i Digitized by tlie Internet Arcliive in 2008 witli funding from IVIicrosoft Corporation littp://www.arcli ive.org/details/collegealgebraOOfinericli COLLEGE ALGEBRA Br HENEY BURCHARD PINE PaoFESsoR OF Mathematics in Princeton University GINN AND COMPANY BOSTON • NEW YORK • CHICAGO • LONDON ATLANTA • DALLAS • COLUMBUS • SAN FRANCISCO ..v'^^ f ¥ ENTERED AT STATIONERS HALL COPYRIGHT, 1901, 1904, BY HENRY B. FINE ALL RIGHTS RESERVED PRINTED IN THE UNITED STATES OF AMERICA 532.10 '-r Wi)t atbenxam ]$ttes PREFACE In this book I have endeavored to develop the theory of the algebraic processes in as elementary and informal a manner as possible, but connectedly and rigorously, and to present the processes themselves in the form best adapted to the purposes of practical reckoning. The book is meant to contain everything relating to algebra that a student is likely to need during his school and college course, and the effort has been made to arrange this varied material in an order which will properly exhibit the logical interdependence of its related parts. It has seemed to me best to divide the book into two parts, a preliminary part devoted to the number system of algebra and a principal part devoted to algebra itself. I have based my discussion of number on the notion of cardinal number and the notion of order as exhibited in the first instance in the natural scale 1, 2, 3, •••. There are con- siderations of a theoretical nature in favor of this procedure into which I need not enter here. But experience has con- vinced me that from a pedagogical point of view also this method is the best. The meaning of the ordinal definition of an irrational number, for example, can be made clear even to a young student, whereas any other real definition of such a number is too abstract to be always correctly understood by advanced students. My discussion of number may be thought unnecessarily elaborate. But in dealing with questions of this fundamental character a writer cannot with a good conscience omit points which properly belong to his discussion, or fail to give proofs IV PREFACE of statements which require demonstration. I hope the details of the discussion will interest the more thoughtful class of students ; but all that the general student need be asked to learn from it is the ordinal character of the real numbers and of the relations of equality and inequality among them, and that for all numbers, real and complex, the fundamental operations admit of definitions which conform to the commu- tative, associative, and distributive laws. In the second or main part of the book I begin by observ- ing that in algebra, where numbers are represented by letters, the laws just mentioned are essentially the definitions of the fundamental operations. These algebraic definitions are stated in detail, and from them the entire theory of the algebraic processes and the practical rules of reckoning are subsequently derived deductively. I shall not attempt to describe this part of the book mi- nutely. It will be found to differ in essential features from the text-books in general use. I have carefully refrained from departing from accepted methods merely for the sake of nov- elty. But I have not hesitated to depart from these methods when this seemed to me necessary in order to secure logical consistency, or when I saw an opportunity to simplify a matter of theory or practice. I have given little space to special devices either in the text or in the exercises. On the other hand, I have constantly sought to assist the student to really mas*ier the general methods of the science. Thus, instead of relegating to the latter part of the book the method of undetermined coefficients, the principal method of investigation in analysis, I have introduced it very early and have subsequently employed it wherever this could be done to advantage. This has naturally affected the arrange- ment of topics. In particular I have considered partial frac- tions in the chapter on fractions. They belong there logically, and when adequately treated, supply the best practice in ele- mentary reckoning that algebra affords. PREFACE V Again, I have laid great stress upon the division transfor- mation and its consequences, and in connection with it have introduced the powerful method of synthetic division. The earlier chapters on equations will be found to contain a pretty full discussion of the reasoning on which the solution of equations depends, a more systematic treatment than is custom- ary of systems of equations which can be solved by aid of the quadratic, and a somewhat elaborate consideration of the graphs of equations of the first and second degrees in two variables. The binomial theorem for positive integral exponents is treated as a special case of continued multiplication, experience having convinced me that no other method serves so well to convey to the student the meaning of this important theorem. I have introduced practice in the use of the general binomial theorem in the chapter on fractional exponents, but have deferred the proof of the theorem itself, together with all that relates to the subject of infinite series, until near the end of the book. In the chapters on the theory of equations and determinants there will be found proofs of the fundamental theorems regard- ing symmetric functions of the roots of an equation and a discussion of the more important properties of resultants. These subjects do not belong in an elementary course in algebra, but the college student who continues his mathemat- ical studies will need them. The like is to be said of the chap- • ters on infinite series and of the chapter on properties of continuous functions with which the book ends. The ideas which underlie the first part of the book are those of Rowan Hamilton, Grassmann, Helmholtz, Dedekind, and Georg Cantor. But I do not know that any one hitherto has developed the doctrine of ordinal number from just the point of view I have taken, and in the same detail. In preparing the algebra itself I have profited by sugges- tions from many books on the subject. I wish in particular to acknowledge my indebtedness to the treatises of Chrystal. vi PREFACE The book has been several years in preparation. Every year since 1898 the publishers have done me the courtesy to issue for the use of the freshmen at Princeton a pamphlet containing what at the time seemed to me the most satisfac- tory treatment of the more important parts of algebra. With the assistance of my colleagues, Mr. Eisenhart and Mr. Gillespie, I endeavored after each new trial to select what had proved good and to discard what had proved unsatisfactory. As a consequence, much of the book has been rewritten a number of times. No doubt subsequent experience will bring to light many further possibilities of improvement ; but I have hopes that as the book stands it will serve to show that algebra is not only more intelligible to the student, but also more inter- esting and stimulating, when due consideration is given to the reasoning on which its processes depend. HENRY E. FINE Princeton University June, 1905 K' CONTENTS PAET FIRST — NUMBERS PAGE I. The Natural Numbers — Counting, Addition, and Mul- tiplication 1 II. Subtraction and the Negative 16 III. Division and Fractions 27 IV. Irrational Numbers S9 V. Imaginary and Complex Numbers 70 PART SECOND — ALGEBRA I. Preliminary Considerations 79 II. The Fundamental Operations . > 93 III. Simple Equations in one Unknown Letter 110— IV. Systems of Simultaneous Simple Equations .... 127—— V. The Division Transformation 155-— VI. Factors of Rational Integral Expressions .... 176 VII. Highest Common Factor and Lowest Common Multiple 196 V^III. Rational Fractions 213 IX. Symmetric Functions 245 X. The Binomial Theorem 252— _ XI. Evolution 260 XII. Irrational Functions. Radicals and Fractional Exponents 271 XIII. Quadratic Equations 298——' XIV. Discussion of the Quadratic Equation. Maxima and Minima 304..^ XV. Equations of Higher Degrek which can be solved by Means of Quadratics 309.- vii vm CONTENTS PAGE XVI. Simultaneous Equations which can be solved by Means of Quadratics 317 XVII. Inequalities 340 XVIII. Indeterminate Equations of the First Degree . 342 XIX. Ratio and Proportion. Variation 347 XX. Arithmetical Progression 3.54 XXI. Geometrical Progression 357 XXII. Harmonical Progression 362 XXIII. Method of Differences. Arithmetical Progres- sions OF Higher Orders. Interpolation . . . 364 . XXIV. Logarithms 374 XXV. Permutations and Combinations 393 XXVI. The Multinomial Theorem 408 XXVII. Probability 409 XXVIII. Mathematical Induction 424 XXIX. Theory of Equations 425 XXX. Cubic and Biquadratic Equations 483 XXXI. Determinants and Elimination 492 XXXII. Convergence of Infinite Series . . . . y . . 520 XXXIII. Operations with Infinite Series 539 XXXIV. The Binomial, Exponential, and Logarithmic Series 553 XXXV. Recurring Series 560 XXXVI. Infinite Products 564 XXXVII. Continued Fractions 666 XXXVIII. Properties of Continuous Functions 577 INDEX 591 A COLLEGE ALGEBRA PART FIRST— NUMBERS I. THE NATURAL NUMBERS — COUNTING, ADDITION, AND MULTIPLICATION GROUPS OF THINGS AND THEIR CARDINAL NUMBERS Groups of things. In our daily experience things present themselves to our attention not only singly but associated in groups or assemblages. The fingers of a hand, a herd of cattle, the angular points of a polygon are examples of such groups of things. We think of certain things as constituting a group, when we distinguish them from other things not individually but as a whole, and so make them collectively a single object of our attention. For convenience, let us call the things which constitute a group the eleynents of the group. Equivalent groups. One-to-one correspondence. The two groups of letters ABC and DEF are so related that we can combine all their elements in pairs by matching elements of the one with elements of the other, one element with one element. Thus, we may match A with D, B with E, and C with F. Whenever it is possible to match all the elements of two groups in this manner, we shall say that the groups are equivalent; and the process of matching elements we shall call bringing the groups into a one-to-one relation, or a relation of one-to-one correspondence. 1 2 \ A \ (JOjiXtGE ALGEBRA TIxGOi e>n., ;,^_ (wa ^/y;/^,^.^ (^rc equivalent to the same third group, they are equivalent to one another. For, by hypothesis, we can bring each of the two groups into one-to-one correspondence with the third group. But the two groups will then be in one-to-one correspondence with each other, if we regard as mates every two of their elements which we have matched with the same element of the third group. Cardinal number. We may think of all possible groups of things as distributed into classes of equivalent groups, any two given groups belonging to the same class or to different classes, according as it is, or is not, possible to bring them into one-to-one correspondence. Thus, the groups of letters ABCD and EFGH belong to the same class, the groups ABCD and EFG to different classes. The property which is common to all groups of one class, and which distinguishes the groups cf one class from those of another class, is the number of things in a group, or its cardinal number. In other words, The number of things in a group, or its cardinal number, is that property which is common to the group itself and every group which may he brought into one-to-one correspondence with it. Or we may say : " The cardinal number of a group of things is that property of the group which remains unchanged if we rearrange the things within the group, or replace them one by one by other things " ; or again, " it is that property of the group which is independent of the character of the things themselves and of their arrangement within the group." For rearranging the things or replacing them, one by one, by other things will merely transform the group into an equiva- lent group, § 2. And a property which remains unchanged during all such changes in the group must be independent of the character of the things and of their arrangement. THE NATURAL NUMBERS 3 Part. We say that a first group is a part of a second group ■when the elements of the first are some, but not all, of the elements of the second. Thus, the group ABC is a part of the group ABCD. From this definition it immediately follows that If the first of three gt'oups be a jjart of the second, arid the second a j^^f^'t of t/ie third, then the first is also a part of the third. _glnite and infinite groups, Wa,-sa,y-that a group or assem- blage is finite when it is equivalent to no one of its parts ; infinite when it is equivalent to certain of its parts.* Thus, the group ABC is finite; for it cannot be brought into one-to- one correspondence with BC, or with any other of its parts. But any never-ending sequence of marks or symbols, the never-ending sequence of numerals 1, 2, 3, 4, • • • , for example, is an infinite assemblage. We can, for instance, set up a one-to-one relation between the entire assemblage 1, 2, 3, 4, • • • and that part of it which begins at 2, namely, between 1, 2, 3, 4, 5, • • • (a) and 2,3,4,5,6,..., (b) by matching 1 in (a) with 2 in (b), 2 in (a) with 3 in (b), and so on, — there being for every numeral that we may choose to name in (a) a corresponding numeral in (b). Hence the assemblage (a) is equivalent to its part (b). Therefore (a) is infinite. Less and greater cardinal numbers. Let M and iV denote any 8 two finite groups. It must be the case that 1. M and N are equivalent, or 2. M is equivalent to a part of N, or 3. N is equivalent to a part of M. * Of course we cannot actually take account of all the elements, one by one, of an infinite group — or assemblage, as it is more often called. We regard such an assemblage as defined when a law has been stated which enables us to say of every given thing whether it belongs to the assemblage or not. 4 A COLLEGE ALGEBRA In the first case we say that M and N have the same cardinal number, § 4, or equal cardinal numbers ; in the second case, that the cardinal number of M is less than that oi N ; in the third, that the cardinal number of M is greater than that of N. Thus, if M is the group of letters c6c, and N the group de/gr, then M is equivalent to a part of N, to the part de/, for example. Hence the cardinal number of M is less than that of N, and the cardinal number of N is greater than that of M. 9 Note. It follows from the definition of finite group, § 7, that there is no ambiguity about the relations "equal," "greater," and "less" as here defined. Thus, the definition does not make it possible for the cardinal number of M to be at the same time equal to and less than that of N, since this would mean that M is equivalent to N and also to a part of N, therefore that N is equivalent to one of its parts, § 3, and therefore, finally, that N is infinite, § 7. 10 Corollary. If the first of three cardinal numbers be less than the second, and the second less than the third, then the first is also less than the third. For if Jf, N, P denote any groujis of things of which these are the cardinals, M is equivalent to a part of N, and JV to a part of P; therefore M is equivalent to a part of P, §§ 3, G. 11 The system of cardinal numbers. By starting with a group which contains but a single element and repeatedly " adding " one new thing, we are led to the following list of the cardinal numbers : 1. The cardinal number of a " group " like I, which contains but a single element. 2. The cardinal number of a group like II, obtained by adding a single element to a group of the first kind. 3. The cardinal number of a group like III, obtained by adding a single element to a group of the second kind. 4. And so on, without end. We name these successive cardinals "one," " two,"*' three," •••, and represent them by the signs 1, 2, 3, • • • . THE NATURAL NUMBERS ' 5 Observations on this system. Calling the cardinal number 12 of any finite group a finite cardinal, we make the following observations regarding the list of cardinals which has just been described. First. Every cardinal contained in this list is finite. For the group I is finite, since it has no part to which to be equiva^ " lent, § 7 ; and each subsequent group is finite, because a group obtained by adding a single thing to a finite group is itself finite.* Thus, II is finite because I is ; III is finite because II is ; and so on. Second. Every finite cardinal is contained in the list. For, by definition, every finite cardinal is the cardinal number of some finite group, as M. But we can construct a group of marks III • • • I equiva- lent to any given finite group if, by making one mark for each object in M. And this group of marks must have a last mark, and therefore be included in the list of § 11, since otherwise it would be never-ending and therefore itself, and with it M, be infinite, § 7. Third. No two of these cardinals are equal. This follows from the definition in § 8. For, as just shown, all of the groups I, II, III, • • • are finite ; and it is true of every two of them that one is equivalent to a part of the other. * We may prove this as follows (G. Cantor, Math. Ann., Vol. 46, p. 490) : If M denote a finite group, and e a single thing, the group Me, obtained by adding e to M, is also finite. For let G = H denote that the groups G and H are equiA'alent. If Me is not finite, it must be equi^^^lent to some one of its parts, § 7. Let P denote this part, so that Me = P. (1) Suppose that P does not contain e. Let/ denote the element of P which is matched with e in Me, and represent the rest of P by Pi. _ Then sinoe Mp = P\ f and e = /", we have M= P\. But tliis is inipossibie, since M is finite and Pi is a part of M, § 7. (2) Sujipfisc tliat P does contain e. It cannot be tliat e in P is matched with e in Me, for then the rest of P, which is a part of M, would be equivalent to M. But suppose that e in P is matched with some other element, as g in Me, and that e in Me is matched with/ in P. If Me = P oe true on this hypothesis, it must also be true if we recombine the elements e, /, g so as to match r in P with e in Me, and / in P with g in Me. But, as "ins't shown, we should then have a part of P equivalent to M. Hence this hypothesis albo is impossible A COLLEGE ALGEBRA THE NATURAL SCALE. EQUATIONS AND INEQUALITIES 13 The natural numbers. We call the signs 1, 2, 3, oi their names " one," " two," " three," positive integers or natural numbers. Hence A natural number is a sigyi or symbol for a cardinal number. 14 The natural scale. Arranging these numbers in an order corresponding to that already given the cardinals which they represent, § 11, we have the never-ending sequence of signs 1,2,3,4, 5, ••, or "one," "two," "three," "four," "five," •••, which we call the natural scale, or the scale of the natural numbers. 15 Each sign i?i the scale indicates the number of the signs in that part of the scale which it terminates. Thus, 4 indicates the number of the signs 1, 2, 3, 4. For the number of signs 1, 2, .3, 4 is the same as the number of groups I, II, III, llli, and this, in turn, is the same as the number of marks in the last group, llll, § 8. And so in general. 16 The ordinal character of the scale. The natural scale, by itself considered, is merely an assemblage of different signs in which there is a first sign, namely 1 ; to this a definite next follow- ing sign, namely 2 ; to this, in turn, a definite next following sign, namely 3 ; and so on without end. In other words, the natural scale is merely an assemblage of different signs which follow one another in a definite and knoivn order, and having a first but no last sign. Regarded from this point of view, the natural numbers themselves are merely marks of order, namely of the order in which they occur — with respect to time — when the scale is recited. 17 It is evident that the scale, in common with all other assem- blages whose elements as given us are arranged in a definite , and known order, has the following properties : THE NATURAL NUMBERS 7 1. We may say of any two of its elements that the one "precedes" and the other "follows," and these words "pre- cede " and " follow " have the same meaning when applied to any one pair of the elements as when applied to any other pair. 2. If any two of the elements be given, we can always deter- mine ivhich precedes and which follows. 3. If a, b, and c denote any three of the elements such that a precedes b, and b precedes c, then a precedes c. An assemblage may already possess these properties when given us, or we may have imposed them on it by some rule of arrangement of our own choosing. In either ease we call the assemblage an ordinal system. Instances of the first kind are (1) the natural scale itself ; (2) a sequence of events in time ; (3) a row of points ranged from left to right along a horizontal line. An instance of the second kind is a group of men arranged according to the alphabetic order of their names. An assemblage may also have " coincident " elements. Thus, 18 in a group of events two or more may be simultaneous. We call such an assemblage ordinal when the relations 1, 2, 3 hold good among its 7io7i-co incident elements — it being true of the coincident elements that 4. If a coincides with b, and b with c, then a coincides with c. 5. If a coincides with b, and b precedes c, then a precedes c. It is by their relative order in the scale that the natural 19 numbers indicate relations of greater and less among the cardinal numbers. For of any two given cardinals that one is greater whose natural number occurs later in the scale. And the relation : " if the first of three cardinals be less than the second, and the second less than the third, then the first is less than the third," is represented in the scale by the relation : " if a precede b, and b precede c, then a precedes c." 8 A COLLEGE ALGEBRA In fact, we seldom employ any other method than this for comparing cardinals. We do not compare the cardinal numbers of groups of things directly, by the method of § 8. On the contrary, we represent them by the appropriate natural numbers, and infer which are greater and which less from the relative order in which these natural numbers occur in the scale. The process causes us no conscious effort of thought, for the scale is so vividly impressed on our minds that, when any two of the natural numbers are mentioned, we instantly recognize which precedes and which follows. Thus, if we are told of two cities, A and B, that the population of A is 120,000, and that of B, 125,000, we immediately conclude that B has the greater number of inhabitants, because we know that 125,000 occurs later in the scale than 120,000 does. 20 Numerical equations and inequalities. In what follows, the word " number " will mean natural number, § 13 ; and the letters a, h, c will denote any such numbers. 21 When we wish to indicate that a and b denote the same num- ber, or " coincide " in the natural scale, we employ the equation a = b, read " a equals b." 22 But when we wish to indicate that a precedes and J follows in the natural scale, we employ one of the inequalities a a, read " b is greater than a." 23 Of course, strictly speaking, these words, "equal," "less," and " greater," refer not to the signs a and b themselves, but to the cardinals which they represent. Thus, the phrase, " a is less than b," is merely an abbreviation for, "the cardinal which a represents is less than the cardinal which b represents." But all that the inequality a rop- erty of the group because of which we shall arrive at the same natural number in whatever order we count the group. This is the definition of cardinal number to which we are naturally led if we choose to make the natural scale, defined as in § 16, our starting point in the discussion of number. ADDITION 29 Definition of addition. To add 3 to 5 is to find what number occupies the third place after 5 in the natural scale. We may find this number, 8, by counting three numbers forward in the scale, beginning at 6, thus : 6, 7, 8. We indicate the operation by the sign +, read ''plus," writing 5 + 3 = 8. And in general, to add b to a is to find what number occupies the ^th place after a in the natural scale. Since there is no last sign in the scale, this number may always be found. We call it the sum of a and b and represent it in terms of a and b by the expression a -\- b. 30 Note. The process of finding a + b hy counting forward in the scale corresponds step for step to that of adding to a group of a things the elements of a group of b things, one at a time. Hence (1) the result of the latter process is a group oi a + b things, § 8, and (2) if a and b denote finite cardinals, so also does a + b. See footnote, p. 5. THE NATURAL NUMBERS 11 Since a + 1, a + 2, and so on, denote the 1st, 2d, and so 31 on, numbers after a, the sequence a + 1, a -\- 2, • • • denotes all that portion of the scale which follows a. Hence any given number after a may be expressed in the form a + d, where d denotes a definite natural number. The process. To add large numbers by counting would be 32 very laborious. We therefore memorize sums of the smaller numbers (addition tables) and from these derive sums of the larger numbers by applying the so-called "laws" of addition explained in the following sections. The laws of addition. Addition is a " commutative " and an 33 " associative " operation ; that is, it conforms to the following two laws : The commutative law. a -{- b — b -{- a, 34 The result of adding h to 2i is the same as that of adding a to b. The associative law. a + (1) -\- c) = (a -\- b) + c, 35 The result of first adding c ^o b amd then adding the sum so obtained to a, is the same as that of first adding b ^o a and then adding c to the sum so obtained. Note. In practice, we replace the expression (a -)- 6) + c by a + 6 + c, 36 our understanding being tliat the expression a + 6 + c + • • ■ represents the result of adding 6 to a, c to the sum so obtained, and so on. Proofs of these laws. We may prove these laws as follows. 37 Pirst. The co7nmutative law : a -{- b = b -{- a. Thus, the sums 3 + 2 and 2 + 3 are equal. For 3 + 2 represents the number found by first counting off three numbers, and after that two numbers, on the natural scale. Thus, The group counted 1, 2, 3, 4, 5, (a) the counters, 1, 2, 3, 1, 2. (b) But as there is a one-to-one relation between the groups of signs (a) and (b), and every one-to-one relation is reciprocal, § 2, we may interchange the roles of (a) and (b) ; that is, if we make (b) the group counted, (a) will represent the group of counters. 12 A COLLEGE ALGEBRA Hence finding 3 + 2 is equivalent to counting the group of signs 1, 2, 3, 1, 2. (b) In like manner, finding 2 + 3 is equivalent to counting the group 1, 2, 1, 2, 3. (c) But as (b) and (c) consist of the same signs and differ only in the manner in w^hich these signs are arranged, the results of counting them are the same, § 27 ; that is, 3 + 2 = 2 + 3. Similarly for any two natural numbers, a and h. Second. The associative laiv : a -\-(h -\- c) = (a -\- b) -j- c. For in counting to tlie 6th sign after a, namely to a + b, and then to the cth sign after this, namely to (a + 6) + c, we count 6 + c signs all told, and hence arrive at the (6 + c)th sign after a, namely at a + (6 + c). The notion of cardinal number is involved in the proofs just given. But addition may be defined and its laws established independently of this notion, as is shown in the footnote below.* * The Italian mathematician Peano has defined the system of natural num- bers without using the notion of cardinal number, by a set of "postulates" which we may state as follows — where " number " means " natural number." 1. The sign 1 is a number. 2. To each number a there is a next following number — call it a +. 3. This number a + is never 1. 4. If a += 6 + , then a= 6. 5. Every given number a is present in the sequence 1, 1 +, (1 +)+,••• . The numerals 2, 3, • • • are defined thus : 2=l + ,3=2 + ,---. The sum a+6 is to mean the number determined (because of 5) by the series of formulas a + 1 = a +, a + 2= (o + 1) +, • • • . The series of formulas just written is equivalent to the single formula 6. a + (6 + l)= (a + b) + l. From 6, by " mathematical induction," we may deriA'e the laws of addition: 7. a + (b + c)= (a+b) +c. 8. a + b=b + a. First. If 7 is true when c= ^, it is also true when c= A; + 1. For, by 6 and 7, a + [b + {k + l)]--a + [{b + k) + l]=[a + {b + k)] + l = [{a + b) + k] + l= {a+b)+(k + l). But, by 6, 7 is true when c= 1. Hence 7 is true when c= 2, .-. when c= 3, .•.•■• when c= any number, by 5. Second. We first prove 8 for tlie particular case : 8'. a + 1 = 1 + a. If 8' is true for a= k, then {k + 1) + \= (1 + k) + 1= 1 + (k+ 1), by G. Hence if 8' is true for a= k, it is also true for a= k+1. Hence, since 8' is true fora= 1, it is true fora-^ 2, .-. for a= 3, • ■ •. Finally, if 8 be true for b=k, it is true for b=k + l. For, by 7 and 8', a + (k + 1) = {a + k) + 1= \ + (a + k) = l + (k + a)=(l + k)+ a={k + l) + a. Hence, since 8 is true (by 8') when 6 = 1, it is true for 6=2, .-. for 6 = 3, .-. ■ • • . See Stolz and Gmeiner, Theoret'iarhc Arithtnrtik, pp. 13 ff., and tlio refer- ences to Peano there given; also Huntington in Bulletin of the American Mathematical Society, Vol. IX, p. 40. H. (irassmanu {Lehrbuch der Arith- metik) was the first to derive 7 and 8 from tJ. THE NATURAL NUMBERS 13 General theorem regarding sums. By making repeated appli- 38 cation of these laws, §§ 34, 35, it can be shown that The sum of any finite number of numbers will be the same, whatever the order in which we arrange them, or whatever the manner in which we group them, when adding them. Thus, a + b + c + d = a + c + b + d. For a + b + c + d = a + (b + c) + d §35 = a + {c + b) + d § 34 z=a + c + b + d. §36 Rules of equality and inequality for sums. First. From the 39 definition of sum, § 29, and the rules of § 24, it follows that 1. If a = b, then a -{- c = b -]- c. 2. If a < b, then a + c < b + e. 3. If a > b, then a -{- c > b -^ c. Here 1 is obvious, since it a = b, then a and b denote the same number. We may prove 3 as follows, and 2 similarly. Ifa>5, let a = 6 + cZ, §31. Then a + c = (6 + d) + c = {& + c) + d, §§ 34, 35, .-, >6 + c. Second. From 1, 2, 3 it follows conversely that 4. li a -\- c = b -\- c, then a = b. 5. If a + c < b + c, then a < b. 6. U a -\- c> b -\- c, then a > b. Thus, ii a + c = b + c, then a = b. Foi- otherwise we must have either ab and therefore a + Ob + c (by 3). Third. It also follows from 1, 2, 3 that 7. If a = b, and c = d, then a -\- c = b -\- d. 8. If a < b, and c < d, then a -]- c < b -{- d. 9. If a > b, and c > d, then a -\- c > b -\- d. Thus, if a = b, then a + c = 6 + c, and if c = d, then b + c = b + d. Hence a + c = b -^ d. 14 A COLLEGE ALGEBRA MULTIPLICATION 40 Definition of multiplication. To multiply a by 5 is to find the sum of b numbers, each of which is a. We call this sum the product oi a hj b and express it in terms of a and ^ by a x b, oi a- b, or simply ab. Hence, by definition, 41 ab = a -\- a ■• -to b terms. 42 We also call a the multiplicand, b the multiplier, and a and b the factoi's of ab. 43 The process. To find products by repeated addition would be very laborious. We therefore memorize products of the smaller numbers (multiplication tables), and from these derive products of the larger numbers by aid of the laws of addi- tion and the laws of multiplication explained in the following sections. 44 The laws of multiplication. Multiplication, like addition, is a commutative and an associative operation, and it is "dis- tributive " with respect to addition ; that is, it conforms to the following three laws : 45 The commutative law. ab = ba. The result of multiplying a % b is the same as that of mtd- tiplying b by a. Thus, 2-3 = and 3 • 2 = G. 46 The associative law. a (be) = (ah) c, The result of multiplying a by the jiroduct be is the same as that of multiplying the product ab by c. Thus, 2 (3 • 4) = 2 • 12 = 24 ; and (2 • 3)4 = 6 • 4 = 24. In practice we write abc instead of {ab)c. Compare § 36. 47 The distributive law. a(b -\- c)= ab + ac, The result of first adding h and c, and then multiply- ing a by the sum so obtained, is the same as that of first THE NATURAL NUMBERS 15 uiultiplymcj a hy b and a by c, and then adding the joroducts so obtained. Thus, 3 (4 + 5) = 3 • 9 = 27 ; and 3 • 4 + 3 • 5 = 12 + 15 = 27. Proofs of these laws. We may prove these laws as follows : 48 First. T\\Q distributive law : ab -{- ac = a(b + c). (1) For ah + ac = {a + a-\r • • ■ toh terms) + (a + «+■•• to c terms) § 41 = a + a + a + • • • to (6 + c) terms = a (6 + c). §§ 35, 41 Hence a (1) -\- c + ■■■)= ab -\- ac -{-■■ -. (2) Thus, a{b + c -\- d) = a{h ^- c) + ad = ah + ac + ad. by (1) and § 35 We also have ac + he = (a. + b) c. (3) For ac + hc = {a + a + ■ ■ ■ to c terms) + (h + b + ■ ■ ■ to c terms) = (a + 5) + (a + &') + ••• to terms = {a + b)c. § 38 Second. The coniniutative law: ab = ba. a6 = (1 -f 1 + . . . to a terms) b = 1 . 6 + 1 • 5 H to a terms by (3) = 6 + 6 + • • • to a terms = ba. § 41 Third. The associative law: (ab)c =^ a(bc). (ab) c = ah + ab + ■■• to c terms § 41 = a{h + b + ■■■ toe terms) = a (be). by (2) and § 41 General theorem regarding products. These laws can be 49 extended to products of any finite number of factors. Thus, The j)voduct of any finite number of factors is independent of the order in tvhich the factors are multiplied tor/ether. Rules of equality and inequality for products. These are : 50 1. If a = b, then ac = be. 4. If ac = be, then a = b. 2. If a < b, then ac < be. 5. If ac < be, then a b. then ac > be. 6. If ac > be, then a > b. 16 A COLLEGE ALGEBRA Here 1 is obvious, since ii a = b, then a and b denote the same number. We may prove 3 as follows, and 2 similarly. If a>b, let a = b + d. Then ac = {b + d)c = bc + dc, .: >bc. The rules 4, 5, 6 are the converses of 1, 2, 3 and follow from them by the reasoning used in § 39. From 1, 2, 3, by the reasoning employed in § 39, it follows that If a — b and c = d, then ac = hd. li a b and c> d, then ac > bd. II. SUBTRACTION AND THE NEGATIVE THE COMPLETE SCALE 51 Subtraction. To subtract 3 from 5 is to find what number occupies the 3d place before 5 in the natural scale. We find this number, 2, by counting three numbers backward in the scale, beginning at 4, thus : 4, 3, 2. We indicate the operation by the sign, — , read "minus," writing 5 — 3 = 2. And, in general, to subtract b from a is to find what number occupies the ^th place before a. We call this number the remainder obtained by subtracting b from a, and represent it in terms of a and b by the expres- sion a — b. We also call a the mimierid and b the subtrahend. 52 Addition and subtraction inverse operations. Olearly the third number before 5 is also the number from which 5 can be obtained by adding 3. And, in general, we may describe the remainder a — b either as the ^'th number before a, or as the number from which a can be obtained by adding b, that is, as the number which is defined by the equation 53 (^a-b)-\-b = a. SUBTRACTION AND THE NEGATIVE 17 Again, since saying that 7 occupies the 3d place after 4 is equivalent to saying that 4 occupies the 3d place before 7, we have 4 + 3 — 3 = 4. And so, in general, {a -\- b) - b =^ a. 54 Since a + ^ — ^ = a, § 54, subtraction undoes addition ; and 55 since a — b -\- b — a, § 53, addition undoes subtraction. We therefore say that addition and subtraction axQ inverse opera- tions. The complete scale. The natural scale does not fully meet 56 the requirements of subtraction ; for this scale has Si first num- ber, 1, and we cannot count backward beyond that number. Thus, on the natural scale it is impossible to subtract 4 from 2. But there are important advantages in being able to count backward as freely as forward. And since the natural scale is itself merely a system of signs arranged in a definite order, there is no reason why we should not extend it backward by placing a new ordinal system of signs before it. We therefore invent successively the signs : 0, which we place before 1 ; — 1, which we place before ; — 2, which we place before — 1 ; and so on. In this manner we create the complete scale . . ., _ 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, • . •, which has neither a first nor a last sign or '' number," and on which it is therefore possible to count backward, as well as forward, to any extent Avhatsoever. Observe the sijmmetry of this scale with respect to the sign 57 0. As 3 is the third sign after 0, so — 3 is the third sign before ; and so in general. Meaning of the new numbers. One of these new signs, 58 namely 0, may be said to have a cardinal meaning. Thus, counting backward from 3 corresponds to the operation of removing the elements of any group of 3 things, one at a time= This operation may be continued until all the elements have 18 A COLLEGE ALGEBRA been removed, and we may call the sign of the cardinal num- ber of the resulting " group " of no elements. We therefore often regard as one of the natural numbers. But — 1, — 2, — 3, • • • have no cardinal meaning whatsoever. On the other hand, all these new signs have the same ordinal character as the natural numbers. Every one of them occupies a definite position in an ordinal system which includes the natural numbers also. And we may consider it definedhy this position precisely as we may consider each natural number defined by its position in the scale. We regard this as a sufficient reason for calling the signs — 1, — 2, — 3 • • • numbers. 59 Positive and negative. To distinguish the new numbers — 1, — 2, — 3, • • • as a class from the old, we call them negative, the old jjositive. The numbers of both kinds, and 0, are called integers to distinguish them from other numbers to be considered later. 60 Algebraic equality and inequality. Let a, b, c denote any num- bers of the complete scale. According as a precedes, coincides with, or follows b, we write a < b, a = b, or a > b. 61 Since by definition the complete scale is an ordinal system, § 17, the rules of § 24 apply to it also ; thus, If a < b and b < c, then a < c. 62 When a < b, that is, when a precedes b in the complete scale, it is customary to say that a is algebraically less than b, or that b is algebraicalli/ greater than a. Observe that the words "less" and "greater," as thus used, mean "precede" and "follow," in the complete scale — this and nothing more. Thus, " — 20 is less than — 18 " means merely " — 20 precedes — 18." 63 Absolute or numerical values. We call 3 the numerical value of — 3 or its absolute value, and use the symbol | — 3 1 to represent it, writing | — 3 1 = 3. Similarly for any negative number. The numerical value of a positive number, or 0, is the number itself. Thus|3| = 3. SUBTRACTION AND THE NEGATIVE 19 Numerical equality and inequality. Furthermore we say of 64 any two numbers of the complete scale, as a and b, that a is numerically less than, equal to, or greater than b, according as |a| <, =, or > |b|. Thus, while - 3 is algebraically less than 2, it is numerically greater than 2, and while - 7 is algebraically less than - 3, it is numerically greater than — 3. OPERATIONS WITH NEGATIVE NUMBERS New operations. We also invent operations by which the 65 negative numbers and may be combined with one another and with the natural numbers, as the latter are themselves combined by addition, multiplication, and subtraction. We call these operations by the same names, and indicate them in the same way, as the operations with natural numbers to which they correspond. Employing a, as in § 60, to denote any number of the com- plete scale, but a and b to denote natural numbers only, we may define these new operations as follows : Definitions of addition and subtraction. These are : 66 1. a + & is to mean the hth. number after a. 2. a — i is to mean the hi\\ number before a. 3. a + and a — are to mean the same number as a. 4. a -f (— V) is to mean the same number as a — &. 5. a — (— b) is to mean the same nmnber as a + 6. In other words, adding a positive number h to any number a is to mean, as heretofore, counting b places forward in the scale ; subtracting it, counting b places backward : while adding and subtracting a negative number are to be equiva- lent respectively to subtracting and adding the corresponding positive number. 20 A COLLEGE ALGEBRA Thus, byl, -3 + 2 = — 1, since — 1 is the 2d number after — 3. by 2, 2 — 5 = — 3, since — 3 is the 5th number before 2. by 4, - 5 + (- 2) = - 5 - 2 = - 7 (by 2). by 5, - 6 - (- 2) = - 6 + 2 = - 4 (by 1). 67 Definition of multiplication. This is : 1. • a and a ■ are to mean 0. 2. a (— b) and (— a) h are to mean — ah. •3. (— a) (— b) is to mean ah. In other words, a product of two factors, neither of which is 0, is to be positive or negative according as the factors have the same or opposite signs. And in every case the numerical value of the product is to be the product of the numerical values of the factors. Thus, by 2, 3 x - 2 = - 6, and - 3 x 2 = - 6. by 3, - 3 X - 2 = 6. 68 The origin and significance of these definitions. Observe that the statements of §§66 and 67 are neither assumptions nor theorems requiring demonstration, but what we have called them — definitions of neiv operations. Thus, it would be absurd to attempt to prove that 2 (— 3) = — 2 • 3 with nothing to start from except the definition of multiplication of natural numbers, § 40, for the obvious reason that — 3 is not a natural number. The phrase "2 taken — 3 times" is meaningless. But why should such operations be invented ? To make the negative numbers as serviceable as possible in our study of relations among numbers themselves and among things in the world about us. The new operations have not been invented arbitrarily ; on the contrary, they are the natural extensions of the old opera- tions to the new numbers. In dealing with the natural numbers, we first defined addi- tion as 2i process — coimting forward — and then showed that SUBTRACTION AND THE NEGATIVE 21 the results of this process have two properties ivhich are inde- pendent of the values of the numbers added, namely : 1. a + b=^b -\- a. 2. a + (Ij + c) = {a + b) + c. Similarly we proved that products possess the three general properties : 3. ab = ba. 4. a (be) = (at) c. 5. a (b -\- c) = ab -{- ac. When we employ letters to denote numbers, these properties 1-5 become to all intents and purposes our tvorkinr/ definitions of addition and multiplication ; for, of course, we cannot then actually carry out the processes of counting forward, and so on. Clearly if corresponding operations with the new numbers are to be serviceable, these " definitions " 1-5 must apply to them also. And §§ 66, 67 merely state the solution of the problem : To make such an extension of the meanings of addition, multi- plication, and subtraction that sums and products of any num- bers of the comjilete scale may have the properties 1-5, and that subtraction may contimie to be the inverse of addltioii. Thus, (1) when we define adding a positive number & to a as counting forward, and subtracting it as counting backward, we are merely repeating the old definitions of addition and subtraction. (2) From this definition of addition it follows that —6 + 6 = 0. But if the commutative law a + 6 = 6 + aisto hold good, we must have — 6 + 6 = 6 + (— 6), and therefore 6 + (— 6) = ; or, since 6 — 6 = 0, we must have b + ( — 6) = 6 — 6. This suggests the definition a + ( — 6) = a — 6. (3) If our new addition and subtraction are, like the old, to be inverse Operations, we must also have, as in § 66, 5, a — (— 6) = a + 6. (4) Again, to retain the old connection between addition and multipli- cation, § 41, we must have, as in § 67, 2, ( — a) 6 = — a + ( — a) + • • ■ to 6 terms = — a — a — ■ ■ ■ to^ lisrms = — ah. (6) If the commutative law ah = ba is to hold good, we must also have a(- 6) = (- 6)a = -6a = -a6, as in §67, 2, 22 A COLLEGE ALGEBRA (6) Similarly, + + ■ • • to a terms = 0, and this fact together with our wish to conform to the law ab = ba leads to the definitions of § 67, 1, namely, • a = and a • = 0. (7) Finally, it follows from (6) that {- a) {- b + b) = - a -0 = 0. But if the distributive law is to hold good, we must also have {- a){- b + b) = (- a) { ~ b) + {- a)b = (- a) (- b) - ab, by (4). "We therefore have {— a) {— b) - ab = 0. And since also ab — ab = 0, we are thus led to define (— a) (— b) as ab, as in § 67, 3. 69 The operations just defined conform to the commutative, asso- ciative, and distributive laws. It remains to prove that the new- operations are in complete agreement with the laws which suggested them. To begin with, we have a + (6 + o) = a + & + c, (1) a-(b + c) = a-b-c, (2) a + b-b=SL-b + b = ai, (3) as follows from the definitions of addition and subtraction as counting forward and backward, by the reasoning in §§ 37, 52. I. The commutative law, a + b = b + a. First, -a + b = b+(-a). Forifa>6, let a = d + b. §§31,34 Then -a + b = -{d + b) + b =,^d-b + b = -d; by (2) and (3) and b + {-a) = b-{b + d), § 66, 4 = b-b-d = -d. by (2) Proceed in a similar manner when b>a. Second, - a + (- b) = - b -{- (- a). For -o if {-?>) = -(a + &) = -(& + «) = -6 + (-«), by (2) and §66,4. II. The associative law, a + (b + c) = (a + b) + c. First, 8i+[b+(-c)^ = a + b+(- c). Forif6>c, let b = d + c. §§31,34 SUBTRACTION AND THE NEGATIVE 23 Then a + [6 + (-c)] = a +[d + c +(- c)] = a + d, and a + fe + (- c) = a + d + c + (- c) = a + d. by (3) and §66,4 Proceed in a similar manner when c>b. Second, a + [(- &) + c] = a + (- 6) + c. This follows from I and the case just considered. Third, a + [- Z- + (- c)] = a + (- i) + (- c). This follows from (2) and § 66, 4, since - 6 + (- c) = - (6 + c). III. The commutative law, ab = ba. First, (- a)b^b(- a). For {-a)b = -ab = -~ba==b{-a). §45; §67,2 Second, (- a) (- b) = (- b) (- a). For {-a){-b) = ab = ba^{~b){-a). §45; §67,3 IV. The associative law, a (be) = (ab)c. First, (- a) [(- b) (- c)] = [(- a) (- ^)] (- c). For {-a)[{-b){~c)] = {-a)-bc = -abc, § 46 ; § 67, 2, 3 and [(_a)(_6)](_c) = aft. (- c) =-a&c. §67,2,3 Second, the other cases may be proved in the same way. V. The distributive law, a (b + c) = ab + ac. First, a[b +(— c)] = ab + a (— c). For [b +{- c)]a = [b + {- c)] + [b + {- c)] + ■■■ to a terms = 6 + 6-1 to a terms + ( — c) + ( - c) + • • • to a terms = ba + {-c)a. § 41 ; § 67, 2 ; II and III Hence a[b + {- c)] =ab + a(- c) by III Second, from this case the others readily follow. Thus, (-a)[6+(-c)]=-a[6 + (-c)] = -Qab + a(- c)] = (- a)b+{- a) (- c). 24 A COLLEGE ALGEBRA 70 The general result. As has already been observed, § 68, in literal arithmetic or algebra, the laws a + b = b + a, and so on, are equivalent to dejinitions of addition and multiplication, even when the letters a, b, c denote natural numbers. And we have now shown that these definitions apply to all numbers of the complete scale. By means of these laws we may change the form of a literal expression without affecting its value, whatever numbers of the complete scale the letters involved in the expression may denote. Thivs, whether a, 6, c, d denote positive or negative integers, we have (a + 6) (c + d) = (a + 6) c + (a + f") d = ac + hc + ad -\- bd. 71 Rules of equality and inequality for sums. We may prove by reasoning similar to that in § 39 that According as a < , = , or > b, so is a + c < , = , or > b + c ; and conversely. Hence it is true for positive and negative numbers alike that 72 An equation remains an equation and the sense of an inequality remains unchanged when the same number is added to both sides, or is subtracted from both. 73 Rules of equality and inequality for products. Observe that changing the signs of any two numbers a and b reverses the order in which they occur in the complete scale, § 57. Thus, we have - 3 < - 2, but 3 > 2 ; - 5 < 2, but 5 > - 2. From this fact and the reasoning of § 50 it follows that According as a <, = , or > b, so is ac < , = , or > be, but a(— c) >, =, or < b(— c); and conversely. Hence SUBTRACTION AND THE NEGATIVE 25 Mxdtiphjing both sides of an equation by the same number, 74 positive or negative, leaves it an equation. Multiplying both sides of an inequality by the same positive number leaves its sense unchanged. But multiplying both sides of an inequality by the same nega- tive number changes its sense, from <.to>, or vice versa. From the first of these rules and the definition of multipli- cation by 0, namely a • = 0, we derive the following important theorem : 1. If a = b, then ac = be. 75 2. If ac = be, then a = b, uitless c = 0. The exceptional case under 2 should be carefully observed. Thus, from the true equation, 2 • = 3 • 0, of course it does not follow that 2 = 3. Zero produets. If a j^roduct be 0, one of its factors must be 0. 76 Thus, if ab = 0, either a = or b = 0. For, since ■ b is also equal to 0, we have ab = • b, and therefore a = 0, unless b = 0. § 75 Numerieal values of products. The numerical value of a prod- 77 uct of two or more factors is the product of the numerical values of the factors. Thus, |(-2)(- 3)(-4)| = |~24|=24; and|-2|.|-3|.|-4| = 24. Numerical values of sums. The numerical value of a sum of 78 two numbers is the stmt of their numerical values when the numbers are of like sign, but the numerical difference of these values when the numbers are of contrary sign. Thus, |-3 + (-5)| = |-8| = 8;and|-3|-M-5| = 3-)-5 = 8. But |2-l-(-5)| = |-3| = 3; and|-5|-2 = 3. 26 A COLLEGE ALGEBRA THE USE OF INTEGRAL NUMBERS IN MEASUREMENT 79 Measurement. We use numbers not only to record the results of counting groups of distinct things, but also to indi- cate the results of vieasur'ing magnitudes, such as portions of time, straight lines, surfaces, and so on. 80 We measure a magnitude by comparing it with some particu- lar magnitude of the same kind, chosen as a unit of measure. 81 If the magnitude contains the unit a certain number of times exactly, we call this number its measure. In particular, we call the measure of a line segment the length of the segment. Thus, we may measure a line segment by finding how many times we can lay some chosen unit segment, say a foot rule, along it. If we find that it contains the foot rule exactly three times, we say that it is three feet long, or that its length — that is, its measure — is 3. 82 The usefulness of the natural numbers in measurement is due to the fact that, by their relative positions in the natural scale, they indicate the relative sizes of the magnitudes whose measures they are. 83 Application of the negative numbers to measurement. We often have occasion to make measurements in opposite " directions " from some fixed " point of reference." Thus, we measure time in years before and after the birth of Christ, longitude in degrees west and east of Greenwich or Washington, tempera- ture in degrees helow and above zero. We may then distinguish measurements made in the one direction from those made in the other by the simple device of representing the one by positive numbers, the other by negative numbers. 84 Thus, consider the following figure : -P_4 P-s P-i /•_! O Pi P.^ Ps DIVISION AND FRACTIONS 27 Here the fixed point of reference, or origin, is 0, the unit is OPi, and the points P^, -P3, •••, P-i, -P_2j ••• are such that OPi = P1P2 = P2P3 = ■•• = P^iO = P_2P-i = •■■• Above these points we have written in their proper order the numbers of the complete scale, so that comes over 0. The distance of each point P from 0, — that is, the length of the segment OP, — is then indicated by the numerical value of the number written above it ; and the direction of P from O is indicated by the sign of that number. Thus, — 3 over P_ 3 indicates that P-3 is distant 3 units to the left of O. Moreover, the order in which the points occur on the line is indicated by the order in which the corresponding numbers occur in the scale. Points used to picture numbers. Inasmuch as there is a one- 85 to-one relation, § 2, between the system of points • • •, P_2, P_i, 0, Pi, Po, • • • and the system of numbers ■••, — 2, —1, 0, 1, 2, •••, either system may be used to represent the other. In what follows we shall frequently use the points to picture the numbers. III. DIVISION AND FRACTIONS DIVISION REPEATED SUBTRACTION The two kinds of division. There are tivo operations to which 86 the name division is applied in arithmetic and algebra. The one may be described as repeated subtraction, the other as the inverse of multiplication. There is a case in which the two coincide. We call this the case of exact divisio7i. Division repeated subtraction. To divide 7 by 3 in the first 87 of these senses is to answer the two questions : 1. What multiple of 3 must we subtract from 7 to obtain a remainder which is less than 3 ? 2. What is this remainder ? 28 A COLLEGE ALGEBRA We may find, the answer to both questions by repeatedly subtracting 3. Thus, since 7 — 3 = 4 and 4 — 3 = 1, we must subtract 3 twice, or, what comes to the same thing, we must subtract ^ X 2. And the remainder is 1. This kind of division, then, is equivalent to repeated suhtrao tion. Its relation to subtraction is like that of multiplication to addition. Observe that the four numbers 7, 3, 2, 1 are connected by the equation 7 = 3-2 + 1 And so in general, if a and b are any two natural numbers, to divide a by b, in the sense now under consideration, is to find two natural numbers, q and r, one of which may be 0, such that 88 a = bq -{- 7' and r < b. 89 We call a the dividend, b the divisor, q the quotient, and r the remainder. 90 Note. When a and h are given, two numbers q and r satisfying § 88 may always be found. Thus, if a < 6, we have q' = and r = a. If a>6, it follows from §§ 31, 35 that we can continue the sum 6 + 5 + • ■ • until it either equals a or will become greater than a if we add another b. And if q denote the number of terms in this sum, we shall have, § 41, either a = hq, ov a = bq -{- r, where rtraction ; that is, a b a + b .a b a — b — I — = } and • - = c c c c c c For if a = qc, and b = q'c, we have a + b = qc + q'c = {q + q') c. §§ 39, 47 a -\- b , a b „ ^, Hence = g + (^' = - + . § 94 c c c And similarly for subtraction. Thus, ^ + ^ = 6 + 3 = 9; and 1^ + ^ = 2-1 = 9. 3 3 3 3 Formulas for adding and subtracting quotients. These are 98 a c ad -\- be a c ad — be ■ b^d^ bd ' b~d^ ~~bd 30 A COLLEGE ALGEBRA And similarly for subtraction. 18 ,10 a , o Q . 18- 5 + 10 -3 120 ^ Thus, = 6 + 2 = 8; and = = 8. 3 5 8-5 15 99 Formula for multiplying quotients. This is a c ac h'd^bd' For if a = §&, and c = g'd, we have ac = qq' ■ bd. §§ 50, 45, 46 ^ o. c , ac „ ^ . Hence -.- = g.g=--. §94 a bd 15 6 ^ ^ ,^ ^. 15-6 90 ,^ Thus, = 5-3 = 15; and = — = 15. 3 2 '3-2 6 100 Formula for dividing one quotient by another, when this division is exact. This is a c ad ad h ' d he be For if a = 5&, c = q'd, and also q = q"q\ we have t ^ - = c/d, according as ad <, =, or > be. 1. For if - = -, then -6d = -d6. §50 d b d But -6 = a, and -d = c. §§93,94 Hence ad = 6c. 32 A COLLEGE ALGEBRA And we can show in a similar manner tliat K a/hc/d, tlien ad>bc. 2. But from all this it follows, conversely, that If ad = 6c, then a/h = c/d. For otherwise we should have either (1) a/hc/d, and therefore ad>bc. And we can show in the same way that If adbc, then a/b>c/d. 105 Enlarging the ordinal number system. But the relative ordei of ad and be in the scale is known, whether the values assigned a, b, c, d be such as make a divisible by b, and c by d, or not. Therefore, take any two natural numbers, a and b, of which b is not 0, and with them form the expression j, or a/b. If a is exactly divisible by b, let a/b denote, as heretoforCj the natural number which is the quotient of a by i ; but if not, regard a/b for the moment merely as a new symbol, read " a over &," whose relation to division is yet to be given, § 122. Then give to all such symbols a/b, c/d, and so on, the property of order already possessed by those which denote natural numbers, by supposing them arranged in accordance with the rule: a/b shall precede, coincide with, or follow c /d^ according as ad precedes, coincides with, or follows be. Or, employing the signs <, =, >, as heretofore, to mean " precede," '' coincide with," " follow," — 106 Let a/b <, =, or > c/d, according «s ad <, =, or > be. Thus, 4/5 is to precede 7/8, that is, 4/5<7/8, since 4-8<7-5. Again, 2/3 is to lie between and 1, or 0<2/3a/6. since 1 ■ 6>a-0. Again, 1/0, 2/0. and so on, will occupy the same place in our ordinal system. For 1/0=2/0, since 10= 2 0. But the rule will give no definite position to the symbol 0/0. For whatever the values of a and b, we should have 0/0= a/b, since • 6 = a • 0. adcfrecede Q. 3. Negative fractions shall be arranged tvith respect to one another {and negative integers) in accordance with the rule: — -<,=, or > — T' according as — ad <, =, or >— be. 112 The system of rational numbers. To distinguish integers and fractions alike from other numbers which we have yet to con- sider, we call them rational numbers. And we call the system which consists of all these numbers the rational system. This system possesses an important property which does not belong to its part, the integral system, namely: 113 The rational system is dense; t/iat is, betiveen every two unequal rational numbers there are other rational numbers. For let - and - be any two fractions, such that < - • We can prove h d d as follows that the fraction — lies between - and - • 2bd b d Since - <-, we have ad 38 10 Thus, between - and - we have — — TE,~ W^i' 4 6 5s • 4 • o 4o Zi DIVISION AND FRACTIONS 35 Hence, when speaking of rationals, one must carefully avoid 114 such expressions as the " next number greater or less " than a given number ; for no such number exists. To each integer there is such a next integer, but between any rational and a rational assigned as the next, there are always other rationals. Operations with fractions. In what follows let a, b, c, d denote 115 any given integers, positive or negative. In §§ 98-102 we proved that, when a/h and c/d denote mtegers, we have ^ a ^ c ad + he ' b^ d~ bd 2 -- b e ad — be d bd a c ae h d he' -r e ad . f- - = — , whe d be ad . — is an integer. But the second member of each of the equations 1, 2, 3, 4 has a meaning even when a/b and c/d are not integers. Each of them is a definite fraction of the kind defined in §§ 110, 111. Hence 1, 2, 3, 4 at once suggest an extension of the mean- ings of addition, subtraction, multiplication, and division which will make these operations applicable to fractions, namely : The sum of two fractions a/b and c/c? is to mean the fraction 116 {ad + he)/hd. The difference obtained by subtracting the fraction e /d from 117 the fraction a/b is to mean the fraction (ad — bc)/hd. The product of two fractions a/b and c/d is to mean the 118 fraction ac /bd. The quotient resulting from dividing the fraction a/b by the 119 fraction c/d is to mean the fraction ad /he Observe that these definitions are equivalent to tlie rules for reckoning with fractions given in elementary arithmetic. The commutative, associative, and distributive laws control 120 these generalized operations. Thus, a c^oc^ca^c a b d hd dh d h ^^ ' 36 A COLLEGE ALGEBRA 121 The rules of equality and inequality, §§ 71, 73, also hold good for these operations. Thus, if = -; • ^' then v = :^ • b f d f b d For if ^.f = :^.^ then aedf=cebf. §§118,100,111 b f d f Hence ad = cb, and therefore - = - • §§73,106,111 a 122 Definition of a fraction as a quotient. The fraction a/b may now be described as the number which multijjlied by b tc ill pro- duce a, that is, as the number which is defined by the equation 123 b ■b = a. a b b-- a ~b b _ 1 " ab " \-b a T b b For 'l.b^l-- = ^ = ^- = a. §§106,111,118 b 6 1 l-b \ b 124 Division the inverse of multiplication. From §§ 118, 119, it follows that area ^ a c c a 7 -^ - X - = 7 and 7 X - h- - = 7 ; b d d b b d d b in other words, that multiplication and division, as defined in §§ 118, 119, are inverse operations. Compare § 55. For, by §§ 118, 119 and §§ 100, 111, we have a c c _ad c _ adc _a dc _a a c . c _ac c _ acd _n rd _a b d d be d bed b cd b' b d d bd d bdc b dc b Hence we may describe the kind of division now before us as the inverse of multijMcation and say 125 To divide a/b Inj c/d is to find a number ivhich multiplied by c/di ivill produce a/b. By introducing fractions into our number system, we have made it possible always to find such a number, except when the divisor c/d is 0. DIVISION AND FRACTIONS 37 This is the usual meaning of division in arithmetic and algebra. It is the generalization of exact division, § 93. Reducing a fraction to its lowest terms. Irreducible fractions. 126 If the numerator and denominator of a fraction have a common factor, we can remove it from both without changing the value of the fraction. For — = -, since am ■ b = a- bm, § 106. bin b When all such common factors have been removed, the frac- tion is said to be in its lowest terms, or to be irreducible. Theorem. If a/b be an irreducible fraction, and a'/b' any 127 other fraction which is equal to it, then a' and b' are equimul- tijjles of di and b resjJectively. For since a' /b' = a/b., and therefore a'b = ab\ a is a factor of a'b. But, by hypothesis, a has no factor in common with b. Hence a must be a factor of a', § 492, 1. We therefore have a' = ma, where m is some integer. But substituting ma for a' in a'b = ab', we have mab — ab', and therefore b' = mb, § 50. Corollary. If tivo irreducible fractions are equal, their numer- 128 ators must be equal, and also their denoniinato->'S. THE USE OF FRACTIONS IN MEASUREMENT Fractional lengths. The definition of length given in § 81 129 only applies to such line segments S as contain the unit segment s a certain number of times exactly. But even if S does not contain s exactly, it may still be commensurable with s; that is, it may contain the half the third, or some other aliquot part of s exactly. In that case we define its length as follows : If a given line segment contains th e hth part of the unit segment 130 a times exactly, we say that its length is the fraction a/b. 38 A COLLEGE ALGEBRA Thus, if S contains the 10th part of s exactly 7 times, the length of S (in terms of s) is 7 / 10. 131 Note. Observe that if a/h is the length of S in terms of s according to this definition, so also is every fraction of the form ma/mh. For if S contains the 6th part of s exactly a times, it will contain the j/i6th part of s exactlj' ma times. 132 Fractions are useful in measurement for the same reason that integers are useful : namely, hy their relative positions in the rational si/stem, they indicate the relative sizes of the segments whose lengths they are. For if a/6 and c/d are the lengths of S and T in terms of s, so also are ad/bd and bc/bd, § 131 ; that is, the bdth part of s is contained in S exactly ad times, in T exactly 6c times. Hence S<, =, or >T, according as ad<, =, or >6c, that is, S<, =, or >T, according as a/b<, = , or >c/d. § 106 133 Note. It hardly need be said that the definition of length here given is equivalent to the definition oi fraction given in elementary arithmetic, and that greater or lesser fractions are there defined as fractions v^hich correspond to greater or lesser line segments or other magnitudes. 134 Rational numbers pictured by points. Fractions, as well as integers, may be pictured by points on an indefinite straight line, § 85. _4 _3 _|_2 -1 1 2 ? 3 4 >' 6 A p Thus, to construct a point, P, which will picture 7/3 in the same way that A pictures 1, we have only to start at the origin and lay off the third part of the unit OA seven times to the right. P', the corresponding point to the left of 0, is the picture of - 7/3. We proceed in a similar manner in the case of any given fraction, positive or negative. 135 All such points are arranged along the line in an order corresponding to that of the rationals which they picture. With this in mind we often s])eak of one rational as lying to the left or right of another rational, or as lying between two other rationals. IRRATIONAL NUMBERS 39 IV. IRRATIONAL NUMBERS PRELIMINARY CONSIDERATIONS Definitions. The product aa is represented by a^, read "a 136 square " ; the product aaa, by a^ read " a cube " ; the product aaa ... to n factors, by a", read " the nth power of a." In the symbols a^, a^, a", the numbers 2, 3, n are called exponents; a itself is called the base. Finding a^ from a is called squaring a; finding a^, cubing b.) finding a", raising a to ^A«3 /ith power. The operation which consists in raising a given number to a given power is also called involution. Roots and logarithms. If, as we are supposing, a is a rational 137 number, and n a positive integer, a" is also a rational number. Call this number b ; then a" = b. This equation suggests two new problems : First. To assign values to n and b, and then find a. Second. To assign values to a and b, and then find n. Thus, (1) let n = 2 and 6 = 9. The equation then becomes and we find that a = 3 or - 3 ; for both 32 = 9 and (- 3)2 = 9. Again, (2) let a = 2 and 6 = 8. The equation then becomes 2" = 8, and we find that n = 3 : for 2^ = 8. When a» = b, 138 1. a is called the nth root of b, and is expressed in terms of n and b by the symbol v^, the simpler symbol V^, read " square root of &," being used when n = 2. 2. n is called the logarithm ofh to the base a, and is expressed in terms of a and b by the symbol log„ b. 40 A COLLEGE ALGEBRA Thus, since 3^ = 9 and (— 3)- = 9, both 3 and — 3 are square roots of 9, and both may be written V9 ; but see § 139. Again, 2 is the logarithm of 9 to the base 3 ; that is, 2 = logs 9- 139 Note. Instead of representing both the square roots of 9 by the symbol V9, we may represent the positive one, 3, by v9, and the negative one, — 3, by — V9. This is the usual method of representing square roots in elementary algebra, and we shall follow it. 140 Evolution and finding logarithms. The operation by which V6 is found, when n and b are given, is called extracting the nth root of h, or evolution. The operation by which log^b is found, when a and b are given, is called Jinding the logarithm o/b to the base a. Both these operations are inverses of involution, §§ 55, 124. 141 Note. The reason that involution has two inverses, while addition and multiplication each has but one, will be seen by comparing the three equations \. a -\-h = c. 2. ab = c. 3. a^ = c. Since a + b = b -{- a, and ab = ba, the problem : Given c and 6 in 1 or 2, find a, is of the same kind as the problem : Given c and a, find b. But since a* is not equal to 6", the problem : Given c and b in 3, find a, is wholly different in kind from the problem : Given c and a, find 6. 14J? New numbers needed. We shall subsequently study these new operations in detail ; for in algebra they are second in importance to the four fundamental operations only. But the point which now concerns us is this: Tlie// necessitate further extensions of the number system. In fact, it is at once evident that vo can denote a rational number in exceptional cases only. Thus, to cite the simplest of illustrations, neither V^ nor V2 can denote a rational number. For 1. Since the square of every rational num ber is positive, no rational exists whose square is — 1. HeuCe V— 1 cannot denote a rational number. 2. No rational number exists who.se siiuare is 2. For clearly 2 is not the square of any integer, and we can show as follows that it is not the square of any fraction. IRRATIONAL NUMBERS 41 Suppose p/q to he a fraction in its lowest terms, such that (p/g)2 = 2, or p'^/q-^ = 2/1. But since p^/q~ is in its lowest terms, § 492, 2, it would follow from \his, § 128, that jj- = 2, wliich is impossible, since p is an integer. Therefore V2 cannot denote a rational number. It can be shown in the same way that if a/b be any fraction in its lowest terms, ^a/b cannot denote a rational number, unless both a and b are nth powers of integers. We are to make good this deficiency in our number system by creating two new classes of numbers : the irrational num- bers, of which v2 is one, and the imaginary mimbers, of which V— 1 is one. We shall treat the irrational numbers in the present chapter and the imaginary numbers in the chapter which follows. THE ORDINAL DEFINITION OF IRRATIONAL NUMBERS In the present chapter the letters a, b, c, and so on, will denote any rational numbers, whether positive or negative, integral or fractional. General properties of the rational system. The rational num- 143 bers constitute a system which has the following properties : 1. It is an ordinal system. 2. It is dense ; that is, between every two unequal numbers of the system, a and b, there lie still other numbers of the system. 3. The sum, difference, product, and quotient of every two numbers of the system are themselves numbers of the system, the quotient of any number by excepted. By the definitions which follow, we shall create a more extended system which possesses these same three properties, and which includes the rational system. Separations of the first kind. 1. The number J separates the 144 remaining numbers of the rational system into two classes : 42 A COLLEGE ALGEBRA the one class consisting of all rationals wliicli jyrecede (are less than) ^, the other of all rationals which folloiv (are greater than) I. Let us name these two classes of numbers C^ and C'2 respectively. Cj I ^ In the figure, the half line to the left of the point \ contains the point- pictures of all numbers in the class Ci, and the half line to the right the point-pictures of ail numbers in the class C2, § 134. From §§ 109, 111, and 113, it immediately follows that 1. Each number in C\ precedes every number in Co. 2. There is no last number in C\, and no first in C\. Thus, were there a last number in Ci, there would be numbers between it and 1/3, § 113, which is impossible since, by hypothesis, all rationals less than 1 /3 are in Ci. 145 2. Instead of thus separating the rational system into the three parts Cj, \, Co, we may join \ to C'l, so forming a class C'l' made up of C\ and \, and then say: The number 1 separates the entire rational system into two parts, C'l' and C\, such that : 1. Each number in C'/ precedes every number in Cg. 2. There is a last number in C/, namely \, but there is no first number in (\. 146 3. Or we may join \ to Co, call the resulting class Co', and then say : The number \ separates the entire rational system into two parts, Ci and Cj', such that : 1. Each number in Ci precedes every number in C2'. 2. There is no last number in Cj, but there is z. first number in Co', namely \. It is evident that each of the rational numbers defines similar separations of the rational system. 147 Conversely, if we are able, iijany way, to separate the entire rational system into two parts, Bx and B^, such that each IRRATIONAL NUxMBERS 43 number in B^ precedes every number in B2 and that there is either a last number in B^ or a first in B^, the separation will serve to distinguish this last or first number from all other numbers and, in that sense, to define it. Thus, let us assign the negative rationals to Bi and the remaining rationals to B^- There is then no last number in Bi, but is the first number in B^. And zero is distinguished from all other numbers when called the first number in B2, as perfectly as by the symbol 0. Note. Obviously there cannot be both a last number in Bi and a first 148 in B2. For there must then be rationals between these two numbers, § 113, whereas, by hypothesis, every rational belongs either to Bi or to B^. Separations of the second kind. But we can also, in various 149 ways, separate the eiitire rational system into a part A^ in which there is no last number, and a part Ao in which there is no first number. Thus, since no rational exists whose square is 2, § 142, every rational is either one whose square is less than 2, or one whose square is greater than 2. Let A2 consist of all positive rationals whose squares are greater than 2, and let Ai consist of all the other rational numbers. Then 1. Each number in A^^ precedes every number in A^. For let a\ be any number in ^1, and ai any number in An. Evidently ai a. 3. If among the rationals which precede a there are some which follow b, then a itself must follow b (or b precede a). We indicate this by the formula : a > b or b < a. 157 It thus appears that when any two different real numbers are given, we can at once infer which precedes and which follows; also, that we may always draw the following con- clusions with respect to three given real numbers, a, b, c : If a = b, and b = c, then a = c. If a < b, and b < c, then a < c. If a = b, and b < c, then a < c. 158 The real system is dense. For there are rational numbers not only between any two unequal rationals, § 113, but also between any two unequal irrational numbers, and between any two numbers one of which is rational and the other irrational, § 15 a 159 The real system is continuous. The real system, therefore, possesses the first and second of the properties of the rational IRRATIONAL NUMBERS 47 system enumerated in § 143. But it possesses an additional property not belonging to the rational system, namely : If the entire real system he separated into tivo parts, R^ and R25 such that each number in E-i precedes every number in Rjj there is eitJier a last nutnber in Ri or a, Jirst in Rj, but not both. For in separating the real system into the parts i?i and E^, we separate the rational system into two parts, A^ and A^, the one part consisting of all the rationals in i?i, the other of all the rationals in i?2. Every rational belongs either to ^1 or to A^, and each rational in Ai precedes every rational in A^. Let a be the umnber which the separation A^, An defines, §§ 147, 154. Then either a is a rational — namely the last number in ^ 1 or the first in A2, § 147, —or, if there be no last number in Ai and no first in A^, a is an irrational lying between Ai and An, § 154. 1. If a is the last number in Ai, it is also the last in jRi; for were there any number in Bi after a, there would be rationals between it and a, that is, rationals in Ai after a, which is impossible. 2. Similarly, if a is the first number in A^., it is also the first in R^. 3. If a is irrational, it must, by hypothesis, belong either to Ei or to i?2- If a belongs to Ei, it is the last number in Ri ; for were there any number in i?i after a, there would be rationals between it and a, § 158, that is, rationals in Ai after a, which is impossible. And, in like manner, if a belongs to E2, it is the first number in E^. Finally, there cannot be both a last number in Ei and a first in E2, since there would be rationals between these two numbers, § 158, that is, rationals belonging neither to Ai nor to A^ ; which is impossible. To indicate that the real system is dense and at the same time possesses the property just described, we say that it is conti7iiioiis. Theorem. A real number, a, either rational or irtiational, is 160 defined whenever a law is stated by means of ivhich the entire real system m,ay be separated into tivo parts, Rj, R2, such that each number in Ri precedes every number in R2 ; this number, a, tlien being either the last numher in Rj or the first in Rj.' This is an immediate consequence of §§ 147, 159. 48 A COLLEGE ALGEBRA APPROXIMATE VALUES OF IRRATIONALS 161 Given any irrational number, a, defined as in § 154. By the method illustrated below we can find a pair of rationals, the one less and the other greater than a, which differ from each other as little as we please. Such rationals are called approximate values of a. Let a be the irrational, V2, which lies between all positive rationals whose squares are less, and all whose squares are greater than 2. 1. We may find between what pair of consecutive integers a lies by computing the squares of 1, 2, 3, ■ • ■ successively, until we reach one which is greater than 2. We see at once that 1^ < 2 and 22 > 2. Hence a lies between 1 and 2, or 1 2; for 1.42= i.ge^ 1.52=: 2.25. Hence a lies between 1.4 and 1.5, or 1.4d', and -«i>-d, §§73,121 and therefore as - ai >d' - d, §§ 39, 121 which is impossible, since it contradicts 3. 50 A COLLEGE ALGEBRA Nor can there be two numbers, one or both of which are irrational^ Ij'ing between every ai and a^ ; for between these two numbers there would be two rationals also lying between every ai and as, § 158, which we have just shown to be impossible. 164 Note. This theorem differs from the definition of an irrational number, § 154, in that it is not here a part of the hypothesis that every rational lies either in Ai or in A2. 165 Addition. Let a and b denote any two ffiven real numbers, rational or irrational, and let aj, a2, h^ h^, denote any rationals whatsoever such that «! < a < 02 and ij < b < b^- (1) Observe that there is no last number of the kind denoted by rti or ^1, and no first number of the kind denoted by a^ or Z'a ; and that if any positive number, as 8, be assigned, it matters not how small, we can alw^ays choose «i, a^, and h^, h», § 162, so that both a, - «i < 8 and h., - h, < 8. (2) When both a and b are rational, say a = a and b = ^, we can find their sum, a + /3, by the rule of § 116 ; and it follows from (1), by § 121, that «! + ^1 < « + /3 < «2 + ^^2- Moreover, whether a and b are rational or not, it follows from (1), by § 121, that «! + Z>i < «2 + K (3) These considerations lead us to define the sum of a and b, when one or both are irrational, as follow^s : 166 The Slim of & and b, vriften a + b, is to mean that number 'which lies between all the numbers a, + bi and all the numbers a2 + bg. In other words, it is the number defined by the formula ai + bi < a 4- b < a., + b.,, where aj, aj, bj, bo denote any rationals whatsoever such that ai < a < ao and bi < b < bj. IRRATIONAL NUMBERS 61 To justify this definition we must show that there is one and but one such number a + b. This follows from § 163 ; for 1. Each ai + 6i is less than every aai and 6i>6i', and therefore ai + 61 > a/ + &i'. 3. If any positive rational, 5, be assigned, we can choose ai, aa, &i, 62, so that a2-ai<5/2 and 62 -6i<5/2, §102 and therefore {ao + 62) - (ai + bi) <8. § 121 Definition of — a. Let a, a^, a^ have the same meanings as in 167 § 165. Considerations like those in § 165 lead us, when a is irrational, to define — a as follows : The symbol — a is to mean the number defined by the formula 168 — aa < — a < — ai, where ai, a2 denote any i-ationals whatsoever sxich that ai < a < a2. It follows from § 163 that there is one and but one such number — a ; for 1. Each — ai is less than every — ai, since ai: - ^1) + &i {do — tti), and we can choose ai, a-i, h\, h-2, § 16'2, so that &2 - 61 < 5/2 ffl2 and no - ffj < 5/2 ^^i, (1) and therefore ao (62 — &i) + ^^i {(lo - ai) < S. (2) ■ We may make such a choice of ai, ao, /*i, ft.,, as follows : First take any particular number of the kind bo, as 62', and then choose Oi) Ozj so that as- ai<5/2 62'. (3) IRRATIONAL NUMBERS 53 Next, using the ao thus found, choose 61, 62, so that 62 — f'i<5/2a2, as in (1). Since 61 < h-z and therefore 5/2 62' < 5/2 61, it follows from (3) that a-z - ai<5/2 6i, as in (1). Multiplication, one or both factors negative or 0. Let a and b 172 denote any two given positive numbers. Then 1. a (— b) and (— a) b are to mean — ab. 2. (— a) (— b) is to mean ab. 3. a ■ and • a are to mean 0. Definition of 1/a. Let a be any given positive number, and 173 cii, Go, any positive rationals vk^liatsoever such that rti < a < og- Considerations like those of § 165 lead us, when a is irra- tional, to define 1/a as follows : The symbol 1/a is to viean the number defined by the formula 174 l/fl2< l/a< 1/ai, ivhere ai, ao denote any 2^ositire rationals whatsoever such that «! < a < ao. It follows from § 103 that there is one and but one such number 1/a; for 1. Each l/tto is less than every l/oi, § 106. 2. Tliere is no last l/a2 and no first 1/ai. (Compare proof, § 168, 2.) 3. Any positive 5 being given, we can choose ai, ao so that 1/ai- l/a2<5. For 1/ffi -l/«2<5, if a2-ai<5-airt2. §§106,117 But if ai denote any particular number of the kind «!, we can choose ai, ao so that ai > ai and an — ai< 5ai'-, and therefore < 5aiOo. Definition of l/(— a). Let a denote any given positive 175 number. Then l/(— a) is to mean —1/a. 54 A COLLEGE ALGEBRA 176 Division. The quotient of a by b (b not 0) is to mean the number a • 1 /b, that is, a 1 b^^b- The meaning of a • 1/b itself is known from tlie preceding definitions. When a and b are positive, it follows from §§ 171, 174 that we may also define a/b by the formula ai/62 b, so is a -f c <, =, or > b + c; also, ac <, =, or > be, if c > 0, but ac >, =, or < be, if c < 0. IRRATIONAL NUMBERS 55 Thus, if a < b, then a + c < b + c. Tor let d and d + a he any two rationals between a and b, and choose Ci so that Ci < c < Ci + a. Then, since a0, then acp such that 7"' < p"' + 5, or (/'" — p"' < 5 ; for we shall then have p'" < 7'" < b, so that p is not the last rti. But (/'" - p"' = {q- p) (7'"-' + r/'-'-^p + • • ■ + qp'"-' + p"—') §'308 <(q — p)ma2.'"'~^, if 02' be any particular a^, <5, if q -p + d/viuo' "'-''■. We ean show in a similar manner that there is no first 03. This established, it may readily be proved that a = v b. IRRATIONAL NUMBERS 57 For, since ai we have ai™ < a™ < a2'". §§ 171, 181 But b is the only number between every a^"' and every a^'". Hence a"' = b, tliat Is, a = A^b. Rules of equality and inequality. Let a and b denote any 184 positive real numbers, and 7u any positive integer. Then According as a <, =, ov > b, so is a"' <, =, or > b"', (1) and Va<,=^, or > V^. (2) We may prove (1) by repeated use of § 179. Thus, if a Vb, we should have a < or > b. Rules of exponents. Let a and b denote any two real num- 185 bers, and rn and 7i any two positive integers. Then 1. a'" • a" = a'" + ". 2. («'")" = «'"". 3. (ab)"' = a"'b"\ Thus, a^ ■ a^ = aaa ■ aa = aaaaa = a^ — a^ + - §177 (a2)3 = o2.a2.a2 =n2 + 2 + 2 =cfi-5 by 1 {abf = abab-ab = aaa ■ bbb =a^.b^ § 177 And similarly for any other positive integral values of m and n. A theorem regarding roots. Let a and b denote any positive 186 real numbers, and 7n any positive integer. Then Vrt V6 = VaZ. For ( Va • 'v^)'« = ( v'a)™ • (v^)-" = ab §§ 182, 185, 3 and (\^b)'» = ab. § 182 Hence (Va • \/6)"' = (v'^)"' and therefore 'Va-"Vb= Vab. § 184, (2) 58 A COLLEGE ALGEBRA VARIABLES AND LIMITS 187 Variables. We say that a never-ending sequence of num bers, such as a I, ito, a 3, • • • , «„, • • • , is (/ive7i or known, if the value of every particular term «„ is knowTi, or can be computed, when the index n which shows its position in the sequence is given. We often have occasion to consider variables which are supposed to be running through such given but never-ending sequences of values. Thus, I, f, |, • • •, -^, • • • is such a given never-endinfj sequence, and X is such a variable if we suppose it to be running through this sequence, that is, to be talking successively the values |, f , |, • • • • 188 Limits. As x runs through the sequence i, §, |, •••, it con- tinually approaches the value 1, and in such a manner that if we assign any positive number, as 8, it matters not how small, the difference 1 — x will ult ately become and remain less than the number so assigned. Thus, after x reaches the 100th term of the sequence, 1 — x will remain less than .01. We express all this by saying that, as x runs through the sequence \, §, |, ■ • •, it approaches 1 as limit. And in general 189 A variable x, tvhi/cJi is stqjposed to be running through a given nevei'-ending seque?ice of values, is said to approach the number a as limit, if the difference a — x tvill tdtimately become and remain numericallij less than evenj jjositive number 8 tliat we may assign. Observe that it is not enough that a - x become less than 5 ; it must also remain less, if x is to approach a as limit. Thus, if X run through the sequence i, 0, f , 0, |, 0, • ■ • , the difference 1 - X will become less than every 5 that we can assign, but it will not remain less than this 5, and x will not approach 1 as limit. In particular, a - x may become 0; that is, x may reach its limit a. IRRATIOXAL NUMBERS 59 To indicate that x is approaching the limit a, we write 190 either a: = a, read ''ic approaches a as limit/' or lim x — a, read " the limit of x is a." Whether a variable x approaches a limit or not depends 191 entirely on the character of the sequence of values through which it is supposed to be running. Thus, while x approaches a limit when it runs through the sequence i» f » f ) • • • > plainly it does not approach a limit when it runs through the sequence 1, 2, 3, 4, • • -, or the sequence 1, 2, 1, 2, ■ ■ ■. Hence the importance of the following theorems : Theorem 1. If the variable x continually increases, bid, on 192 the other hand, remains always less than some given 7iumber c, it approaches a limit. And this limit is either c or some number which is less than c. For by hypothesis there are numbers which x will never exceed. Assign all such numbers to a class i?2, and all other numbers, that is, all numbers which x will ultimately exceed, to a class '7?i. We thus obtain a separation of the entire system of real numbers into two parts, B.-^, R^, so related that each number in ill is less than every number in Ro. Obviously there is no last number in R^. Hence, § 160, there is a first number in R^. Call this number a. As x increases, it will approach a as limit. For however small 8 may be, if only positive, a — 8 belongs to the class of numbers Ri, which x will ultimately exceed. Hence x will ultimately remain between a — 8 and a, and therefore differ from a by less than 8. In the same manner it may be demonstrated that If the variable x continually decreases, but, on the other hand, 193 re7)iains always greater than some given number c, it approaches a limit. And this limit is either c or some number which is greater than c. 60 A COLLEGE ALGEBRA 194 Regular sequences. It is not necessary, however, that x should always increase or always decrease, if it is to approach a limit. Thus, X is sometimes increasing and sometimes decreasing, as it runs through the sequence — i, |, — |, 1-5, • • • ; but it approaches as hmit. We shall prove that x will or will not approach a limit, according as the sequence of values (Xi, a^, ■■•, a„, ••• through which it runs, has or has not the character described in the following definition : 195 The sequence aj, aj, • • • , a„, • • • is said to he regular, if for every positive test number 8 that may he assigned a corresjiond- ing term d^ can he found, which ivill differ numerically from every subsequent term by less than S. 1. Thus, the sequence 1.4, 1.4], 1.414, • • • (1), § 161, is regular. For the difference between the first term, 1.4, and every subsequent term is less than 1/10; that between the second term, 1.41, and every subsequent term is less than 1 / 10^ ; that between the nth term and every subsequent term is less than 1/10". Now, however small 5 may be, we can give n a value which will make 1/10" smaller still ; and if k denote such a value of n, the A;th term of 1.4, 1.41, • • • will differ from every .subsequent term by less than 5. Thus, if we assign the value 1/500000 to 5, we have 1/10^ < 5, so that the sixih term of 1.4, 1.41, • ■ • will differ from every subsequent term by less than this value of S. 2. The following sequences are also regular : h h h if, •••, (2) h h h H, •••, (3) -h tV, ••• , (4) 2, 1, 1, 1, •••. (5) Observe that in (2) each term is followed by a greater term, in (3) by a lesser term, in (4) sometimes by a greater term, sometimes by a lesser. We sometimes encounter regular sequences like (5), all of whose terms after a certain one are the same. Evidently a variable which runs through such a sequence will ultimately become constant, that is, will reach its limit. 3. The following sequences are not regular : 1,2,3,4,..., (6) hhhh"-- (7) IRRATIONAL NUMBERS 61 For in (6) the difference between a term and a subsequent one may always be' indefinitely great, and in (7) it may always be |, and therefore not less than every number, 1 for instance, that we can Formulas for regular sequences. 1. We may indicate the 196 relation between the term u,. and every subsequent term, a , by the formula, § 63 : |a^ — a^l < 8 for every p> k. (1) 2. Again, since any terms a^ there may be which are > a^. will lie between a^. and a^. + S, and any which are < a^ will lie between a,. — 8 and a^, we may also write Oi — S < flj, < % + 8 for every 2> > k. (2) 3. It follows from (2) that if some of the terms a^ are less, and some are greater than a^., the difference between two of these terms may exceed 8, but not 2 8. But we can always find a term, a„ which corresponds to 8/2 as a/, corresponds to 8. The difference between every two terms after a, will then be numerically less than 2(8/2), or 8; that is, the relation between every two of these terms will be that indicated by the formula \^P ~ *-?! < ^ ^°^' every 2^ > q > I. (3) Theorem 2. The variable x will ajxproach a limit if the 197 sequence of values ai, a2, • • •, a^, • • •, tliroiigh which it is supposed to run, is a regular sequence. For there are numbers to whose right x will ultimately remain as it runs through the sequence aj, a^, ■■ ■, a,^, ■ ■ -. (1) Thus, if 5 and at have the meanings above explained, x will remain to the right of a^. — 5 after it reaches the value a/,, § 196 (2). Assign all such numbers to a class, Ri, and all other num- bers • — that is, all numbers to whose right x will not remain — to a class, R^. 62 A COLLEGE ALGEBRA We thus obtain a separation of the entire system of real numbers into two parts, Ri, R^, so related that each number in Ri is less than every number in R^. By § 160, a definite number, a, exists at which this separation occurs. Thus, if the sequence be — h h ~ h i\' " ' '' ^^^ negative rationals constitute Ri, but and the positive rationals, E^ ; and a itself is 0. As X runs through the sequence (1), it will approach this number a as limit. For assign any positive test number, 8, it matters not how small. Since (1) is regular, we can find a term, «„„ § 196 (3), such that [dp — a^] < 8/2 for every j) > q > m. (2) But since a — 8/2 belongs to 7?i, all the values of x after a certain one will lie to the right of a — 8/2. And since a + 8/2 belongs to R^, among these values there will be some after a„, which lie to the left of a + 8/2 ; for otherwise a + 8/2 would belong to Ri, since x would ultimately remain to its right. Thus, if the sequence be — J, i, — |, j-\, • • •, and 5 = jL, all values of X after the/our^ft, Jg^ Ue between a — 5/2 and a + 5/2, that is, between - ^V and ^V- Let a^ denote such a value of x. Then a-8/2<< m, it follows, §§ 78, 178, that I a — a^\ < 8 for every 2^ > q'. In other words, after x reaches the value a^ the difference & — X remains numerically less than 8. Therefore x approaches a as limit, § 189. 198 Conversely, if x is a}ij)roaching a limit, a, the sequence of values ai, a,^, ■■■, a^, •••, through wliich it is supjjosed to run, must be regular. IRRATIONAL NUMBERS 63 For since the difference a — x will ultimately become and remain numerically less than every assigned positive number, 8, § 189, we can choose a^. so that 1^ — "t| < 2/2 and |a — a^^] < 8/2 for every p > k; whence [a^^ — w^.] < 8 for every j? > /,;. Hence the sequence a^, a^, ■■■, «„, • • • is regular, § 196 (1). We may combine §§ 197, 198 in the single statement: The sufficient and necessary condition that a variable aj^proach 199 a limit is that the sequence of values througli which it is sup- posed to run he a regular sequence. SOME IMPORTANT THEOREMS REGARDING LIMITS In the present section a and h will denote any given real numbers, and x and y variables which are supposed to run through given never-ending sequences of values. The limit 0. From the definition of limit, § 189, it imme- 200 d lately follows that 1. If the variable x will ultimately become and remain numerically less than every positive number, 8, that may be assigned, then x approaches as limit ; and conversely. 2. If X approaches a as limit, then a — x approaches as limit ; and conversely. Thus X approaches the limit 0, as it runs through the sequence \^ \^\^l•••\ and 1 — X approaches the limit 0, as x runs through the sequence |, f , |, • • • . A variable whose limit is is called an infinitesimal. Theorem 1 . Tf x = and y = 0, and A and B remain nwner- 201 ically less than some fixed number, c, as x and y vary, then Ax -I- By = 0. For assign any positive number, 5, it matters not how small. Since x = 0, x will ultimately remain numerically < 3/2 c. § 200, 1 64 A COLLEGE ALGEBRA Since y = 0,y will ultimately remain numerically < 5/2 c. § 200, 1 Hence Ax + By will ultimately remain numeiically < 2c — , .■• < 5, 2c and therefore approaches as limit, § 200, 1. Thus, if X d= and y == 0, then {xy — o)x + 2y ~0. 202 Note. This theorem may readily be extended to any finite number of variables. Thus, if X = 0, 2/ = and 2 = 0, then Ax + By + Cz = 0. 203 Theorem 2. The limit of the sum, difference, product, quotient of tivo variables which approach limits is the sum, difference, product, quotient of these limits : that is, if x and y apjyroach the limits a, and h respectively, then 1. x + 7j~a + b. Z. ory^zah. 1. x — y^a — h. 4. .r/y = a/i, unless J = 0. Tor, since a - x = and 6 - ?/ = 0, § 200, it follows from § 201 that ^(a_x) + B(6-2/) = 0. (1) The formulas 1, 2, 3, 4 may be derived from (1). Thus, 1. a + 6 - (X + y) = (a- X) + (6 - y) .: = 0, by{i) that is. x + y = a + b. § 200, 2 2. a - b - {X - y) ^ {a - x) - {b - 7/) .: =0, by(i) that is, X — y = a - b. § 200, 2 3. ab - xy ^ {a - x)b + {b - y)x .: = 0, by(i) that is, xy = ab. § 200, 2 -\- X V ' ^hlhil-lh^'-^-^^-y^k- . -0, by(l) that is, x/y = a/b. § 200, 2 204 Corollary. 7/* X = a, then x" = a". 205 Theorem 3. The limit of the nth root of a var 'able which 7/" X = a, then vx =L Va IRRATIONAL NUMBERS 65 1. When a = 0. Assign any positive number, 5. Since x = 0, x will ultimately remain numerically < 5". § 200, 1 Hence Vx will ultimately remain numerically < 5. § 184 Therefore V^ = 0. § 200, 1 2. AVhen a is notO. It follows from a later section, § 308, that x — a is always exactly divisible by Vx — Va, and that the quotient Q does not approach the limit when x = a. It therefore follows from § 203 (1), by setting ^ = 1 / Q and i? = 0, that Vx - Va =:: (X - a) / Q £= 0, that is, Vx = Va. § 200, 2 RELATION OF THE IRRATIONAL NUMBERS TO MEASUREMENT Length of a line segment incommensurable with the unit. If a 206 line segment S be incomviensurable with the unit segment s, • — • that is, if, as when S and s are diagonal and side of the same square, we can prove that no aliquot part of s, however small, is contained in S exactly — the definition of length given in § 130 does not apply to S. But there is then a definite irrational number, a, which stands in the following relation to S : The segments which are commensurable with the unit s fall into two distinct classes, those which are less than S and those which are greater than S. The rational numbers which are their lengths, § 130, fall into two corresponding classes, which we may call A^ and A 2. Every positive rational belongs either to A^ or to A^, each number in A^ precedes every number in A^, and, finally, there is neither a last number in A^ nor a first in A.^* There is, then, § 154, a definite irrational number, a, which lies between all numbers in A^ and all in ^2- We call this * For were there a last number in A^, then among the segments commensur- able with s and less than S there would be a greatest, say S'. But no svich sesrnient exists. For according to the Axiom of Archimedes, explained in the following footnote, we could find an aliquot part of s which is less than S — S' ; and the sum of S' and this part of s would be commensurable with s, less than S and greater than S'. 66 A COLLEGE ALGEBRA number a the length of S. We therefore have the following definition : 207 The length of any segment, S, incommensurable loith the unit, s, is that irrational number, a., ivhich lies between all rationals which are lengths of segments less than S and all rationals which are lengths of segments greater than S. Thus, V2 is the length of the diagonal of a square in terms of the side. 208 If the length of S in terms of s is a, we write S = as, and that whether a is rational or irrational. 209 Real numbers pictured by points. As in the figure of § 134, take any right line and on it a fixed point as origin; also some convenient unit, s, for measuring lengths. And by the distance from of any point P of the line, understand the length of the segment OP in terms of s, §§ 130, 207. We choose as the picture of any given number, a, that point, P, of the line whose distance frorn is the nutnerical value of a., the point being taken to the right or left of 0, according as a is positive or negative. If a is a rational number, we can actually construct P, § 134, On the contrary, if a is irrational, we usually cannot con- struct P. We then assume that P exists, in other words, that on the line there is a single point, P, lying between all points which picture rationals less than a and all which picture rationals greater than a.* * This is not the place for a disoussion of the axioms of geometry ; but we may mention the following beoanse of their relation to the subject of measurement now under consideration. 1. Axiom of Archimedes. 1/ s and S denote two line segments such that sS. 2. Axiom of continuity. Tf all the points of a right line be separated into two classes, R, and K,, siicli that mrl, j,nint in'R^ lies to the left of every point i?iR,, thej-e is either a last jmiiit in K, nrniirst in R,. (i) The Axiom of Avchiincdcs is invnlved in the assumption that every line segment can be measunid. For the hrst step in measuring S in terms of s is to find an integer, m, such that (/)X- I) ss, and therefore PQ>s/m. But this is impossible. For we can select from B a number b and from a number c such that c—ba. For, by §§ 108, 176, 178, we can choose an integer, m, such that m>a/b and therefore ??ib>a. The real system would not possess this property — at least not without a sacrifice of some of its other properties — were we to invent more than one irrational for a separation of the rational system of the kind described in § 154. Thus, if every rational is either an o, or an a^, and a^<\)5. For it would then follow that c — b>6/m, which is impossible since c-bTi, and Ti = M, we have, by case 1, Si>ai6is. Similarly S2 < 02628- Hence ai6is + i) = (a . - 6 . 0) + (a • -f 6 • 0) z = 0. Division. We define the quotient of a. + hi by c + di as the 228 complex number which multiplied hy c -\- di will give a + hi. When c + di is not 0, there is one and but one such number, namely, that in the second member of the equation a + hi ac 4- hd he — ad . + c + di c' + d-' ' c^ + d"" But when c + di is 0, no determinate quotient exists. For the product of the right member of this equation by c + di is a + 6t, as the reader may easily verify by aid of § 226. We discover that this number is the quotient as follows : If a number exists which multiplied by c + di will give a + 6i, let it be « + 2/i. Then (x + yi) (c + di) = a + hi. (1) or {ex — dy) + {dx + cij)i = a + hi. (2) and therefore ex — dy = a and dx + aj = b. (3) § 225 Solving this pair of equations for x and y, we obtain ac + bd be — ad . ., , „ „ ,,, X — , y — J unless c^ + d^ = 0. (4) 74 A COLLEGE ALGEBRA Moreover, since (4) is the only pair of walues of x and y which will satisfy (o), the corresponding number x + yi is the only number which multiplied by c + di will give a + hi. It is evident from (4) that when (fl -f # = our definition of quotient is meaningless. But if c- + d^ = 0, both c = and d — 0, since other- wise we should have a positive number equal to 0. And if c — and d = 0, the divisor c + diis 0. 229 The commutative, associative, and distributive laws. The oper- ations just defined evidently include the corresponding opera- tions with real numbers. Like the latter, they conform to the commutative, associative, and distributive laivs. Thus, (a + a'i) (b + b'i) = ab - a'b' + (ah' + a'b) i, (1) and (6 + h'i) {a + a'i) =ba - b'a' + (b'a + ha') i. (2) But the second members of (1) and (2) are equal, by § 177. Hence (a + a'i) (b + b'i) = (6 + b'i) (a + a'i). And the remaining laws may be established similarly. 230 Rules of equality. Let a, b, c denote any complex numbers. 1. If a = b, then a + c = b + c. 2. If a + c = b -f c, then a = b. 3. If a = b, then ac = be. 4. If ac = be, then a = b, unless c = 0. 1. For let a = a + a'i, b = ?> + b'i, and c = c + c'i. If a + a'i = b + b'i, then a = b and a' = h'. § 225 Hence a -{■ c = h -^ c and a' -\- c' = h' + c', § 178 and therefore (a -t- c) -f- (a' + c') t = (?) + c) + (6' + c')i, §225 that is, a + c = b + c. § 226 2. If a + c = b + c, we have a + c + (— c) = b + c + (— c), byl and therefore ' a = b. § 226 8 and 4. The proofs of these rules are similar to those of 1 and 2 respectively. / IMAGINARY AND COMPLEX NUMBERS 75 Corollary. If a product vanish, one of its factors must vanish. 231 This follows from § 230, 4, by the reasoning of § 76. Absolute value of a complex number. The positive real number 232 Vo.- + b' is called the absolute or numerical value of a -{- bi and is represented by \a + bi\. Hence, by definition, Thus, \2 -\- i\ = V'4 + 1 =: Vs. When 6 = 0, this definition of numerical value reduces to that already given for real numbers, § 63. For a geometrical interpretation of this definition see § 239. We also say of two complex numbers that the first is numer- 233 ically less than, equal to, or greater than the second, according as the absolute value of the first is less than, equal to, or greater than that of the second. Thus, 2 + 3 lis numerically greater than 3 + i. For |2 + 3i| = VT3 and |3 + i| = VlO, and Vl3>VlO. Theorem 1 . The absolute value of a product of two complex 234 numbers is equal to the product of their absolute vcClues. Let the numbers be a = a + a'i and b = 6 + h'i. "Since ab = a6 - a'b' + {ah' + a'b) i, § 226 we have [ab| = V(a& - a'b')^ + {ah' + a'h)"^. § 232 But on carrying out the indicated operations it will be found that {ah - a'h')^ + {ah' + a'6)2 = (a^ + a'2) (62 + b"^). Hence ^{ab - a'b')^ + {ah' + a'by^ = Va2 + a'^ ■ Vb'^ + 6'2. § 186 That is, |abl = la|.lbl. Theorem 2. The absolute value of a sum of two complex 235 numbers cannot exceed the sum of their absolute values. Employ the same notation as in § 234. Then Va2 + a'2 + Vfts + 6'2^ V(a +^6)2 + {a' + b')^ (1) if a2 -f- o'2 + 62 + 5'2 + 2 V(a2 + a'2) (62 + 5'2) > a2 + 62 + a'2 + 6'2 + 2 (a6 + a'6') § 184 76 A COLLEGE ALGEBRA .-. if V'(a2 + a'2) (62 + b'-^) ^ab + a'b' § 178 .-. if d^b'^ + a'26'2 + a26'2 + a'262 > a262 + a'26'2 + 2 aba'b' § 184 .-. if a26'2 + a'262 > 2 aba'b' § 178 .-.if (a6'-a'6)2>0. (2) §178 But (2) is always true since tlie square of every real number is positive (or 0). Hence (1) is always true — which proves our theorem. Thus, 12 + t| = V5 and |l + 3i| = VlO. But |(2 + i) + (l + 3i)| = 5, and 5arallelo- ^C gram OACB. Then C is the graph o/ a + b. For, draw the perpendic- ulars BB, AE, CF, AG. Then a, a', h, b' are the lengths of OE, EA, OD, BB respectively, and the trian- gles OBB and AGO are congruent. - OE + EF = OE + OB = a + b, in length, -. FG + GC = EA + BB = a' + b', in length. ;raph oi a + b + (a' + b') i, or a + b, § 220, 1. D E F Y' Hence OF - and EC -- Therefore C is the When O, A, B are in a straight line (and always) C may be found by drawing AC equal in length and direction to OB. Since OC + «. 2. a + (/> + c) = (a + b)+c. 3. ab = ba. 4. a (be) = (ab) c. 5. a(b + c)= ab -[- ac. It may therefore be said that the formulas 1-5 are practi^ cally all the definition of addition and multiplication that we either need or can use when combining literal expressions ; and the like is true of the remaining operations of arithmetic. PRELIMINARY CONSIDERATIONS 81 Thus, to add 2x + Sy and 4 a; + 5 ?/ merely means to find the simplest form to which the expression 2x + Sy + {4:X + 6y) can be reduced by applying formulas 1-5, and adding given numbers. We thus obtain 2a: + .3?/ + (4x + 5?/) = 2x + Sy + ix + ^y by 2 = 2 X + (3 2/ + 4 X) + 5 ?y = 2 X + (4 X + 3 y) + 5 ?/ by 2 and 1 = 2x + 4x + 3y + r,y = (2x + 4x) + (3?/ + 5^) by 2 = (2 + 4) X + (3 + 5) y = Qx + Sy, the sum required, by 3 and 5 THE FUNDAMENTAL RULES OF RECKONING In accordance with what has just been said, we may regard 246 addition, subtraction, multiplication, division, involution, and evolution as defined for algebra by the following rules and formulas, which we shall therefore call the fundamental rules of reckoning. In these formulas — which have been established for num- bers of all kinds in the first part of the book — the letters a, b, c denote any finite numbers whatsoever, and the sign of equality, =, means " represents the same number as." Addition. The result of addinrj 6 to a is the expression 247 a + b. We call this expression the sum of a and b. It has a value, and but one, for any given values of a and b. In particular, a + = + a = a. Addition is a commutative and an associative operation ; that 248 is, it conforms to the two laws, §§ 34, 35 : a + b = b + a, a-\-(b -\-c) = (a + b)+c. The following rules of equaUty are true of sums, § 39 : 249 If a = b, then a -\- c = b -\- c. If a -\- c — b -\- c, then a = b* Subtraction. 'Y\\\'S> \?>\)ci% inverse of addition, %b^. Given any 250 two numbers, a and b, there is always a number, and but one, * I,ater we shall fiud that this rule does not hold good when c is infinite. 82 A COLLEGE ALGEBRA from which a can be obtained by adding b. We call this num- ber the remainder which results on subtracting b from a, and we represent it by the expression a — b. Hence, by definition, (a — b) -\- b = a. In particular, we represent — b by — b. 251 Multiplication. The result of multiplying a by & is the expression ab. We call ab the •product of a by b. It has a value, and but one, for any given values of a and b. In particular, a • = • a = 0, whatever finite value a may have. When 5 is a positive integer, ah — a ■{- a ^ •■ • tob terms. 252 Multiplication is ^.commutative and an associative operation, and it is distributive with respect to addition ; that is, it con- forms to the three laws, §§ 45-47 : ab = ba, a (be) = (ab) c, a (b + c) = ab -\- ac. 253 The following rules of equality are true of products, §§ 75, 76 : If a — b, then ac = be. If ac = be, then a = b, unless c = 0* If ac = 0, then a = 0, or c = 0. 254 Division. This is the inverse of multiplication, § 124. Given any two numbers, a and b ; except when b is 0, there is always a number, and but one, from which a can be obtained by multiplying by b. We call this number the quotient which results on dividing a by b, and we represent it by the expres- sion - or a/b. Hence, by definition, {j\b = a, except when i = 0. 255 Involution. This is a case of repeated mnlflplicatio7i. We represent the continued product a ■ a ■ ■ ■ to n factors by a" and call it the 7ith power of a. * Later we shall find that this rule does uot hold good when c is infinite. PRELIMINARY CONSIDERATIONS 83 In the symbol a", n is called the exponent, and a the hase. Involution, or raising to a power, conforms to the following 256 thi-ee laws, called the laws of exponents, § 185 : a™ • a" = a"' + «, (a'")" = a""*, (ciby = a"b\ The following i-ules of equaliti/ are true of powers, § 184 : 257 If a = b, then a" = ft". If a" = Zi-, then a = h, or a = — b. The second of these rules and the general rule of which it is a particular case will be demonstrated later. Evolution. This is one of the inverses of involution, §§ 138, 258 140. Given any ^jostVtiJe number a, there is a positive number, and but one, whose ?^th power is equal to a. We call this number the 2)ri7ic ip a I nth root of a, and we represent it by Va, or, when w = 2, by Va. Hence, by definition, (>/«)" = a. But this positive number, -Va, is not the only number whose 7ith. power is equal to a. On the contrary, as will be shown subsequently, there are n different numbers whose nth powers are equal to a ; and this is true not only when a is positive, but also when a is any other kind of number. When a is positive and n is odd, the principal nth root of — a is — va. On the reversibility of the preceding rules. We have called 259 certain of the rules just enumerated rules of equaliti/ ; we may call the rest rules of combinatio7i. Observe that all the rules of combination and the rules of equality for sums are reversible, but that the rules of equality for products and powers are 7iot completely reversible. Thus, according to the distributive law, a{b + c) = ab + ac, which is one of the rules of combination, we can replace a{b + c) by ab + ac, or reversely, ab + ac by a (6 + c). 84 A COLLEGE ALGEBRA Again, if a = 6, we may always conclude that a + c t=b + c, and reversely, that, if a + c = 6 + c, then a = b. But while, if a = 6, we may always conclude that ac = bc; on the con- trary, if ac = be, we can conclude that a = b only when we know that c is not 0. And while from a = 6 it always follows that a^ = b^, from a^ = 62 jt only follows that either a = b or a = — b. 260 The rules of inequality. The formula a ^ b means " a is not equal to b." Of two given unequal real numbers, a and b, the one is alg/e- hraically the greater, the other algebraically the lesser, § 62. If a is the greater and b the lesser, we write a > b or b < a. In particular, we have a > or a < 0, according as a is positive or negative. 261 For any given real numbers a, b, c, we have the rules, §§178, 184: 1. If a = b and b = c, then a = c. If a = b and b < c, then a < c. If a < b and b < c, then a <. c. 2. According as a <, =, or > Z», so is a + c <, =, or > ^> -f- r, and ac <, =, or > be, if c>0; but ac >, =, or <. be, if c < 0. 3. When a and b are j)ositive, according as a <, =, or > J, so is a" <, =, or > h"; and Va <, =, or > V^. As has already been pointed out, the rules under 2 and 3 which involve only the sign = hold good of imaginary num- bers also. This is also true of the rule : If a = 6 and b = c, then a = c, which we may call the general rule of equality. PRELIMINARY CONSIDERATIONS 85 ADDITIONAL ALGEBRAIC SYMBOLS Besides the symbols whose meanings have been explained 262 in the preceding sections, the following are often employed in algebra : 1. Various signs of aggregatio7i, like the parentheses () employed above, and [ ], ^ ^, to indicate that the expression included by them is to be used as a single symbol. 2. The double signs ±, read " plus or minus," and q:, read " minus or plus." Thus, in a ± 6 =F c, which means a -{-h — c or a — h -\- c, the upper signs being read together and the lower signs similarly. 3. The symbol .'. for hence or therefore. 4. The symbol • • • for and so on. 5. Also, •.• for since ; > for not greater than; <^ for not less than ; \ for greater or less than. ALGEBRAIC EXPRESSIONS Any expression formed by combining letters, or letters and 263 numbers, by the operations just described, is called an algebraic expression. Note. The number of times that an operation is involved in such 264 an expression may be limited, as in 1 + x + x^, or unlimited, as in 1 + X + x2 + • • •, supposed to be continued without end. In the one case we say that the expression is finite, in the other, infinite. For the present we shall have to do with finite expressions only. It is customary to classify algebraic expressions as follows, 265 according to the manner in which the variable (or unknown) letters under consideration occur in them : An expression is called integral if it does not involve an 266 indicated division by an expression in which a variable letter occurs ; fractional, if it does. 86 A COLLEGE ALGEBRA Thus, ifx and y are the variable letters. , but a, b, c constants, en ax'- -\-bx + c and l^v, are integral. t y + 1 X and 2 + x 1 -X are fractional. 267 An expression is called rational if it does not involve an indicated root of an expression in which a variable letter occurs; irrational, if it does. Thus, a + Vbx is rational, but Vy + Vy -x is irrational. 268 Notes. 1. In applying these terms to an expression, we suppose it reduced to its simplest form. Thus, Va;2 + 2 xy + y'^ is rational, since it can be reduced to the rational form x + y. 2. The terms integral, rational, and so on, have nothing to do with the numerical values of the expressions to which they are applied. Thus, X + 2 is a rational integral expression, but it represents an integer only when x itself represents one. It represents a fraction for every fractional value of x, and an irrational number for every irrational value of X. 269 When an algebraic expression A is made up of certain parts connected by + or — signs, these parts with the signs imme- diately preceding them are called the terms of A. Thus, the terms of the expression I + m a + a^c-{b + c) + [d + e]~ {f + g] + h + i+j\- -^ -\-k\ ^ '^P are a, a-c, — (b + c), and so on, those of the terms which themselves consist of more terms than one being enclosed by parentheses or some equivalent sign of aggregation, §262, 1. 270 Integral expressions are called monomials, binomials, trino- mials, and in general polynomials, according to the number of their terms. 271 In any monomial, the product of the constant factors is called the coefficient of the product of the variable factors. Thus, in 4 ab'^xhj^, 4 ab'^ is the coefficient of x^y*. At the same time, it is proper to call any factor the coefficient of the rest of the product. PRELIMINARY CONSIDERATIONS 87 In every monomial the coeflBcient should be written first. When no ooefBcient is expressed, it is 1. Thus, 1 is the coefficient of x'^y. Like terms are such as differ in their coefficients at most. 272 Thus, — 2 z-y and hxP^y are like terms. The degree of a monomial is the sum of the exponents of such 273 of the variables under discussion as occur in it. Thus, if the variables are x and y, the degree of 4 ah'^x^y* is seven ; that of ax3, tkree ; that of 6, zero (see § 595). The degree of a polynomial is the degree of its term or terms 274 of highest degree ; and the degree of any integral expression is that of the simplest polynomial to which it can be reduced. Thus, the degree of az^ + hz-y + cy^ + dx- + ey +/ is three; and the degree of (x — 1) (x — 2) is two. It is convenient to arrange the terms of a polynomial in the 275 order of their degrees, descending or ascending, and if there are several terms of the same degree, to arrange these in the order of their degrees in one of the variables. This order is observed in the polynomial given in § 274. A polynomial is said to be homogeneous when all its terms 276 are of the same degree. Thus, 5 x^ — 2 x-y + 4 xy^ + y"^ is homogeneous. Polynomials in a single variable. Rational integral expres- 277 sions in a single variable, as a-, are of especial importance. They play much the same role in algebra as integral numbers in arithmetic. In fact we shall find that they possess many properties analogous to those of integral numbers. They can always be reduced to the form of a polynomial in x, that is, one of the forms : UqX -t «!, a^pc!^ + a-iX + a.,, a^fc^ + a^x"^ + a^x + ctg, • • •, or, as we shall say, to the form : ttfyX" + a^x"-^ -f- a^x"-^ 4 h a„_iX + a„, 88 A COLLEGE ALGEBRA where n denotes the degree of the expression, and the dots stand for as many terms as are needed to make the entire number of terms n + 1. The coefficients Uo, cii, ■■■, denote constants, which may be of any kind. In particular, any of them except ao may be 0, the polynomial then being called incomplete. Observe that in each term the sum of the subscript of a and the exponent of x is the degree of the polynomial. Thus, in 5 x^ — x* + 2 x2 + X — 3, we have n = 6, a© = 5, Ui = 0, a-i = 0, Cs = — 1, 04 = 2, as = 1, tte = — 3. 278 Functions. Clearly an algebraic expression like cc + 2 or x^ + y, which involves one or more variables, is itself a vari- able. We call X + 2 o, function of x because its value depends on that of x in such a way that to each vahie of x there corresponds a definite value of x -\- 2. For a like reason we call x^ + y a function of x and y and, in general, we call every algebraic expression a, function of all the variables which occur in it. 279 What we have just termed integral or fractional, rational or irrational expressions in x, x and y, and so on, Ave may also term integral or fractional, rational or irrational functions of X, X and y, and so on. 280 We shall often represent a given function of x by the symbol /(a;), read "function of x." We then represent the values of the function wliich correspond to x = 0, 1, b, hj /(O), /(I), f(h). Thus, a fix) = X + 2, we have /(O) = 2, /(I) = 3, f(b) = 6 + 2. And, in general, if /(x) represent any given expression in x, f(b) represents the result of substituting b for x in the expression. When dealing with two or more functions of x, we may represent one of them by f(x), the others by similar symbols, asF(x),(x),yf;(x). In like manner, we may represent a function of two vari- ables, x and y, by the symbol f(x, y), and so on. PRELIMINARY CONSIDERATIONS 89 1. What is the degree of x-yz^ + 2 x^y*z^ + 3 x'^y'^z^ with respect to X, y, and z separately ? with respect to y and z jointly ? with respect to X, 2/, 2 jointly ? 2. What is the degree of (x + 1) (2 x^ + 3) (x^ - 7) ? 3. Given 3 x'^ + x^ — 4 x* + x^ — 12 ; what are the values of n, ao, Oi, • • in the notation of § 277 ? 4. If /(x) = 2x3 - x2 + 3, find/(0), /(- 1), /(3), /(8). 5. If /(x) - {x2 - 3x + 2)/(2x + 5), find/(0),/(- 2),/(6). 6. If /(x) = X + V^ + 3, find /(I), /{4), /{5). 7. If /(x) = 2 X + 3, what is /(x - 2) ? /(x^ + 1) ? 8. If /(x, y) = x3 + X - ?/ + 8, find the following : /(O, 0), /(I, 0), /(O, 1), /(I, 1), /(_2, -3). IDENTICAL EQUATIONS OR IDENTITIES If A denotes the very same expression as B, or one which 281 can be transformed into B by the rules of reckoning, §§ 247- 258, we say that A is identically equal to B. The notation A = B means ".4 is identically equal to 5." Thus, X (X + 2) + 4 is identically equal to x2 + 2 (x + 2). For X (x + 2) + 4 = (x^ + 2 x) + 4 = x2 + (2x + 4) = x2 + 2(x + 2). §§248,252 We call A = 5 an identical equation, or identity. Hence An identical equation A = B is a statement that a first expres- 282 sion, A, can be transfornied into a second expression, B, by means of the rules of reckoning. In particular, an identical equation like 283 3-8 + 2 = 4 + 7-14 in which no letters occur, is called a numerical identity. 90 A COLLEGE ALGEBRA The following very useful theorem is implied in § 282. 284 Theorem. If two polynomials in x are identically equal, their corresponding coefficients are equal ; that is, If aox" + aix°-^ -\ h a„ = boX" + bjx"-' H [- b„, tlien ao = bo, Ei = bi, • • •, a„ = b„. For were these coefficients different, the polynomials would be different as they stand and the first could not be transformed into the second by the rules of reckoning. Thus, if ax2 + 3 X - 3 = 2 x2 + 6x + 0, then a = 2,b = S, c = - S. If, instead of being constants, the coefficients ao, Oi, • • ■ , bo, 6i, • • • denote algebraic expressions which do not involve x, it follows from the identity aoX" + UiX"-^ + ■ • • = boX" + bix"-^ + • • • tliat ao = 6o, ai ~bi, ■ ■ •, in other words, that the expressions denoted by corresponding coefficients, ao and bo, and so on, are identically equal. 285 A similar theorem holds good of two identically equal poly- nomials whose terms are products of powers of two or more variables with constant coefficients. Thus, if a + 6x + cy + dx2 + exy -\- fy- + ■ ■ ■ = a' + b'x + c'y + d'x'^ + e'xy ■\-f'y'^ + • • • , then a = a', b = b', c = c', d = d', e = e', f = /', • • • . 286 Properties of identical equations. In algebraic reckoning we make constant use of the following theorems : Theorem 1 . If A = B, then B = A. For the process by which A may be transformed into B is reversible since it involves only rules of combination, § 259. But the reverse process will transform B into A. Thus, we may reverse the transformation in the example in § 281. For x2 + 2 (x + 2) = x2 + (2 X + 4) = (x2 + 2x)-f-4 = x(x + 2) + 4. §§248,252 Theorem 2. If A = C and B = C, then A = B. For since B=C, we have C = B. by Theorem 1 Hence A = C and C = B, and therefore A = B. PRELIMINARY CONSIDERATIONS 91 Thus, since x (x + 2) + 4 = a;2 + 2 a; + 4, §§ 248, 252 and x2 + 2 (X + 2) = x2 + 2 X + 4, §§ 248, 252 we have . x (x + 2) + 4 = x^ + 2 (x + 2). Theorem 3. A71 identity remains an identity when the same operation is ijerformed on both its members. This follows from the rules of equality, §§ 249, 253, 257. Thus, a A = B, then A + C = B + C, and so on. On proving identities. To prove of two given expressions, 287 A and B, that .1 = B, it is not necessary actually to transform A into B. As § 286, 2, shows, it is sufficient, if ive can reduce A and B to the same form C. The following theorem supplies another very useful method. If from a supposed identity, A = B, a known identity, C = D, 288 can be derived by a reversible process, the stipposed identity A = B is true. For since the process is reversible, A = B can be derived from C = D. Therefore, since C = D is true, A = B is also true. Example. Prove that a + b — b is identically equal to a. If we suppose a + b — b = a (1) it will follow that [{a + b) - b] + b = a + b. (2) § 249 But (2) is a known identity, § 250, and the step (1) to (2) is reversible. Therefore (1) is true. That it is not safe to draw the conclusion A = B unless the process from ^ = J5 to C = D is reversible may be illustrated thus : If we suppose x = — x (1) it will follow that x^ = {- x)2. (2) Here (2) is true, but it does not follow from this that (1) is true, since the step (1) to (2) is not reversible, § 259. And in fact, (1) is false. Identity and equality. It is important to remember that 289 identity is primarily a relation of form rather than of value. At the same time. If A and B are finite expressions, and A = B, then A and B have equal values for all valves of the letters which may occur in them. 92 A COLLEGE ALGEBRA For, by hypothesis, we can transform A into B by a limited number of applications of the rules a + b — b + a, and so on. But a -\- b and b + a have equal values vi'hatever the values of a and b ; and so on. The reason for restricting the theorem to finite expressions will appear later. Conversely, if A and B have equal values for all values of the letters in A and B, then A = B. This will be proved subsequently. Hence in the case of finite expressions we may always replace the sign of identity of form, = , by the sign of equality of value, =, and when A = B, write A = B. We shall usually follow this practice. This use of the sign = is to be carefully distinguished from that described in § 325. ON CONVERSE PROPOSITIONS 290 Consider a proposition which has the form If A, then B, (1) or, more fully expressed : If a certain statement, A, is true, then a certain other statement, B, is also true. Thus, If a figure is a square, then it is a rectangle. If a; = 1, then x - 1 = 0. 291 Interchanging the hypothesis, A, and the conclusion, B, of (1) we obtain the converse proposition If B, then k* (2) Thus, the converses of the propositions just cited are : If a figure is a rectangle, then it is a square. If z - 1 = 0, then z = \. 292 As the first of these examples illustrates, the converse of a true proposition may he false. * A proposition like If A and B, then C, which has a flnvble hypothesis, has tv)0 converses: n.imely, If C aixl B, then A, and If A and C, then /?. Simi- larly, if there be a triple hypothesis there are three converses; and so on. THE fu:ndamental operations 93 But the converse of a true proposition : li A, then B, is 293 always true when the process of reasoning by which the con- clusion, B, is derived from the hypothesis, A, is reversible; for by reversing the process we may derive A from B, in other words, prove If B, then A. The method of proving a proposition by proving its converse by a reversible process is constantly employed in algebra. An illustration of this method has already been given in § 288. When a proposition : li A, then B, is true, we call A a suffi- 294 cient condition of B, and B a necessary condition of A. Tims, the proposition If x — 1, then (a; — 1) (a; — 2) = is true. Hence x = 1 is a sufficient condition tliat (x — 1) (x — 2) = 0, and (x — 1) (x — 2) = is a necessary condition that x = 1. When both the proposition If A, then B, and its converse 295 If B, then A, are true, we say that A is the sufficient and neces- sary condition of B ; and vice versa. Thus, both (1) If X = 1, then x - 1 = 0, and (2) If x - 1 = 0, then X = 1 , are true. Hence x = 1 is the sufficient and necessary condition that X — 1 = ; and vice versa. II. THE FUNDAMENTAL OPERATIONS ADDITION AND SUBTRACTION Sum and remainder. Let A and B denote any two algebraic 296 expressions. By the sum of A and B, and by the remaiyider to be found by subtracting B from A, we shall mean the simplest forms to which the expressions A -\- B and A — B can be reduced by aid of the rules of reckoning, §§ 247-258. Some useful formulas. In making these reductions the fol- 297 lowing formulas are very serviceable, namely : 1. a + 1) — c = a — c -\-h. 2. a — (b + c) = a ~ b — c. 3. a + (b — c) = a -\- b — c. 4. a — (b — c) = a — b -[- c, 5. a (b — c) — ab — ac. 94 A COLLEGE ALGEBRA These formulas may be described as extensions of the com- mutative, associative, and distributive laws to subtraction. We may prove 1 and 2 by aid of the rule, § 249 : Two exjyressions are equal if the resxdts obtained by adding the same expressio7i to both are equal. 1. a-\-h — c = a — c-\-b. For the result of adding c to each member is a + Z». Thus, [(a + 6) - c] + c = a + 6, § 250 and (a-c)Jrb + c^{a-c) + c + b=:a-\-h. §§248,250 2. a — (b -{- e) = a — b — c. For the result of adding b + c to each member is a. Thus, [a - (b + c)] + (b + c) = a, § 250 and a~b — c + (b + c) = a~b — c + c + b = a-b + b = a. §§ 248, 250 We may prove 3, 4, 5 as follows : Since h=(b-c)+c, § 250 we have, 3. a + b — c = a +[(b — c)-\- c"]— c = a+(b-c) + c-c § 248 = a+(b-c). by 1 and § 250 4. a — b + c = a— [(b — r) + f ] + c = a—(b — c)—c + c by 2 = a-(b-c). §250 5. ab — ac = a [(i — c) -\- c']— ac = a(b-c)+ ac - ac § 252 ^a(b-c). by 1 and §250 Observe that it follows from § 248 and the formulas 1-4 that a series of additions and subtractions may be performed in any order tvhatsoeuer. THE FUNDAMENTAL OPERATIONS 95 Thus, a-b + c~d + e = a + c-b-d + e, byl = a + c - (b + d) + e - a + c + e - (b + d), by 2 and 1 = a + c + e — b — d. ' by 2 Rules of sign. The " rules of sign " which follow are par- 298 ticular cases of the formulas 3, 4, 5 just established. 1. a -\- (— c) = a — c. 2. a — (— c)= a -\- c. 3. a(— c) = — ac. 4. (— a) (— c) = ac. We obtain 1, 2, 3 at once by setting 6 = in § 297, 3, 4, 5 respectively. We may prove 4 as follows : (-a)(-c) = (-a)(0-c) = (-a)0-(-a)c §297, 5 = — {— ac) = ac. by 2 and 3 Rule of parentheses. From the formulas § 248 and § 297, 299 2, 3, 4, we obtain the following important rule : Parentheses preceded by the + sign may he removed ; paren- theses preceded by the — sign may also be removed, if the sign of every term within the parentheses be changed. Parentheses may be introduced in accordance with the same rule. Thus, a + 6 — c — d + e = a + 6 — (c + d — e). To simplify an expression which involves parentheses within parentheses, apply the rule to the several parentheses successively. Thus, a -\b - [c - {d - e)']\ = a - h + [c - {d - e)] = a — b-\-c — {d — e) = a — h + c — d-\-e. Of course the parentheses may be removed in any order; but by beginning with the outermost one (as in the example) we avoid changing any sign more than once. 96 A COLLEGE ALGEBRA 300 Rules for adding and subtracting integral expressions. From the formulas of §§ 248, 252, 297 we derive the rules : To odd (or subtract) two like terms, add {or subtract) their coefficients, and affix the common letters to the result. To add two or more polynomials, write all their terms in succession with their signs unchanged, and then simplify by combining like terms. To subtract one polynomial from another, change the sign of every term in the subtrahend and add. Example 1. Add 4 ab"^ and — bah- ; also subtract — 5 alfi from 4 aV^. We have 4 aft'^ + ( - 5 ah"^) = (4 - 5) at^ = - 0*2 ; § 248 and 4a62-(-5a62) = [4-{-5)]rt62 = 9a52. §297,5 Example 2. Add x^ + ax"y + 2 ah^ and hx-y — 5 ah^. We have x^ + ax"y + 2 a&3 + (6x2?/ _ 5 aW) = x3 + ax^y + 2 aft' + bx^^y -6ab^ § 299 = x^ + ax'^y + bx'^y + 2ab^-5 ab^ § 248 = x3 + (a + 5) x-'-y - 3 a/A §§ 252, 297, 5 Example 3. Subtract 2 a"b - ab- + b^ from a^ + a-b + b^. We have a' + a"b + b^ - (2 a"b - ab" + b^) = a' + a:-b + b^-2a"-b + ab"- - b^ § 299 = a^ - a-b + ab-. §§ 252, 297 When the polynomials to be added (or subtracted) have like terms, it is convenient to arrange these terms in columns and then to add (or subtract) by columns. Example 4. Add a* + a^b - 2 a-b"- - ¥ and ab"^ + 3a262 - a% and subtract 5 a^b^ — ab^ from the result. We have a* + a^?) - 2 a%"- - -¥ -a^b + S aW- + a65 - ba"-b"- + a63 0*^ - 4 a262 -1- 2a65- ^ THE FUNDAMENTAL OPERATIONS 97 EXERCISE n 1. Addiax-y, —Qax-y, iJbx'^y, and — Sbx-y. 2. Add 7 a2 + 2 a - 62, 3a + b'^-2a^, and b^-4a-ia^. 3. Add 3x2 _ 5x + 6, x^ + 2x - 8, and - 4x2 + 3x - 7. 4. Add 4 a-3 + a^b - 5 6% 5 a^ - 6 a62 - a26, i a^ + 10 63, and 6 6' - 15 a62 - 4 a26 - 10 a^. 5. Subtract 4a — 26 + 6c from 3 a + 6 — c. 6. Subtract 2 x2 - 5 x + 7 from x^ + 6 x2 + 5. 7. What must be added to a^ + 5 a26 to give a^ + b^? 8. From x^ + y^ — 6x + 5y take the sum of -2x2-Gx + 7?/-8 and x^ + 2x'^ - by + 9. 9. Simplify - (a + 6) + ^ - a - (2 a - 6) ^ - 6 (a - 4 6). 10. Simplify 6 X - ^ 4 X + [2x- (3x + 5x + 7 - l) + 3] -8^. 11. Simplify 2 a - [4 a - c + ^ 3 a - (4 6 - c) - (6 + 3 c) ^ - 6 c]. 12. Subtract x - (3 ?/ + 2 z) from z - [3 x + (y + 52)]. 13. To what should x^ + 8 x + 5 be added to give x^ — 7 ? 14. To what should x< - 9 x2 + 3 y be added to give y'^ + x - 7 ? MULTIPLICATION Product. By the product of two algebraic expressions, A 301 and B, we shall mean the simplest form to which the expression AB can be reduced by means of the rules of reckoning. Of especial importance in such reductions are : 1. The commutative, associative, and distributive laws. 2. The law of exponents »"*•«" = a"'+". 3. The rules of sign : a(— b) = (— a)b =— ab; (— a) (— b) = ab. Rules for multiplying integral expressions. 1. To find the 303 product of two monomials, multiply tJte product of the numerical 98 A COLLEGE ALGEBRA factors hy that of the literal factors, simplifying the latter by adding exponents of powers of the same letter. Give the result the + or — sign, according as the monomials have like or unlike signs. 2. To find the product of a polynomial hy a moyiomial or polynomial, mnltiply each term of the multiplicand by each term of the multiplier and add the products thus obtained. The first rule follows from the commutative and associative laws and the law of exponents. The second rule follows from the distributive law ; thus, (a + 6 + c) (??i + n) = (a + b + c) ??i + (a + 6 + c) n = am + hm + cm + an + 6n + en. The first rule applies also to products of more than two monomials. When an odd jiiimber of these monomials have — signs, the sign of the product is — ; otherwise it is +• A product of 7}iore than two polynomials may be found by repeated applications of the second rule. Example 1. Find the product of - 4 a"h-x^, 2 hx'^, and - 3 a^x. We have - 4 a^b'^x^ • 2 6x* • - 3 a^x = 24 aWx'-hx^a'^x = 24 a^b^xs. Example 2. Find the product of a — 2 6 and ab — b^ + a^. For convenience we arrange both factors in descending powers of a, and choose the simpler factor as multiplier. We then have (a2 + ab- b-) (a - 2 6) = a^ + a"-b - ab"- -2a%-2 ab"- + 2 63 = a3 - a'^b - 3 ab"^ + 2 b^. 303 The degree of the product Avith respect to any letter (or set of letters) is the sum of the degrees of the factors with respect to that letter (or set of letters). This follows from §302, 1, and the fact that the term of highest degree in any product is the product of the terms of highest degree in the factors. Thus, the degrees of x2 + 1 and x^ - 1 are two and three respectively, and the degree of the product (x^ + 1) (x^ - 1), or x^^ + x^ - x^ - 1, is five. THE, FUNDAMENTAL OPERATIONS 99 When both factors are homogeneous, § 276, the product is 304 homogeneous. For if all the terms of each factor are of the same degree, all the products obtained by multiplying a term of the one by a term of the other are of the same degree. Hence the sum of these products is a homogeneous polynomial. Arrangement of the reckoning. When both factors are poly- 305 nomials in x or any other single letter, or when both are homogeneous functions of two letters, it is convenient to arrange the reckoning as in the following examples. Example 1. Multiply 2x3 - a;2 + 5 by a; - 3 + x^. 2 3 _ 2 4. i^ ^^ arrange both factors in descend- 2 , a- _ Q ^°S (or ascending) powers of x and 2^5 _ — --^ '- rJi place multiplier under multiplicand. 2 4 _ 3 ' c ^6 tli^n write in separate rows the _„3„2 _ir; "partial products" corresponding to o . , — I _ , , Q >, , , Tz the several terms of the multiplier, 2 x^ + x* - 7 x3 + 8 x2 + 5 X — 15 , . , , , , ^ ' placing them so that like terms, that is, terms of the same degree, are in the same column. Finally we add these like terms by columns. Example 2. Multiply x- — ?/2 + 2 xy by 2 y + x. x2 + 2x?/ —y" In this case both factors are homogeneous X 4- 2 ?/ functions of x and y. x3 + 2 Oo^y — xy'^ We arrange them both in descending powers 2 x^y + 4 xy'^ — 2 y^ of x, and therefore in ascending powers of y x^ + 4x-2/ + 3x2/^ — 2 y3 and then proceed as in Ex. 1. Detached coefficients. In the reckoning illustrated in § 305, 306 Ex. 1, the terms are so arranged that their positions suffice to indicate Avhat powers of x occur in them. We may make use of this fact to abridge the reckoning by suppressing x and writing the coefficients only, and it is always worth while to do this when the given polynomials have numerical coefficients. If either polynomial is incomplete, care must he taken to indicate every niissinrj term hij a () coefficient. 100 A COLLEGE ALGEBRA . Example. Multiply x^ - 3 12 + 2 by x^ + 3 a;2 - 2. We arrange the reckoning as in § 305, 1 — d + + j,^ . ^^^ write the coefficients only, indi- X -f o -r u _ eating the missing terms by coefficients. 1 — d + + 2 ^g ^^j^ jj^g partial product correspond- 3 - 9 + + 6 ing to the term of the multiplier. Insert- — — — — — — — ing the appropriate powers of x in the final 1 + — 9 + U+ — — result — beginning with x^ since the sum of the degrees of the factors is six — we obtain the product required, namely, x6-9x-* + 12x2-4. The degree of the product, .six, is also indicated by the number of terms, seven, in the result 1+0-9 + + 12-0-4, § 277. This is called the method of detached coefficients. It applies not only to polynomials in a single letter, — both arranged in descending or ascending powers of that letter, — but also to homofjeneous polynomials in two letters. For in arranging two such polynomials in descending powers of one of the letters, we at the same time arrange them in ascending powers of the other letter, so that the position of any coefficient will indicate what powers of both letters go with it. 307 Formulas derived by the method of detached coefficients. Con- sider the following examples. Example 1. Prove the truth of the identity (a* + a^b + am + ah^ + 6*) (a - 6) = a^ - fts. We perform the multiplication indicated in the first member by detached coefficients, and r so obtain the coefficients of the product arranged 1 ^" descending powers of a and in ascending powers of h. 1+0 + + + 0-1 ^^ ^^^^ .^ advance that the degree of the product is five, which is also indicated by the number of terms, six, in the final result. Hence the product is a^ + • a*6 + • aW + • a26» + • a6* - U", or a^ - U'. Example 2. Prove the truth of the identities (a2 - o6 + &2) (a + 6) = a8 + 68. (1) (cfi - cC^h + a62 - 63) ^a^-h) = a* - 6*. (2) THE FUNDAMENTAL OPEIlATlCNS 101 Proceeding precisely as in Ex. 1, we have 1 - 1 + 1 (1) 1 - 1 + 1 - 1 (2) 1 +1 1 + 1 1-1+1 1-1+1-1 1-1+1 1-1+1-1 1 + + + 1, i.e. 0=5 + 63. 1+0 + + 0-1, i.e. a' - ¥. By the method illustrated in these examples we may prove the truth of the following identities, of which the examples are special cases, namely : For eve7'i/ positive integral value of 7i we have 308 (a"-i + a"-26 -\ h ab"-^ + ft" "i) (a - ft) = a" — ft", For every positive odd value of 7i, we have 309 (a«-i - a'-zft -\ aft"-2 + ft"-') (a + ft) = a" + ft". And for every positive even value of 7i, we have 310 (a«-i - a^-H 4 h aft"~^ - ft""') (a + b) = a" - ft". Powers of a binomial. We can compute successive powers 311 of a + ft by repeated multiplications. These multiplications are readily performed by detached coefficients. As the coefficients of the multiplier are always 1 + 1, it is only necessary to indicate for each multiplication the partial products and their sum. We thus obtain (1) 1 + 1 i.e. a + b. 1 + 1 (2) 1 + 2 + 1 i.e. a2 + 2a6 + 62 = (a + 6)2. 1+2 + 1 (3) 1 + 3 + 3 + 1 i.e. a' + 3 a26 + 3 a62 + 63 = (a + 6)3. 1+3+3+1 (4) 1 + 4 + 6 + 4 + 1 i.e. a* + 4 d^b + 6 a262 + 4 06' + 6* = (a + 6)*. Observe that in each multiplication the coefficients of the second partial product are those of the first shifted one place to the right. Hence when we add the coefficients of the two 102 A COLLEGE ALGEBRA pai'tiAl 'products and so' obtain the coefficients of the next power, we are merely applying the rule : To any coefficient in a power already found add the coefficient which precedes it ; the sum will be the correspondiny coefficient in the next power. All the coefficients of this next poiver, except the first and last, can be found by this rule ; these are 1 and 1 Thus, the coefBcients of (4) which correspond to 3, 3, 1 in (3) are 3 + 1 or 4, 3 + 3 or 6, 1 + 3 or 4. Applying the rule to (4), we obtain 4 + 1 or 5, 6 + 4 or 10, 4 t- 6 or 10, 1 + 4 or 5. Hence (a + 6)5 = a5 + 5 a*6 + 10 a%^ + 10 a%^ + 5 a¥ + h^. Evidently the coefficients of any given power of a + ^ can be obtained by repeated applications of this rule. Example. Find successively (a + 6)^, (a + by, (a + h)^. Products of two binomial factors of the first degree. The student should accustom himself to obtaining products of this kind by insjiection. We have {x -{- a) {x + b) = X- + {a + b)x -\- ab. (1) (aoX + a{) (boX + b{) =-- a^boX^ + (ao^i + Oi^o) ^ + a A. (2) In the product (1) the coefficient of x is the snm and the final term is the product of a and b. In the product (2) the first and last coefficients are products of the first coefficients and of the last coefficients of the factors, and the middle coefficient is the sum of the '' cross-products " a^x and tti^o- Example 1. Find the product (x + 5) (x — 8). (X + 5) (X - 8) = x2 4 (5 - 8) X - 40 = x2 - 3 X - 40, Example 2. Find the product (x + 3y) (x + 10?/). (X + 3?/) (X + lOy) = x2 + (3 + 10)X7/ + SOy^ = x2 + \Zxy + Z^y\ THE FUNDAMENTAL OPERATIONS 103 Example 3. Find the product (2 x + 3) (4x + 7). (2x + 3)(4x + 7) = 2-4x2 + (2.7 + 3-4)x + 3-7 = 8x2 + 26x + 21. Example 4. By the methods just explained find the products (X - 10) (X - 15), (3a + 46) (5 a -66), {7 x - y)(5x - 3y). Product of any two polynomials in x. Consider the product 314 (aoX^ + aix'^ + a„x + a^ {1)qX- + h-^x + h<^ + (Cfi^>2 + ajji + aj)^) X'^ + («2^''2 + «3^l) ^ + «3^2- The product is a polynomial in x whose degree is the sum of the degrees of the factors. And the coefficient of each term may be obtained by the following rule, in which a^ denotes one of the numbers a^, a^, a^, a^, and bf. one of the numbers ^0) ^1) ^2- Find the difference between the degree of the product and the degree of the term, and then form and add all the products aj^b^. in which h + k equals this difference. Thus, to obtain the coefficient of x-, we find the difference 5 — 2, or 3, and then form and add 0162, 0261, 0360, these being all the products ahbk in which h + k = Z. This rule applies to the product of any two polynomials in X of the form a^x"' + • ■ • + «,„ and b^yX'^ -\ + ^„. It also indi- cates how to obtain any particular coefficient of the product when the factors have numerical coefficients. Example 1. Find the coefficient of x^"" in the product {aoa:'5 + a^x'^^ + ■■■ + a^x + 075) (60x0 + 6ix59 + • • • + 659X + beo). The degree of the product is 75 + 60 or 135 ; and 135 - 100 = 35. Hence the coefficient of x^'"' is 00635 + 01634 + • • • + 03461 + 03560. Similarly the coefficient of x^ is O40660 + O41659 + ■ • • + O74626 + 075635. Example 2. Find the coefficient of x^ in the product (3 X* - 2 x3 + x2 - 8 X + 7) (2 x3 + 5 X- + 6 X - 3). The required coefficient is (- 2) (- 3) + 1 • 6 + (- 8) 5 + 7 • 2, or - 14. 104 A COLLEGE ALGEBRA Example 3. In the product of Ex. 1, find the coefficients of x"* and of x23_ Example 4. In the product of Ex. 2, find separately the coefficients of x^, x^, X*, x2, and x. Products found by aid of known identities. The following formulas or identities are very important and should be care- fully memorized. (a + by = a'' + 2ab + b\ (1) (a -hf = 0^-2 ah + h'. (2) (a + h) {o -b) = a'- h\ (3) To this list may be added the formulas given in §§ 308, 309, 310, and the following, § 311 : ' {a^hf = a?^?> a^b ■{- ^ ab'' + b^. (4) Inasmuch as the letters a and b may be replaced by any algebraic expressions whatsoever, these formulas supply the simplest means of obtaining a great variety of products. The following examples will make this clear. Example 1. Find the product (.3x — 5?/)2. (3x - 52/)2 = (3x)-i - 2 ■ 3x • 5?/ + (5 (/)2 = 9x2 - 30x2/ + 25t/2. by (2) Example 2. Find the product {x2 + xy + y-) {x2 - xy + y"). (x2 + x?/ + 2/2) (x2 -xy + 2/2) = [(a;2 + ^2) + ^y^ |-(a;2 + yi) _ ^y] = (x2 + 2/2)2 _ y.2yi = a;4 + x^^ + y*. by (3), (1) Example 3. Explain the step.s in the following process. (X + 2/ + 2) (X - 2/ + 2) (X + 2/ - 2) (X - 2/ - 2) = [x + (2/ + z)-] [X - (2/ + 2)] . [X + (2/ - 2)] [x - (2/ - 2)J = [X2 - (2/ + 2)2] . [X2 - (2/ - 2)2] = [(X2 - 2/2 - 22) -2yZ-\- [(X2 _ 2/2 _ 22) + 2 2/2] = [X2 - (2/2 + 22)]2 _ 4 y2z2 = X* - 2 X2 (2/2 + 22) + (2/2 + 22)2 _ 4 ^2^2 = X* + 2/* + 2* - 2 x22/2 - 2 2/222 _ 2 22x2, THE FUNDAMENTAL OPERATIONS 105 Observe in particular that by this method we may derive from (1) and (4) the square and cube of any polynomial. Thus, we have (a + 6 + c)2 = [(a + 6) + c]2 = (o + 6)^ + 2 (a + 6) c + c2 = a- + 62 ^ c2 + 2 a6 + 2 ac + 2 6c. (a + 5 + c)3 = (a + 6)3 + 3 (a + b)-c + 3 (a + 6) c2 + c^ = a3 + 63 + c3 + 3a'i6 + 362a + 362c + 3c26 + 3c2a + 3o2c + 6a6c. Generalizing the first of these results we have the theorem : The square of any polynomial is equal to the sum of the 316 squares of all its terms together with twice the products of every two of its terms. Example 1. Find the product (a-6 + 2c-3 d)2. Example 2. Find the product (1 + 2x + 3x2)2. Example 3. Find the product (x^ — o:;hj + xy- — y^)^. Powers of monomial products. By the ?2th power of any 317 algebraic expression, A, we shall mean the simplest form to which the expression .4" can be reduced by the rules of reckoning. From the laws of exponents (a'")" = a'"" and (ab)" = a"!/" we derive the following rule : To raise a monomial expression A to the nth power, raise its 5(^18 numerical coefficient to the nth power a?id multiply the exponent ^~~ of each literal factor by n. If the sign of A be — , gii'e the result the sign + or —, accord- ing as n is even or odd. Thus, { - 2 ax2(/7)4 = ( - 2Ya*z«y^-» = 16 a*x«y^«. For by repeated applications of the law {ab)" = a"6« we have (- 2 ax2?/7)4 =, (_ 2)ia* (x^)* {y-)\ and by repeated applications of the law {a'")" = a™" we have (- 2)4a4(x2)4 (2/7)4 = l6a*x82/2^. 106 A COLLEGE ALGEBRA EXERCISE m In the following examples perform each multiplication by the most expeditious method possible. In particular employ detached coefficients where this can be done with advantage ; also the identities of § 315. 1. Multiply 3x5 - 2x1 - x3 + 7x2 - 6x + 5 by 2a;2 - 3x + 1. 2. Multiply 5 x3 - 3 ax2 + 2 a-'x + a^ by 3 x2 - ax - 2 a^. 3. Multiply x5 - x^t/ + x^y- - x-y^ + xy* - y^ hy x + y. 4. Multiply 3 x3 - 2 x2 + 7 by 2 x3 - 3 X + 5. 5. Multiply 7 X — 2 2/ by 4 X — 5 ?/, by inspection. 6. Multiply a2 - ax + 6x — x2 by 6 + x. 7. Multiply X* — 2 X + 5 x2 - x^ by 3 + x2 — x. 8. Multiply 2 x" - 3x«-2 + 5x»-3 by x»-2 - x^-s. 9. Multiply a2 _ a& + 3 62 by a2 + a6 - 3 62. 10. Multiply x + 32/-2zbyx-3y + 2z. 11. Multiply x2 + xy + 2/2 + X — y + 1 by X — ?/ — 1. 12. Multiply a2 + 62 + c2 + 6c + ca - a6 by a + 6 - c. 13. Multiply 3x-22/ + 5byx-4y + 6. 14. Multiply x + 7?/-32:by2x + ?/-8 2. 15. Find the product (6 + x) (6 - x) (62 + x2). 16. Also (x2 + X + 1) {x2 - X + 1) (X* - x2 + 1). 17. Also {x + y + z) {- X + y + z) {x - y + z) {x + y — z). 18. Form a table of the coefficients of the first four powers of x2 + x + 1, 19. Continue the table of coefficients of successive powers of a + 6 as far as the tenth power. 20. Find (4 x - 3 y)- and (4 x - 3 y)^ 21. Find (x + 2?/ + 3z-4 m)2. 22. Find (x + 2y + Sz)^; also (x + 2y -3 2)8. 23. Multiply (a + 2 6)2 by (a - 2 6)2. 24. Find the coefficients of x29 and of x^^ in the product (aoX2T + aiX26 + . ■ • + 02635 + Uii) (6oxi9 + biX^» + ••' + bi»x + 619). THE FUNDAMENTAL OPERATIONS 107 25. Find the coefficients of x^, x^, and x* in the product (2x6 - 3x5 + 4x* - 7x3 + 2x - 5) ^g^^ _ x3 + 2x2 + 3x - 8). 26. Verify the following identities : 1. (x + y + 2)3 - (x3 + ys^z3) = 3iy + z) {z + x) (x + y). 2. {a^ + bf^) (x2 + 2/2) = (ax + byf + (6x - ay)\ 3. (a2 - 62) (x2 _ yi) = (ax + by)^ - (bx + ay^. 4. (a + 6 + c)3 = a3 + 63 + c3 + 3a2{6 + c) + 362(c + a) + 3c2(a + 6) + 6a6c. VH. Simplify the following powers : (2 a2x32/'')5, (-x^y^zy, (a26"'c3)2», (a^ftncS")". 28. Simplify the following products : ( - a62c3) (a36)2 ( _ acZ)h^ ( _ 2 x'^y^f (axSyiija. DIVISION Quotient. Let A and B denote any two algebraic expressions 319 of wliich B is not equal to 0. By the quotient of A divided by B^ we shall mean the simplest form to which the fraction A j B can be reduced by the rules of reckoning. Formulas. In making such reductions the following formu- 320 las are especially useful, namely, 1 ^ _ ^ n *'" 1 a"* 1 , z. — = a™"", when m> n\ — = > when n> m. a" a" a"~"' 3. 4. — a a a a b ~~ V -h~~V a + b a b d ~ d^ d — a a We may prove 1, 3, and 4 by aid of the rule, § 253 : Tivo expressions are equal if their products by any third expression (not 0) are equal. 108 A COLLEGE ALGEBRA For in 1 the product of each member by be is ac. Thus, — 6c = ac : and - be = ~ b ■ c - ac. §§ 254, 252 be b b Again, in 3 the product of each member of the first equa- tion by h is —a, and the products of each member of the second and third equations hj — b are a and — a respectively. Thus, -^6 = -a;and {--\b = -~b=-a. §§298,254 6 \ b / b Finally, in 4 the product of each member hj d i?, a -\- b. Thus, "^^d = a + b; f" + -V^ = "d + ^ cZ = a + 6. §§254, 252 d \d d/ d d The formula 2 is a particular case of the formula 1. Thus, if m> n, a™ = a"'-" ■ a". § 256 am a"'—" -a" , , Hence — = = a™-". by 1 a" a" 321 Rules for simplifying A/B. The formulas 1, 2, and 3 give us the following rules for simplifying A/B. 1. Cancel all factors common to numerator and denominator. 2. When numerator and denominator involve different powers of the same letter (or expression) as factors, cancel the lower power and subtract its exponent from that of the higher power. 3. Give the quotient the -\- or — sign, according as the numerator and denominator have the same or opposite signs. Thus, = 6a'^-2 = ba^, and = = car —a' a' -2 a* 322 Rules for dividing by a monomial. From the definition of division and § 320, 4, we derive the following rules. 1. To divide one monomial by another, form a fraction by writing the dividend over the divisor, and simplify. THE FUNDAMENTAL OPERATIONS 109 2. To divide a 'pol ynoviial hy a monoviial, divide each term of the dividend by the divisor, and add the quotients so obtained. — 8 a^b'^c 4 a^c Thus, — 8 a^b-c -^ 6 a¥'d = = , by cancelling the Gab^d '6¥d ^ common factor 2 ab- and applying the rule of signs. Again, {az^ — 4 a-z-) -^ ax = = x' — i ax. ax ax But when ci lias no factor in common with a and. b, we regard {a-{-b)/d as a simpler form of the quotient than a/d + b/d. Division of a polynomial by a polynomial. If A and B are 323 polynomials which have common factors, the quotient is the expression to which A/D reduces when these factors are cancelled. Thus, \i A=x'^ — y"-, B = x--{-2xy + y", the quotient is {x — y)/(x + y). A x2 - 2/2 (xj^y)(x-y) x-y For B x'^ + -2xy + y- (x + y)'^ x + y In another chapter we shall give methods for finding the factors which are common to two polynomials. The process called, long division is considered in Chapter V. Complex expressions. Observe that a -i- b x c means j c, 324 while a -=- be, like a -^ (b x c), means a [be. In the chapter on fractions we shall consider complex expressions in which a number of indicated multiplications and divisions occur. In particular we shall find that ax(hxc^d)=axbxc^d. (1) a^{bxc-T-d)=a-^b-^cxd. (2) In (1) the signs x and -f- within the parentheses remain unchanged when the parentheses are removed ; but in (2) each X is changed to -i-, and each -t- to x . 110 A COLLEGE ALGEBRA EXERCISE IV 1. Divide 15 a^bc- by 10 ab^c'^ 2. Divide 75x^2" by - 100 ax^z^. 3. Divide - 35x2'"^" by 28x"'2/'" + ". 4. Divide - 54 \ {ab^-)'^c 1 5 by - 1 8 ^ a (l^c)^ ] ^ 5. Divide x-y — xy- by x^ — |/2_ 6. Divide (x^ - y^) (x^ + y^) by (x - y) {x- - xy + y^). (a - 6)2(6 _ c)3(c - a)* 7. Simplify 8. Simplify 9. Simplify (6 - a)(c- bY{a - c)3 30 a263c4 - 25 a^ftScS + 20 a*¥c^ 3(x_2/)4_2(x-2/)3 + 5(x-2/)2 (2/ - xY 10. Simplify 4 a^ x (3 a63c2)2 ^ (a6c)2 - 6 6c. 11. Simplify the following (1) by performing the operations In the order indicated, (2) by first removing the parentheses. a? ^\a^ ^ (a* ^ ofi y. a) y. (a^ x a ^ a"^)]. 12. By what must 2 a {x-y^Y ^e multiplied to give - 4 a2 {z^y'^Y ? III. SIMPLE EQUATIONS IN ONE UNKNOWN LETTER CONDITIONAL EQUATIONS 325 The expressions 3 a; — 4 and a; + 6 are not identically equal, § 281, and therefore are not equal in value for all values of x. If asked, " For what value or values of x are the values of these expressions the same ? ", we begin by supposing them to be the same, and state the supposition thus : 3a;-4 = a: + 6. The expression thus formed is called a conditional equatio?i, or an equation of condition, because it states a condition which SIMPLE EQUATIONS 111 the " unknown letter " x is to satisfy. It serves the purpose of restricting x to values which satisfy this condition, being true when the values of 3 a; — 4 and a; + 6 are the same and then only. Similarly a: + y = is a conditional equation in the two unknown letters x and y, and, in general. When the exjjresslons A ayul B are not identically equal, 326 A = B is a conditional equation. This equation means : '' A and B are sujj/josed to hai^e equal values.'' And it restricts the variable letters in A and B to values for tvhich this sujjposition is true. The letters whose values the equation A = B thus restricts are called the unknoxcn letters of the equation. In what follows, the word " equation " will mean " con- ditional equation." If the only letters in an equation are the unknown letters, 327 as X, y, z, we call it a numerical equation ; but if there are also known letters, as a, h, c, we call it a literal equation. Thus, 2x — Sy = 5isa, numerical, but ax + by = c is a. literal equation. A literal equation does not restrict the values of the known letters. If both A and B are rational and integral with respect to 328 the unknown letters, the equation .1 = B is said to be rational and integral. But if A or B is irrational or fractional, the equation is said to be irrational or fractional. No account is taken of numbers or known letters in this classification. Thus, V2x + y/b — c is both rational and integral. In the case of a rational integral equation reduced to its 329 simplest form, § 340, the degree of the term or terms of highest degree is called the degree of the equation itself. Thus, the degree of ax'^ -\-hx = cis two ; that of xH^ -\- y* = b is five. The degree is measured with respect to all the unknown letters, but these letters only. 112 A COLLEGE ALGEBRA 330 Equations of the first degree are often called simple or linear equations ; those of the second, third, fourth degrees are called quadratic, cubic, biquadratic equations respectively. 331 An equation in one unknown letter, as x, restricts oc to a finite number of values. We say that these values of x satisfy the equation, or that they are its solutions or roots. Hence 332 A root of an equation in x is any number or knoxcn expression which, if substituted for a:, will make the equation an identity. Thus, 1 and — 2 are roots of the equation z^ ^ x = 2 ; for 12 + 1 = 2 and {-2)2 + (-2)ee2. Again, a — 6 is a root of z + 6 = a ; for (a — 6) -f 6 = a. 333 Notes. 1. An equation may have no root ; for it may state a condi- tion which no number can satisfy. Thus, no finite number can satisfy the equation x + 2 = x + 3. 2. In every equation in x which /i«s roots, x is merely a sym.bol for one or other of these roots. In fact the equation itself is merely a disguised identity, a substitute for the several actual identities obtained by replacing X by each root in turn. Thus, z2 ^ X = 2 is merely a substitute for the two identities 12 + 1=2 and (-2)2 + (-2) = 2. ON SOLVING EQUATIONS 334 To solve an equation in one unknown letter is to find all its roots, or to prove that it has no root. The reasoning on which the process depends is illustrated in the following examples. Example 1. Solve the equation 3z — 4 = x + 6. Starting with the supposition that x has a value for which this equation is true, we may reason as follows : If 3x-4 = x + 6, then 3x-4 + (-x + 4) = x + 6 + (-x + 4), or 2x = 10, and therefore x = 5. Hence, if 3x — 4 = x + 6, then x = 5. (1) (2) §249 (3) §300 (4) §253 (a) SIMPLE EQUATIONS 113 The proposition (a) thus proved states that if (1) is ever true, it is when a: = 5, in other words, that the only number which can be a root of (1) is 5; but it does not state tliat 5 is a root of (1). Tliat statement would be If X = 5, then 3 x - 4 = x + 6. (6) And (6) is not the same as (a) but its converse, § 291. We may prove that 5 is a root of (1) by substituting 5 for x in (1) ; for we thus obtain the true identity 3-5 — 4 = 5 + 6. But this step is not necessary, except to verify the accuracy of our reckoning. For when a true proposition has been proved by a reversible process, we may always conclude that its converse is true, § 293. And this is the case with (a), since the process by which (4) was derived from (1) is made up of reversible steps and is therefore reversible as a whole. Thus, If x = 5, (4) then 2x = 10, (3) §253 or 3x - -4 + (-x + 4) = x + G + (- ■x + i), (2) §300 and therefore 3 X - 4 = X + 6. 0) §249 Hence, in proving the proposition (o) by a reversible process, we have at the same time proved the converse proposition (6), that is, we have not only proved that no other number than 5 can be a root of (1), but also that 5 is itself a root of (1). Example 2. Solve the equation x^ = 9. If X- = 9, (1) then either x = 3, (2) or x = - 3. (3) § 257 Hence (1) can have no other roots than 3 and — 3. But both 3 and — 3 are roots of (1), since the step by which each of the equations (2) and (3) has been derived from (1) is reversible. Thus, (1) follows from (2) and also from (3), § 257. These examples illustrate the following general principles: In seeking the roots of an egtiation in x, we treat the eqiia- 335 tion as if it were a known identity, and endeavor to find all the equations of the form x = c which necessarily follow from it, ivhen thus regarded, by the rules of reckoning. If the process by which one of these equations x = c has been derived is reversible when x has the value c, we may at once 114 A COLLEGE ALGEBRA conclude that c is a root ; and the process is reversible if it is made up of reversible steps. 336 It is important to remember that the mere fact that a certain value of x has been derived from an equation by the rules of reckoning does not prove it to be a root. The process must be reversible to Avarrant this conclusion. Thus, from x - 2 = 0, (1) it follows that (a; - 2) (x - 3) = 0, (2) § 253 and hence that either x = 2, or x = 3. (3) § 253 But we have no right to draw the absurd conchision that 3 is a root of (1). For when x = 3 we cannot reverse the process, that is, divide both members of (2) by x — 3, since the divisor x — 3 is then 0. On the other hand, when x = 2 we can reverse tlie process, since x — 3 is then not but — 1 ; and 2 is a root of (1). TRANSFORMATION THEOREMS 337 In the light of what has just been said we may regard any correct application of the rules of reckoning to an equation as a legitimate transformation of the equation ; and if such a transformation is reversible, we may conclude that it leaves the roots of the equation unchanged. Hence the following theorems. 338 Theorem 1. The following transformations of an equation leave its roots unchanged, namely : 1. Applying the rules of combination, § 259, to each member 2. Adding any expression xvhich has a finite value to both members, or subtracting it from both. 3. Multiplying or dividing both members by the same constant {7lOt 0). For all the rules of reckoning involved in these transformations are reversible, § 259. We may also state the proofs of 2 and 3 as follows : SIMPLE EQUATIONS 115 If A and B denote expressions in a;, the roots of the equation A = 3 are numbers which substituted for xin A and B make A = B^% 332. But any value of z which makes A^B and C finite will make A -y C = B -\- C^ and conversely, § 249 ; hence the roots of ^ = i? are the same as those oiA-\-C — B-\-C. Again, if c denote any constant except 0, any value of x which makes A^B will make cA = cB, and conversely, § 253 ; hence the roots oi A = B are the same as those of cA = cB. Thus, in § 334, Ex. 1, the equations 3x-4 = z + 6, (1) 3x-4 + (-x + 4) = x + 6 + (-a; + 4), (2) 2x = 10, (3) X = 5. (4) all have the same root, 5. Here (2) is derived from (1) by the transformation 2, (3) from (2) by the transformation 1, and (4) from (3) by the transformation 3. Corollary. The following transformations of an equation 339 leave its roots unchanged, namely: 1. Transposing a term, with its sign changed, from one member to the other. 2. Cancelling any terms that may occur hi both 7nembers. 3. Changing the signs of all terms in both members. For 3 is equivalent to multiplying both members by — 1. And 1 and 2 are equivalent to subtracting the term in question from both members of the equation. Thus, if from both members oix — a-\-b=c+b (1) we subtract — a + b = — a + h we obtain x = c + a. (2) The effect of the subtraction is to cancel b in both members of (1) and to transpose — a, with its sign changed, from the first member to the second. By aid of these transformations, §§ 338, 339, every rational 340 integral equation in a- may, without changing its roots, be reduced, to the standard form UqX" + ttio;""^ + ■ • • + «„-i^ + «„ = 0. 116 A COLLEGE ALGEBRA We suppose such an equation reduced to this form when its degree is measured, § 329. The like is true of rational integral equations in more than one unknown letter. Thus, x2 + 3x + 5 = x2 — 4x + 7 can be reduced to the form 7 x — 2 = 0. Its degree is therefore one, not two. 341 Theorem 2. Whe?i A, B, aiid C are integral, the equation AC = BC has the same roots as the two equations A = B and C = 0. For any value of x which makes AC = BC must make either A = B or C = ; and, conversely, any value of x which makes either ^ = i.' or C = will make ^C=BC, §§251, 253. In this proof it is assumed that A, B, C have finite values for the values of x in question. This is always true when, as is here supposed, A, B, C are integral ; but it is not always true when A, B, C are fractional. In particular, ivhen A and C are integral, the equation AC = has the same roots as the equations A = and C = jointly. Thus, the roots of the equation x^ := 3 x are the same as those of the two equations x = 3 and x = 0, that is, 3 and 0. Similarly the roots of (x — 1) (x — 2) = are the same as those of the two equations x — 1 = and x — 2 = 0, that is, 1 and 2. 342 Hence the effect of multiplying both members of an inte- gral equation A = B hy the same integral function C is to introduce extraneous ^voots, namely, the roots of the equation C = 0. Conversely, the effect of removing the same integral factor C from both members of an integral equation AC = BC, is to lose certain of its roots, namely, the roots of C = 0. On the other hand, in & fractional equation, it is usually the case that no extraneous roots are introduced when both members are multiplied by the lowest common denominator of all the fractions. Thus, if the equation be l/x= l/(2x — 1), and we multiply both members by x(2x — 1), we obtain 2x — 1 = x, whose root is 1. As 1 is not a root of x(2x — 1) = 0, we have introduced no extraneous root. SIMPLE EQUATIONS 117 Corollary. The integral equation A^ = B^ has the same roots 343 as the equations A = B and A = — H jointly. For A'^ = B^ has the same roots as A^ — B- = 0, § 339. And since A^ -B' = {A- B) (J. + B), the equation A'^ - B- = has the same roots as the two equations A — B — and A + B = 0, %Ml, and therefore the same roots as the two equations A — B and A = — B, § 339. Thus, the roots of the equation (2x — 1)2 = (x — 2)2 are the same as those of the two equations 2x — l=x — 2 and 2 x — 1 = — (x — 2), that is, — 1 and 1. Hence the effect of squaring both members of the equation 344 A = B is to introduce extraneous roots, namely, the roots of the equation A = — B. Conversely, the effect of deriving from A"^ = B^ the single equation A — B is to lose certain of the roots, namely, the roots of the equation A = — B. Since A" - £" = (A - B) (^"-^ + A^'-'B H \-B^-^),% 308, 345 it follows by the reasoning of § 343 that the roots of .4" = ii" are those of A = B and A"-^ + A"--B H \- B"-^ = jointly. Thus, since x^ — 1 = (x — 1) (x2 + x + 1), the equation x^ = 1 has the same roots as the equations x = 1 and x- + x + 1 = jointly. The theorems just demonstrated, §§ 338-345, hold good for 346 equations in more than one unknown letter if the word root be replaced by the word solution, § 355. Thus, by § 339, the equation x + 2?/ — 3 = (1) has the same solu- tions as the equation x = — 2y + 3 (2), that is, every pair of values of X and y which satisfy (1) will also satisfy (2), and conversely. Equivalent equations. When two or more equations have the 347 same roots (or solutions), we say that they are equivalent. Thus, § 3§§, the equations A = B and A + C = B + C zxe. equivalent. Again, §341, the equation AC = BC is equivalent to the two equations ^ = B and C = 0. But_x?^ = 9 (1} and x = 3 (2) are not equivalent although both have the root 3. For (1) also has the root — _3, which (2) does not have. 118 A COLLEGE ALGEBRA SOLUTION OF SIMPLE EQUATIONS 348 From the transformation theorems of §§ 338, 339 we may at once derive the following rule for solving a simple equation in one unknown letter, as x. To solve a simple equation in x, reduce it to the form ax = b, Then 1. If a.=f^ 0, the equation has the single root b/a. 2. 7/" a = 0, and b g^ 0, the equation has no root. 3. i/" a = 0, and b = 0, the equation is an identity. If the equation has fractional coefficients, it is usually best to begin by multiplying both members by the lowest common denominator of these fractions. This process is called clearing the equation of fractions. We then reduce the equation to the form ax = b hj trans- posing the unknown terms to the first member and the known terms to the second, and collecting the terms in each member. To verify the result, substitute it for x in the given equation, 2 'r X ■ 2 X Example 1. Solve — — = - - (4 - x). To clear of fractions, multiply both members by the Led., 6. Then 4x - 3(x - 2) = z - 6(4 - x), or 4x-3x + 6:=x-24 + 6x. Transpose and collect terms, — 6 x = — 30. Therefore x = 5. ■r. .. s- 2-5 5-2 5,,^, Verification. — — = - — (4 — 5). Example 2. Sol-'e mx + n = px -\- q. Transpose and collect terms, [m — p)x = q - n. Hence if m :^p, the equation has the single root {q — n)/{m — p). If m =p and q 7^ n, it has no root. If m = p and q — n, it is an identity and every value of X satisfies it. SIMPLE EQUATIONS 119 Example 3. Solve (x + a)(x + b) = {x - af. Expand, x^ + (a + 6) x + a6 = x- - 2 ax + a^. Cancel x^, and transpose and collect terms. Then (3 a + 6)x = a2 - a6, a2-a6 and therefore 3a + 6 Sometimes a root of an equation can be fomid by inspectioji. 349 The equation is then completely solved if it be a simple equa- tion, for it can have no other root than the one thus found. Example. Solve (x - a)2 - (x - 6)2 = (a - b)^ Evidently this is a simple equation, and when x = 6 it reduces to the identity (6 — a)2 = (a — 6)2. Hence its root is b. The roots of an equation of the form AB = 0, in which A 350 and B denote integral expressions of the first degree in x, can be found by solving the two simple equations ^ = and B = 0, § 341. In like manner, when A, B, C are of the first degree, the roots of ABC = 0, AC = BC and A'^ — B^ may be found by solving simple equations, §§ 341, 343. Example 1. Solve (x - 2) (x + 3) (2 x - 5) (3 x + 2) = 0. This equation is equivalent, § 347, to the four equations x-2 = 0, x + 3 = 0, 2x-5 = 0, 3x + 2 = 0. Hence its roots are 2, - 3, 5/2, - 2/3. Example 2. Solve 4 x2 - 5 x = 3 x2 + 7 x. This equation has the same roots as the two equations X = and 4x — 5 = 3x + 7. Its roots are therefore and 12. EXERCISE V Solve the following equations. 1. 15- (7-5x) = 2x + (5-3x). 2. X (x + 3) - 4 X (X - 5) = 3 X (5 - x) - 16. 3. (X + 1) (X -f 2) -■ (X 4- 3) (X + 4) = 0. 120 A COLLEGE ALGEBRA X X X X ' + 2 + 4 + 8 + 1^ 5. z-2[x-3(x + 4)-5] = 3^2x-[x-8(x-4)]^ -2. 6. 2^3[4{5x-l)-8]-20^ -7 = 1. 7. Ui[Hi*-l)-6] + 4^=l- 5X-.4 1.3 X - .05 _ 13.95 - 8x ^^ 'J~^ 2 ~ 1.2 11. 3 ex - 5 a + 6 - 2 c = 6 6 - (a + 3 6x + 2 c). 12. (6 - c) (a - X) + (c - a) (6 - X) + (a - 6) (c - X) = 1 - x. 13. [- - = 2, by inspection. a + 1 a 1 A x + 1 a + 6 X — + — a - 1 _ 2a a^ - 6-^ 15. m n 2x. 16. (2x-l)(3x-l)(4x + l)(5x + 2) = 0. 17. (x2 - x) (2 X - 5) = (x2 - x) (x + 9). 18. (X + 2)3 - (X - 2)3 = 32 X + IG. 19. [(a + 6)x - c]-^ = [(a - 6)x + c]2. 20. (x2 - 2 X + 1)2 - (x - 1)2 (X - 3)2 = 0. PROBLEMS 351 On solving problems. In the following problems it is required to derive the values of certain unknown numbers from given relations, called the conditions of the j)roblem, connecting these numbers with known numbers and one another. In each case we represent one of the unkno^vn numbers by a letter, as x. The given conditions then enable us to express SIMPLE EQUATIONS 121 the remaining unknown numbers in terms of x and to form a single equation connecting the expressions thus obtained. This equation is the statement of the problem in algebraic symbols. We solve it for x. If the problem have any solution, it will be the value thus found for x, together with the corresponding values of the other vmknown numbers. It may happen, however, that the value thus found for x is not an admissible solution of the problem. For the problem may be one which imposes a restriction on the character of the unknown numbers, as that they be integers, and the equa- tion in X into which the statement of the problem has been translated does not express this restriction. Having solved the equation in x, therefore, we must notice whether the result is a number of the kind reqiiired before we accept it as a solution of the problem. If it is not, we conclude that the problem is an impossible one. Example 1. The sum of the digits of a certain number of two digits is 12. If we reverse the order of the digits we obtain a number which is 4/7 as great. What is the number ? Here there are four unknown numbers, namely, the tens digit, the units digit, the value of the number as it stands, and the value when the digits are reversed ; but all four can be readily expressed in terms of either units or tens digit. Thus, let X = the tens digit. Then 12 — x = the units digit, lOx + (12 — x) = value of required number, 10 (12 — x) + X = value with digits reversed. By the remaining condition of the problem, we have 10(12 - X) + x = K10« + (12 -X)]. (1) Solving this equation we obtain x = 8, which being an integer less than 10, is an admissible solution of the problem. The like is true of 12 — X or 4. Hence the required number is 84. Notice that with a slight modification the problem becomes impossible. Thus, if we require that reversing the digits shall double the value of the number, we have, instead of (1), the equation 10(12 -x) + x = 2[10x + (12- X)]. (2) 122 A COLLEGE ALGEBRA And solving (2) we obtain x = 32/9, which being fractional is not an admissible solution of the problem. "When dealing with a problem which has to do with certain magnitudes, as intervals of time, remember that the letters used in stating the problem algebraically are to represent not the magnitudes themselves, but the numbers which are their measures in terms of some given unit or units. Care must also be taken to express the measures of all magnitudes of the same kind, whether known or unknown, In terms of the same unit. Example 1. A tank has a supply pipe A which will fill it in 3 hours, and a waste pipe B which will emiJty it in 3 hours and 40 minutes. K the tank be empty when both pipes are opened, how long will it be before the tank is full ? Let X denote the number of hours required. Then 1 /x is the part filled in one hour when both A and B are open. But were A alone open, the part filled in one hour would be 1/3. And were B alone open (and water in the tank) the part emptied in one hour would be l/3j or 3/11. XT 113 2 Hence - = , or — X 3 11 33 Therefore x = 33/2 hours, or 16 hours 30 minutes. Example 2. A crew can row 2 miles against the current in a certain river in 15 minutes ; with the current in 10 minutes. What is the rate of the current ? And at what rate can the crew row in dead water ? Let X = rate of current in miles per minute. As the rate of the crew against the current is 2/15 in miles per minute, in dead water it would be 2/15 + x. And as the rate of the crew with the current is 2/10. or 1/5 in miles per minute, in dead water it would be 1 /5 — x. Hence VV + ^ = i ~ s;, whence * = t^ (miles per minute), and .j^ 4- X = ^ (miles per minute). Example 3. At what time between two and three o'clock do the hour and minute hands of a clock point in opposite directions ? Let X = number of minutes past two o'clock at the time required. SBIPLE EQUATIONS 123 Since the minute hand starts at XII it will then have traversed x minute spaces. The hour hand starts at II, or 10 minute spaces in advance of the minute hand, but it moves only 1/12 as fast as the minute hand. Therefore when the minute hand is at x minute spaces past XII, the hour hand is at 10 + x/12 minute spaces past XII. But by the conditions of the problem, at the time required the minute hand is 30 minute spaces in advance of the hour hand. Hence x = Ao + — ) + 30, or solving, x = 43 J'y minute spaces. Therefore the hands point in opposite directions at 43j7j- minutes after two o'clock, or 16/j- minutes before three o'clock. Sometimes in the statement of a problem the known num- 354 bers are denoted by letters, as o, b, c. The value found for x will then be an expression in a, b, c which may represent an admissible solution of the problem for certain values of these letters, but not for others. The discussion of the follow- ing problem, known as the problem of coiiriers, will illustrate this point. Example. Two couriers A and B are traveling along the same road in the same direction at the rates of m and n miles an hour respectively. B is now d miles in advance of A. Will they ever be together, and if so, when ? Let X = the number of hours hence when they will be together. A will then have traveled mx miles, and B nx miles ; and since B is now d miles in advance of A, we have mx = nx + d. (1) whence {m — n)x = d. (2) and therefore X = hours hence. (3) 1. If A is to overtake B, this value of x must be positive ; and since by hypothesis d, m, n all denote positive numbers, this requires that 711 > n. Which corresponds to the obvious fact that if A is to overtake B, he nmst travel faster than B does. 2. At the same time we can interpret the negative value which x takes if we suppose m < n as meaning that A and B vtere together d/(n — m) hours ago. 124 A COLLEGE ALGEBRA 3. If m = n, we cannot, properly speaking, derive (3) from (1), since the process involves dividing by in — n which is 0. But we can derive (3) from (1) if m differs at all from ?i, it matters not how little. And if in (3) we regard m as a variable, which while greater than n is continually approaching equality with n, the fraction d/{m — n) becomes a variable which continually increases, and that without limit, § 510. All of which corresponds to the obvious fact that the smaller the excess of A's rate over B's, the longer it will take A to overtake B, and that A will jiever overtake B if his rate be the same as B's rate. 4. Finally, if we suppose both m = n and d = 0, the equation (1) is satisfied by every value of x. Which corresponds to the obvious fact that if A and B are traveling at the same rate and are now together, they ■will always be together. EXERCISE VI 1. The sum of the digits of a certain number of two digits is 14. If the order of the digits be reversed, the number is increased by 18. What is the number ? 2. By what number must 156 be divided to give the quotient 11 and the remainder 2 ? 3. There are two numbers whose difference is 298. And if the greater be divided by the less, the quotient and remainder are both 12. What are the numbers ? 4. The tens digit of a certain number of two digits is twice the units digit. And if 1 be added to the tens digit and 5 to the units digit, the number obtained is three times as great as if the order of the digits be first reversed and then 1 be subtracted from the tens digit and 5 from the units digit. What is the number ? 5. If 2 be subtracted from a certain number and the remainder be multiplied by 4, the same result is obtained as if twice the number and half a number one less be added together. What is the number ? 6. A father is now four times as old as his son. If both he and his son live 20 years longer, he will then be twice as old as his son. What are the present ages of father and son, and how many years hence will the father be three times as old as the son ? 7. A tank can be filled by one pipe in 3 hours, and emptied by a second in 2 hours, and by a third in 4 hours. How long will it take to empty the tank if it start full and all the pipes are opened ? SIMPLE EQUATIONS 125 S. A and B can do a certain piece of work in 10 days ; but at the end of the seventh day A falls sick and B finishes the piece by working alone for 5 days. How long would it take each man to do the entire piece, working alone ? 9. At what time between eight and nine o'clock do the hands of a watch point in the same direction ? in opposite directions ? 10. How soon after four o'clock are the hands of a watch at right angles ? 11. In a clock which is not keeping true time it is observed that the interval between the successive coincidences of the hour and minute hands is 66 minutes. What is the error of tlie clock (in seconds per hour) ? 12. Four persons, A, B, C, D, divide $1300 so that B receives f as much as A, C f as much as B, and D f as much as C. How much does each receive ? 13. A man leaves J his property and .f 1000 besides to his oldest son ; I of the remainder and $1000 besides to his second son ; h of the sum still remaining and $1000 besides to his youngest son. If $3500 still remain, what is the amount of the entire property ? ^^ 14. If 2 feet be added to both sides of a certain square, its area is increased by 100 square feet. What is the area of the square ? 15. The height of a certain flagstaff is unknown ; but it is observed that a flag rope fastened to the top of the staff is 2 feet longer than the staff, and that its end just reaches the ground when carried to a point 18 feet distant from the foot of the staff. What is the height of the staff ? 16. A purse contains a certain number of dollar pieces, twice as many half-dollar pieces, and three times as many dimes. If the total value of tlie pieces is $11.50, how many pieces are there of each kind ? 17. A man invests $5000, partly at 6% and partly at, 4%, so that the average rate of interest on the entire investment is [>]%. What sum does he invest at each rate ? 18. In what proportions should two kinds of coffee worth 20 cts. and 30 cts. a pound respectively be combined to obtain a mixture worth 26 cts. a pound ? 19. A pound of a certain alloy of silver and copper contains 2 parts of silver to 3 of copper. How much copper must be melted with this alloy to obtain one which contains 3 parts of silver to 7 of copper ? 126 A COLLEGE ALGEBRA 20. If a certain quantity of water be added to a gallon of a given liquid, it contains 30% of alcohol ; if twice this quantity of water be added, it contains 20% of alcohol. How much water is added each time, and what percentage of alcohol did the original liquid contain ? 21. A train whose rate of motion is 45 miles per hour starts on its trip from Philadelphia to Jersey City at 10 a.m., and at 10.30 a.m. another train whose rate is 50 miles an hour starts on its trip from Jersey City to Philadelphia. Assuming that the two cities are 90 miles apart, when will the trains pass each other, and at what distance from Jersey City ? 22. If two trains start at the times mentioned in the preceding exam- ple and pass each other at a point half way between Jersey City and Philadelphia, and if the slower train moves | as fast as the swifter one, what are their rates, and when do they pass each other ? 23. A rabbit is now a distance equal to 50 of her leaps ahead of a fox which is pursuing her. How many leaps will the rabbit take before the fox overtakes her if she takes 5 leaps while the fox takes 4, but 2 of the fox's leaps are equivalent to 3 of her leaps ? 24. If 19 ounces of gold weigh but 18 ounces when submerged in water, and 10 ounces of silver then weigh 9 ounces, how many ounces of silver and of gold are there in a mass of an alloy of the two metals which weighs 387 ounces in air and 351 ounces in water ? 25. A traveler set out on a journey with a certain sum of money in his pocket and each day spent \ of what he began the day with and $2 besides. At the end of the third day his money was exhausted. How much had he at the outset ? 26. The base of a certain pyramid is a square, and the altitude of each of the triangles which bound it laterally is equal to an edge of the base. Were this edge and altitude each increased by 3 inches, the area of the pyramid would be increased by 117 square inches. What is the area of the pyramid ? 27. The sum of the digits of a certain number of two digits is a. If the order of the digits be reversed, the number is increased by b. What is the number? Show that the solution is admissible only when 9a>b and when both 9 a + 6 and 9a — b are exactly divisible by 18. 28. Two persons A and B are now a and 6 years old respectively. Is there a time when A was or when A will be c times as old as B, and if so, when ? Discuss the result for various values of a, b, c, as iu § 354. SIMULTANEOUS SIMPLE EQUATIONS 127 IV. SYSTEMS OF SIMULTANEOUS SIMPLE EQUATIONS SIMULTANEOUS EQUATIONS A conditional equation in two unknown letters, as x and y, 355 will be satisfied by infinitely many pairs of values of these letters. We call every such pair a solution of the equation. The like is true of an equation in more than two unknown letters. Thus, the equation 2 x + y = 3 (1) is satisfied if we give any value what- soever to X and tlie corresponding value of 3 — 2x to y. For in (1) substitute any number, as b, for x and 3 — 26 for y, and we have the true identity 2 6 + (3 - 2 6) = 3. Thus, x = 0, ?/ = 3;x = l, 2/ = l;x = 2, ?/ = — 1; •••are solutions of (1). Note. When two unknown letters, x, y, are under consideration, the 356 equation x = 2 means that x is to have the value 2, and y any value what- soever; in other words, the equation x = 2 then has an infinite number of solutions. And the like is true of any equation which involves but one of the unknown letters. It is therefore natural to inquire whether there may not be 357 pairs of values of x and y which will satisfy two given equa- tions in these letters. Such pairs usually exist. Thus, both the equations 2 x + ?/ = 3 and 4 x + 3 ?/ = 5 are satisfied when X = 2 and 2/ = - 1 ; for 2 ■ 2 + (- 1) = 3, and 4 • 2 + 3(- 1) = 5. Simultaneous equations. Two or more equations involving 358 certain unknown letters are said to be siynultaneous when each unknown letter is supposed to stand for the same number in all the equations. Thus, the equations 2x + i/ = 3 (1) and 4x + 3?/ = 5 (2) are simul- taneous if we suppose x to denote the same number in (1) as in (2), and y similarly. It is not necessary that all the unknown letters occur in every one of the equations. Thus, x = 2, y = 3 constitute a pair of simultaneous equations in X and y. 128 A COLLEGE ALGEBRA 359 Generally speaking, the supposition that certain equations are simultaneous is allowable only when the number of equations is equal to, or less than, the number of unknown letters. Thus, the two equations cc = 2 and x = 3 cannot be simultaneous, since x must denote different numbers in the two. 360 A solution of a system of simultaneous equations is any set of values of the unknown letters which will satisfy all the equations of the system. Thus, x = 2, ?/ = -lisa solution of the system 2x + 2/ = 3, 4x + 3t/ = 5. 361 To solve a system of simultaneous equations is to find all its solutions or to prove that it has no solution. 362 The reasoning on which the process depends is similar to that described and illustrated in §§ 334, 335. Thus, in the case of a pair of equations in x and y we begin by supposing that x and y actually have values which satisfy both equations. On this supposition the equations may be treated like identities and the rules of reckoning applied to them. By aid of these rules w^e endeavor to transform the equations into one or more pairs of equations of the form X = a, y = b. If the process by which such a pair x = a, y =b has been derived is reversible when x, y have the values a, b, we may at once conclude that a, b is one of the solu- tions sought; and the process is reversible if it consists of reversible steps. The only new principle involved in all this is the following: 363 Principle of substitution. Jf from, the supposition that all the given equations are actually satisfied it follows that the values of a certain pair of expressions, A and B, are the same, the one expression may be substituted for the other in any of the equatiotis. SIMULTANEOUS SIMPLE EQUATIONS 129 Example. Solve the pair 2x + y = 12, (1) 2/ = 8. (2) From the supposition that x and y actually have values v?hich satisfy both equations it follows that the value of ij in (2) and therefore in (1) is 8. Substituting this value, 8, for y in (1), we obtain 2a;+8^12, (3) whence x = 2. (4) Therefore, if (1), (2) have any solution, that solution is x = 2, y = 8. But conversely, x = 2, ?/ = 8 is a solution of (1), (2), inasmuch as the process from (1), (2) to ^4), (2) is reversible. Thus, (3) follows from (4), and then (1) from (3), (2). Note 1. This principle of substitution is a consequence of the several 364 rules of equality, §§ 249, 253, 257, and of the general rule of equality, If a = 6, and 6 = c, then a = c, ^ 261. Thus, we may prove our right to make the preceding substitution as follows : liy = 8, then ?/ + 2x = 8 + 2x, or2x + 8 = 2x + 7/, § 249. And if 2 X + 8 = 2 X + 2/, and 2x + y = 12, then 2 x + 8 = 12, § 261. Note 2. Of course this principle can be applied only when we have 365 a right to suppose the given equations to be simultaneous. Thus, from x = 2 and x = 3 we cannot draw the absurd conclusion 2 = 3, because we have no right to suppose x = 2, x = 3 simultaneous. TRANSFORMATION THEOREMS In view of what has just been said we may regard any 366 correct application of the rules of reckoning to a pair of equations as a legitimate transformation of the pair ; and if such a transformation be reversible, we may conclude that it leaves the solutions of the pair unchanged. Hence the following theorems, which hold good for equa- tions in any number of unknown letters. Theorem 1. The solutions of a pair of equations remain 367 unchanged when the transformations o/ §§ 338, 339 «r« applied to the equations separately. 130 A COLLEGE ALGEBRA For the solutions of the individual equations remain unchanged by such transformations. Thus, the pair of equations 3 x — 2 ?/ = 1 and ?/ — 2 x = 5 has the same solutions as 3 x — 2 y = 1 and y = 5 + 2x. 368 Theorem 2. The j^air of equations y = X, f(x,y)=0 has the same solutio)is as the pair y = X, f(x,X)=0. Here X denotes any expression in x alone (or a constant), /(x, y) any expression in x and y, and/(x, X) the result of substituting X for y in /(x, y), §280. The theorem is merely a special case of the principle of substitution. Thus, the pair of equations y = x + 2 and 3 x — 2 ?/ = 1 has the same solutions as ?/ = x + 2 and 3 x — 2 (x + 2) = 1. 369 Theorem 3. The pair of equations A = B, C = D has the same solutions as the pair A + C = B + D, C = D. For A = B, C = I) has the same solutions as A + C = B + C, C = D, % 338, and A + C = B + C, C = D has the same solutions as A + C = B + n, C = D, ^ 3(53. Thus, the pair x + y = 5 and x — ?/ = 1 has the same solution as x + ?/ + (x — ?/) = 5 + 1 and x — // = 1, and therefore as 2x = (i and x - y = 1. 370 Corollary. Before applying the theorem of § 369 we may, without changing their solutions, multiply both the given equations — that is, both members of each equation — by any constants we please, except 0. Hence If k and 1 denote amj constants except 0, the pair of equations, A = B, C = D has the same solutions as the pair kA ± IC = kB ± ID, C = D. SIMULTANEOUS SIMPLE EQUATIONS 131 Theorem 4. When A, B, and C are integral, the pair of 371 equations AB = 0, C = has the same solutions as the tivo pairs A = 0, C = and B = 0, C = 0. For AB — has the same solutions as the two equations ^ = and B = jointly, § 341. Hence the solutions of the pair AB = 0, C = are the same as those of the pairs ^ = 0, (7 = and ^ = 0, C = jointly. Thus, the solutions of xy = and x + y = 2 are that of x = and x + y = 2, together with that of y = and x + y = 2. Equivalent systems. Two systems of simultaneous equations 372 are said to be equivalent when their solutions are the same. Thus, the pair of equations x + 2z/ = 5, 2x + 2/ = 4is equivalent to the pair 3x + ?/ = 5, 4x + 32/ = 10, both pairs having the same solution 1, 2. Again, the pair xy = 0, x + ?/ = 2 is equivalent to the two pairs x = 0, X + y = 2 and y = 0, x + y — 2. ELIMINATION. SOLUTION OF A PAIR OF SIMPLE EQUATIONS Elimination. To eliminate an unknown letter, as x, from a 373 pair of equations is to derive from this pair an equation in which X does not occur. We proceed to explain the more useful methods of elimi- nating X 01 y from a pair of simple equations in x and y, and of deriving the solution of the equations from the result. Method of substitution. This method is based on the theorem 374 of § 368. Example. Solve x + 3 ?/ = 3, (1) 3x + 5y = l. (2) Solving (1) for x in terms of ?/, x = 3 — 3 ?/. (3) Substituting 3 - 3 ?/ for x in (2), 3 (3 - 3 2/) + 5 ?/ = 1. (4) Solving (4), y = 2. (5) Substituting 2 for 2/ in (3), x = - 3. (6) 132 A COLLEGE ALGEBRA Hence the solution, and the only one, of (1), (2) is x = — S, y — 2. For, by §§ 367, 3(j8, the following pairs of equations have the same 3olution, namely: (1), (2); (3), (2); (3), (4); (3), (5); (6), (5); and the solution of (6), (5) is x = — 3, ?/ = 2. The same conclusion may be drawn directly from § 362. For the process from (1), (2) to (5), (6) is reversible. Verijicatioji. -3 + 3-2 = 3, (1) 3 ( - 3) + 5 • 2 = 1. (2) Here (4) w^as obtained by eliminating x by substitution. To eliminate an unknown letter, as x, from a pair of equations by substitution, obtain an expression for x in terms of the other ' letter (or letters) from one of the equations, and then in the other equation replace x by this expression. 375 The following example illustrates a sjDCcial form of this method, called elimination by comparison. Example. Solve x + 5?/ = 7, (1) a; + 6 2/ = 8. (2) Solving both (1) and (2) for x in terms of ?/, x = 7-5y, (3) x = 8-67/. (4) Equating these two expressions for x, 7 — 5 ?/ = 8 — 6 y. (.5) Solving (5) y=\. (0) Substituting 1 for y in (3), x = 2. (7) Hence the solution of (1), (2) is x = 2, ?/ = 1. 376 Method of addition or subtraction. Tliis method is based on the theorem of §§ 3G9, o70. Example. Solve 2x-G?/ = 7, 3x + 4?/ = 4. (1) (2) Multiply (1) by 3, Gx- 18?/ = 21. (3) Multiply (2) by 2, Gx+ 8v= 8. (4) Subtract (4) from (3), -26y = 13. (5) Whence, 2/ = -1/2. (6) Substitute - 1/2 for y in (1) 2x -G(-l/2) = 7. (7) Whence, x = 2. (8j Hence the solution of (1), (2) i&x^ = 2, 2/ =-1/2. SIMULTANEOUS SIMPLE EQUATIONS 133 For by §§ 367, 368, 370, the following pairs of equations have the same solution, namely : (1), (2) ; (1), (5) ; (1), (6) ; (7), (6) ; (8), (6) ; and the solution of (8), (6) is x = 2, ?/ = - 1/2. Verification. 2 • 2 - 6(- 1/2) = 7, (1) 3 • 2 + 4(- l/2) = 4. (2) Here x was eliminated by subtraction. We can also find the value of z directly from (1), (2) by eliminating y by addition. Thus, Multiply (1) by 2, 4x-12y = 14. (9) Multiply (2) by 3, 9x + 12y = 12. (10) Add (9) and (10), 13 x =26. (11) Whence, as before, x =2. (12) To eliminate an nnknoivn letter, as x, from a pair of simple equatiotis by addition or subtraction, multiply the equations by nuinbei's which will make the coeffiycients of x in the resulting equatio7is equal numerically. Then subtract or add according as these coefficients have like or xinlike signs. Exceptional cases. Let A =i), B = denote a pair of simple 377 equations in x and y. Tlie preceding sections, §§ 374, 376, show that this pair ^ = 0, J5 = has one solution and but one, unless the expressions A and B are such that in eliminating x we shall at the same time eliminate y. This can occur in the following cases only. 1. If the expressions .1 and B are such that A = kB, where k denotes a constant, we say that the equations ^1 = and B = ^ are not independent. Evidently if J. = kB, every solution of £ = is a solution of J. = 0, and vice versa, so that the pair ^ = 0, i? = has infinitely many solutions. Thus, let ^ = 2 X + 6 ?y - 10 = (1), and B = X + 3 ?/ - 5 = (2). Here ^ = 2 /J, so that yl = and i? = are not independent. Observe that if to eliminate x we multiply (2) by 2 and subtract the result from (1) we at the same time eliminate y. 2. If A and B are such that A = kB -\- I, where k and I denote constants, I not 0, we say that the equations ^ = and .6 = are not consistent. 134 A COLLEGE ALGEBRA In this case the pair A = 0, B = has no solution ; for any values of x, y that make B = will make A = I, not ^ = 0. Thus, \QtA = 2x + Qy -9 = (3), andB = a; + 32/-5 = (4). Here J. = 2 B + 1, so that A = and B = are not consistent. If we eliminate x from (3), (4) we shall at the same time eliminate y. 378 Formulas for the solution. We may reduce any given pair of simple equations in x, y to the form ax -{- by = c, (1) a'x + b'y = e', (2) where a, b, c, a', b', c' denote known numbers or expressions. By § 377, the pair (1), (2) has one solution, and but one, unless a constant k can be found such that a' = ka and b' = kb, and therefore ab' — a'b =k(ab — ab)= 0. To obtain this solution, eliminate y and x independently by the method of subtraction, § 376. The results are (ab' — a'b) x = b'c ~ be', (3) (ab' — a'b) y = ac' — a'c. (4) Therefore, if ab' — a'b ^ 0, the solution of (1), (2) is b'c — be' ae' — a'e ab' — a'b ab' — a'b These formulas are more easily remembered if written ^ _ y ^ -1 be' — b'c ca' — c'a ab' — a'b (5) (6) Did we not know in advance that the pair (1), (2) has a solution when ab' — a'b ^ 0, the argument here given would only prove that if the pair (1), (2) has any solution, it is (5). EXERCISE VII Solve the following pairs of equations for x and y. j'x + y = 62, j'6x-52/ = 2r), j-45x-137/ =161, ^' |x-2/=12. ■ l4x-3y = 10. ' 1 18x + ll y = 32. x-3 = 7-x, j'12x = 9- lOz/, ('22/-3x = 0, 8x-32/-61 = 0. ■ t82/ = 7-9x. ' t5x-32/-2 = 0. SIMULTANEOUS SIMPLE EQUATIONS 135 fx/.S + by = S\, (2{2x + Sy) = S{2x-3y) + 10, L5x + 32/ = 1.65. ■ l4x-3?/ = 4(6y- 2x) + 3. r{x + 2){y + l) = (z-5){y-l), ( ax + by = a'^ + 2a + b^, L X (4 + ?/) = - 2/(8 -X). ■ lbx + ay = a^ + 2b + b'^. Cax + by::^c, f {a-b)x + (a + b)y = 2(a^-b^), \(a + b)x+(a~b)y = 2{a:^ + b'^). ^' r ^-y _ x + 2?/-5 _ y-S _ y + 2x-5 13. "I" " 14. J 4 6 ~ 4 6 ' 5x-2y + 6 = 0. ' ax + by — c, \px = qy. fx + yy- 3 2 X x x + y ^2"^ 9 7 {^1 = 1 a b c ' x y _\ a b' c' 15. ^ 16. - + ^ = l+x, a b X V , V + - = 1 + 2/- \^b a 17. Show that the following equations are inconsistent. llx -23 2/ = 10, 6z-10y = 15. 18. In Ex. 15 assign values to a, 6, c, a', 6', c' for which the equations are (1) not consistent, (2) not independent. PAIRS OF EQUATIONS NOT OF THE FIRST DEGREE WHOSE SOLUTIONS CAN BE FOUND BY SOLVING PAIRS OF SIMPLE EQUATIONS A pair of equations which are not of the first degree with 379 respect to x and y may yet be of the first degree with respect to a certain pair of functions of x and y. We can then solve the equations for this pair of functions, and from the result it is ofiten possible to derive the values of x and y themselves. 1 , r. , 2 5 , 9 10 , Example 1. Solve -H = 1, -H = 5. X 3y X y Both equations are of the first degree with respect to 1 /x and 1/y. Solving for 1/x and 1/y, we find l/x = 1/3, 1/2/ = 1/5. Hence x = 3, 2/ = 5. 136 A COLLEGE ALGEBRA Example 2. Solve 3x + - = 6, 7x = 1. Solving for x and y/x, we find x = 1, rj/x = 3. Hence x = 1, y = 3. 380 Given a pair of equations reducible to the form AB =^ 0, A'B' = 0, where A, B, A', B' denote integral expressions of the first degree in x and y. It follows from the theorem of § 371 that all the solutions of this pair can be obtained by solving the four pairs of simple equations A = 0, ^' = 0; A =0, B' = 0; B = 0, A' = 0; B = 0, B' = 0. Example. Solve x^ — 2xy = 0, (1) (x + y-l){2x+y-S)=0. (2) This pair is equivalent to the four pairs X = 0, X + y - 1 = 0, (3) X = 0, 2 X + ?/ - 3 = 0, (4) X - 2 2/ = 0, X + 7/ - 1 = 0, (5) x-22/ = 0, 2x + y -5 = 0. (6) Solving these four pairs (3), (4), (5), (0) we obtain the four solutions of (1), (2), namely: x, y = 0, 1 ; 0, 3 ; 2/3, 1/3; G/5, 3/5. 381 And, in general, if ABC ■ ■ ■ and A'B'C" ■ ■ ■ denote products of m and n integral factors of the first degree in x and y, all the solutions of the pair of equations ABC • • • = 0, A'B'C • • • = can be found by solving the mn pairs of simple equations obtained by combining each factor of the first product equated to with each factor of the second likewise equated to 0. If all these pairs of simple equations are both independent and consistent, we thus obtain mn solutions, that is, the number of solutions of the given equations is the prod^ict of their degrees. EXERCISE Vra Solve the following pairs of equations. 2x 3?/ X y 2. 10x + 5 = 5, 15x + — = 8. y y SIMULTANEOUS SIMPLE EQUATIONS 131 y 2(3-y) ^ 3 X X 2' X X + 2)-- + 1. = 0. 4. 12/ = 0, (a; + 2 2/- l){3x 5. xy - y = 0, Sx -8y + 5 = 0. 6. X (cc - ?/) (X + ?/) = 0, X + 2 2/ - 5 = 0. 7. (X - 1) (y - 2) = 0, (X - 2) (2/ - 3) = 0. 8. 2/2 ={x- 1)2, 2x + 32/-7 = 0. 9. (2x + 2/)2=(x-3 2/ + 5)2, (x + 2/)2 = 1. 10. (x-52/ + 8)(x + 32/ + 5) = 0, (2x + 2/ + 5){5x + 2?/ - 14) = 0. GRAPHS OF SIMPLE EQUATIONS IN TWO VARIABLES Graph of a pair of values of x and y. It is convenient to represent /^a/?-*' of values of two variables, as x and y, by points in a plane. In the plane select as axes of reference two fixed straight lines, X^OX and Y'OY, which meet at right angles at the point 0, called the origin ; and choose some convenient unit for measuring lengths. Then if the given pair of val- ues be a; = a, y = h, proceed as follows : On X'OX and to the right or left of O, according as a is positive or negative, measure off a segment, 0.4, whose length is \a\, the numerical value of a. Similarly on Y'OY and above or below 0, according as h is positive or negative, measure off a segment, OB, whose length is|/.|. Then through A and B draw parallels to Y'OY and X'OX respectively. We take P, the point in which these parallels sY 382 138 A COLLEGE ALGEBRA intersect, as the point-picture, or graj^li^ of the pair of values X ^^ a, y = h. It is convenient to represent both the value-pair x = a, y = b and its graph P by the symbol (a, b). We call the number a, or one of the equal line segments OA or BP, the abscissa of P ; and b, or one of the equal seg- ments OB or AP, the ordinate of P. And we call the abscissa and ordinate together the codrdi?iates of P. We also call X'OX the x-axis or the axis of abscissas, and Y'OY the y-axis or the axis of ordi^iates. Observe that this method brings the value-pairs of a;, y into one-to-one correspondence, § 2, with the points of the plane ; that is, for each value- pair (a, b) there is one point P, and reciprocally for each point P there is one value-pair (a, 6) found by measuring the distances of P from Y'OY and X'OX respectively, and giving them their appro- priate signs. In particular, the graph of (0, 0) is the origin, that of (a, 0) is a point on the z-axis, and that of (0, h) is a point on the y-axis. Example. Plot the value- pairs (4, 4), (-3, 3), (-4,0), (- 5, - 4), (3, - 2). Carrying out. the construc- tion just described for each value-pair in turn, we obtain their graphs as indicated in the accompanying figure. Notice particularly how the position of the graph depends on the signs of the coordinates. 383 The graph of an equati^ in x and y. If, as is commonly the case, a given equation m x and y has infinitely many real solutions, there will usually be a definite curve which con- tains the graphs of all these solutions and no other points. We call this curve the (jraph of the equation. r (-3,3) 1 (4,4) • 1 1 i 1 ., ^ (-4,0) 1 1 ' • ' ' • ■— T ' — :Y (3,-2) (-0 -4) SIMULTANEOUS SIMPLE EQUATIONS 139 But the graph of an equation may consist of more than one curve. Observe that we here include straight lines among curves. Theorem. The graph of every simple equation in one or both of the letters x and y is a straight line. On this account simple equations are often called linear equations. The student may readily convince himself of the truth of the theorem by selecting some particular equation and " plotting " a number of its solutions. (-1 X- 6) (0,4) ^IV) (2^0) (3,-2) Thus, take the equation ?/ = — 2 x + 4. When x=:0, 1, 2, 3, ••• we have ?/ = 4, 2, 0, - 2, • • • And plotting these value-pairs (0, 4), (1, 2), (2, 0), (3, — 2) • • • as in the accompanying figure, we find that their graphs all lie in the same straight line. We Ta?iY prove the theorem as follows : 1. When the equation has the form X = a, or y = b. Example. Find the graph of x = 2. This equation is satisfied by the value 2 of x and every value of y, § 356. Hence the graph is a parallel to the 2/-axis at the distance 2 to its right. For this line con- tains all points whose abscissas are 2, and such points only. And so, in general, the graph oi x = a is a parallel to the ^/-axis at the distance \a\to the right or left according as a is positive or negative; and the graph of 2/ = i is a pa lei to the x-axis at the distance |^>| above or below according A'^ (x = 2) as b is positive or negative. In particular, the graph of y ic = is the y-axis. is the X-axis, and that of 384 140 A COLLEGE ALGEBRA 385 2, When the equation has the form y = mx. Example. Find the graph of y = 2 x. The graph is the right line which passes through the origin (0, 0) a,nd the point (1, 2) ; for this line contains every point whose ordinate is twice its abscissa, and such points only. And so, in general, the graph of y = mx is the right line which passes hrough the origin and the point (1, m). 3. When the equation has the form y = mx + c. Example. Find the graph of y = 2 x + 3. Evidently we shall obtain the graph of this equation if we increase the ordinate of every point of the graph of y = 2x by 3. But that comes to the same thing as shifting the line y = 2x upward parallel to itself until its point of intersection with the {/-axis is 3 units above the origin. And so, in general, the graph of y = mx + c is a right line parallel to the graph of y = mx and meeting the y-axis at the distance \c\ from the origin, above or below, according as c is posi- tive or negative. To find this line. As any two of its points suffice to determine a right line, we may find the graph of any equation, ax -\- by + c =z 0, as in the following example. Example. Plot the graph of3a; + y — 6 = 0. First, when y = 0, then x = 2. Second, when x = 0, then y = 6. Hence we have only to plot the points (2, 0) and (0, 6), that is, the points where the line will meet the axes, and draw the line which these points deternnne (see figure in § 380). This method fails when the equation has one of the forms x = a, y = b, y - mx. We then find the line by the methods explained in § 384, 1 and 2. SIMULTANEOUS SIMPLE EQUATIONS 141 \ r \ <5' \ V- \^ Xe/^^'-' ^J>^S,4) M\^,3) \ ^^;> V' \ :?'^5, 1) ,'' \ Example. Find the graphs of the equations {4x + 2/-7)(3x + 2y-17) = 0, (X - 2 y + 6) (2 X - 3 2/ - 7) = 0, and the graphs of their solutions. 387 Graph of the solution of a pair of simultaneous simple equations. 386 This is the point of intersection of the two lines which are the graphs of the equations themselves ; for this point, and this point only, is the graph of a solution of both equations. Thus, the solution of2x-3y + 7 = (l),and3x + 2/-6 = 0(2)isx = l, y = 3. And, as the figure shows, the graphs of (1) and (2) intersect at the point (1, 3). When the given equations are not consistent, § 377, 2, their graphs are A' lines which have no point in common, thatis,j9araZZenines; when theequa- tions are not independent, § 377, 1, their graphs are lines which have all their points in commonj, that is, coincident lines. Thus, the equations ?/ = 2 x, y = 2 X + 3 are not consistent, and the graphs of these equa- tions, § 384, 3, are parallel lines. Again, the equations?/ = 2x, 3 ?/ = 6 X, which are not inde- pendent, have the same graph. The graph of an equa- 388 tion of the form AB = consists of the graphs of ^ = and B = jointly ; for the solutions oi AB = are those of J = and B = jointly, §§ 341, 346. (1) (2) 142 A COLLEGE ALGEBRA The graph of (1) consists of the lines PQ and ES which are the graphs of4x + y — 7=0 and 3x4-2?/ — 17=0 respectively. The graph of (2) consists of the lines PS and QR which are the graphs ofx — 2?/ + 5 = and 2x — 3?/ — 7 = respectively. The points P, Q, E, S in which the pair PQ, RS meets the pair PS, QR are the graphs of the solutions of (1), (2), namely, (1, 3), (2, - 1), (5, 1), (3, 4). 389 Graph of an equation of higher degree in x and y. We find a number of the solutions of the equation, plot these solutions, and then with a free hand draw a curve which will pass through all the points thus found. By taking the solutions " near " enough together, we can in this way obtain a curve which differs from the true graph as little as we please. In work of this kind it is con- venient to use paper ruled into small squares, as in the accom- panying figure. Example. Find the graph of the equation y = x-. When x = 0, 1, 2, 8, 4, ■ • • we have y = 0, 1, 4, 9, 16, • • • And when - 1, - 2, - 3, - 4, •• • we have 1, 4, 9, 10, • • • Taking the side of a square as the unit of length, plot the corresponding points (0, 0), (1, 1), (2, 4) ■ • •(— 1, 1), (— 2, 4)- • •. A few of them suffice to indicate the general character of the graph, the curve in the figure, except between x = — 1 and x = + 1. It lies wholly above the x-axis, extending upward indefinitely ; and it lies symmetrically with respect to the y-axis, the same value of y corre- sponding to X = a and x = — a. When x = ±l/2, ±1/4,... we have 2/ = 1/4, 1/16, ••• and plotting one or two of the corresponding points, we find that the graph touches the x-axis. ' i \ \ \ \ 1 \ 1 \ \ / , k y SIMULTANEOUS SIMPLE EQUATIONS 143 EXERCISE IX 1. Plot the following pairs of values of x and y. (0, 0), (5, 0), (0, - 7), (6, 2), (- 7, - 1), (- 4, 3), (5, - 9). 2. Find the graphs of the following equations. x = 0, y = 0, 2 2/ + 7 = 0, 3(/ + x = 0, x + 2/ + 5 = 0, 7x4-3?/- 18 = 0, 3x-4y = 24. 3. Find the graphs of the following. X2/ = 0, (X 4- y - 3) (X - 2 2/) = 0, x2 - 1 = 0, x2 = 4 y"^, x"^ -\- y"^ = 0. 4. Find the solutions of the following pairs of equations by the graphi- cal method and verify the results algebraically. rx + 2/-3=:0, r32/ + 2x + 19 = 0, ^'\ x-2?/ = 0. ^ ' t22/-3x + 4 = 0. 5. Do the same with each of the following pairs. r'(x-4y + 6)(x + 32/ + 6) = 0, r-(y_x-2)x = 0, t(3x + 22/-10)(2x-2/ + 5) = 0. I (y - x + 2)2/ = 0. 6. Find the graphs of the following two equations. 2/ =-(x+ 1)2, 2/ = x\ SYSTEMS OF SIMPLE EQUATIONS WHICH INVOLVE MORE THAN TWO UNKNOWN LETTERS Method of solving a system of n simple equations in n unknown 390 letters. A pair of equations in three unknown letters will ordinarily have infinitely many solutions. Thus, the pair x = 2 2;, y = z + \ has infinitely many solutions ; for both equations are satisfied if we assign any value whatsoever, as 6, to z and the values 2 h and & + 1 to x and 2/- But a system of three simple equations in three unknown 391 letters ordinarily has one, and but one, solution, which may be obtained as in the following example. Example. Solve the system of equations 3x-22/ + 4z = 13, (1) 2 X + 5 ?/ - 3 z = - 9, (2) 6x + 3y + 23 = 7. (3) 144 A COLLEGE ALGEBRA Elimiuate z between two pairs of these equations, thus : Multiply (1) by 3, 9 x - 6 y + 12 2 = 39 (4) Multiply (2) by 4, 8x + 20?/- 122 =-36 (5) Add ITx+Hy =3 (6) Again, (1) is 3x-2y + 42=13 (7) Multiply (3) by 2, 12x + 6y + 42 = 14 (8) Subtract (7) from (8), 9x + 8y =1 (9) Eliminate y between the resulting equations (6), (9), thus: Multiply (6) by 4, 68 x + 56 ?/ = 12 (10) Multiply (9) by 7, 63x + 56i/= 7 (11) Subtract (11) from (10), 5x =5 (12) Hence x = 1. Substituting x = 1 in (9), we find y — —\. Substituting x = 1, y = — 1 in (1), we find 2 = 2. Therefore, § 362, if (1), (2), (3) has any solution, it is x = 1, y = - 1, 2 = 2. But the process by which we have dei-ived x = l,y = — 1,2 = 2 from (1), (2), (3) is reversible. In fact, it may readily be traced back- ward step by step. Hence x = 1, y = - 1, 2 = 2 is the solution of (1), (2), (3). We may also prove as follows that x = 1, y = — 1, 2 = 2 is the solution of (1), (2), (3). It is evident by § 368 that x = 1, y = - 1, 2 = 2 is the solution of (12), (9), (1). We therefore have only to prove that the system (12), (9), (1) has the same solution as the given system (1), (2), (3). Let us represent the equations (1), (2), (3), with the known terms trans- posed to the first members, thus : ^ = 0, (1) B = 0, (2) C = 0. (3) It will then follow from the manner in which (9) and (12) were derived, that we may express the equations (1), (9), (12) thus: ^=0, (1) -^ + 2C = 0, (9) 19^ + 16B-14C = 0. (12) Evidently any set of values of x, y, z that makes A=0, B = 0, C = will make^^O, -A +2C = 0, 19 A + 16B-UC = 0. Conversely, when A=0 and —A +2C = 0, then C = 0; and when also 19 ^ + 1(5 B-14C = 0, then B = 0. Hence the system (1), (2), (3) has the same solution as the system (1), <9), (12), namely, x = 1, y = - 1, 2 = 2. SIMULTANEOUS SIMPLE EQUATIONS 145 In the case just considered, from the given system of three 392 equations in the three unknown letters, x, y, z, we derived a system of tivo equations in two letters, x, y, and then from this system, a single equation in one letter, x. And, in general, if we start with a system of n simple equa- tions in n unknown letters and take n — 1 oi these steps, wg shall arrive at a single equation in one of the letters, as x, of the form ax — h = 0. Then, ujiless a = 0, the system has one, and hut one, solution, in which the value of x is, b /a and the values of the other unknown letters may be found by successive siibstitutions in the equations obtained in the process. This may always be proved as in the example. On the other hand, if a = the system ordinarily has infi- nitely many solutions when b = 0, and no solution when b ^ 0. This will be proved in § 394. A much less laborious method of solving a system of simple 393 equations is given in the chapter on determinants. In certain cases labor may be saved by special devices. Example. Solve x + 1/ + 2 = 8, (1) X + 2/ + ?t = 12, (2) X + 2 + tt = 14, (3) 7J + z + u = 14, (4) Add (1), (2), (3), (4), 3x + 3?/ + 32 + 3u = 48. Hence x -\- y + z + u = 16. (5) And subtracting each of the equations (4), (3), (2), (1) in turn from (5), we obtain x = 2, y = 2, 2 = 4, m = 8. Exceptional cases. Let A=Q, B = 0, C = denote a system 394 of simple equations in x, y, z, and, as in § 392, let ax — b — ^ denote the equation obtained by eliminating y and z. 1. If a = and J = 0, it will be found that one of the func- tions A,B,C may be expressed in terms of the other two, thus : A^kB -\- IC, where k and I denote constants. We then say that the equations A = d, B — 0, C — are not Independent. 146 A COLLEGE ALGEBRA From the identity A = IB + IC it follows that every solu- tion oi B = and C = is a solution of ^ = 0. Hence if ^ = and C — are consistent, § 377, 2, the three equations A = 0, B = 0, C = will have infinitely many solutions. Thus, consider the system of equations ^ = 3a;-2y + 42-13 = 0, (1) £=:2x + 5y-3z+9 = 0, (2) C=7x + 8 7/-2z+5 = 0. (3) Eliminating z between (1) and (2), , 3^ +4B=17x + 14 2/-3 = 0. (4) Eliminating z between (1) and (3), ^ + 2Cee17x + 142/-3 = 0. (5) Eliminating y between (4) and (5), 24 + 4B-2C= 0-x -0 = 0. (6) Here the final equation ax — 6 = has the form • x — = 0, and in deriving it, we find that the expressions A, B, C are connected by the identity 2 A + i B - 2 C ~0, or C = A + 2 B. And, in fact, we see on examining (1), (2), (3) that C may be obtained by multiplying 5 by 2 and adding the result to A. Hence the system (1), (2), (3) has infinitely many solutions. 2. If a = and b ^ 0, it will be found that one of the func- tions A, B, C may be expressed in terms of the other two, thus : A = IcB + IC + m, where k, I, vi denote constants, m not 0. We then say that the equations A = 0, B = 0, C = are 7iot consistent. t'rom the identity A = kB + IC + m it follows that ^ = 0, B = Q, C = have no solution. For any values of x, y, z that make B = and C = will make A = m, not A = 0. Thus, consider the system of equations A = 3x-2y + iz-lS = 0, (1) B = 2x + 5y -Sz+ = 0, (2) C = 7x-l-8y-2z+ 6 = 0. (3) SIMULTANEOUS SIMPLE EQUATIONS 147 Eliminating z and y as above, we obtain 2^+4B-2C = 0x-2 = 0. Hence the final equation ax — 6 = has the form • x — 2 = 0, and A, B, C are connected by the identity C = ^ + 2^+L And, in fact, on examining (1), (2), (3) we find this to be the case. Hence the system (1), (2), (3) has no solution. Systems of simple equations in general. From the preceding 395 discussion we may draw the conclusion : Ordinarily a system of m siynple equations in n unknown letters has one solutio7i tvJien m = n, infinitely many solutions when m < n, no solution ivhen m > n. JVhenever exceptions to this rule occur, two or more of the equa- tions are connected hy identical 7'elations of the kind described in §§ 377, 394. In particular, a system of three simple equations, ^ = 0, B = 0, C = 0, in tivo unknown letters, oc, ?/, has a solution when, and only when, A, B,C are connected by an identity of the form A = kB -{- IC and B—0, C = are consistent. Thus, the system x - y = 1 (1), x + y = 1 (2), 3x - y = 10 (3) has no solution ; for the solution of (1), (2), namely, x = 4, y = 3, does not satisfy (3). On the other hand, the system x-?/ = l(l),x + y = 7 (2), 3 x - y = 9 (4) has a solution ; for (4) is satisfied by x = 4, y — S. But observe that 3x - y - 9 = 2 (X - 2/ - 1) + (X + 2/ - 7). Let the student draw the graphs of (1), (2), (4). He will find that they meet in a common point. EXERCISE X Solve the following systems of equations. (x + y=n, 1. J 2/ + z = 13, [z +x = 12. fx-f22/-3z = 3, 3. J 3x-52/ + 7z = 19, Ux-Sw- llz=:- 13. x+y+z= 1, x + 2y + Sz = 4, x + 3?/ + 72 = 13. 5x-22/ = - -33, X + 2/ - 7 z - = 13, x+3y=- 10. 148 A COLLEGE ALGEBRA X + 2y — 4z 2 X - 3 y = 0, y - 4 z = 0. 11, (3x-5 = 2(x-2), (x+l)(y-l)-(x + 2)(y-2) + 5, 2x + Sy + z = 6. 1 1 -+ - 1 _ 1 X y 3 2 = 4. by + z — 4:U 3y + u = \2, x + 2y + Su = 8, 17, 10. 2/ + 2 + u = 4, X + Z + M = 3, X + 2/ + M = 1, X + y + 2 = 10. CX + &?/ = I, by +,az — m, 11. Ix = my = nz, r2x = 3y = 6z, _ax + by + cz = d. ' 1 (x+22/+2-16){3x-2y + 20)=0. Show that the following systems are not independent and find the identity which connects the equations of each system. 3, fSx-Sy + 7z = lO, 13. 2/ - z = - 5, 14. } 2x + 52/-32 = 12, 2-X = 2. I6x + 9y-z = 80 PROBLEMS 396 The following problems can be solved by means of simple equations in two or more unknown letters, as a;, y, •••. How many of these letters it is best to employ in any case will depend on the conditions of the problem. But when a choice has been made of the unknow^n numbers of the problem which X, 1/, • ■ • are to represent, and the remaining unknown numbers, if any, have been expressed in terms of these let- ters, it will V)e found that the conditions of the problem still unused will yield just as many independent and consistent equations connecting x, ?/,••• as there are letters x, i/, ■■■. In fact, if they gave more than this number of equations, the SIMULTANEOUS SIMPLE EQUATIONS 149 problem would have no solution ; if less, an infinite number of solutions, § 395. The remark of § 352 on the restrictions which the nature of the problem may impose on the character of the unknown numbers applies here also. Example 1. In a certain number of three digits, the second digit is equal to the sum of the first and third, the sum of the second and third digits is 8, and if the first and third digits be interchanged, the number is increased by 99. Find the number. Let X = hundreds digit, y = tens digit, z = units digit. Then the number is 100 x + lOy + z. But, by the conditions of the problem, we have X + z = y, (1) 2/ + z = 8, (2) 1002 + 10?/ + x = 100a;+ 10?/ + z + 99. (3) Solving (1), (2), (3) we find x = 2, 7/ = 5, 2 = 3. Hence the number is 253. Example 2. After walking a certain distance a pedestrian rests for 30 minutes. He then continues his journey, but at | of his original rate, and on reaching his destination finds that he has accomplished the entire distance, 20 miles, in 6 hours. If he had walked 4 miles further at the original rate and then rested as before, the journey would have taken 5& hours. What was his original rate, and how far from the starting point did he rest ? Let X = original rate in miles per hour, and let y = number of miles from starting point to resting place. Expressing in terms of x and y the number of hours taken by (1) the actual journey, (2) the supposed journey, we have X 2 7x/8 ' ^' X 2 7x/8 ^ ^' Solving (1), (2) for y/x and 1/x, we find y/x = 3/2, ] /x = 1/4. Hence x — 4, y = 6. Example 3. Two vessels, A and B, contain mixtures of alcohol and water. A mixture of 3 parts from A and 2 parts from B will contain 40% of alcohol ; and a mixture of 1 part from A and 2 parts from B will 150 A COLLEGE ALGEBRA contain 32% of alcohol. What are the percentages of alcohol in A and B respectively ? Let X and y denote the percentages of alcohol in A and B respectively. Then by the given conditions we shall have 3x 2,^4^ - + ^ = ^. (2) 6 5 100 ^ ' 3 3 100 ^ ' Solving (1), (2) we find x = 52/100, or 52%, y = 22/100, or 22%. EXERCISE XI 1. Find three numbers whose sum is 20 and such that (1) the first plus twice the second plus three times the third equals 44 and (2) twice the sum of the first and second minus four times the third equals — 14. 2. The sum of three numbers is 51. If the first number be divided by the second, the quotient is 2 and the remainder 5 ; but if the second number be divided by the third, the quotient is 3 and the remainder 2. What are the numbers ? 3. Find a number of two digits from the following data : (1) twice the first digit plus three times the second equals 37 ; (2) if the order of the digits be reversed, the number is diminished by 9. 4. A owes $5000 and B owes $3000. A could pay all his debts if besides his own money he had |- of B's ; and B could pay all but $100 of his debts if besides his own money he had \ of A's. How much money has each ? 5. Find the fortunes of three men. A, B, and C, from the following data : A and B together have p dollars ; B and C, q dollars ; C and A, r dollars. What conditions nuist p, q, and r satisfy in order that the solution found may be an admissible one ? 6. A sum of money at simple interest amounts to $2556.05 in 2 years and to $2767.10 in 4 years. What is the sum of money, and what the rate of interest ? 7. A man invested a certain sum of money partly in 4% bonds at par, partly in 5% bonds at 1 10, and his income from the investment was $650. If the 4% bonds had been at 80 and the 5% bonds at 110, his income from the investment would have been $100 greater. How much did he invest ? 8. Find the area of a rectangle from the following data : if 6 inches be added to its length and 6 inches to its breadth, tlie one becomes | of the other, and the area of the rectangle is increased by 84 square inches. SIMULTANEOUS SIMPLE EQUATIONS 151 9. A gave B as much money as B had ; then B gave A as much money as A had left ; finally A gave B as much money as B then had left. A then had $16 and B $24. How much had each originally ? 10. A and B together can do a certain piece of work in 5} days ; A and C, in 44 days. All three of them work at it for 2 days when A drops out and B and C finish it in Ij^^ days. How long would it take each man separately to do the piece of work ? 11. Two points move at constant rates along the circumference of a circle whose length is 150 feet. When they move in opposite senses they meet every 5 seconds ; when they move in the same sense they are together every 25 seconds. What are their rates ? 12. It would take two freight trains whose lengths are 240 yards and 200 yards respectively 25 seconds to pass one another when moving in opposite directions ; but were the trains moving in the same direction, it would take the faster one 3| minutes to pass the slower one. What are the rates of the trains in miles per hour ? 13. Two steamers, A and B, ply between the cities C and D which are 200 miles apart. The steamer A can start from C 1 hour later than B, overtake B in 2 hours, and having reached D and made a 4 hours' wait there, on its return trip meet B 10 miles from D. What are the rates of A and B ? 14. In a half-mile race A can beat B by 20 yards and C by 30 yards. By how many yards can B beat C ? 15. A and B run two 440-yard races. In the first race A gives B a start of 20 yards and beats him by 2 seconds. In the second race A gives B a start of 4 seconds and beats him by 6 yards. What are the rates of A and B ? 16. Two passengers together have 500 pounds of baggage. One pays $1.25, the other $1.75 for excess above the weight allowed. If the bag- gage had belonged to one person, he would have had to pay $4. How much baggage is allowed free to a single passenger ? 17. Given three alloys of the following composition : A, 5 parts (by weight) gold, 2 silver, 1 lead ; B, 2 parts gold, 5 silver, 1 lead ; C, 3 parts gold, 1 silver, 4 lead. To obtain 9 ounces of an alloy containing equal quantities (by weight) of gold, silver, and lead, how many ounces of A, B, and C must be taken and melted together ? 18. A and B are alloys of silver and copper. An alloy which is 5 parts A and 3 parts B is 52% silver. One which is 5 parts A and 11 parts B is 42% silver. What are the percentages of silver in A and B respectively ? 152 A COLLEGE ALGEBRA 19. A marksman who is firing at a target 500 yards distant hears the bullet strike 2| seconds after he fires. An observer distant 000 yards from the target and 210 yards from the marksman hears the bullet strike 2^5 seconds after he hears the report of the rifle. Find the velocity of sound and the velocity of the bullet, assuming that both of these velocities are constant. 20. A tank is supplied by tv70 pipes, A and B, and emptied by a third pipe, C. If the tank be full and all the pipes be opened, the tank will be emptied in 3 hours ; if A and C alone be opened, in 1 hour; if B and C alone be opened, in 45 minutes. If A supplies 100 more gallons a minute than B does, what is the capacity of the tank, and how many gallons a minute pass through each of the pipes ? PROBLEMS ILLUSTRATING THE METHOD OF UNDETERMINED COEFFICIENTS 397 We proceed to consider one or two simple problems relating to the subject matter of algebra itself. The inquiry may arise with regard to some particular func- tion of the variables under consideration, Can this function be reduced to a certain specified form and, if so, what are its coefficients when reduced to this form? The following example will illustrate the method of attack- ing a problem of this kind. Example. Can the expression x^ + 4x + 6 he reduced to the form oi a polynomial of the second degree in z + 1, and, if so, what are its coef- ficients when reduced to this form ? The most general expression of the form in question may be written a(x + 1)2 + 6(x + 1) + c, where a, b, c denote constants. Hence, if the reduction under consideration is possible, we must have a;2 + 4x + 6 = a(x + l)2 + ?,(x + l) + c (1) or x"^ + Ax + (j = ax^ + {2 a + h)x + (a + b + c). (2) By § 284, (2) and hence (1) will be an identity when, and only when, the coefficients of like powers of x in (2) are equal, that is, when a = 1, 2a + b = 4, a + ?> + c = 6, or, .solving for a, b, c, when a = l, b = 2,c~3 Hence x2 + 4x + 6 = (x + 1)2 + 2(x + 1) + 3- SLMLLTANEOCS SIMPLE EQUATIONS 153 Observe tliat we set the given expression equal to an expres- sion of the required form but with undetermined coefficients. We then find, that to make this supposed identity true, the coefficients must satisfy certain conditional equations. And by solving these equations we obtain the values of the coefficients. The following is a more general kind of problem including that just considered. Certain conditions are stated and the question is then asked, Does any function of a certain specified form exist which will satisfy these conditions, and, if so, what are its coefficients ? To solve such a problem, we construct an expression of the form in question with undetermined coefficients. These coeffi- cients are the unknown numbers of the problem and the given conditions yield the system of equations which they must satisfy. If this system of equations has a single solution, we obtain a single function satisfying the given conditions ; if the system has no solution, no such function exists ; if the system has infinitely many solutions, the problem is indeter- minate, there being infinitely many functions satisfying the given conditions. It is here supposed that the function under discussion is definite expression, § 264. Example. If possible, find a polynomial in x, of the second degree, which has the value when x = \ and when x = 3, and the value 6 when x = 4. The polynomial in question must have the form ax^ + hx + c. And by the conditions of the problem a-|-6 + c = 0, 9a + 3 6 + c = 0, 16a + 46 + c = 6. Solving for a, 6, c, we find a = 2, 6 = — 8, c = 6. Hence the required polynomial is 2 x^ _ 8 x + 6. Had the problem been to find a polynomial of the^rs^ degree satisfy- ing the given conditions, there would have been no solution ; had it been to find one of the third degree, there would have been infinitely many somtions. 154 A COLLEGE ALGEBRA 399 The method illustrated in the preceding sections is called the method of xmdetermined coefficients. It is the principal method of investigation in algebra and we shall often have occasion to apply it as we proceed. EXERCISE XII 1. Express 3 x" - x^ + 2 x - 5 as a polynomial in x - 2. 2. Express 4x2 + 8x + 7asa polynomial in 2 x + 3. 3. rind/(x) = ax2 + 6x + c such that /(-l) = n, /(l) = -5, /(5) = 6. 4. Find f(x) = ax^ + ftx^ + ex + d such that /(O) = 5, /(- 1) = 1, /(I) = 9, /(2) = 3L 5. Find/(x, y) = ax + hy -\- c such that /(O, 0) = 4, /(4, 4) = 0, /(I, 0) = 6. 6. Find a simple equation ax + 6y + 1 = two of whose solutions are X = 3, y = 1 and x = 4, 2/ = - 1. 7. Can a simple equation ax -\- hy -\- c = be found which has the three solutions x = 3, y = l; x = 4, 2/=-l; x = l, 2/ = l? 8. Find the simple equation whose graph is the straight line deter- mined by the points (2, 3), (- 4, 5). 9. Determine c so that the graph of3x + 2/ + c = will pass through the point (- 2, 3). 10. Find two simple equations, ax + fey + 1 = (1), a'x + ft'y + 1 =0 (2), such that both are satisfied by x = 2, y = 3 and also (1) by x = 7, y = 5 and (2) by x = 3, ?/ = 7. Plot the graphs of these equations. 11. Find the equation x^ + 6x2 + ex + (Z = whose roots are - 2, 1, and 3. 12. Find an equation of the form x^ + hxy + ex + % = which haa the solutions x=l, ?/ = 0; x = 2, 2/ = l; x = — 2, 2/ = l. 13. Express 3 x + 2 ?/ — 3 in the form a(x + 2/- l) + 6(2x-2/ + 2) + c{x + 2y-3), -where a, 6, and c denote constants. THE DIVISION TRANSFORMATION 155 V. THE DIVISION TRANSFORMATION THE GENERAL METHOD Preliminary considerations. In § 319 we defined the quotient 400 of .i by B as the simplest form to which the fraction A/ B can be reduced by the rules of reckoning. We are now to give a general method for finding the quotient as thus defined, when A and B are polynomials in the same letter, as x, and the degree of A is not less than that of B. 1. It is then possible that i? is a factor of A, in other words, that A can be reduced to the form A = QB, (1) where Q is an integi^al function of x. We then have — = Q, B that is, the quotient of ^ by £ is the integral function Q; and we say that A is exactly divisible by B. Thus, ifJ. = «3 + 4a;2_2x-5 and i? = x2 + 3 x - 5, it will be found that x^ + 4x2 — 2x — 5 = (x + 1) (x^ + 3x — 5), an identity of the form (1), q being x + 1. ^ x3 + 4x2-2x-5 Hence — = = x + 1. B x2 + 3x-5 2. But it will usualli/ happen that B is not a factor of A. We cannot then reduce A to the form QB; but, as we shall show, § 401, we can reduce it to the form A = QB + R, (2) where both Q and R are integral functions of x, and the degree of R is less than that of B. A R We then have — = Q -\- —, 156 A COLLEGE ALGEBRA that is, the quotient of A by B is the sum of an integral func- tion, Q, and a fraction, B/B, whose numerator is of lower degree than its denominator. In this case we call Q the integral part of the quotient, and R the remainder. Thus, if J. = a;3 + 2 x2 + 3 X + 3 and B = x2 + 2 X + 2, we can at once reduce A to the form (2) by writing it x3 + 2x2 + 3a;+ 3 = x(x2 4-2x + 2) + (x + 3), where Q is x and E is x + 3, which is of lower degree than B. „ ^ x3 + 2x2 + 3x + 3 , x + 3 ^^'^'^^ ^^ X2 + 2X + 2 =^ + xm:¥^T^- 401 The division transformation. It remains to show how to reduce A to the form QB + R, where R is of lower degree than B and may be 0. The process by which this is usually accom- plished is called the division transformation, or " long division." It is illustrated in the following example. Let .4 = 2 X* + 3 ic' + 4 a;2 + a; - 2 and £ = a-2 - a; + 1. Here the degree of B is two, and the problem is to find an integral function, Q, such that the remainder, iJ, obtained by subtracting QB from A, shall be of \\\q first degree at most and may be ; for if such a function, Q, be found, we shall have A - QB = B, and therefore .1 = QB + R. Since the degree of A is four and that of R is to be not greater than one, Q must be such that the first three terms of A are cancelled when we subtract QB. This suggests the follow- ing method for finding Q. A = 2x' + 3x8 + 4a-2-l- x-2 -2x3 + 2x2 x^- x + l=B 2x-'B = 2x* 2x^ + 5x+7=Q A-2x'B = 5x« + 2x2+ ^_2 (1) 5xB = 5:r8-5x2 + 5x A-(2x' + 5x)B = 7x2-4.T-2 (2) 7B = 7a,2_7^^7 A-i2x^ + 5x + 7)B = 3a:-9 = R (3) THE DIVISION TRANSFORMATION 157 Evidently we shall cancel the leading term of A if we sub- tract any multiple of B which has the same leading term that A has. The simplest multiple of this kind is 2x'^B, where the multiplier 2 x^ is found by dividing the leading term of A, namely 2 a'*, by the leading term of B, namely x^. Subtracting 2x-B from A, as above, we have A -2x^B^5x^ + 2x^ + x-2. (1) We may cancel the leading term of the remainder (1) thus obtained, and with it the second term of .4, by a similar process. The quotient of 5 a;^ by x^ is 5 a; ; and multiplying £ by 5 x and subtracting we have ^ - (2 a;2 + 5 cc) 5 = 7 a;2 - 4 X - 2. (2) Finally, we shall cancel the leading, term of the remainder (2), and with it the third term of A, by subtracting 7 B, where the multiplier, 7, is found by dividing 7 x^ by x'^. The result is A - (2 a--' + 5 a- + 7) B = 3 X - 9. (3) The remainder (3) is of the first degree, and we obtain it by subtracting (2 x'^ -\- 5 x -\- 1) B from .1. Hence the polynomials Q and R which we are seeking are Q = 2x- -\-^x + l and /? = 3 x - 9. And writing the identity (3) thus, .4 = (2a-2 + 5 x + 7) £ + (3 a: - 9), we have A in the form QB + E, where the degree of R is less than that of B. We therefore have the following rule for finding Q and R when A and B are given. Arrange hoth A and B according to desceiiding powers ofx. Divide the leading term of A by the leading term of B ; the quotient will be the first term of Q. Multiply B by this first term of Q, and subtract the product from A. 158 A COLLEGE ALGEBRA Proceed in a similar inaniier with the remainder thus obtained, dividing its leading term by the leading term, of B, and so on. Continue the process until a remainder is reached which is of lower degree than B. We shall then have found all the terms ofQ, and the final remainder will be K — QB or E,. It is customary to arrange the reckoning in the manner illustrated above. We can then use detached coefficients, as in multiplication. Example 1. Given vl = 2 a;5 - 6 x* + 7 x^ + 8 x^ - 19 x + 20, and ^ = x2-3x + 4; find Q and R. 2-6 + 7 + 8-19 + 2011 -3 + 4 2-6+8 2+0-] -1 + 8- -1 + 3- -19 4 5- 5- -15+20 -15 + 20 Hence Q = 2x3-x+5 and E = 0. Observe that instead of tlie first remainder, —1 + 8 — 19 + 20, we write only that part, —1 + 8—19, which is involved in the next subtrac- tion ; also that the second coefBcient of Q is 0, because the tioo leading terms of A are cancelled by the first subtraction. Example 2. Given ^ = x* + 2 x^ + 3 x2 + 2 x + 4, B = x- + 2x, find q and R. 402 Remarks on this method. 1. Observe that in the division transformation each intermediate remainder plays the role of a new dividend ; also that if R^ denote any such remainder, C, the part of Q already obtained, and (22 the rest of Q, we have A = Q^B + 7^1 and R^ = QnB + R. 2. The process by which Q and R are found is not itself the division of A by B, but a preliminary operation consisting of multiplications and subtractions, the aim of which is to reduce A to the form A = QB + R. The division of ,4 by B does not occur until we pass from the identity A = QB -{- R to the identity A/B=Q + R/B. THE DIVISION TRANSFORMATION 159 At the same time it is customary to call the operation by which Q and R are found "division," and also to call Q the " quotient " instead of the " integral part of the quotient " even when R is not ; and we shall usually follow this practice. But " dividing A by £ " does not then mean, as in § 254, finding an expression which multiplied by B will produce A, but finding, first, what multiple of B we must subtract from A to obtain a remainder which is of lower degree than B, and, second, what this remainder is. Compare § 87. 3. The steps by which the integral expression A is reduced to the integral form QB -\- R may be taken whatever the value of X. Hence A and QB + R have equal values for all values of X, ereii those values for which B is eqnal to 0. On the other hand, neither A / B nor Q + R/B has any meaning when B = 0. Thus, if A =x- + x + 1 and B = x - 1, we find, by § 401, x2 + x + 1 = (x + 2) (x - 1) + 3, (1) x^ + X + 1 3 and therefore — - — i~ = x + 2 + • (2) X — 1 X — 1 ^ ' Here B = when x = 1. Substituting 1 for x in (1) and (2), we have 3 = 3, which is true, but 3/0 = 3 + 3/0, which is meaningless. 4. The transformation of A to the form QB + R is U7iique, that is, there exists but one pair of integral functions Q and it (of which it is of a lower degree than B) such that A = QB + R. For were there a second such pair, say Q', R', we should have QB + R=Q'B + R' and therefore {Q - Q') B = R' - R. But this is impossible, since R' — R would be of lower degree than B but (Q — Q')B not of lower degree than B. The effect of multiplying the dividend or divisor by a constant. 403 The following theorems will be of service further on. 1. If 7re multiply the dividend by any constant, as c, we multiply both quotient and remainder by c. F.or if A = QB + R, then cA = cQ-B + cR. 160 A COLLEGE ALGEBRA 2. If we multiply the divisor hy c, we divide the quotient by c, but leave the remainder unchanged. For if A = QB + R, then A = ^-cB + E. 3. If ive multiply both dividend arid divisor by c, tve multiply the remainder by c, but leave the quotient uiichanged. For if A = qB + R, then cA = Q,-cB + cR. 4. If at any stage of a division transformation we multiply an intermediate remainder or the divisor by c, the final remainder, if changed at all, is merely multiplied by c. This follows from 1 and 2 and § 402, 1. The student Avill do well to verify these theorems for some particular case. Thus, we may verify the second theorem by dividing A =4x2 + 6a; + l first by B = 2 X - 1, and then by 2 5 = 4 x - 2. 4 + 6 + 11 2-1 4 + 6 + 1 14-2 4-2 12 + 4 4-2 |l + 2 8+1 8+1 8-4 .-. Q = 2x + 4, 8-4 .-. Q = x + 2, 5 E = 5. 5 R = b. 404 Division by the method of undetermined coefficients. We may also find Q and R when .1 and B are given, as follows : Example 1. Divide ^ = 2 x< + 3 x^ + 4 x2 + x - 2 by B = x^ - x + 1. Since the degree of A is four and that of B is Uoo, we know in advance that the degree of Q is two and that the degree of R is one at most. Hence let Q = CqX^ + cix + C2 and R — dox + dj, where such values are to be found for the coefiBclents Cq, Ci, c^, do, di that we shall have 2X* + 3X3 + 4X2 + X - 2 = (cox2 + CiX + C2) (x2 - X + 1) + cfox + di CqX* + - Co j -r' + Co + Ci I - Ci + C2 x2 + ci X -t- r2 1 -Col +dl\ (1) + do - Co + Cl = 3, Co- Ci+ C2 = 4, Ci- C2+do = 1, C3+di^- -2, THE DIVISION TRANSFORMATION 161 But to make (1) an identity, we must have, § 284, Co= 2. •. Ci = 3 + Co = 3 + 2 = 5. ,-. Co = 4 - Co + Ci = 4 - 2 + 5 = 7. •. rfo = 1 - ci + C2 = 1 - 5 + 7 == 3. •. (ii = - 2 - C2 = - 2 - 7 = - 9. Hence Q = 2x2 + 5x + 7 and i? = 3x - 9, as in § 401. Example 2. Divide 6x5 + 13x* - 12x3 - 11x2 + llx-2 by 2x2 + a;-2. Exact division. Let A and B denote polynomials in x with 405 literal coefficients, and suppose the degree of B to be m. For the division of .4 by B to be exact, the remainder R must equal identically. This requires that all the coefficients of R be 0. Since the degree of R is m — 1, it has 7n coefficients, § 277, and evidently these coefficients are functions of the coefficients of A and B. Hence In order that a jjoli/nomial A viai/ be exactbj divisible by a polynominl of the vath deyree, I>, tlic coeffieients of A and B must satisfy m conditions. The following example will illustrate this fact. Example 1. For \Yhat values of a and f> is x^ + 3x- + 6x + 2 exactly divisible by x- + ax + 1 ? We have x^ + 3 x2 + 6x +2 I x^ + ax + 1 x^ + ax' + X |x + (3 - a) (3-a)X'i + (6 - l)x +2 (3 - a)x2 + (3a - a'^)x + (3 - a) (6 - 1 - 3a + a2)x + (a - 1) Hence a and h must satisfy the turn conditions ft_l_3a + a2 = 0, a — 1 = 0; whence a = 1 and 6 = 3. Example 2. Determine I and m so that 2x3 + 3x2 + te + ?n ^ay jj^ exactly divisible by x- + x — 6. Dividend and divisor arranged in ascending powers of x. Let 406 A and B denote dividend and divisor arranged in ascendiny powers of x, and suppose that A does not begin with a term 162 A COLLEGE ALGEBRA of lower degree than B begins with. We may then obtain an integral expression for A in terms by B by the process of cancelling leading terms described in § 401. If A is exactly divisible by B, this result is the same as when A and B are arranged in descending powers ; but if A is not exactly divisible by B, the result is entirely different. The following examples will make this clear. H-3x + 3x2 + x3 1 + x 1 -2x+ x2 1 + X 1+ X 1 +2x + x2 1+ X l-3x + 4x- ! 2x + 3x2 2 X + 2 x2 (1) -3x+ X2 -3x-3x2 (2) X2 + X3 X2 + X3 4x2 4x2 + 4x3 -4x3 From this reckoning it follows by the reasoning of § 401 that l + 3x + 3x2 + x3 = :{1+2X + X2) (1 + X) (1) 1 - 2 X + x2 :(1 -3X + 4x2)(l + x)-4x3. (2) The result (1) is the same as that obtained when the given dividend and divisor are arranged in descending powers of x. This must always be the case when, as here, tlie division is exact, as follows from § 402, 4. But the result (2) is entirely different from that obtained when we arrange 1 — 2 x + x2 and 1 + x in descending powers of x. We then get x2 - 2 X + 1 = (x - 3) (X + 1) + 4. (3) Both (2) and (3) are true identities, but they give us different expres- sions for x2 — 2 X + 1 in terms of x + 1 and lead to different expressions for the quotient of x2 — 2 x + 1 by x + 1 , namely : l-2x + x2 4x3 = l-3x + 4x2 , 1 + X 1 + X *i:il^±i= x-3 4-^-. X + 1 X + 1 407 Observe that in the arrangement according to ascending powers the degrees of the leading terms of the successive remainders increase, and, except when the division is exact, the process has no natural termination. By taking steps enough we can obtain, as the integral part of the quotient, a THE DIVISION TRANSFORMATION 163 polynomial of as many terms and therefore of as high a degree as we please. Hence If A. and B denote jjoli/tioniials arranged in ascending poivers of X, A not being exactly divisible by B nor beginning with a term of lower degree than B begins with, we can reduce the quotient of A by B to the form B-^ +B' ivhere Q' and R' are integral functions arranged in ascending j\owers of x, Q' ending xoith as high a power of x as we please and E' beginning with a still higher poiver. If the number of terms in Q! is n, we call Q' the quotient of A by li to n terms, and R' the corresponding remainder. When the value of x is small (how small will be shown subsequently) we can make the value of R' / B as small as we please by taking n great enough ; that is, we can find a polynomial Q' whose value will differ as little as we please from that of A / B. On this account the polynomials Q' are sometimes called approximate integral expressiotis for the frac- tion A/B. Thus "dividing" 1 by 1 — x to n "steps," we obtain 1 x" :. = 1 +x + x2 + ...+x''-i + ; I — X 1 — X If we give x any value numerically less than 1, we can choose n so tliat 1 + a; 4- • • • + X" — 1 will differ in value as little as we please from 1/(1 -x). Thus, if x=: 1/3, then xV(l - x) = 1/18, so that 1 + x + x2 differs from 1/(1 — x) by only 1/18. Similarly 1 + x + x- + x^ differs from 1 /(I — x) by only 1/54 ; and so on. Quotients to n terms found by the method of undetermined coefficients. We proceed as in the following example. Example 1. Find the quotient (3 — x)/(l — x + 2x2) to four terms. 3 — X Let = ao + aix + a^x" + asx^ + • • • . (1) 1 — X 4- 2 x2 164 A COLLEGE ALGEBRA Multiplying both members by 1 — z + 2 x^ and collecting terms, we have x3 + ... (2) ao + ai x + 02 x^ + as - ao - ai — a^ + 2ao + 2ai But to make (2) an identity, we must have, § 284, ao = 3, ai — ao = — 1, .-. Oi = — l + ao=2. 02 — tti + 2 ao = 0, .-. a2 = ai — 2 ao = — 4. as — a2 + 2 ai = 0, .-. as = a2 — 2 ai = — 8. Hence (3 - x) / (1 - x + 2 x2) = 3 + 2 a; - 4 a;'^ - 8 x^ + . ■ • . Example 2. Find (2 + x + 3 x-) / (1 + x - x-) to five terms. 409 Polynomials involving more than one variable. Let two poly^ nomials, A and B, be given which involve more than one variable. Unless ^4 is of lower degree than B with respect to some one of the variables, it is 2)ossihle that .4 is exactly divisible by B, in other words, that an integral function Q exists such that A /B = Q. We may discover whether or not this is so, and if it is find Q, by first arranging both .1 and B as polynomials in some one of the variables and then applying the method of § 401. Example 1. Divide x^ + j/S ^ 2-3 _ 3 xyz hy x -\- y -\- z. x3 -?,yz.x^ (2/3 + Z-) \x + (y + z) x^ + {y + z)x-^ |x2 -(y + z)x + {yi -yz + z'^) - {y + z)x- - Syz-x -(y +Z)x2-(y + 2)2x {y'^-yz + z-^)x + {y3 + z^) {y^-yz + z''^)x + {y^ + z^) Hence (x^ + y^ + z^ - 3 xyz) / (x + y + z) = x"^ + y"- + 2- ~yz-zx- xy. Example 2. Divide 2 x- + 5 xy + 3 y^ _|. 7 ^ + n y - 4 by x + ?/ + 4. If .4 is not exactly divisible by B, this method will yield an expression for ./I//? of the form A/B= Q + R / B, where Q and R are integral with respect to the letter of arrangement, THE DIVISION TRANSFORMATION 165 and R is of lower degree than B with respect to that letter. But the form of this exjjression lu'dl depend on what choice is made of the letter of arrangement. Example. Divide 'ix'^ + Q xy -\- y'^ hy 2 x -\- y. (1) Choosing x as the letter of arrangement, we have Ax'^ + 'Ixy |2x + 2y Hence 4xy+2y'^ ^ =2x+2y- _ y. 2x + 7y • 2x + y (2) Choosing y as the letter of arrangement, we have y2 + 6 yx + 4 x2 l y + 2x y'^ + 2yx ly + 4 X Hence 42/X + 4X2 2/2 + 6-(/x + 4x2 , ^ 4x2 4yx + 8x2 — = y + ix — — • _4a;2 y + 2^ 2/ + 2X EXERCISE Xm 1. By the method of § 401 and using detached coefficients, divide 6x4-7x3-3x2 + 24x-20by 3x2 + x- 6. 2. Also 3x* - 2x3 - 32x2 + 66x - 35 by x2 + 2x - 7. 3. Also 2 x5 - 5 X* + 13 x^ - 15 x2 + 22 x by x2 - 2 x + 4. 4. Also 4 x^ - 3 x5 + 19 X* + 2 x3 + 4 x2 - 4 X + 7 by x3 - x + 5. 5. By the method of undetermined coefficients, § 404, divide 2x'' - 3x2 + X - 5 by x2 - 3x + 2. 6. Also 2x5 - 3x* + x2 - 5 by x^ - 3x + 2. 7. Given 4 = Sx^ - 5x2 - 7x + 12 and jB = 3 x2 + x - 5, reduce A to the form A = QB + R, where R is of lower degree than B. Also write down the corresponding expression for A/B. 8. Determine a and b so that x* + ax' + x2 + te 4- 1 may be exactly divisible by x2 — 2 x + 1 . 9. For what values of a and b is (x* + 2 x'' + 3 x2 + ax + 6) / (x2 + 3 x + 5) reducible to an integral expression ? 10. Divide x^ + x^ + x^ + x + 1 + 2 (x* + x2) by x2 + x + 1. 1G6 A COLLEGE ALGEBRA 11. Divide 2x2 + 5x2/ -3?/2-5x + 13?/- 12by x + 3?/-4. 12. Divide 2 a"^ - b- - 6 c^- - ab + ac + bbc hy 2 a + b - 3 c. 13. Divide a^ (6 + c) + b- {c + a) - c^- {a + b) + abc by ab + be + ca. 14. Divide x* + (a - 3)x3 + (4 - a)x2 - 2ax + 8a by x2 - 3x + 4. 15. Divide 8 x^ - 27 j/^ by 2 x - 3 y, using detached coefficients. 16. Also X* - 4 x?/3 + 3 2/* by X - y. 17. Also 6a^ + a^b - a^b^ + 11 a%^ - 5a¥ + 4b^ hy 2a"- - ab + b\ 18. Tlie dividend being 2 x^ + xy^ + 2/^ and the divisor being 2 x + y, find Q and B, first, when x is taken as "letter of arrangement," second, when y is taken as that letter. 19. Arranging dividend and divisor in ascending powers of x, find quotient to three terms and remainder when dividend is 1 — 3 x + 5 x"- and divisor 1 + x + 3 x^. 20. Also when dividend is 1 + x + 3 x^ and divisor 1 — 3 x + 5 x^. 21. By the method of undetermined coefficients, § 408, find to four terms the quotient 1 / (1 — 2 x). 22. Also the quotient (2 + 3 x + 4 x^) / (1 - x + 2 x'^) to /our terms. SYNTHETIC DIVISION AND THE REMAINDER THEOREM 410 Synthetic Division. We proceed to explain a very expedi- tious method of making the division transformation, § 401, when the divisor has the form x — b, that is, is a binomial of the first degree whose leading coefficient is 1. Consider the result of dividing aoX^ + "i-c^ + anX + Og by X — h. a^x^ + aiX^ + «2a^ + «3 U — ^ L gpa:8 - a^bx'' \a^x;^ + (nj> + ffi^r + jaj''' + ^i^^ + ^2) {ajb + ai)«^ + a^x (a J) + ax)x^ — {ay + a^b)x (a.Jj^ + ajj -\- a^)x + a^ (a^b^ + a,h + a„)x -(a,h' + aj>^ + aJ>) UqO^ + Uib'^ +a2b + as = R THE DIVISION TRANSFORMATION 167 The coefficients of Q and R are ao, ttob + «!, aoP + a^b + a^, ajj^ + a-^b'^ + aj) + a^. Observe t hat the first of these coefficients is the leading coefficient of the_ dividend and that the rest may be obtained one after ^the other by the follo wing rule : Multijdy the coefficient last obtained by b and add the next unused coefficient of the dividend. Thus, ao&2 + ai& + a2 = {a^h + ai) 6 + a^, and aolfi + axh- + a^h + as = {aoh" + aih + 02) 6 + as. This rule applies whatever the degree of the dividend may be. For since the coefficient of the leading term of the divisor is 1, each new coefficient of Q will always be the same as the leading coefficient of the remainder last obtained. Like that coefficient, therefore, it is found by multiplying the preceding coefficient of Q by ft and adding a new coefficient of the divi- dend. And for a like reason we shall obtain R, if we multiply the last coefficient of Q by J and add the last coefficient of the dividend. Hence, when the divisor has the form x — b and the divi- 411 dend the form aoX" + Oja;""^ -f • • • + «„, we can find Q and R as follows, where Cq, c^, ■■• c„_i denote the coefficients of Q. ao «! 052 •• ■ «„-i a„ \b Cpb cj) ■ ■ Co Ci C2 R We first write down the coefficients of the dividend in their proper order and b at their right. Under ao we write Cq, which we know to be the same as ao- We then multiply Cq by b, set the product Cob under ai, add, and so obtain Cj. In like manner we multiply c^ by b, set the product c^b under a^, add, and so obtain c^. 168 A COLLEGE ALGEBRA And we continue thus, multiplying and adding alternately, until all the coefficients Uq, a^, ■ ■ ■ a„ are exhausted. Example. Divide 3 x* - 5 x^ - 4 x^ + 3 x - 2 by x - 2. We have 3_5-4+3 -£[2 6 2-4-2 3 1-2-1,-4 Hence Q = 3 x^ + x^ - 2 x - 1 and E = - 4. This very compact method is called syntJietic division. The student should accustom himself to employ it whenever the divisor has the form x — h. 412 Remarks on this method. 1. In dividing synthetically when the dividend is an incomplete polynomial, care must be taken to indicate the missing powers of a; by coefficients. 2. Since x + b — x — (— b), we may divide synthetically by a binomial of the form x -{- b. It is only necessary to replace b by — b in the reckoning just explained. Example 1. Divide x* — 1 by x + 1. Here x 4- 1 = x - (—1), and dividing by x — (— 1), we have 1+0+0+0 -1 I- 1 _ ni ±i ^ id 1-1+1-1, Hence Q = x^-x- + x-l and R = 0. 3. To divide by a binomial of the form ax — /?, write it thus : a(x — (S/a). Then divide synthetically by a; — /3/n-, and let Q and R represent the quotient and remainder so obtained. The quotient and remainder corresponding to the divisor ax- ft will he Q/a and It, § 403, 2. Example 2. Divide 3 x^ - 11 x^ + 18 x - 3 by 3 x - 2. Here 3x — 2 = 3(x — 2/3), and dividing by x — 2/3, we have 3-11+18 -3 [2/3 J -_6 _8 3-9 12, 6 THE DIVISION TRANSFORMATION 169 Hence the required quotient is (3a;2 _ 9x -f 12)/3, or x* — 3x + 4 and the remainder is 5. Example 3. Divide 5 x^ _ x^ + x + 2 by x - 3. Example 4. Divide x^ + 6 X'^ + 11 x + 6 by x + 3. Example 5. Divide 2 x^ - 3 x2 + 8 x - 12 by 2 x - 3. The Remainder Theorem, When a polynomial in x is divided 413 by X — h, a remainder is obtained which is equal to the result of substituting b for x in the dividend ; so that if f (x) denote the dividend, f (b) will denote the remainder. The demonstration of this theorem is given in § 410 ; for it is there shown that if we divide ao^;^ + «ia;^ + a^x + a^ by X — b, we obtain the remainder aj)^ + «ii^ + cL^b + dz, and, in general, that if we divide f(x) = ao^^" + cfi^""^ + • • • + «„ by X — b, the remainder will be Oq^" + aji""^ + • • • + «„> ot f(b). The theorem may also be proved as follows : If /(x) be the dividend, x — & the divisor, (x) the quotient, and R the remainder, then, §401, /(x) = (X) {x-b) + R, where R, being of lower degree than x — 6, does not involve x at all and therefore has the same value for all values of x. The two members of this identity havs equal values whatever the value of X. In particular they have equal values when z = b. Hence f{b) = ! (x), where ^1 (x) is integral. (1) If in (1) we set x = ^, we have f(b) = (b-a)Mb)- (2) But by hypothesis f(b) = 0, and b — a ^ 0. Therefore, since when a product vanishes one of its factors must vanish, § 253, it follows from (2) that <^i (b) = 0. But if ^1 (b) = 0, then ^1 (x) is exactly divisible hj x — b, § 415, and if we call the quotient i(x) we have <^i (ic) = (a; — b) (f)2 (x), where o (x) is integral. (3) Substituting this expression for (fti^x) in (1), we have f(x) = (x-a)(x-b).cl>,(x), (4) which proves that f(x) is exactly divisible by (x — a)(x — b). Continuing thus, we may prove the more general theorem If f(x) vanishes for x = a, b, c, •••, the7i f(x) is exactly 418 divisible by (x — a) (x — b) (x — c) • • •. Thus, 2 x3 + 3 x2 - 2 X - 3 vanishes when x = l, for 2 + 3-2-3 = 0, and when x = - 1, for - 2 + 3 + 2 - 3 = 0. Elence 2x3 + 3x2 — 2x — 3 jg exactly divisible by (x — 1) (x + 1), or x2 — 1, as may be verified by actual division. 172 A COLLEGE ALGEBRA Example L Find a polynomial /(x), of the second degree, which will take the value when x = 2 and when x = S, and the value 6 when » = 4. Since /(x) is of the second degree and is exactly divisible by (x — 2)(x — 3), § 417, it may be expressed in the form/(x) = ao(x - 2) (x - 3), where Uq denotes a constant. And since /{4) = 6, we have 6 = ao(4 - 2) (4 - 3), whence Uq - 3. Hence /(x) = 3(x - 2) (x - 3) = 3x2 _ I5x + 18. Example 2. Find a polynomial /(x) of the third degree which will vanish when x = 2 and when x = 3, and will take the value 6 when x = 1 and the value 18 when x = 4. Reasoning as before we have /(x) = (aox + ai) (x - 2) (x - 3) where Oo, ai are constants. Again, since /(I) = 6, and /(4) = 18, we have 6 = (ao + ai) (i - 2) (1 - 3), or ao + ai = 3, (1) 18 = (4ao + ai)(4-2)(4-3), or 4 ao + ai = 9. (2) Solving (1) and (2), we obtain ao = 2, ai = 1. Hence /(x) = (2x + 1) (x - 2) (x - 3) =:: 2x3 - Ox^ + 7x + 6. 419 ■' Theorem. A pol>/nomial £(x); whose degree is n, cannot vanish for more than n values of x. For if f(x) could vanish for more than n valites of x, it ■would be exactly divisible by the product of more than n factors of the form x — a, § 41 8, which is evidently impossible since the degree of such a product exceeds 7i. 420 Theorem. If we know of a certain polynomial /(x), ivhose degree cannot exceed n, that it will vanish for more than n values of x, we may conchide that all its coefficients are 0. For if the coefficients were not all 0, the polynomial could not vanish for more than n values of x, § 419. We say of such a polynomial that it vanishes identically. 421 Theorem. If two polynomials of the nth degree, f (x) and <^(x), have eAjual values for more than n values of x, their corresponding coefficients are equal. THE DIVISION TRANSFORMATION 173 For let f(x) = aoa;" + ai^;""^ H h «„ and <^ (cr) = h^xf + b-ipc"-'^ -\ \-b„\ also let i// (x) = /(x) — ^ (a;) = (ao - ^-o) ic" + («! - ^-i) aj"-^ + • • • + (a„ - *„). Then yj/ (x) is whenever the values of f(x) and ^ (x) are the same, and by hypothesis these values are the same for more than n values of x. Hence the polynomial \p (x) = (oq — ^o) ^" + •••+(«■„ — ^„), whose degree does not exceed n, vanishes for more than n values of x, and therefore, § 420, all its coefficients are 0. Therefore aQ — bo = 0, a^ — b^ — 0, ■■•, a„ — i„ =0, whence a^ = b^, a^ = J^, • • •, a„ = b„, that is, the corresponding coefficients oif(x) and <^ (x) are equal. Thus, if /(x) = 2 x2 + 6x + 5 and (x) = ax^ + 3x + c have equal values wheu x = 2, 4, 6, we must have a = 2, 6 = 3, and c = 5. EXERCISE XIV 1. Divide x* — 3 x^ — x^ — 11 x — 4 by x — 4 synthetically. 2. Also 5x5 - 6x* - 8x3 + 7 x2 + 6 X + 3 by X - 3. 3. Also 3x* + x2 - 1 by X + 2. 4. Also 3x3+ 16x2 - 13x - 6 by 3x + 1. 5. Also 3 x3 - 6 x2 + X + 2 by 3 X - 1. 6. Also x3 — (a + 6 + c) X- + {ah + ac -|- hc)x — abc by x — a. 7. Also 2 x^ - x3y - 7 x2?/2 + 7 xy3 _ 10 y4 by X - 2 ?/. 8. Given /(x) = 2x3 - 5x + 3. By the method of § 414, find /(I), /(2), /(5), /(- 1), /(- 3), /(- 6). 9. By aid of the remainder theorem, determine »n so that x3 + ?nx2 - 20X + 6 may be exactly divisible by x — 3. 10. In a similar manner, determine I and m so that 2 x3 — x^ + Zx + m may be exactly divisible by (x + 2) (x — 4). 174 A COLLEGE ALGEBRA 11. By § 416, show that Sbm + am — 2an — 6bn is exactly divisible hy m — 2 n, also by a + 3 b. 12. By §§416, 417, show that a{b - c)^ + b{c - a)^ + c{a - b)^ is exactly divisible by (a — b){b — c) (c — a). 13. Find the integral function of x of the third degree which vanishes when a; = 1, 4, — 2, and takes the value — 16 when x = 2. 14. Find the integral function of x of the third degree which vanishes when X = 2, 3, and takes the value 6 when x = and the value 12 when x = l. 15. Show that 2 x"* — ax -i^- 1 and x3 + 5x + 2 cannot have equal values for four values of x. EXPRESSION OF ONE POLYNOMIAL IN TERMS OF ANOTHER 422 Let A and B denote two polynomials in x, A of higher degree than B. Divide Ahj B and call the quotient Q, the remainder R ; then A = QB + R. (1) If Q is not of lower degree than B, divide Qhy B and call the quotient Qi, the remainder R^ ; then Q=Q,B + R,. (2) Similarly, if Q^ is not of lower degree than B, divide Qi by B and call the quotient Q^, the remainder R^; then Qi = Q,B + /?2- (3) Suppose that Q2 is of lower degree than B. We then have A = QB + R by (1) = \Q,B + R,\B + R by (2) = \(Q,B + R,)B + R,\B + R by (3) = QoB^ + R.B'' + R^B + R, where all the coefficients Q2, Ri, Ri, R are of lower degree than B. THE DIVISION TRANSFORMATION 175 And, in general, if any polynomial A be given which is of higher degree than B, and we continue the process just described until a quotient is reached which is of lower degree than B, we shall have A = Q^_iB'- + R^_iB'—^ -\ h R^B + R where R, Ri, ••-, Rr-i, Q,.-i denote the successive remainders and the final quotient, all being of lower deyree than B. Example. Reduce x^ _ 4 x* + 3 x^ - -/ "-^ x + 4 to the form of a poly- nomial in x^ + X + 1 with coefficients whose "degrees are less than two. Using detached coefficients, we may arrange the reckoning thus : 1-4 + 3 1 + 1 + 1 -5+2-1 -5-5-5 7 + 4 + 1 7 + 7 + 7 -3-6+4 - 3 - 3 - 3 ?1 = « - ' 12 + 3 .-. Ei = 12x + -3 + 7 .-. E = -3x + 7. Hence x^ - 4 x* + 3 x^ - x^ + x + 4 = (x - 6) (x2 + X + 1)2 + (12x + 3) (x2 + X + 1) - (3x - 7). In particular, this method enables us to transform any poly- 423 nomial in x into a polynomial of the same degree va. x — b with constant coefficients. Example. Transform 2x^ — x2 + 4x — 5 into a polynomial in x — 2. We may perform the successive divisions synthetically and arrange the reckoning as follows : 2 - 1 +4 - 5[2 4 6 20 2+3+10, 15 .-. E = 15 4 14 2+7, 24 .-. Bi = 24 4 2, 11 .-. i?2 = 11 and Q2 = 2. Hence 2 x^ - x2 + 4 x - 5 = 2 (x - 2)^ + 11 (x - 2)2 + 24 (x - 2) + 15. 176 A COLLEGE ALGEBRA EXERCISE XV 1. By the method of § 422 express x* + x^ — 1 iu terms of x^ + 1. 2. Also 4 X* + 2 x3 + 4 x2 + X + 6 in terms of 2 x'^ + 1. 3. Also 2x'^ - 3x6 + 2x5 4- 5x* - x2 + 6 in terms of x^ - x2 + x + 3. 4. Also x5 + x^y^ + x^y^ + y^ in terms of x^ + xy H- y"^. 5. By the method of § 423 express 2 x^ - 8 x2 + x + in terms of x - 3 6. Also x5 + 3 X* - 6 x* + 2 x2 - 3 X + 7 in terms of x + 2. 7. Also x3 + 9 x2 + 27 X in terms of x + 3. 8. Also x' + 3 x2 4- X — 1 in terms of x + 1. VI. FACTORS OF RATIONAL INTEGRAL EXPRESSIONS PRELIMINARY CONSIDERATIONS 424 Factor. Let A denote a rational integral function of one or more variables. Any rational integral function of these variables which exactly divides A is called 2^ factor of .-i. Hence in order that a given function, F, may be a factor of ^, it is sufficient and necessary 1. That F be rational and integral with respect to the variables of which ^ is a function. 2. That A be reducible to the form A = GF, where G also is integral. Example 1. Since 2x2 - 2x?/ = 2x(x - ?/), both x and x-y are factors of 2 x2 — 2 xy. Example 2. Since 3x2 - 2^2 = (Vsx + v'2y) ( V3x - V22/), both V3 X + V2 2/ and V3 X — V2 2/ are factors of 3 x2 — 2 ?/2. Example 3. Althouj;h x - y = ( Vx + Vy) ( Vx - V^), we do not call Vx + v^ and Vx — Vy factors of x — y, because they are not rational with respect to x and y. FACTORS OF INTEGRAL EXPRESSIONS 177 Note 1. The coefficients of a factor need not be either integral or 425 rational. On the contrary, they may be numbers or expressions of any kind. In Ex. 2 they are irrational. Therefore, since x^ — y = {x + \y) (x — V^), the expression x^ — y, regarded as a function of both x and y, cannot be factored ; but regarded as a function of x alone, it has the factors x + V^ and x - Vy. And the like may be said of other expressions which involve more than one letter. Note 2. Except when dealing exclusively with functions having 426 integral coefficients, it is not customary to include a mere "numerical factor," like 2 in Ex. 1, in a ''st of the factors of a given integral func- tion, A ; for if, as here, we do not require the coefficients of an integral function to be integers, any mere number (or constant) whatsoever may be said to divide A exactly. For a like reason, if F is a factor of A, and c is any constant (not 0), cF is also a factor of A ; but we regard F and cF as essentially the same factor and include but one of them in a list of the factors of A. Thus, in Ex. 1, it would be equally correct to say that 2x and x — y, or that — 2 X and y — x, are the factors. Theorem. If F is a factor of B, and ^ is a factor of A, then 427 F is a factor of A. For, by § 424, A and B are reducible to the forms A = GB and B = HF, where G and H are integral. Hence A = GHF=GH-F, that is, F is a factor of .4, § 424. Prime, composite. An integral function may have no other 428 factor than itself (or a constant). In that case we call it prime. But if it have other factors, we call it composite. Thus, X + 2/2 and x — 2 y are prime, but x^ — y"- is composite. A composite function, A, of the nth. degree, is the product 429 of not less than two nor more than n prime functions, B, C, ■ ■ ■. These prime functions are called the prime factors of A. 178 A COLLEGE ALGEBRA 430 In what follows we shall assume that 1. Any given function A has hut one set of prime factors. 2. All other factors of A are products of these prime factors. 3. Two or more of these pirime factors may be equal, but A can he expressed in only one way as a product of powers of its different prime factors. These theorems, of which 2 and 3 are corollaries to 1, will be proved in §§ 484, 485 for the case in which A is a function of a single variable, and they can be proved generally. Thus, since x^y^ - 2 x^y* = xxyyy (x - 2 y), the prime factors of x3y3 _ 2x2i/* are x, x, y,y,y,x-2 y. Its other factors, as x^, xy, and so on, are products of two or more of these prime factors. Its different prime factors are x,y,x-2y, and it can be expressed in but one way as a product of powers of these factors, namely thus: x'^y^{x - 2 y). 431 Factoring. To factor a given function, A, completely is to "resolve it into its prime factors," that is, to reduce it to the form A = BCD--, where B, C, Z», ••• denote prime functions. But ordinarily we do not attempt to discover these prime factors at the outset. We endeavor first to resolve A into a product of some ttvo of its factors, as F and G, next to resolve F and G, and so on, until the prime factors are reached. And even the first step in this process may be called "factoring" .4. Factoring is the reverse of multiplication. A multiplication usually involves two main steps : (1) a number of applications of the distributive law, in order to replace {a + b)c by ac + 6c, and so on ; (2) the combina- tion of like terms in the result thus obtained. To reverse the process, we must (1) separate the terms thus combined — it is in doing this that the difficulty of factoring consists — and then (2) apply the distributive law in order to replace ac + be by (a + b) c, and so on. It must not be supposed that every composite function can be actually factored. Thus, while it can be proved that x& + ax^ + bx- + ex + d is composite, it can also be proved that the factors of this expression cannot be found by algebraic methods, that is, by applying, a finite number of times, the various algebraic operations. FACTORS OF INTEGRAL EXPRESSIONS 179 EXPRESSIONS WHOSE TERMS HAVE A COMMON FACTOR Expressions whose terms all have a common factor, mono- 432 mial or polynomial, can be factored by a single application of the distributive law, namely : ab -{- ac -\- ad -\- • • ■ = a (b -\- c -\- d -{- ■ • ■). Example 1. Factor 2 a^c + 2 abc + 4 ac^ — 6 acd. All the terms have the factor 2ac. " Separating " it, we have 2a~c + 2abc + 4ac- - Gacd = 2ac{a -{- b + 2c - 3 d). Example 2. Factor a{c — d) + b{d — c). Both terms have the factor c — d. Separating it, we have a{c - d) + b{d - c) = a(c - d) - b{c - d) = {a - b) {c - d). Factors such as these should be separated at the outset. Some expressions which are not in this form as they stand 433 can be reduced to it by combining such of their terms as have a common factor. Example 1. Factor ac + bd + ad + be. Combining ac and ad, also be and bd, we obtain a(c + d) + b{e + d), a binomial whose terms have the common factor e + d. Hence ae + bd + ad + be = {a + b) (c + d). Observe that the parts into which we separate the expres- sion in applying this method must all have the same number of terms. jExample 2. Factor a"^ -{■ ab — bd — ad -\- ae — ed. We must have either two groups of three terms, or three groups of two terms. Four of the terms involve a, namely, a^, a6, — ad, ac, and the remaining two involve d, namely, — bd and — cd. To obtain groups which have the same number of terms we combine the term — ad with the d-terms, and have a2 + a6 + ac - od - 5d - cd = a (a + 6 + c) - d (a + & + c) = {a - d) {a -Y b + c). 180 A COLLEGE ALGEBRA EXERCISE XVI Factor the following expressions. 1. 6 z*j/322 _ 12 x2y*z + 8 x2y3. 2. 2 n2 + (n - 3) n. 3. ab - a + b - 1. 4. mx - nx - mn + n"^. 5. 3xy-2x- 122/ + 8. 6. lOxy + 52/2 + 6x + Sy. 7. x^y"^ - x^ys + 2 x22/ - 2 X2/2. 8. x* + x^ + x2 + x. 9. ac + 6d - (6c + ad). 10. a2c - dbd - abc + a^d. 11. ad + ce+bd + ae + cd + be. 12. a'^ + cd - ab - bd + ac + ad. FACTORS FOUND BY AID OF KNOWN IDENTITIES 434 In the second chapter we derived a number of special products, as (a + b) (a — b) = a- — b^. If a given function, A, can be reduced to the form of one of these products, we can write down its factors at once. The following sections will illustrate this method of factoring. 435 Perfect trinomial squares. This name is given to expressions which have one of the forms a"^ ± 2 ab -\- b'\ Such expressions can be factored by means of the formulas : a^-\-2ab-\-b^ = (a + b)(a + b) = (a + by. a''-2ab + b'' = (a-b)(a-b) = (a- bf. Observe that in a perfect trinomial square (properly arranged) the middle term is twice the product of square roots of the extreme terms, and that the factors, which are equal, are obtained by connecting the principal square roots of the extreme terms by the sign of the middle term. To extract the square root of the perfect square is to find one of these equal factors. Example 1. Factor 9 x2 - 12 xy + 4 y^. This is a per fect square, s ince V2xy = 2 v9x2 • V4?/2. And since V9x2 = 3 x, V4y2 = 2 y, and the sign of the middle term is -, we have 9x2 _ \2xy + 42/2 = {3x -2 2/)(3x - 2y) = (3x -2yY. FACTORS OF INTEGRAL EXPRESSIONS 181 Example 2. Factor a- + b" + c- + 2 ab + 2 ac + 2bc. We can reduce this to the form of a trinomial square by grouping the terms thus : a2 + 2 a6 + 62 + 2 ac + 2 &c + c2 = (rt + 6)2 + 2 (a + 6) c + c2 = (a + 6 + c)2. Example 3. Factor the following expressions. 1. x2 + 14x + 49. 2. 9 --60 + 02. 3. 9x'^y^ + 30xy + 25. 4. a;2 - 4a;y + 47/2 4. 6a; - 12?/ + 9. 5. 64 a8 - 48 a* + 9. 6. a2 + 62 + c2 - 2 a6 + 2 ac - 2 6c. A difference of two squares. Expressions of this form, or 436 reducible to it, can be factored by aid of the formula: a^-b-'=(a + b) (a - b). Thus, x2 - 2/2 _ 22 + 2 2/2 = x2 - (2/2 -2yz + z"^) = x^- {y - zf = (X + y - 2) (X - 2/ + Z). A very useful device for reducing a trinomial expression to this form is that of making it a perfect square by adding a suitable quantity to one of its terms and then subtracting this quantity from the resulting expression. Thus, x* + x22/2 + 2/* = a;* + 2 ^2^2 + yi- x^y^^ = (X2 + ^2)2 _ y-lyl = (x2 + 2/2 + a;2/) {x2 + y2 _ a;y). Example. Factor the following expressions. 1. x* - 2/6. 2. 6 a3 - 6 a62. 3. 12 aH^ - 1h axy\ 4. 25x2" -49x2™. 5. 36x^-1. 6. x* - 3 x22/2 + ?/*• A sum of two squares. By making use of the imaginary 437 unit i = V— 1, §§ 218, 220, a sum of squares a^ + b"^ can be reduced to the form of a difference of squares and then factored by § 436, the factors being imaginary. For since %" = -!, we have b'' = -{- b^) = - (iby. Hence a"" + b^ = a^ - {ibf = (a + ib) (a - ib). 182 A COLLEGE ALGEBRA As we have seen, §§ 219, 220, i conforms to all the ordinai-j rules of reckoning. One has only to remember when employ- ing it that r = — 1. 438 Sums and differences of any two like powers. In §§ 308, 309, 310 we proved that First. Whether n is odd or eve7i, a--b-=^{a- b) {a—' + a^-^ + • • • + ah"~^ + Z-""')- (1) Second. When n is even, a--b-={a + b) (a"-! - a^-'b + •■- + ab'^~' - b"-'). (2) Third. When n is odd, «" + &"=(« + b) (a"-' - a^-H + ai"-2 + b^-'). (3) Hence the following theorems : 1. a" — b" is always dunsible by 2i — h. 2. a° — b" is divisible by a. + h when n is even. 3. a" + b" is divisible by a, + 10 when n is odd. 4. In every case the quotient consists of the terms ^n-i a"-2b---ab"-2 b"-^ connected by sigyis which are all + when a — b is the divisor, but are alternately — and + ivhen a + b w the divisor. Thus, 1. X6 - 1 = (X - 1) (X5 + X* + X3 + X2 + X + 1). 2. a;6 - 1 = (X + 1) (x^ - X* + x3 - x2 + X - 1). 3. 8 a3 + 27 6^c3 = (2 aY + (3 bcf = (2 a + 3 6c) [(2 af - (2 a) (3 he) + (3 6c)2] = (2 a + 3 6c) (4 a2 - 6 a6c + 9 62c2). Example. Factor the following expressions. 1. 04x3 - 125 2/\ 2. 27 x^ + 1. 3. 16 x* - Sly*. 439 When n is composite. The following theorems are an imme- diate consequence of § 438, (1), (2), (3) and § 436. \. Tfn is a multiple of any integer, p, then a" — b" is exactly divisible by aP — b^. FACTORS OF INTEGRAL EXPRESSIONS 183 Thus, x^ -y^ = (a:2)3 - (y2)3 = (X2 - y2) {X* + x2?/2 + yi), 2. If n is an even multiple of any integer, p, then a" — h" is exactly divisible by zP + b^. Thus, «6 - 2/6 = (X3)2 - (2/3)2 = (X3 + 2/3) (X3 - 2/3). 3. 7/* n is a?i odd multiple of any integer, p, then a" + b° is exactly divisible by aP + b^, ivhether n itself is odd or even. Thus, X6 + 2/6 = (X2)3 + (2/2)3 = (X2 + 2/2) (X* - x22/2 + 2/*). 4. 7/" n zs a ^^^ovt^er o/ 2, ("Ae/i a" + b" can be resolved into factors of the second degree by repeated use of the device exjdained in § 43 C. Thus, x8 + 2/8 = x8 + 2 x42/* + 2/8-2 x^y* = («* + 2/*)2-2x-'2/^ = (x* + 2/* + V2 x22/2) (x4 + 2/* - V2 x22/2). Again, X* + y* + V2x22/2 = X* + 2 x22/2 + 2/* _ (2 - V2)x2y2 = (x2 + 2/2)-^- (2-V2)x22/2 = (x2 + 2/2 + V2 - V2x2/)(x2 + 2/2 - V2 - V2X2/), and so on. As each of these " quadratic " factors can be resolved into two (imagi- nary) factors of the first degree by § 444, the complete factorization of rt" + 6" is always possible when n is a power of 2. "When n is composite, it is best to begin by resolving a" + i", or a" — b", into two factors whose degrees are as nearly equal as possible. It will always be possible to factor at least one of the factors thus obtained. Thus, the factorization of x^ — 2/6 given under 2 is the best. Continu- ing, we have X6 - 2/6 = (X3 + 2/3) (X3 - 2/3) = (X + 2/) (x2 - X2/ + 2/2) (X - y) (x2 + X2/ + 2/2). 184 A COLLEGE ALGEBRA Example. Factor the following expressions. 1. X* + y^. 2. x8 - 2/8. 3. a;9 + ^9. 440 The theorems of §§ 438, 439 also apply to expressions of the form a'" ± V' when m and n are multiples of the same integer p. Thus, X6 - yl5 = (X2)3 - (i/5)3 = (X2 - 2/5) (X* + x2l/5 + 2/10). EXERCISE XVII In the following examples carry the factorization as far as is possible without introducing irrational or imaginary coefficients. 1. 4 z^y - 20 x22/2 + 25 X2/3. 2. 28 te'^ - 63 «2/2. 3. x2 ^ 4 y2 + 9 z2 _ 4 a;2/ - 12 2/2 + 6 zx. 4. (7 a2 + 2 62)2 _ (2 a2 + 7 62)2. B. (7x2 + 4x-3)2-(x2+4x + 3)2. 6. 4(1 - 62 - a6) - a2. 7. X* + x2 + 1. 8. a* - 6 a262 + 6<. 9. a« + 4 a2 + 16. 10. 9 x* + 15 x2y2 + 16 y^. 11. 4 (a6 + cd)2 - (a2 + 62 - c2 - cZ2)2. 12. 570x62/3 -9 2/1^ 13. x» - 2/9. 14. xi2 - 2/I2. 15. xw + 2/10. 16. x5 _ 32. 17. x' + 2/". FACTORS FOUND BY GROUPING TERMS 441 Sometimes the terms of a polynomial in x can be combined in groups, all of which have some common factor, as F. This common factor F is then a factor of the entire expression. Compare § 433. Example 1. Factor x^ + 3 x2 - 2 x - 6. Noticing that the last two coefficients are equimultiples of the first two, we have x8 + 3x2-2x-0 = x2(x + 3) - 2{x + 3) = (x2 - 2) (z + 3) = (X + V2) (x - V2) (X + 3). FACTORS OF INTEGRAL EXPRESSIONS 185 Example 2. Factor x^ + 2x^ -\- 2x + I. Combining terms whiich have like coefficients, we have :=(X2_X+ 1)(X + 1) + 2X(X + 1) = (X2 - X + 1 i- 2 X) (X + 1) = {X2 + X + 1) (X + 1). Sometimes this can be accomplished by first separating one of the given terms into two terms. Example 3. Factor x^ + 4 x^ + 5 x + 6. We have x3 + 4 x2 + 5 X + G = x3 + 3 x2 + x2 + 3 X + 2 x + 6 = x2 (X + 3) + X (X + 3) + 2 (X + 3) = (x2 + X + 2) (X + 3). Consider also the following example. Example 4. Factor x* + 2 x^ -f- 3 x^ + 2 x + 1. We have x* + 2x3 + 3x2 + 2x + l = x* + 2x3 + x2 + 2x2 + 2x + l = (x2 + x)2 + 2 (x2 + x) + 1 = (X2+X + 1)2. EXERCISE XVIII Factor the following expressions. 1. x*-x3 + x-l. 2. x^i -x5-8x2 + 8. 3. x4- 2x3 + 2x-l. 4. x'5- 7x2-4x + 28. 5. x6 - x*y- - x-y* + ?/. 6. x^ + 2 x2 + 3 x + 2. yf. x5 + 2x< + 3x3 + 3x2 + 2x + l. 8. x* + 4x3 + 10x2 + 12x + 9. FACTORIZATION OF QUADRATIC EXPRESSIONS The quadratic x- + px + q, factored by inspection. This is 442 sometimes possible, when p and q are integers. Since (x + a)(x + b) = x"" + (a + b) x + ab, we shall know the factors of x"^ + px + 5- if we can find two numbers, a and h, such that a + I — v and ab = q. 186 A COLLEGE ALGEBRA Two such numbers always exist, § 444, though they are seldom rational. But when rational they are integers, § 454, and may be found by inspection, as in the following examples. Example 1. Factor x2 + 13 x + 42. We seek two integers, a and 6, whose product is 42 and sum 13. As both ah and a + 6 are positive, both a and h must be positive. Hence among the positive integers whose product is 42 — namely, 42 and 1, 21 and 2, 14 and 3, 7 and 6 — we seek a pair whose sum is 13, and find 7 and 6. Hence x2 + isx + 42 = (x + 7) (x + 6). Example 2. Factor x2 - 13 x + 22. Here both a and h must be negative ; for their product is positive and their sum negative. Hence, testing as before the pairs of negative inte- gers whose product is 22, we find — 11 and — 2 ; for — 11 — 2 = — 13. Hence x2 - 13x + 22 = (x - 11) (x - 2). Example 3. Factor x2 - 9 x - 22. Here, since ah is negative, a and h must have opposite signs ; and since a + 6 is negative, the one which is numerically gi-eater must be negative. Hence we set — 22 = — 22 x 1 = — 11 x 2, and, testing as before, find a = - 11 and 5 = 2 ; for - 11 + 2 = - 9. Hence x2 - 9 x - 22 = (x - 11) (x + 2). Example 4. Factor the following expressions. 1. x2 + 3x4-2. 2. x2-lGx + 15. 3. x2-4x-12. 4. x2 + X - 30. 5. x2 + 20 X + 96. 6. x2 - 21 x + 80. 443 The quadratic ax^ -f bx + c factored by inspection. This is sometimes possible, when a, h, and c are integers. By multiplying and dividing by a, we may reduce ax"^ -{-hx-\-c to the form \_(axY + h (ax) -{- ac] / a, and then factor the bracketed expression with respect to ax by the method just explained, namely, by finding two integers whose product is ac, and sum b. Example 1. Factor 2 x^ + 7 x + 3. Wehave 2xC + 7x + 3 = <^^li±lia^bJ ^(2x+^)(2x+2) = (x + 3)(2x-fl). i FACTORS OF INTEGRAL EXPRESSIONS 187 Example 2. Factor abz^ + {a- + b~}x + ah. ah _ {abx + a-) {ahz + b") ~ ^6 = (6x + a) (ax + h). Example 3. Factor IQx"^ + 12x- 63. In this case it is not necessary to multiply and divide by 16, for we have 16 x2 + 72 X - 63 = (4 xf + 18 (4 x) - 63 = (4x + 21)(4x-3). Example 4. Factor the following expressions. 1. 6x2-13x + 6. 2. 5x2 + 14x-3. 3. 14 x2 + X - 3. 4. 18 x2 + 21 X + 5. 5. 49 x2 + 105 X + 44. 6. abx"^ - {ac - b^-) x - be. The quadratic x^ + px + q or ax^ + bx + c factored by com- 444 pleting the square. While the preceding methods apply in particular cases only, the following is perfectly general. «... (..?)•=...,..?, we can make x^ + px a perfect square by adding ^^ that is, the square of half the coefficient of x. This process is called covijiletlng the square of x^ + px. 1, We shall not affect the value of x^ -^ px -\- q, if we both add and subtract jt»^/ 4. But by this means we can transform the expression into the difference between two squares and then factor it by § 436. Thus, x"^ +px + q = x^+px + - ■J-'^S' =(.+iy-^ 4 4 =(..f + ^^)(.,|_vS).,, 188 A COLLEGE ALGEBRA 2. Since ax^ + bx -{- c = ai x^ + - x -{- -\f we may obtain the factors of this expression by substituting b /a for 2^ and c/a for q in (1). Simplifying the result, we have 'b''-Aac\ G + lc- V62-4«c' 2a ) 2a ax-' + bx-^c = a[ X + — + x + -- r—— • (2) Example 1. Factor x^ - 6 x + 2. We have x2-6x + 2 = z2_6x + 32-32 + 2 = (X - 3)2 _ 7 = (X - 3 + V?) (X - 3 - V7). Example 2. Factor x2 + 8 x + 20. We have x^ + 8 x + 20 = x^ + 8 x + 42 - 42 + 20 = (X + 4)2 + 4 = (X + 4)2 - 4 i2 =r (X + 4 + 2 (x + 4 - 2 1). Here we first obtain a sum of squares, (x + 4)2 + 4, and then transform this sum into a difference by replacing 4 by — 4 i-, § 437. The factors are imaginary. Examples. Factor 3x2 — 5x + 1. ;x2 -^ 5x + 1 = 3 rx2 _ -X + -l '5\2 /5\2 r We have 5\2 13- 36. „/ 5 V]3\/ 5 Vl3\ Example 4. Factor the following expressions. 1. x2 + 10x + 23. 2. x2-10x + 24. 3. x2 - 12 X + 45. 4. x2 + X + L 5. 2 x2 + 3 X + 2. 6. x2 - 4 ax - 4 62 -I- 8 a6. FACTORS OF INTEGRAL EXPRESSIONS 189 Homogeneous quadratic functions of two variables. The methods 445 of §§ 442-444 are applicable to quadratics of the form ax^ + hxy + cy'^. Example 1. Factor x^ — 8xy + Wy'^. We have x2 - 8 xy +Uy^ = x'^-8xy + 16 1/2 - 2 ?/2 = (X - 4 2/)2 - 2 2/2 = [X - (4 + V2) y] [X - (4 - V2) y]. Example 2. Factor the following expressions. 1. x2 + 5xy + 4 y~. 2. x2 — x?/ + y^. Non-homogeneous quadratic functions of two variables. Such 446 functions are ordinarily jirime. But when composite, they may be factored as in the following example. Example 1. Factor ^ = x2 + 2x2/ - 8 2/2 + 2 x + 14 ?/ - 3. If A is composite, it is the product of two polynomials of the first degree. Moreover its terms of the second degree, x^ + 2xy — 8 2/2, must be the product of the terms of the first degree in these polynomials. We find by inspection that x2 + 2 xy — 8 2/2 = (x + 4 ?/) (x — 2 y). Hence, if A is composite, there must be two numbers, I and m, such that we shall have x2 + 2x2/-8 2/2 + 2x + 14y-3 = {x + Ay + l){x-2y + m) = x2 + 2 X2/ - 8 2/2 + (Z + m)x + {im-2l)y + Im. (1) But to make (1) an identity, we must have, § 285, l + m = 2 (2), - 2 Z + 4 m = 14 (3), Im = - S (4). From (2) and (3) we find I = — I, m = 3. And these values satisfy (4); for - 1 . 3 = - 3. Therefore x2 + 2 x?/ - 8 ^2 + 2 x + 14 2/ - 3 = (x + 4 2/ - l)(x - 2 ?/ + 3). Note. The example shows how exceptional these composite functions are. If, leaving A otherwise unchanged, we replace the last term, — 3, by any other number, A becomes prime ; for (4) will then not be satisfied by Z = - 1, m = 3. This method is also applicable to homogeneotis quadratic functions of three variables. 190 A COLLEGE ALGEBRA Thus, to factor x"^ + 2xy - 8y'^ + 2xz + liyz - Sz^, we set «2 + 2 xy - 8 1/2 4- 2 xz + 14 7/z - 3 22 = (X + 4 2/ + ;«) (X - 2 2/ + mz) and then proceed as above, again finding I = — I, m — S. Example 2. Factor 2 x2 - 7 xy + 3 ?/ + 5 a;^ _ 5 ^/z + 2 z2. Example 3, Show that x2 — 2/2 + 2 x + y — 1 is prime. 447 Polynomials of the nth degree. We have shown that every polynomial of the second degree, a^yx" + a^x + a^, is the product of factors of the first degree. The like is true of polynomials in X of every degree, though no general method exists for finding these factors; in other words, Theorem. Every ■polynomial in x, of the nth degree, f (x) = aox" + aiX"-^ -\ h an_iX + a„ is the product ofn factors of the first degree; that is, there are n binomials, x — ^1, x — )82, • • •, x — j8^ sucli that f(x) = ao(x-/?0(x-A)---(x-^„). The proof of this theorem will be given later. 448 Corollary. A homogeneous polynomial in two variables, x and y, of the nth degree, is the product of n homogeneous factors in X and y, of the first degree. Thus, the homogeneous polynomial aox^ + aix^y + 02X^2 + azy^ may be derived from aox^ + aix2 + a^x + as by substituting x/y for x and multiplying the result by y^. But by § 447, aoX^ + aix2 + otox + a.3 = ao (x - ft) (x - /So) (x - /Sg), and if we substitute x/y for x in this identity and then multiply both members by y^, we obtain aox3 + aix2y + a^xy"^ + 03?/^ = ao(x - ^^y) (x - fty) (x - /Ssy). EXERCISE XIX Factor the following expressions. 1. x2 _ 14 jc + 48. 2. x2 - 21 X - 120. .3. 6x2_63x-22. 4. 16x2 + 64x + 63. 5. 54x2-21x + 2. 6. 12 x^ + 20 xy - 8 2/'. FACTORS OF INTEGRAL EXPRESSIONS 191 7. X* - 13 x2 + 36. 8. x-^y - 3 x2y2 _ 18 xy^ 9. x2 - 3 X + 3. 10. 3 x2 + 2 X - 3. 11. x2- 4x2/ -22/2. 12. x2-6ax-962_i8a6. 13. a6x2-(a2 + 62)x-(a2-62). 14. x2 + M + dx + 6x + cx2 + cdx. 15. x2'-8x;/ + 152/2 + 2x-4?/-3. 16. x2+3xt/ + 22/2 + 32X + 5?/2 + 222. APPLICATIONS OF THE REMAINDER THEOREM AND SYNTHETIC DIVISION On finding factors by aid of the remainder theorem. Let f(x) 44S denote a polynomial in x. By the remainder theorem, § 415, if b denote a number such that f(b) — 0, then x — b is & factor of f(x). We can sometimes find such a number b by inspection. Example. Factor /(x) = x^ — 5x + 4. Since /(I) = 1 — .5 + 4 = 0, x — 1 is a factor of /(x). l + 0-5 + 4[l Dividing /(x) by x - 1, we obtain the quotient 1 1— X2 + X-4. 1 1 - 4, jjp^^g ^^^^ = (X - 1) (x2 + X - 4). Note. Observe that vrhenever, as in this example, the algebraic sum of the coefficients of /(x) is 0, x — 1 is a factor of /(x). Polynomials with integral coefficients. If asked to factor a 450 polynomial, /(a*), with integral coefficients, it is usually best to look first for any factors of the first degree with mtegral coefficients that it may have. These may always be found by aid of the following principles, §§ 451, 452. A polynomial f(x) = aoX" + ais;""' + • • • + «„, with integral 451 coefficients, may have a factor of the form x — b, where b is an integer. But if so, b must be a factor of a„, the constant term o/f(x). Thus, let f{x) = OoX^ + aix2 + UnX + as. If x - 6 is to be a factor of /(x), we must have, § 415, f{b) = aob^ + ai62 4- a^b + as = 0, and therefore (aob^ + aib + 02) & = - as. Therefore, since ao62 + aib + 02 denotes an integer, 6 is a factor of 03. 192 A COLLEGE ALGEBRA Hence all such factors x — b may be found as in the follow- ing example. Example. Factor /(x) = 3 a;^ - 3 x* - 13 x^ - 1 1 a;2 _ iq x - 6. The factors of the constant term, — 6, are ±1, ±2, ±3, ±6, and b must have one of these values if x - 6 is to be a factor of f(x). We test these values of b as follows by synthetic division. 3_3_13_11_10 — 61— 1 Since /(I) ?i 0, x — 1 is not a factor. -3+ 6+ 7+ 4 + 6 We therefore begin by testing X- (- 1) 3 3~6^^ T^ 4^ 6 01—1 <5r ^ + 1- The division proves to be _3, g_ 24. 6* exact, the quotient being Qi = 3 x* 3 ^Tq II 2 '^ 6 13 — 6 x^ — 7 x2 — 4 X — 6, and the remainder 9 Q _(. g' 0. Hence x + 1 is one factor of /(x), 3 Q 2 ^"^^ ^^ '^ ^^^ product of the remaining factors. We have next to factor Qi. It also proves to be exactly divisible by X + 1, the quotient being Q2 = Sx' - Ox^ + 2x - 6. To factor Q2, vi^hose constant term is also — 6, we test successively X + 1, X — 2, X + 2, but in each case obtain a remainder which is not 0. Hence none of these are factors. But testing x — 3, we find that it divides Q2 exactly, the quotient being Q3 = Sx^ + 2. Therefore /(x) = (x+l)2(x-3)(3x2 + 2). 452 A polynomial f{x) = a^"" + ayx''-^ H 1- o,„ with integral coefficients, may have a factor of the form ax — /?, where a and /? denote integers which have no common factor. But if so, a must be a factor of ao, and (3 a factor of a„. This theorem includes that of § 451. Thus, let /(x) = aox' + aiX^ + a^x + 03. If ax - ;3, or a{z - /3/cr), is to be a factor of /(x), we must have, § 415, K',) 8\ efl /32 B and therefore aoiS^ + ai/32a + a2/3a2 + aaa^ = 0. (1) From (1) we obtain aojS' = - (ai/32 + oaiSa + aza"^) a. (2) Therefore, since a-[^ + OnPa + 030-2 is an integer, a- is a factor of ao,8^. But a has no factor in common with /3'', § 492, 2. Hence a is a factor of ao, § 492, 1. Again from (1), {ao^ + a^^a + a,2a2)^ = - aacr^, (3) whence, reasoning as before, we conclude that ^ is a factor of Og. FACTORS OF INTEGRAL EXPRESSIONS 193 Hence all siich factors ax — /3 niay be found as in the following example. Example. Factor f(x) = 6 x* + 5 x' + 3 x2 - 3 x - 2. If crx — /3 is to be a factor of /(x), a must have one of the values ± 1, ±2, ±3, ±6, and /3 one of the values ±1, ± 2 ; therefore /3/a: must have one of the values ±1, ±2, ±1/2, ± 1/3, ± 2/3, ± 1/6. We may test az — ^ for these various values of ^/a by dividing /(x) by X — /3/a synthetically. If the division is exact and Q denotes the quotient, then ax — ^ is a factor of f{x) and Q/a is the product of the remaining factors, § 412, 3. Testing x — 1, x + 1, x — 2, x + 2, successively, we find that none of them divides /(x) exactly. But x+ 1/2 6 + 5 + 3-3- 2- |- 1/2 ^^gg^ ^^^ quotient being Qi = 6 x^ + 2 x^ - -^^^-^ — — — + 2 X — 4. Hence 2 x + 1 is one factor of "*" " "^ ~ ' fix) and the product of the remaining factors rf + i + i -[-/^ is Qi/2 = 3x=' + x2 + x-2. ?±^±-? We next proceed to factor Qi/2. If "^ "^ ' ax — /3 is to be a factor, /3/(:ir must have one ■^ "^ ^ + ^ of the values ±1, ±2, ±1/3, ± 2/3. But we already know that x — 1, x + 1, x — 2, x + 2 are not factors, since they are not factors of f{x). Testing x — 1/3, x + 1/3, we find that neither of them divides Qi/2 exactly; but x — 2/3 does, the quotient being Q2 = 3x- + 3x + 3. Hence 3x — 2 is a factor of Qi/2, and the product of the remaining factors is (^.2/ 3 = x- + x + 1. Therefore /(x) = (2x+ l)(3x-2)(x2 + x+ 1). Note. It often becomes evident before a division by x — b or x — ^/a 453 is completed that the division cannot be exact. Thus, the reckoning here given suffices to prove that x — 2 will not divide 5x^ — ix^ + x + 8 exactly ; for since the ~ + + L "divisor " 2, the last coefficient of Q already found, namely 6, and the unused coefficients of the divi- 5 6 dend, namely 1 and 8, are a\\ positive, the remaining coefficients of Q and R must be positive, that is, E cannot be 0, Similarly we may conclude from the reckoning ^ + '^ + '^ - ^ [1/3. here given that x - 1 /3 will not divide Sx^ + x2 - — - + X — 2 exactly. For the number which occurs next in the reckoning, namely, 2 • 1 /3, or 2/3, is a fraction, and this will cause the remaining coefficients of Q and R to be fractional, so that R cannot be 0. 194 A COLLEGE ALGEBRA 454 It follows from § 452 that a ^polynomial f (x) = x" H \- a^, whose leading coefficient is 1, the rest being integers, cannot vanish for a rational fractional value of x. For if /(/3/a) - 0, then/(x) must be exactly divisible by ax - ^ pnd therefore a must be a factor of 1, which is only possible when cr = j^, 1. 455 Factoring polynomials and solving equations. It follows from § 350 that the problem of resolving a polynomial f{x) into its factors of the first degree is essentially the same as that of solving the equation f{x) = 0. Example 1. Solve f{x) = 2 x* + z^ - 17 x'^ - 16 x + 12 = 0. By § 452 we find that /(x) = (2 x - 1) (x + 2)2 (x - 3). Hence the equation /(x) = is equivalent to the four equations 2x-l = 0, x + 2 = 0, x + 2 = 0, x-3 = 0. Therefore the roots of f(x) = are 1 /2, - 2, - 2, and 3. ■ Example 2. Solve x^ + 3 x2 = 10 x + 24. Transposing, x^ + 3 x2 - 10 x - 24 = 0. Factoring, (x + 2) (x - 3) (x + 4) = 0. Hence the required roots are — 2, 3, and — 4. EXERCISE XX Factor the following expressions. 1. x3-7x + 6. 2. x3 + 6x2 + llx + 6. 3. x4 - 10x3 + 35x2 -50x + 24. 4. x*-2x2 + 3x-2. 5. 6x^- 13x2-14x-3. 6. 2x3 -5x2y- 2x^2 + 22/8. 7. 2x*-x3-9x2 + 13x-5. 8. 4x6 - 41 x" + 46x2 - 9. 9. 6x5 + 19x* + 22x3 + 23x2 ^ i6x + 4. 10. 5x6 - 7 x5 - 8x* - x3 + 7 x2 + 8 X - 4. Solve the following equations. 11. x2-4x-12 = 0. 12. 6x2-7x + 2 = 0. 13. x2_5x = 14. 14. x2 + 6x = 2. 15. x8-9x2 + 26x = 24. 16. x* + 2x3 - 4x2 - 2x + 3 = 17. x«-l=0. 18. 10x3-9x2-3x4-2 = 0. FACTORS OF INTEGRAL EXPRESSIONS 195 EXERCISE XXI The following expressions can be factored by methods explained in the present chapter. Carry the factorization as far as is possible without introducing irrational or imaginary coefficients. 1. 6 X2/ + 15 X - 4 2/ - 10. 2. a'^bc - ac^-d - abH + bcd^. 3. a3 (a - 6) + b^ {b - a). 4. a^ - 81 abK 5. a*b - aP-b^ + a^fia _ a6*. 6. 3 abx'^ -Qaxy + bxy - 2 2/2, 7. 3x6-192 2/6. 8. (x2 + x)3-8. 9. 64x62/3-2/^5. 10. x2-(a - &)x-a6. 11. x2«_3a;«_18. 12. a;_a;2 + 42. 13. 3x« + 3x3-24x-24. 14. x^ - 9x3 + 8x2 - 72. 15. 2 xc - a2 + x2 - 2 a6 + c2 - 62, iq ^i ^^z _ qq) + 64. 17. a2 - 2 a& + 62 - - 5 « + 5 6 + G. 18. x* - 10 x"y'- + 9 y*. 19. 6x2 - 7x2/- 52/2 -4x- 22/. , 20. x* - (a2 + 62) x2 + a262. 21. 4(XZ + M2/)2-(x2-2/2 + 22_u2)2. 22. 14 x2 + 19 X - 3. 23. 1 + 19 y - 66 2/2. 24. x2/3 + 55 x:^y^ + 204 x^y. 25. a* - 18 a262c2 + 81 b*c^. 26. (x2 - 7 x)2 + 6 x2 - 42 x. 27. 8 (X + ?/)''- 27 (X - 2/)3. 28. (x - 2 2/) x^ - (2/ - 2 x) j/S. 29. x2 + a2 - 6x - a6 + 2 ax. 30. x^ - 2/^ - (x - 2/)^. 31. x5 - x4 - 2 x3 + 2 x2 + X - 1. 32. 6-' + 62 + 1. 33. 2 x2 + 7 X2/ + 3 2/2 + 9 X + 2 2/ - 5. 34. a* + 4. 35. x2-X2/-22/2+4xz-52/z + 3z2. 36. 4a* + 3 a262 + 96*. 37. x2 - 8 ax - 40 a6 - 25 62. 38. x^ + x* + 1. 39. (x2 + 2x-l)2-(x2-2x + l)2. 40. (ox + 62/)2 - (6x + ay)2, 41. x3 - ax2 - 62x + a62. 42. x* + 6x3 _ aH - a^b. 43. a2-962+ 126c-4c2. 44. 8a3 + 12 a2 + 6a + 1. 45. X* - 2 x3 + 3 x2 - 2 X + 1. 46. (ax + byf + {bx - ay)^. 47. 4x5+4 x*-37x3-37x2+9x+9. 48. x*-4x + 3. 49. x2 + 5ax + 6a2-a6-62. 50. 15x3 + 29x2 - 8x - 12. 51. a6cx2 + (a262 + c^) x + abc. 52. 2 x^ - ax2 - 5 a*x - 2 a^. 196 A COLLEGE ALGEBRA 53. (a - 5) x2 4- 2 ax + (« + b). 54. x^^ - y'^\ 55. X* -6x^ + 1 X- + 6x -8. 56. 4 x^ - 3 x - 1. 57. 3x5 - lOx* - 8x3 - 3x2 + lOx + 8. 58. 5x* + 24x3-15x2- 118x + 24. 59. a'^bc + ac- + acd - abd - cd - d"^. 60. X* + y^ + z* -2 x-y- - 2 yH'^ - 2 zH'^. VII. HIGHEST COMMON FACTOR AND LOWEST COMMON MULTIPLE HIGHEST COMMON FACTOR 456 Highest common factor. Let .1 , B, • •• denote rational, inte- gral functions of one or more variables, as x, or x and y. If A, B, ■ ■ ■ have no factor in common, we say that they are prime to one another. But if tliey have any common factor, they have one whose degree is highest ; we call it their highest common factor' (H.C.F.). Thus, x2 + 2/2 and x + ?/ are prime to one another. But 4x2/z5, 8x2*, and 'kx'^yz^ have the common factors x, z, z"^, z', xz, xz2, xz^, and their highest conunon factor is xz^. 457 Notes. 1. We here ignore common numerical factors. 2. It is sometimes said of two or more functions which are prime to one another that their H.C.F. is L 3. The numerical value of the H.C.F. of A and B is not necessarily the greatest common divisor of integral numerical values of A and B. Thus, the H.C.F. of (2x + l)x and (x - l)x is x. But when x = 4, the values of (2 X + 1) x and (x — 1 ) x are 36 and 12, and the greatest common divisor of 3() and 12 is not 4, but 12. 458 Theorem 1 . The H.C.F. o/" A, B, • • ■ is the product of all the different common prime factors of K, B, •••, each raised to the lowest 2wwer in which it occurs in any of these functions. The truth of this theorem is obvious if we suppose each of the functions A, B,--- expressed in the form of a product of HIGHEST COMMON FACTOR 197 powers of its different prime factors and, as in § 430, assume that there is but one such expression for each function. Thus, the different common prime factors of xyz^, xz*, and x^yz^ are X and z, and the lowest powers of x and z in any of these functions are x and zK Hence the H.C.F. is xz^ Observe tliat if it were possible to express a given function in more than one way in terms of its prime factors, the process described in the theorem might lead to various results corresponding to the various ways of expressing A, B, ■ • -, and there might be more than one common factor of highest degree. Applications of this theorem. When the given functions can 459 be completely factored, their H.C.F. may be written down at once by aid of the theorem of § 458. Example 1. Find the H.C.F. ofx^y"^ -6x*y^ + 9x^y^ and x*y - 9xV- We have x^y"^ - 6 x*?/3 + 9x^y* = x^y^ (x - 3 y)^ and x*y-9x^y^ = x^y{x-Sy){x + 3y). Hence the H.C.F. is x^y (x - 3 y). Example 2. Find the H.C.F. of the following. 1. 2x'^y-z^, Sx^y% and ix^y*. 2. x^ — y'^, x^ + 2 xy + y-, and x^ + y^. 3. x2 - X - 6, x2 + 6 X + 8, and x^ + 5 x + 6. 4. x3-6x2 + llx - 6 and 2x3 - 9x2 + 7x + 6. If the prime factors of one of the functions A, B, ••• are 460 known, we can find by division or the remainder theorem which of them, if any, are factors of all the other functions. The H.C.F. may then be obtained by aid of § 458. Example 1. Find the H.C.F. of /(x) = x2 - 3 X + 2 and (x) = x* - 3 x^ + 5 x2 - 8 x + 5. By inspection we have f{x) = (x - 1) (x — 2). Testing x = 1 and x = 2 in ^(x), we find (1) = 0, but 'od>(ct, B ■ C • D • • •. For, as just demonstrated, A is prime to BC. And since A is prime to BC and also to D, it is prime to the product BCD ; and so on. 484 Theorem 3. A composite function has one and but one set of jjrime factors. For let P denote the given function and n its degree. If P is composite it has some factor A. If ^4. in turn, is composite, it has some factor B. Continuing thus, we must ultimately come upon a prime function ; for the degrees of the successive functiims P, A, B, ■■• begin with the finite number n, decrease, and cannot fall below 1. Let F denote this prime function. It is one of the prime factors of P, § 427, and we have P= FM, where M is integral. Similarly if M is composite, a prime function F' exists such that M= F'M', and therefore P = FF'M', where M' is integral. Continuing thus, we reach the conclusion that a .series of prime func- tions F, P', P", ■ • ■ exists, whose number cannot exceed n, such that P = F-F'F" ■■■. Hence P has at least one set of prime factors. Moreover P can have but one such set of factors. For, suppose that P = FF'F"-- = G-G'- G" ■■■ where G, G', G", • • ■ also denote prime functions. Then G cannot be prime to all the functions P, P', P", • • ■ , for, if so, it would be prime to their product P, § 483, whereas it is a factor of P. Suppo.se, therefore, that G is not prime to P, for example. Then G and P have a common factor. But G and P are prime functions, and two prime functions can have no factor in x but themselves in common. Hence G differs from P by a numerical factor, as c, at most, and we have G = cF. HIGHEST COMMON FACTOR 211 But substituting this value of G in the identity FF'F" •• ■ = GG'G" • • •, and dividing both members by F, we have F'F"---^cG'G"---, from which it follows by a mere repetition of our reasoning that G' differs from one of the functions F\ F", • • • by a numerical factor at most. Continuing thus, we reach the conclusion that the set of fvmctions G, G\ G", • • • differs from the set F, F', F", • ■ • at most by numerical factors or in the order in which they are arranged. Corollary. A composite function can be expressed in only one 485 way as a product of powers of its different prime factors. This follows at once from the identity P = i^- F' • F" • • • , if we replace each set of equal factors in the product F ■ F' ■ F" ■ ■ • by the correspond- ing power of one of these factors. Irreducible factors. By the irreducible factors of an integral 486 function with rational coefficients, we usually mean the factors of lowest degree with rational coefficients. Thus, while the prime factors of (x — 1) (x^ — 2) are x — 1, x — v^, X + V2, the irreducible factors are x — 1 and x^ — 2. From the theorems just demonstrated and the theorem of 487 § 469, it follows that A reducible integral function with rational coefficients can be expressed in only one way as a product of powers of its different irreducible factors. DIGRESSION IN THE THEORY OF NUMBERS Theorems analogous to those just demonstrated hold good 488 for integral numbers. We shall employ the letters a, b, and so on, to represent integers, positive or negative (not 0), and shall mean by a factor of a any integer which " exactly divides a. A prime number is an integer which has no other factors 489 than itself and 1. 116|325[2 232 ri= 93|ll6[l 93 r2= 23|tt3 [i 92 n= 1 Hence, starting w 212 A COLLEGE ALGEBRA 490 If two integers, a and b, have no common factor except 1, a is said to be prime to b. 491 Theorem, i/'a is prime to b, two integers, m and n, can always be found such that ma + nb = 1. For since a is prime to 6, if we apply the usual method for finding the greatest common divisor, we shall obtain 1 as the final remainder. We may deduce the theorem from this fact by the reasoning of § 479. Thus, let a - 325, b = 116. Applying the method for finding G.C.D., we have 325 = 2-110 + 93, or 93 = 325-2- 116 (1) 116 = 1 - 93 + 23, or 23 = 116 - 1 • 93 (2) 93 = 4 • 23 + 1, or 1 = 93 - 4 - 23 (3) fith (3), and substituting first the value of 23 given by (2), and then the value of 93 given by (1), we have 1 = 93 - 4 . 23 =.5.93 -4.116 = 5-325- 14- 116. Therefore 5 . 325 + (- 14) - 116 = 1. Hence we have found two integers, m = 5 and ?i = — 14, such that ?u-325 + n- 116 = 1. And similarly in every case. Example. Find integers m and ?i such that 223 m + 125 n = 1. 492 Corollaries. From this fundamental theorem we may derive for integral numbers theorems analogous to those derived for integral functions in §§ 481-485, and by the same reasoning. In particular we may prove that 1. If Si is ])ri?ne to b, a/id the product ac is divisible by b, then c is divisible by b. 2. If a, is prime to b and c, then a is prime to be. 3. A composite number can be expressed in one ivay, and hut one, as a product of powers of its different prime factors. RATIONAL FRACTIONS 213 VIII. RATIONAL FRACTIONS REDUCTION OF FRACTIONS Fractions. Let A and B denote any two algebraic expres- 493 sions, of which B is not 0. The quotient of A by B, expressed in the form A / B, is called a fraction ; and A is called the numerator, B the denominator, and .1 and B together the terms of this fraction. When both A and B are rational, A ] B is called a rational 494 fraction. When both A and B are integral, A / B \s called a simple 495 fraction ; but if .4 or B is fractional, A / B is called a complex fraction. A simple fraction is called a proper or an improper fraction, 496 according as the degree of its numerator is or is not less than that of its denominator. ^, x-y ,2z2_3 2a;2+i x^ - 3 . Thus, and are proper, and improper, X2 + 2/2 x3 + 1 ^ ' X2 + 1 x2 + 1 An improper fraction whose terms are functions of a single 497 variable can be reduced to the sum of an integral expression and a proper fraction, § 400. This sum is called a mixed expression. T,, 2x2+1 1 x3-3 x + 3 Thus, = 2 — - , = x . X2 + 1 X2 + 1 X2 + 1 X2 + 1 Allowable changes in the form of a fraction. These depend on 498 the following theorem, § 320, 1. The value of a fraction remaij^s unchanged ichen its mimer- ator and denominator are multiplied or divided by the same expression (not 0). In particular, we inay change the signs of both numerator aiid denomina- tor, this being equivalent to multiplying both numerator and denominator by — 1. Changing the sign of the numerator or of the denominator alone will change the sign of the fraction itself, § 320, 3. 214 A COLLEGE ALGEBRA If the numerator or denominator be a polynomial, changing its sign is equivalent to changing the signs of all its terms. a+h—c c—a—b c—a—b a+b—c Thus, = 7 = r = -7 a — b + c b — c — a a — b -j- c b — c — a K the numerator, denominator, or both, are products of certain factors, we may change the signs of an even number of these factors ; but chang- ing the signs of an odd number of them will change the sign of the fraction. Thu (a -b){c- d) ^ (b - a) {c - d) ^ _ (b -a){d- c) ''^' (e-f){g-h) (e-f){h-g) (f-e){g-h)' 499 Reduction of fractions. To simjdifi/ a fraction is to cancel all factors which are common to its numerator and denominator, this being a change in the form of the fraction which will not affect its value, § 498. When this has been done the fraction is said to be in its lowest terms, or to be iri'educihle. We discover what these common factors are, or show that there are none, by the methods of Chapter VII. W"e look first for common monomial factors and other common factors which are obvious by inspection or which can be foimd by aid of the remainder theorem, and when these simpler methods fail we apply the general method of § 465. The following examples will illustrate some of these methods. Example 1. Simplify (aec — ade) / {bde — ebc). „„ , aec — ade ae (c — d) a(c — d) a We have = — ^^ '- = ^ - = bde — ebc be {d — c) b{c — d) b Example 2. Simplify {x^ + x" + x + 0)/{x- + 3 x + 2). By inspection, the factors of the denominator are x + 1 and x + 2. Hence if numerator and denominator have any common factor, it must be one of these. Testing by synthetic division, we find that the numer- ator is not divisible by x + 1, but is divisible by x + 2, the quotient being x2 - X + 3, x^ + x^ + x + 6 x2-x-f-3 Hence x2-l-3x + 2 x + 1 RATIONAL FRACTIONS 215 Example 3. Simplify (x^ + 7 x + 10) / (x^ + 5 x + 6). Subtracting denominator from numerator, we have x3 + 7 X + 10 - (x3 + 5 X + G) = 2 (X + 2). Hence, if the numerator and denominator have any common factor, it must be X + 2, § 461. But the numerator does not vanish when x = — 2. Hence, § 415, the fraction is already in its lowest terms. Example 4. Simplify — ^^ '—^ ^^ '— !^ '-. {a -b)(b- c) (c - a) Here the only possible common factors are a — b, b — c, and c — a. Setting a = 6 in the numerator, we have b^(b ~ c) + b-{c — b), or 0. Hence, § 417, the numerator is divisible by a — 6. And we may show in the same way that it is divisible by 6 — c and c — a. Therefore the numerator is exactly divisible by the denominator. But the two are of the same degree, namely three, in a, b, c. Their quotient must therefore be a mere number ; and since the a^ terms in the two, when arranged as polynomials in a, are a- (6 — c) and — a^ (6 - c) respec- tively, this number is — 1. Hence the given fraction is equal to — 1. Example 5. Simplify (2x3 + 13x2-0x + 7)/{2x* + 5x-5 + 8x2-2x + 5). By §465, we find that the H.C.F. of numerator and denominator is 2x2 — x + 1. And dividing both , numerator and denominator by 2 x2 _ X + 1, we obtain 2x3 + 13x2-6x + 7 x + 7 2x4 + 5x3 + 8x2- 2x + 5 x'^ + 3x + 5 EXERCISE XXIV Reduce the following fractions to their lowest terms. x5y3 _ 4 x^y^ _ (x6 - y^) (X + y) X^y^ - 2 X^^ (X3 + y3) (x4 _ ^4) x2_4x-21 . 3x2 -8x- 3 x2 + 2x-63 3x2 + 7x + 2 3x2- 186x + 27b2 5x2 + 6ax + a' 2x2-1862 * ■ 5x2 + 2ax-3a2' (x2-25)(x2-8x + 15) g 15x2-46x + 35 {x2-9)(x2-7x + 10) ' ■ 10x2-29x + 21* 216 A COLLEGE ALGEBRA a;*4-a;V + 1/* -^q x^ - tj^ + z"^ + 2 xz (a;3 4. ^3) (a;3 _ ^^3) x:^ ^ yz _ z''^ + 2 xy (1 + xyY - (X + vY ,2 2»ix - my - 12?ix + 6ny 1 — x2 ' 6 mx — 3 »i2/ — 2 Tix + nj/ 2 x3 + 7 x^ - 7 X - 12 x3-8x2 + 19X-12 15. 2x3 + 3x2-14x- 15 2x3-13x2 + 17x4-12 x* + x3 + 5x2 + 4x + 4 ,„ x3-2x2-x-6 2x* + 2x3 + 14x2 + 12x + 12 x* + 3x3 + 8x2 + 8x + 8 (X2 + C2)2 - 4 &2x2 (g - 6)3 + (& - c)3 + (C - X* + 4 6x3 + 4 {,2x2 _ c4 ■ ■ (a - 6) (6 - c) (c - a) OPERATIONS WITH FRACTIONS 500 Lowest common denominator. To add or subtract fractious, we first reduce them to equivaleut fractions having a commoa denominator. Evidently the lowest common multiple of the given denomi- nators will be the common denominator of lowest degree. It is therefore called the lowest common denoviinator (L.C.D.) of the given fractions. Example. Reduce —, — , and — to a lowest common denominator, be ca ab The L.C.M. of the given denominators is abc. To reduce a /be to an equivalent fraction having the denominator abc, we must multiply both its terms by a. Similarly we must multiply both terms of b/ca by b, and both terms of c/ab by c. a _ a2 6 _ ^ c _ c2 be abc ca abc ab abc 501 To reduce two or more fractions to equivalent fractions having a lowest common denominator, find the lowest common miilti- ple of the given denominators. Then in each fraction replace the denominator by this lowest common multiple, and multiply the numerator by the new factor thus introduced in the denominator RATIONAL FRACTIONS 217 Addition and subtraction. For fractions which have a common 502 denominator the rule of addition and subtraction is contained in the formula, § 320, 4, a b c _a -\- b — c d d d d Hence to find the algebraic sum of two or more fractioit^, If necessary, reduce them to a lowest common de?iommato7\ Co7inect the numerators of the resulting fractions by the signs which connect the given fractions, and tvrite the result over the common deriominator. Finally, simjdify the result thus obtained. This rule applies when integral expressions take the place of one or more of the fractions ; for such an expression may be regarded as 2^. fraction 7vhose denominator is 1. It is best to reduce each of the given fractions to its lowest terms, unless a factor which would thus be cancelled occurs in one of the other denominators. Care should bo taken that the expression selected as the lowest common denominator actually is this denominator. A frequent mistake is to treat factors like a — b and b — a, which differ only in sign, as distinct, and to introduce both of them in the lowest common denominator. It is often better to combine the given fractions by pairs. Example 1. Simplify a + b 1 1 26 1 26 a-h a'^-b'^ minator is a- — 6^, and we have a-b a + b 26 a2_62 ' a'-i_62 a^ - 62 a-6 + a + 6-26 2a-26 2 a-2 - 62 a2 ~ 62 a + b a + b a-b a^ - b^ Observe that the denominator of the sum, when reduced to its lowest terms, may be but a factor of the lowest common denominator. 218 A COLLEGE ALGEBRA 1 X3_3x + 1 Example 2. Simplify x 1 — X X"- — 1 Since the first denominator is 1, and the second is — (x — 1), the lowest common denominator is x^ — 1. We therefore have 1 x3 - 3x + 1 _ xs - X X + 1 x3 - 3x + 1 ^ ~ 1 -X X2- 1 ~ X-2 - 1 X2 - 1 X2 - 1 _ x^-x + x + l-x3 + 3x-l _ 3x ~ X2-1 ~X2-1' Example 3. Simplify -i- + _A_ _ _J_ _ _±- . Here it is simpler to combine the fractions by pairs. Thus, 1 1 X + 2 - (X - 2) ^ 4 X - 2 X + 2 ~ X- - 4 x2 - 4 _^ 2 ^o X-l-(x + l) ^ 4 x+1 x-1 x2-l x2-l* _^ 4 _ ^ x2 - 1 - (x2 - 4) ^ 12 x2-4 x-i-l" (x2-l)(x2-4) x*-5x2 + 4' x2-l 2x2 + 3x-2 Example 4. Simplify x^ + x2 - 2 X 2 x3 + x2 + 3 X By aid of the remainder theorem, § 415, we find that x — 1 is a common factor of the two terms of the first fraction, but is not a factor of the second denominator. We therefore simplify the first fraction by cancelling x-1 in both terms, thus obtaining (x + 1) /(x^ + x2 + 2 x). By § 465, the H.C.F. of x^ + x^ + 2x and 2x3 + x2 + 3x - 2 is x2 + x + 2 ; and x3 + x2 + 2x = (x2 + x + 2)x, 2x3 + x2 + 3x - 2 = (2x - 1) (x2 4 x + 2). Before reducing to a common denominator, we inquire whether 2 x — 1 is also a factor of the numerator 2 x2 + 3 x - 2. AVe find that it is, and cancelling it, reduce the second fraction to (x + 2)/(x2 -f x + 2). x2-l 2x2 + 3x-2 Hence x4 + x2-2x 2x3 + x2 + 3x-2 x+1 . x+ X(x2 + X+2) X2 + X + 2 x + l + x' + 2x ^ x^ + sx^.! " «3 4. x2 + 2 X ~ X3 + X2 + 2 X 503 Multiplication. The product of two or more fractions may be feund by applying the following theorem. RATIONAL FRACTIONS 219 The product of two or more fractions is the fraction whose numerator is the product of the numerators of the given fractions, and its denominator the product of their denominators. mi- a c ac For the product of each member by hd is ac (see § 253). Thus, — 6d = ac; and -• '^■6d = '^6-d = ac. §§252,254 hd b d b d In particular, to multiply a fraction by an hitegral expres- sion, multiply its numerator by that expression. The fraction ac /hd should always be reduced to its lowest terms before the multiplications indicated in its numerator and denominator are actually performed. Example 1. Multiply (x-^ - i) / (x3 ^ i) by (^ + 1) / (a; - 1). Wehave 5^:^ x + 1 ^ (x^ - l)(a; + 1) ^ x^ + x + 1 X3 + 1 X - 1 (X3 + 1) (X - 1) X2 - X + 1 Example 2. Multiply 1 - (x - 2) / (x^ + x - 2) by (x + 2) /x. _, / x-2\x + 2 x2 x + 2 X We have ( 1 ) • = = \ x2 + X-2/ X X2 + X-2 X x-1 Involution. From § 503 we derive the rule : 504 To raise a fraction to any given power, raise both numerator and denominator to that power. Thus, ■^-^ ^^ .6/ b" o b) ~b"b For (-1 = •••ton factors to n factors a» .6/ b b b-b- • -to n factors 6" Example. Find the cube of — ab"c^/efg*. a62c3\3 a^b'^c^ We have ( I = V efg* ) e^Pg^"^ Division, To invert a fraction, as a Jh, is to interchange its 505 numerator and denominator. The fraction h / a thus obtained is called the reciprocal of a Jh. 220 A COLLEGE ALGEBRA To divide one fraction hy another, multiply the dividend by the reciprocal of the divisor. a c a d ad Thus, _ ^ _ = _ . - = — b d b c be For the product of each member by c/d is a/6 (see § 253). ™, a c c a J ad c adc a ofnco nn mo Thus, 7-^3X:; = r; and— •- = —- = -. §§252,254,503 b d d b be d bed b In particular, to divide a fraction by an integral expression, multiply its denominator by that expression. Example 1. Divide (X2 - xy + 2/2) /(x2 _ yi) by (X* + x2y2 + yi)/{x^ - y*). x2 - xy + 2/2 X* + x2?/2 + 2/* _ x2 - X2/ + 2/*^ X* - y* We have X2 - 2/2 X* - 2/* X2 - 2/2 X* + x22/2 + J/* X2 + W2 X2 + XT/ + 2/2 Example 2. Divide (x^ + 5 x + 6) / (x + 1) by x2 + 6 x + 8. We have »^ + ^^ + ^ (X2 + 6X + 8) -^ + 5x4-6 x + 1 ' (x + l)(x2 + 6x + 8) _ (x + 2) (X + 3) _ X + 3 ~ (X + 1) (X + 2) (X + 4) ~ x2 + 5x + 4' 506 Rational expressions in general Complex and continued frac- tions. We have shown that the sum, difference, product, and quotient of two fractions are themselves fractions. It therefore follows that every rational expression, § 267, can he reduced to the form of a simple rational fraction. No general rule can be given as to the best method for redu- cing a given rational expression to its simplest form. Seek to avoid all unnecessary steps. In particular be on the watch for opportunities to reduce fractions to their lowest terms, and perform no multiplications of polynomials until the reduction can be carried no further without so doing. 507 As has already been said, § 495, the fraction A / B is called complex when A or B, either or both, is fractional. RATIONAL FRACTIONS 221 In simplifying a complex fraction, A I B^ it is sometimes better to divide A by B at the outset by the rule of § 505, sometimes better first to multiply both A and B by the L.C.M. of all the denominators in A and B. Before taking either of these steps it is often best to simplify A and B separately. a + & , ,\ I / a — h E.amp.el. Simplify (»^;+l)/("^° + l). We have a + h a + h + a — h 2a a — b a — b a — b a-b a -b + a + b 2a a+6 a+b a+6 _ 2a a + 6 _ a + b ~ a — b 2a a — b Observe that when the terms of a complex fraction are simple fractions we may cancel any factors which are common to the numerators or to the denominators of these fractions. Thus, in the third expression above, we may cancel 2 a. Example 2. Simplify (^^ - ^^) / (-^^ + -A_ ) . We may proceed as in Ex. 1 ; but a simpler method is to begin by multiplying both terms by (a + b){a-b). We thus obtain a b a - b~ a + b a{a + b) - b{a - b) _ a"^ + ab - ba + b"^ _ ^ "T~ "~& a{a - b) + b{a + b) ~ a^ - ab + ba + b^ a + b a — b Example 3. Simplify ■ "■^^ Working from the bottom upwards, we have a a a(d/ + e) adf + ae c , , cf b(df+e) + cf bdf+be + cf Complex fractions like that in Ex. 3 are called continued fractions. 222 A COLLEGE ALGEBRA EXERCISE XXV Simplify the following expressions. 2a-Sb 2a + 3b ia^-9b^ X + 1 X2 - 1 X3 + 1 1 1 x2-3x + 2 x2-5x + 6 x2-4x + 3 x+1 x+2_ x+3 (X - 1) (X - 2) (2 - X) (X - 3) (3 - X) (1 - x) 5. -i ^ + -i '-. X + b X + c X — b X — c (a -b){a- c) {b -c){b- a) (c -a){c- b) ^ yz (x + g) ^ zx {y + a) ^ xy (z + a) {x-y){x-z) {y-z){y-x) (z - x) (z - j/) 1 8x*-33x 2x + 6 x + 3-2x 8x3- 27 4x2 + 6x + 1\2 (.,i)%(,,i)-,(.,,i)-_(.,^X,,>)(.,,i) (a + 6)^-c3 {6 + c)3-a3 (c + a)^ - b' a + 6 — c 5 + c — a c + a — 6 x2-4 3x2-14x-5 11. x3-3x2-x + 6 3x3-2x2-10x 12. . ' + ' + ' x4_4x2_a; + 2 2x*-3x3-5x2 + 7x-2 2x* + 3x3-2x2-2x + l -(--i)-(''4)- -a-^D<'"->- ,^ «2_ 6x + 6 x2 + 7x + 12 X2 + X-6 15. x2 + 3x-4 x2-8x + 15 x2-4x-5 ifi 1 Ji rx-1 1/x-l (x-2)(x-3) \-t1 RATIONAL FRACTIONS 223 ^^_ ax + x^_26x^3_ ^^_ (.._^._,. + 2,z)^^-^ + ^ 19. 2 6 - ex (a + x)2 " ' X - y - z / a + b a^ + b^\/a + b a"^ + b^ \ 1 1_ 1 1_ a^b J__j_ 20 ^ y + zy x + z ^^ b a a* 6* 1 1 1 1 a b /I . 1\2 z y + z y x + z b („-.) 22. ^^=^ 23. x + ^ x-2- x + 1 x-1 « + !_ x-2" »' INDETERMINATE FORMS Limits. Suppose the variable x to be taking successively 508 the values 1/2, 3/4, 7/8, 15/16, and so on without end; then evidently x is approaching the value 1, and in such a manner that the difference 1 — x will ultimately become and remain less than every positive number that we can assign, no matter how small that number may be. We indicate all this by saying that as x runs through the never-ending sequence of values 1/2, 3/4, 7/8, 15/16, •••, it apjwoaches 1 as limit. And in general, if x denote a variable ivhich is supposed to be rimning through some given but never-ending sequence of values, and if there be a 7iumber a s^ich that the difference a — x will ultimately become and remain numerically less than every posi- tive number that we can assign, we say that x approaches this number a as limit. To indicate that x is approaching the limit a, we write X = a, or a = lim x. It will be noticed that the word variable has here a more restricted meaning than in § 242. Whether or not a variable of the kind here under consideration approaches a limit depends on the sequence of values through which 224 A COLLEGE ALGEBRA it is supposed to be running. Thus, if the sequence be 1, 2, 1, 2, ■ • • , the variable will not approach a limit. A full discassion of variables and limits will be found in §§ 187-205, which the student is advised to read, at least in part, in this connection. 509 Theorems respecting limits. In § 203 it is proved that if the variables x and y approach limits, then their sum, difference, product, and quotient also approach limits, and lim (x -\- y) = lim x + lim y. lim (x — y) = lim x — lim y. lim xy = lim x ■ lim y. lim - = t: , unless lim ?/ = 0. y hm y From these theorems it follows that if F(x) denote any given rational function of x, and F(a) its value when x = a, then F(x) will approach F(a) as limit ivhenever x approaches a as limit, that is, limF(a-) = F{a), where ^^^^ F(.r) is read "limit of F(x'), as x approaches a." Thus, ^™ (2x2 - 3x + 1) = 2 a2 _ 3 a + L 510 Infinity. Evidently if x be made to run through the never ending sequence 1, 2, 3, 4, • • • , it will ultimately become and remain greater than every number that we can assign. A variable x %vhich will thiis ultimately become and remain numerically greater than every number that we can assign is said to approach infinity. For the word infi7iity we employ the symbol oo, and, to indicate that x is approaching infinity, write x = (x) is not divisible by x — a, we assign the value /(a) /^ (a) when x = a, and so have (x - a) fix) ^ f(x) (x — a) (f> (x) (x) for all values of x, the value a included. 515 The form a/0. The fraction 1/x takes the form 1/0 when x = 0. While we cannot divide 1 by 0, we can divide 1 by a value of X which differs as little as we please from 0. Moreover we have shown, § 512, that if x be made to approach as limit, then 1/x will approach oo. We therefore assign to 1/0, and in general to a/0, when a=^ 0, the " value " oo, writing a ^ = 00. And for a like reason, to every fraction of the form /(x)/(x-«)<^(,r), where /(cr) and (x) are integral and /(.t) is not divisible by X — a, we assign the "value" oo when x = a; our meaning RATIONAL FRACTIONS 227 being that the fraction will always approach oo when x is made to approach a as limit. Conclusion as to the values of a fraction. From §§ 514, 515 516 we draw the following conclusions regarding a simple fraction of the form f{x)l^ix). 1. If /(x)/<^(a:) is in its lowest terms, it will vanish for values of x which make its numerator f (x) vanish, and become infinite for values of x which make its denoniinator <^ (x) vanish. For all other finite values of x it has a value different from both and co. 2. But if f(x) and <^ (x) have the common factor x — a, and /(ic) contains this factor m times and <^(x) contains it n times, then f(x)/cf)(x) will vanish for x = a when m > n, become infinite when in < u, and have a value different from both and co when m — n. Thus when x = 2, we have a;-2 ^^ ^±2 = 00 ^^^z31 = o ^^-^^ (X - 2)2 ^1 x + 1 'x-2 '^'x(x-2) 'x(x-2)2 '^' x(x-2)2 2' The form 00/00. It is often important to know what limit 517 the value of a fraction /(aj)/^ (a;) approaches when x is indefi- nitely increased, that is, when x = . Consider the following examples. We have shown, § 512, that 1/x, 1 /x^, • ■ = 0, when x = 00. XT u . X2-X + 3 1 - 1/X + 3/x2 . ^ ^^ ,,^ Hence, when x = oo, — = ■ — - — = 1/2, (1) 2x2 + x-4 2 + 1/X-4/X2 ^' X + 2 1 + 2/x x2 + x + 5 X + 1 + 5/X x2 + x-7 X + 1-7/X = 0, (2) = 00. (3) 2x + 3 2 + 3/x And in general, when x = , the fraction {a^x^ + aia:"'-^ + • • • + a,„) / {b^x- + Wx^-^ + . . . + ^, J approaches the limit ao/bo, if, as in (1), the degrees of numer- ator and denominator are the same ; the limit 0, if, as in (2), 228 A COLLEGE ALGEBRA the degree of the denominator is the greater ; the limit oo, if, as in (3), the degree of the numerator is the greater. And in each case the limit is called the "value which the fraction takes when x = oo," that is, when the fraction itself assumes the indeterminate form oo/co. 518 The forms • oo and oo — oo. A rational function of x may take one of the indeterminate forms • oo or oo — oc for some particular value of x. But the expression can then be reduced to one which will take one of the forms 0/0, a/0, or oo/oo already considered. 1. Thus, (x2 — 1) • takes the form • co when x= I. But, except "^ " ^ 1 x2 - 1 when X = 1, we have (x^ — I) = , and therefore X — 1 X — 1 lim r/a;2 _ 1) . _J_'\ = lim ^^^ll = lim (x + 1) = 2. Hence we assign to the given expression the value 2, when x = 1. 1 2 2. Again, takes tlie form oo — oo when x = 0. But, X x(x + 2) ^ 2 except when x = 0, we have = , and therefore X X (X + 2) X (x + 2) lim n ?_"1 = lim '^^Olx X(X + 2)J =^=0; lim 1 _1 x(x + 2)J =^-0x{x + 2) ^-"x + 2 2 Hence we assign to the given expression the value 1/2 when x = 0. 519 General conclusion. Therefore, if a given function of a single variable, as F(x), assumes an indeterminate form when x = a, proceed as follows : Reduce the exj^ression to its simj)Iest form, and then find what limit its value approaches when x is made to approach a as limit. Call this limit the value which the function has when X = a. 520 Note. This method is restricted to functions of a single variable, as F{x). For the reason that the method yields definite results is this : the value of ''™ F (x) depends solely on the value of a and not on the values which X may take in approaching a ; and the like is not true of functions of more than one variable. RATIONAL FRACTIONS 229 Thus, suppose that x and y are unrelated variables, and consider the fraction x/y when x = and y = 0. The limit, if any, which x/y will approach when x = and y = 0, depends on the sequences of values through which x and y may run. For example, a variable will approach as limit if it runs through either of the following sequences : 1/2, 1/3, 1/4, ■•• (1); 1/2^ 1/3-:, 1/42, ••• (2) If X runs through (1), and y through (2), then x/y will run through the sequence 2, 3, 4, • ■ • , and approach oo. But if x runs through (2), and y through (1), then x/y will run through the sequence 1/2, 1/3, 1 /4, • • • , and approach 0. Therefore, if x and y are unrelated variables, we regard x/y as abso- lutely indeterminate when x = and y = 0. And so in general. Infinity in relation to the rules of reckoning. If we take 521 infinite values of the letters into account, we must state the rules of §§ 249, 251, 253 as follows : 1. a • = 0, unless a. = oo. 2. If ac = he, then a = b, unless c = or oo. 3. li a -{- c = b + c, then a = b, unless c = oo. It is important to keep these exceptional cases in mind when applying the rules to the solution of equations. Thus, consider the product x — 1. When x = 1, the second x^ — 1 factor, X — 1, is ; but as the first factor, 1 /(x^ — 1), is then oo, it does not follow that the product is 0. The product is 1/2 in fact, § 518. Infinite roots of equations. Instead of saying, as we have 522 been doing, that the equation x -\- 2 = x -\- S and other simple equations which will reduce to the form 0-x = b, have no root, we sometimes say that the^/ have the root oo. For however small a may be, if not actually 0, ax = b has the root b/a. And if, keeping b constant and different from 0, we make a approach as limit, b/a will approach oo, § 512. In other words, a,s ax = b approaches the form Ox — b, its root b/a approaches the value oo. It is therefore quite in 230 A COLLEGE ALGEBRA agreement with the practice explained in § 515 to say that when ax = b has the form Ox = h, it has the root oo. Observe that if we regard a; + 2 — x + 3asa true equation whose root is 00, we are not driven to the absurd condusion that 2 = 3. For since a; = CO we have no right to infer that the result of subtracting x from both members is a true equation, § 521, 3. 523 Infinite solutions of simultaneous equations. In like manner, instead of saying of a system of inconsistent simple equations, § 377, 2, § 394, 2, that it has no solution, we sometimes say that it has an infinite solution; for from such a s^^stem we can obtain by elimination a single equation of the form Ox = b, and, by § 522, this equation has the root oo. Thus, we may say that the pair of inconsistent equations y — x = (1), y — a; = 1 (2) has an infinite solution. Observe that this pair (1), (2) is the limiting case, as m = 1, of the pair y - mx = (3), y-x = l (2). The solution of the pair (3), (2) is X = l/(m — 1), y = m/ {in — 1), and when m = 1, both l/(m — 1) and mf {m — \) approach infinity. The same thing may be shown by the graphical method, §§ 386, 387. For, when m = 1, the graph of (3) approaches parallelism with that of (1), and the point of intersection of the two graphs recedes to an infinite distance in the plane. EXERCISE XXVI Assign the appropriate values to the following expressions. 2. 2. 1 when x = 1. x3 - 2 X + 1 x2-2ax + a2 1. 4. 1 when x = a. x2-(a+6)x + a6 when X = — 2. when X = 1. x3-3x2 + 3x-l X2- 5x + 6 X2- 6x + 8' X 2-1 X2- 2x + l (3 X + 1) (X + 2)2 (X2- - 4) (x2 + 3x + 2) X3 -X2-X + 1 . RATIONAL FRACTIONS 231 ^ Sz^-x + 5 ^ x^ + 1^ _3x_^ (2.^ + l)(x3-5)^ when x = oo. * 2 x2 + 6 X - 7 X x2 + 1 (x* + 1) (X - 6) -X — 1 X — 2 , „ 8. 1 when x = 3. x2 - 9 X (X - 3) 9. 1 5 when x = 1. X — 1 X (x — 1) X2 + ^ ^-^ 10. ^ , whenx = 2. 11. . when x = oo. 2 X- 1 3x + l "^ ^ X - 2 x2 + 1 FRACTIONAL EQUATIONS On solving a fractional equation. Any given fractional eqiia- 524 tion may be transformed into one which is integral by multi- plying both its members by D, the lowest common denominator of all its fractions. We call this process clearing the equation of fractions. It follows from §§ 341, 342 that the integral equation which is thus derived will have all the roots of the given equation, and, if it has any roots besides these, that they must be roots of the equation D = and so may readily be detected and rejected. Example 1. Solve ? + — ^^^ = 0. (1) xx-lx(x-l) Clear of fractions by multiplying by D = x (x — 1). We obtain 3(x - 1) + 6x - (x + 13) = 0. (2) Solving (2), X = 2. (3) Therefore, since 2 is not a root of D = x (x - 1) = 0, it is a root of (1), and the only root. Example 2. Solve ? + \ ?^^,, = 0- X x-1 x{x-l) (1) Clearing of fractions, 3 (x - 1) + 6 x - (x + 5) = 0. (2) Solving (2), a; = 1- (3) 232 A COLLEGE ALGEBRA As 1 is a root of D = a;(x - 1) = 0, it is not a root of (1). In fact, when X = 1, the first member of (1) has the form 3 + 6/0-G/O; and by § 518 we find that its value is 8, not 0. Hence (1) has no root. We may sum vip the method thus illustrated in the rule : 525 To solve a fractional equation, clear of fractions hy multiply- ing both members by D, the lowest common denominator of all the fractions. Solve the resulting integral eqxiation. The roots of this equation — except those, if any, for which D vanishes — are the roots of the given equation. 526 Note. We may also establish this rule as follows : Let N /D = represent the result of collecting all the terms of the given equation in one member and adding them ; then N = will be the integral equation obtained by clearing of fractions. 1. If iV/D is in its lowest terms, the roots of iV/D = and i\r = are the same ; for a fraction in its lowest terms vanishes when its numerator vanishes, and then only, § 516. 3 6 x+, 13 8{x-2) N Thus, in § 524, Ex. 1, - + ^ = ~ .{ = 7:- ' ^ 'x x-1 x(x-l) x(x-l) D Here N/D is in its lowest terms and the root of iV/D = is the same as the root of N — 0, namely, 2. 2. If N/J) is not in its loioest terms, iV = will have roots which N/D = does not have, namely, the roots for which the factor common to N and D vanishes. 3 6 x + 5 8(x-l)iV Thus, in § 524, Ex. 2, - + = -^ ^ = -• ' * 'x x-1 x(x-l) x(x-l) D Here N/D is not in its lowest terms, and the root of iV = 0, namely 1, is not a root of N/D = 0; for when x = 1, iV/D = 8, § 514. Evidently 1. is the general case and 2. is exceptional. 3 6 X + a ^ Thus, consider the equation -\ ; — = v. X x-1 x(x-l) Here N/D = [8x - (a + 3)]/x(x - 1), and this is in its lowest terms except when u = 5 or — 3. RATIONAL FRACTIONS 233 3. It must not be inferred from what has just been said that the given equation will never be satisfied by a root of iV" = which is also a root of X» = 0. X 2 1 Thus, consider the equation = 0. X — 1 X X (x — 1) Here N/D = (x — l)2/x(x — 1) and, by § 516, this expression vanishes when X = 1. But observe that N = (x — 1)^ = has the root 1 a greater number of times than D = x{x — 1) — has this root. In applying the rule of § 525, care must be taken not to 527 introduce extraneous factors in the expression selected as the lowest common denominator. If any fraction in the equation is not in its lowest terms, begin by simplifying this fraction, unless the factors thus cancelled occur in other denominators. Before clearing of fractions it is sometimes best to combine certain of the fractions, or to reduce certain of them to mixed expressions. 6 X + 5 x2 11 Example 1. Solve !x + 15 6x-2x2 5 Here the terms of the first fraction have the common factor x — 5, and those of the second the common factor x. Cancelling these factors, we have x-1 x 11 x-1, X 11 x-3 6-2x" 5 x-3 2(x-3)" 5 Clearing of fractions. lOx- -10 + 5x: = 22x- Solving, X -- = 8. Example 2. Solve ^ + V x + 2 ^x + 6 x + 7 x + 2 x + 3 x + 5_ x + 6 Reducing ; each fraction to a mixed expression and simplifying. .^- 1 x + 7 ^- 1 x + 6 Transposing so that the terms in each member may be connected by minus signs, _J 1__^ 1_ x + 2 x + 3~x + 6 x + 7* 234 A COLLEGE ALGEBRA Combining the terms of each member separately, 1 1 X- + bx + 6 x^ -f 13 X + 42 ■ Clearing of fractions, x^ + 13 x + 42 = x^ + 5 x + 6. Solving, x = -9/2. The given equation may be solved by clearing it of fractions as it stands, but that method is much more laborious. 528 Simultaneous fractional equations. The general method of solving such a system is to clear the several equations of frac- tions and then to find, if possible, the solutions of the result- ing system of integral equations. The solutions thus found — except those, if any, which make denominators in the given equations vanish — are the solutions of these equations, §371. But if the equations are of the form described in § 379, or if they can be reduced to this form, they should be solved by the method explained in that section. Example 1. Solve the following pair of equations for x, y. y — 2 2/ — 4 xy — 2x iy — 2y'^ xy Clearing both equations of fractions and simplifying, v^'e obtain x-?/ + l = 0, X + 2y - 8 = 0. Solving X = 2, 2/ = 3. Therefore, since none of the denominators in the given equations vanish v?hen x = 2 and y = 3, these equations have the solution x = 2, 2/ = 3. Example 2. Solve the follow^ing system for x, y, z. ^-±1=^^, -y^ = -l 2iz + x) + xz = 0. xy G y + z 2 ^ ' These equations can be reduced to the form, § 379, 1 1_5 1 1__2 1 1__1 x y 6 y z 3 z x 2 Solving for 1 /x, l/?y, 1/z, we find 1/x = 1/2, 1/?/ = 1/3, 1/z = - 1. Hence the required solution is x = 2, y — 3, z = — 1. RATIONAL FRACTIONS 236 EXERCISE XXVn Solve the following equations for x. 1. 6X-1 4x-7^^ 3x + 2 2x-5 ? 6x 8 1 5x-l 3-15X 6 s •4 1 4 x-2 x-4 x2-6x + 8 + ' 2x + 3 x-5 2x2-7x-15 1 2 (x + l)(x-3) (x-3)(x + 2) (x + 2)(x + l) 2 2 3 x2-l x2 + 4x-5 x2 + 6x + 5~ x+ 1 2x 5 = 0. 3x+l 5-6x 5 + 9x-18x2 g _^+ g ^ x + 6 ^ a + 6 g x3 + 1 a;^ - 1 _ " 6 (X + &) a (X + a) a6 ' ' x + 1 x - 1 ~ 10 ^'' + ^^ + 1 + =^ - 1 x2 + 5x + 4 x2 + 3x-4 ^^ x-8 x-9 _ x + 7 x + 2 'x-3 x-4~x + 8 x + 3* 12 ^ + '^ . ^ + 9 _ a: + 10 x + 6 "x + e x + S^x + O x + 5* „ x3 + 2 x3-2 15 16. = 4 X. x-2 x + 2 x2-4 14 1 ^-^ I ^^' + ^^ ^0 ■ X - 1 X2 - 1 1 - X* 15. -J- + -1^±A L_^o x3-8 2x2 + 4x + 8 x-2 ax + c bx + d . 16. 1 = a + b. X — p X — q 236 A COLLEGE ALGEBRA 17. X- 1 X — 8 X- + X X + 2 2 x2 - X + 7 „ x2-ax + 26x-2a6 62-x2 3 c2 18. ; 1 — — + X — 26 X — 2c (x - 6)2 (X - c)2 19 (^ - '^y- I (X - 6) (X - c) (X - c) (X - a) ■ (X - a) (x - 6) 20 ^^^±^-^iiA_^il2^^ X2 + X X2 - 1 X2 - X 21. -^ + -^-^+1 = 0. x + 2 x-2 x2-4 Solve the following for x and y, or for x, ?/, and z. 22. 3x + y - 1 _ 6 X-2/ + 2 ~ 7' X + 9 _x + 3 [2/ + 4~^3" 24. xy X + y ' + z a, 23. 25. <^ PARTIAL FRACTIONS 529 It follows from § 506 that every rational function of a single variable, as x, can be reduced to the form of an inte- gral function, or a proper fraction, or the sum of an integral function and a proper fraction. For certain purposes it is useful to carry this reduction further and, whe7i a proper fraction A/B is given, find the simplest set of fractiovs of which A/B is the sum. The method depends on the following theorems in wMch the letters A, B, P, Q, and so on, denote rational integral func- tions of X. RATIONAL FRACTIONS 237 Theorem 1. The sum and the difference of two proper frac- 530 tions A/B and C/D are themselves proper fractions. A , C AD±BC For — ± — = B D ED Since A is of lower degree than B, AD is of lower degree than BD. And since C is of lower degree than D, BC is of lower degree than BD. Hence AD ± BC is of lower degree than BD. Theorem 2. Let 1 and V denote integral f mictions, and A./ 'B 531 and A.' /W projier fractions. If I + A/B = I' + A'/B', then 1 = 1' and A/B = A'/B'. For, by hypothesis, I — f = A'/B' — A/B. But I — I' denotes an integral function (or 0), and, § 530, A'/B' — A/B denotes a proper fraction (or 0). Therefore, since an integral function cannot be identically equal to a proper fraction, we have / - /' = and A'/B' - A/B= 0, or I =r and A/B = A '/B'. Theorem 3. Let A/ FQ denote a proper fraction who.ie denom- 532 inator has been separated into two factors, P a7id Q, ivhich are jjrime to one another. This fraction can be reduced to a sum of two proper fractions of the forms, B/P and C/Q. For, since Q is prime to P, we can find, § 479, two integral functions M and N, such that 1 = il/Q + NP, and therefore A = AMQ + ANP. A AMQ + ANP AM AN Hence — = = (i) PQ PQ P Q ^ ^ If AM/P and AN / Q are proper fractions, our theorem is already demonstrated. 238 A COLLEGE ALGEBRA If AM I F and AN / Q are not proper fractions, reduce them to sums of integral functions and proper fractions, and let the results be 4 If R A N (' (2) AM B AN r ■ 1 + and ■: A + P P Q Q Substituting these expressions for AM J P and AN / Q in (1), we have But since A/PQ, B/P, and C /Q are proper fractions, it follows from (3), by §§ 530, 531, that 7 + A' = and A _B C PQ~ P q' as was to be demonstrated. 533 Note. The fraction A/PQ can be reduced to but one such sum B/P+C/Q. A B C B' C lor suppose = — = \-^i (1) PQ p q p q ^ ' where B' /P and C / Q also denote proper fractious. Then ^^^ = ■^-^, and therefore ^^ ~ ^'^ ^ =C'-C. (2) But (2) is impossible unless B — E' = and C — C = 0. For other- wise (2) would mean that (B — B') Q is exactly divisible by P, and this cannot be the case since Q is prime to P and B — B' is of lower degree than P, § 481. 534 Partial fractions. We call the fractions B/P and C/Q, whose existence we have just T^xovedi, partial fractions of A/PQ. To resolve a given fraction of the form A / PQ into its par- tial fractions B/P and C / Q, it is not necessary to carry ouL the process indicated in § 532 ; we may apply the method of undetermined coefficients, § 397, as in the following example. Example 1. Resolve {2 x"^ + \) / {z^ — V) into a sum of two partial fractions. This is a proper fraction, and its denominator is a product of two factors, X — 1 and x^ + x + 1, which are prime to each other. RATIONAL FRACTIONS 239 Hence, § 532, (2x2 + l)/(x^ — 1) is equal to a sum of two proper frac- tions whose denominators are a: — 1 and x^ + x + 1 respectively. The numerator of the first of these fractions must be a constant, that of the second an expression whose degree is one at most. Hence we must have 2x2+l_ a bx + c x^-1 "" x-1 "^ x2 + X + 1 ^^^ where a, 6, and c denote constants. To find a, b, c, clear (1) of fractions. We obtain 2 x^ + 1 = a (x- + x + 1) + (6x + c) (x - 1), or 2x2 + 1 = (a + 6)x2 + (a - 6 + c)x + (a - c). (2) As (2) is an identity, the coefficients of like powers of x are equal, § 284. Hence a + b-2, a-b + c = 0, a-c = l, (3) or, solving (3), a — I, 6 = 1, c = 0. Therefore ^^!±i ^ ^A^ + ? '. X3 - 1 x-1 X2 + X + 1 Example 2. Resolve (5x + 4)/(x'* + x^ + x^ — x) into a sum of two partial fractions. General theorem regarding partial fractions. From the theorem 535 of § 532 we may draw the following conclusions. 1. Let A / PQR denote a proper fraction in which the three factors of the denominator P, Q, R are prime to one another. This fraction can be reduced to a sum of the form A _B D E PQR ~ P Q R where B fP, D / Q, and E / R denote proper fractions. For since P is prime to QR, § 482, A / PQR is the sum of two proper fractions of the form B/P+ C / QR, §532; and since Q is prime to R, C J QR is itself the sum of two proper fractions of the form D/Q + E/R,% 532. The like is true when the denominator is the product of any number of factors all prime to one another. 2. Consider the proper fraction A / PQ^ in which P is prime to Q. By § 532 it can be resolved into the sum A _B C PQ^~ P q^' 240 A COLLEGE ALGEBRA We cannot apply the theorem of § 532 to the fraction C I Cl\ since the factors Q, Q, Q are not prime to one another. But since C is of lower degree than Q^ it can be reduced, § 422, to a polynomial in Q of the form C = CiQ2 + C2Q + C3, where Ci, C2, and Tg are of lower degree than Q. Dividing each member of this identity by Q^ we have Hence the given fraction can be reduced to the sum A _ ^ Ci C'2 C3 where B is of lower degree than P, and C^, C^, C3 are of lower degree than Q. And so in general when a factor, as Q, occurs more than once in the denominator. We therefore have the following theorem. Suppose that the denominator of a given proper fraction has been separated into factors — some occurring once, some, it may he, more than once — which are all prime to one another. The fraction itself can then be resolved into one and but one sum of proper fractions in which (1) for each factor, P, which occurs but once, there is a single fraction of the form B/P, and (2) for each factor, Q, tvhich occurs r times, there is a group of r fractions of the form Ci/Q + Cj/Q'^ H h C,/Q^ ivhere Ci, C2, • • •, C, are all of lower degree than Q. 536 Simplest partial fractions. It can be proved that ever}^ poly- nomial in X with real coefficients is the product of factors of one or both the types x — a and a;^ + px + q, where a, p, and q are real, but where the factors of x"^ + />x + q have imagi- nary coefficients. Moreover it follows from §§ 469. 532 that, if the numerator of a given proper fraction and the factors into which its RATIONAL FRACTIONS 241 denominator has been separated have real coefficients, so will the numerators of the corresponding partial fractions. Hence, by § 534, Every 'proper fraction whose mimerator and denominator have real coefficients is equal to a definite sum of partial frac- tions related us follows to the factors x — a and x^ + px + q o/" its denominator. 1. For every factor x — a occurring once there is a single fraction of the form A/(x — a), xoliere A is a real constant. 2. For every factor x — a occurring r times there is a group of r fractions of the form Ai/(x - a) + k,/{x _ a)2 + • ■ ■ + A,/(x - a)^ where Aj, Ag, • • • A^ are real co?istanfs. 3. For every factor x^ -|- px + q occurring once there ts a, single fraction of the form (Dx + E)/(x'^ -f- px + q), ivhere D and E are real constants. 4. For every factor x^ + px + q occurring r times there is a group of r fractions of the form (D,x 4- E0/(x2 + px + q) + • • • + (D,x + E,)/(x2 + px + q)^ where Dj, Ej, Do, Ej, • • • D^, E^ denote real co)istants. The fractions here described are usually called the simplest 537 partial fractions of the given fraction. They are best found by the method of undetermined coefficients. ifl -\- X — 3 Example 1. Resolve into its simplest partial fractions. By § 536, we have ^^^ = -^ + -^ + -^— (1) (X - 1) (X - 2) (x - 3) X - 1 X - 2 X - 3 ^ ' where A, B, C are constants. Clearing (1) of fractions, we obtain x2 + X - 3 = 4 (X - 2) (x - 3) + B(x - 3) (X - 1) + C(x - 1) (X - 2). (2) We may find A, B, C by arranging the second member of (2) accord- ing to powers of x and equating coefficients of like powers ; but, since A, B, C are constants, the same results will be obtained by the following method, which is simpler. 242 A COLLEGE ALGEBRA In (2) let a; = 1, and we have - 1 = ^ ( - 1) ( - 2), .-.A =-1/2 next let x = 2, and we have 3 = 5 ( - 1) • 1, .-. B = - H; finally let a: = 3, and we have 9=C-2-l, .-.(7 = 9/2. Hence = h (X - 1) (I - 2) (x - 3) 2 (X - 1) X - 2 2 (X - 3) X + 1 Example 2. Resolve into its simplest partial fractions x(x-lf X, o ro^ ,. a; + l A B C B By § 536, we have = — + x(x-l)3 X x-1 (x-l)^ {x-lf and therefore x + 1 = J. (x - 1)'' + J5x (x - 1)- + Cx (x - 1) + Dx. (1) In (1) let X = 0, and we have 1 = J. (- 1)3, .■.A=-l; next let x = 1, and we have 2 = J), .-. D = 2. Substitute these values of A and D in (1), transpose to the first mem- ber the terms thus found, namely — (x — 1)^ and 2x, and simplify the result. We obtain x3 - 3 x2 + 2 X = 5x (X - 1)- + Cx (X - 1). (2) Dividing both members of (2) by x(x — 1), we have x-2 = B(x-l)+C. (3) Equating coefiicients of like powers of x in (3), we have 1 = B and - 2 = - B + C, .-. B = 1 and C = - 1. x+1 11 1,2 Hence = 1 \- x(x-l)3 X x-1 (x-l)2 (X - 1)3 Or we may arrange (1) according to powers of x, obtaining X + 1 = {A + B)x^ ~ {S A + 2 B - C)x^ + {3 A + B - C + D)x - A. Equating coeflQcients of like powers of x, we have ^ + -5 = 0, 3A + 2B-C = 0, 3A + B - C + D = -[, - A = I. And from these equations we find, as before, A=-l, B = l, C = - 1, D = 2. 5 a;2 — 4 X -I- 16 Example 3. Resolve — — —- into its simplest partial \X — X + 1 ) - (x — 3) fractions. The factors of x^ - x + 1 being imaginary, we have, § 536, 5x2 -4x -f 16 ^ Ax + B Cx + D E (X2 - X + 1)'^ (X - 3) ~ (X2 - X + 1)2 X2 - X + 1 "*" X-S' where A, B, C, D, E are constants. RATIONAL FRACTIONS 243 Clearing of fractions, 6x2 -4x + W^{Ax + B){x-S) + {Cx + D){x^-x + l){x~3) + E{x^-x + 1)2. (1) "We may find A, B, C, B, E by arranging (1) according to po^Yers of X and then equating coefficients of like powers ; but the following method is simpler. In (1) let X = 3, and we have 49 - 49^, .-. E ~\. Substitute this value of E in (1), transpose the term (x^ — x + 1)2 thus found to the first member, simplify, and divide both members b^ X - 3. We obtain - (x3 + x2 + X + 5) = ^x + i? + (Cx + D) (x2 - X + 1). (2) Next divide both members of (2) by x2 — x + 1. We obtain 2x + 3 Ax + B - X - 2 - ^— -- = ^— + CX + I). (3) X2-X+Ix2-X + 1 ^ By § 531, the fractional parts and the integral parts in (3) are equal. Hence - x - 2 = Cx + D, and therefore C = - 1, Z) = - 2, and -2x — 3=J.x+ B, and therefore J. = - 2, j5 = - 3. ^, ^ 5x2-4x + 16 2x + 3 x + 2 1 Therefore = 1 (x2 - x + 1)2 (X - 3) (x2 - X + 1)2 a;2 - X + 1 X - 3 When the denominator of the given fraction has the form (x — ay, it is best to begin by expressing the numerator in powers of a; — a, § 423. Similarly when the denominator has the form {x^ + px + qY, the factors of x"^ + px + «/ being imagi- nary, we express the numerator in powers of x" + p^ + ?• X* 4- x3 _ 8 x2 + 6 X 32 Example. Resolve into its simplest partial fractions. By § 423, we find x* + x3 - 8 x2 + 6 X - 32 = (X - 2)* + 9(x - 2)3 + 22(x - 2)2 + 18 (x - 2) - 28. Dividing both members by (x — 2)^, we have x^ + x3-8x2 + 6x-32 _ 1 9 22 18 28_ (x-2)5 ~x-2 (x-2)2 (x-2)3 (x-2)* {x-2)5" If given an improper fraction, we may first reduce it to the sum of an integral expression and a proper fraction and then resolve the latter into its partial fractions. 244 A COLLEGE ALGEBRA x3- 2x2 -6a; -21 Example. Apply this method to the fraction x-! — 4 X — 6 ^. , x3-2x2-6x-21 „ 7x-ll We have = x + 2 + x2_4x-5 x2-4x-5 7x-ll X + 2 (X + 1) (X - 5) ^nd proceedlDg as above we find 7x-ll 3 4 (X + 1) (X - 5) X + 1 X - 5 EXERCISE XXVm Resolve the following into the simplest partial fractions whose denomi- nators have real coefficients. 1 2x+ 11 2. 6x-l (X - 2) (X + 3) (2x + l)(3x - 1) 3 4x 4 x2 + 2 X + 3 (X + 1) (X + 2) (X + 3) {x-l){x-2)(x-3)(x-4) 5. x2 + 2 1+X3 6. 8x + 2 X-X3 7 x^-x^-~5x\-i 8 2X3-X2 + 1 x2-3x + 2 (X - 2)* 9 x-1 10. 6 2x3-5x2-12x 2 X* - x2 - 1 n 2x3-3x2 + 4x-5 12 X2 + X + 1 (x + 3)s (.C2+ l)(x2 + 2) n x2 + 6 X - 1 14 3x -1 (X- 3)2 (x-1) (X - 2) (X2 + 1) 1*) 2 x6 - X + 1 16 2x2 -x + 1 (X2 + X+ 1)3 (X2 - X)2 17. 3x2- X + 2 (x2 + 2) (X2 - X - 2) 18. x^+px + q (x - a) (X - b) (X - c) 19. 2x2 -3x -2 20. X3 + X + 3 X(X-l)2(x + 3)3 X*+X2+1 SYMMETRIC FUNCTIONS 245 IX. SYMMETRIC FUNCTIONS ABSOLUTE SYMMETRY AND CYCLO-SYMMETRY Absolute symmetry. In the expression x'^ -\- y^ -{- z^ the let- 540 ters a-, y, z are involved in such a manner that if any two of them be interchanged a;^ + ^/^ + z^ is transformed into an iden- tically equal expression, namely, y'^ -\- x^ -\- z"^, or z"^ -\- y"^ + x^, or x^ -{- z^ + \f. To indicate this, we say that x"^ -\- y"^ ■\- z^ is symvietric with respect to x, y, z. And, in general, a function of a certain set of letters is said to be symmetric with respect to these letters when every inter- change of two of the letters will transform the function into an identically equal functioi-. Other examples of symmetric functions are (xy + xz + yz)/{x + y) (x + z) {y + z) with respect to a;, y, z, a-\-h -\- c and (x + a) (x + h) (x + c) with respect to a, 6, c. On the other hand, x -|- ?/ — z is not symmetric ; for if we interchange y and z we obtain x + z — y, which is not equal to x + ?/ — z. We call 2 xhj and 3 yH terms of the savie type with respect 541 to the variables cc, ?/, z, because the variable parts of these terms, namely, x^y and y'^z, can be transformed into one another by interchanges of pairs of the letters x, y, z. And so in general. The sufficient ayid necessary condition that an integral func- 542 tion of certain letters, as x, y, z, he symmetric with respect to these letters is that all its terms of the same type shall have the same coefficie7its. This implies that if a symmetric function contains one term of a certain type, it must contain all terms of that type ; that is, all terms that can be derived from the term in question by making every jwssible interchange of the letters. Thus, if ax- + by- + cz"^ is to be symmetric, we must have a = b = c. Again, if a symmetric function of x. ?/, z contains the term x'^y, it must contain all the terms x'^y -(- y-x 4- x^z + z^x -f y'^z + z'^y. 246 A COLLEGE ALGEBRA 543 This theorem will indicate the general form of a symmetric function of given degree with respect to a given set of letters. Thus, the general form of a symmetric function of the first degree with respect to x,y, z, u i& a{x + y ^ z -[■ u) + h, where a and 6 denote constants. Again, the most general symmetric and homogeneous functions of the first, second, and third degrees with respect to x, y, z are 1. a{x + y+ z). 2. a (x2 + y'i + z^)-\-h {xy + xz + yz). 3. a (x3 + 2/3 + 2-5) + b (xhj + y'-x + x^z + z'^x + y^z + z'^y) + cxyz. 544 On expressing a symmetric function. The notation 'Zx" means the sum of all terms of the same type as x"^ ; that is, if a-, y, z are the letters under consideration, %x- = x'^ + if + z^. Similarly ^x^y = xhj + y'^x + x-'z + z^x + y^z + zhj ; and so on. Any given symmetric function may be represented by select- ing from its terms one of each type, and writing the symbol 2 before their sum. Thus, S (2 X - x3?/2) = 2x + 2y + 2z - x^y^ - t/'^x- - x^z^ - z^x'^ - y^z^ - z^y^. 545 When writing out symmetric functions at length, it is best to arrange the terms in accordance with some fixed rule. The following examples Avill indicate a convenient rule for the arrangement of integral symmetric functions. Suppose that the letters under consideration are a, 6, e, d, and by the normal order of these letters understand the order a, 6, c, d. We shall then write "Lab and Ea6c as follows : 2a6 = a6 + ac + ad + 6c + hd + cd, I,abc = abc + abd + acd + bed. Observe that in each term we write the letters in their normal order. In forming Hah we take each letter a, 6, c in turn and after it write each subsequent letter. The terms of Zabc are derived in a similar manner from those of 2a&. We shall arrange the terms of Sa'"6", when m ^f n, as follows : Sa263 = a253 + 62^3 + a^c^ + c^a^ + • • • + c^d^ + d'^c'^. Observe that we keep the order of the exponents fixed ; then under the exponents we write the letters of the first term of 'Lab in both the orders ab and ba, and so on. SYMMETRIC FUNCTIONS 247 In like manner we may write Sa263c4 = a263c4 + a'^c^b* + b^-c^a^ + b^a^c* + c^a^b* + c'^b^a* + (terms similarly derived from the remaining terms of 'Labc). A general theorem regarding symmetry. It follows from the 546 definition of symmetry, § 540, tliat a symmetric function will remain symmetric when its form is changed by the rules of reckoning. In particular, The sum, difference, jjroduct, and quotient of two symmetric functions are themselves symmetric. By aid of this theorem we may obtain the result of combin- ing given symmetric functions by algebraic operations without actually carrying out these operations. It is only necessary to compute the various typical terms of the result. Example 1. Find (Sa)- = (a + b -{■ c -^ )2. Evidently the required result is a homogeneous symmetric function of the second degree consisting of terms of the two types cfi and 2 ab. Hence (Sa)2 = 2a2 + 2 Sa6. Example 2. Find 2x2 • Sx = (x2 ^ yi j^ z^){x + y + z). Evidently this product is a sum of terms of the two types x^ and x'^y. Hence 2x2 . sx = 2x3 + 2x2?/ ^x^ + y^ + z^ + x^y + 2/2x + x^-z + z^x + y-z + z^y. Example 3. Find (2x)3 = {x + y +zf. The required result is homogeneous, symmetric, and of the third degree with respect to x, y, z. We must therefore have, § 543, {x + y + zY = a (x3 + 2/3 ^ 23) + 6 {x^y + yH + x2z + z2x + y'^z + z'^y) + cxyz. To find the values of the constants a, b, c, assign any three sets of values to x, y, z which will yield equations in a, 6, c, and solve these equations. Thus, putting x = 1, 2/ = 0, z = 0, we have 1 = a. (1) Again, putting x = 1. ?/ = 1, z = 0, we have 8 = 2 a + 26. (2) Finally, putting x = 1, 2/ = 1, 2 = 1, we have 27 = 3o + 6& + c. (3) Solving (1), (2), (3), we obtain a = 1, 6 = 3. c = 6. Hence (2x)3 = 2x^ + 3 2x2?/ + 6 2x2/z. 248 A COLLEGE ALGEBRA 547 Cyclo-symmetry. In the expression x^y + y"z + z^x the let- ters X, y, z are involved in such a manner that if we replace x by y, y by z, and z by x, we obtain an identically equal expres- sion, namely, y'^z + z^x + x-y. To indicate this, we say that x^y + y'^z + z^x is cyclo-symmetric, or cyciJie, with respect to the letters x, y, z, taken in the order x, y, z. And, in general, an expression is said to be cyclo-symmetric, or cyclic, with respect to certain letters arranyed in a given order, if it is transformed into an identically equal expression when we replace the first of these letters by the second, the second by the third, and so on, and the last by the first. Such an interchange of the letters is called a cyclic interchange. 548 Observe that the terms of x'^y + y'^z + z^x are themselves arranged cyclicly ; that is, so that the first changes into the second, the second into the third, and the third into the first, when we replace x hy y, y by z, and z by x. Cyclic expres- sions are of frequent occurrence and reckoning with them is greatly facilitated by arranging them cyclicly. 549 Evidently every symmetric function is cyclic, but not every cyclic expression is symmetric. Thus, x'^y -\- y'^z + z-x, though cyclic, is not symmetric. Its vahie changes if x and y are interchanged. To make it symmetric we must add the group of terms y'^x -\- z'^y -f- x^z. 550 As the example shows, a cyclic function will ordinarily not contain all the terms of a given type, but such of these terms as it does contain will have the same coefficients. 551 Theorem. The sum, difference, product, and quotient of tivo cyclic fujictions are themselves cyclic. This follows at once from the definition of cyclo-symmetry. Example. Find the product {x^y + yH -f z'^x) (x -t- ?/ -|- z). Evidently the product is cyclic but not symmetric. Moreover it con- tains the terms x^y, x'^y-, x"yz, each once, and terms of these types only. Hence the product is x^y + yH Jr z^x + x-y- + yH- -f zH"^ -\- x'^yz ■\- yHx -\- zHy. SYMMETRIC FUNCTIONS 249 EXERCISE XXIX 1. State the letters with respect to which the expression X4 - 2 2/* + 2^ + 4 (X3 - 2/3) (2/3 - 23) (x^ + ^2) is symmetric. 2. Write out in full the following symmetric functions of a, 6, c. Sa262, sa364, 2aV&, 2a263c5, Sa^ft'^c*, S(a + 6)c, S(a + 62)c3, S(a + 26 + 3c). 3. Show that (a — b) (b — c) (c — a) is cyclic but not symmetric with respect to a, b, c ; also that (a — b)^{b — c)^{c — a)^ is symmetric. 4. Is (a — 6)2 (6 — c)2 (c — d)'^ (d — a)^ symmetric with respect to a, 6, c, d? 5. Arrange the following sets of expressions cyclicly. y- — X", 2^ — 2/2, x~ — 2- ; a-bc, abd'^, acH, b'^cd ; {a - c) (6 - a), (a -c)(c- b), {a -b)(b- c). 6. Write out in full the cyclic functions of a, 6, c, d whose first terms are a63c2, a{b- c), (b + 2c){a + d), a"- /{a -b){a- c). 7. Prove the truth of the following identities. Sa3 • Sa = 2a* + ^a^b ; Zab ■ 2a = 2a-'6 + 3 2a6c. FACTORIZATION OF SYMMETRIC AND CYCLIC FUNCTIONS By aid of the remainder theorem and the principles just 552 explained it is often possible to factor a complicated symmetric or C3^clic function with comparatively little reckoning. Example 1. Factor x^ {y — z) + y^ {z — x) -\- z^ (x — y). This function vanishes when y = z^ for X3 (2 - 2) + 23 (2 - X) + z3 (x - 2) = 0. Hence the function is exactly divisible by ?/ — 2, § 416 ; and for a like reason it is exactly divisible by z — x and by x — 2/, and therefore by the product (2/ — z){z — x) (x — y). Both dividend and divisor are cyclic and homog»neous, and their degrees are four and three respectively. Hence the quotient must be 250 A COLLEGE ALGEBRA a cyclic and homogeneous function of the first degree and therefore have the form k(x-i- y + z), where k denotes some constant. Hence x^ {y-z)+ 2/3 (z-x)+z-{x-i/)~k{y- z) {z -x){x ~y,{x + y ^- z) (1) To find k, assign to x, y, z any set of values for which the coefficient of k will not vanish. Thus, putting x = 2, y = 1, z = 0, we have 6 = — 6A;, orA: = — L Or we may find k by equating the coefficients of like powers of x in the two members of (1) arranged as polynomials in x. Thus, the x^ term in the first member is x^ {y — z), and in the second member it is — kx^{y — z), whence as before fc = — 1. We therefore have X3 {y -Z) + 2/3 (Z - X) + Z3 (X _ y) = _ (y - z) (z - X) (X - ?/) (x + ?/ + z). Example 2. Factor {x + y + zY — x^ - y^ — z^. This function vanishes when x = — y, for {- y + y + zY + y^ - y^ -z^ = 0. Hence the function is exactly divisible by x + 2/ ; and for a like reason it is divisible by 2/ + z and by z + x, and therefore by (x + y) {y + z) (z + x). As the dividend and divisor are symmetric and homogeneous and of the fifth and third degrees respectively, the quotient must be of the form k {x2 + 2/2 + z2) + i (xy + yz-\- zx), § 543. Hence {x + y + zf - x^ - y^ - z^ = (x-Vy){y + z) (z + X) [k (x^ -f y'^ + z'-^) + Z (X2/ + 2/2 + zx)]. Putting X = 1, 2/ = 1) 2 = 0, we obtain 15 = 2 fc + L Putting X = 2, 2/ = 1) z = 0, we obtain 35 = 5 A: + 2 i. Solving for A; and i, we find fc = 5, Z = 5, and therefore have (x-\-y + z)^ -x^ -y^ - z5 = 5(x + 2/) (2/ + z) (z + X) (x2 + 2/2 + z2 + a;y + 2/z + zx). Example 3. Factor {x + y -\-zY -{y -\- z-xY - {z + x-yY - {x +y - zY- This function vanishes when x = 0, for (2/ + zY - (2/ + z)3 - (z - yY -(y- zY = 0. Hence the function is exactly divisible by x - or x ; and for a like reason it is divisible by y and by z and therefore by xyz. Since both dividend and divisor are of the third degree, the quotient is some constant, k. Hence (x + y + zY-{y + z-xY-(z + x-yY-ix + y- zY = kxyz. Putting x = 1, 2/ = 1, z = 1, we find k = 24, and therefore have {x + y + zY-{y + z-xY-{z + x-yY-{x + y-zY = 2ixyz. SYMMETRIC FUNCTIONS 251 The method just explained is sometimes useful in simplifying 553 cyclic fractional expressions. a^ h^ c3 Example. Simplify 1- {a-h){a-c) {b-c)(b-a) (c-a){c-b) This expression is cyclic with respect to a, 6, c. The lowest common denominator is [b — c) (c — a){a — b). On reducing the fractions to this lowest common denominator we obtain as the first numerator — a^ (b — c). Hence, by cyclo-symmetry, the second and third numerators are —b^(c — a) and — c^{a — b). Adding these three numerators and factoring the result, § 552, Ex. 1, we have (ct + & + c) (6 — c) (c — a) {a — b). Hence the given expression reduces to a + 6 + c. EXERCISE XXX Factor the following expressions. 1. x2(y - 2) + r/-2 (z - x) + z'^ (X - y). 2. yz {y - z) + zx {z - x) + xy (x - y). 3. {y - zY + {z- x)3 + (X - y)\ 4. x{y - z)3 + ^ (z - x)3 + z (x - ?/)3. 6. x2 (y - zf + ?/ (z - x)3 + z2 (x - y)\ 6. X* (2/2 - 22) + 2/« (Z2 - X2) + Z* {x2 - 2/2). 7. (x + 2/ + 2)3 — x3 — 2/3 — 2^. 8. {y - 2)6 + (2 - x)5 + (X - 2/)5. 9. (x + 2/ + z)5 - (2/ + 2 - x)5 - (z + X - 2/)^ - (X + 2/ - zf. 10. (2/ - 2) (2/ + 2)3 + (z _ X) (2 + X)3 + (X - 2/) (X + 2/)"- 11. x{y + 2)2 + 2/(z + X)2 + 2(X + 2/)2 - 4X2/2. 12. X5 (2/ - 2) + 2/^ (Z - X) + Z5 (X - ?/). Simplify the following fractional expressions. a* b* c* 13 14. (a - 6) (a - c) (6 - c) (6 - a) (c - a) (c - b) X + a X + b x + c ^a"^^^6Ha^^ (6 - c) (6 - a) (c - a) (c - 6) ° 252 A COLLEGE ALGEBRA a* — 6c b" — ca c^ — ab 16. : — - + (a - 6) (a - c) (b -c)(b - a) (c - a){c - b) 16. (& + c)2 ^ jc + a)^ ^ (a + 6)2 17. (o - 6) (a - c) (6 - c) (6 - a) (c - a) (c - 6) a2 62 c2 (a - 6) (a - c) (X - a) (6 - c) (6 -a){x- b) (c - a) (c - 6) (x - c) X. THE BINOMIAL THEOREM 654 Structure of continued products. To obtain the product (a + b i- c + d)(e +/+ g)(h + k) we may multiply each term of a -\- b -\- c -{- d by each term of e +f+ ff, then multiply every product thus obtained by each term of h + k, and finally add the results of these last multiplications. Hence Ave shall obtain one term of the product if we select one term from each of the three given factors and multiply these terms together. And we shall obtain all the terms of the product if we make this selection of terms from the three given factors in all possible ways. Thus, selecting b from the first factor, g f'-om the second, and k from the third, we have the term bgk of the product ; and so on. Since we can select a term from a-\-b-{-c-\-d in four ways, a term from e +f+ g in three ways, and a term from h + k in two ways, the number of terms in the complete product is 4 • 3 • 2, or 24. And so, in general, The product of any number of 2'>ohjnomials is the sum of all the products that can be obtained by selecting one term from each factor and multiplying these terms together. And if the first factor has m terms, the second n, the third p, and so on, the number of terms in the complete j^^oduct — before like terms, if any, have been collected — is mnp • • •. THE BINOMIAL THEOREM 258 This theorem supplies a useful check on the correctness of a 555 multiplication. It may be applied to a product in which like terms have been collected, provided its terms represent sums of terms of like sign and without numerical coefficients, the coefficient of a term then indicating how many uncollected terms it represents. Thus, by our theorem, the product {a + b + c) {a + b + c) should have 3 • 3 or 9 terms, which are all of the same sign. But, as we have shown, (a + 6 + c) (a + 6 + c) = a2 + 62 _|. c2 -f 2 a6 + 2 ac + 2 6c, and this repre- sents an uncollected product of 1 + 1 + 1 + 2 + 2 + 2 or 9 terms, as it should. Similarly, the product (a + b) (a + b) (a + b) should have 2 • 2 • 2 or 8 terms. But this product when simplified is a^ + 3 a-b + Sab- + b^, which means, as it should, an uncollected product of 1 + 3 + 3 + 1 or 8 terms. One should bear this theorem in mind when reckoning with 556 symmetric functions by the methods of the last chapter. Thus, the student has proved, p. 249, Ex. 7, Sa6 • 2a = Za^b + 3 :Eabc. To test this formula, suppose that only the letters a, b, c are involved. Then Za6 has 3 terms, Sa has 3 terms, Sa^ft has 6 terms, and 2a6c has 1 term ; and 3-3 = 6 +3-1, as it should. Products of binomial factors of the first degree. The theorem 557 of § 554 enables one to obtain the product of any number of factors of the form a; +- ^ by inspection. Thus, (x + h,) (x + ^-,) (x + b,) = x^ + (bi + b^ + b,) x"" + (^1^.2 + ^1^3 + Kh) X + bAK For, selecting x from each factor, we have the term x^. Selecting in all possible ways x's from two of the factors and a b from the third, we have the terms bix"^, b^x"^, b^x^. Selecting in all possible ways x from one of the factors and b's from the other two, we have the terms bib2X, h^b^x, b^b^x. Selecting ft's from all three factors, we have the term b^bj)^. Observe that when the terms of the product are collected, as in the formula, the coefficient of a-'^ is the sum of the three letters ^i, b^, b^, the coefficient of x is the sum of the products of 254 A COLLEGE ALGEBRA every two of these letters, and the final term is the produci of all three. Hence these coefficients are symmetric functions of h^, b^, b^, as was to be expected since (x + bi) (x + b^) (x -f- b^ is itself symmetric with respect to b^, b^, b^. Observe also that since (x + b-^) {x + ^g) {^ + ^s) is sym- metric with respect to b^, b^, b^, we may obtain the product by finding one term of each type, as x^, b^x^, b^b-^x, b^b^b^, and then writing out all the terms of these several types. 558 By the same reasoning we may prove the general formula (x + b,) (x + b.) (x + b,)--- (x + b„) = x"-^ B^x"-^ + Box"-^ H + 5„, where Bi = 'Sbi = bi + b.^ + b^ ^ h b„, B^ = ^b^b^ = iA + bA + • • • + hh + ■■■ + K-iK, Bs = 2&1M3 = hhh + bAh + •••+ K-2K-l^n, B„= hhh---b,/, that is, Bi is the sum, and B^ is the i^oduct of all the letters *i> ^2> • • • ^n) ^^^ '^^ intermediate coefficients are : iJo, the sum of the products of every txoo of these letters ; B^, the sum of the products of every three ; and so on. Thus, we obtain one term of the product each time that we select ?/s from three of the factors and o-'s from the rest. Making the selection in all possible ways, we obtain the terms bib^bsX""^, b^b^biX"'^, ■ ■ -, and their sum is Bsx""-^. Observe that, as indicated above, the coefficients Bi, B^, ■•; B„ are symmetric fimctions of the letters bi, bo, •••, h^. 559 In like manner, we have (x — bi) (x - J2) (x - ^^3) ■■■(x- I>„) = x"- B.x"--' + B^x—^ + (- 1YE„, where Bi, B^, •■• B„ have the same meanings as in § 558. and THE BINOMIAL THEOREM 255 the signs connecting the terms are alteiuately — and +, the last sign, that of (— l)"i?„, being + when n is even and — when n is odd. We obtain this formula by merely changing the signs of all the letters b^, h^, ••• &„ in the formula of § 558. For this leaves unchanged every B whose terms are products of an even number of Z»'s, and merely changes the sign of every B whose terms are products of an odd number of i's. Example. By the method of §§ 557-559 find the following products. 1. (X + 1) (X + 2) (a; + 3). 2. (x+ 2)(x - 3) (x + 4). 3. (x4-a)(a; + 6)(x + c)(x + d). 4. (x - 2/)(x + 22/)(x - 32/){x + 4?/). The number of terms in the sums 2bi, Sbibo, • • •. Let ??i, Wr,, • • • 560 denote the number of terms in 2/^i, 26A> • • • respectively. 1. Since 2&i =l h-^-\-h.^ -{-•■• A^h^, evidently Wi = n. 2. If we multiply each of the n letters h^, h^, ■•■, i„ by each of the other w — 1 letters in turn, we obtain n (ji — 1) products all told. But these n (n — 1) products are the terms of l.hil>2, each counted twice. Hence /Zg, the number of terms in 2&i62» is 71 (71 — l)/2, or 71—1 71 ()l — 1) Thus, we have &1&25 &1&3) • • -1 bibn; b^bi, h^z-, • • •» bibn ; • • ■ ; &«&!> b„b2, • • •, 6„6„_i. There are n groups of these products, and n — 1 products in each group, hence n(n — 1) products all told. But the product 6162 here occurs twice, namely once in the form &162 and once in the form 62&1 ; and so on. 3. Again, if we multiply each of the ??2 terms of ^b^h^ hy each of the n — 2 letters which do not occur in that tern\ we obtain Wg (w — 2) products all told. But these ti^ (w — 2) products are the terms of ^b-fij)^, each cotmted three times. Hence 713, the number of terms in ^b^h.jb^, is n2{7i — 2)/3, or 71-2 n (71-1) {71 -2) 256 A COLLEGE ALGEBRA Thus, we have 6i&2&3) bibibi, • ■ •, bib2bn; &1&3&2) bib^bi, • • •, ^i^s^nj • • • ; b„-ib„bu bn-ib„b2, ■ ■•, 6a_i6A-2. There are n^ groups of these products, and n — 2 products in each group, hence n^ (n — 2) products all told. But the product 616263 here occurs three times, namely in three forms, 6162 • 63, 6163 • 62, 6263 • 61. And similarly every term of S616263 here occurs three times, once for each of the three ways in which a product of three letters may be obtained by multiplying the product of two of the letters by the remaining letter. 4. By the same reasoning we can show that n-3 _ n(n-l)(n- 2) (n - 3) '''-''' 4 ~ 1.2-3.4 and, in general, that n(n — l)(n — 2) ■•• to r factors "'-'- 1.2.3...,- Thus, the numbers of products of four letters 6i, 62, 63, 64, taken ov,e, two, three, four at a lime, are 4-3 ^ 43 -2 , 4-3.2-1 . „i = 4, n2 = — = 6, n3 = j^:^ = 4, n, = ^-^-^ = L 561 Binomial theorem. If in the formula of § 558, namely, (x + bi) (x + b^)--- (x + b„) = ic" + Bix"-^ + B^x''-- -\ \- B„, we replace all the n different letters bi, b^, ■ ■ ■ b„ by the same letter b, and x by a, the first member becomes (a + b)". Again, since each of the 71 terms of B^ becomes b, and each of the Tiz terms of B^ becomes b"^, and so on, we have, § 560, n -nb B - ^^("-1) /,. n _n( n-l)(n-2-) ■Di — no, n^— J ,^ 0, li^— 103 <>,•••. Our formula therefore reduces to the following : (a + &)» = a" + J a^-'b + '^^^^ a-^' + 1-2.3 « 6 + •. where THE BINOMIAL THEOREM 257 1. The, 7iumber of terms on the right is n -)- !• 2. The exponents of a, decrease by one and those ofh increase by one from term to term, their sum in each term being n. 3. The first coefficient is 1, the second is n, and the rest of them may be found by the following rule: Multiply the coefficient of any term by the exponent of a in the term and divide by the exponent of b increased by 1 ; the result will be the coefficient of the next term. This formula is known as the binomial theorem, the expres- sion on the right being called the expansion of {a + ^)" by this theorem. •Thus, (a + 6)3 = a3 + 3a26 + — a62 + ?_L?ji53 ^ ' 1-2 1.2-3 = a3 + 3 a?-h + 3 a62 + W. Since a + ^ is symmetric with respect to a and b, it follows 562 from § 542 that the terms of the same type in the expansion of {a + Z»)" — namely those involving a." and i", a"~'6 and ab"~^, and so on — must have the same coefficients. But these are the first and last terms, the second and next to last terms, and in general every two terms which are equally removed from the beginning and the end of the expansion. Hence the last term is b", the next to last is nab"~^, and so on. Since the number of terms is n + 1, there will be one middle term when n is even, two when n is odd. When there are two middle terms, they are of the same type and have the same coefficients. And by what has just been said, the coeffi- cients of the terms which follow the middle term or terms are the same as those which precede them but in reverse order. It may also readily be shown that the coefficients increase 563 up to the middle term or terms and then decrease, so that the middle coefficient or coefficients are the greatest. This follows from the rule of coefficients, § 561, 3, since in the terms which precede the middle term or terras the exponent of a is greater than the exponent of b increased by 1, while in the terms which follow it is less. 258 A COLLEGE ALGEBRA 564 By changing the sign of h in the preceding formula and simplifying, we obtain (a - hy = a" - na^-'b + ^^^^^ a-'b' n(n-l)(n-2) 1-2.3 « ^ + » where the terms which involve odd powers of b have — signs^ and those which involve even powers have + signs. Example. Find the expansion of (2 x - y^)^. Substituting 2 x for a, and y^ for 6, in the formula, and remembering that the last three coefficients are the same as the first three in reverse order, § 562, we have (2 X - 2/3)6 ^ (2 x)6 - 6 (2 x)^y^ + ^ (2x)* (t/3)2 _ ^lllA (2 x)^ (2/3)3 + . . . = (2 x)6 - 6 (2 x)52/3 + 15 (2 x)* {yY - 20 (2 x)3 {y^ + 15 (2 X)'^ (7/3)4 _ 6 (2 X) (2/3)5 + (y3)6 = 64x6 - 192x^2/3 + 240x*2/e - 160x3j/9 + mx'^y^'^ - 12x2/^5 + y". 565 The general term. From § 561 it follows that the (r -f l)th term in the expansion of {a + by is n(n — V) (n — 2) ■ • ■ to r factors ■ — ^ -^ ' a^~^b^. 1 • 2 • 3 • • • r This, with a minus sign before it when r is odd, is also the (r + l)th term in the expansion of {a — by. Example 1. Find the eighth term in the expansion of (x — yy^. Here n = 16 and r + 1 = 8, or r = 7. Hence the required term is 16 -15 -14 13 12 11 10 1.2.3-4-5-6-7 x^z/T = - 11440x92/'- Example 2. Does any term in the expansion of (x3 + l/x)i2 contain x2f» ? If so, find this term. Let r + 1 denote the number of the term. Then, since n = 12, a = x*, and b = 1/x, vre mu.st have a'>-'-b'- = (x3)i2-'-/x'- = x^-*'- = x2o. THE BINOMIAL THEOREM 259 This condition is satisfied if 36 — 4 r = 20, or r = 4. Hence the fifth term contains x^o, and substituting in the formula we find this term to be lUijJ^a;2o = 495 x2o. 1-2. 3-4 EXERCISE XXXI Expand the following by aid of the binomial theorem. 1. (3x + 2 2/)3. 2. (a -6)8. 3. (1 + 2x2)7. 4. (2 + l/x)4. 5. {x-3/x)6. 6. (x/y-y/z)^ 7. (1 - X + 2 x2)*. 8. (a2 + ax - x2)3. 9. Find the sixth term in (1 + x/2)". 10. Find the eighth term in (3 a - 4 6)i2, 11. Find the middle term in (a^ - 2 bcy°. 12. Find the two middle terms in (1 — x)^. 13. Find the coeflicient of x^ in (1 + x)^. 14. Find the coefficient of x* in (3 - 2 xy. 15. Find the coefficient of x^ in (1 — x-)^. 16. Find the coefficient of x^ in (1 + 2 x)^ + (1 - 2«)". 17. Find the constant term in (x + l/x)i2. 18. Find the coefficient of x"^ in (2 x — 1 /x) is. 19. Find {x + 2y) {x — Sy) {x — 5y) by inspection. 20. Find (x + 2) (x + 3) (x - 4) (x - 5) by inspection. 21. What is the number of terms in the product {a + b + c + d){f+g + h){k + l) {m + n+p + q)? 22. Find the sum of the coefficients in the following products. 1. (i + X2 + X3 + X4)3. 2. (1 + 2 X + X2)2 (1 + X + 3 X^)^. 23. What is the sum of the coefficients in the following symmetric functions of four letters a. 6, c, d when expanded ? 1. Sa2 • 2a. 2. 2,a^ ■ "Slabc. 3. Sa6 • Zabc. 24. Show that the sum of the coefficients in the expansion of (a + 6)" is 2". 25. Show that in the expansion of (a — 6)" the sum of thg positive coefficients is numerically equal to the sum of the negative coefficients. 260 A COLLEGE ALGEBRA XL EVOLUTION 566 Perfect powers. Given a rational function P. It is possible that P is a lyerfeot nth power; in other words, that a second rational function Q exists such that P — Q". If so, this rational function Q will be an nth root of P. In the present chapter we consider the problem : A rational function P being given, it is required to determine whether or not P is a perfect nth power, and, if it is, to find its 7ith root Q. We suppose n to denote a given positive integer. 567 Roots of monomials. Let P denpte a rational monomial reduced to its simplest form. If P is a perfect nth power, an 7ith root of P may be obtained by the following rule. Divide the exj)onents of the several literal factors of P by n, and multiply the result by the principal nth root of the numeri- cal coefficient ofP. This follows at once from the rule for involution, § 318. Thus, (aW/c'«)" = ai''6'''/t"'", §318. Hence a'^b'/C" is an nth root of a'-"V"/c""\ § 566, and it is obtained by dividing the exponents of 568 The root thus obtained is called the principal nth root of P (compare § 258). We shall mean this root wh^n we speak of the nth root of P, or when we use the symbol VP. Example 1. Find the cube root of - 8 a^h^/21 x^y^. We have a/ - 8 a%^ _ 2 ab^ Example 2. Find the following roots, 1 04 a*66 ^^Ixhfz^K 3. 6/32^ 100 c8di2 569 Roots of polynomials. Consider the following examples. Exainp^le 1. Determine wliether or not 4 a;* — 4 a;^ + 13 x2 — 6 a; + a perfect square, and, if it is, find its square root. EVOLUTION 261 If this is a perfect square, evidently it must have a square root of the form 2 x'^ +px + q, where p and q are constants. We must therefore have Ax* - 4x^ + rSx^ - 6x + 9 = (2x'^ + px + qf = 4x* + 4jjx3 + (p- + ■iq)x'^ + 2pqx + q'^, which requires, § 284, that p and q satisfy the equations : 4p = -4 (1), 2)2 + 4g = 13 (2), 2pg = -6 (3), q^ = 9 (4). From (1) and (2) we find p = - 1, q = 3, and these values of p and q satisfy (3) and (4) ; for 2 (- 1) 3 = - 6, and 3^ = 9. Hence 4 x'* - 4 x-^ + 13 x- - 6 x + 9 is a perfect square and 2 x^ - x + 3 is its square root. Example 2. Find the cube root of x6 + 6x5 + 21x* + 44 x^ + 63x2 f 54x + 27. If this be a perfect cube, it will have a cube root of the form x^+px + q. We must therefore have x6 + 6 x5 + 2 1 X* + 44 x3 + 63 x2 + 54 X + 27 = (x2 + px + qf = x6 + 3px6 + 3(p2 + q)xi + (p3 + epq)x^ + 3 {p-q + g2) x2 + 3 J992^ + q^, which requires, § 284, that p and q satisfy the six equations : 3p = 6 (1), 3(p2 + g) = 21 (2),-.. gs ^ 27 (6). From (1) and (2) we obtain p = 2, g = 3. And these values of p and q will be found to satisfy the remaining equations (3)- • -(6). Hence x^ + 6x5 + • • • + 54x + 27 is a perfect cube, and its cube root is x2 + 2 X + 3. By the method illustrated in these examples it is always possible to determine whether or not a given polynomial in x is a perfect nth. power, and, if it is, to find its nth. root. Let the polynomial be aox"' + aja;"'"^ + • • • + «„,. If this be a perfect wth power, its degree m must be a multiple of n so that m = kn, where k is an integer ; and it must have an nth root of the form ax'' + A^x'-'^ -\- ■ ■ • + A^., where a denotes the principal wth root of ao, and -4i, • • ■, Ai. are unknown constants. We call this root the principal nt^i root. To determine whether OoX"' + • ■ • + a,„ has any such root, and to find this root if it exists, we set aoX'" + a^x'"-' 4- . . . + a„ = (ax' + A,x'-' + ■■■ + A,)". 262 A COLLEGE ALGEBRA / reduce the second member to the form of a polynomial in x, and then equate its coefficients to those of the like powers of X in the first member. We thus obtain a system of nk equa- tions in A^, Ao, ■■■, A^. The first k of these equations will give a single set of values for A^, An, ■■■, A^\ and this set of values must satisfy the rest of the equations if a^x™ + • ■ ■ -\- a^ is to be a perfect ?ith power. Example 3. Find the cube root of 8 x6 - 12 x5 + 18 x« - 13 x3 -f 9 x2 - 3 X + 1. 570 Square roots of polynomials. If a polynomial P is a perfect square, its square root may also be obtained by the following method. As in the preceding section, let P denote a polynomial in x of even degree and arranged in descending powers of x. Let us suppose that P is a perfect square and that a, b,c, ■•• denote the terms of its square root arranged in descending powers of x, so that P = (a+J + c + -- -y. The problem is, knowing P, to find a, b, c, ■■ ■. Now, whatever the values of a, b, c, ■■• may be, we have (a + by = a^ + 2ab + b^ = a^ + (2a + b)b, {a + b + cY={a + by + 2 {a + b)c + c"" = a'+{2a + b)b+[_2{a + b) + c-]c, {a + b -{- c -\- dy == a^ + {2 a + b)b +[2{a + b) + c-]c + [2{a + b + c)+cqd, and so on, a new group of terms being added on the right with each new letter on the left, namely, a group formed by addimj the tieio letter to twice the sum of the old letters and multijjlying the result by the new letter. Therefore, since by hypothesis P = [a -{- b + c + • • -y, we have P = a'+(2a + b)b+{_2{a + b)+ (f\c + [2(a + 6 + c) + o?](^ + ..., EVOLUTION 263 where the leading terms of the several groups on the right, namely, d\ 2 ah, 2, ac, 2ad,---, are all of higher degree in x than any of the terms which follow them. From this identity we may find a, b, c, ■ ■ ■ as follows : 1. Evidently a is the square root of the leading term of P. 2. Subtract a^ from P. As the leading term of the remainder, Ri, must equal 2 ab, we may find b by dividing this term by 2 a. 3. Having found b, form {2a + b)b and subtract it from Ri. As the leading term of the remainder, R^, must equal 2 ac, we may find c by dividing this term by 2 a. 4. Continue thus until a remainder of lower degree than a is reached. If this final remainder is 0, P is, as was supposed, a perfect square and its square root is a + Z> + c + • ■ •. If this final remainder is not 0, P is not a perfect square ; but we shall have reduced P to the form P=(a + b + c + .--y+R, that is, to the sum of a perfect square and an integral function which is of lower degree than a. It is convenient to arrange the reckoning just described as in the following example. Example. Find the square root of 4 x* - 4 z^ + 13 a;2 _ g ^ + 9. P = 4x*-4x3 + 13x2-6x + 9 |2x'^-x + 3 = a + 6 + c a^ = 4x* 2a + 6 = 4x2-x - 4x3 + 13x2 - 6x + 9 = El = P - a2 - 4 x3 + x2 =i2a + b)b 2(a + 6) + c = 4x2- -2x + 3 12x2 - 6x + 9 = E2 = P - (a + 6)2 12 x2 - 6 X + 9 = \2{a + b) + clc = E = P - (a + 6 + c)2 Since the final remainder is 0, P is a perfect square and its square root is 2x2 - X + 3. Compare § 569, Ex. 1. Observe that as each new remainder R^, R^, ■•• is found we divide its leading term by 2 a and so get the next term of the root. Then at the left of the remainder we write twice the part of the root previously obtained plus the new term of 264 A COLLEGE ALGEBRA the root. We multiply this smn by the new term of the root, subtract the result from the remainder under consideration, and thus obtain the next remainder. Example. Find the square root of 25 x* — 40x3 + 46 x'^ - 24x + 9. 571 This method is applicable to a polynomial P which involves more than one letter, provided it be a perfect square. We first arrange P in descending powers of one of the letters, with coefficients involving the rest, and then proceed as in § 570, it being understood that x now denotes the letter of arrangement. 572 Approximate square roots. We may also apply this method to a polynomial in x arranged in ascending powers of this letter. But a, b, c, ■ ■■ will then be arranged in ascending powers of x, and the degrees of the successive remainders will increase. Hence, § 570, 4, if P is not a perfect square but has a constant term, we can reduce it to the form that is, to the sum of a perfect square and a polynomial, R', whose lowest term is of as high a degree as we jjlease. For small values of x we can make the value of it' as small as we choose by carrying this reckoning far enough. Hence in this case we call a + i, a + ^ + c, • • • the approximate square roots of P to two terms, three terms, and so on. It should be added that these approximate roots are found more readily by the method of § 569. Example 1. Find the square root of 1 + x to four terms. By § 569, we write Vl + x = 1 + px + ^x^ + rx^ + • • ■ . Squaring, 1 + x = 1 + 2px + (p^ + 2 q) x"- + 2{j>q + r) x^ + . • • Hence, §284, 2p=l, ^2 + 2^ = 0, pg + r = 0, or solving, p = l/2, q'=— 1/8, r=l/16. Therefore the required result is 1 + x/2 — x'-/8 + x^/lO. Let the student verify this by the method of §§ 570, 57 L Example 2. Find the square root of 4 — x + x^ to three terms. EVOLUTION 265 Square roots of numbers. From the formulas in § 570 we 573 also derive the ordinary method of finding the square root of a number. Example. Find the square root of 533(31. Let a denote the greatest integer with but one significant figure, whose square is contained in 5336 L Its significant figure will be the leading figure of the root and its remaining figures wiil be O's. We find a as follows : Remembering that for each at the end of a there will be two O's at the end of a-, we mark off in 53361, from right to left, as many periods of two figures as we can, thus : 5'33'61. Each of the periods 61 and 33 calls for one at the end of a, and the remaining period, 5, calls for the initial figure 2, since 2 is the greatest integer whose square is less than 5. Hence a — 200. Having found a, we proceed quite as when seeking the square root of a polynomial. This is indicated in the scheme below at the left, where b denotes the second figure of the root multiplied by 10, and c the units figure. The scheme at the right gives the reckoning as abridged in common practice. a + b + c '5' 33' 61 1 200 + 30+1 4 00 00 = a^ 2a = 400 2a + 6 = 430 1 33 61 = Ri 1 29 00 = (2 a + 6) 6 2 (a + 6) = 460 4 61 = R2 2 (a + ?>) + c = 461 14 61 = [2(a + b) + c]c = R We first subtract a^, then find the significant figure of b by dividing the remainder i?i by 2 a, next find Ro by subtracting {2a + b)b from Ri, and finally c by dividing R2 by 2 (a + b). The simplest way of accomplishing all this, as indicated in the abridged scheme at the right, is to omit final O's and to bring down one period at a time. Then, as each new remainder is obtained, we write at its left twice the part of the root already found as a "trial divisor," obtain the next figure of the root by dividing the remainder by this trial divisor, and complete the divisor by affixing this figure to it. We then multiply the complete divisor by the new figure of the root, subtract, and so obtain the next remainder. If too large a figure is obtained at any stage •n the process, that is, a figure which makes the product just described greater than the remainder in question, we try the next smaller figure. 266 A COLLEGE ALGEBRA 574 Approximate square roots of numbers. The method just explained also enables us to obtain approxhnate values of the square roots of numbers which are not perfect squares. Example. Find an approximate vahie of the square root of 7.342 cor- rect to the third place of decimals. 7.34'20'00I2.709 Evidently for each decimal fig- ' ure m the root there are two such . _ - o figures in the number. Hence we „ 90 Hence separate the decimal part of the 54^^ 20 00 V7:M = 2.709... ™^'' into periods of two fig- . ures, proceednig from the decimal 1 — —— point to the right. The integral part of the number we separate into periods, as in § 573, proceeding from the decimal point to the left. Observe that a decimal number cannot be a perfect square if it has an odd number of decimal figures. 575 Cube roots of polynomials. There is also a special method for finding the cube root of a polynomial P, when P is a perfect cube, analogous to that just given for finding a square root. Let P denote a polynomial in x whose degree is some multi- ple of 3 and which is arranged in descending powers of x. Let us suppose that Pisa perfect cube and that a,b,c,-'- denote the terms of its cube root, arranged in descending powers of x, so that P = (a-\-h-\-c-{--- -y. The problem is, knowing P, to find a, b, c, ■■ -. Now, whatever the values of a, l>, c, ■■■ may be, we have (a + by= a^ + (3 fl2 + 3 ai + b"") b, (a + i + c)3 = a" + (3 «2 + 3ab + b^)b + [3(a + by + 3(« + b)c + c2]c, and so on, a new group of terms being added on the right with each new letter on the left, namely, a group formed bij adding together three times the square of the sum of the old letters, tJiree times the product of the sum of the old letters by the new letter, and the square of the new letter, and then multiplying the result by the new letter. EVOLUTION 267 Therefore, since by hypothesis P = (a + 6 + c + -- •)^, we have P = a^ + (Sa"" + 3 ab + b^)b + [3 (a + by + 3 (a + b)c + c^^c+ • ■ ; where the leading terms of the several groups on the right, namely, a^, 3 a^b, 3 a'^c, • • •, are all of higher degree in x than any of the terms which follow them. From this identity we may find a, b, c, ••• as follows : 1. Evidently a is the cube root of the leading term of P. 2. Subtract a^ from P. As the leading term of the remain- der, Ri, must equal 3 a^, we may find b by dividing this term by 3 a^. 3. Having found b, form (3 a^ -\- 3 ab + b^) b and subtract it from Ri. As the leading term of the remainder, R2, must equal 3 a^c, we may find c by dividing this term by 3 a\ 4. Continue thus until a remainder is reached which is of lower degree than a"^. If this final remainder is 0, then P is, as was supposed, a perfect cube and its cube root is a + ^ + c + • ■ • . If this final remainder is not 0, P is not a perfect cube, but we shall have reduced it to the form p = (a + b + c-] y -h R, where R is of lower degree than a^. It is convenient to arrange this reckoning as follows : Example. Find the cube root of x6 + 6x6 + 21x* + 44x3 + 63x2 ^ 54 3; + 27. |x2 + 2x + 3 3a2 = 3x* x6 + 6x5 + 21x^ + 44x3 + 63x2 + ,54x + 27 3(x2)2 = 3x* 3x2-2x + ( 2x)2z=6x3 + 4x2 Ix* + 6x3 + 4x2 6 x5 + 21 X* + 44 x3 + 63 x2 + 64 X + 27 6x5 + 12x4+ 8x3 3{x2+ 2x)2 = 3x* + 12x3+ 12x2 3(x2 +2x)3 + 32= 9x2 + 18x + 9 ;x* + 12x3 + 21x2 + 18x + f)x* + 36x3 + 63x2 + 54x + 27 = Rj Qx" + 36x3 + 63x2 + 54x + 27 =R 268 A COLLEGE ALGEBRA Since the final remainder is 0, x^ + 6 x^ + • • • + 54 x + 27 is a perfect cube and its cube root is x- + 2 x + 3. Compare § 569, Ex. 2. Observe that as each new remainder R^, Ri, • •• is found, we divide its leading term by 3 a^ and so get the next term of the root. Then at the left of the remainder we write the sum of three times the square of the part of the root previously obtained, three times the product of this part by the new term, and the square of the new term. We multiply this sum by the new term, subtract the result from the remainder under consideration, and thus obtain the next remainder. 576 This method is also applicable to a polynomial which involves more than one letter, if it be a perfect cube (compare § 571). The method may also be applied to a polynomial in x arranged in ascending powers of this letter, — if it does not lack a constant term. If the polynomial is not a perfect cube, we thus obtain approximate cube roots (compare § 572). 577 Cube roots of numbers. We may also find the cube root of a number by aid of the formulas of § 575. Example. Extract the cube root of 12487168. a + 6 + c N = 12' 487' 168 [200 + 30 + 2 = 232 8 000 OOP 3a2 = 120000 Zah= 18000 62 = 900 138900 4 487 108 = Ri = 2f 4 167 OOP = (3 a2 + 3 a6 + b^-) b 1 (a + 6)2 = 158700 l(a + 6)c= 1380 c2 = 4 160084 320 168 = E2 = iV - (a + bf 320 168 = [3 (a + 6)2 + 3 (a + 6) c + c"-] c =R = N-{a + b + c)\ In order to find a, the greatest number with one significant figure whose cube is contained in N, we begin by marking off periods of three figures in N from right to left (also from the decimal point to the right when there are decimal figures in N), thus : 12' 487' 168. Each of the periods 168 and 487 calls for one at the end of a, and the remaining EVOLUTION 269 period, 12, calls for the initial figure 2, 2 being the greatest integer whose cube is contained in 12. Hence a = 200. The rest of the reckoning is fully indicated above. Observe that each new figure of the root is found by dividing the remainder last obtained by three times the square of the part of the root already found ; thus, we find the significant figure of b by dividing jRi by 3 a-, and c by dividing 2?2 by 3 (a + b)'^. If too large a figure is thus obtained, we test the next smaller figure. The process may be abbreviated in the same way as that for finding the square root of a number. Approximate cube roots of numbers which are not perfect cubes may also be found by this process (compare § 574). Higher roots of polynomials. The fourth root of a polynomial 578 which is a perfect fourth power may be obtained by finding the square root of its square root ; similarly the sixth root of a polynomial which is a perfect sixth power may be obtained by finding the cube root of its square root. It is also possible to develop special methods, analogous to those of §§ 570, 575, for finding any root that may be required. But the general method of § 569 makes this \mnecessary. In fact we have given the special methods for square and cube roots explained in §§ 570, 575 only because of their historic interest and their relation to the problem of finding square and cube roots of numbers. EXERCISE XXXn Simplify the following expressions. 3/ 27xe^_ /5i9^_ ^^_^^_^^ \ 125a92i2 \625c2d8 ^ ^ y ^ y/ By § 569 or § 570 find the square roots of the following. 4. X* -2x3 + 3x2 -2x+ 1. 5. x2 - 2 X* + 6 x3 - 6 X + x6 + 9. 6. 4 x6 + 12 x^y + 9x4?/2 - 4 x^yS _ 6 xV + 2/®- 7. 4»2_20x + 13 + 30/x + 9/a;2. 270 A COLLEGE ALGEBRA 8. 49 -84a; -34x2 + 60x8 + 25x*. 9. x8 + 2 x^ - x6 - X* - 6 x3 + 5 x2 - 4 X + 4, 10. (X2 + 1)2_4X(X2-1). 11. 4x* + 9x2y2 _ 12x32/ + 16x2 - 24x2/ + 16. 12. X2/2/2 + 2/2/X2 + 2 + 2 X2 + 2 y2 + x22/2. Find approximate square roots to four terms of the following. 13. l-2x. 14. 4-x + 3x2. By § 569 or § 575 find the cube roots of the following. 15. x6 + 3 x5 + 6 x* + 7 x3 + 6 x2 + 3 X + 1. 16. 27 xi2 + 27 xio - 18 x8 - 17 x6 + 6 X* + 3 x2 - 1. 17. 8 x6 - 36 ax5 + 90 aH* - 135 aH^ + 135 a*x2 - 81 a^x + 27 a^. 18. x^/y^ + 2/Va;3 + Zx'-/y^ + 3y2/a;2 + Q^/y + 62//X + 7. 19. Find the approximate cube root to three terms of the expression 1 - X + X2. 20. By § 569 or § 578 find the fourth root of x8 -4x^ + 10x6- 16x5+ 19x*- 16x3 + 10x2 -4x + 1. 21. By § 569 find the fifth root of a;io + 5a;9 4. i5a;8 + sox^ + 45x6 + 51 xS + 45x* + 30x3 + 15x2 + 5x + 1. 22. To make x* + 6x3 + 11 x2 + ^j. _|_ 5 a perfect square, what values must be assigned to a and h ? Find the square roots of the following numbers. 23. 27889. 24. 2313.01. 25. 583.2225. 26. 4149369. 27. .00320356. 28. 9.024016. Find approximate square roots of the following numbers correct to the third decimal figure. 29. 2. 30. 55.5. 31. 234.561. Find the cube roots of the following numbers. 32. 1860867. 33. 107284.151. 34. 1036.433728. IRRATIONAL FUNCTIONS 271 XII. IRRATIONAL FUNCTIONS. RADICALS AND FRACTIONAL EXPONENTS REDUCTION OF RADICALS Roots. In what follows the letters a, b, ■■ • will denote posl- 579 tive numbers or literal expressions supposed to have positive values. Again, ">/« will denote the principal nih. root of a, that is, the positive number whose wth power is a ; in other words, the positive number which is defined by the formula (Vo)" = a. Finally, when n is odd, -^ — a will denote the principal nth. root of — a, namely — va. And when we use the word root we shall mean principal root. Note. This is a restricted use of the word root; for any number whose 580 nth power equals a is itself an nth root of a, and there are always n such numbers, as will be proved subsequently. Thus, since 22 = 4 and (- 2)2 = 4, both 2 and - 2 are square roots of 4. We shall indicate the principal root 2 by Vi, the other root - 2 by -Vi. When n is odd and a is real, one of the nth roots of a is real and of the same sign as a, and the rest are imaginary. When n is even and a is positive, two of the nth roots of a are real, equal numerically, but of contrary sign, and the rest are imaginary. When n is even and a is negative, all the nth roots of a are imaginary. In the higher mathematics Va usually denotes any nth root of a, not, as here, the principal root only. Radicals. Any expression of the form Va or b Va is called 581 a radical ; and n is called the index, a the radicand, and h the coefficient of the radical. When both a and b are rational numbers or expressions, h 'Va is called a simple radical. Thus 5 Vi is a simple radical whose index is 3, its radicand 4, and its coefl&cient 5. 272 A COLLEGE ALGEBRA 582 Formulas for reckoning with radicals. The rules for reckoning with radicals are based on the following formulas, in which m, n, p denote positive integers. 1. Va^ = "-^'aF\ 2. ^b = Ta.-Vb. 3. ^~±^^. 4. {Vay = VoF\ 5. VVa = '>/«. Observe in particular that, by 1, the value of a radical is not changed if its index and the exponent of its radicand are multiplied by the same positive integer or if any factor common to both is cancelled ; thus, Vo^ = Vo^. The similarity of this rule to the rule for simplifying a fraction is obvious. These formulas may be proved by aid of the definition (Va)" = a, the laws of exponents («.'")"=«""', (aby = a"b", and the rule of equality, § 261, 3, Two positive numbers are equal if any like powers of these numbers are equal. Thus, 1. Vo^ = Vrt'"^, since their «^;th powers are equal. For ( Va^'yp = a"'P ; and (va'")"-^' = (a'")'' = a"v. 2. 'Vab = V a • V^, since their nth. powers are equal. For (V^)« = ab ; and (V^ ■ %/6)" = (V^)" • ( v/fe)" = ab. 3 \/- = > since their nth. powers are equal. ^^ ]" = [(y/a)"]'» - a'". IRRATIONAL FUNCTIONS 273 5. V V a = Va, since their mnth powers are equal. For ( Vfl)"»« = a ; and ( V Va)'"'" = ( Va)" = a. The following examples will show the usefulness of these formulas. 1. -v/s = V^ = V2. 2. VSab^ = Vi^ . V2ab = 26 V2ab. 3. 3/ 3c Vsc Vsc . 5/ , 10, ; V#^ = I^ = ^- '• ^^32x'V = V32x:^.5 = V2^ 5. ( V2^)2 = V(2 x2/-^)2 = V4xV = V VT^. - On simplifying radicals. That form of a radical is regarded 583 as simplest in w^hich the radicand is the simplest integral expression possible. Hence for simplifying radicals we have the following rules, which are immediate consequences of the formulas just demonstrated. 1. If the radicand he a -power whose exponent has a factor in common with the index, cancel that factor in both exponent and index. Thus, -^27 x-V = \/(3x^ = v'3xy2. 2. // aiuj factor of the radicand be a power ivhose exponent is divisible by the index, divide the exponent by the index and then remove the factor from under the radical sign. Thus, vie x"?/9 = V2*x'»x37/82/ = 2 xy"^ ^xHj. 3. If the radicand he a fraction, multipdy its numerator and denominator by the simplest expression which ivill render it pos- sible to remove the de^iominator from under the radical sign. „, s/xw 3 U xyz 1 ^ n Thus, -V — ^ = \\-~ = 7^^^ ^y^- ' \2 22 \ 8 23 2 2 Similar radicals. Radicals which, when reduced to their 584 simplest forms, differ in their coefficients only are said to be similar. Thus, yf\x^y and VsTx^ are similar ; for their simplest forms, namely 2 x Vx^ and 9 x^y "^'^i differ in their coefficients only. 274 A COLLEGE ALGEBRA 585 On bringing the coefficient of a radical under the radical sign. Since b V^ = V'6"a, the coefficient of a radical may be brought under the radical sign if its exponent be multiplied by the index of the radical. EXERCISE XXXm Reduce each of the following radicals to its simplest form. 1. 5. Vl8. 2. V3/2. 6. 4^25a56iOci5d6. V588. 3. V-272. 4. V-IOOO. V3/2. 7. V3/4. 8. \/3/16. 9. 10. Vl28 a26*c8. 13. ■s/a"62"c3". 11. V8X<52/9215. 12. "^25a26*c6. 14. ■v^a2« + '63« + V». 15. ^X6 - X32/3. 16. V(x2- 2/-) (•« + 2/)- 17. 18. ^/a^^>l - - 2 a-'65 + a--;6«. 19. 3 a3 + 63 \ 32 a?)2 3/ a3 -■ ^^9w■ 22. , ,„.3 - Vf -^-f -i -«*• Va3«63„ + 2 Bring the coeiiicients of the following under the radical sign. a -\-b \a — b a - 6 \a + b 25. 3 a V3 a. 26. '-^-^^ "V/ '^'' — , • 27. 3 ax % i / 27 a3x3 Show that the following sets of radicals are similar. 28. V18, V50, and V\J%. 29. V^, ^fm, and ^8/9. 30. V(x3 - 2/3) (x _ y) and Vx*y2 _i- xHj^ + x-y*. OPERATIONS WITH RADICALS 586 Addition and subtraction. We have the rule : To reduce the algebraic sxim of two or more radicals to its simplest form, simplify each radical aiid then combine such of them as are similar by adding their coefficients. IRRATIONAL FUNCTIONS 275 Example. Add Vl6 a^b, - Vg a?b, 3 Vi, and - 2 Vi/2. We have VWa^ - V9a26 + 3 V^ - 2 Vl72. = 4aV6-3aV6 + 3V2-V2 = aV6 + 2V2. Observe that a sum of two dissimilar radicals cannot be reduced to a single radical. Thus, we cannot have Vx + Vy^ Vx + y except when x or ?/ is ; for squaring, we have x + y + 2 y/xy = x + y, .-.2 Vxy = 0, .-. xy — 0, .-. either x = or y = 0. Reduction of radicals to a common index. It follows from the 587 formula Va'" = Va™^ that we can always reduce two or more radicals to equivalent radicals having a common index. The least common index is the least common multiple of the given indices. Example. Keduce va^ and Vfts to their least common index. The least common multiple of the given indices, 6 and 8, is 24. And , Va^ = Va2o and VP - Vfts. Comparison of radicals. We make the reduction to a common 588 index when we wish to compare given radicals. K. 20^ 6 /— Example 1. Compare vl6, v6, and V3. The least common multiple of the given indices, 15, 10, 6, is 30 ; and 15 , 30 , 30 , 10 ,— 30 , 30 , 6 ,— 30 , 30 , . V 16 = V162 = V256 ; V6 = V63 = V216; V3 = V35 = V2i3. Therefore, since 256 > 243 > 216, we have V16 > V3 > V6. Example 2. Compare 2 V3 and ViT. Bringing the coeflBcient of the first radical under the radical sign, § 585, and then reducing both radicals to the common index 6, we have 2 V3 = V12 = v^ = \/l728 ; ^/41 = V^ = V168I. Therefore, since 1728 > 1681, we have 2 Vs > ViT. Multiplication and division. From the formulas 589 Va • V^ = -y/ab and we derive the following rule : 276 A COLLEGE ALGEBRA To multiphj or divide one radical hy another, reduce them, if necessary to radicals having the least common index. Then find the product or quotient of their coefficients and radicands separately. Example 1. Multiply 4 >/xy by 2 Vx-y^. We have 4 Vxy • 2 \/x2^ = 8 V^ • \/^ = 8 VxV = Sxy Vx?/. Example 2. Divide 6 Vxy by 2 vxy. We have 6 V^/2 Vxy = 3 V^/ Vxy = 3 v^. 590 Involution. From the formulas (V^)'» = y38 • 2 a Vft = 3 ^ VT^ = 3 vT^. IRRATIONAL FUNCTIONS 277 Simple radical expressions. By a simjjle radical expression 592 we shall mean any expression which involves simple radicals only. Thus, Va + V^ is a simple radical expression. We call such an expression integral when it involves no fraction with a radical in its denominator. By the rules just given, sums, differences, products, and jMivers of simple integral radical expressions can be reduced to algebraic sums of simple radicals. In § 607 we shall show that the like is true of quotient s. But ordinarily a root of a simple radical expression, as \a + V^, cannot be reduced to a simple radical expression. Example 1. Multiply 3V6 + 2V5by2V3- VlO. We hare (3 Ve + 2 V5) (2 Vs - VlO) = 6 vTs + 4 vTs - 3 Veo - 2 V50 = 8 V2 - 2 VT5. Example 2. Square V2 + Vi. We have (V2 + \/4)2 = 2 4-2 V2 V^ + VT6 = 2 + 4^2 + 2-^. EXERCISE XXXIV Reduce the following to their least common index. 6/- 10/— IS/— 3, i. 6; 1. V3, V3, and V3. 2. Va^ V2 a^b^ andVTffS. Compare the following. 3. 3 Vi and 2 ^3. 4. V3, v^, and ^/l. Eeduce each of the following to a simple radical in its simplest form. 5. V35 -=- Vfjl. 6. 10 -4- V5. 7. 4 - ■\/2. 8. Ve • VTo • VT5. 9. ^/m • ^/m • -v/Ts. lo. 2 V3 •- 3 v^. 11. Vi-V^-v^. 12. -v/3-i-V^. 13. 2V35. V65H-V91- 14. ^a^b^c'' ■ -x/osb^cs. 15. "'V^ ■ '"v^. 16. V^ - V^. 17. \/a26c2 . 4/^&2^. A COLLEGE ALGEBRA 18. Va-'v^. 19 20. Va62- V'a65^(v'^- Vai26"). 21. (Vl2)3. 22. (\/a2)6. 24. Vv^- 25. ^^/Vs. 27. V v^. 28. V2V2. 31. '•v^V;?'. 23. (2Vx2/223)6. 26. Vv'a^bVc^. 29. V2V2. 32. (V Va) . 30. VV2-.V'2. Simplify each of the following as far as possible. 33. V12 + V75 - Vis + Vl47. 34. Vl25 + VT75 - V28 + Vl/20, 35. "v^ - VTo8 + ^^T/i. 36. ^^a/hc + y^h/ca + ^c/ah. 37. V50 _ Vii + ^- 24 + ^. 38. V(^r+6J% - V^ - V62^. Vax3 + 6 ax2 + 9 ax - Vax^ - 4 a-x^ + 4 a^x. 40. (x + y)^f—^-{x-y)■ 41. (V2 + V3+V6)- Ve. 43. (V6+V5)(V2 + Vl6). 45. (1+V3)3. — y \ x2 — ?/" 42. ( V(5 + VTo + VTi) -- V2. 44. Vo + 2 V2 • V5 - 2 V2. 46. (Va + v'a + l)(Va- Va + 1). FRACTIONAL AND NEGATIVE EXPONENTS 593 In many cases reckoning with radicals is greatly facilitated by the use oi fractional exponeyits. Thus far we have attached a meaning to the expression a" only when n denotes a positive integer. The rules for reckon- ing with such expressions, namely, 1. a"' -a" = a'" + ", 2. (a"")" = a"'", 3. (ab)" = a''b", are among the simplest in algebra. It is therefore natural to inquire : Can we find useful meanings for a'\ in agreement with these rules, when n is not a positive integer ? IRRATIONAL FUNCTIONS 279 The definition aP/i = Va^. Tak£ _fl.^ f orj natance. We wish, 594 if possible, to find a meaning for this symbol which will be in agreement with the rules 1, 2, 3. But, to be in agreement with 1, we must have (a^y = a^.a^ = J^'^ = a^ = a, that is, a^ must mean either \a or — Va. We choose the more convenient of these two meanings, and define a^ as Va. We thus find that one of the conditions which we wish a^ to satisfy suffices to fix its meaning. Similar reasoning leads us to define a} as Va, a' as Va^, and in general a« as '\aP, that is, as the principal qth root of aP. p pm p Observe that since a'' = Va** = 'Vo^ = a«"', the value of a« is not changed when ^9/3- is replaced by an equivalent fraction. Thus, a^ — a^ = a^\ also a"^ = a^ = a^. The definition a° = 1. Again, to be in agreement with 1, 595 we must have «<>»"' = ftO + m^^m^ and therefore a° = «"'/«"' = 1. We are therefore led to define a" as 1. 596 The definition a~® = 1/a®. Finally, to be in agreement with 1, we must have, § 595, ar'-a' = a-' + ' = a" = 1, and therefore a~* = 1/a'. We are therefore led to define a~^ as 1/a'. Thus, by definition, a-s = 1 /a^, oT^ = \/a^ = \/ V^. p Jt remains to prove that the meanings thus found for a', a°, and a~^ are in complete agreement with the rules of exponents. Theorem 1 . The Iww a" a° = a'"+° holds good for all rational 597 values of m. and n. 280 A COLLEGE ALGEBRA Let 2h <1j ^5 * denote any positive integers. Then 1. When m =p j q and n = r/s, we have, § 582, - - Q I s/ 9*/ 9-'/ a'^.a' = -^JaP ■ V a'" = Va^' • Va«'" qs / -+- 2. When ??i = — p/q and ?i = — r/^, we have, by Case 1, 3. When m =p /q and n =■ — r / s, and p/q > r / s, we have 4. When m=p/q and n = — r/s, and plq■■"« = a™. 2. When n = p / q, where ^ and (7 are positive integers, we have, by Case 1, (a"')9 = V(a"')" = Va"* = a « = a «. 3. When n = — s, where s is any positive rational, we have, by Cases 1, 2, ^ ^ (a"')" a"" IRRATIONAL FUNCTIONS 281 Theorem 3. The law (ab)" = a"b° holds good for all rational 599 values of -a. 1. Let 7i = p/q, where p and q denote positive integers. Then p p p (aby = -Vjaby = -^a^b" = VaJ' ■ Vbi' = a^b^. 2. Let n = — s, where s denotes any positive rational, whether integral or fractional. Then, by Case 1, ^ ^ (aby a'b" Applications. The following examples will illustrate the 600 use of fractional and negative exponents. A complicated piece of reckoning with radicals often becomes less confusing when this notation is empl-^yed. Example 1. Simplify \ a/^a. We have ^ a/^a -. {aa~^^- = («*)i = a^ = Va. 4, 6. Z. Example 2. Simplify Va^» • \'cv>h ^ ^a-b^. We have va^^ • va'^fe -^ Va-6^ = a^b^ ■ a^^ ■ a~h~^ Example 3. Expand {x^ + y~^^. We have (x' + y-')^ = (x^^ + 3(j;3)2y-§ + 3x?(y-i)2 + (y-^f = X- + .3 x'y~ ^ + 3 x%~ ^ + y~\ Example 4. Divide x — ?/ by x* + x^y^ + y^. Arranging the reckoning as in § 401, we have X — y [x^ + x'y^ + y^ X + x^y^ + x^y^ I x» — y^ - x^j/^ - x^2/^ - y Hence the quotient — x^y^ — x^2/5 — y isz^ — y^. 282 A COLLEGE ALGEBRA EXERCISE XXXV Express as simply as possible without radical signs 1. '-vV. 2. V^. 3. aV^. 4. bVb*- V65. Express without negative or fractional exponents 5. a»^. 6. c-i-5. 7. ((Z')-6. 8. (e-«^)-^. Express with positive exponents and without radical signs 9. a-V6-°c-2. 10. x-^V^. 11. (l/V^)-4. 12. a;-2 Vr-^/?/-2 V^. Express as simply as possible without denominators 13 ^ _ ^ _ a~M^~' + c~') 6 + c be c-2 a-2(6 + c) 6- +c-i' Reduce each of the following to its simplest exponential form. 14. (3i)l 15. 8li. 16. (-27)1 17. 8-5. 18. a?aM. 19. akr^arA 20. (a^6)5a^6l 21. ab-^/a-^b. 22. (a')l 23. (a-i6-2c3)-2. 24. (-32aio)l 25. (- a66-9)-*. 26. 6-iVF5^6-iVFi. 27. (a-^ V6^)3. 28. (8a-i5/^125a3)-J. 29. Va^ (fjc- 1)-2. 30. Va-iV^. 31. ■V'a? V^/V-^^ • V^. 32. [(X*):*]:*. 33. (x^' + a^J';/!'' + ^)a: + ». 34. (x' - r)/(x-i + 2/--S). 35. Multiply x* + x"y^ + y^ by x* - x^y^ + y*. 36. Divide a2 _ h^ by a^ - 6'. 37. Expand (x^ - yh^)*. 38. Simplify [(e^ + e-^)2 - 4)]*. 39. Find square root of x2 + 4x'y^ + Axy -\- Qx^y^ + 12 2/2 4. Qx-^yK 40. Find cube root of x^ + 3x2 + 6x + 7 + 6x-i + 3x-2 + xr^. IRRATIONAL FUNCTIONS 283 THE BINOMIAL THEOREM FOR NEGATIVE AND FRACTIONAL EXPONENTS If in the binomial expansion, § 561, (a + by = a" + na"-'b + ''^'^~^^ a'^-'b^ + • • • we assign a fractional or negative value to w, we shall have 601 on the right a never-ending, or infinite, series ; for none of the coefficients n, n(n — l)/2, • • • will then be 0. It will be shown further on that if b < a the sum of the first m terms of this series will approach the value of (a + by as limit when m is indefinitely increased ; in other words, that, by adding a sufficient number of the terms of this series, we may obtain a result approximating as closely as we please to the value of (a + by. This is what is meant when iu is said that the binomial theorem holds good for (a + by when n is fractional or nega- tive and b <. a. Example 1. Expand (8 + x"')' to four terms. Putting n = 1/3, a = 8, 6 = x~^ in the formula, we have (8 + x"')' = 8s + ^ • 8-3X-* + 5__^8-§(x-^)2 2 l(-t)(-D 8-S(x-^)3 + ' 5x-? 12 288 20 Example 2. Find the sixth term in the expansion of 1 / (a' -f- x^)^ or (a^ -I- x^)-2. Putting 71 = — 2, a = a^, & = X3, r = 5 in the formula for the (r -|- l)th term, § 565, we have (-2)(-3)(-4)(-5)(-6) . , = _ 6 a-ix^. 1-2-3-4.6 ^ ' ^ ' 284 A COLLEGE ALGEBRA Example 3. Expand Vl + x to four terms. Since Vl + x = (1 + x)^ we have n = h, a = 1, b = x. 2 2-3 2 8 16 The result is the same as that obtained in § 572, Ex. 1. Example 4. Find an approximate value of VlO. We have VlO = (32 + i)^ = 3 (1 + i}K + ■ 6 216 3888 = 3 + .16666 - .00462 + .00025 + • • • = 3.1623 nearly EXERCISE XXXVI Expand each of the following to four terms. 1. (1 + x)^. 2. (a^ + x'^~K 3. 4^(27 -2x)2. 4. (a^ + x)^. 5. (a-i-6~5)-4. q (^^ + ^y)-6. _ 1 o 1 \ vTXsv^y 2 + 3x ^(1 + x)2 ^ Vl +3Vx' 10. Find the tenth term in (1 + x)-3. 11. Find the seventh term in (x-2 — 2 ?/*)i 12. Find the term involving x^ in (1 — x')». 13. Find the term involving x-2 in x~^ (2 + x~s)-3_ 14. By the method illustrated in § 601, Ex. 4, find approximate values of the following. 1. Vgo. 2. V^. 3. Vsi. IRRATIONAL FUNCTIONS 285 RATIONALIZING FACTORS Rationalizing factors. When the product of two given radi- 602 cal expressions is rational, each of these expressions is called a rationalizing factor of the other. Thus, (Va + V6) ( Va - Vb) = a-h. Hence Va + Vd is a rational- izing factor of Va — Vft, and vice versa. It can be proved that every finite expression which involves simple radicals only has a rationalizing factor. The following sections will serve to illustrate this general theorem. Rationalizing factors of functions of square roots. Every expres- 603 sion which is rational and integral with respect to Va? can be reduced to the form A + B V^, where A and B are rational and integral with respect to x ; and A -]- B V^ has, with respect to x, the rationalizing factor A — B V^, obtained by merely changing the sign of -\x. Thus, 2(Vx)* + 3x(Vx)3 may be written 2x- +_3a;2 Vx. Hence this expression has the rationalizing factor 2 a;^ _ 3 ^1 Vx. We may obtain a rationalizing factor of an expression which is rational and integral with respect to any finite num- ber of square roots, as Va;, Vy, V^, • • ■ , by repetitions of the process just explained. For we shall obtain a result which is completely rational if we multiply the given expression by its rationalizing factor with respect to V^, the product by its rationalizing factor with respect to Vy, and so on. Example. Find the rationalizing factor ofl+Vx + v^ + 2 Vxy. We have l+Vy + \^(l+2 V^). (1) Multiply (1) by 1 +^^-^^(1 + 2 Vy) . (2) We obtain (1 + V^Y - x (1 + 2 V^)% or 1 -x + 2/-4xy + 2V?/{l -2x). (3) Multiply (3) by l-x + 7/-4xy-2Vy(l-2x) . (4) We obtain (1 - x + ?/ - 4 xy)"- - 4 y (1 - 2 xf. (5) Therefore, since (5) is completely rational, the product of (2) and (4) is the rationalizing factor of (1). 286 A COLLEGE ALGEBRA 604 Rationalizing factors of binomial radical expressions. The rationalizing lactor of an expression of the form Va ± V6 may be found as in the following example. Example. Find the rationalizing factor of v a + v6. We have Va + Vb = a^ + b^ = (a"^)^ + {b^)K (1) But, §438, (a2)B + (a^)^ will exactly divide the rational expression a* - 63, the quotient being (a^)^ - (a-')^(65)6 + (53)i. (2) Hence (2) is the rationalizing factor of (1). 605 On rationalizing the denominator of a fraction. Any irrational expression of the form A /B, in which B involves simple radi- cals only, may be reduced to an equivalent expression having a rational denominator by multiplying both A and B by the rationalizing factor of B. Example 1. Rationalize the denominator of 1/ Va^. TTT t. 1 1 a* a^ */- We have -j — — — = — — ^ = — = y/a/a. V X- + Cl^ + 'Vx- Example 2. Rationalize the denominator of Vx2 + a2 - Vx2 _ (£1 We have V^:jr^ + Vx2 -a? _ (Vx2 + a2 + Vx2-a2)2 Vx2 + a2 - Vx2 - a2 ( Vx2 + a2 - Vx2 - a2) ( Vx2 + 02 + Vx2 - a2) _ x2 + Vx* - a* ~ ^2 606 In computing an approximate value of a fractional numeri- cal expression which involves radicals, one should begin by rationalizing the denominator. Much unnecessary reckoning is thus avoided. Example. Find an approximate value of (1 + Vs) / (3 — Vi) which is correct to the third decimal figure. Wehave i±^ = il±l^^^lii±^ = 1+ V^ . 2.414- .- . 3 - V^ (3 - V2) (3 -f- v^) IRRATIONAL FUNCTIONS 287 Division of radical expressions. To divide one radical expres- 607 sion by another, we write the quotient in the form of a fraction and then rationalize the denominator of this fraction. Example. Divide 4 + 2V5byl-V2+ V5. We have 4 + 2 V5 (4 +2V5){1 + V2-V5) . - 3 + 2V2 -V5 + V1O 1 4. V2 + V5 (1 + V2 + V5) (1 + V2 - VS) V2 - 1 ^ (-3 + 2V2-V5 + VlO)(V2 + l) ^ J _ V2 + V^. (V2-1)(V2 + 1) General result. It follows from § 592 and § 607 that every expression which involves simple radicals only can be reduced to an algebraic sum of simple radicals. EXERCISE XXXVn Find rationalizing factors of the following. 2. ^^o^VftS. 3. x? + x^ + x'. 5. Vx + Vy + Vz. 6. Vxy+Vyz+V^. - Vu. 8. Vx + Vx + 1. 10. ^ - "^62. 11. x^ _ yk^ 13. 1 + xV'. 14. X? + x^ + 1. 16. 1 + V2 + Vs. 17. 1 + ^/2. 19. v'li + ^/6+^/3. Reduce each of the following to a fraction having a rational number or expression for its denominator. 20. -J—. 21. « + ^^ - ^-^ 1. v^. 4. V^ + V6c. 7. y/z + y/y -yfi 9. x^^yK 12. xJ + yl 15. 3-V5. 18. V9 + V3 + I. 23. 25. V^V62 1 6 + V62 - a2 1 +V2 +V3 I-V2+V3 I+V2 + V3 + V6 288 A COLLEGE ALGEBRA 27. "^ + ^^ ■ 28. ^_l_ + ^^_. Vx + Vy + Vx + y V3 - 1 V3 + 1 Find approximate values of tlie following expressions correct to the third decimal figure. 29. -A-. 30. ' + ^ - '^^"^ V125 V? V2 + Vs IRRATIONAL EQUATIONS On solving an irrational equation. The general method of solving an irrational equation is described in the following rule. First, rationalize the equation. Next, solve the resulting rational equation. Finally, test all the solutions thus obtaiiied in the given equa- tion and reject those which do not satisfy it. For, let P = denote the given equation, and PR = the rational equation obtained by multiplying both members of P = by iZ, the rationalizing factor of P. By § 341, the roots of PR ^0 are those of P = and R = jointly. We dis- cover which of them are the roots of P = by testing them in this equation. Example. Solve x — 7 — Vx - 5 = 0. Multiplying both members by the rationalizing factor x — 7 + Vx — 6, we obtain (X _ 1)2 _ (X - .5) = 0, or simplifying, x2-15x + 54 = 0. Solving, by § 455, x = 9 or 6. Substituting 9 for x in x - 7 - Vx- 5 = 0, we have 9 - 7 - V9-5=0, which is true. Hence 9 is a root. But substituting 6, we have 6 - 7 - V6 - 5 = 0, which is false. Hence 6 is not a root. But observe that 6 is a root of the equation x - 7 + Vx '- 6 = 0, obtained by equating the rationalizing factor to ; for 6 — 7 + Ve — 5 = is true. IRRATIONAL FUNCTIONS 289 An equation which involves the radical V J may be ration- 610 alized with respect to this radical by collecting the terms which involve \A in one member and the remaining terms in the othei*, and then raising both members to the «th power. By repetitions of this process an equation which involves square roots only may be completely rationalized. It follows from § 345 that this method is equivalent to that described in § 609, but it involves less reckoning. Example 1. Solve V Vx + a = Vft. Cubing both members, Vx + a = 6^. Transposing and squaring, x = (IP — a)2. Substituting this result in the given equation, we -find it to be a root. Example 2. Solve Transposing, Vx — 4 Squaring, x — 4 = 81 Simplifying, vx + 5 = 5. Squaring, x + 5 = 25. Solving, X = 20. Substituting 20 for x in the given equation, we have V25 + Vi6 = 9, which is true. Hence 20 is a root. Notes. 1. Observe, as in Ex. 1, that we rationalize an equation with 611 respect to the unknown letter only and make no attempt to rid it of radicals which do not involve this letter. 2. Observe also that a n irra tion al equ ation may have no root. Thus, the equation Vx + 5 — Vx — 4 = 1) has no root. For if we attempt to solve it we shall merely repeat the reckoning in Ex. 2 and shall again obtain the result x = 20 ; and V25 — Vl6 = 9 is false. 3. We may add that the simplest method of rationalizing an equation of the form Va + V^ + Vc + Vb = (or Va + Vb + Vc + E = 0) is to begin by writing it thus : Va+Vb = -Vc -Vd (otVa -\-Vb = ~Vc -E) and then to square both members. The resulting equation will involve but two radicals and it may be rationalized as in Ex. 2. 290 A COLLEGE ALGEBRA 612 Simultaneous irrational equations. To solve a system of such equations we may first rationalize each equation, then solve the resulting rational system, and finally test the results thus obtained in the given system. But if the equations are of the form described in § 379, they should be solpd by the method there explained. Exampk soiFea D} 1. I Solve 5 + V^ + 5 = Vx + V^, x + 2y=\-i. Squaring (1), x-5+?/ + 5 + 2 Vxy + ox- by •Zo ■= X -\- y ■]- 2 Vary, or y/xy ■\- bx - by - -lb = Vxy. (3) Squaring (3) and simplifying, x — y = b. (4) Solving (4), (2), x = 9, y = 4. (5) Substituting x = 9, ?/ = 4 in (1), we have Vi + V9 = V9 + Vi, whicli is true. Hence x = 9, ?/ = 4 is tlie solution of (1), (2). Example 2. Solve Vx + 6 + 2/Vy = 4, (1) 2 Vx + 6 + 6/%/?/ = 9. (2) Solving for Vx + 6 and 1 /Vy, we find Vx + 6 = 3, 1 /Vy = 1 /2. (3) And from (3) we obtain x = 3, 2/ = 4, which is the solution of (1) and (2). EXERCISE XXXVm Solve the following equations for x. 1. x* = 4. 2. x-^ = 3. 3. x? = 8. V3. d. --7 4. {V2x- 1)^ 6. Vox + Vftx + Vex : 8. Vx + 4 + Vx -t 11 V ^ + VS + Vx = 2. 7. V4x2 + x+ 10 = 2x+l. 9. V4X + 5+ Vx+1- V9x+10=0. 10. 12. 13. 'x + 1 + = 0. Vx + 2 /x + V + Vx - 2 = Vx + 2 + Vx - 1. 11. Vx2 + 3x - 1 - Vx'-^-X- 1: 'x + 3 + Vx = 2. 14. J. ^x+1 ' + ' ^X-1 Vx2-1 IRRATIONAL FUNCTIONS 291 Solve the following for x and y. Vx + 17 + Viy - 2 = Vx + 5 + 16, r_^- __ 3, + 2 V X + 2/ - 4 17. Show that Vx + a + Vx + 6 + Vx + c + Vx + d = will reduce to a rational equation of the first degree. 18. Show that Vax + h + vcx + d - Vex +/ = will reduce to a rational equation of the first degree if Va + Vc — Ve = 0. QUADRATIC SURDS Surds. Numerical radicals like V2'and Vd, in Avhich the 613 radicaud is rational but the radical itself is irrational, are called surds. A surd is called quadratic, cubic, and so on, according as its index is two, three, and so on. Theorem 1. The product of two dissimilar quadratic surds 614 is a quadratic surd. Suppose that when the surds have been reduced to their simplest forms, their radical factors are Va and Vi, The product of va and V ^ is Va6, and this is a surd unless ah is a perfect square. But ab cannot be a perfect square, since by hypothesis a and b are integers none of whose factors are square numbers, and at least one of the factors of a is different from every factor of b. Thus, V2 ■ Vs = Ve, Ve • Vis = V90 = 3 Vio. Theorem 2. The sum and the difference of two unequal quad- 615 ratic surds are irrational numbers. This is obvious when the surds are similar. Hence let Va and V^ denote dissimilar surds. 292 A COLLEGE ALGEBRA Suppose, if possible, that Va + V^ = c, (1) where c is rational. Squaring both members of (1) and transposing, 2 Vo^ = 0^ -a-h, (2) which is impossible since 2 Vo^ is irrational, § 614, while c^ — a — h is rational. Theorem 3. i/" a + Vb = c + Vd, ivhere Vb and Vd are surds, then a = c and b — (\.. For, by hypothesis, V^ — V(/ = c — a. But this is impossible unless Vft — V^ = and c — a = 0, since otherwise V^ — Vc^ would be irrational, § 615, and equal to c — a, which is rational. Hence b = d and a = c. Square roots of binomial surds. We have ( V^ ± V?/)2 = x + ij±2 Vxy. Hence if a + 2 V^ denote a given binomial surd, and we can find two jMsiti a e rational numbers x and 3/ such that X -\- 1/ = a and cry = b, then Vo; + Vy will be a square root of a + 2 V^ and Vx — Vy will be a square root oi a — 2 Va, and both these square roots will be binomial surds. When such numbers x, y exist they may be found by inspection. Example 1. Find the square root of 37 — 20 V3. Reducing to the form a - 2 %^, 37 - 20 \^ = 37 - 2 VSOO. But 300 = 25-12 and 37 = 25 + 12. Hence V37 -2 V3OO = V'25 - Vl2 = 5 - 2 V3. Example 2. Find the square root of 13/12 + V5/6. _ , 13 \l 13 , V30 13 + 2V3O We have 12 + Vc =^ I^ + "^ = — 1^ IRRATIONAL FUNCTIONS 29S Since 30 = 10 • 3 and 13 = 10 + 3, we have Vi3 + 2 Vio = VlO + V3. Hence Note. We may sis (4), ^13 + 2 V30 12 obtain formv Vx + v^ = V-^-V-y. )y (2), x-y-- x + y- X - V10 + V3 > V12 lias for X and y ; /12O+V36 1 12 2 as follows : a — Va2 _ ^~ 2 V30 By hypothe = Va + 2 A (1) and = Va-2V6. (2) Multiplying But = Va2-45. = a. (3) (4) Solving (3), a + Va'- -4b 2 -45 Observe that these values are rational only v^hen a^ - 4 6 is a perfect square. Hence in this case only is the square root of a + 2 Vft a bino- mial surd. EXERCISE XXXIX Find square roots of the following. 1.9 + V56. 2. 20 + 2 V96. 3. 32-2 Vl75. 4. 1+- — -. 5. 7-3V5. 6. 8V2 + 2V3O. 5 7. 2 (a + Va2 - 62). 8. 6 - 2 Vab - a^. Simplify the following. 618 9. Vn + 12 V2. 10. V9 + 4 V4 + 2 V3. IMAGINARY AND COMPLEX NUMBERS Complex numbers. Since all even powers of negative num- 619 bers are positive, no even root of a negative number can be a real number. Such roots are hnag'mary numbers. Definitions of the imaginary numbers and of the operations by "which they may be combined are given in §§ 217-228, which the student shoiild read in this connection. According to these definitions, 1. The symbol i = V— 1 is called the unit of imaginaries. 294 A COLLEGE ALGEBRA 2. Symbols of the form at, where a is real, are called pure imaginaries. 3. Symbols of the form a + bi, where a and b are real, are called complex numbers. 4. Two complex numbers are equal when, and only when, their real parts and their imaginary parts are equal, so that If a -\- bi = c -^ di, then a = c and b = d. 5. The siun, difference, jyroduct, or quotient of two complex numbers is itself a complex number (in special cases a real number or a pure imaginary) which may be found by applying the ordinary rules of reckoning and the relation i^ = — 1. The like is true of any positive integral power of a complex number, since by definition (a + biy = (a. + bi) (a + bi) ■ ■ ■ to n factors. Example 1. Add 5 + 3 i and 2 — 4 i. We have 5 + 3 i + (2 - 4 i) = (5 + 2) + (3 - 4) i = 7 - i. Example 2. Subtract 6 + 2 i from 3 + 2 i. We have 3 + 2 i - (6 + 2 i) = (3 - 6) + (2 - 2) i = - 3. Example 3. Multiply 2 + 3 i by 1 + 4 i. We have (2 + 3 i) (1 + 4 i) = 2 + 3 i + 8 i + 12 ^2 = 2 + 3i + 8i-12 = - 10+ lit. Example 4. Expand (1 + i)2. We have (1 + i)2 = 1 + 2 i + i2 = 1 + 2 i - 1 = 2 i. Example 5. Find real values of x, y satisfying the equation (x + yi) i - 2 + 4 i = (x - 2/x) (1 + i). Carrying out the indicated operations, we have - (y + 2) + (X + 4) i = (X + 2/) + (X - 2/)i. Equating the real and the imaginary parts, § 619, 4, - (y + 2) = X + y and x + 4 = x - y, or, solving, x = 6, y = — 4. In §§ 238-241 we have given a method for representing complex numbers by points called their graphs, and rules for obtaining from the graphs of two complex numbers the graphs IRRATIONAL FUNCTIONS 295 of their sum and product. Let the student apply these rules to Exs. 1, 3, 4. Conjugate imaginaries. Two complex numbers like a + hi 620 and a — bi, which differ only in the signs connecting their real and imaginary parts, are called conjugate imaginaries. The product of two conjugate imaginaries is a positive real 621 number. Thus, (a + hi) (a - hi) = a^ - hH'^ = a2 + 62, Hence a fraction, as (a + hi) / (c + di), may be reduced to 622 the form of a complex number by multiplying both its terms by the conjugate of its denominator. Example. Divide 54-7iby2 — 4z. We have M^jj ^ (5 + 7^) (2 + 4 Q 2 - 4 i (2 - 4 i) (2 + 4 i) - 18 + 34 i _ _ ^ 17 , 20 10 10 The powers of i. From the equation p = — 1 it follows that 623 the even powers of i are either — 1 or 1, and the odd powers either i or — i. Thus, i3 = z2 . i = — i ; i< = i3 . ^ _ _ i . j — _ ^2 _ j . ^^^ go on. To find the value of i" for any given value of n, divide n by 4. Then, according as the remainder is 0, 1, 2, 3, the value of i« is 1, i, — 1, — i. Thus, i'^ — (i*)8 = 1 ; i^ = i-^ ■ i = i; and so on. Even roots of negative numbers. The number — 4 has the 624 two square roots 2i and —2i; for (2iy = 2^i^ = — 4, and (— 2iy = (— 2y^i^ = — 4. We select 2 i a s the principal square root, and write V— 4 = 2i and — V— 4 = — 2 i. Similarly the principal square root of any given negative number — a is Vai, that is, V— a = ^/ai. From this definition of principal square root it follows that 625 it — a and — b are any two negative numbers, then V— a V— b — — -Vab. For V-a • V- 6 = Vai • Vbi = i^VaVb = - Vab. 29G A COLLEGE ALGEBRA Thus, while the product of the principal square roots of two negative numbers — a, — b, is 07ie of the square roots of their product ab, it is not, as in the case of real numbers, the principal square root of this product. When reckoning with imaginaries, it is impo£tant to bear in mind this modification of the rule Va Vb = ^(ib. All chance of confusion is avoided if at the outset we replace every symbol V— a by Vai. Example 1. Simplify V^ • ( V^)5 • (^^5)7. We have V^ . ( V^)5 . ( V^)" = V2 i • ( Vs j)M^ 0^ = V2- (V3)5- {V5)Tii3 = 1125 VsOi. Example 2. Multiply 2 + V^ by 1 + V^l. We have (2 + V~9) (i + y^^) = (2 + 3 /) (1 + i) = - 1 + 5 i. 626 The higher even roots of negative numbers are complex numbers. This will be proved subsequently. Thus, one of the fourth roots of — 4 is 1 + i ; for (1 + j)4 = 1 + 4 i + G i^ + 4 (3 + i4 = 1 + 4 i _ 6 - 4 i + 1 = - 4. 627 Square roots of complex numbers. As will be proved farther ■ on, all roots of complex numbers are themselves complex num- bers. We may find their square roots as follows. We have (v^ ± i Vy)- = x — y ± 2 t Vxy. Hence, if a + 6J denote a given complex number in which h is positive, and we can find two positive numbers x and y such that X - y = a, (1) and 2 Vxy = b, (2) then Vx + t Vy will be a square root of a + bi, and Vx — iVy will be a square root of a — bi. We may find such numbers x and y as follows. By hypothesis Vx + iVy = Va + bi, (3) and Vx - iV^ = Va - bi. (4) Multiplying (3) by (4), x + y = VoM^. (6) IRRATIONAL FUNCTIONS But, by (1), x-y ■ Hence, solving (5) and (6), x 297 (6) Va2 + 62 and a+Va2 + 62 And both these values are positive since V a2 + 62 > a. Example. Find a square root of — 1 + 4 Vs t. Here a = -l and Va2 + 62 = V(_ 1)2 + (4 y/l)i Hence x = (-l+9)/2 = 4 and 2/ = (1 + 9)/2 = 5. Therefore V_i+4 V5i- 2 + V5 ». EXERCISE XL Simplify the follovring. 1. V-49. 2. V- 18. 3. V-8-V-12. 4. V-22. 5. (V^)2. 6. ii2. 7. i-'. 8. ii5. 9, y/x-y-y/y -X. 10. (2+V-3){l+V-2). 11. (V-2)7(V-3)9. 12. (l + 2z)3 + (l -2iY. !'■ A— %• . V-a2 iV62 14. 4 + 6i 4-6i^ 1+i 1-i 15. ( V3 + 4 i + V3 - 4 i)2. 16. (l + i3)/(l+i). ,^ a + 6i 9 + 3V2i a-6i (3 + V2i)(l +V2i) 19. Divide 4 by 1 + V3^. 20. Find a fourth root of - 16. 21. Show that (- 1 + V3i)/2 is a cube root of 1. 22. Show that (1 + i)/V2 is a fourth root of - 1. 23. Find real values of x and y satisfying the equation 3 + 2 i + X (i - 1) + 2 2/i = (3 i + 4) (x + J/). Find square roots of the following. 24. 5 + 12t. 25. 2i 26. 4a6 + 2(o2 - 68)i 298 A COLLEGE ALGEBRA XIII. QUADRATIC EQUATIONS 628 General form of the equation. Every quadratic equation in one unknown letter, as x, may be reduced to the form ax? + hx + c = 0, where a, h, and c denote known numbers. If, as may happen, 6 = 0, the equation is called a pxire quadratic ; if & ^ 0, it is called an affected quadratic. 629 The roots found by inspection. The roots of the equation ax^ -^ bx -\- c = are those particular values of x for which the polynomial ax'^ -\- hx + c vanishes, § 332. There are two of these roots. If the factors of ax^ + bx -{- c are known, the roots of ax^ + bx -{- c = are also known, for they are the values of x for which the factors of ax" + bx -\- c vanish, §§ 253, 341. If the factors are x — a and x — /3, the roots are a and /3. Example 1. Solve the equation a;^ _|_ ^ — 6 = 0. We have x2 + x - 6 = (x + 3) (x - 2). The factor x + 3 vanishes when x = — 3, and the factor x — 2 vanishes when X = 2. Hence the roots are — 3 and 2. . Example 2. Solve abz^ — (a^ + b'^) x + (a^ — b-) = 0. Factoring, by § 443, [ax - (a + b)] [bx - {a - b)] = 0. Hence the roots are (a + b)/a and (a — b)/b. In particular, since x'^ — q =(x — Vy) (x + V^), the roots of the pure quadratic x^ — q =0 are Vy and — Vy. Again, since ax^ -i- bx = (ax + b)x, the roots of a quadratic of the form ax^ -\- bx = are —b I a and 0. Thus, the roots of 4 x2 = 9 are 3 /2 and - 3 /2 ; the roots of 2 x2 - x = are and 1 /2 ; the roots of 5x2 = are and 0. 630 Conversely, to obtain tlie quadratic whose roots are two given numbers, as a and y8, we form the product (.r — a) (x — ft) and equate this product to 0. Thus, the quadratic whose roots are - 2 and 1 /3 is (x + 2) (x - 1/3) = 0, or3x2 + 6x-2 = 0. QUADRATIC EQUATIONS 299 Example 1. Solve the following quadratics. 1. x2 + 2x-8 = 0. 2. 2x2-7x + 3 = 0. 3. (2x-l)(x-2) = x2 + 2. 4. (x-l)(x-3) = {2x-l)2. Example 2. Find the quadratics whose roots are 1. -2/3,-3/2. 2. a, -a. 3.1/4,0. General formula for the roots. But ax^ -\- bx + c may always 631 be factored ; for, as was shown in § 444, ■ ax^ -}- bx + c ^ a [x - - ^> + ^^' - 4 ac l f _ -b--^b^-A ac l Therefore, since the roots of ax"" + bx + c = (1) are the values of x for which the factors of ax^ -\- bx -{- c vanish, these roots are -b + Wb^-4ac , _ J _ V^»2 - 4 ac X — — ■ and X = — — — - — ■ j 2 a 2 a or, as we usually write them, -b± V^/2 _ 4 ac X = (2) 2 a ^ '^ This formula (2) should be carefully memorized, for ,it enables one by mere substitution to obtain the roots of any given quadratic which has been reduced to the form (1). Example. Solve 4 x2 + 105 x = 81. Reducing to the form (1), 4x2 + 105 x - 81 = 0. • Here a = 4, 6 = 105, and c = - 81. - 105 ± V1052 + 4-4.81 ,, , . 3 Hence x = = » that is, - or - 27. 8 4 When b is an even integer, a and c also being integers, it is 632 more convenient to use the formula -b/2±^(b/2r-ac ^ x = , (3) which is obtained by dividing the numerator and denominator of (2) by 2. 300 A COLLEGE ALGEBRA Example. Solve 3 x^ + 56 x - 220 = 0. Here 6/2 = 28, and substituting in (3) we have - 28 ± V282 + 3.220 ^, , . 10 x= = ) that IS, — or -22. 3 3 633 Any given quadratic may also be solved by applying directly to it the process of completing the square, § 444, as in the following example. But since this method involves needless reckoning, it is not to be recommended — except when the formula of § 631 has been forgotten. Example. Solve 3i2-6x + 2 = 0. Transposing the known term and dividing by the coeflBcient of x^, x2-2x = -2/3. Completing the square of the first member, x2-2x + 1 = 1/3. Extracting the square root of both members, a; _ 1 = ± V3/3, whence x = (3 ± V3)/8. 634 The methods just explained enable one to solve oxij fractional equation which yields a quadratic when cleared of fractions. But see §§ 524-527. Example 1. Solve 1 = ^ • x+1 x+2 x+3 x+4 Clearing of fractions and simplifying, 2 x^ + 10 x + 11 = 0. c , . _5±V3 Solvmg, x = Both of these values of x are roots of the given equation, for they cause none of its denominators to vanish. Example 2. Solve ^^^ + ^^ + ""^ = 0. ^ X2-1 X2-X X2 + X Clearing of fractions by multiplying by the lowest common denominator x(x2 — 1), and simplifying, we obtain 3x2 + 2x-5 = 0, whence x = 1 or - 5/3. But 1 cannot be a root of the given equation, since its first two denominators vanish when x = 1. Hence — 5/3 is the only root of this equation. QUADRATIC EQUATIONS 301 EXERCISE XLI Solve the following equations. 1. a;2 + 2x = 35. 3. x2 = 10 X - 18. 5. 2x2 + 3x- 4 = 0. 7. x2 + 9x- 252 = 0. 9. 8x2 -82x + 207 = 0. 11. x2-3x- 1 -f V3 = 0. 13. (X - 2)2(x _ 7) = (X + 2) (X - 3) (x - 6). 14. ^:^ + ^+-2 = 2. X + 2 2x Q ^ 16. 2. 4x2-4x=3. 4. 9x2 + 6x + 5 = 0. 6. (2x-3)2 = 8x. 8. 12 x2 + 56x- 255 = 0. 10. 15x2 -80x- 64 = 0. 12. x2-(6 + i)x + 8 + 2i = 0. ?)(x -6). 15. X x-\ 17 3 1 2x 2(x2-l) 4x + 4 8 2x + l 4x-2 1-4x2 8 18. 2x-1^3x+^^5x-U_ ^g _^+i L_ + i. = o. x-2 x-3 x-4 x(x-2) 2x-2 2z 20. -A ^ = ^_._^_. X — 1 4 — X X — 2 3-x 2^ x + 3 , 2x + l 17a; + 7 ,„ 4(x + 2)(3x-l) 3(3x-l)(x + 4) 6(x + 4)(x + 2) 22. ^ + ^ + ^— + ^ + ^ =0. 2 x2 - 7 X + 3 x2 - 2 a: - 3 2 x2 + a; - 1 23. 3 x2 + (9 a - 1) X - 3 a = 0. 24. x2 - 2 ax + a^ - 6^ = 0. 25. c2x2 + c (a - 6) X - a6 = 0. 26. x2 - 4 ax + 4 a2 - 62 = q. 27. x2 - 6 acx + a2(9 c2 - 4 62) = o. 28. (a2 - 62)a;2 - 2 (a2 + 62)^ + a2 _ 52 = 0. 29. l/(x-a) + l/(x-6) + l/(x-c) = 0. 30 {x-aY-{x-W ^^^^ ^^ (X - a) (x - 6) a2 - W- < 302 A COLLEGE ALGEBRA EXERCISE XLU 1. Find two consecutive integers whose product is 506. 2. Find two consecutive integers the sum of whose squares is 481. 3. Find two consecutive integers the difference of whose cubes is 91. 4. Find three consecutive integers the sum of whose products by pairs is 587. 5. Find a number of two digits from the following data : the product of the digits is 48, and if the digits be interchanged the number is diminished by 18. 6. The numerator of a certain fraction exceeds its denominator by 2, and the fraction itself exceeds its reciprocal by 24/35. Find the fraction. 7. A cattle dealer bought a certain number of steers for $1260. Having lost 4 of them, he sold the rest for $10 a head more than they cost him, and made $260 by the entire transaction. How many steers did he buy ? 8. A man sold some goods for $48, and his gain per cent was equal to one half the cost of the goods in dollars. "What was the cost of the goods ? 9. If $4000 amounts to $4410 when put at compound interest for two years, interest being compounded annually, what is the rate of interest ? 10. A man inherits $25,000, but after a certain percentage has been deducted for the inheritance tax and then a percentage for fees at a rate one greater than that of the inheritance tax, he receives only $22,800. "What is the rate of the inheritance tax ? 11. A man bought a certain number of $50 shares for $4500 when they were at a certain discount. Later he sold all but 10 of them for $5850 when the premium was three times the discount at which he bought them. How many shares did he buy ? 12. The circumference of a hind wheel of a wagon exceeds that of a fore wheel by 8 inches, and in traveling 1 mile this wheel makes 88 less revolutions than a fore wheel. Find the circumference of each wheel. 13. A square is surrounded by a border whose width lacks 1 inch of being one fourth of the length of a side of the square, and whose area in square inches exceeds the length of the perimeter of the square in inches by 64. Find the area of the square and that of the border. QUADRATIC EQUATIONS 303 14. The corners of a square the length of whose side is 2 are cut off in sucli a way that a regular octagon remains. What is the length of a side of this octagon ? 15. A vintner draws a certain quantity of wine from a full cask con- taining 63 gallons. Having filled up the cask with water, he draws the same quantity as before and then finds that only 28 gallons of pure wine remain in the cask. How many gallons did he draw each time ? 16. A man travels 50 miles by the train A, and then after a wait of 5 minutes returns by the train B, which runs 5 miles an hour faster than the train A. The entire journey occupies 2i hours. What are the rates of the two trains ? 17. A pedestrian walked 6 miles in a certain interval of time. Had the time been 1/2 hour less, the rate would have been 2 miles per hour greater. Required the time and rate. 18. A pedestrian walked 12 miles at a certain rate and then 6 miles farther at a rate 1/2 mile per hour greater. Had he walked the entire distance at the greater rate, his time would have been 20 minutes less. How long did it take him to walk the 18 miles ? 19. From the point of intersection of two straight roads which cross at right angles, two men, A and B', set out simultaneously, A on the one road at the rate of 3 miles per hour, B on the other at the rate of 4 miles per hour. After how many hours will they be 30 miles apart ? 20. If A and B walk on the roads just described, but at the rates of 2 and 3 miles per hour respectively, and A starts 2 hours before B, how long after B starts will they be 10 miles apart ? 21. If from a height of a feet a body be thrown vertically upward with an initial velocity of b feet per second, its height at the end of t seconds is given by the formula h = a + bt — 16 1^. The corresponding formula when the body is thrown vertically downward is ^ = a — 6^ — 16 1^. (1) If a body be thrown vertically upward from the ground with an initial velocity of 32 feet per second, when will it be at a height of 7 feet? of 16 feet ? Will it ever reach a height of 17 feet ? (2) A body is thrown from a height of 64 feet vertically downward with an initial velocity of 48 feet per second. When will it reach the height of 36 feet ? (3) If a body be dropped from a height of 36 feet, when will it reach the ground ? 304 A COLLEGE ALGEBRA XIV. A DISCUSSION OF THE QUADRATIC EQUATION. MAXIMA AND MINIMA 635 Character of the roots. The discriminant. Let a and y8 denote the roots of ax" + ^x + c = 0, so that, § 631, h 4- V/y- ^A,ic ^ -b - Vft^ - 4 ac 2 a 2 a The radicand i^ — 4 ac is called the dlscrimbiant of ax'^ -\-hx -\- c — Q. When the coefficients a, b, c are real, the character of the roots a, p is indicated by the sig7i of the discriminant. Thus : 1. When b^ — 4 ac is jjosifive, the roots are real aiid distinct. 2. When b^ — 4 ac is 0, the roots are real and equal. 3. When b'^ — 4 ac is negative, the roots are conjugate imagi- naries. It should also be observed that 1. When i^ — 4 ac = 0, then ax"^ + &a; + c is a perfect square. 2. When a is positive and c is negative, the roots are always real, since b- — Aac is then positive. 3. If a, b, c are rational, the roots are rational when, and only when, i^ — 4 ac is a perfect square. Example 1. Show that the roots of x^ - Gx + 10 = are imaginary. They are imaginary since 6'^ — 4 ac = (— 6)- — 4 ■ 1 ■ 10 = — 4. Example 2. For what value of ??i are the roots of mx- + 3 x + 2 = equal ? We must have S^ - 4 • m ■ 2 = 0, that is, ?«, = 9/8. Example 3. If possible, factor y^ + X7j - 2x^ + Ux + y - 12. Arranging the polynomial according to powers of y and equating it to 0, we have y^ + (x + l)y - (2x2 - 11 x + 12) = 0. ^ , . _ (X + 1) ± V<)x2-42x + 49 Solvmg, y = — '^ ■, that is, y = X - i, or y - - 2 X + S. A DISCUSSION OF QUADRATIC EQUATIONS 305 Hence, § 681, 2/2 + xy - 2 x2 + 11 x + ?/ - 12 = (y - x + 4) (?/ + 2 x - 3). Observe that the factorization is possible only because the radicand 9 x2 — 42 X + 49 is a perfect square. Relations between roots and coefficients. If a and /3 denote 636 the roots of ax^ -\- bx + c — 0, we have, § 631, ax'^ -{- bx -\- c = a[x — a) (x — f3). Dividing both meniljers of this identity by a and carrying out the multiplication iu the second member, we have b c x^ + - X + - = x- - (a + B)x + aB. Since this is an identity, the coefficients of like powers of x in its two members are equal, § 264, that is, a -^ /3 = — b I a and a^ = c ] a. This may also be proved by adding and multiplying the values of a and ^ given in § 631. Therefore, since a, ^ are the roots oi x'^ -^ bx I a -^ c j' a — ^, we have the theorem : In any quadratic of the form x^ + px + q = the coefficient of X with its sign changed is equal to the sum of the roots, and the constant term is equal to the product of the roots. Thus, in the quadratic Gx^ + x = 2, that is, x^ + x/6 - 1/3 = 0, the sum of the roots is —1/6, and their product is — 1 /3. Example 1. Solve Ox^ - lOx + 1 = 0. Obviously one of the roots is 1, for 9 — 10 + 1 = 0. Therefore, since the product of the roots is 1/9, the other root is 1/9 -^ 1 or 1/9. Example 2. Find the equation whose roots are three times those of 3x2 + 8x + 5 = 0. Let a and j3 denote the roots of 3 x^ + 8 x + 5 = 0. Then a + /3 = - 8/3 and a/3 = 5/3. Hence the required equation is x2_ (3a + 3/3)x + 3rt-3i3 = x2- 3(a + /3)x + 9a/3 = x2 + 8x + 15 = 0. Symmetric functions of the roots. The expressions a -\- (3 and 637 afi are symmetric functions of the roots a, jS, § 540. All other rational symmetric functions of a and can be expressed 306 A COLLEGE ALGEBRA rationally in terms of these two functions, a + ft and aft, and therefore rationally in terms of the coefficients of the equation. For every such function can be reduced to the form of an integral sym- metric function or to that of a quotient of two such functions. If an integral symmetric function contains a term of the type Zca''/3i' + ', it must also contain the term kaP + i^p, § 5427lmtiUierefore kaP^P {ai + /3'?). But cxP^p = (cr/S)?, and it may readily be showl^■fe^jisuccessive applications of the binomial theorem that ai + /S"? can be expresSed>4n^ terms of powers of a + ^ and a/3. Thus, since {a 4-/3)2=^2 + 2 n-/3 + /32, we have a^ + p^ = (a + /3)2 - 2^/3. Similarly we find a^ + /33 = (a + /3)3 - 3 a/3 (a + /3). Example. The roots of x^ + p,x + g = being a, j3, express 1 / a + 1 / /S and a3/3 + a/33 jn terms oi p and q. We have l/a + 1/^ = (a + /3)/a/3= -p/q, and a3|3 + a/33 = ap(a^ + ^2) =a/3[(a + /3)2-2a^]= g (p2-2g). 638 Infinite roots. Suppose that, instead of being constants, the coefficients of ax"^ -\- bx + c = are variables. We can then show that if a approaches as limit, one of the roots will approach oo ; and if both a and b (but not c) approach 0, both roots will approach x. For the formulas for the roots are - b + V62 - 4 ac ^ -b - Vb- - 4 ac 2 a Multiply both terms of the fraction a by — 6 — V6- — 4 ac and both terms of the fraction /3 by - 6 + V^"- - 4 ac. We obtain 2c „ 2c a = — 5 + Vb^ -4ac b - V 62 _ 4 ac By §§ 203, 205, if a = 0, then V52 - 4 ac = b. Therefore if a = 0, then a == — c/b and /3 = 00, and if a = and 6 = 0, then a == 00 and /3 = 00. It is customary to state these conclusions as follows, § 519: One root of ax^ -f- bx + c = becomes injinite when a van- ishes, and both roots become infinite when a and b {Imt not c) vanish simultaneously. A DISCUSSION OF QUADRATIC EQUATIONS 307 Maxima and minima. Let ?/ be a function of x^ § 278. It 639 may happen that, as x increases, y will increase to a certain value, m, and then begin to decrease, or that y will decrease to a certain value, m', and then begin to increase. We then call m a maximum value of y and m' a minimum value. Thus, y = {x — 1)2 — 4 has a minimum value when x = 1, this value being — 4. For if x start from a value less than 1 and increase, (x - 1)2 will first decrease to and then increase. Similarly y = 4 — (x — 1)^ has a maximum value, 4, when x = 1. Every quadratic trinomial ax^ -^hx -\- c with real coefficients 640 has either a maximum or a minimum value, which may be found as in the following examples. Example 1. Find the maximum or minimum value of ?/ = a;'- + 6 x — 7. By completing the square, x'- + 6x - 7 = (x + 3)2 — 10. Hence when x = — 3, ?/ has a minimum value, namely — 16. Example 2. Divide a given line segment into two parts whose rectangle shall have the greatest possible area. Let 2 a denote the length of the given segment, x and 2 a — x the lengths of the parts, y the area of their rectangle. Then y = x (2 a - x) = 2 ax - x2 = a2 - (a - x)2. Hence y has a maximum value when x = a, that is, when the given seg- ment is bisected and the rectangle is a square whose area is a'. The maximum and minimum values of quadratic trinomials 641 and of certain more complex functions may also be found by the following method. Example. Find the maximum and minimum values, if any, of 2/ = (4x2-2)/(4x-3). Clearing of fractions and solving for x, we have ± Vy2 _ 3 y ^- 2 2/ ± V(y - 1) (2/ - 2) 2 2 By hypothesis, x is restricted to real values. Hence y can only take values for which the radicand {y — 1) {y — 2) is positive (or 0), that is, the value 1 and lesser values and the value 2 and greater values. It follows from this that 1 is a maximum and 2 a minimum value of y. 308 A COLLEGE ALGEBRA For observe that as y increases to 1, the two values of x, namely 2/ + 2)/2 and (y + V?/2 _ 3 ^ + 2) / 2, respectively increase Hence, conversely, as x increases through 1/2, (2/-V2/2 and decrease to 1/2 y first increases to 1 and then decreases 642 Variation of a quadratic trinomial. Given y = ax^ -\-hx -{■ c, where a is positive. By completing the square, we obtain Hence 3/ has a minimum value when x =— b /2 a, this mini- mum value being (4ac — b^)/4:a. (T As X increases from — 00 to + 00, ?/ will first decrease from + 00 to (i ac — b^) / A a and then increase to + 00. Thus, let 2/ = x2 _ 2 X - 3 = (a; - 1)2 - 4. As X increases from — 00 to + , y first decreases from 00 to — 4 and then increases from — 4 to 00. Moreover ?/ = when x^ _ 2 x — 3 = 0, that is, when x = — 1 or 3. Until X reaches the value — 1, y is posi- tive ; it then remains negative until x = 3, when it again becomes positive. When x= 3,-2, -1, 0, 1, 2,3,4, 5,-.. we have ?/=••• 12, 5, 0,-3, -4, -3,0,5,12,. ••. We may obtain the graph of ?/ = x^ — 2 x — 3 by plotting these pairs of values and passing a curve through them, as in § 389. '(■\'_A\ ^ ' } Observe that to the zero values of y there correspond the points where the graph cuts the x-axis, and that to the minimum value of y there corresponds the lowermost point of the graph, which is also a turning point of this curve. EXERCISE XLin 1. For what valuesof 7?i are the roots of (?«+2)x2- 2 >?ix + l=0 equal? 2. What are the roots of {m'^ + m) x'- + 3 mx -2 = when m = - 1 ? when ?w = ? EQUATIONS SOLVABLE BY QUADRATICS 309 3. If possible, factor Sx- + 5xy - 2y- - 5x + 'ty - 2. 4. For what values of m can x- - y'^ + mx -{- 5y — 6he factored ? 5. The roots of x'^ + px + q = being a and /3, express {a - /3)2, a* + /3*, and a/^ + /3/a in terms of p and q. 6. The roots of2x2-3x + 4 = being a and /3, find the values of a/p^ + p/a^ and a^^ + ap^ 7. The roots of a;- + x + 2 = being rr and /3, find the equations whose roots are - a, - )3 ; 1/a, 1/^; 2 a, 2/3; a + 1, /3 + 1. 8. Find the maximum and minimum values of the following. 1. a;2_8x + 3. 2. 2x2 -X + 4. 3. l + 4x-x2. 4. x/(x2 + l). 5. l/x + l/(l-x). 6. (x + 1)/ (2x2-1). 9. Find the greatest rectangle that can be inscribed in a given circle ; also the rectangle of greatest perimeter. 10. A man who is in a boat 2 miles from the nearest point on the shore wishes to reach as quickly as he can a point on the shore distant 6 miles from that nearest point. If he can row 4 miles an hour and walk 5 miles an hour, toward what point should he row ? 11. "What height will a body reach if thrown vertically upward from the ground with an initial velocity of 48 feet per second, and when will it reach this height? See p. 303, Ex. 21. XV. EQUATIONS OF HIGHER DEGREE WHICH CAN BE SOLVED BY MEANS OF QUADRATICS Equations which can be factored. Given an integral equation 643 in the form A = 0. If we can resolve A into factors of the first or second degrees, we can find all the roots of ^1 = by equating the several factors of A to zero and solving the result- ing equations. For if ^ = BC •■■, then ^1 = is equivalent to B=zO, C = 0, ■•■, jointly, § 341. Example 1. Solve x* + x2 + 1 = 0. By § 436, X* + x2 + 1 = (x2 + X + 1) (x2 - x + 1). 310 A COLLEGE ALGEBRA Hence x* + x^ + 1 = is equivalent to the two equations x2 + X + 1 = and x2 - x + 1 = 0. „ , . ^ . -1 ± iVs- 1 ± j. Vs Solving these equations, x = or Example 2. Solve x* - x^ - 5 x^ - 7 x + 12 = 0. Factoring by the method of § 451, we find that x* - x3 - 5x2 - 7 X + 12 = (X - 1) (X - 3) (x2 + 3x + 4). Hence x* — x'' — 5 x^ — 7 x + 12 = is equivalent to the three equations X - 1 = 0, X - 3 = 0, and x2 + 3 X + 4 = 0, whose roots are 1, 3, and (- 3 ± i V7)/2. Example 3. Solve the following equations. 1. 6x3 - 11x2 + 8x -2 = 0. 2. X* -5z3 + x2 + llx + 4 = 0. 644 Equations of the type au^ + bu + c = 0, where u denotes some function of x. If the roots of au'^ + bu -{- e = when solved for u are a and fi, this equation is equivalent to the two equations m — a and u — /?, for, § 631, au^ + bu -\- c = a(u — a) (u — (3). Hence to solve au^ + bu + c — for x, we have only to solve the two equations u = a and u = (3 for x. Example 1. Solve 3 x< + 10 x2 - 8 = 0. Solving for i2, x^ = 2/3 or - 4. Hence x = ±Vt5/3 or ±2i. Example 2. Solve x^ + 3 - 10x~^ = 0. Multiplying by x', x^ + 3 x^ - 10 = 0. Solving for x', x^ = 2 or — 5. Hence x = ±2V2or ±5iV5. Example 3. Solve (x^ + 3 x + 4) (x2 + 3 x + 5) = 6. "We may reduce this equation to the form (x2 + 3x)2 + 9 (x2 + 3x) + 14 = 0. Solving for x2 + 3x, we obtain the two equations x2 + 3 X = - 2, and x2 + 3 x = - 7, whose roots are — 1, — 2, and (— 3 ± i Vl9)/2. EQUATIONS SOLVABLE BY QUADRATICS 311 Example 4. Solve (x + 1) (x + 2) (x + 3) (x + 4) = 120. By multiplying together the first and fourth factors, and the second and third, we reduce the equation to the form (x2 + 5 X + 4) (X- + 5 X + 6) = 120, which may be solved in the same way as the equation in Ex. 3. Example 5. Solve x^ + 10 x^ + 31 x^ + 30 x + 5 = 0. By completing the square of the first two terms, we obtain {x2 + 5x)2 + 6 (x2 + 5x) + 5 = 0. Solving for x^ + 5 x, we obtain the two equations x2 + 5 X = — 5 and x^ + 5 x = — 1, whose roots are (- 5 ± V6)/2 and {- 5 ± V2T)/2. Example 6. Solve 8 ?i+l^ + 3 ^inl _ n = 0. x2 - 1 x2 + 2 X Observing that the second fraction is the reciprocal of the first, we mul- tiply both members of the equation by the first fraction, thus obtaining 0. Solving for (x^ + 2x) /(x^ — 1), we obtain the two equations x2 + 2x , ,x2 + 2x 3 = 1 and = - , x2 - 1 x2 - 1 8 v7hose roots are —1/2 and —3, —1/5. All the values of x thus found are roots of the given equation since they cause none of its denominators to vanish. Example 7. Solve the following equations. 1. 3x*- 29x2 + 18 = 0. 2. x''-6x3 + 8x2 + 3x = 2. 3. (X - a) (X + 2 a) (x - 3 a) (x + 4 a) = 24 a*. 4. (4x2 + 2x)/(x2 + 6) + (x2 + G)/(2x2 + x) _ 3 = 0. Reciprocal equations. These are equations which remain 645 unchanged when we replace x hy 1/x and. clear of fractions. If we arrange the terms of such an equation in descending powers of x, the first and last coefficients will be the same, also the second and next to last, and so on ; or each of these pairs of coefficients will have the same absolute values but contrary signs. 312 A COLLEGE ALGEBRA Thus, 2x* + 3x3 + 4x2 + 3x4-2 = and x°-2x* + 4x3-4x2 + 2x-l = are reciprocal equations. Eeciprocal equations of the fourth degree may be reduced to the quadratic form and solved as follows. Example 1. Solve 2x* - Sx^ + 4x2 - 3x + 2 = 0. Grouping the terms which have like coeflBcients and dividing by x2, we reduce the given equation to the form <-'+^)-K'^+i)+^=''- X2. Since x"^ -\-\ /x^ = (x + 1 /x)2 - 2, we may reduce this equation to the torn, J ("= + i)-K^ + i) = ''- Solving for x + 1/x, we obtain the two equations X + - = and x + - = -, X X 2 whose roots are i, — i, and (3±iv7)/4. Every reciprocal equation of odd degree has the root 1 or — 1 ; and if the corresponding factor a; — 1 or x 4- i be sepa- rated, the "depressed" equation will also be reciprocal. Hence reciprocal equations of the third and Jifth degrees can be solved by aid of quadratics. Example 2. Solve 2x5-3x2-3x + 2 = 0. Grouping terms, 2 (x" + 1) - 3 (x2 + x) = 0. Since both terms of this equation are divisible by x + 1, it is equivalent to the two equations X + 1 = and 2-x2 - 5 x + 2 = 0, whose roots are —1, and 2, 1/2. Example 3. Solve x^ - 5 x* + 9 x' - 9 x2 + 5 x - 1 = 0. Grouping terms, (x^ — 1) — 5 x (x^ — 1) + 9 x2 (x — 1) = 0. Dividing by x — 1, we find that this equation is equivalent to X - 1 = and x* - 4 x^ + 5 x2 - 4 x + 1 = 0, whose roots are 1, and (1 ± i V3)/2, (3 ± V5)/2. EQUATIONS SOLVABLE BY QUADRATICS 313 Example 4. Solve the following equations. 1. x3 - 2X-2 + 2x - 1 = 0. 2. X* - 4x3 + 5z2 _ 4a; + 1 = q. 3. x5 + x* + x^ + x2 + X + 1 = 0. Binomial equations. This name is given to equations of the 646 form x" 4- a = 0. They can be solved by methods already given when cc" + a can be resolved into factors of the first or second degrees. Example 1. Solve x^ — 1 = 0. Since x^ - 1 = (x - 1) (x^ + x + 1), the equation x^ - 1 = is equivalent to the two equations X - 1 = and x- + x + 1 = 0. Solving, x = 1 or (- 1 ± i V3)/2. Example 2. Solve x^ - 32 = 0. From x5 — 32 = 0, by setting x = V32 y = 2 y, we obtain ?/5 — 1 = 0. By §§ 438, 043, t/^ — 1 = is equivalent to the two equations 2/ - 1 = and 2/4 + 2/3 + 2/-2 + y + 1 = 0. Solving, 2/ = 1, (- 1 -£ V5 + i Vio ± 2 V5)/4, or (- 1 ± V5 - z ViO ± 2 V5)/4. Hence x = 22/ = 2, (_l±V5 + £ ViO ± 2 V5) /2, or (-1 ±V5-i ViO± 2 V5)/2. By the method here employed every binomial equation x" ± a = can be reduced to the reciprocal form y" ± 1 = 0. Example 3. Solve the following equations. 1. x3 + 8 = 0. 2. X* + 1 = 0. 3. x6 + 1 = 0. These examples illustrate the theorem : Every number has 647 n nth roots. Thus, a cube root of 1 is any number which satis- fies the equation x^ = 1\ and in Ex. 1, we found three such numbers, namely 1, (— 1 + t V3)/2, and (— 1 — i V3)/2, Irrational equations. If asked to solve an irrational equation, 648 we ordinarily begin by rationalizing it, § 609. But, as will be illustrated below, certain equations admit of a simpler treat- ment than this. Whatever method is used, care must be taken to test the values obtained for the unknown letter before accepting them as roots of the given equation. 314 A COLLEGE ALGEBRA Example I. Solve V2x -3 - Vox - 6 + V8~ Transposing, v 2 x — 3 + v 3 x — 5 = v 6 x — 6. Squaring and simplifying, V(2 x — 3) (3 x — 5) = 1. Squaring and simplifying, Gx^ — 19x + 14 = 0. Solving, X = 2 or 7/6. Testing these values of x in the given equatinn, we find that 2 is a root but that 7/6 is not a root. We may also rationalize the given equation by the method of § 603. "We thus discover that — 4 (6x'^ — 19x + 14) is identically equal to (V2x-3-V5x-6 + V;^x- 5)(V2x -3+V5x-6 -V3x-5). ( V2x'^ + V5X-6 + V3x-5)(V2x-3 - Vsx -6 - V3x-5). Tliere are but two values of x for which the product 6x2 — 19 x + 14 can vanish. The first factor on the right vanishes for one of these values, 2; the second, for the other, 7/6. Hence there is no value of x for which the third or fourth factor can vanisli. Example 2. Solve V4x + 3 + Vl2x + 1 = V24 x + 10. Some equations which involve a single radical can be reduced to the form of a quadratic with respect to this radical. We then begin by solving for tlie radical. Example 3. Solve 2 x2 - 6 x - 5 Vx^ - 3 x - 1 -5 = 0. Observing that the x terms outside the radical are twice those under the radical, we are led to write the equation in the form 2 (x2 - 3 X - 1) - 5 Vx--! - 3 X - 1 -3 = 0. Solving for v'x2 — 3x — 1, we obtain the two equations Vx2-3x-l =3, and Vx2 - 3x - 1 = - 1/2. The second of these equations must be rejected since according to the convention made in § 579 a radical of the form ^/a cannot have a negative Talue. Squaring the first equation, we obtain x2-3x-l = 9, whose roots are 5 and — 2. Testing 5 and — 2 in the given ecjuation, we find that both of them are roots. Example 4. Solve 2 x2 - 14 x - 3 Vx2 - 7 x + 10 + 18 = 0. EQUATIONS SOLVABLE BY QUADRATICS 315 Sometimes an equation may be reduced to a form in which, both members are perfect squares or one member is a perfect square and the other is a constant. Example 5. Solve Ax" + x + 2x VSxM-x = 9. We may write this equation in the form 3 x2 + X + 2 X V3 X- + X + x2 = 9. The first member is a perfect square, and extracting the square root of both members, we obtain the two equations Va x2 + X + X = 3, and V3 x- + x + x = - 3. Solving these equations, x = 1, — 9/2, or (5 ± V97) /4. Testing these results, we find that only 1 and — 9/2 are roots of the given equation. Sometimes all the terms, when properly grouped, have a common irrational factor. Example 6. Solve Vx^— 7 ax + 10 a- — Vx"- + ax — 6 a^ = x — 2 a. Here the first two terms and also x — 2 a have the factor Vx — 2 a. Separating this factor, we find that the equation is equivalent to Vx — 2a = and Vx — 5a — Vx + 3a = Vx — 2 a. Solving these equations, x = 2a, — 10 a /3, or 6a. Testing, we find that only 2a and —10 a/3 are roots of the given equation. Example 7. Solve V3 x'^ - 5 x - 12 - V2 x^ - 11 x + 15 = x - 3. If one or more of the terms of the equation are fractions with irrational denominators, it is often best to rationalize these denominators at the outset. Example 8. Solve ( Vx + Vx -3) / ( Vr - Vx - 3) = 2 x - 5. Rationalizing the denominator in the first member and simplifying, we have Vx2 - 3 X = 2 X - 6. Solving, X = 3 or 4. Testing, we find that both 3 and 4 are roots of the given equation. Example 9. Solve ( Vx - 1 - Vx + 1) / ( Vx - 1 + Vx + 1) = x - 3. 316 A COLLEGE ALGEBRA EXERCISE XLIV Solve the following equations. 1. 4a;* -17x2 + 18 = 0. 2. 3x^-4x^ = 7. 3. (x2 - 4) (x2 - 9) = 7 x2. 4. (2x2 -x-3)(3x2 + x -2)2=0. 5. x< + x5 + x2 + 3 X - 6 = 0. 6. X* - 2 x3 + x2 + 2 X - 2 = 0. 7. (3x2-2x + I)(3x2-2x- 7) + 12 = 0. 8. x4 - 12 x3 + 33 x2 + 18 X - 28 = 0. 9. 4x* + 4x3 - x2 - X - 2 = 0. 10. X* - 2 x3 + 2 x2 - 2 X + 1 = 0. 11. x< + x3 + 2 X2 + X + 1 = 0. 12. x5 - 11 X* + 36 x3 - 36 x2 + 11 X - 1 = 0. 13. x5- 243 = 0. 14. (2x-l)s = l. 15. (1 + x)3 = (1 - x)3. 16. (X - 2)* - 81 = 0. 17. (a + x)3 + (6 + x)3 = (a + 6 + 2 x)^. 18. (a - x)* - (6 - x)4 = (a - &) (a + 6 - 2 x). 1 ^g x2 + 3x + l g4x2 + 6x 4x2 + 6x „ 1 x2 + 3x + l -2 21. 3x2 22. 4x2 2x-5V3x2-2x + 3 + 9 = 0. 2 X - 1 = V2 x2 - X. 23. V-r-^ 31. V5x2-6x + l - V5x2 + 9x-2 SIMULTANEOUS QUADRATIC EQUATIONS 817 32. ^^^±^ + 3:.0. 33. ^ + 4^2^ = 2. V2X-1 - V3x 34. (X + a)5 + (X + b}^ + (X + c)' = 0. 35. X (X - 1) (x - 2) (X - 3) = C • 5 • 4 . 3. 36. (x + a)2 + 4 (X + rt) Vx =: a- - 4 a Vx. XVT. SIMULTANEOUS EQUATIONS WHICH CAN BE SOLVED BY MEANS OF QUADRATICS A PAIR OF EQUATIONS IN X, Y, ONE OF THE FIRST DEGREE, THE OTHER OF THE SECOND A pair of simultaneous equations of the form 649 f(x, y) = ax^ + bxi/ + ci/' -{- dx + ei/ +f = 0, (f) (ic, y) = a'x + ^>'// + c' = may be solved as in the following example. Example. Solve y^ - x- + 2x + 2y + 4 =0, (1) 2x-y-7 = 0. (2) From (2), ?/ = 2 x - 7. (3) Substituting (3) in (1), 3 x2 - 22 x + 39 = 0. (4) Solving (4), X = 13/3 or 3. (5) Substituting in (.3), y = 5/3 or - 1. (6) The solutions of (1), (2) are the pairs of values x= 13/3, i/ = 5/3; x = 3, y = -l. (7) For, §§368, 371, the following pairs of equations are equivalent: (1), (2) to (4), (2); (4), (2) to (5), (2); (5), (2) to (7). We may indicate the solutions (7) thus: 13/3, 5/3; 3, —1. Care must always be taken to group corresponding values of x and y. Ordinarily such a pair of equations will have tico finite solu- 650 tions. But if the group of first-degree terms in ^ {x, y), namely 318 A COLLEGE ALGEBRA a'x + Vy, is a factor of the group of second-degree terms in fix, 7j), namely ax^ + bxy + cy"; while (^ (x, y) itself is not a factor of f{x, y), there will be only one finite solution or no such solution. And if ^(x, y) is a factor of /(x, y), there will be infinitely many solutions. Example 1. Solve ?/'- - x^ + 2 a; + 2 y + 4 = 0, (1) y -mx = 0. (2) Eliminating y, {rrfi - 1) x2 + 2 (m + 1) x + 4 = 0. (3) If 11 fi -1^0, (3) has two finite roots, and (1), (2) two finite solutions. But \iy - mx is a factor of 2/2 _ ^2, that is, if in = ± 1, then wi2 -1=0 and (3) does not have two finite roots. Thus, if jn = \, (3) reduces to x + 1 = 0, which has but one finite root, the other being infinite, § 638. And if m = - 1, (3) reduces to 4 = 0, which has no finite root, both roots being infinite, § 638. Hence if (2) has the form ?/ - x = 0, the pair (1), (2) has but one finite solution, the other being infinite. And if (2) has the form ?/ + x = 0, the pair (1), (2) has no finite solution, both solutions being infinite. Example 2. Solve y- ~ x- + 2x + 2y = Q, (1) y + x = 0. (2) Eliminating y, x^ - x^ + 2x - 2x = 0. (3) But (3) is an identity and is satisfied by every value of x. Hence every pair of numbers x = a, y = -aisa solution of (1), (2). The reason for this result is that y + xisa, factor of y^ - x^ + 2x + 2y. 651 When A, B, C are integral functions, the pair of equations AB ==0, C = is equivalent to the two pairs A = 0, C = and 5 = 0, C = 0, § 371. This principle and § 649 enable us to solve two integral equations f(x, y) = 0, (f> (x, y) = when- ever f(x, y) can be resolved into factors of the first or second degrees and <^ {x, y) into factors of the first degree. Example. Solve x^ + xy- — bx = 0), (1) {2x-y)(x + y-\) = 0. (2) This pair of equations is equivalent to the four pairs X = 0, 2 X - 2/ = 0, (3) X = 0, X + y - 1 = 0, (4) x2 + y2_5^0, 2x- j/ = 0, ' (5) x2 -I- 2/2 _ 5 = 0, X -I- 2/ - 1 = 0. (6) SIMULTANEOUS QUADRATIC EQUATIONS i 319 Solving the pairs (3), (4), (5), (6), we find the soUitions of (1), (2) to be 0, ; 0, 1 ; 1,2; - 1, - 2 ; 2, - 1 ; -1,2. A pair of integral equations in x, y can be solved by means 652 of quadratics only when it has one of the forms described in §§ 649, 651 or when an equivalent pair which has one of these forms can be derived from it. Tims, the pair of equations of the second degree, y^ — x + 1 = 0, y = x% cannot be solved by quadratics. For there is no simpler method of solving this pair than to eliminate y, which gives x* - x + 1 = 0, an equa- tion of the fourth degree which cannot be solved by quadratics. The preceding sections illustrate the truth of the following 653 important theorem : A pair of integral equations f (x, y) = 0, <^ (x, y) = 0, whose degrees are m and n respectively, has mn solutions. Thus, the pair x^ + x?/^ - 5 x = 0, (2 x - ?/) (x + ?/ - 1) = has 3 • 2, or 6, solutions, §651. See § 381 also. It should be added, however, that if the groups of terms of 654 highest degree in f{x, y) and (x, y), but not f{x, y) and <^(.r, //) themselves, have a common factor, there are less than mn finite solutions. Thus, for every factor of the first degree which is common to the groups of terms of highest degree in f{x, y) and (x, y) there is at least one infinite solution ; for every such factor which is also common to the groups of terms of next highest degree there are at least two infinite solutions; and so on. lif(x, y) and ^(x, y) themselves have a common factor, there are infinitely many solutions. Thus, the pair x^ - xy2 + xy _ 2/2 - y = (1), x2 - yS _ i = o (2) cannot have more than three finite solutions ; for there are 3 • 2, or 6, solutions all told, and at least one of these is infinite since x + ?/ is a common factor of the groups of terms of highest degree in (1) and (2), namely x^ — xy2 and x"^ — ?/'-, and at least two others are infinite since x — y is a common factor of the groups of highest and next highest degree in (1), (2), namely x^ — xy-^ x2 — ?/2 and xy — 2/2, (x — ?/)• j^ r.x2 + y2-8 = 0, ^^ p2_.ry_2^/2 + y^0, 320 A COLLEGE ALGEBRA EXERCISE XLV Solve the following pairs of equations. ^ r7x2_6xy = 8, ^ ixy^l, ^ rx2 + x = 4y2, ■l2x-3?/ = 5. ■ L3x-5y = 2. ' [Sx + dy^l. ^ rSx2-Sxy-y"--ix-Sy + 3 = 0, L3x -?/-8 = 0. rx2 + 5y2_8x-72/ = 0, r2x2-x?/-3y = 0, ■tx + 3?/ = 0. ■\7x-6y-4 = 0. rx2 + 3x?/ + 2;/2_ 1 = 0, J2x + 3y = 37, lx + ?/ = 0. ■ ll/x + l/y = 14/45. g rl/z/-3/x = l, ■ ^^ rx2 + xy + 2 = 0, L7/XZ/- 1/2/2 = 12.. ■ i(3x + 2/)(2x + 7/-l) = ( rx2 (x + 1)2 = (2/ -1)2. \(x-22/){x + 2/-3) = 0. 13. Determine m so that the two solutions of the pair ?/2 + 4 x + 4 = 0, y — mx shall be equal. 14. Determine m and c so that both solutions of the pair ic2 4- .ry - 2 2/2 + X = 0, y = mx + c shall be infinite. 15. By the method of § 650, Ex. 2, show that 2x — 2/ + 4 is a factor of 2x2 + X2/ - 2/2 + 10 X + 2/ + 12. 16. Show that the pair x?/ = 1, X2/ + x + 2/ = has not more than two finite solutions, and that the pair x'~y + xy = \, x-y + y"^ = 2 has not more than four finite solutions. PAIRS OF EQUATIONS WHICH CAN BE SOLVED BY FACTORIZATION, ADDITION, OR SUBTRACTION 655 When both equations are linear with respect to some pair of functions of x and y. We begin by solving the equations for this pair of functions by tlie methods of §§ 374-376. Example 1. Solve 2x2 - 3^/2 = — 58, (1) 3x2 + 2/- =111. (2) SIMULTANEOUS QUADRATIC EQUATIONS 321 Solving for x^, 2/2, we obtain x"- = 25, 7/2 = 36, whence, x = ± 5, y = ±G. By §§ 367-372, the pair (1), (2) is equivalent to the four pairs x = 5, y = 6; x = — 5, ?/ = 6; x = 5, ?/= — 6: x = — 5, ?/ = — 6. Hence the solutions of (1), (2) are 5, 6 ; - 5, 6 ; 5, - 6 ; - 5, - 6. Example 2. Solve the following pairs. J ax2 + by" = a, jSx^ -l/y^ = 2, 16x2 -ay'- = b. " 1 5x2 + 3/^2 = 120. When one of the equations can be factored. This is always 656 possible when the equation in question has the form ax^ -\- bxy + cy"^ — 0, and, in general, when it is reducible to the form air -\- bu. -{- c = 0, where u denotes a function of x, y. Example 1. Solve x2 + y2 4. x - 11 ?/ - 2 = 0, (1) x2- 5X2/ + 02/2 = 0. (2) Factoring (2) by solving for x in terms of j/, x = 2y, (3) or ' x — Zy. (4) Solving (1), (3) and (1), (4), we obtain all the solutions of (1), (2), namely 4, 2; -2/5, - 1/5; 3, 1 ; -3/5, - 1/5. Example 2. Solve 2x2 + 4x?/ + 2y2 _|_ 3^ + 3 y - 2 = 0, (1) 3 x2 - 32 (/2 + 5 = 0. (2) We may write (1) thus : 2 (x + ?/)2 + 3 (x + y) - 2 = 0. Solving, x + 2/=:]/2, (3) or X + 2/ = - 2. (4) Solving (2), (3) and (2), (4), we obtain all the solutions of (1), (2), namely 1, - 1 /2 ; 3/29, 23/58 ; - 3, 1 ; - 41 /29, - 17/29. Example 3. Solve the following pairs. ^ rx2 + x2/-6 = o, ^^ rx + .^x-.^^^ U2_2X2 = 1. 322 A COLLEGE ALGEBRA 657 When the given equations may be combined by addition or sub- traction so as to yield an equation which can be factored. This is always possible when both the given equations are of the form ax^ + i>^y + cy"^ = d. Example 1. Solve 6 x^ - x?/ - 2 ?/2 = 56, (1) 5 x2 - xy - y2 = 49. (2) We combine (1) and (2) so as to eliminate the constant terms. Multiply (1) by 7, 42 x2 _ 7 xy - 142/2 = 392. (3) Multiply (2) by 8, 40 x2 - 8 xy - 8 2/2 = 392. (4) Subtract (4) from (3), 2 x2 + x?/ - 6 2/2 = 0. (5) Solve (5) for X, x = 3 2//2, (6) or X = — 2 2/. (7) Solving (2), (6) and (2), (7), we obtain all the solutions of (1), (2)* namely ±3V36/10, ±^^35/5; ±2V2T/3, tV^/3. And, in general, we obtain an equation which can be factored when the given equations are of the second degree, and can be combined by addition or subtraction so as to eliminate (1) all terms of the second degree ; (2) all terms except those of the second degree ; (3) all terms which involve x (or y) ; or (4) all terms which do not involve x (or y). Example 2. Solve 2 x2 + 4 X2/ - 2 x - ?/ + 2 = 0, (1) 3x2 + Gx2/-x + 32/ = 0. (2) Here we can eliminate all terms of the second degree by multiplying (1) by 3, and (2) by 2, and subtracting. We thus obtain 4x + 92/-6 = 0. (3) Solving (2), (3), we obtain - 3. 2 ; - 2, 14/0, and these are all the finite solutions of (1), (2). See § 654. Example 3. Solve x2 - 3 x?/ + 2 2/2 + 4 x + 3 ?/ - 1 = 0, (1) 2 x2 - 6 X2/ + 2/- + 8 X + 2 y - 3 = 0. (2) Here all the terms which involve x can be eliminated by nmltiplying (1) by 2 and then subtracting (2). We thus obtain 32/2 + 4 2/ + 1 =0. (3) Solving (1), (3), wo obtain all the four solutidns of (1), (2), namely 1/3, - 1/3; - 1(5/3, - 1/3; (- 7 ± V57)/2, - 1. SIMULTANEOUS QUADRATIC EQUATIONS 323 Consider the following example also. Example 4. Solve x- + 2xy + 2y^ + ^x = 0, (1) xy + y^ + 2,y + \ = 0. (2) Multiply (2) by 2 and add to (1). We obtain (x + 22/)2 + 3{x + 22/) + 2 = 0. (3) Solving (3), x + 22/ = -l, (4) or a; + 2 y = - 2. (5) Solving (2), (4) and (2), (5), we obtain all the solutions of (1), (2), namely - 3 ± 2 V2, 1 T ^2 ; - 3 ± ^5, (1 =F "^5) /2. Example 5. Solve the following pairs of equations. ^ r2x2 + xy + 52/=0, rx2 +7/2 _ 13=0, ■ 1x2 + ?/2 + 10 ?/ = 0. ■ t xy + 2/ - X = - 1. "When the equation obtained by eliminating x or y can be 658 factored. From any pair of equations of the second degree we can eliminate x or y by the following method. The result- ing equation will ordinarily be of the fourth degree and not solvable by means of quadratics. But if we can resolve it into factors of the first or second degrees, we can solve it and so obtain the solutions of the given pair. Example 1. Solve 10 x2 + 5 ?/2 _ 27 x - 4 y + 5 = 0, (1) x2 + ^2 _ 3 X _ y = 0. (2) First eliminate ?/2 by multiplying (2) by 5 and subtracting the result from (1). We obtain 5 x2 - 12 X + 2/ + 5 = 0, or ?/ = - 5 x2 + 12 X - 5. (3) Substituting (3) in (2), 5x* -24x3 + 40x2 -27x + 6 = 0. (4) Factoring, by § 45l, (X - 1) (X - 2) (5 x2 - 9 X + 3) = 0. (5) Solving (5), § 643, x = 1, 2, (9 ± V21)/10. (6) Substituting (6) in (3), ?/ = 2, - 1, (7 ± 3 \^)/10. (7) The pairs of corresponding values (6), (7) are the solutions of (1), (2). Example 2. Solve x2 - 3 xy + 2 ?/2 + 3 x - 3 y = 0, 2 x2 + a-y - y2 + X - 2 y + 3 = 0. 324 A COLLEGE ALGEBRA EXERCISE XL VI Solve the following pairs of equations. 17x2-22/2 = 10. ■ tl/?/2_4/j.2 = 8. ^ ry^ + xy + 6 = 0, ^ rx'^ + y'^ - 3x + 2y - 39 = 0., ' \y2-y -2 = 0. ' I3x2_i7a:y + 102/2 = 0. j'y2_x2-5 = 0, l4x2 + 4x2/ + 2/2 + 4x + 2i x2+5x2/-2x + 32/ + l = 0, 3x2 + 15x2/-7x + 82/ + 4 = 0, x2 - 15 X2/ - 3 2/2 + 2 X + 9 2/ = 9 5x2/ 4- 2/- - 3 2/ = - 21. 1x2 + 3x2/ -42/2 = 25, fx{x + 3y) r2x^ + -6xy -iy^ = 'Zb, r:x(x + 32/) = l 1 15x2 + 24x2/ -31 2/2 = 200. ' \x^- ^y- = 4. fx- - 3 x?/ + 3 2/2 = x22/2, ja;2 + X2/ + 2/- = 38, 1 7x2 -10x2/ + 4 2/2 =12x22/2. ' 1x2 - x;/ + ?/2 = 14. j'x2-x2/ + y2 = 21(x-2/), rx2 + 2/-8 = 0, 1x2/- = 20. ■ 1 2/2 + 15 X - 40 = 0. 10. PAIRS WHICH CAN BE SOLVED BY DIVISION 659 In solving a pair of equations it is sometimes advantageous to combine them by multiplication or division ; but care must then be taken not to introduce extraneous solutions nor to lose actual ones (see §§ 362, 342). 660 If given a pair of the form AB =. CD (1), B = D (2), where A, B, C, D denote integral functions of x, y, we may replace B by D in (1), thus obtaining the pair AD = CD, B = D which is evidently equivalent to the two pairs .4 = C, B = D and D = 0, J5 = 0. We may obtain the pair A = C, B = D by dividing each member of (1) by the corresponding member of (2); but if we then merely solve this pair A =^ C, B — D, we lose some SIMULTANEOUS QUADRATIC EQUATIONS 325 of the solutions of (1), (2), except, of course, when either B or Z) is a constant, so that the pair iJ = 0, Z> = has no solution. Example 1. Solve x* -\- x-y- -\- y* = 2\, (1) x2 + X!/ + 7/2 = 7. (2) Dividing (1) by (2), x"^ - xy + y^ = 3. (3) Solving (2), (3), we obtain all the finite solutions of (1), (2), namely 2, 1 ; - 2, - 1 ; 1, 2 ; - 1, - 2. See § 654. Example 2. Solve x^ - ?/3 ^ - 3 (x + 1) y, (1) x2 + xy + y-^ = X + 1. (2) Dividing (1) by (2), x-y = -2,y. (3) The pair (1), (2) has the same finite solutions as the pair (2), (3) and the pair x- + x?/ + t/^ = o (4), x + 1 = (5) jointly. And the solutions of (2), (3) and (4), (5) are 2,-1; - 2/3, 1/3; - 1, (1 ± i V3)/2. Example 3. Solve (x + y)" = x, (1) x2 - t/2 = _ e y. (2) Dividing (1) by (2), (x + 2/)/(x - ?/) = - x/6?y. (3) Clearing of fractions, x"^ + bxy + Qy"^ = 0. (4) The pair (2), (4) has the four solutions 0, ; 0, ; 4, - 2 ; 9/4, - 3/4. The process by which (4) was derived from (1), (2) is reversible when a;, y have the values 4,-2 or 9/4, — 3/4, but not when they have the values 0, 0. Hence this reckoning only proves that 4,-2 and 9/4, -3/4 are solutions of (1), (2), §362. It is obvious by inspection that 0, is a solution of (1), (2); but it should be counted only once as a solution, not twice as in the case of (2), (4). This follows from the fact that (1), (2) can have but three finite solutions, § 654. It may also be shown thus : In (1), (2) make the substi- tution y = tx (5). We obtain (1 + ty^x"^ = x (U), (1 - t"^)^^ = - Qtx (2'). And (5), (1'), (2') yield x — 0, y = once and but once. EXERCISE XLVn ^ rx3-y' = 63, 2 rx + y = g8, rx* + x22/2 4-^4 = 931, ^ r(x + 2/)(x2-2i I x2 + x?/ + ?/2 = 49. ■ \(X-7J) (x2 - 2 \ , r(x + 2/)2(x-?/) = 3xy+6j/, ^ j'x2-3xy + 2y' Ix*-- j^2-a; + 2. " lx2-2/2 = -5y, 326 A COLLEGE ALGEBRA SYMMETRIC PAIRS OF EQUATIONS 661 A pair of equations in x, y is said to be symmetric if it remains unchanged when x and y are interchanged. • Thus, the following pairs, (a) and (6), are symmetric. IxV + xy + 1 = 0. ^'\y^ = 2y + Zx. Symmetric pairs are of two types, those like (a) in which the individual equations remain unchanged when x and y are interchanged, and those like (b) in which the two equations change places when x and y are interchanged. 662 Symmetric pairs of the first type. The simplest pair of sym- metric equations is x -{- y = a, xy =^h. This pair may be solved as in § 649, but the following is a more symmetric method. Example. Solve x + y = 5, (1) xy = 6. (2) Square (1), x'^ + 2xy + y^ = 25. (3) Multiply (2) by 4, 4xy = 24. (4) Subtract (4) from (3), x"^ - 2xy + y'^ = \. (5). Hence x — y — 1, (6) or X - 2/ = - 1. (7) From (1), (6), x = 3, y = 2 ; and from (1), (7), x = 2, 2^ = 3. 663 If given a more complicated pair of symmetric equations, we may transform each equation into an equation in x + y and xy, § 637, and then solve for these functions ; or in the given equations we may set x = ti + v, y = u — v and then solve for u and v. The second method is essentially the same as solving the given equations for x -{- y and x — y; for, since X = u -{- V, y = u — t^, we have « = (a; -f- y) /2, ^) =(x — y) /2. Example 1. Solve 2x- + r^ xy + 2y'^ + x + y + \ = 0, (1) x2 + 4x2/ + 2/2 + 12x + 12?/ + 10 = 0. (2) In (1) and (2) for x^ + y"^ substitute (x + y)^ - 2xy. 0, (3) 0. (4) 0. (5) 4, (6) -2/3. 0) -37, (8) -11/9. (9) SIMULTANEOUS QUADRATIC EQUATIONS 327 Collecting terms, 2{x + y)~ + xy + (x + y) + 1 {X + y)^ + 2x!/ + 12 {x + y) + lO: Eliminating xy, S(x + y)^ - 10 (x + ?/) - 8 : Solving, X + y or x + y ■ Hence, from (3), , xy ■■ or xy ■■ And solving the pairs (6), (8) and (7), (9) for x, y, v?e have X, 2/ = 2±Vil, 2t^^41; (-1 ±2 V3)/3, (-1 t2 V3)/3. Example 2. Solve x* + y* = 97, (1) x + y = 5. (2) In (1) and (2) set x = ii + v, y = u — v. We obtain (u + ?))* + (u - v)* = 97, (3) and 2 M = 5. (4) Eliminating u, 16 u* + 600 v"- - 151 = 0. (5) Solving, v=:±l/2 or ±iVl5T/2. (6) Substituting m = 5 / 2 (4) and the four values (6) of v in the formulas X = u + V, y = u ~ V, we obtain X, 2/ = 2, 3; 3, 2; (5 ±iVT51)/2, (5 ^F i vl51)/2. Evidently if a: = cr, ?/ = /? is one solution of a symmetric pair, X = 1^, y = ix is another solution. Unless a = y8, these two solutions are different ; but xt/ and x + i/ have the same values for x = a, y = ^ as for x = (3, y — a, and the corre- sponding values oi X — y, namely a — jB and jS — a, differ only in sign. Hence the values of xy or x -{- y derived from a symmetric pair will be less numerous than the values of x or y, that is the degree of the equation in xy or x + y derived from the pair by elimination, as in Ex. 1, will be less than the degree of an equation in x ov y similarly derived would be. As for the equation in x — y, if c is one of its roots, — c must be another root. Hence this equation will involve only even powers of x ~ y, as in Ex. 2, or only odd powers with no constant term. 328 A COLLEGE ALGEBRA 664 Note. The methods just given are applicable to pairs of equations which are symmetric with respect to x and - y or some other pair of functions of x and y. Tlius, x* + y* = o, x-y -h may be written X* + {-yY = a, x + {-y) = h. 665 Symmetric pairs of the second type. Such a pair may some- times be solved as follows. Example. Solve x'^Tx + Sy, (1) 2/3 = 7 2/ + 3x. (2) Adding (1) and (2), x^ + ?/3 = ]0 (x + y). (3) Subtracting (2) from (1), x/ - 2/3 = 4 (x - y). (4) By § 341, (3) is equivalent to the two equations X + 2/ = 0, (5) and X- - xy ^ y- = 10. ^6, Similarly (4) is equivalent to the two equations X - 2/ = 0, (7) and X- + xy -\- y- = 4. (8) And solving (5), (7) ; (5), (8) ; ((1)^(7) ; (6), (8), we obtain 0, ; 2^- 2 ; -2, 2; ±VlO, ±Vl0; (1 ± Vl3)/2, (1 T ^^)/2 ; (-l±Vl3)/2, (_l^Vr3)/2. EXERCISE XLVIII Solve the following pairs of equations. rx + ?/ = 5, rx2 + 2/2 = 200, ^ rx2 + 2/2 = 293, \xy + m = 0. ■ tx + 2/ = 12. ■ 1x2/ = 34. fx2 + 2/2 = 85, rx'5 + 2/^ = 513, ^ r x^ + J/'' = 468, ■ Ix - 2/ = 7. Ix + y = 9. 1x22/ + x2/2 = 420 (-27x3 + 04 2/3 = 65, r x" + 2/* = 82, ^ rx5 + 2/^ = 32, ■ l3xf42/ = 5. ■ lx-2/ = 2. ' tx + 2/ = 2. rx + 2/ = l/2, .a.y + ^4.y + 19 = 0, ■^^- I 56('? + I) + 113 = 0. ■ 1x22/ + X2/2 + 20 = 0. x< + 2/* - (-i;- + 2/-) = 72, ^g r x22/ + X2/2 = 30, 2/2 + 2/2 = 10. ■ ll/x+ 1/2/ = 3/10. j2 rx* + 2/*-(.x2 + 2/2) = 72, ^g (X'y + : ■ 1x2 + j.22/2 + ,y2 ^19. • \ 1 /x + rx2 + 3x2/ + 2/2 + 2x + 2 2/=8, ^^ r x3 = 5 2/, l2x2 + 22/2 + 3x + 32/ = 14. " 1.2/3= 5x. SIMULTANEOUS QUADRATIC EQUATIONS 329 SYSTEMS INVOLVING MORE THAN TWO UNKNOWN LETTERS A system of three equations in three unknown letters can be solved by means of quadratics when one of the equations is of the second degree and the other two of the first degree ; also when it is possible to reduce the system to one or more equiva- lent systems each consisting of one equation of the second degree and the rest of the first degree. The like is true of a system of four equations in four unknown letters, and so on. If A, B, C are integral functions of degrees m, n^ji in x, y, z, and no two of them have a common factor, the equations A = 0, B = 0, C = will have vmp solutions. But some of these solutions may be infinite. Example 1. Solve z- — xy — 1 = Q, (1) X + y + z = 0, (2) 3 X - 2 2/ + 2 z + 2 = 0. (3) Solving (2), (3) for x and y in terms of z, x = -(4z + 2)/5, (4) 2/ = (-z + 2)/5. (5) Substituting in (1) and simplifying, 7 22 + 2 z - 57 = 0. (6) Solving (6), Hence, from (4), (5), x, Example 2. Solve the system Dividing (2) by (1), Substituting (4) in (3), From (5), (4), (1) we obtain 2 =-3 or 19/7. 2/, z = 2, 1, -3; -y, - 1 19 7' 7 ' X2/ = 6, (1) yz = 12, (2) 2X = 8. (3) z/x = 2 or 2 = 2x. (4) X- = 4, whence x = ± 2. (5) 2/, z = 2, 3, 4 ; - 2, - 3, - -4. EXERCISE XLIX Solve the following systems of equations- rx + y = S, iy + z = 2. -x(y + z) = 12, 2. - 2/{2 + x) = 6, 1x2 _ 2/2 ^ 19. .2(X + 2/) = 10 r{y + b){z + c) = a'^, «! (2 + c)(x + a) = 62, l(x + a)(t/ + 6)=c«. 330 A COLLEGE ALGEBRA EXERCISE L Solve the following systems of equations by any of the methods of the present chapter. \2z-3y = 5. "' \x-y = l. ' \xy ^ {b'^ - a'-) / 'i. ( \ 3 (a b ^ , ,., = 1, (x + y = a + b, [x2 + 2/2 = 0. — - ~ = 12. [x + b y + a Ixy 2/2 Lx y 5 Lxy t 4 10. a{x-\-y) = b(x-y) = xy. 11. 40x2/ = 21(x2-?/2) = 210(x + ?/). ■4x2-252/2 = 0, rx2+3x2/-92/2 = 9, 10 2/2 -3 2/ = 4. ■ tx2- 13xy + 2l2/2 = -9. 12 r4-^-25.'^ = 0, ^3_ rx^- 12x2-10 2/2 -3 2/ = 4. 1x2 rx2-72/2-29 = 0, rx/y + 2//x = 65/28, lx2-6xy + 92/2-2x + 62/ = 3. " 1 2 (x2 + 2/2) + (x 16. y) = 34. rx22/ = a^ j'x22/ + X2/2 = a, rx = a(x2 + 2/2), ^^ rx22/ + X2/2 = a, ^g re 1x2/2 = &. ■ 1x22/ - X2/2 = 6. ■ 12/ -&{a;2 + 2/2), r(x + y)/(x-2/) = 5/3, ^^ r3(x3- y^) ^ isx^,, l(2x + 32/)(3x-22/) = 110a2. ' ix-y^l. 19. , .(2x + 32/)(3x-2 2/) = 110a2. 2j ,x* + 2/* = aS 22 f21(x + 2/) = 10x2A rx* + 1x4-: y = a. U + 2/ + a;2-f 2/2 = 68. 23. x2 + 2/2 = X2/ = X + 2/. 24. x2 - X2/ + 2/^ = 3 a2 = x2 - 2^ rx2 + x2/ + 2/- = 21, 26 r4x2-32/2 = 12(x-2/), Ix + Vxy -f 2/ = 7. ' 1x2/ = 0. j'x2 + 2/2 = x + ?7-f20, 28 ra;- + 4x-32/ = 0, lx2/ + 10 = 2(x + 2/). ■ li 2/2 + lOx - 92/ = 0. X2/ — x/y = a, y = 2. Ixy - y/x = l/a. r28(x6 + 2/5) = 61{x3 + y3), ^^ rx2/ - x/2/ = a, ■ lx + 2/ = 2. ■ I SIMULTANEOUS QUADRATIC EQUATIONS 331 f{x + \x + 3,_ (x + 1)3 + (2, -2)3 = 19, ^^ y^ - xy - yz = 6, X + 4 2/ + z = 14, 34. x-y + 2Z-0. C{y + z){x-{-y + z) = 10, 35. J (z + x) (X + ?/ + z) = 20, 36. l(x + ?/)(x + 2/ + z) = 20. EXERCISE LI 1. The difference of the cubes of two numbers is 218 and the cube cf their difference is 8. Find the numbers. 2. The square of the sum of two numbers less their product is 63, and the difference of their cubes is 189. What are they ? 3. Tlie sum of the terms of a certain fraction is 11, and the product of this fraction by one whose numerator and denominator exceed its numerator and denominator by 3 and 4 respectively is 2/3. Find the fraction. 4. Separate 37 into three parts whose product is 1440 and such that the product of two of tliem exceeds three times the third by 12. 5. The diagonal of a rectangle is 13 feet long. If each side were 2 feet longer than it is, the area would be 38 square feet greater than it is. What are the sides ? 6. The perimeter of a right-angled triangle is 36 inches long and the area of the triangle is 54 square inches. Find the lengths of the sides. 7. The hypotenuse of a right-angled triangle is longer than the two perpendicular sides by 3 and 24 inches respectively. Find the sides of the triangle. 8. Find the dimensions of a room from the following data : its floor is a rectangle whose area is 224 square feet, and the areas of two of its side walls are 126 and 144 square feet respectively. 9. A rectangle is surrounded by a border whose width is 5 inches. The area of the rectangle is 168 scjuare inches, that of the border 360 square inches. Find the length and breadth of the rectangle. 332 A COLLEGE ALGEBRA 10. In buying coal A gets 3 tons more for -$135 than B does and pays $7 less for 4 tons than B pays for 5, Kequired the price each pays per ton. 11. A certain principal at a certain rate amounts to $1248 in one year at simple interest. "Were the principal $100 greater and the rate 1| times as great, the amount at the end of 2 years would be $1456. What is the principal and what is the rate ? 12. A man leaves $60,000 to his children and grandchildren, seven in all. The children receive ^ of it, .which is $2000 more apiece than the grandchildren get. How many children are there and how many grand- children, and what does each receive ? 13. At his usual rate a man can row 15 miles downstream in 5 hours less time than it takes him to return. Could he double his rate, his time downstream would be only 1 hour less than his time upstream. What is his usual rate in dead water and what is the rate of the current ? 14. Three men A, B, C together can do a piece of work in 1 hour, 20 minutes. To do the work alone it would take C twice as long as A and 2 hours longer than B. How long would it take each man to do the work alone ? 15. Two bodies A and B are moving at constant rates and in the same direction around the circumference of a circle whose length is 20 feet. A makes one circuit in 2 seconds less time than B, and A and B are together once every minute. What are their rates ? 16. On two straight lines which meet at right angles at the points A and B are moving toward at constant rates. A is now 28 inches from and B 9 inches ; 2 seconds hence A and B will be 13 inches apart, and 3 seconds hence they will be 5 inches apart. At wliat rates are A and B moving ? 17. Three men A, B, and C set out at the same time to walk a certain distance. A walks 4\ miles an hour and finishes the journey 2 hours before B. B walks 1 mile an hour faster than C and fiuislies tlie journey in 3 hours less time. What is the distance ? 18. Two couriers A and B start sinmltaneously from P and Q respec- tively and travel toward each otlier. When they meet A has traveled 12 miles farther than B. After their meeting A continues toward Q at the same rate as before, arriving in 4i; hours. Similarly B arrives at P in 7^ hours after the meeting. What is the distance from P to Q ? SIMULTANEOUS QUADRATIC EQUATIONS 333 GRAPHS OF EQUATIONS OF THE SECOND DEGREE IN X, Y Examples of such graphs. The graph of any given equation 667 of the second degree in .r, y may be obtained by the method illustrated in the following examples. Example 1. Find the graph of y- — \x. (1) Solving for y, 2/ = ± 2 Vx. (2) From (2) it follows that when x is negative, y is imaginary ; when x is 0, 2/ is ; when x is positive, y has two real values which are equal numeri- cally but of opposite signs. Hence the graph of (1) lies entirely to the right of the y-axis, passes through the origin, and is symmetric with respect to the x-axis. When x = 0, 1/4, 1/2, 1, 2, 3, 4,..., we have ?/ = 0, ± 1, ± ^2, ± 2, ± 2 V2, ± 2 Vs, ±4, • • • . We obtain the part of the graph given in the figure by plotting these solutions (0, 0), (^, 1), (], — 1), • • • and passing a curve through the points thus found. Com- pare § 389. It touches the ^ y-axis. y/^/ ^-^9.6) This curve is called a parabola. It consists of one "in finite branch, "here extending indefinitely to che right. Example 2. In what pohits is the graph of 2/- = 4 X (1) met by the graphs of y = x — S (2), ?/=x + l(3), y = x + 3(4)? 1. The solutions of (1), (2) are 1, - 2 ; 9, 6. Hence, § 386, the graphs of (1), (2) intersect in the points (1, — 2), (9, 6), as is .shown in the preceding figure. 2. The solutions of (1), (3) are equal, namely 1,2; 1, 2. Hence the graph of (3) meets the graph of (1) in two coincident points at (1, 2). Tliis means that the graph of (3) touches the graph of (1) at (1, 2), as is indicated in the figure. 3. The solutions of (1), (4) are imaginary, namely — l±2V2i, 2 ± 2 V2 i. Hence, as the figure shows, the graphs of (1), (4) do not meet. 334 A COLLEGE ALGEBRA Example 3. Find the graph of 2/2 -2x2/ + 2x2- Solving for y, we have y = x - 1X + 22/ + 1 -l±V4x-x2. (2) The values of y given by (2) are real when the radicand 4 x — x2, or x(4 — x), is positive (or 0), that is, vfhen x lies between and 4 (or is or 4). Hence the graph of (1) lies between the lines x = and x = 4. When X = and when x = 4 the values of y given by (2) are equal: namely — 1,-1 when x = 0, and 3, 3, when x = 4. Tliis means that the graph of (1) tmches the line x = at the point (0, — 1) and the line x = 4 at the point (4, 3). See Ex. 2, 2. The line y — X ~\ passes through these points of tangency, for when 4x — x- vanishes, (2) gives the same values for y that y = X — 1 gives. For each value of x between and 4 the equation (2) gives two real _Y and distinct values of y, obtained by increasing and diminishing the value of X — 1 by that of V4 x — x2. Hence for each of these values of x there are two points of the graph of (1). They ^ are most readily obtained by drawing the line y = x — \ and then increasing and diminis hing its ordinate fur the value of x in question by tiie value of V4 x — x^. Thus, when x = 0, 1, 2, 3, 4, we have for the line ?/ = — 1, 0, 1, 2, 3, and for the graph of (1) 2/ = - 1, ± V^, 1 ± 2, 2 ± V3, 3. The figure shows the oval-shaped curve which the points thus found determine. It is called an ellipse. By solving (1) for x and applying the method of § 641, we may show that the highest and lowest points of the curve are (2 + V2, 1 + 2 V2) and (2 - V2, 1 - 2 y.^2). Example 4. Find the graph of t/'^ - x- + 2 x + 2 ?/ + 4 = 0. (1) Solving for y and factoring the radicand in the result, 2/ = - 1 ± V(x + 1) (X - 3). (2) The radicand (x + 1) (x — 3) vanishes when x = — 1 and when x = 3, and in both cases (2) gives equal values of y, namely — 1, — 1. This SIMULTANEOUS QUADRATIC EQUATIONS 335 means that the graph of (1) touches the line x = — 1 at the point (—1, — 1) and the line x = 3 at the point (3, — 1). The line y = — 1 passes through these points of tangency. The radicand (x + 1) (x — 3) is positive when and only when x < — 1 or X > 3. For every such value of x the equation (2) gives two real and distinct values of y and therefore two points of the graph of (1). These points may be obtained by drawing the line y — — 1 and then increasing and diminishing its con- st ant ordinate — 1 by the value of V(x + l)(x-3). Hence, as is indicated in the figure, the graph of (1) consists of two infinite branches, the one touch- ing the line x = — 1 and extending indefinitely to the left, the other touching the line x = 3 and extend- ing indefinitely to the right. This curve is called an hyperbola. There are two straight lines called asyinptotes, which the infinite branches of this hyperbola tend to touch, and which they are said to touch at infinity. These lines are the graphs of the equations ?/ = x — 2 and y = — X, which we obtain as follows. Compare § 650, Ex. 1. Eliminating y between (1) and the equation y = nix + c, (3) we obtain (??i2 _ i) x2 + 2 {mc + m + 1) x + (c^ + 2 c + 4) = 0. (4) Both roots of (4) are infinite, § 638, if m- — 1 = and mc + m + 1 = 0, that is, if m = 1, c = — 2, or if ?n = — 1, c = 0. Hence both solutions of (1), (3) are infinite if (3) has one of the forms y = x-2 (3') or y = - x. (3") Therefore the graphs of (3') and (3") each meet the graph of (1) in tiuo injiniiely distant coincident points. Example 5. Find the graph ot y\- 4x1/ + 3x^ + 6x - 2y ^ 0. (1) Solving for y, we have y = 2x + 1 ± Vx2 - 2x + 1, (2) that is, ?/ = 3 X or ?/ = X + 2. Hence the graph of (1) consists of the pair of right lines y = 5x and y = x + 2. Except when the radicand vanishes, that is, when x — 1 = 0, the equa- tion (2) gives two real and distinct values of y. But when x — 1 = it 336 A COLLEGE ALGEBRA ' (-4,JA^'^ 1) \A J ^X^_ ::::::: ^^-3) gives two equal values of y, namely 3, 3. Hence the line x — 1 = meets the graph of (1) in two coincident points at (1, 3). Of course this cannot mean that the line x - 1 = touches the graph of (1) at (1, 3). It means that the points coincide in which the line x — 1 = meets the two lines, y = Zx and y = x + 2, which together constitute the graph of (1). Example 6. Find the graphs, and their intersections, of x2 + if- = 25, (1> xV16 + 2/V9==2. (2) The graph of (1) is a circle whose center is at the origin, 0, and whose radius is 5. The graph of (2) is an ellipse. These curves intersect at the four -" points (4, 3), (-4, 3), (-4, -3), (4, — 3), these points being the graphs of the solutions of (1), (2). Example 7. Find the graphs, and their intersections, of xy - S y - 2 = 0, xy + 2y + 5 = 0. From (1) we obtain Here there is one real value of y for each real value of x. When x = - M, -1, 0, 1, 2, 2] ■we have y = 0, - i, - f , - 1, - 2, - 8, ± <», 8, 2, 0. And plotting these solutions, we obtain an hyperbola whose two infinite branches are indicated by the unbroken curved lines in the figure, and whose asymptotes are y = (found as in Ex. 4), and x — 3 = (since, when X = 3, then y = oo). In a similar manner, we find the graph of (2) to be the hyperbola indi- cated by the dotted curve and having the asymptotes y = and x + 2 = 0. The equations (1), (2) have the single finite solution x = 1, ?/ = — 1, the remaining tliree solutions being infinite, § 654. The hyperbolas which are the graphs of (1), (2) meet in the single finite point (1, 0) (2) ^ = 2/(x-3). (3) d hence one point of the graph, H, But since they SIMULTANEOUS QUADRATIC EQUATIONS 337 have the common asymptote y = 0, they are regarded as having two infinitely distant coincident intersections at (oc, 0) ; and since they have the parallel asymptotes a; — 3 = and x + 2 =r 0, they are regarded as having one infinitely distant intersection at (0, co). General discussion of such graphs. Generalizing the results 668 obtained in the preceding exaui])les, we reach the following conclusions. Suppose any equation of the second degree in x, y given, with real coefficients, as ax" + 2 hxxj + hif + 2 ryro^ + 2/y + c = 0. (1) If h is not 0, and we solve for ?/, we have hy = -{lix^f)±-jR, (2) where R = (A^ - ab) x' + 2 (hf - brj) x + (f^ - be). From (2) we obtain two real values of y for each value of x for which the radicand R is positive. Corresponding to these two values of y there are two points of the graph which may- be found by drawing the line by^-ihx+f) and then increasing and decreasing its ordinate for the value of X in question by the value of -y/lt/b. See § 667, Exs. 1, 3, 4. The form of the graph depends on the character of the factors of R. 1. When (hf - byf - (A^ - ah) {f - be) = 0. In this case jR is a perfect square, § 635, and the first member of (1) can be resolved into factors of the first degree, § 635, Ex. 3, If these factors have real coefficients, the graph of (1) is a pair of right lines. See § 667, Ex. 5. 2. When {hf -bg^- {h^ - ab) (f - be) > 0. In this case, unless h^ — ab = 0, the radicand R can be reduced to the form R = (h^ — ab) (x — a) (x — /3) (3) where (I and (3 are real and a < ^, § 635. 338 A COLLEGE ALGEBRA If h^ — ab < 0, the product (3) is positive when and only when X lies between a and (3. Hence the graph of (1) will be a closed curve lying between the lines x — a = and x — /3 = 0, which it touches. It is therefore an ellijjse (or circle). See § 667, Ex. 3. If 7^2 _ (jj > 0, the product (3) is positive when and only when x ^. Hence the graph will consist of two infinite branches, the one touching the line x — a = and extending to its left, the other touching the line x — (3 = and extending to its right. It is therefore an hyperbola. See § 667, Ex. 4. If h^ -ab = 0, we have R = 2 {hf - bg) x+{p- be), where kf—bg4^ 0, and this is positive when and only when we have x-> — (f^ — be) /2(hf— bg). Hence the graph will consist of one infinite branch lying entirely to one side of the line 2 (hf— bg) X + {f'^ — be) = 0, which it touches. It is therefore ^ 2)arabola. See § 667, Ex. 1. 3. When (hf-bg)^-(h^-ab)(f^-be)<0. In this case the roots of R = are conjugate imaginaries, § 635, and if we call them X + fii and A — /xi, we can reduce R to the form R = (h^ - ab)l(x - X)^ + fx^^ (4). If h^ — ab>0, the product (4) is positive for all values of x. Hence the graph of (1) will consist of two infinite branches which lie on opposite sides of the line bi/ — — (hx +/)• It is therefore an hi/jwrbola. Thus, i/"^ — x" — 1. If K^ — ab< 0, the product (4) is negative for all values of X. Hence the graph of (1) will be wholly imaginary. Thus, x^ + y' + 1 = 0. In the preceding discussion it is assumed that bi^O. But if ^ = 0, while a ^ 0, and we solve (1) for x instead of y, we arrive at similar conclusions. If both a = and b = 0, the graph of (1) is an hyperbola, as in § 667, Ex. 7, or a pair of straight lines of which one is parallel to the cc-axis, the other to the y-axis. SIMULTANEOUS QUADRATIC EQUATIONS 339 EXERCISE LII Find the graphs of the following equations. 1. 2/2 = - 8 X. 2. x2 + 2/2 = 9. 3. (y _ x)"^ = z. 4. x2 + 2x2/ + 2y2 = 8. 5. y2 - 4x?/ + 3x2 + 4x = 0. 6. y'^ -2xy + 1-0. 7 . y'^ - 2xy - 1 = 0. 8. 2 x2 + 3 2/2 _ 4 X + 6 2/ = 0. 9. 2/^ - x2 - 3 x + 2/ - 2 = 0. 10. 2 x2 + 4 X2/ + 4 2/2 + X + 4 2/ - 5 = 0. 11. 4 x2 - 12 X2/ + 9 2/2 + 3 X - 6 y = 0. Find the graphs of the following pairs of equations and their points of intersection. 12. 15. 16. rx2/ = l, ^3_ rx2- 2/2 = 1, ^^ rx2 + 2/^ l3x-52/ = 2. Ix2-x2/ + x = 0. 12/2= 2z j'2/2-X2/-2x2-2z-22/-2 = L2/2-X2/-2x2 + 2 = 0. • (X - 2 2/) (X + 2/) + X - 3 2/ = 0, :2/)(x-2/) + 2x-62/ = 0. r(x-2i l(x-2. 17. Find the graph of x2 + 2/2 — 6x — 22/ + l=0 and its points of intersection with the axes of x and y. 18. Show that the graph of {x — y)- - 2{x + y) + I = touches the x and 2/ axes. 19. Show that the line y = Sx + 5 touches the graph of Wx" + y" — \G = at the point (-3/5, 16/5). 20. For what values of m will the line y = mx + 3 touch the graph of x2 + 2 y2 = G ? 21. For what values of c will the line 7x — 4y + c — touch the graph of 3 x2 — 2/2 + X = ? 22. Show that the lines y = and x — 2 y + 1 = are the asymptotes of the graph of X2/ — 2 2/2 + 2/ + 6 = 0. 23. Find the asymptotes of the graph of the equation 2 x2 + 3 X2/ - 2 y2 ^ a; + 2 2/ + 2 = 0. 24. For what values of X is the graph of x2 + \xy + y'^ = x an ellipse ? a parabola ? an hyperbola ? 340 A COLLEGE ALGEBRA XVII. INEQUALITIES 669 Single inequalities. An absolute inequality is one like ^' + y' + 1 > which holds good for all real values of the letters involved ; a conditional inequality is one like x — 1 > which does not hold good for all real values of these letters, but, on the contrary, imposes a restriction upon them. 670 The principles which control the reckoning with inequalities are given in § 261. From these principles it follows that the sign > or < connecting the two members of an inequality will remain unchanged if we transpose a term, with its sign changed, from one member to the other, or if we multiply both members by the fiduiwe positive number ; but that the sign > will be changed to <, or ince versa, if we multiply both members by the same negative number. Example 1. Prove that a2 + 6^ > 2 ah. We have (a - 6)2 > 0, that is, a2 - 2 a6 + tfi > 0, and therefore cfi + lf->2 ah. Example 2. Prove that a"^ + h'^ + c^ > ah + he + ca. We have cC^ + h'^>2ah, 62 + c2 > 2 be, c2 + a2 > 2 ca. Adding the corresponding members of these three inequalities and dividing the result by 2, we have a'^ -\- b^ + c'^>ah + be + ca. Example 3. Solve the inequality 3x + 5>x + 11, that is, find what restriction it imposes on the value of x. Transposing terms, 2 x > 6, whence x > 3. Example 4. Solve x2 - 2 x - 3 < 0. Factoring, (x + 1) (x - 3) < 0. To satisfy this inequality one factor must be positive and the other negative. Hence we nmst have x > — 1 and < 3, that is, — 1 < x < 3. 671 Simultaneous inequalities. A system of one or more ine- qualities of the form ax -{- hy + c > INEQUALITIES 341 may be solved for the variables x, y hj a, simple graphical method which is based upon the following consideration : Draw the straight line which is the graph ot ax + bi/ + c = 0, § 385. Then for all i)airs of values of x, y whose graphs lie on one side of this line we shall have ax + by -\- €> 0, and for all pairs whose graphs lie on the other side of the line we shall have ax -\- by -^ c <. 0. Thus, let (xi, yi) be a point on the graph oi y — (mx + c) = so that yi — {mxi + c) = 0. Then, if 1/2 < Vi so that the point (xi, 2/2) lies below the line, we have yo — (mxi + c) < 0, and if y^ > 2/1 so that the point (xi, 2/3) lies above the line, we have v/3 — {mxi + c) > 0. Example. Solve the simultaneous inequalities ki = x-2y + K0, ^o = x + y-5<0, k3 = 2x-y - 1>0. Find the graphs of ki = 0, ko = 0, ks = 0, as indicated in the figure. The inequality A:2 < is satisfied y by those pairs of values of x, y whose graphs lie on the side of the line A;2 = toward the origin ; for when X = 0, 2/ = 0, we have A;2 = — 5, that is < 0. It may be shown in a similar manner that the inequalities ki<0 and A;3>0 are .satisfied by the pairs of values of x, y whose graphs lie on the sides of the lines ki = 0, k^ = remote from the origin. Therefore the given inequalities ^1 < 0, A:2 < 0, ^"3 > are satisfied by the pairs of values of x, y whose graphs lie within the triangle formed by the three lines. EXERCISE LIII In the following examples the letters a, b, c are supposed to denota unequal positive numbers. 1. Trove that a/6 + 6/a> 2. 2. Prove that (a + b) (a^ + b"^) > (a2 + 62)2. 3. Prove that a^ + 6^ > a'^b + ab". 4. Prove that a"b + b"a + W-c + d^b + c'^a + a^o 6 abc. 342 A COLLEGE ALGEBRA 5. Prove that a^ + 63 + c^ > 3 ahc. 6. Solve X + 7>3x/2 - 8. 7. Solve 2 x2 + 4 X > x-^ + 6 X + 8. 8. Solve (X + 1) (X - 3) (X - 6) > 0. 9. Solve 7/-x-2<0, x-3<0, y + l>Oby the graphical method. 10. Also ?/-x>0, 2/-2x<0. 11. Also x + ?/ + 3>0, ?/-2x-4<0, y + 2x + 4>0. 12. Prove that x^ + 2 x + 5 > is true for all values of x. 13. Solve x2 + 2/2 _ 1 < 0, 2/2 _ 4 x < by a graphical method. XVIII. INDETERMINATE EQUATIONS OF THE FIRST DEGREE 672 Single equations in two variables. Given any equation of the form ax -\-oy = c where a, b, c denote integers, of which a and b have no common factor. We seek an expression for all pairs of integral values of X and y which will satisfy this equation ; also such of these pairs as are positive as well as integral. 673 Theorem 1. All equations ax + by = c of the Idnd just described have integral solutions. For since a and h are prime to one another, by the method explained in § 491 we can find two integers p and 7, positive or negative, such that ap -\-hq — \ and therefore a (pc) + h (qc) = c, and this proves that x = pc, y — qc is a, solution of az + by — c. 674 Theorem 2. If x = x,,, y = y^ he one integral solution of such an equatio7i ax + by = c, all of its integral solutions are given by the formulas X = Xo + bt, y = yo - at whe7i all possible integral values are assigned to tc INDETERxMiNATE EQUATIONS 343 First, X = Xo + bt, y = yo — at is always a solution of ax + by — c. (1) For, substituting iu (1), a (Xq + bt) + b(yo — at) = c, or, simplifying, axo + byo = c, which is true since, by hypothesis, x = Xo, 2/ = 2/o is a solution of (1). Second, every integral solution of (1) is given hy x = Xo + bt, y = yo — at. For let X = Xi, y = yi denote any second integral solution. Then axi + byi — c and axo + byo = c, whence, subtracting, b (yi — yo) = — a (Xi — Xq). (2) From (2) it follows that & is a factor of the product of the integers a and Xi — Xo. Therefore, since b is prime to a, it must be exactly con- tained in Xi — Xo, § 492, 1, and if we call the quotient f, we have Xi — Xo = bt', or Xi = Xo + bt\ (3) And substituting (3) in (2) and simplifying, we also have 2/1 = 2/0- at\ (4) From § § 673, 674 it follows that every equation ax -\-hy = c 675 of the kind just described has infinitely many integral solu- tions. When a and h have contrary signs there are also infinitely many positive solutions ; but when a, and b have the same sign there is but a limited number of such solutions or no such solution. Thus, one solution of 2 x + 3 y = 18 is x = 3, ?/ = 4. Hence the general solution is x = 3 + 3 <, i/ = 4 — 2 i. The positive solutions correspond to < =: — 1, 0, 1, 2 and are x, y = 0, 6 ; 3, 4 ; 6, 2 ; 9, 0. The theorem of § 674 enables one to write down the general 676 integral solution of an equation of the kind under considera- tion as soon as a particular solution is known. A particular solution may often be found by inspection. Thus, one solution of 10 a; + 3 2/ = 12 is a; = 0, ?/ = 4. A particular solution may always be found by the method indicated in § 673 ; also by the method illustrated in the following example. Example. Find the integral solutions of 7x + 19 y = 213. (1) Solving for the variable with the smaller coefficient, here x, and reducing, we have x = ^li:=i^ = 30-2y + ^-^. (2) 344 A COLLEGE ALGEBRA Hence if x is to be an integer when y is one, (3 - 5?/)/ 7 must be an integer. Call this integer m, so that (3 - 5 ?/) / 7 = u. Then 5 y + 7 u = 3. (3) Treating (3) as we have just treated (1), we have 3-7M 3-2« .,. 2/ = 5 Set (3 - 2 m)/ 5, which must be integral, equal to v. Then 2 w + 5 « = 3. (5) Treating (5) as we have already treated (1) and (3), we have „ = ^Jil-"==i-2. + l:^. (6) 2 2 When v-1 the fractional term (l-r)/2 vanishes and u has the integral value — 1. Substituting it = - 1 in (4), we obtain ?/ = 2. Substituting y = 2 in (2), we obtain x = 25. Hence the general integral solution of (1) is x = 25 + 19«, 2/ = 2-7<. There are two positive solutions corresponding to t — -\ and t = respectively, namely : x, y = 6, 9 ; 25, 2. Observe that in the fractional terms of (2), (4), (6) the numerical values of the coefficients of y, u, v, namely 5, 2, 1, are merely the successive remainders occurring in the process of finding the greatest common divisor of the given coefficients 7 and 19. We finally obtain the remainder, or coefficient, 1, because 7 and 19 have no common factor. The like will be true if we apply the method to any equation ax + by = c in which a and b have no common factor. Hence the method will always yield a solution of such an equation. But in practice it is seldom necessary to complete the reckoning above indicated. Thus, having obtained (4), we might have observed that m = - 1 will make (3 - 2u)/5 integral, which would have at once given us y = 2 and therefore x = 25, by (2). 677 Observe that an equation ax -{-by = c with integral coeffi- cients of which a and b have a common factor, as d, can have no integral solution unless d is also a factor of c. For if x and y were integers, d would be a factor of ax + by and therefore of c. Thus, 4 a; + 6 y = 7 has no integral solution. INDETERiMINATE EQUATIONS 345 Simultaneous equations. The following example will illus- 678 trate a method for tindiug the integral solutions, if there be any, of a pair of simultaneous equations in three variables with integral coefficients. Example. Find the integral solutions of Sx + Gy-2z = 22, (1) 5x + 8y -(3Z = 28. (2) First eliminate z and simplify the resulting equation. We obtain 2 x + 5 ?/ = 19. (3) Next find the general solution of (3), as in § 676. We obtain x = 7 + [,t, y = 1 - 2t. (4) Next substitute (4) in (1) and simplify the result. We obtain 2z-St = 5. (5) Next find the general solution of (5). We obtain 2 = 1 - 3 m, « = - 1 - 2 u, (6) where u denotes any integer whatsoever. Finally substitute i = — 1 — 2m in (4) and simplify. We obtain x = 2 — lOu, y — 3 + 4^u, z = l — Su, (7) which is the general solution required. The only positive solution is that corresponding to m = 0, namely x = 2, 2/ = 3, 2 = 1. Observe that the given equations will have no integral solu- tion if either of the derived equations in two variables has none, § 677. We proceed in a similar manner if given three equations in four variables, and so on. Single equations in more than two variables. The following 679 example illustrates a method of obtaining formulas for the integral solutions of a single equation in more than two variables with integral coefficients. Example. Find the integral solutions of 5a; + Sy + 19 2 = 50. (1) Solving for X, x = 10 - y - 3 2 - 'l^jtJ^. (2) 5 Set {3y + iz)/5, which must be integral, equal to u. 346 A COLLEGE ALGEBRA Then 3 y + 4 2 = 5 u. (3) Solving for 2/, y = u-z+ "~^ - (4) Set (2 u — 2)/3, which must be integral, equal to V. Then z-2u - 3v. (5) Substituting (5) in (4), y — — u + iv. (6) Substituting (5) and (6) in (2), x = 10-6« + 5u. (7) The formulas (5), (6), (7), in which u, v may have any integral values whatsoever, constitute the general solution required. Substituting u = 2, v = 1 in the formulas (5), (6), (7), we obtain a positive solution of (1), namely x = 3, y = 2, z = 1. EXERCISE LIV Find the general integral solutions of the following ; also the positive integral solutions. 1. 6x-172/ = 18. 2. 43x-12y = 158. 3. 16x + 59y = l. 4. 72x + 23 2/ = 845. 5. 49x-27y = 2S. 6. 47 x - 97 y = 50L ^ r2x + 52/-8z = 27, r5x + 2y = 42, ' L3x + 22/ + z = ll. '13^-72 = 2. 9. 4x + 3y = 2z + 3. 10. 2x + 32/ + 4z = 17. 11. Find the number of the positive integral solutions of the equation 3x + 7 2/ = 1043. 12. Reduce the fraction 41/35 to a sum of two positive fractions whose denominators are 5 and 7. 13. A man buys calves at $1 a head and lambs at $6 a head. He spends in all $110. How many does he buy of each'.' 14. Separate 23 into three parts such that the sum of three times the first part, twice the second part, and five times the third part will be 79. 15. Find the smallest number which when divided by 5, 7, 9 will give the remainders 4, 6, 8. 16. Two rods of equal length are divided into 250 and 253 equal divi- sions respectively. If one rod is laid along tl>e other so that their ends coincide, which divisions will be nearest together? RATIO AND PROPORTION 347 XIX. RATIO AND PROPORTION. VARIATION RATIO AND PROPORTION Ratio. In arithmetic and algebra it is customary to extend 680 the use of the word ratio, § 215, to numbers ; and, if a and b denote any two numbers, to define the ratio of a to i as the quotient a /b. . (Compare § 216.) The ratio of a to i is denoted by a /b or by a:b. In the ratio a : Z» we call a the atitecedent and b the co?iseqtient. Properties of ratios. Since ratios such as a : 6 are fractions, 681 their properties are the properties of fractions. Hence The value of a ratio is not clianged when both of its terms are ?nultijjlied or divided by the same number. Thus, a lb = ma : mb = a/n : b/n. On the other hand, except when a = b, the value of a:b is changed when both terms are raised to the same power, or when the same number is added to both. In particular. If a,,h, and m are positive, the ratio a:b is increased by adding m to both a and b wheii a < b ; decreased, when a > b. „ a + m a m(h — a) For , = ~ > 6 + m b b{b + m) * and 7n{b — a)/b{b + ?«) is positive ornegative according as ab. Proportion. When the ratios a : b and c : d are equal, the 682 four numbers a, b, c, d are said to be in projjortion, or to be jiroportional. This proportion may be written in any of the ways a fb = c / d, or a\b — c:d, or a\b::c:d. It is read '' a is to i as c is to f/." In the proportion a : b = c : d, the terms a and d are called the extremes, and b and c the means. Again, d is called the fourth proportional to a, b. and c. 348 A COLLEGE ALGEBRA 683 Theorem. In any proportion the product of the extremes is equal to that of the means ; that is, If a : b = c : d, then ad = be. For from a/b = c/d we obtain ad = be by merely clearing of fractions. Example. The first, second, and fourth terms of a proportion are 1/2, — 3, and 5 respectively; find the third term. Calling the third term x, l/2:-3 = x:5. Hence 5-l/2=: — 3x, or, solving for X, x = — 5/6. 684 Conversely, if the product of a first pair of numbers be equal to that of a second pair, the four numbers tvillbe in p)ropjortion when arranged in any order which makes one of the pairs means and the other extremes. For, let ad = be. Dividing both members by bd, we have a/b = c/d. Hence a-.b =: c :d (1) and c ■.d = a-.b. (2) Similarly by dividing both members of ad = be by cd, ab, and ae in turn, we obtain a:c = b:d (3) and b:d = a:c, (4) d -.b = c -.a (5) and c la^ d -.b, (6) d:c = b:a (7) and b:a = d:c. (8) 685 Allowable rearrangements of the terms of a proportion. From §§ 683, 684 it follows that if a, b, c, d are in proportion when arranged in any one of the orders (l)-(8), they will also be in proportion when arranged in any other of these orders. In pcirticular, 1. In any 2)ro}mrtion the terms of both ratios may be inter- changed. Thus, if a-.b = c\d, then b : a = d : c. 2. I?i auy j)r()j)(>rtion either the means or the extremes may be interchanged. Thus, if a : 6 = c : d, then a : c = 6 : d. RATIO AND PROPORTION 349 The transformations 1 and 2 are called inversion and alter- nation respectively. Other allowable transformations of a proportion. 686 If we know that a : b = c : d, we may conclude that 1. a + b:b = c + d:d. 2. a - b :b =z c - d:d. 3. a -\- b : a — b = c -{- d : c — d. 4, ma : inb = nc : nd. 5. ma : nb = mc : nd. 6. a" : i" = c" : d". For in 1 take the product of the means and extremes and we have ad-\^bd = be + bd, that is ad = be, which is true since a:b = c :d. Hence 1 is true, § 288. The truth of 2-6 may be proved in a similar manner. The transformations of a:b = c: d into the forms 1, 2, 3 are called composition, division, and comjjositio7i and division respectively. Example. Solve x2 + 2x + 3 : x^ _ 2a; - 3 = 2x2 + x - 1 :2x2-x + 1. By 3, 2x2:2(2x + 3) = 4x2:2(x-l). Hence x2 = 0, (1) or by 4, 5, 1 : 2x + 3 = 2 : x - 1. (2) Solving (1) and (2), x = 0, 0, or - 7/3. Theorem. In a series of equal ratios any antecedent is to its 687 consequent as the sum of all the antecedents is to the sum of nil the consequents. Thus, if a\:bi — a-2 : b-2 = a^ : 63, then Ui : hi = ai + a2 + as : 61 + 62 + ^3. For let r denote the common value of the equal ratios. We then have ai/bi = r, ai/b<2-r, 03/63 = r. Hence Oi = rhx, a-i = rb-2, «j = rbg, or, adding, a\ + a^ + as = r(bi + 60 + 63). 61 + 6a + &3 350 A COLLEGE ALGEBRA Example 1. If x:{b - c)yz = y:{c - a)zx = z -.{a -h) xy, then x2 4. y2 + 22 = 0. For multiplying the terms of the first ratio by x, those of the second by y, and those of the third by 2, and then applying our theorem, we have X2 __ 7/2 _ Z2 ^x- + y^ + Z2 {b — c) xyz (c — a) xyz {a — b) xyz which evidently requires that x'^ + y^ + z- = 0. The device employed in the proof just given will be found useful in dealing with complicated problems in proportion. Example 2. Prove that if a :b = x:y, then flS + 2 6^ : a62 = x3 + 2 y3 . xy^. Set a/b = x/y = r, so that a = rb and x = ry. Then (a^ + 2 5^) / a62 = (rSfts + 2 63) / r&3 = (r-3 + 2)/r, and (x3 + 2 y 3) / a;?/2 = (r^y3 + 2 ?/3) / ryS ^ (^3 + 2) / r. 688 Continued proportion. The numbers a, b, c, d, ■ ■ ■ are said to be in continued proportion if a -.b = b ic = c:d := • • ■. If three numbers a, b, c are in continued proportion, so that a : b = b : c, then b is called a mean jnoportional to a and c, and c is called a third proportional to a and ft. T/'a, b, c are in continued proportion, then b^ = ac. For since a-.b = b -.c, we have b- — ac, § 683. EXERCISE LV 1. Find a fourth proportional to 15, 24, and 20; a third proportional to 15 and 24 ; a mean proportional between 5 aW and 20 ab^ ; a mean proportional between Vl2 and V75. 2. IfSx — 2y = x — 5y, find x : y ; also x + y : x — y. 3. If 2 x2 — 5xy — 3 2/2 = 0, find x :y; also y : x. 4. If ax + by + cz = and a'x + 6';/ + c'z = 0, then X : y : z = be' — b'c : ca' — c'a : ab' — a'b. 5. If a : 6 = c : d, then 06 + cd is a mean proportional between a^ 4- c* and 62 4. d2. 6. If (a2 + 62) cd - (c2 + d2) a^^ then either a:6 = c:dora:6 = d:C. RATIO AND PROPORTION. VARIATION 351 7. Ua:b -c:d, then V^ -\-Vb-Wa + b a& 63 c3 (a + 6 + c)^ 9. If the numbers ai, tto, •••, a„ ; bi, b^, ■ ■ ■ ■, b„ ; ^i, Z21 •••) ^i are all posi- tive, the ratio liai + hn2 + • • • + 'n«« : ^1^1 + ^2^2 + • ■ • + l,,bn is intermediate in value to the greatest and least of the ratios oi : 61, a^ : 62, ■ • • , a„ : 6„. 10. li a — b -.k = b — c :l = c — a : m, and a, b, c are unequal, then k + I + m = 0. 11. If X : mz — ny = y -.nx — Iz = z -.ly — mx, then Ix + my + nz = and x2 + ?/2 + 22 = 0. 12. If ai : bi = Uo : bi = as : 63, then each of these ratios is equal to 1 1 {kal + Z2«2 + '3«3)" : ('1^1 + l^b^ + kb^r- 13. By aid of § 68G solve each of the following equations. x^ + ax - a 2x2 + rt (2) x^ — ax + a 2x- — a 2x3-3x2 + 2x4-2 3x3-x2 + 10x-26 2 x" - 3 x2 - 2 X - 2 3 x3 - x2 - 10 X + 26 14. Separate 520 into three parts in the ratios 2:3:5. 15. Two casks A and B are filled with two kinds of sherry mixed in A in the ratio 3 : 5, in B in the ratio 3 : 7. What amount must be taken from each cask to form a mixture which shall consist of 6 gallons of one kind and 12 gallons of the other kind ? VARIATION One independent variable. If two variables y and x are so related that however their values may change their ratio remains constant, then y is said to vary as x, or 1/ and x are said to vary proportionally. More briefly, y is said to vary as x when y / x = c, ot y = ex, where c denotes a constant. The notation y cox means " y varies as x." If given that y varies as x, we may at once write y = ex ; and if also given one pair of corresponding values of x and 352 A COLLEGE ALGEBRA y, we may find c. The equation connecting y and x is then known, and from it we may compute the value of y which corresponds to any given value of x. Example. If y varies as x, and y = 12 when x = 2, what is the value of y when x = 20 ? We have y = ex, and, by hypothesis, this equation is satisfied when y = 12, x = 2. Hence 12 = c • 2, that is c = 6. Therefore y = x. Hence when x = 20 we have y = (3 • 20 = 120. 691 Instead of varying as x itself, y may vary as some function of x, for example as x'^, or as a; + 1, or as 1 1 x. In particular, if y varies as 1/x, we say that y varies inversely as x. Example. Given that y is the sum of a constant and a term which varies inversely as x ; also that y = \ when ic = — 1, and y = b when x = 1. Find the equation connecting x and y. By hypothesis, y = a + b/x, where a and b are constants. Since this equation is satisfied byx = -l, y = l, and by x = 1, y = 5, ^^ ^^^^ 1 = a - 6 and 5 = a + b. Hence a = 3, 6 = 2, and the required equation is y = 3 + 2/x. 692 More than one independent variable. Let a? and y denote variables which are independent of one another. If a third variable z varies as the product xy, so that z = cxy, we say that z varies as x and y jointly ; and if z varies as the quotient X /y, so that « = c • x/y, we say that z varies directly as x and ijiversely as y. Thus, the area of a rectangle varies as the lengths of its base and alti- tude jointly ; and the length of the altitude varies directly as the area and inversely as ihe length of the base. 693 Theorem. If irlu-n x is constant z varies as y, and token y is constant z varies as x, then ichen both x and y vary, z varies as the product xy. For select any three pairs of values of x and ?/, such as Xi, ?/i ; Xn, yi ; Xi, yi ; and let Zi, Z2, Za denote the corresponding values of z, so that VARIATION 363 xi, 2/1, zi, (1) Xl, 2/2, 23, (2) X2, 2/2, Z2 (3) are sets of corresponding values of the three variables. Then since the value of x is the same in (1) as in (2), and, by hypoth- esis, for any given value of x, z varies as y, we have, § 689, 21/2/1 = 23/2/2- (4) Similarly since yn is common to (2) and (3), we have Zs/a-l =22/X2. (6) Multiplying together the corresponding members of (4) and (5), 2i/a;i2/i = Z2/a;22/2. (6) Therefore corresponding values of z and xy are proportional ; that is, z varies as xy, § 689. EXERCISE LVI 1. If 2/ varies as x, and y = ~2 when x = 5, what is the value of y when X = 7 ? 2. If y varies inversely as x^, and y — 1 when x = 2, for what values of X will 2/ = 3 ? 3. Given that y is the sum of a constant and a term which varies as X- ; also that 2/ = 1 when x = 1, and y = %vhen x = 2. Find the equation connecting x and y. 4. If y varies directly as x- and inversely as 2^, and y = 1 when x = — 1 and 2 = 2, what is the value of y when x = 3 and 2 = — 1 ? 5. If y varies as x, show that x^ — y- varies as xy. 6. If the square of y varies as the cube of z, and z varies inversely as X, show that xy varies inversely as the square root of x. 7. The wages of 3 men for 4 weeks being $108, how many weeks will 5 men work for $135? 8. The volume of a circular disc varies as its thickness and the square of the radius of its face jointly. Two metallic discs having the thicknesses 8 and 2 and the radii 24 and 36 respectively are melted and recast in a single disc having the radius 48. What is its thickness? 9. A right-circular cone whose altitude is a is cut by a plane drawn parallel to its base. How far is the plane from the vertex of the cone when the area of the section is half that of the base ? How far is the plane from the vertex when it divides the cone into two equivalent parts ? 354 A COLLEGE ALGEBRA XX. ARITHMETICAL PROGRESSION 694 Arithmetical progression. This name is given to a sequence of numbers which may be derived from a given number a by repeatedly adding a given number d, that is, to any sequence which may be written in the form a, a + d, a + 2d, • • • , a. + (?i — 1) d. (I) Since d is the difference between every two consecutive terms of (I), it is called the common difference of this arith- metical progression. Thus, 2, 5, 8, 11 is an arithmetical progression in which d = S, and 2, — 1, — 4, — 7 is an arithmetical progression in which d = — ,3. 695 The nth term. Observe that in (I) the coefficient of d in each of the terms a, a + d, a -]- 2 d, ■ ■ ■ is one less than the number of the term. Hence the general or mih term is a -\-(m — l)d; and if the entire number of terms is n and we call the last term I, we have the formula l = a + (n-l)d. (II) Example. The seventh term of an arithmetical progression is 15 and its tenth term is 21 ; find the first term a and the common difference d ; and if the entire number of terms is 20, find I. We have ^ a + 6 d = 15 and a + 9 d = 21. Solving for a and d, o = 3, d = 2. Hence l-S + 19.2 = 41. 696 The sum. Evidently the next to the last term of (I) may be written I — d, the term before that, I — 2d, ■ • • , the first term, I —(71 — 1) d. Hence, if S denote the sum of the terms of (I), we have S = a+(a + d) + (a + 2d)-] \-[a+(?i- !)(/], S = I -^-(l - d) + (I - 2 d)+ ■ ■ ■ +11 - {n - l)d2. ARITHMETICAL PROGRESSION 355 Adding the corresponding members of these two equations, we obtain 2 S = n(a + I). Therefore S = ^(a + l). (Ill) Example. Find the sum of an arithmetical progression of six terms ■whose first term is 5 and whose common difference is 4. Since n = 6, we have Z = 5 + 5 • 4 = 25. Hence S = « (5 + 25) = 90. Applications. If in an arithmetical progression any three 697 of the five numbers a, I, d, n, S are given, the formulas (II) and (III) enable us to hnd the other two. The only restric- tion on the given numbers is that they be such as will lead to positive integral values of >i. Example. Given d = 1/2, 1 = 3/2, S = -15/2; find a and n. Substituting in (11), (HI), 3 2 ■ = a + n-1 2 (1) - 15 ' 2 ■ n/ -2V -D- (2) Eliminating a, n2- - J n — 30: = 0. (3) Solving (3), n : = 10 ( 3r - 3. - The value n = -.3 is inadmissible. Substituting ; n = :10 in (1). we obtain a = — 3. Hence ?i = 10, a = — 3, and the arithmetical progression is - .3, - 2.V, - 2, - li, - 1, - i, 0, I, 1, 11. Arithmetical means. If three numbers form an arithmetical 698 progression, the middle number is called the arithmetical mean of the other two. The arithmetical mean of any two numbers a and b is one half of their sum. For if X be the arithmetical mean of a and 6, then the sequence a, x, b is an arithmetical progression. Hence x — a — b — x, and therefore x — {a + b)/2. 356 A COLLEGE ALGEBRA In any arithmetical progression all the intermediate terms may be called the arithmetical means of the first and last terms. It is always possible to insert or " interpolate " any number of such means between two given numbers a and h. Example. Interpolate four arithmetical means between 3 and 5. We are asked to find the intermediate terms of an arithmetical progres- sion in which a = 3, I — b, and ?i = 4 + 2 or 6. Substituting i = 5, a = 3, ?i = 6 in (II), we have 5 = 3 + 5 d, whence d = 2/5. Hence the required means are 3§, 3f , 4i, 4f. EXERCISE LVII 1. Find the twentieth term and the sum of the first twenty terms of 3, 6, 9, •••; of -3, -1\, 0, • • • . 2. Find a formula for the sum to n terms of 1, 2, 3, • • • ; of 1, 3, 5, • • • ; of 2, 4, G, ••-. 3. Find the sum of the first n numbers of the form 6r + 1, where r denotes or a positive integer. 4. Find the arithmetical progression of ten terms whose fifth term is 1 and whose eighth term is 2. 5. Insert five arithmetical means between — 1 and 2 6. Given n = IG, a = 0, d = 4/3 ; find Z and S. 7. Given n = 7, Z = - 7, (Z = - 5/3 ; find a and S. 8. Given n = 12, a = - 5/3, Z = 31 1 ; find d and S. 9. Given a = 2, Z = - 23^, .S = - 559 ; find n and d. 10. Given n = 7, n = 3/7, .S = 45 ; find d and /. 11. Given a = 4, cZ = 1/5, Z = 9s ; find n and S. 12. Given n = 9, d = - 4, .S = 135 ; find a and I. 13. Given n = 10, I = - 2, S = Ui>; find a and d. 14. Given d = 5, Z = - 47, S = - 357 ; find n and a. 15. Given a = - 10, d = 7, .S = 20 ; find n and I. 16. Show Ihat if «'-, h'-, c- are in arithmetical progression, so also are 1/(6 + c), l/(c + «), 1/(a + b). GEOMETRICAL PROGRESSION 357 17. Show that the sum of any n consecutive integers is divisible by n, if n be odd. 18. Find an arithmetical progression such that the sum of the first three terms is one half the sum of the next four terms, the first term being 1. 19. Three numbers are in arithmetical progression. Their sum is 15 and the sum of their squares is 83. Find these numbers. 20. Find the sum of all positive integers of three digits which are multiples of 9. 21. If a person saves $\oO a year and at the end of the year puts this sum at simple interest at 4%, to how much will his savings amount at the end of 11 years? 22. Two men A and B set out at the same time from two places 72 miles apart to walk toward one another. If A walks at the rate of 4 miles an hour, while B walks 2 miles the first hour, 2?^ miles the second hour, 3 miles the third hour, and so on, when and where will they meet ? XXI. GEOMETRICAL PROGRESSION Geometrical progression. This name is given to a sequence 699 of numbers which may be derived from a given number a by repeatedly multiplying by a given number r, that is, to any sequence which may be written in the form a, ar, ar'^, •■-, ar"~^. (I) We call r the common ratio of the geometrical progression (I) and say that the progression is increasing or decreasing according as r is numerically greater or less than 1. Thus, 1, 2, 4, 8 and 1, - 2, 4, - 8 are increasing geometrical progres- sions in which r = 2 and — 2 respectively; while 1, 1/2, 1/4, 1/8 is a decreasing geometrical progression in which r — \ /2. The nth term. Observe that the exponent of r in each term 700 of (I) is one less than the number of the term. Hence in a geometrical progression of n terms whose first term is a and whose ratio is r, the formula for the last term I is I = ar-\ (II) 358 A COLLEGE ALGEBRA 701 The sum. Let S denote the sum of the geometrical pro- gression (I). Then S = a -{- ar + ar^ -] + ar"-^ + ar"-^ and rS = ar -{- ar^ + • • • + a?*""^ + ar"~^ + ar\ Subtracting the second of these equations from the first, we obtain (1 — r)S = a — ar'\ Therefore ^^a^,-^ (III) In applying this formula to an increasing geometrical pro- gression we may more conveniently write it thus: S = a{i^-l)/ir-l). From (II) we obtain rl = ar". Hence (III) may also be written thus : S = (a-rl)/{l - r), or 5 = (rl - a) / (r - 1). Example. In the geometrical progression 2, — 4, 8, — 16, • • • to eight terms, find I and S. Here a = 2, r = — 2, and n = 8." Hence, by (II), l = 2{-2y = - 256, ..,.„„,, « = .L^._-. 702 Applications. If in a geometrical progression any three of the five numbers a, I, r, n, S are given, the formulas (II) and (III) determine the other two. Moreover these two numbers can be actually found by methods already explained, except when the given numbers are a, n, S or I, n, S. If one of the unknown numbers is n, it must be found by inspection ; but this is always possible if admissible values have been assigned to the given numbers, since n will then be a positive integer. Example 1. Given r = 3, n = 6, S = 728 ; find a and I. Substituting the given values in (II) and (III), we have i = a • 35 =: 243 a, and 728 = a^-^ = 364a. 3-1 Solving these equations, a = 2, 1 = 486. GEOMETRICAL PROGRESSIOX 359 Example 2. Given a = 6, n = 5, 1 = 2/27 ; find r and S. By (II), 2/27 = 6r*, whence r* = 1/81, or r = ± 1/3. Therefore, by (III), if r = 1 /3, then S = 6 ^-(^/^)^ = ^ , ' "^^ '' 1-1/3 27 and if r = -l/3, then S = e^-llil/^ = '^. l-(-l/3) 27 Hence there are two geometrical progressions in which a = 6, n = 5, and i = 2/27. Example 3. Given a = - 3, Z = - 46875, S = - 39063 ; find r and n. Substituting in the formula S = (a — rl) / (I — r), § 701, we have _ 39063 = - 3 + 46875 r ^ ^^^^^^ r = -b. 1 — r Therefore, by (II), _ 46875 = - 3(- 5)"-i, or (- 5)"-! = 15625. But by factoring 15625 we find 15625 = 5^ = (- 5)6. Hence n — 1 = 6, that is n = 7. Example 4. Given a = 3, n = 5, -S = 93 ; find r and I. By (III), 93 = 3^^^ = 3 (H- r + r2 + r3 + ^4). Hence r* + r* + r^ + r - 30 = 0. Thus this problem involves solving an equation of the fourth degree ; and, in general, when a, n, S are the given numbers, to find r we must solve an equation of the degree n — 1. In this particular case, however, we may find one value of r by the method of §455. It is 2. Substituting r = 2 in (II), we have Z = 3 • 2* = 48. Geometrical means. If three numbers form a geometrical 703 progression, the middle number is called the geometrical mean of the other two. The geometrical mean of any two numbers a and b is a square root of their x>rod;uct. For if X be the geometrical mean of a and 6, the sequence a, x, h is a geometrical progression. Hence x/a = h/x and therefore x = ± Va6. In any geometrical progression all the intermediate terms may be called the geometrical means of the first and last terms. 360 A COLLEGE ALGEBRA We may insert any number of such means between two given numbers a and b, as in the following example. Example. Insert four geometrical means between 18 and 2/27. It is required to find the intermediate terms of a geometrical progres- sion in which a = 18, i = 2/27, and n = 4 -f 2 = 6. Substituting the given values in (II), we have 2/27 = 18 r5, whence r = 1/3. Hence the means are 6, 2, 2/3, 2/9. 704 Infinite decreasing geometric series. We call an expression of the form ^ ^ ^^. ^ ^,., ^ . . . _^ ^,.„_i _^ . . .^ ^^^ supposed continued without end, an infinite geometric series. By the formula (III), the sum of the first n terms of (1) is a{l-r")/(l~r). Suppose that ?• < 1 numerically. Then, as n is indefinitely increased, y" will approach as limit, § 724, and therefore a(l — ?•")/(! — /•) will approach a /(I — r) as limit. We call this limit the sum of the infinite series (1). Hence, if S denote the sum of (1), we have Example 1. Find the sum of 1 + 1/2 -l- 1/4 + 1/8 H . Here a = 1 and r = 1 /2. Hence S = 1/[1 - 1/2] = 2. Example 2. Find the value of the recurring decimal .72323 ■ • • . 2.3 23 The part wliich recurs may be written 1 1 , and, by (2), 1000 100000 ' J V /' 023 23 the sum of this infinite series is -^—^ — or Adding .7, the part which l-.Ol 990 & . 1 358 does not recur, we obtain for the value of the given decimal 495 EXERCISE LVIII 1. Find the fifth term and the sum of the first five terms of the geometrical progre.ssion 2, — 6, 18, ■ • ■. 2. Find the fourth term and the sum of the first four terms of the geometrical progression 4, 6, 9, • • •. GEOMETRICAL PROGRESSION 361 3. Find the sums of the following infinite geometric series : 12_6+3-...;l-i + i----;f + i+i^ + ---. 4. Find the values of the following recurring decimals : .341341 ••-, .0567272..., 8.45164516 •• =. 5. Given a = - .03, r = 10, n = 6 ; find I and *\ 6. Given n = 7, a = 48, i = 3/4 ; find r and S. 7. Given a = 1/16, r = 2, Z = 8 ; find n and S. 8. Given n = 5, r = -3, l = Sl; find a and S. 9. Given a = 54, r = 1/3, 6' = 80| ; find n and I. 10. Given n = 4, a = - 3, S = - 408 ; find r and I. 11. Given a =-9/16, Z = -16/9, S = - 781/144; find n and r. 12. Given n = 0, r = - 2/3, S = 665/216 ; find a and I. 13. Given r = 3/2, I = mi, S = 83J ; find n and a. 14. Given 7i = 4, Z = 54/25, S = 544/25 ; find a and r. 15. Given n = 5, Z = 48, S = 93 ; find a and r. 16. Find the positive geometrical mean of 0^/6 and h^/a. 17. Insert three geometrical means between 5 and 405. 18. The third term of a geometrical progression is 3 and the sixth term is — 3/8. Find the seventh term. 19. Find a geometrical progression of four terms in which the sum of the first and last terms is 133 and the sum of the middle terms is 70. 20. Find three numbers in geometrical progi'ession such that their sum shall be 7 and the sum of their squares 91. 21. Three numbers whose sum is 36 are in arithmetical progression. If 1, 4, 43 be added to them respectively, the results are in geometrical progression. Find the numbers. 22. There are four numbers the first three of which are in arithmetical progression and the last three in geometrical progression. The sum of the first and fourth is 16 and the sum of the second and third is 8. Find the numbers. 23. What distance will an elastic ball traverse before coming to rest if it be dropped from a height of 15 feet and if after each fall it rebounds to 2/3 the height from which it falls? 362 A COLLEGE ALGEBRA XXII. HARMONICAL PROGRESSION 705 Harmonical progression. This name is given to a sequence of numbers whose reciprocals form an arithmetical progression, that is, to any sequence which may be written in the form 1/a, \l{a^d), l/(a + 2fi), •••, l/[a+(;i-l)rf]. Thus, 1, 1/2, 1/3, 1/4 and 3/2, 3/4, 3/6, 3/8 are harmonical progressions. Example. Prove that if a, 6, c are in harmonical progression, then a:c=ia — 6:6 — c. Since 1/a, 1/6, 1/c is an arithmetical progression, we have l/6-l/a = 1/c- 1/6. Hence c (a — 6) = a (6 — c), that is a : c = a — 6 : 6 — c. 706 To find any particular term of an harmonical progression, we obtain the term which occupies the same position in the corresponding arithmetical progression and invert it. Example. Find the tenth term of the harmonical progression 3/5, 3/7, 3/9, •■■. By § 695, the tenth term of the corresponding arithmetical progression 5/3, 7/3, 9/3, •• • is 23/3. Hence the tenth term of 3/5, 3/7, 3/9, ••• is 3/23. 707 Harmonical means. If three numbers are in harmonical pro- gression, the middle number is called the harmonical mean of the other two. Again, in any harmonical progression all the intermediate terms may be called the harmonical means of the extreme terms. Example 1. Find the harmonical mean of a and 6. If this mean be x, then 1 /a, 1 /x, 1 /6 is an arithmetical progression. Hence l/x - 1/a = 1/6 - 1/x, or 2/x = 1/a + 1/6. Therefore x = 2 a6/ (a + 6). Example 2. Prove that the geometrical mean of two numbers a and 6 is also the geometrical mean of their arithmetical and harmonical means. Let A, G, and ZT denote respectively the arithmetical, geometrical, and harmonical means of a and b. HARMONICAL PROGRESSION 363 Then ^ = ^+^, G = V^, iT^li^. 2 a + b J rr « + ft 2 a6 , ^, Henco AH = = ab = G^. 2 a + 6 Example 3. Prove that when a and 6 are positive, A> G>H. „, , . „ a + b 2ab {a-b^ We have A - H = = -i '—. 2 a + b 2 {a + b) Therefore, since (a - 6)V2 (a + 6) is positive, we have A>H. And since, by Ex. 2, G is intermediate in value to A and H, we have A>G>H. EXERCISE LIX 1. Continue the harmonical progression 3/5, 3/7, 1/3 for two terms. 2. Find the harmonical mean of 3/4 and 5. 3. Insert four harmonical means between 10 and 15. 4. The second and fourth terms of an harmonical progression are 4/5 and — 4. Find the third term. 5. The arithmetical mean of two numbers is 4 and their harmonical mean is 15/4. Find the numbers. 6. The geometrical mean of two numbers is 4 and their harmonical mean is 16/5. Find the numbers. 7. Show that if a, b, c are in harmonical progression, so also are tt/(6 + c), b/{c + a), c/{a + b). 8. Three numbers are in harmonical progression. Show that if half of the middle term be subtracted from each, the results will be in geomet- rical progression. 9. Show that if x is the harmonical mean between a and 6, then l/(x-a) + \/{x-b) = \/a + l/b. 10. The bisector of the vertical angle C of the triangle ABC meets the base AB at D, and the bisector of the exterior angle at C meets AB produced at E. Show that AD, AB, AE are in harmonical progression. 11. The point P lies outside of a circle who.se center is 0, and the tangents from P touch the circle at T and T. If tlie line PO meets the circle at A and B and TT at C, show that PC is the harmonical mean between PA and PB. 364 A COLLEGE ALGEBRA XXIII. METHOD OF DIFFERENCES. ARITH- METICAL PROGRESSIONS OF HIGHER ORDERS. INTERPOLATION ARITHMETICAL PROGRESSIONS OF HIGHER ORDERS 708 Differences of various orders. If in any given sequence of numbers we subtract each term from the next following term, we obtain a sequence called the first order of differences of the given sequence; if we treat this new sequence in a similar manner, we obtain the second order of differences of the given sequence ; and so on. Thus, if the given sequence be 1^, 2\ .S^, • • • , we have given sequence 1, 8, 27, 64, 125, 216, • • • , first differences 7, 19, 37, 61, 91, ••-, second differences 12, 18, 24, 30, • • • , third differences 6, 6, 6, • • • . The fourth and all subsequent differences are 0. 709 Arithmetical progression of the rth order. This name is given to a sequence whose rth differences are equal, and whose subsequent differences are therefore 0. Thus, 1^, 2^, 33, 4-^, • • • is an arithmetical progression of the third order, for, as just shown, its third differences are equal. An ordinary arithmetical progression, § 694, is of the first order, each of its first differences being the common difference d. 710 The nth term of an arithmetical progression of the rth order. Given any arithmetical progression of the rth order «1, ^2, «3J ^4, •••» «„, «„ + !' '■■, (1) and let d^, d^, ■■■, d,. denote the first terms of its successive orders of differences. We are to obtain a formula for a„ in terms of a^, r/,, d«, ■■■, d,., and 71. The first order of differences of (1) is METHOD OF DIFFERENCES 365 The first term of (2) is di, and the first terms of its first, second, • • • differences are iL, d^, ■ • • ; for the first differences of (2) are the second differences of (1), and so on. Hence when we have found an expression for any particular term of (1), we can derive from it an expression for the corre- sponding term of (2) by applying the rule : Replace a^, d^, d^, • • • by d^, d^, d^ ••■. (3) Now since d^ — a» — a^, we have a^ = Oi -{- dy. Starting with this formula for a^, we may reckon out a 3, 04, • • • as follows : We have aj = ai + dy Hence, by (3), ^3 - - a2 = C?i + (^2 Adding, ^3 = f/i + 2 r/i + f/a Hence, by (3), 04- -«3= dy + 2d._-\-d^ Adding, a^ = ay + 3(/i + 3f/o + (/3 and so on indefinitely, the reckoning, so far as coefficients are concerned, being precisely the same as that given in § 311 for finding the coefficients of successive powers of a + h. There- fore, by § 561, we have the formula «« = «i + (" — l)<^^i ^ (..-l)(..-2) ,^ ^ . . . + (.-^V..(.-.) ^ ^j^ Example. Compute the fifteenth term of 1^, 2-^, • • • by this formula. Here, § 708, a^ = 1, (h = 7, d. = 12, ds = dr = G. 14 13 14 ■ 1.3 • 12 Hence ais = 1 + 14 ■ 7 + — ^j— ■ 12 + — -^-^ 6 = 3375. Sum of the first n terms of an arithmetical progression of the 711 rth order. Let S„ denote this sum, the sequence being «!, aa, as, ■■-, a„, a„^j, •••, (1) and di, d^,---, d^ having the same meanings as in § 710. 366 A COLLEGE ALGEBRA Form the sequence of which (1) is the first order of differ- ences, namely : 0, «!, rti + «2) «1 + «2 + ^3, • • •, «! + «2 + • • • + Cini ' ' "• (2) Then S„ is the (?i + l)th term of (2), and since (2) is an arithmetical progression of the (« + l)th order whose first term is 0, and the first terms of its several orders of differences are «i, di, cL, ■ ■ ■ d,., we have, by (I), Example. Find the sum of the first fifteen terms of 1^, 2^, 3^, • • •. Here, § 708, n = 15, ai = 1, ch = 7, dz = 12, ds = d,. = 6. nr 15-14 „ 15 14 13 ,„ 15 14 13 12 ^ ,,,^^ Hence Si5 = 15+ 7 + 12 + = 14400. 2 2-3 2.3-4 712 Piles of spherical shot. 1. To find the number of shot when the pile has the form of a triangular pyramid. The top course contains 1 shot, the next lower course 1 + 2 shot, the next 1 + 2 + 3, and so on. Hence, if there are n courses, the number of shot is the sum of n terms of the sequence 1, 3, 6, 10, 15, • • •. The first differences of this sequence are 2, 3, 4, 5, • • • , and the second differences are 1, 1, 1, • • • . Hence 1, 3, 6, ■ • • is an arithmetical progression of the second order in which ai = 1, di = 2, (fj = 1. Therefore, by (11), .S„ = n + ''S'l^l^ . 2 + n(n~l){n-2) 1-2 1.2.3 _ n (n + 1) (n + 2) 1.2-3 Thus, in a pile of twenty courses there are 20 . 21 . 22/6 = 1540 shot. 2. To find the number of shot when the pile has the form of a pyramid with a square base. Enumerating the shot by courses as before, we obtain the sequence P, 22, 32, 42, ... . The first differences are 3, 5, 7, . . • , and the second are 2, 2, ■ • • . Hence 12, 22, 32, ... is an arithmetical progression of the second order in which ui = l,di = 3, dz = 2. METHOD OF DIFFERENCES 367 Therefore, by (II), 5„:=n + ^^-3 + ^<--;^^(;-^) .2 ^ n(n + l)C2n + l) . 1-2-3 Thus, when n = 20, the pile contains 20 • 21 • 41/6 = 2870 shot. 3. To find the number of shot when the pile has a rectan- gular base and terminates at the top in a row of ^^ shot. Again enumerating the shot by courses, we obtain the sequence p, 2{p+l), 3(p + 2), 4(p + 3),.... The first differences are p + 2, p + i, p + 6, ■ ■ ■ , and the second differ- ences are 2, 2, • • • . Hence p, 2 (p + 1), 3 (p + 2), • • • is an arithmetical progression of the second order in which ai = p, di — p + 2, and dz = 2. Therefore, by (II), -S„ = np + "^^^ (p + 2) + ^n -IHn^- 2) _ ^ _ n{n + l) { Sp + 2n-2) Thus, when ?i = 20 andp = 5, the number of shot is 20 • 21 • 53/6 = 3710. A theorem respecting arithmetical progressions. An examina- 713 tion of the formula for the nth term of an arithmetical progression of the rth order, § 710, (I), will show that if we carry out the indicated multiplications and arrange the result according to descending powers of n, we can reduce it to the form where the coefficients b^, b^, ■• -, b^ are independent of n. Thus, when r = 2, we have (n-l)(n-2), a„ = ai + (n - 1) di + ^^ ^2 = I n2+ {di - ^ d2) n + (ai - d, + d^). Therefore the terms of any arithmetical progression of the rth order, Oj, a^, a^, ■ ■ ■, are the values for 71 = 1, 2, 3, • ■ ■ of a certain polynomial b^n'' + bin'"^ + ■ ■ ■ + b^ whose degree with respect to n is r. We are going to show conversely that 368 A COLLEGE ALGEBRA 714 Theorem. If (x.) denote any rational integral function of the vth degree, as the sequence of numbers ^(1), <^(2), <^(3), •••, obtained by setting x = 1, 2, 3, • • • successively in <^(x), is an arithmetical progression of the rth ojrler. Here the given sequence of numbers is (!), <^(3)-<^(2), cA(4)-<^(3)..., (2) are the values of ^(x + 1) — <)!> (x) for a; = 1, 2, 3, • • •. But <^(.r + 1) — 4>(^) '^i^y be reduced to the form of a poly- nomial in X. Call this polynomial <^i (x). Its degree is r — 1. For, by the binomial theorem, § 561, we have or'"+ 6iX''-i + • •) ::= rboX'-^ + •■■ Similarly, if we write <^l (X + 1) - ! (X) = 2 (X), <^2 (X + 1) - <^2 (X) = (f>Z (X), and so on, the values of ^3(x), cf>3(x), ■•■, r(x) for x = 1, 2, 3, ■ ■ will be the second, third, • • •, rth differences of (1). But <^,.(x) is a constant and the rth differences of (1) are therefore equal. For the degree of ^gC^) is (r — 1)— 1, or r — 2; that of 3(x) is r — 3; and finally that of ^^(x) is r — r, or 0. For example, if (x) = 2 x^ — x + 1, we have 01 (x) == 2 (X + 1)3 - (X + 1) + 1 - (2 x3 - X + 1) = 6x2 + 6x + 1, 02(x) = 6(x + l)2 + 6(x + 1) + 1 - (Gx2 + 'jx + l) = 12x + 12, ,x + b^x^ + ■■■ + Kx% (1) find ht^, bi, ■■■, b,. by the method just illustrated, and then employ (1) as a formula for computing y for values of x inter- mediate to Xi, X2, ■ ■ ■, a^r + l- 719 Method of differences. When x^, Xo,---, x.+i are consecutive integers, the formula (1) of § 718, may be reduced to the form (X — CCi) (X — Xo) -, , y = yr+ix-x,)ch+^ Y^ -d, + --- (x-x,)...(x-x,) ^ 1 . 2 • • • r '■' ^ ^ where di, d^, ■■-, d, denote the first terms of the successive orders of differences of ?/i, y-i,---, y, + i. For since Xi, X2, • • •, a;,. + i are consecutive integers, the corresponding values oibo + biX-\ 1- ^x*", namely yi, 2/2, • • ■ , y,- + u form an arithmet- ical progression of the rth order, § 714. Hence we may also obtain 2/1, 2/2, •• • by substituting n = 1, 2, • • • in the formula, § 710, (I), namely y = yi + {n-l)di + ^-^ ch + ■■■ + i.2...r But setting n = 1, 2, 3, • • • in this formula will give identically the same results as setting x = Xi, X2, X3, • • • = Xi, Xi + 1, Xi -f- 2, • • • in (2). Therefore the second member of (2) and that of (1), § 718, have equal values for r + 1 values of x. But both are of the rth degree. Hence they are identically equal, § 421. Thus, as in § 718, for x = 2, 3, 4, 5, let y = 5, 4, - 7, - 34. We have 2/1, 2/2, ^3, 2/4 = 5, 4, - 7, - 34. First differences -1, -11, -27. Second differences - 10, - 10. Third difference - 6. Substituting in (2), Xi = 2, Xa = 3, X3 = 4, 2/1 = 5, di = - 1, ''2 = - 10, dg = - 6, we have 2/ = 5 - (x - 2) - 5 (x - 2) (x - 3) - (x - 2) (x - 3){x - 4), which may be reduced to 2/ = 1 - 2 x + 4 x^ _ x^, as in § 718. METHOD OF DIFFERENCES 373 Example. Given v^jjo = 3.1072, v'31 = 3.1414, -v^^ 3.1748, and \/33 = 3.2075; find VsLC. yi, 2/2, 2/3, 2/4 = 3.1072, 3.1414, 3.1748, 3.2075. First differences .0342, .0334, .0327. Second differences - .0008, - .0007. Third difference .0001. Substituting in (2) Xi = 30, x^ = 31, x^ = 32, yi = 3.1072, di = .0342, da = - .0008, ds = .0001, and x = 31.6, we have 3.1072 + (1.6) (.0342) + 11:M^(_ .0008) + (^-^H'^) (~ '^^ (_ .0001) = 3.1072 + .05472 - .000384 + .0000064 = 3.1615 +. Lagrange's formula. The formula (1) of § 718 may also be 720 reduced to the following form, due to Lagrange : _ (.r — a-o) (x — Xs) ■ ■ ■ (x — x,.^^) ^ ~ ^' (^1 - x.^{xi - a-3) • • ■ (X, - x,^,) (x-x,)(x-x,)---(x-x,.^,) ^' i^-, - ^i)(-^2 - ^3) • • • (^2 - *. + i) {x-x,){x-x^)---{x-x:) For the right member of (3) is an integral function of x of the ?-th degree and its values for ic = a-^, Xo, ■■■, x,.,^^ are ?/i, Z/2) ■■-, l/r+i- Thus, if we set x = x^, every term except the first vanishes and the first term reduces to yi. Hence, § 421, the right member of (3) and that of (1), § 718, are equal for r + 1 values of x and are therefore equal identically. Thus, as in § 718, for x = 2, 3, 4, 5, let 2/ = 5, 4, - 7, - 34. Sub- stituting in (3), we obtain (x_3)_(x-_4)_(x-_5) (2 - 3) (2 - 4) (2 - 5) ^ (x-2)(x-4)(x-5) _ (x-2)(x-3)(x-5) _ ^^ (x-2) (x-3) (x-4) (3_2)(3-4)(3-5) (4-2) (4-3) (4-5) " (5-2) (5-3) (5-4)' which will reduce to ?/ = 1 - 2 x + 4 x^ - x^, as in § 718. 374 A COLLEGE ALGEBRA EXERCISE LXI 1. For X = - 3, - 2, - 1, it is known that ?/ = - 20, 6, 0, 4 ; find y when X — — 5/2, also when x = — 1 /2. 2. Given that /(4) = 10, /(6) = - 12, /(7) = - 20, /(8) = - 18 ; find /(x) and then compute /(12). 3. Giventhat252 = 625, 262 = 676, 272 = 729; find 26.542 by the method of differences. 4. Given that 23 = 8, 3-3 = 27, 43 = 64, 53 = 125; find 4.83 by the method of differences. 5. Given that 1/22 = .04546, 1/23 = .04348, 1/24 = .04167, and 1/25 = .04; find 1/23.6 by the method of differences. 6. Given that V432 = 20.7846, V43 3 = 20 .8087, V434 = 20.8327, V435 = 20.8566, V436 = 20.8806 ; find V435.7 by the method of differ- ences. 7. By aid of Lagrange's formula find the polynomial of the third degree whose values for x = — 2, 0, 4, 5 are 5, 3, — 2, — 4. XXIV. LOGARITHMS PRELIMINARY THEOREMS REGARDING EXPONENTS 721 Theorem 1. If a, denote any real numher greater than 1, p and p, q denote jmsitive integers, then a^ > 1. For a > 1, .-. aP > 1, .-. ^ai' > 1, .-. a'/ > 1, § 261. 722 Theorem 2. If a, denote any real numher greater than 1, and V, s any two rationals such that r > s, then dJ > a^ For r - s>0, .-. a''-''>l, .-. a'--'- a'>a', .-. a'->a% §§ 721, 261. 723 Theorem 3. 7/"a > 1 and n be integral, then ^^^"^ a" = 00. Fur since a > 1, we may write a = 1 -\- d, where d is positive. Then a" = (1 + d)", and since (1 + d)">l + nd, § 561, wehave a" >l + nd. Therefore, since 1™ (1 + nd) = 00, we have 1"" a" = 00. LOGARITHMS 375 Theorem 4. If < a. < 1, and n be integral, lim a" = 0. 724 For let a = 1 /6, where 6 > 1, since a < 1. Then li^" a« = 1 /J™ ^" = 0, § 512, since Jii^ 6» = oo, § 723. 1 Theorem 5. If u ^»e integral, lii^i V^ = JiiJ a" = 1. 725 1. When a > 1, we have a" > 1, § 721, so that a" = 1 + d„, where d„ is some positive number dependent on n. Tlien a = (1 + d«)», .-. a>l + ?id«, .-. d„ <(a - \) / n. Tlierefore, since 1™ (a - l)/n = 0, § 512, we have Jij"^ cZ„ = 0. Hence Ji|^ a" = jij" (1 + d„) = 1. 2. When < a < 1, let a = 1/6, where &>1, since a1 or a < 1, but to fix the ideas we shall suppose that a > 1. There are infinitely many sequences of rational values through which X may run in approaching h as limit. From among them select some particular increasing sequence, and represent x by x' when supposed to run through this sequence. Then as x' == b, the variable a-^' continually increases. § 722, but remains finite — less, for instance, than a"^, if c denote 376 A COLLEGE ALGEBRA any rational greater than b. Hence a-^' approaches a limit, § 192. Call this limit L. It only remains to prove that a' will approach this same limit L it x approach b through any other sequence of rationals than that through which x' runs. But a^ = a^ • a^ - ^' and therefore lim a"" - lim a^' ■ lim a*- ==' = L, since lim a^-^' — 1, § 726. 728 Irrational exponents. We employ the symbol a^ to denote the limit which a^ will approach when x is made to approach b through any sequence of rational values. Hence by a'', when b is irrational, we shall mean ^^]^^ a*. 729 Having thus assigned a meaning to a^ when x is irrational, we can readily prove that li]^^ a"" = a* when x approaches b through a sequence of irrational values. For let x', x, x" denote variables all of which approach b as limit, x' and x" through sequences of rational values and x through a sequence of irra- tional values, and such that x' ■+ s'> -a^ + <=. §§ 203, 728 2. {a>'y = a'"'. For (a^)y = fF". Hence lim (ax)w - jii"^ a'"", or (a}>)y = aPv. §728 Hence lim (aby — lini (jb.-/^ or {a!>Y = aK §§ 728, 729 3. (aby = a'=¥. For (ab)?' = W'bv. Hence lim (a/*)" = lim a«by = lim a" • lim 6v. §203 That is, (a6)'- = a'-^-. §728 LOGARITHMS 377 LOGARITHMS. THEIR GENERAL PROPERTIES Logarithms. Take a, any positive number except 1, as a base 731 or number of reference. We have shown that every real power of a, as a''^, denotes some definite positive number, as m. In ,a subsequent section we shall show conversely that every positive number, m, may be expressed in the form a*^, where fi is real. If a'^ = m, we call /a the loijaritlim of m. to the base a and 732 represent it by the symbol log„?>i. Hence the logarithm of m. to the base a is the exponent of the power to which a must be raised to equal ?», that is a'""""' = m. Thus, 3* = 81, .-. 4 = log381; 2-3 = 1/8, .-. -3 = log.2l/8. Since a" = 1, we always have log„l = 0; and since a^ ■= a, 733 we always have log,, a = 1. When a > 1, it follows from a^ = m, by § 722, that to any 734 increase in the number m there corresponds an increase in its logarithm p. ; also that if m is greater than 1, its loga- rithm /A is positive, and that if m lies between 1 and 0, its logarithm /u, is negative. Again, when a > 1, we have, § 723, 735 lim a*^ =. CO, and liii^ xr^ = lim 1 /a'^ = 0. We therefore say, when a > 1, that log„oo = co, and log„0 = — oo. Theorem 1. The logarithm of a product to any base is the 736 su)n of the logarithms of the factors to the same base. For let m = a*^, that is ^u = log„ ??;, and n — ay^ that is v = log„ji. Then 9?in = a'^a'' = a'-' + '', that is, loga»i?i= ^ + V = log,, m + logaW. Theorem 2. The logarithm of a quotient is the logarithm of 737 the dividend minus the logarithm of the divisor. For if m = a*^ and n = a", we have rn/n = af^ /a"" = a'^-'', that is, logam/n = fj. — v =■ loga»i — log«ft. 378 A COLLEGE ALGEBRA 738 Theorem 3. The logarithm of any power of a number is the logarithm of the number multiplied by the exponent of the power. For if m-a)^, we have nv = {af^y = a/^^, that is, loga m'' = r/j. = r loga m. 739 Theorem 4. The logarithm of any root of a mimher is the logarithm of the number divided by the index of the root. For if m = a", we have V//i = Va*^ = a% that is, loga >/w = fi/s = (loga m)/s. 740 The practical usefulness of logarithms is due to the proper- ties established in §§ 736-739. Logarithms of numbers to the base 10 have been computed and arranged in tables. If we avail ourselves of such a table, we can find the value of a product by an addition, of a quotient by a subtraction, of a power by a multiplication, and of a root by a division. Thus, log— 1^ = log v^5 + log Ve - log 325 §§ 736^ 737 = (log 5) /7 + (log 6) /8- 25 log 3. §§738,739 Hence, to obtain the value of v5 ViJ/O-s, we have only to look up the values of log 5, log 6, and log 3 given in the table, then to reckon out the value of (log 5) /7 + (log 0) /8 - 25 log 3, and finally to look up in the. table the number of which this value is tlie logarithm. EXERCISE LXII 1. Find log24, log42, logv'gS, log5025, loga 729, logio.OOl, log2l/64, log2.125, logaVa^, log8l28, log„2a3. 2. If logio2 = .3010 and logio3 = .4771, find the logarithms to the base 10 of 12, 9/2, v^, Vg. 3. Express loga 600^ in terms of loga 2, loga 3, and loga5. LOGARITHMS 379 4. Express the logarithms of each of the following expressions to the base a in terms of loga&, logaC, logad. (1) 6^c~Vf^'- (2) ^/a-2v^**-^&Wa"^• ■3V81'\^729-9-' = 31/] 5. Prove that logs V 81 V 729 • 9" ^ = 31/18, X 4- vx^TTT 6. Prove that log„ = 2 loga (x + Vx^ - 1). COMMON LOGARITHMS Computation of common logarithms. For the purposes of 741 numerical reckoning we employ logarithms to the base 10. These are called, common logarithms. In what follows log m will mean logi„?». We have 10° = 1, .-. log 1 = ; 10^ = 10, .-. log 10 = 1 ; 742 102 ^ 100, .-. log 100 = 2, ■ • • ; also 10"' = .1, .-. log .1 = - 1 ; 10-2 = .01, .-.log .01 =-2, •••. Hence, for the numbers whose common logarithms are iyite- gers we have the table : The numbers • • • .001, .01, .1, 1, 10, 100, 1000, • eir logarithms • • -3, _ 9 -1, 0, 1, 2, 3,- Observe that in this table the numbers constitute a geomet- rical progression in which the common ratio is 10, and the logarithms an arithmetical progression in which the common difference is 1. The numbers in this table are the only rationals whose 743 common logarithms are rational, for all fractional powers of 10 are irrational. But, as we proceed to show, every positive number has a common logarithm, and the value of this loga- rithm may be obtained correct to as many places of decimals as may be desired. If we extract the square root of 10, the square root of the result thus obtained, and so on, continuing the reckoning in 380 A COLLEGE ALGEBRA each case to the fifth decimal figure, we obtain the- following Jable : 10^=3.16228, 10^^=1.07461, 10"' = 1.00451, 10^ = 1.77828, W' = 1.03663, 10"^* = 1.00225, 10^ = 1.33352, 10"« = 1.01815, 10=^" = 1.00112, 10^"* = 1.15478, 10"« = 1.00904, 10"'" = 1.00056, and so on, the results obtained approaching 1 as limit as we proceed (compare § 725). The exponents 1/2, 1/4, • • • on the left are the logarithms of the corresponding numbers on the right. 744 By aid of this table we may compute the common loga- rithm of any number between 1 and 10 as in the following example. Example. Find the common logarithm of 4.26. Divide 4.26 by the next smaller number in the table, .3.16228. The quotient is 1.34719. Hence 4.26 = 3.16228 x 1.34719. Divide 1.34719 by the next smaller number in the table, 1.33352. The quotient is 1.0102. Hence 4.26 = 3.16228 x 1.33352 x 1.0102. Continue thus, always dividing the quotient last obtained by the next smaller number in the table. If q,, denote the quotient in the nth division, we shall obtain by this method an expression for 4.26 in the form of a product of n numbers taken from the table and q„, the result being 4.26 = 3.16228 x 1.33352 x 1.00904 x • • • x g„ = 10* • 10^ . lO'^s . . • g„ = 10^ + s + =^= '° " '"'"' g„. As n increases, the exponent 1/2 + 1/8 + 1/256 + • • • to n terms also increases. But it remains less than 1, since it is always a part of the infi- nite series 1/2 + 1/4 + 1/8 + • • • who.se sum is 1, § 704, Ex. 1. Hence it approaches a limit which is some number less than 1, § 192. Represent this limit thus : 1/2 + 1/8+1 /256 + • • • . Again, as n increa.ses, q„ approaches 1 .as limit. For each quotient lies between the divisor used in obtaining it and 1, and as the process is continued the divi.sors approach 1 as limit. Hence 4.26 = 1*/" 10= + s + ^Js + • • • ""' '""■»(,„ = 10^ +k + '.h + ---^ and there- fore log 4.26 = I'/V+l/B + 1/256 + • • • = .6294 • • •• LOGARITHMS 381 From the common logarithms of the numbers between 1 and 745 10 we may derive the common logarithms of all other positive numbers by the addition of positive or negative integers. Example. Find the common logarithms of 42.6 and .426. 1. We have 42.6 = 10 x 4.26. Hence log 42.6 = log 10 + log 4.26 = 1 + log 4.26 = 1.6294. 2. Again, .426 = 4.26/10 = 10-1x4.26. Hence log .426 = log lO-i + log 4.26 = -1 + log 4. 26 = 1.6294. Similarly log 426 = 2.6294, log .0426 = 2.6294, and so on ; that is, we may obtain the logarithm of any number which has the same sequence of figures as 4.26 by adding a positive or negative integer to log 4.26. Characteristic and mantissa. In a logarithm expressed as in 746 the preceding example we call the decimal part the mantissa and the integral part the characteristic. Thus, log 42.6 = 1.6294 and log .426 =1.6294 have the same mantissa .6294 and the characteristics 1 and — 1 respectively. As has been shown in § 745, if n denote a positive number 747 expressed as an integer or decimal, the inaiitissa of log n depends solehj on the sequence of figures in n, a7id the charac- teristic on the 2^08 it ion of the decimal point in n. If n lies between 1 and 10, that is, if it has but one figure 748 in its integral part, the characteristic of log 7i is 0, § 744. If we shift the decimal point in such a number n one place to the right, that is, if we multiply 7i by 10, we add 1 to the characteristic of log n. Similarly if we shift the decimal point two places to the right, we add 2 to the characteristic of log n, and so on, § 745. Hence the rule : If n>l, the characteristic of log n is one less than the member offgures in the integral part of n. Thus, log 426000 = 5.6294, log 42600000 = 7.6294, and so on. 382 A COLLEGE ALGEBRA 749 In like manner, if in a number n which has but one figure in its integral part we shift the decimal point yx places to the left, that is, if we multiply n by 10"'^, we add — /a to the char- acteristic of log n. Thus, log .426 = - 1 + log 4.26 = 1.6294, log .0426 = 2.6294, and so on, § 745. In practice we find it convenient to write these negative characteristics 1, 2, • • ■ in the form 9 — 10, 8 — 10, • • •, and to place the positive part 9, 8, • ■ • before the mantissa, and the - 10 after it. Thus, instead of 1.6294 we write 9.6294 - 10. Hence the rule : Ifnr, ()(■)( )5 6675 6684 6693 6702 6712 47 6721 6730 6739 6749 ()7r]S 6767 6776 6785 6794 6803 48 6812 6821 6830 6S.",9 6S4K 6857 6866 6875 6884 6893 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093 7101 7110 7118 7126 7135 7143 7152 62 7160 7168 7177 7185 7193 7202 7210 7218 7226 7235 53 7243 7251 7259 7267 7275 7284 7292 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7372 7380 7388 7396 LOGARITHMS 385 N 1 a 3 ^ 5 6 7 8 9 55 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 755!) 7566 7574 7582 7589 7597 7604 7612 7619 7627 58 7684 7642 7649 7667 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 7760 7767 7774 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 7973 7980 7987 63 7993 8000 8007 8014 8021 8028 8035 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8625 8531 8537 8543 8549 8555 8561 8567 72 8573 8579 8585 8591 8597 8003 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9730 9741 9745 9750 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 386 A COLLEGE ALGEBRA 752 To find a number when its logarithm is given. We have merely to reverse the process described in the preceding section. Example 1. Find the number whose hjgarithm is 5.9552 — 10. We find the mantissa 9552 in the table in the row marked 90 and iu the column marked 2. Hence the required sequence of figures i.s 902. But since the characteristic is 5 — 10, the number is a decimal with 9 — 5, or 4, O's at the right of the decimal point, § 749. Hence the required number is .0000902. Example 2. Find the number whose logarithm is 7.5520. Looking in the table we find that the given mantissa 5520 lies between the mantissas 5514 and 5527 corresponding to 356 and 357 respectively; The lesser of these mantissas, 5514, differs from the greater, 5527, by 13 and from the given mantissa, 5520, by 6. Thus, if we add 13 to the mantissa 5514, we add 1 to the number 356, Hence if we add 6 to the mantissa 5514, we should add 6/13 of 1, or .$ approximately, to 356. Hence the required sequence of figures is 3565, and therefore by the rule for characteristic, § 748, the required number is 35650000. We therefore have the following rule for finding the sequence of figures corresponding to a given mantissa which is not in the table : Fi7id front the table the next lesser mantissa m, the three corresponding figures, and d, the difference between m and the next greater mantissa. ; Subtract va. from the given mantissa and divide the remainder by d, an7iexi7ig the resulting figure to the three figures already obtained. 753 Cologarithms. The cologarithm. of a number is the logarithm of the reciprocal of the number. Since colog m = log l/m=z log l — \ogm = — log m, §§ 733, 737, we can find the cologarithm of a number by merely changing the sign of its logarithm. But to avail ourselves of the table we must keep the decimal parts of all logarithms positive. We therefore proceed as follows : LOGARITHMS 387- Example 1, Find colog 89.2. We have log 1 = 10 — 10 and log 89. 2= 1.9504 Hence colog 89.2= 8.0496-10 Example 2. Find colog. 929. We have log 1 = 10 - 10 and log .929 = 9.9680 - 10 Hence colog. 929= .0320 Hence we may find the cologarithm of a number from its logarithm by beginning at the characteristic and subtracting each figure from 9 until the last significant figure is reached, which figure must be subtracted from 10. To this result we do or do not afiix — 10 according as — 10 is not or is affixed to the logarithm. In this way when the number has not more than three figures we may obtain its cologarithm directly from the table. Computation by logarithms. The following examples will 754 serve to show how expeditiously approximate values of prod- ucts, quotients, powers, and roots of numbers may be obtained by aid of logarithms (compare § 740). Example 1. Find the value of .0325 x .0425 x 5.26. Log (.0325 X .6425 x 5.26) = log .0325 + log .6425 + log 5.26. But log. 0325= 8.5119-10 log .6425= 9.8079-10 log 5.26 = .7210 Hence log of product = 19.0408 - 20 = 9.0408 - 10 Therefore the product is .1099. Example 2. Find the value of 46. 72/. 0998. Log (46. 72/. 0998) = log 46.72 - log. 0998. But log 46.72 = 11.6695 - 10 log .0998 = 8.9991 - 10 Hence log of quotient = 2.6704 Therefore the quotient is 468.2. 388 A COLLEGE ALGEBRA We write log 46.72, that is 1.6G95, in the form 11.6695 - 10 in order to make its positive part greater than that of 8.9991 — 10 which is to be subtracted from it. Example 3. Find the value of 295 x .05631 -^ 806. Log (295 X .05631 -- 806) = log 295 + log .05631 + colog 806. But log 295= 2.4698 log .05631 = 8.7506-10 colog 806= 7.0937 -10 Hence log of required result = 18.3141 - 20 = 8.3141 - 10 Therefore the required result is .02061. Example 4. Find the sixth power of .7929. Log (.7929)6 = 6 X log .7929. But log .7929 = 9.8992 - 10 Hence log (.7929)6 = 59.3952 _ 60 = 9.3952 - 10 Therefore (.7929)6 ^ .2484. Example 5. Find the seventh root of .( Log V.00898 = (log .00898) -4- 7. But log .00898 = 7.9533 - 10 7 )67.9533-70 Hence log V.00898 = 9.7076-10 Therefore v.00898 = .510. Observe that when as here we have occasion to divide a negative loga- rithm by some number, we add to its positive and negative parts such a multiple of 10 that the (luotient of the negative part will be — 10. Negative numbers do not have real logarithms to the base 10 since all real powers of 10 are positive numbers. If asked to find the value of an expression which involves negative factors, we may first find the absolute value of the expression by logarithms and then attach the appropriate sign to the result. Thus, if the given expression were 456 x ( - 85.96), we should first find the value of 450 x 85.96 by logarithms and then attach the — sign to the result. LOGARITHMS EXERCISE LXni Find approximate values of tlie following by aid of logarithms. 1. 79 X 470 X .982. 2. (- 9503) X (- .0086578). 3. 1375600 X 8799000. 4. .0356 X (- .00049). n 8075 ^- 364.9' ^• .00542 24617 .04708 - .00054 .643 X 7095 ■ 67 X 9 X .462 9097 X 5.4086 - 225 X 593 X .8665 10. (2.388)5. 11. (.57)-". 12. (19/11)9. 13. (1.014)26. 14. V67.54. 15. V- .30892. 16. 8l 17. (.001)3. 18. (29t\)'. 19. V| X -v/ll. 20. VjL- (.009)1 21. (.00068)-^ 22. (6|)3-'. 23. (-9306)'. 24. (.0057)2-5. 25. (5648)^ X (- .94)3. 26. 289273 - (.8)1 V.0476 X V222 „„ V943 X 7298 ^5059 X .0088 V.00006 X .99 ^g / 854 X V.042 \ 7.9856 X V.0005 3Q 3/7*x92^x(.01)J \ (.00026)5 X 5968i 389 SOME APPLICATIONS OF COMMON LOGARITHMS Logarithms to other bases than 10. From the logarithm of a 755 number to the base 10 we can derive its logarithm to any- other positive base except 1 by aid of the theorem : The logarithms of a number m to two different bases, a and b, are connected by the formula log,, m = log^ m/log^ b. For let 7/1 = a**, that is /i = log„ 7?i, and let b = a'', that is ;/ = log„6. 1 Since a" = b, we have a = b^. 1 M. Hence 7?i = a*^ = (6'')'^ = 6", that is, logi, m — fx/v — loga m / loga 6. 390 A COLLEGE ALGEBRA Example. Find the logarithm of .586 to the base 7. Tna ^«fi logio .586 _ 9.7679- 10 _ 2321 _ _ We reduce 9.7679 - 10 to the form of a single negative number, namely — .2.321, and perform the final division by logarithms. 756 When m = a the formula gives log^a = l/log^b. 757 The only base besides 10 of which any actual use is made is a certain irrational number denoted by the letter e whose approximate value is 2.718. Logarithms to this base are called natural logarithms. We shall consider the;m in another connection. 758 Exponential and logarithmic equations. Equations in which the unknown letter occurs in an exponent or in a logarithmic expression may sometimes be solved as follows. Example 1. Solve the equation IS^^+s = 14^+7. Taking logarithms of both members, (2 x + 5) log 13 = (x + 7) log 14. o , . 7 log 14 - 5 log 13 2.4532 „ _„ Solving, X = 2 5 — = = 2.268. ^' 2 log 13 - log 14 1.0817 Example 2. Solve the equation log Vx — 21 + | log x = 1. By §§ 736, 739, we can reduce this equation to the form log Vx(x-21) = 1 = log 10. Hence x2 _ 21 x = 100. Solving, X = 25 or — 4. Example 3. Solve the equation x-^°s^ = 10 x. Taking logarithms, 2 (log x)2 = log x + 1. Solving for log X, logx = l or —1/2. Hence x = 10 or 1/VlO. 759 Compound interest. Suppose that a sum of P dollars is put at compound interest for a period of n years, interest being compounded annually and the interest on one dollar for one year being r. Then the amount at the end of the first year will be P + Pr or P (1 + r), at the end of the second year it will LOGARITHMS 391 be P(i + r) • (1 + r) or P(l + ?•)^ and so on. Hence, if A denote the amount at the end of the nth year, we have I A^P(l-\-ry. If interest be compounded semiannually, A = P (1 -{- r/2)^"; if quarterly, A = P (I + r /AY" ; and so on. We call P the 2irese7it worth of A. If A, n, and r be given, we can find P by means of the formula P = A(\ + r)'". Example 1. Find the amount of §2500 in eighteen years at 4% com- pound interest. We have log A = log 2500 + 18 log 1.04 = 3.7039. Hence A = $5057, approximately. Example 2. At the beginning of each of ten successive years a premium of $120 is paid on a certain insurance policy. What is the worth of the sum of these premiums at the end of the tenth year if computed at 4% compound interest? The required value is 120 [1.04 +(1.04)2 + . .. ^(i.o4)io], that is, by § 701, 120 x 1.04 x ^^ 1.04 — 1 By logarithms, (1.04)1" - 1.479. Hence the required value is 120 x 1.04 x .479 h- .04; and this, com- puted by logarithms, gives $1494, approximately. Annuities. A sum of money which is to be paid at fixed 760 intervals, as annually, is called an annuity. It is required to find the present worth of an annuity of A dollars payable annually for n years, beginning a year hence, the interest on one dollar for one year being r. The present worth of the first payment is ^ (1 + r)~S that of the second payment is A (1 + r)"^, and so on. Hence the present worth of the whole is, § 701, If the annuity hejierpetual, that is, if n = oo, then (1 + r)" = oo, and the formula for the present worth reduces to A /r. 392 A COLLEGE ALGEBRA Example. What sum should be paid for an annuity of $1000 payable annually for twenty years, money being supposed to be worth 3% per annum ? The present worth, P, is 1 • .03 L (L03)2oJ By logarithms, we find that (1.03)2o = 1.803. „ „ 1000 r, 1 1 1000 X .803 ^,,„,, Hence P = 1 = = $14845, approximately. .03 L 1.803J .03 X 1.803 , h-f ^ J EXERCISE LXIV 1. Find logs 555, logy. 0463, logioo47. 2. Solve the following exponential equations. (1)3^ = 729. (2) a^= + 2 = a3^. (3)213^ = 516-^+4. 3. Solve the following logarithmic equations. (1) log X + log (X + 3) = 1. (2) log x2 + log X = 2. (3) log (1-2 x)3 - log (3 - x)3 = 6. (4) x'^g^ = 2. 4. Find the amount of $7500 in thirty-five years at 5% compound interest, the interest being compounded annually. 5. Find the amount of $5500 in twenty years at 3% compound interest, the interest being compounded semiannually. 6. Show that a sum of money will more than double itself in fifteen years and that it will increase more than a hundredfold in ninety-five years at 5% compound interest. 7. What sum will amount to $1250 if put at compound interest at 4% for fifteen years? 8. A man invests $200 a year in a savings bank which pays S},% per annum on all deposits. What will be the total amount due him at the end of twenty-five years ? 9. What sum should be paid for an annuity of $1200 a year to be paid for thirty years, money being supposed to be worth 4% per annum ? What sum should be paid were this annuity to be perpetual ? 10. If c denote the length of the hypotenuse of a right-angled triangle and a, b the lengths of the other two sides, b = V(c + a)(c — a). Given c = 586.4, a = 312.2, find b and the area of the triangle, using logarithms. TERMUTATIOXS AND COMBIXATIONS 393 11. If ft, 6, c denote the lengths of the sides of a triangle and s = {a + b + c)/2, the area of the triangle is \''s{s — a){s — b) {s — c). Find the area of the triangle in which a = 410.8, b — 424, c = 25.68. 12. Find the area of the surface and the volume of a sphere the length of whose radius is 23.6 by aid of the formulas S — 4i7tr'\ F=4 7rr*/3, assuming that jr = 3.1416. XXV. PERMUTATIONS AND COMBINATIONS Definitions of permutation and combination. Suppose a group 761 of n letters to be given, as a, b, c, ■■■, k, denoting objects of any kind. Any set of r of these letters, considered without regard to order, is called a combination of the n letters t at a time, or, more briefly, an v-covibinution of the n letters. We shall use the symbol C" to denote the number of such combinations. Thus, the 2-combinations of the four letters a, &, e, d are ab, ac, ad, be, bd, cd. There are six of these combinations, that is, Ct = 6. On the other hand, any arrangement of r of these n letters in a definite order in a row is called a permutation of the n letters, r at a time, or, more briefly, an T-perynutation of the 71 letters. We shall use the symbol P". to denote the number of such permutations. Thus, the 2-permutations of the four letters a, b, c, d are ab, ac, ad, be, bd, cd, ba, ca, da, cb, db, dc. ^ There are twelve of these permutations, that is P^ = 12. ~^ Observe that while ab and ba denote the same combination, they denote different permutations. In what has just been said it is assumed that the letters a. b, ■ ■ ■, k are all different and that the repetition of a letter 394 A COLLEGE ALGEBRA within a permutation or combination is not allowed. This will be the understanding throughout the chapter except where the contrary is stated. 763 A preliminary theorem. AVe have already had occasion to apply the following principle, § 554 : If a certain thing can he done in m waijs, and if, vlien it has been done, a certain other thing can he done in n loays, the entire number of ways in which both things can be done in the order is mn. We reason thus : Since for each way of doing the first thing there are n ways of doing both things, for m ways of doing the first thing there are mn ways of doing both things. More generally, if a first thing can be done in m ways, then a second thing in n ways, then a third in p ways, and so on, the entire number of ways in which all the things can be done in the order stated is 7n -n-jy ■ •■• Example. How many numbers of three different figures each can be formed with the digits 1, 2, 3, • • •, 9 ? We may choose any one of the nine digits for the first figure of the number, then any one of the remaining eight digits for its second figure, and finally any one of the seven digits still remaining for its third figure. Kence we may form 9 ■ 8 • 7, or 504, numbers of the kind required. 763 The number of r-permutations of n different letters. By the reasoning employed in the preceding example we readily prove that this number P'^ is given by the formula P'^ — n(n — 1) (n — 2) ■•■ to r factors. (1) For in forming an ?'-permutation of n letters we may choose any one of the n letters for its first letter, then any one of the remaining n — 1 letters for its second letter, then any one of the n — 2 letters still remaining for its third letter, and so on. Hence, § 762, the entire number of ways in which we may choose its first, second, third, ■ • • , rth letters, in other word? PERMUTATIONS AND COMBINATIONS 395 the entire number of ways in which we may form an /--permu- tation with the n letters, is n (« — 1) (w — 2) • • • to r factors. Thus, the numbers of permutations of t^^ letters a, 6, c, d, e one, two, three, four, five at a time are P\ = b, P| = 5 • 4, P-] = 5 ■ 4 • 3, P] = 5 • 4 • 3 • 2, P^ = 5 • 4 • 3 • 2 • 1. Evidently the rth factor in the product, n {n — 1) (?i — 2) • • • is ?i — (r — 1), or n — r + 1. Hence the formula (1) may be written p» = n (n - 1) (^ -x2) ...(«- r + 1). (2) When r = n, the factor n — r -\- 1 is n — n + 1, or 1, and we have P;; = n (n — 1) • • • 2 • 1, or 1 ■ 2 ■ • • (/i — 1) n. The con- tinued product 1-2 ■ • ■ n is called factorial n and is denoted by the symbol n ! or [n^. Hence the efi^ire number of orders in which we can arrange n letters in a row, using all of them in each arrangement, is given by the formula Pl = nl (3) For a reason which will appear later, § 775, the meaningless symbol 0! is assigned the value 1. Example 1. How many different signals can be made with four flags of different colors displayed singly, or one or more together, one above another ? There will be one signal for each arrangement of the flags taken 1, 2, 3, or 4 at a time. Hence the number is Pj + P| -(- P3 + P^, or 64. Example 2. Of the permutations of the letters of the word fancies taken all at a time, (1) How many begin and end with a consonant ? The first place may be filled in 4 ways, then the last place in 3 ways, then the intermediate places in 5 ! ways. Hence the required number is 4 • 3 • 5 !, or 1440. (2) How many have the vowels in the even places ? The vowels may be arranged in the even places in 3 ! ways, the con- sonants in the odd places in 4 ! ways, and each arrangement of vowels may be associated with every arrangement of consonants. Hence the required number is 3 ! • 4 !, or 144. 396 A COLLEGE ALGEBRA (3) How many do not have c as their middle letter? Evidently c is the middle letter in ! of the permutations, for the remaining letters may be arranged in all possible orders. Therefore the number of the pernmtatious in which c is not the middle letter is 7 ! — 6 !, or 4320. Example 3. Show that P^ = 4 ■ Pl, and that P^^ -2- P^. Example 4. If P\" = 127 Pl", find n. Example 5. How many passenger tickets will a railway company need for use on a division on which there are twenty stations? Example 6. In how many of the permutations of the letters a, e, i, o, u, y,. taken all at a time, do the letters a, e, i stand together ? .. Example 7. With the letters of the word numerical how many arrange- ments of five letters each can be formed in which the odd places are occupied by consonants? Example 8. Show that with the digits 0, 1, 2, • • • , 9 it is possible to ( form P'^" — Pg numbers, each of which has four different figures. Example 9. How many numbers all told can be formed with the digits 3, 4, 5, 7, 8, all the figures in each number being different ? Example 10. In how many ways can seven boys be arranged in a row if one particular boy is not permitted to stand at either end of the row ? 764 Circular permutations. The number of different orders in which n different letters can be arranged about the circum- ference of a circle or any other closed curve is (n — 1)!. For the relative order of the n letters will not be changed if we shift all the letters the same number of places along the curve. Hence Ave shall take account of all the distinct orders of the n letters if we suppose one of the letters fixed in posi- tion and the remaining n — 1 then arranged in all possible orders. But these n — 1 letters can be arranged in {ii — 1)! orders, § 763, (3). Thus, eight persons can be seated at a round table in 7 !, or 5040, orders. Exam])le 1. Show that the number of circular r-permutations of n different letters is P'^/r. PERMUTATIONS AND COMBINATIONS 397 Example 2. Taking account of the fact that a circular ring will come into coincidence with itself if revolved about a diameter through an angle of 180°, show that (n — 1) !/2 different necklaces can be formed by stringing . together n beads of different colors. Example 3. In how many ways can a party of four ladies and four gentlemen be arranged at a round table so that the ladies and gentlemen may occupy alternate seats? Permutations of different letters when repetitions are allowed. 765 With n different letters we can form n'' arrangements or per- mutations of r letters each, if allowed to repeat a letter within a permutation. For in forming a permutation of this kind we may choose any one of the n letters for its first letter, and then again, since repetitions are allowed, any one of the n letters for its second letter, and so on. Hence, § TG2, the entire number of ways in which we can form tlie permutation is » • « • n • • • to r factors, or if. Thus, with the digits 1, 2, .3, • • -, 9 we can form all told 93, or 729, numbers of three figures each. - — Example 1. How many numbers of one, two, or three figures each can be formed with the characters 1, 2, 3, 5, 7 ? Example 2. In how many ways can three prizes be given to seven boys if each boy is eligible for e\ ery prize ? The number of n-permutations of n letters which are not all 766 different. Let us inquire how many distinguishable permuta- tions can be formed with the letters a, a, a, b, c(l), three of which are alike, all the letters being used in each permutation. Compare these permutations with the corresponding permu- tations of the letters a, a', a", b, c(2), all of which are different. If we take any one of the permutations of (1), as abaca, and, leaving b, c undisturbed, we interchange the a's, we get nothing new. But if we treat the corresponding permutation of (2), namely aba'ra", in a similar manner, we obtain 3 ! distinct per- mutations, namely aba'ca", aba"ca', a'ba"ca, a'baca", a"haca', a"ba'ca. Hence to each permutation of (1) there correspond 398 A COLLEGE ALGEBRA 3 ! permutations of (2). The number of the permutations of (2) is 5 !, § 763, (3). Therefore the number of the permutations of (1) is 5! -3!. By the reasoning here employed we can prove in general that the number of distinguishable ?i-permutations of n letters of which p are alike, q others alike, and so on, is given by the formula , 2)\q\--- Example 1. In how many different ways can the letters of the word independence be arranged ? Of the 12 letters in this word 4 are e's, 3 are n's, 2 are d's. Hence the required result is 12 !/4 ! • 3 ! • 2 !, or 1,663,200. Example 2. In how many ways can the letters of the word Antioch be arranged without changing the relative order of the vowels or that of the consonants ? From the proof just given it is evident that the required number of arrangements is the same as it would be if the three vowels were the same and the four consonants were the same. Hence it is 7 !/3 ! • 4 !, or 35. Example 3. How many terms has each of the following symmetric functions of five variables x, y, z, u, v, namely, Sx^y^z, Zx^y'^z^, Zx^yzu, and Xx^y^z^u^ ? We shall obtain all the terms of Hx^y'^z once each if, leaving the expo- nents 3, 2, 1 fixed in position, we write under them every 3-permutation of the letters x, y, z, u, v. Hence the number of the terms is P^, or 60. If we apply the same method to :^x^y-z'^, we obtain the term x^y-z^ twice, once in the form x^y^z^ and once in the form x^z-y'^. Similarly every term is obtained twice, once for each of the orders in which its letters under the equal exponents can be written. Hence the number of terms in 'Lx^y-z'^ is Pi/ 2, or 30. Similarly -Zx^yzu has P]/3 !, or 20, terms, and Sx^^/V-it^ has P^/2 !2 !, or 30, terras. Example 4. In how many ways can five pennies, six five-cent pieces, and four dimes be distributed among fifteen children so that each may receive a coin ? Example 5. In a certain district of a town there are ten streets run- ning north and south, and five running east and west. In how many ways can a person walk from the southwest corner of the district to the northeast corner, always taking the shortest course ? PERMUTATIONS AND COMBINATIONS 399 The number of r-combinations of n different letters. This niim- 767 ber, C", is given by the forinvila For evidently if wq were to form all the r-combinations and were then to arrange the letters of each combination in turn in all possible orders, we should obtain all the ^--permutations. But since each combination would thus yield r! permuta- tions, § 763, (3), all the combinations, C" in number, would yield r ! x C" permutations. Hence r ! x C" = P", and therefore C" = P" h- r !. Thus, the numbers of combinations of the letters a, 6, c, d, e one, two, three, four, five at a time are ^ r ^ 1.2' ^ 1.2-3' * 1.2.3-4' ' 1.2.3.4-5" This expression for C" is the coefficient of the (r -f l)th term in the expansion of (a -f- by by the binomial theorem, § 565. This was shown in § 560 by an argument which is merely another proof of the formula (1). If in the expression just obtained for C" we multiply both 768 numerator and denominator by (>i — r) !, we obtain the more symmetrical formula From this formula (2) it follows that the number of the 769 r-combinations of n letters is the same as the number of the {11 — ?-)-combinations. n ! n ! or C,\_^ = ^;;;3-;yr^^^;3-^^^3^^ = ^^737^177 ^ ^"• This also follows from the fact that for every set of ?'-things taken, a set of /i — r things is left. Thus, C\t - CV = 14 • 13/1 • 2 = 91. Observe how much more readily C}j is found in this way than by a direct application of (1). 400 A COLLEGE ALGEBRA Example 1. There are fifteen points in a plane and no tlu'ee of these points lie in the same straight line. Find the number of triangles which can be formed by joining them. Evidently there are as many triangles as there are combinations of the points taken three at a time. Hence the number of triangles is C'3 , that is, 15 -14 -13/ 1-2. 3, or 455. Example 2. In how many ways can a committee of three be selected from ten persons H) so as always to include a particular person A ? (2) so as always to exclude A ? (1) The other two members of the committee can be chosen from the remaining nine persons in C^, that is, 9 • 8/1 • 2, or 36, ways. (2) The entire committee can be chosen from the remaining nine persons in C^, that is, 9 • 8 • 7/1 • 2 • 3, or 84, ways. Example 3. With the vowels a, e, i, and the consonants 6, c, d, /, g how many arrangements of letters can be made, each consisting of two vowels and three consonants? The vowels for the arrangement can be chosen in C\ ways, the con- sonants in Cgways; then each selection of vowels can be combined with every selection of consonants and the whole arranged in 5 ! ways. Hence the required result is C| • C3 • 5 !, or 7200. Example 4. In how many ways can eighteen books be divided equally among three persons A, B, C ? A's books can be selected in C^,? ways, then B's in C^^ ways, then C's in C;i, or 1, way. Hence, §702, the required result is C^^ ■ C'J • C% or 18!/(6!)3, To find in how many ways the 18 books can be distributed into three sets of 6 books each, we must divide the result just obtained by 3 !, which gives 18 !/(() !)3 3 ! ; for liere the order in which the three sets may chance to be arranged is immaterial. Example 5. "With the letters of the word mathematical how many different selections and how many different arrangements of four letters each can be made ? As the letters are not all different wp cannot obtain the required results by single applicatii)ns of the fornuila.s f(u- C" and P'^. The letters are a, n, a ; ?», m ; t, t; h, e, i, c, I. Hence we may classify and then enumerate the possible selections and arrangements as follows : PERMUTATIONS AND COMBINATIONS 401 1. Those having three like letters. Combining the 3 a's with each of the seven other letters in turn, we obtain 7 selections and 7 • 4 !/3 !, or 28, arrangements. 2. Those having two pairs of like letters. There are 3 such selections and 3 • 4 ! / 2 ! • 2 !, or 18, such arrangements. 3. Those having two letters alike, the other two different. Of such selections there are 3 • C^, or 63 ; of arrangements there are 63-4I/2!, or 756. 4. Those having four different letters. Of selections there are C% or 70 ; of arrangements, 70 • 4 !, or 1680. Hence the total number of selections is 7 + 3 + 63 + 70, or 143 ; of arrangements, 28 + 18 + 756 + 1680, or 2482. Example 6. Find the values of Cj^, C'!*, and Cj;]. Example 7. If C^ = C'-', find n. Example 8. If 2 C] = 5 • q, find n. Example 9. How many planes are determined by twelve points, no four of which lie in the same plane ? Example 10. How many parties of five men each can be chosen from a company of twelve men ? In how many of these parties will a particular man A be included ? From how many will A be excluded ? Example 11. Of the parties described in the preceding example how many will include two particular men A and B ? How many wilL— -^ include one but not both of them ? How many will include neither of them ? Example 12. From twenty Republicans and eighteen Democrats how many committees can be chosen, each consisting of four Republicans and three Democrats ? Example 13. "With five vowels and fourteen consonants how many arrangements of letters can be formed, each consisting of three vowels and four consonants? Example 14. In how many ways can a pack of fifty-two cards be divided equally among four players A, B, C, D ? In how many ways can the cards be distributed into four piles containing thirteen each ? Example 15. How many numbers, each of five figures, can be formed with the characters 2, 3, 4, 2, 5, 2, 3, 6, 7 ? 402 A COLLEGE ALGEBRA 770 Total number of combinations. If in the formula for (a + by, § 561, we set a = b =^1 and then subtract 1 from both mem- bers, we obtain CI+CI + ---+ CI = 2" - 1. Hence the total number of combinations of n different things taken one, two, •••, n at a time, in other words, the total number of ways in which one or more things may be chosen from n things, is 2" — 1. This may also be proved as follows : Each particular thing can be dealt with in one of two ways, that is, be taken or left. Hence the total number of ways of dealing with all n things is 2-2 ■•-.to n factors, or 2", § 762. Therefore, rejecting the case in which all the things are left, we have as before, 2" — 1. Example 1. How many different sums of money can be paid with one dime, one quarter, one half dollar, and one dollar ? Example 2. By the reasoning just illustrated, show that the total num- ber of ways in which one or more things can be chosen from p + q + ■ • • things of which p are alike, q others alike but different from the p things, and so on, is {p + 1) (q + !)• • • — 1. i.^ Example 3. How many different sums of money can be paid with two dimes, five quarters, and four half dollars? 771 Greatest value of C°. In the expression for C", namely n(?i — 1) • • • (?i — r + 1) /rl, the r factors of the numerator decrease while those of the denominator increase. Hence for a given value of w the value of C" will be greatest when the next greater value of r will make (n-r-\-l)/r b, that the odds are a to i in favor of the PROBABILITY 411 event ; when b > a, that the odds are b to a against the event; when a = b, that the odds are even on tlie event. In the first case the probability of the event, namely, a / (a + b), is greater than 1/2; in the second it is less than 1/2 ; in the third it is equal to 1/2. Thus, if a ball is to be drawn from a bag containing five balls, three white and two black, the odds are 3 to 2 in favor of its being white, and 3 to 2 against its being black. Expectation, li jj denote the chance that a person will win 780 a certain sum of money M, the product Mp is called the value of his expectation so far as this sum M is concerned. Thus, the value of the expectation of a gambler who is to win $12 if he throws an ace with a single die is $12 x 1/6, or $2. Examples of probability. In applying the definition of proba- 781 bility, § 776, care must be taken to reduce the possible cases to such as are equally likely. The following examples will illustrate the need of this precaution. Example 1. If two coins be tossed simultaneously, what is the chance that the result will be two heads ? two tails ? one head and one tail ? We might reason thus : There are three possible cases, one favoring the first result, one the second, one the third ; hence tlie chance of each result is 1/3. But our conclusion would be false, since the number of equally likely possible cases is not three but four. For if we name the coins A and B respectively, the equally likely cases are : A head, B head ; A tail, B tail ; A head, B tail; A tail, B head. And since one of these cases favors the result two heads, one the result two tails, and two the result one head and one tail,' the chances of these results are 1/4, 1/4, and 2/4 respectively. Example 2. What is the chance of throwing a total of eight with two dice ? Here the number of equally likely possible cases is 6 • G, or 36, for any face of one die may turn up with any face of the other die. We have a total of eight if the faces which turn up read 2, 6 or 3, 5 or 4, 4. But there are two ways in which 2, 6 may turn up, namely 2 on the 412 A COLLEGE ALGEBRA die A and 6 on the die B, or vice versa. Similarly there are two ways in which 3, 5 may turn up. On the contrary, there is but one way in which 4, 4 can turn up. Hence there are five favorable cases. Therefore the chance in question is 5/36. Example 3. What is the chance of throwing a total of eight with three dice if at least one die turns ace up ? The number of equally likely possible cases is 6 • 6 ■ 6, or 216. We have a total of eight if the faces which turn up read 1, 1, 6 or 1, 2, 5 or 1, 3, 4. But 1, 1, 6 may turn up in 3 !/2 !, or 3, ways, § 766 ; for the numbers 1, 1, G may be distributed among the three dice in any of the orders in which 1, 1, 6 can be written. Similarly 1, 2, 6 and 1, 3, 4 may each turn up in 3 !, or 6, ways. Hence there are 3 + 6 + 6, or 15, favor- able cases. Therefore the chance in question is 15/216, or 5/72. Example 4. An urn contains six white, four red, and two black balls. (1) If four balls are drawn, what is the chance that all are white ? There are as many ways of drawing four white balls as there are 4-com- binations of the six white balls in the urn, namely C^. Similarly, since the urn contains twelve balls all told, the total number of possible drawings is C\^. Hence the chance in question is Cl/C^^, or 1/33. (2) If six balls are drawniwhat is the chance that three of them are white, two red, and one black ? The three white balls can be chosen in C^ ways, the two red balls in Ct ways, the one black ball in Cf ways. Hence the number of ways in which the required drawing can be made is C^- C^- Cf. The total number of possible drawings is C',?. Hence the chance in^uestion is Cg • C| ■ C\/ Cg^, or 20/77. Example 5. Three cards are drawn from a suit of thirteen cards. (1 ) What is the chance that neither king nor queen is drawn ? Aside from the king and queen there are eleven cards. Hence there are Cy sets of three cards which include neither king nor queen. Therefore the probability in question is C\j / C^^, or 15/26. (2) What is the chance that king or queen is drawn, one or both? This event occurs when the event described in (1) fails to occur. Hence the probability in question is 1 - 15/26, or 11/26, § 778. (3) Wliat is the chance that both king and queen arc drawn? We obtain every set of three cards which includes both king and queen if we combine each of the remaining eleven cards in turn with king and queen. Hence the chance in question is ll/C'g, or 1/26. PROBABILITY 413 On the various meanings of probability. 1. The fraction o/m, 782 which we have called the probability of an event, § 776, means nothing so far as the actual outcome of a single trial, or a small number of trials, of the event is concerned. But it does indicate the frequency with which the event would occur in the long run, that is, in the course of an indefinitely long series of trials. Thus, if one try the experiment of throwing a die a very great number of times, say a thousand times, he will find that as the number of throws increases the ratio of the number of times that ace turns up to the total number of throws approaches the value 1/6 more and more closely. 2. There are important classes of events — the duration of life is one — to which the definition of § 776 does not apply, it being impossible to enumerate the ways, all equally likely, in which the event can happen or fail. But we may be able to determine the frequency with which events of such a class have occurred in the course of a very great number of ^>asi^ trials. If so, we call the fraction which indicates this frequency the probability of an event of the class. Like 1/6 in the case of the die, it indicates the frequency with which events of the class may reasonably be expected to occur in the course of a very great number of future trials. Thus, if we had learned from the census reports that of 100,000 persons aged sixty in 1880 about 2/3 were still living in 1890, we should say that the probability that a person now sixty will be alive ten years hence is 2/3. 3. But we also use the fraction a/ m to indicate the strength of our expectation that the event in question will occur on a single trial. The greater the ratio of the number of favor- able cases to the number of possible cases, or the greater the frequency with which, to our knowledge, events of a similar character have occurred in the past, the stronger is our expec- tation that this particular event will occur on the single trial under consideration. 414 A COLLEGE ALGEBRA We may speak of the probability, in this sense, of any kind of future event. Thus, before a game between two football teams, A and B, we hear it said that the odds are 3 to 2 in favor of A's winning, or that the probability that A will win is 3/5. This means that the general expectation of A's winning is about as strong as one's expectation of drawing a white ball from an urn known to contain five balls three of which are white. EXERCISE LXVII 1. The probability of a certain event is 3/8. Are the odds in favor of the event or against it, and what are these odds ? What is the prob- ability that the event will not occur ? 2. The odds are 10 to 9 in favor of A's winning a certain game. What is the chance of his winning the game ? of losing it ? 3. The odds are 5 to 3 in favor of A's winning a stake of $G0. What is his expectation ? 4. The French philosopher D'Alembert said : "There are two possible cases with respect to every future event, one that it will occur, the other that it will not occur. Hence the chance of every event is 1 /2 and the definition of probability is meaningless." How should he be answered ? 5. An urn contains sixteen balls of which seven are white, six black, and three red. (1) If a single ball be drawn, what is the chance that it is white? black? red? (2) If two balls be drawn, what is the chance that both are black? one white and one red ? (3) If three balls be drawn, what is the chance that all are red? none red? one white, one black, one red? (4) If four balls be drawn, what is the chance that one is white and the rest not ? two white and the other two not ? (5) If ten balls be drawn, what is the chance that five are white, three black, and two red ? 6. What is the chance of throwing doublets with two dice ? with > three dice ? 7. What is the chance of throwing a total of seven with two dice ? Show that this is the most probable throw. PROBABILITY 415 8. What is the chance of throwing at least one ace in a throw with two dice ? of throwing one ace and but one ? 9. One letter is taken at random from each of the words factor and banter. What is the chance that the same letter is taken from each ? 10. A box contains nine tickets numbered 1, 2, • • • , 9. If two of the tickets be drawn at random, what is the chance that the product of the numbers on them is even ? odd ? i^'ll. If five tickets be drawn from this box, find the chance (1) that 1, 2, and 3 are drawn ; (2) that one and but one of 1, 2, and 3 is drawn ; (3) that none of these numbers is drawn. 12. If four cards be drawn from a complete pack of fifty-two cards, what is the chance that they are ace, king, queen, and knave ? ace, king, queen, and knave of the same suit ? I J:3. What is the chance that a hand at whist contains four trumps and three cards of each of the remaining suits ? 14. What is the chance of a total of five in a single throw with three dice ? of a total of less than five ? 15. If eight persons be seated at a round table, what is the chance that two particular persons sit together ? COMPOUND EVENTS. MUTUALLY EXCLUSIVE EVENTS Independent events. Two or more events are said to be inde- 783 pendent when the occurrence or non-occurrence of any one of them is not affected by the occurrence or non-occurrence of the rest. In the contrary case the events are said to be interdependent. Thus, the results of two drawings of a ball from a bag are independent if the ball is returned after the first drawing, but interdependent if the ball is not returned. Theorem 1. The prohabiUty that all of a set of independent 784 events will occur is the produc t of the j)robahilities of the single _-> events. '" For consider two such events whose probabilities are «i/mi and ag/wg respectively. 416 A COLLEGE ALGEBRA The number of equally likely possible cases for and against the first event is ?»i, for and against the second vi^, and since the events are independent any one of the m^ cases may occur with any one of the w^ cases. Hence the number of equally likely possible cases for and against the occurrence of both events is inim as was to be demonstrated. The proof for the case of more than two events is similar. The demonstration applies only to events of the kind described in § 776, but for the reasons indicated in § 782 we may apply the theorem itself to any kind of future event, as in Ex. 2 below. Thus, the chance of throwing ace twice in succession with a single die is 1/6 X 1/6, or 1/36. Again, the chance of twice drawing a white ball from a bag which contains five white and four black balls, the ball first drawn being returned before the second drawing, is 5/9 x 5/9, or 25/81. Theorem 2. If the probability of a first event is pi, and if after this event has happened the probahility of a second event is p2, the probability thai both events will occur in the order stated is piP2. And similarly for more than two events. This theorem may be proved in the same manner as Theorem 1. It evidently includes that theorem. Thus, after a white ball has been drawn from a bag containing five white and four black balls, and not replaced, the chance of drawing a second white ball is 4/8. Hence the chance of twice drawing a white ball when the one first drawn is not replaced is 5/9 x 4/8, or 5/18. Example 1. What is the chance that ace will turn up at least once in the course of three throws with a die ? Ace will turn up at least once unless it fails to turn in every throw. The chance of failure in a single throw being 5/6, the chance of failure in all three throws is 5/6 x 5/6 x 5/0, or 125/216. Hence the chance of at least one ace is 1 - 125/216, or 91/216. PROBABILITY 417 Example 2. The chance that A will solve a certain problem is 3/4. The chance that B will solve it is 2/3. What is the chance that the problem will be solved if both A and B attempt it independently ? The problem will be solved unless both A and B fail. The chance of A's failure is 1/4, of B's failure 1/3. Hence the chance that both fail is 1/4 X 1/3, or 1/12. Therefore the chance that the problem will be solved is 11/12. Example 3. There are two purses, one containing five silver coins and one gold coin, the other three silver coins. If four coins be drawn from the first purse and put into the second, and five coins be then drawn from the second purse and put into the first, what is the chance that the gold coin is in the second purse ? in the first purse ? The chance that the gold coin is taken from the first purse and put into the second is C\/ C% or 2/3, § 781, Ex. 5. The chance that it is then left in the second purse is ^ / C\^ or 2/7. Hence the chance that after both drawings it is in the second purse is 2/3 x 2/7, or 4/21. The chance that it is in the first purse is 1 — 4/21, or 17/21. Example 4. If eight coins be tossed simultaneously, what is the chance that at least one of them will turn head up ? Example 5. Four men A, B, C, and D are hunting quail. If A gets on the average one quail out of every two that he fires at, B two out of every three, C four out of every five, and D five out of every seven, what is the chance that they get a bird at which all happen to fire simultaneously ? Example 6. An urn A contains five white and four red balls. A sec- ond urn B contains six white and two black balls. What is the chance of drawing a white ball from A and then, this ball having been put into B, of drawing a white ball from B also ? Example 7. The chance that A will be alive five years hence is 3/4; B, 5/6. What is the chance that five years hence both A and B will be alive ? A alive, B dead ? A dead, B alive ? both dead ? Mutually exclusive events. If two or more events are so 786 related that but one of them can occur, they are said to be mutually exclusive. Thus, the turning of an ace and the turning of a deuce on the same throw of a single die are mutually exclusive events. i -1^ ^ a 418 A COLLEGE ALGEBRA 787 Theorem 3. The probabiliti/ that some one or other, of a set of mutually exclusive events will occur is the sum of the ■probabilities of the single events. For consider two mutually exclusive events A and B. The possible cases with respect to the two events are of three kinds, all mutually exclusive, namely, those for which (1) A happens, B fails ; (2) A fails, B happens ; (3) A fails, B fails. Let the numbers of equally likely possible cases of these three kinds be I, m, and n respectively. Then (a) The chance that either A bv B happens is •; For there are I -\- m -\- n possible and I + m favorable cases- (b) The chance of the single event A is ; r- 1 + (m + n) For since A never happens except when B fails, the I cases in which A happens and B fails are all the cases in which A happens, and the m + n cases in which A fails and B happens or both A and B fail are all the cases in which A fails. (c) Similarly the chance of the single event B is -• ^ ^ "^ ^ 7n-\-(l-\- n) Bt ^ + '>n _ I m I -\- m + n I + {in + n) m + (^ + n) Therefore the chance that either A or B happens is the sum of the chances of the single events A and B. The proof for more than two events is similar. Thus, if one ball be drawn from a bag containing four white, five black, and seven red balls, since the chance of its being white is 1/4, and that of its being black is 5/16, the chance of its being either white or black is 1/4 + 5/16, or 9/16. Of course this result may be obtained directly from the definition of probability, § 776. In fact that definition may be regarded as a special case of Theorem 3. Care must be taken not to apply this theorem to events which are not mutually exclusive. Thus, if asked, as in § 785, Ex. 2, to find the chance that a problem will be solved if both A and B attempt it, A's chance of success being PROBABILITY 419 3/4 and B's 2/3, we cannot obtain the result by merely adding 3/4 and 2/3, since the two events A succeeds, B succeeds are not mutually exclu- sive. The mutually exclusive cases in which the problem will be solved are : A succeeds, B fails ; A fails, B succeeds ; A succeeds, B succeeds. The chances of these cases are, § 784, 3/4x1/3 or 3/12, 1/4x2/3 or 2/12, 3/4 x 2/3 or 6/12; and the sum of these three chances, or 11/12, is the chance that the problem will be solved. Example 1. An urn A contains ten balls three of which are white, and a second urn B contains twelve balls four of which are white. If one of the urns be chosen at random and a ball drawn from it, what is the chance that the ball is white ? We are required to find the chance of one of the following mutually exclusive events : (1) choosing A and then drawing a white ball from it; (2) choosing B and then drawing a white ball from it. The chance of choosing A is 1 / 2, and the chance when A has been chosen of drawing a white ball is 3/10. Hence the chance of (1) is 1/2 X 3/10,or3/20. Similarly tlie chance of (2) is 1 /2 x 4/12,orl/6. Therefore the chance in question is 3/20 + 1/6, or 19/60. Example 2. What is the value of the expectation of a person who is to have any two coins he may draw at random from a purse which contains five dollar pieces and seven half-dollar pieces? The value of his expectation so far as it depends on drawing two dollar pieces is $2 x C'^/C'| = $2 x 5/33 = §.30; on drawing two half-dollar pieces, $1 x C\/ Clf = $1 x 7 /22 = 3-32 ; on drawing one dollar piece and one half-dollar piece, .$1.50 x 5 ■ 7/ Clf = §1.50 x 35/66 = $.80. Hence the total value of his expectation is $.30 + $.32 -|- .§.80, or $1.42. Example 3. Two persons A and B are to draw alternately one ball at a time from a bag containing three white and two black balls, the balls drawn not being replaced. If A begins, what chance has each of being the first to draw a white ball ? The chance that A succeeds in the first drawing is 3 / 5. The chance that A fails and B then succeeds is 2/5 x 3/4, or 3/10, for when B draws, the bag contains four balls three of which are white. The chance that A fails, B fails, and A then succeeds is 2/5x1/4x3/ 3, or 1 / 10, for when A draws, the bag contains three balls all white. Therefore A's total chance is 3/5 -H 1/10, or 7/10, and B's is 3/10. 420 A COLLEGE ALGEBRA Example 4. In the drawing described in Ex. 3 what are the respective chances of A and B if the balls are replaced as they are drawn ? On the first round A's chance is 3/5, B's 2/5 x 3/5, or 6/25; and their chances on every later round, of which there may be any number, will be the same as these. Hence their total chances are in the ratio 3/5 : 6/25, or 5 : 2 ; that is, A's total chance is 5/7, B's 2/7. "^ Example 5. In a room there are three tables and on them nine, ten, and eleven books respectively. I wish any one of six books, two of which are on the first table, three on the second, one on the third. If a friend select a book for me at random from those in the room, what is the chance that it is one of those I wish? ^^f^ Example 6. An owner of running horses enters for a certain race two hor.ses whose chances of winning are 1/2 and 1/3 respectively. What is the chance that he will obtain the stakes ? Example 7. A and B throw alternately with two dice for a stake which is to be won by the one who first throws a doublet. What are their respective chances of winning if A throws first ? 788 Repeated trials of a single event. The following theorems are concerned with the question of the chance that a certain event will occur a specified number of times in the course of a series of trials, the chance of its occurrence on a single trial being known. 789 Theorem 4. If the prohahillty that an event will occur on a single trial is p, the prohahility that it will occur exactly/ v times in the course ofi\ trials is C"p''q"~'', loJiere q = 1 — p. For the probability that it will occur on all of any partic- ular set of r trials and fail on the remaining Ji — r trials is jo''(l — p)"~'', ov ]j''q"-'\ U (J = 1 —J), § 784. But since there are 7i trials all told, we may select this particular set of r trials in C" ways which, of course, are mutually exclusive. Hence the probability in (]uestion is C'^.p'q"''', § 787. Thus, the chance that ace will turn up exactly twice in five throws with a single die, or that out of five dice thrown simultaneously two and but two will turn ace up, is C^ ■ (!)2 • {f)3, or 625/3888. PROBABILITY 421 Observe that C",'.p''q"~'' is the term containing/)'" in the expan- sion of (p + (/)" by the binomial theorem ; lor C" = C,,"^. Theorem 5. The i^fobabiUty that such an event will occur 790 at least r times in the course ofn trials is the sum of the first n — r + 1 terms in the exjjansion o/" (p + cj)", namely, p„ ^ Cjp^-'q + CSp°-2q^ + • • • + C„1,p\l"-^ For the event will occur at least r times if it occurs exactly r times or exactly any number of times g)\ieater than r, and the terms j9", Cij)"~^q, •■■, C„",.^''y"~''repres'^iit the probability of the occurrence of the event exactly n, §i — 1, ■ ■ ■, r times respectively, § 789. ! Thus, the chance that ace will turn up at least twice in the course of five throws with a single die is a)^ + 5(i)* I + 10(DMD^ + 10(i)2(|)3, or 3^8- Example 1. Two persons A and B are playi;ig a game which cannot be drawn and in which A's skill is twice B's. What is the chance that A will win as many as three such games in a se'. of five ? A's chance of winning a single game is 2/3, of losing 1 /3. Hence the chance that A will win as many as three of the five games is the sum of the first three terms of Q + })5, that is, (2)5 + 5(2)4 i + 10(2)3(i)2, or 64/81. Example 2. Under the conditions of Ex. 1 what is the chance that A will win three games before B wins two ? The chance in question is that of A's winning at least three of the first four games played ; and this chance is (5)* + 4(|)'^i, or J^. And, in general, the chance of A's winning m games before B wins n is the same as the chance of A's winning at least m of the first m + m — 1 games played. Example 3. Ten coins are tossed simultaneously. What is the chance that exactly six of tliem turn heads up ? that at least six turn heads up ? Example 4. If four dice be thrown simultaneously, what is the chance that exactly three turn ace up? that at least three turn ace up? Example 5. Under the conditions stated in Ex. 1 what is the chance that A will win at least four of the five games played ? Example 6. Under the same conditions what is the chance that A will win four games before B wins one ? 422 .. COLLEGE ALGEBRA EXERCISE LXVra 1. A bag contailis three white, five black, and seven red balls. On the understanding that one ball is dravi-n at a time and replaced as soon as drav?n, what are the chances of drawing (1) first a white, then a red, then a black ball ? (2) a white, red, and black ball in any order whatsoever ? 2. What is the chance of obtaining a white ball in the first only of three successive drawings from this bag, balls not being replaced ? e 3. What is the v ;lue of the expectation of a person who is allowed to draw two coins at ra. dom from a purse containing five fifty -cent pieces, four dollar pieces, ai (. three five-dollar pieces ? 4. The chance tF'it a certain door is locked is 1/2. The key to the door is one of a ""launch of eight keys. If I select three of these keys at random and go to the door, what is the chance of my being able to open it ? 5. There are three independent events whose chances are 1/2, 2/3, and 3/4 respectively. #hat is the chance that none of the events will occur? that one and but one of them will occur? that two and but two will occur ? that all thret will occur ? 6. Find the odds against throwing one of the totals seven or eleven in a single throw with two dice. 7. What are the odds against throwing a total of ten with three dice ? What are the odds in favor of throwing a total of more than five ? 8. Three tickets are drawn from a case containing eleven tickets numbered 1, 2, • • ■ , 11. What is the chance that the sum of their numbers Is twelve ? What is the chance that this sum is an odd number ? 9. Two gamblers A and B throw two dice under an agreement that if seven is thrown A wins, if ten is thrown B wins, if any other number is thrown the stakes are to be divided equally. Compare their chances. 10. The same two gamblers play under an agreement that A is to win if he throws six before B throws seven, and that B is to win if he throws seven before A throws six. A is to begin and they are to throw alternately. Compare their chances. 11. Three gamblers A, B, and C put four white and eight black balls into a bag and agree that the one who first draws a white ball shall win. PROBABILITY 423 If they draw in the order A, B, C, what are their respective chances when the balls drawn are not replaced ? when they are replaced ? 12. What is the worth of a ticket in a lottery of one hundred tickets having five prizes of $100, ten of §50, and twenty of $5 ? 13. A bag A contains five balls one of which is white, and a bag B six balls none of which is white. If three balls be drawn from A and put into B and three balls be then drawn from B and put into A, what is the chance that the white ball is in A ? 14. The bag A contains ??i balls a of which are white, and the bag B contains n balls b of which are white. Is the chance of obtaining a white ball by drawing a single ball from one of these bags chosen at random the same that it would be if all the balls were put into one bag and a single ball then drawn ? 15. In a certain town five deaths occurred within ten days including January first. What is the chance that none of the deaths occurred on January first? 16. If on the average two persons out of three aged sixty live to be seventy, what is the chance that out of five persons now sixty at least three will be alive ten years hence ? 17. A boy is able to solve on the average three out of five of the prob- lems set him. If eight problems are given in an examination and five are required for passing, what is the chance of his passing ? 18. A person is to receive a dollar if he throws seven at the first throw with two dice, a dollar if he throws seven at the second throw, and so on until he throws seven. What is the total value of his expectation ? 19. In playing tennis with B, A wins on the average three games out of four. What is the chance that he will win a set from B by the score of six to three ? What is the total chance of his winning a set from B, the case of deuce sets being disregarded ? 20. Under the conditions described in the preceding example what chance has A of winning a set in which the score is now four to two against him? 21. Two gamblers A and B are playing a game of chance and each player has staked $32. They are playing for three points, but when A has gained two points and B one they decide to stop playing. How should they divide the $64 ? 424 A COLLEGE ALGEBRA XXVIII. MATHEMATICAL INDUCTION 791 Mathematical induction. A number of the formulas con- tained in recent chapters may be established by a method of proof called mathematical induction. It is illustrated in the following example. Example. Prove that the sum of the first n odd numbers is n^. We are asked to show that l + 3 + 5 + ... + (2n-l) = n2. (1) We see by inspection that (1) is true for certain values of n, as 1 or 2. Suppose that we have thus found it true when n has the particular value ki so that l + 3 + 5 + .-- + (2A;-l) = A;2 (2) is known to be true. Adding the next odd number, naively, 2 (A; + 1) — 1, or 2 A; + 1, to both members of (2) and replacing A;2 + 2 i + 1 by (A: + 1)2, we obtain l + 3 + 5 + ... + (2A; + l) = (A; + l)2. (3) But (3) is what we get if in (1) we replace n by A: + 1. We have there- fore .shown that if (1) is true when n has any particular value k, it is also true when n has the next greater value k + \. But we have already found by inspection that (1) is true when k has the particular value 1. Hence it is true when n = l + l, or 2 ; hence when n = 2 -I- 1, or 3 ; and so on through all positive integral values of n, which is what we were asked to demonstrate. And, in general, if a formula involving n has been found true for n — 1 and we can demonstrate that if true for n = /.• it is also true for w = Ar + 1, we may conclude that it is true for all positive integral values of n. For we may reason: Since it is true when n — 1, it is also true when n = 1 + 1, or 2 ; hence when n = 2 + l,ov 3; and so on through all posi- tive integral values of n. As another illustration of this method we add the following proof of the binomial theorem. For small values of n we find by actual multiplication that (a + 6)" = a" + C'{a"-^b + C?^a"-^b^ + •■• + 0";.^-^ + ■•■. (1) THEORY OF EQUATIONS 426 Multiplying both members of (1) by a + 6, we obtain, § 773, 1, (a + b)" + ^ = a« + i + C'l\a"b+ C'^\a"-^b- f- C;: \a''-'- + ^b'- + •■- + 1 I +^1 +c,ii\ + C" + ' a(" + !)-'■ 6'-+ •••. (2) But (2) is the same as (1) with n replaced by n + I. Hence if (1) is true when n = fc, it is also true when n = k + 1. But (1) is known to be true when n = 1. It is therefore true when n = 1 + 1, or 2 ; therefore when n = 2 + 1, or 3 ; and so on. Since the formula C" + C,."i = C" + ^ can be proved independently of the doctrine of combinations, § 774, the proof of the binomial theorem nere given is independent of that doctrine. EXERCISE LXIX Prove the truth of the following formulas, §§ 701, 712, by the method of mathematical induction. 1. a + ar + ar'^ + ■ ■ ■ + ar" -^ = a (l - r") / (1 - r). 2. 12 + 22 + 32 + . . . + Ti2 = ,1 (,i 4. 1) (2 11 + 1) /G. 3. 13 + 23 + 33 + .. • + n3 = n2(n + l)-/4- 4. 1 + 3 + 6 + ••■ + n(n + l)/2! = n{n + l)(n + 2)/3!. XXIX. THEORY OF EQUATIONS THE FUNDAMENTAL THEOREM. RATIONAL ROOTS The two standard forms of the general equation of the nth 792 degree in x. Every rational integral equation involving a single unknown letter, as x, and of the nth degree with respect to that letter, can be reduced to the standard form a„x" + aiX"~^ + • • • + (^n-i^ + *« = 0. (1) When the coefficients a^, a^, ■■■, a„ are given numbers, (1) is called a numerical equation, but when they are left wholly undetermined, (1) is called the general equation of the wth degree. 426 A COLLEGE ALGEBRA The final coefficient a.,, is often called the absolute term. We call an equation of the form (1) complete or incomplete according as none or some of the coefficients aj, 02> • • •? «„ are 0. Observe that in a complete equation the number of the terms is n + 1. In what follows, when all the coefficients a,,, a^, ■•-, a„ are real numbers, we may and shall suppose that the leading one ttQ is positive, and when they are rational, that they are integers which have no common factor. By dividing both members of (1) by a^, we reduce it to the second standard form x" + b^x"-^-] \-K-i^ + K = Q, (2) in which the leading coefficient is 1, and b^ = Oi/oq, and so on. For many purjjoses (2) is the more convenient form of the equation. In the present chapter it is to be understood that f(x) = denotes an equation of the form (1) or (2). Roots of equations. The roots of the equation f(x) = are the values of x for which the polynomial f(x) vanishes, §§ 332, 333. It is sometimes convenient to call the roots of the equation the roots of the polynomial. From the definition of root it follows that when o„ is one of the roots of /(a-) = is ; also that an equation /(a-) = all of whose coefficients are positive can have no positive root, and that a complete equation f(x) = whose coefficients are alternately positive and negative can have no negative root. Thu.s, 2x3 + x^ + 1 = can have no positive root since the polynomial 2 «* + x2 + 1 cannot vanish when x is positive ; and 2x^ — x^ + Sx — 1 = can have no negative root since 2a;3 - a;2 + 33. _ i cannot vanish when X is negative. Theorem 1 . If h is a root of f (x) = 0, then f (x) is exactly divisible by x — b; and conversely, if f{x) is exactly divisible byx — h, then b is a root of f (x) = 0. THEORY OF EQUATIONS 427 For, by § 413, the remainder in the division of /(a:) by X -b is f{b). But when i is a root of f{x) = this remainder f{b) is 0, § 793, so that f(x) is exactly divisible by x -b; and conversely, when f(x) is exactly divisible by x - b, the remainder f{b) is 0, so that b is &, root of f(x) = 0. Example. Prove that 3 is a root of /(a;) = x^ - 2 x^ - 9 = 0. 1 -2 +0 -9[3 Dividing x^ - 2x2 _ 9 ^y x - 3 synthet- 3 3 9 ically, § 411, we find that the remainder /(3) 1 1 3, 0=/(3) is 0. Hence 3 is a root of /(x) = 0. If 6 is a root of f(x) = 0, so that f(x) is exactly divisible 796 \)y X — b, and we call the quotient cf> (x), we have f(x) = (x-b) (x) vanishes ; in other words, they are the roots of the depressed equation b /c cannot be a root unless c = ± 1, that is, unless b /c denotes the integer ± b. Hence the following theorem, § 454 : An equation of the form x" + ajx""' -f . . . -f a„ = 0, cohere ai, •••, a„ denote integers, cannot have a rational fractional root. It follows from what has just been said that all the rational 803 roots of an equation with rational coefficients can be found by a limited number of tests. These tests are readily made by synthetic division. Example. Find the rational roots, if any, of the equation 3 x5 - 8 X* + x2 + 12 X + 4 = 0. The only possible rational roots are ±1, ±2, ±4, ±1/3, ± 2/3, ± 4/3. We see by inspection that 1 is not a root. Testing 2, we find that it is a root and obtain the depressed equation 3x^-2 x3-4x2-7x-2=0. We find that 2 is a root of this depressed equation also and ob- tain the second depressed equation I x^ + 4 x2 + 4 X + 1 = 0. This equation can have no positive root since -8 6 -2 + -4 -4 + 1 -8 -7 + 12 - 14 - 2, + 4[2 - 4 0(2 6 4 8 4 8 1, 2 -\/?> -1 - 1 -1 3 3, 430 A COLLEGE ALGEBRA all its terms are positive, § 794. Testing —1, we find that it is not a root . Testing — 1/3, we find that it is a root and obtain the third depressed equation x^ + x + 1 — 0. Hence the rational roots of the given equation are 2, 2, —\/Z. Its remaining roots, found by solving x^ + x + 1 = 0, are (— 1 ± i V3)/2. 803 The reckoning involved in making these tests will be lessened if one bears in mind the remark made in § 453 ; also the fact that a number known not to be a root of the given equation cannot be a root of one of the depressed equations, § 796 ; and finally the following theorem : If b is 2}ositive and the signs of all the coefficients in the result of dividing f (x) byx. — b synthetically are plus, f (x) = can have no root greater than b ; ifh is negative and the signs just mentioned are alternately 2)lus and rninus, f(x) = can have 710 root algebraically less tlian b. Eor it follows from the nature of synthetic division that in both cases the effect of increasing h numerically will be to increase the numerical values of all coefficients after the first in the result without changing their signs, so that the final coefficient, that is, the remainder, cannot be 0. Example 1. Show that 2x3+3x2-4x4-5 = has no root greater than 1. 2 +3 — 4 +5|J. Dividing by x — 1, we obtain positive coefli- _ 2 5 1 cients only. Hence there is no root gi'eater 2+5+1, 6 than 1. If we divide by x — 2, we obtain a result with larger coefficients, all positive, namely, 2 + 7 + 10, 25. Example 2. Showthat 3x3+4x2-3x + l = has no root less than -2. 3 +4 - 3 +1 I - 2 Dividing by x + 2, we obtain coefficients _ ~ ^ + 4 — 2 which are alternately plus and minus. Hence 3 — 2 +1,-1 tliere is no root less than — 2. If we divide by x + 3, we obtain coefficients with the same signs as those just found but numerically greater, namely, 3-5 + 12, - 35. 804 We may add that any number which is known to be alge- braically greater than all the real roots of f{x) = is called a THEORY OF EQUATIONS 431 sujjerior limit of these roots, and that any number which is known to be algebraically less than all the real roots oi f(x) — is called an inferior limit of these roots. Thus, we have just proved that 1 is a superior limit of the roots of 2x^ + 3x2 — 4x + 5 = and that — 2 is an inferior limit of the roots of 3x3 + 4x2-3x + l =^0. EXERCISE LXX 1. Form the equations whose roots are (1) a, - 6, a + 6. (2) 3, 4, 1/2, - 1/3, 0. 2. Show that — 3 is a triple root of the equation X* + Sx'' + 18 X- - 27 = 0. 3. Show that 1 and 1/2 are double roots of the equation 4 x5 - 23 x5 + 33 x2 - 17 X + 3 = 0. 4. By the method of § 803 find superior and inferior limits of the real roots of x5 — Sx* — 5x^ + 4 x^ — 7 X — 250 = 0. 5. Show that 2x* — 3x^ + 4x2 — lOx — 3 = has no rational root. Each of the following equations has one or more rational roots. Solve them. 6. x3 - x2 - 14 X + 24 = 0. 7. x^ - 2 x2 - 25 x + 50 = 0. 8. 3 x^ - 2 x2 + 2 X + 1 = 0. 9. 2 x< + 7 x^ - 2 x^ - x = 0. 10. X* 4- 4 x3 + 8 x2 + 8 X + 3 = 0. 11. 2 X* + 7 x3 + 4 x2 - 7 X - 6 = 0. 12. 3x1 + 11x3 + 9x2 + 11X + 6 = 0. 13. x5 - 9 X* + 2 x3 + 71 x2 + 81 X + 70 = 0. 14. 2x5 -8x* + 7x3 + 5x2-8x + 4 = 0. 15. x5 + 3x* -15x3- 35x2 + 54x+ 72 = 0. 16. 12x4 - 32x3 + 13x2 + 8x - 4 = 0. 17. x5 - 7x4 + 10x3 + 18x2 -27x- 27=0. 18. 2x* - 17x3 + 25x2 + 74x- 120 = 0. I 432 A COLLEGE ALGEBRA 19. 4 x» - 9 x-^ + 6 x2 - 13 X + = 0. 20. x5 + 8x-« + 3x3 - 80x2 _ 52x + 240 = 0. 21. 2 x5 + 11 X* + 23 x- + 25 x2 + 10 X + 4 = 0. 22. 6x* - 89x5 + 359x2 _ 254x + 48 = 0. 23. lOx* + 41x'' + 46x2 + 20X + 3 = 0. 24. 30 X* - 108x3 + 107 x2 - 43x + 6 = 0. 25. 12x5 + 20x* + 29x3 + 77x2 + COx + 18 = 0. 26. 2 x6 + 7 x5 + 8 X* + 7 x3 + 2 x2 - 14 x - 12 = 0. 27. 2x6 + 11x5 _^ 24x* + 22x3- 8x2 -.33x- 18 = 0. 28. 5 x6 - 7 x5 - 8 X* - x3 + 7 x2 + 8 X - 4 = 0. RELATIONS BETWEEN ROOTS AND COEFFICIENTS 805 Relations between roots and coefficients. When an equation whose roots are (3i, fS-i, • • •? y8„ is reduced to the second standard form, § 792, (2), the identity in § 798, (3), becomes X" + b^x"-^ + h„x"-- + h^x"-^ H \-b„ = (x - /30 (,r - p,) {X - /3.) • . • {X - p,). Carry out the multiplications indicated in the second mem- ber and arrange the result as a polynomial in x, § 559. Then equate the coefficients of like powers of x in the two members, § 284. We thus obtain the following relations between the coefficients h^^, h,,, ■■■, ^>„ and the roots /?i, /?o, •••, y8„ : -^ = i8i + /8, + i8, + --- + )8,„ (1) b, = p,/3, + 13,(3, + • • • + A/?3 + • • • + f3,.^^/3,„ (2) - ^3 = AA/Sa + A/8.i8, + • • • + /3,_, A,-i/8,„ (3) (-iyb„ = p,p,p,...l3,„ ^n) where the second members of (2), (3), ■•• represent the sum of the products of every two of the roots, of every three, and so on, THEORY OF EQUATIONS 433 and the sign before the first member is plus or minus according as the number of the roots in each term of the second member is even or odd. Hence the theorem : Theorem. In every equation reduced to the form 806 X" + bix"-' + Kx"-2 + . . . + b„ = 0, tlie coefficient bj of the second term, ivith its sign changed, is equal to the sum of all the roots; the absolute term b„, urith its sign changed or not according as n is odd or even, is equal to til e product of all the roots ; and the coefficient \ of each inter- mediate term, with its sign changed or not according as r is odd or even, is equal to the sum of the products of every r of the roots. Before applying this theorem to an equation whose lead- ing coefficient is not 1 we must divide the equation by that coefficient. If the equation be incomplete, it must be remem- bered that the coefficients of the missing terms are 0. Thus, without solving the equation Sx-'^ — Gx + 2 = 0, we know the following facts regarding its roots /3i, p«, (is. Reduced to the proper form for applying the theorem, the eciuation is x^ + Ox^ — 2 x + 2/3 = 0. HcncG ' /3i + /32 + /33 = 0, /3i/32 + /3i/33 + ^2^3 = - 2, ^i^o/Ss = - 2 /3. If all but one of the roots of an equation are known, we can 807 find the remaining root by subtracting the sum of the known roots from — bi, or by dividing b,„ with its sign changed if 71 is odd, by the product of the known roots. Example. Two of the roots of 2 x^ + 3 x^ - 23 x - 12 = are 3 and — 4. What is the remaining root ? The remaining root is - 3/2 - [3 + (- 4)] = - 1/2 ; or again, it is 6-3(-4)=-l/2. When the roots themselves are connected by some given 808 relation, a corresponding relation must exist among the coeffi- cients. To find this relation we apply the theorem of § 806. 434 A COLLEGE ALGEBRA Example 1. Find the condition that the roots of x^ + px^ ^qx + r = shall be in geometrical progression. Representing the roots by a / p, a, a^, we have - + a + a/3 = -p, — + a2 _f, ^2^ = g^ _. a ■ a^ = -r. /3 ^ ^ The third equation reduces to a^ = — r, whence a =\ — r. Dividing the second equation by the first, substituting a = V— r in the result, and simplifying, we have q^ — p^r = 0. Example 2. Solve the equation x^ + 8 x^ 4- 5 x — 50 = 0, having given that it has a double root. Representing the roots by a, a, /3, we have 2 a + ^ = - 8, a:2 + 2 a/3 = 5, a2/3 = 60. Solving the first and second of these equations for a and /3, we obtain a = -5, ^ = 2 and a = -1/3, ^ = -22/3. The values a = — 5, /3 = 2 satisfy the equation a^j3 = 50, but the values a = — 1/3, ^ = — 22/3 do not satisfy this equation. Hence the required roots are — 5, — 5, 2. 809 Symmetric functions of the roots. The expressions in the roots to which the several coefficients are equal, § 805, are si/m- metric functions of the roots, § 540. It will be proved in § 868 that all other rational symmetric functions of the roots can be expressed rationally in terms of these functions, and therefore rationally in terms of the coefficients of the equation. Example 1. Find the sum of the squares of the roots of the equation 2x3-3x2-4x- 5 = 0. Calling the roots a, /3, 7, we have a2 + ^2 + 72 = (a + /3 + 7)2 - 2 (a/3 + ^7 + 7«) = (3/2)2 + 4=61. Example 2. If the roots of x' + px"^ + qx + r = are a, /3, 7, what is the equation whose roots are ^y, ya, a/3 ? If p', g', r' denote the coeflHcients of the required equation, we have - p' = /37 + 7a + a/3 = 7, g' = ^7 . 7a + 7a • a/3 + a/3 ■ ^y = a/37 (or + |3 + 7) = ( - r) ( - i)) = rp, - r' = /37 • 7a • a/3 = (a^7)2 = r"^. Hence the required equation is x^ — qx^ + prx — r2 = 0. THEORY OF EQUATIONS 435 EXERCISE LXXI 1. Two of the roots of 2 x^ - 7 a;^ + lo j - 6 = are 1 ± i ; find the third root. 2. The roots of each of the following equations are in geometrical progression ; find them. (1) 8x3-14x2-21x4-27 = 0. (2) x^ + x2 + 3x + 27 = 0. 3. The roots of each of the following equations are in arithmetical progression ; find them. (1) x3 + 6x2 + 7x-2 = 0. (2) x3 _ 9x2 + 23x- 15 = 0. 4. Show that if one root of x^ + px2 + gx + r = be the negative of another root, pq — r. 5. Find the condition that one root of x^ + px- + qx + r = shall be the reciprocal of another root. 6. Solve X* + 4.x3 + 10x2 + 12x + 9 = 0, having given that it has two double roots. 7. Solve the equation 14x3 — 13x2 — 18x + 9 = 0, having given that its roots are in harmonical progression. 8. Solve the equation x* — x^ - 56 x2 + 36 x + 720 = 0, having given that two of its roots are in the ratio 2 : 3 and that the difference between the other two roots is 1. 9. If a, i3, 7 are the roots of x^ + px- + gx + r = 0, find the equa- tions whose roots are (1) - a, - i3, - 7. (2) fca, k^, ky. (3) 1/a, 1//?, 1/7. (4) a + fc, /3 + A:, 7 + A;. (5) a^, ^, 72. (6) - l/a2, _l//32, - 1/72. 10. If a, i3, 7 are the roots of 2 x^ + x2 - 4 x + 1 = 0, find the values of (1) a2 + ^2 + y2. (2) a3 + ;33 + 73. (3) l/^y+l/ya+l/a^. (4) a/32 + ^,^2 + ^^2 + ^^2 + ^2,^ + a^. 11. If a, /3, 7 are the roots of x^ - 2 x2 + x - 3 = 0, find the values of (1) a/py + ^/ya + y/a^ (2) a^/y + ^y/ a + ycx /^. (3) (^ + 7) (7 + «) (« + ^)- (4) (/32 + 7^) (72 + a"') {a^ + ^). 436 A COLLEGE ALGEBRA TRANSFORMATIONS OF EQUATIONS Some important transformations. It is sometimes advan tageous to transform a given equation f{x) = into another equation whose roots stand in some given relation to the roots of f(x) = 0. The transformations most frequently used are the following : To transform a given equation f (x)= into another tvhose roots are those o/f (x)= tvith their signs changed. The required equation is /(— y) = 0. For substituting any number, as y8, for x in f(x) gives the same result as substitut- ing — /8 for y in /(— ij). Hence, if /(a;) vanishes'when x = /3, y(_ y) will vanish when y =— P; that is, if /S is a root of fix) = 0, - ^ is a root of /(- y) = 0. Therefore every root of f(x) = 0, with its sign changed, is a root of /(— y) = ; and /(—?/)= has no other roots than these, since f(x) = and /(-?/) = are of the same degree. If the given equation is rt^cc" + «ia;"~* + a^x"'^ + • • • -f- a„ = 0, the required equation will be «.,(- t/T + «i (- ?/)""' + a,{- y)"-' + • • • + «^„ = 0, which on being simplified becomes «oy" - «iy""' + o^y"'- + (- !)"«« = O- Hence the required equation may be obtained from the given one by changing the signs of the terms of odd degree when n is even, and by changing the sigyis of the terms of even degree, including the absolute term, when n is odd. We may use x instead of y for the unknown letter in the transformed equation, and write /(— a;) = for /(— y) = 0. THEORY OF EQUATIONS 437 Example. Change the signs of the roots of 4x5 - 9x3 + 6x2 - 18x + 6 = q. Changing the signs of the terms of even degree, we have 4x5 - 9x3 - 6x2 - 13x - 6 = o. In fact, the roots of the given equation are 1 / 2, 3/2, — 2, ± i, and those of the transformed equation are — 1/2, — 3/2, 2, T i- To transfortn a given equation f(x)= into another ichose 812 roots are those o/f (x) — eacli multiplied by some constant, as k. The required equation is f(i//k) = 0. For if /(a*) vanishes when X = P,f{i//k) will vanish when y /k = p, that is, when y — k(i (compare § 811). If the given equation is aoX" + aix"-^ + a^x""-^ -\ 1- a„ = 0, the required equation will be which when cleared of fractions becomes «(,?/" + kaiy"~^ + k'^Ooy"'^ + • • • + k"a,^ = 0. Hence the required equation may be obtained by multiply- ing the second term of the given equation by k, its third term by k^, and so on, taking account of missing terms if any. When k<= — l this transformation reduces to that of § 811. Example. Multiply the roots of x* + 2x3 - x + 3 = by 2. Also divide them by 2. The first of the required equations is x* + 4 x3 — 8 x + 48 = 0, and since dividing by 2 is the same as multiplying by 1/2, the second is a;4 + x3-x/8 + 3/lG = 0, or lOx* + 16x3 - 2 x + 3 = 0. The following example illustrates an important application 813 of the transformation now under consideration. Example. Transform the equation 36x3 + 18x2 + 2x + 9 = into another whose leading coefficient is 1 and its remaining coefficients integers. Dividing by 36, we have x3 + x2/2 + x/18 + 1/4 = 0. (1) 438 A COLLEGE ALGEBRA Multiplying the roots by k, x3 + A-xV2 + fc2x/18 + fcV4 = 0. (2) "We see by inspection that the smallest value of k which will cancel all the denominators is 6. And substituting 6 for k in (2), we have x3 + 3x2 + 2x + 54=0, (3) ■which has the form required. The roots of (3) each divided by 6 are the roots of the given equation (1). 814 To transform a f/ive?i equation f(x) = into another tvhose roots are the reciprocals of those of f (x) = 0. The required equation is /(I/?/) = 0. For \i f{x) vanishes when X = (3, /(1/y) will vanish when 1/y = {3, that is, when If the given equation is CIqX" + (liX"~^ + • • • + <^n-V^ + ^n = 0, the required equation will be «o/y' + «i/y"' + • • • + a„_i/y + «„ = 0, which when cleared of fractions becomes ay + «„_i//'"^ H + aiy + ^0 = 0. Hence the required equation may be obtained by merely reversing the order of the coefficients of the given equation. Example. Replace the roots of 2 x* - x2 - 3 x + 4 = by their recip- rocals. Reversing the coefficients, we have 4x* - Sx^ — x^ + 2 = 0. 815 An equation like 2 x^ -\- 3 x^ — Z x — 2 = 0, which remains unchanged when this transformation is applied to it, that is, when the order of its coefficients is reversed, is called a reciprocal equation, § 645. If /3 is a root of such an equation, l/;8 must also be a root. Hence when the degree of the equation is even, half of the roots are the reciprocals of the other half. The same is true of all the roots but one when the degree is odd; but in this case there must be one root which is its own reciprocal, that is, one root which is either THEORY OF EQUATIONS 439 1 or - 1. Thus, one root of2a;^4-3x2-3cc-2 = 0isl and the otlier roots are — 2 and —1/2. From the nature of this transformation it follows that ivhen 816 an equation has variable coefficients and the leading coefficient vanishes, one of the roots becomes infiiiite; when the tivo leading coefficients vanish, two of the roots become infinite; and so on. Example 1. Show that one of the roots of mx^ + 3x2_2x + l=0 becomes infinite when m vanishes. Applying § 8U to mx^ + 3 x^ - 2 a; + 1 = 0, (1) we obtain x» - 2x2 + 3x + m = 0. (2) If the roots of (2) are ^i, /Sg, /Sg, those of (1) are l/^i, l/iSj, 1//33. By § 806, i3i/32/33 = — m. Hence, if m approach as limit, one of the roots /3i, /3o, /33 must also approach as limit, and if this root be /3i, the corresponding root of (1), namely, 1 /^i, must approach oo, § 512. Example 2. Show that two of the roots of mx^ + iii^x- + x + 1 = become infinite when m vanishes. Applying § 814 to mx^ + m^x'^ + x + 1 ^ 0, (1) we have x^ + x'^ + ni^x + m - 0. (2) If the roots of (2) are ^i, ^2, Pz, those of (1) are l//3i, 1//32, I / ^z- ^1^3 = -m, /3ii32 + ^1^3 + ftft = '«^- (3) It follows from (3) that if m approach 0, two of the three roots ^1, ^2, ^3 must also approach 0, and if these roots be /3i, /Sa, the corresponding roots of (1), namely, l//3i, 1//32, must approach oo. To transform a given eqxiation f (x)= into another whose 817 roots are those of f (x) = 0, each diminished by some constant, as k. The required equation is/(y + ^) = 0. For \ff{x) vanishes when x = /3, f{y + k) will vanish when y + k = jS, that is, when y = /3 — k. If the given equation is f(x) = a^x" + Oiic" -' + [- a„_^x + a„ = 0. the required equation will be f(y + k) = a,(y + ky + a,(y + k)"-' + . . . + a„ = 0, 440 A COLLEGE ALGEBRA which, when its terms are expanded by the binomial theorem and then collected, will reduce to the form where Cq = «„, Ci = nkao + ai, and so on. This method of obtaining <^ (y) from f(x) is usually very laborious. The following method is much more expeditious, at least when the coefficients otf(x) are given rational numbers. It X = i/ + k, then y = x — k, and we have /(^)=/(y + k)=(y)=(x - k), that is, Co(x — k)" -\ h c„_i (x — k) + c„ = a^" H \- a„_iX + a„. If both members of this identity be divided by x — k, if again the quotients thus obtained be divided by x — k, and so on, the successive remainders yielded by the first member, namely, c„, c„_,, • • •, will be the same as those yielded by the second member. Hence we may obtain ^ (ij) from f{x) as follows: Divide f (x) hij x — k, divide the quotient thus obtained hij X — k, and so on. The successive remainders will be c„, Cn_i, •••, Ci, and the final quotient will be % (compare § 423). The divisions should be performed synthetically. Example 1. Diminish the roots of 2 a;3 - 7 x^ _ 3x + 1 = by 4. First method. Substituting ?/ + 4 for x, we have 2 x3 _ 7 x2 _ 3 X + 1 = 2 (y + 4)3 - 7 (y + 4)2 - 3 (y + 4) + 1 = 2 2/3 + 17 2/2 + 37 2/ + 5. Second method. Arranging the reckoning as in § 423, we have 2 - 7 - 3 +1 [4 8 4 _4 2+1+1, 5 .-.03 = 5. 8 _36 2 + 9, 37 .-. C2 = 37. 8 2, 17 .-. cx^ 17 and Co = 2, Hence, as before, we find the required equation to be 2 2/3 + 17 y2 + 37 y + 5 = 0. THEORY OF EQUATIONS 441 Example 2. Increase the roots of x^ + 4 x^ + x + 3 =: by 4. To increase the roots by 4 is the same thing as to diminish them by - 4. Hence the required equation may be obtained either by substitut- ing 2/ — 4 for X or by dividing synthetically by — 4. It will be found to be 2/3 - 8 2/2 + 17 2/ - 1 = 0. By aid of § 817 we can transform a given equation into 818 another which lacks some particular power of the unknown letter. Example 1. Transform the equation x^ — 3x2 + 5x + 6 = into another which lacks the second power of the unknown letter. Substituting x = y + k, we have y'^ + {Sk — 3)y'^ + • ■ ■ . Hence we must have 3 A: — 3 = 0, that is, fc = 1. And diminishing the roots of x3 - 3x2 + 5x + 6 = by i^ ^^e obtain x^ + 2x + 9 = 0. Example 2. Transform the equation x*^ — 5 x'- + 8 x — 1 = into another which lacks the first power of the unknown letter. Substituting x = y + k, we have 2/3 + (3 /^ _ 5) 2/2 + (3 ^2 _ 10 A: + 8) 2/ + • • • = 0. Hence we must have 3 fc^ _ lO A: + 8 = 0, that is. A; = 2 or 4/3. Diminishing the roots of x^ — 5 X'^ + 8 x — 1 = by 2, we obtain x3 + x2 + 3 = 0. If, when we diminish the roots of f(x) = hy k, we obtain 819 an equation ^(a:r)=0 whose terms are all positive, ^ is a superior limit of the positive roots of /(a')= 0, § 804. For in this case (f>(x)=0 has no positive root, §794. Hence any- positive roots that f(x) = may have become negative when diminished by k. They are therefore less than k. The process of synthetic division is such that it is possible by inspection and trial to find the smallest integer k for which all the terms of (x) is rational, we shall obtain an equation F (y) = whose roots are connected with those of f(x) = by tlie relation y = <^ (x), so that if the roots of /(a-) = are ^i, (3.,, ■■■, fi,, those of F{y) = are <^ (ySi), <^(A)^ •••, <^(^„). The transformations of §§812, 814, 817 afford further illustrations of this theorem. In the first of these transformations the equation y = (x) is y = kx, in the second it is ?/ = 1 /a-, and in the third it is y = x — k. When, as in these cases, y =^ cf)(x) can be solved for x, the elimination of X is readily effected. Example 1. Find the equation whose roots are the squares of the roots of x^ + px"^ + qx + r = 0. In this case the relation y = )] [x —(a — ib)]. This product has real coefficients, for, since i^ = — 1, lx-(a + ib)] [x - (a - ib)] = (x- a)2 + b^ = x'--2ax+ (a^ + b^). Since the polynomials f{x) and x^ — 2 ax + (a^ + b'-) have the common factor x —{a + ib), they have a highest common factor. This highest common factor must be either x—{a + ib) ov x^ — 2ax+ (a^ + b^). But it cannot he x — (a + ib), since this has imaginary coefficients, whereas the highest common factor of two polynomials with real coefficients must itself have real coefficients, § 469. Hence the highest common factor oif(x) and x^-2ax+ (a^ + b^) is x^ - 2 ax + (a"" + P) ; in other words, f(x) is divisible hy x^ - 2 ax + (a- + b-), as was to be demonstrated. Example. One root of 2 x^ + 5 x^ + 4G x - 87 = is - 2 + 5 i. Solve this equation. Since - 2 + 5 i is a root, — 2 — 5 i is also a root. But the sum of all the roots is — 5/2, § 806. Hence the third root is -6/2-(-2 + 5i-2 -5i) = 3/2. THEORY OF EQUATIONS 445 Corollary 1. Every polynomial f (x) icith real coefficients is 822 the product of real factors of the first or second degree. For to each real root c of /(a*) = there corresponds the real factor x — c otf(x), § 795 ; and to each pair of imaginary roots a + ib, a — ib of f(x) = there corresponds the real factor a:- - 2 ax + (a^ + b-) oif(x), § 821. Corollary 2. The product of those factor's o/f (x) which corre- 823 sjjond to the imaginary roots of f(x) = is a function of x %iihich is 2JiJsitire for all real values of x. Tor it may be expressed as a product of factors of the form (x — ay + b'^, § 821, and every such factor, being a sum of squares, is positive for all real values of x. Corollary 3. Every equation with real coefficients whose degree 824 is odd has at least one real root. For the number of its imaginary roots, if it have any, is even, § 821, and the total number of its roots, real and imagi- nary, is odd, § 798. Hence at least one root must be real. Tims, the roots of a cubic equation with real coefficients are either all of them real, or one real and two imaginary. By the reasoning employed in § 821 it may be proved that 825 if a -\- 'wb is a root of a given equation with rational coeffi- cients, a — -y/b is also a root; it being understood that a and b are themselves rational, but "vb irrational. Irreducible equations. Let (j>(x) = be an equation whose 826 coefficients are both rational and real. We say that this equation is irreducible if (x) has no factor whose coefficients are both rational and real (compare § 486). Thus, x2 — 2 = and x- + x + 1 = are irreducible equations, but x~ — 4 = is not irreducible. Theorem. Let t(x)=0 be any equation ivhose coefficients 827 ((re both rational and real, and let ^(x)= be an irreducible equation of the same or a lower degree. 446 A COLLEGE ALGEBRA If one of the roots of ^ (x) = he a root of f (x) = 0, then all of the roots of ^ (x) = are roots of f (x) = 0. This may be proved by the reasoning of § 821. For if f(x) = and <^ (x) = have the root c in common, f(x) and (x) have the common factor x — c, ^ 795, and therefore a highest common factor which is either x — c, some factor of (x) which contains a; — c, or ^ (x) itself. But since, by hypothesis, (x)— is an irreducible equa- tion, (x)- in other words, /(x) is exactly divisible by ^(a;). Hence f(x) may be expressed in the form f(x) = Q(x) will have at least one less variation than f (x) has. For since b is positive, it follows from the rule of synthetic division, § 411, that when /(a:) is divided by a; - ^., the coeffi- cients of the quotient are positive until the first negative coefficient of f(x) is reached. If then or later one of them becomes negative or zero, they continue negative until the next positive coefficient off(x) is reached, and so on. Hence (x) can have no variations except such as occur at the same THEORY OF P:QUATI0NS 447 or earlier terms of /(a-). But since the division is, by hypoth- esis, exact, the last sign in <^ (x) must be contrary to the last sign in f{x), and therefore ^ (x) must lack the last variation in/(,r). 1 +1 -2 -10 -1 +12 -4[2 Thus, /(x) = a;6 + a;5 - 2 x* 2 -f6 + 8 - 4 - 10 + 4 - 10 x^ - x2 + 12 X - 4 is ex- 1+3+4-2-5+2, actly divisible by x - 2, tlie quotient being (x) = x^ + 3 x* + 4 x^ — 2 x^ — 5 x + 2. Observe that the first two variations of/ (x) ai-e reproduced in (x), but not the third. 1 -1 +1 -7 +2|^ Again, f(x) = x* - x-'' + x'^ - 7 x + 2 is 2 +2_ +_n —2_ exactly divisible by x — 2, the quotient 1 +1 +3 —1, being 0(x) = x^ + x2 + 3x — 1. In this case only one of the four variations of /(x) is reproduced in 0(x), and we have an illustration of the fact that when intermediate variations of f{x) disappear in 0(x), they disappear in pairs. Theorem (Descartes's rule of signs). A71 eq uation f (x) = cannot 830 have a. greater number of positive roots than it has variations, 7ior a greater number of negative roots than the equation f (— x) = has variations. 4]. For let /3i, yS.j, •■ -, (3,. denote the positive roots of /(.r) = 0. if we divide /(if) by x — /3i, the quotient thus obtained by X — (32, and so on, we obtain a final quotient c{>(x) which has at least r less variations than /(.r) has, § 829. Therefore, since (x) cannot have less than no variations, f(x) must have at least r variations, that is, at least as many as f(x) = has positive roots; 2. The negative roots of/(.r)=0 become the positive roots of /(— x) = 0, § 811. And, as just demonstrated, /(— x) = cannot have more positive roots than variations. Hence /(a-) = cannot have more negative roots than /(— a-)=0 has variations. Thus, the equation /(x) = x^ — x^ — x' + x — 1 = cannot have more than three positive roots nor more than one negative root. For /(x) = has three variations, and /(- x) = 0, that is, x^ + x^ + x^ - x - 1 = 0, has one variation. 448 A COLLEGE ALGEBRA 831 Corollary. A complete equation cannot have a greater nuin^ her of negative roots titan it has j)ernia?ie?ices. For when f(x) = is complete its permanences correspond one for one to the variations of /(— x) = 0, since of every two consecutive like signs in /(x) = one is changed in /(— x) — 0, §811. Thus, if f{x) = is x5 + X* - 6 x3 - 8 X'- - 7 X + 1 = 0, (1) en /( - x) = is x5 - X* - 6 x^ + 8 X- - 7 X - 1 = 0. (2) In (1) we have permanences at the terms x*, — Sx^, — 7x, and at the corresponding terms of (2), namely, — x'*, 8 x'-, — 7 x,| we have variations. Since (1) has two variations and three permanences, /(x) = cannot have more than two positive roots nor more than three negative roots. 832 Detection of imaginary roots. In the case of an incomplete equation we can frequently prove the existence of imaginary roots by aid of Descartes's rule of signs. Let f (x) = be an equation of the nth degree which, has no zero roots, and let v and v' denote the number of variations in f (x) = and f (— x) = respectively. The equation f (x) = must have at least n — (v + v') imaginary roots. For f{x) — cannot have more than v positive roots nor more than v' negative roots, § 830, and therefore not more than w + v' real roots all told. The rest of its n roots must therefore be imaginary. This theorem gives no information as to the imaginary roots of a complete equation, since v + v' is equal to n in such an equation. Example. Show that x^ + x^ + 1 = lias four imaginary roots. In this case /(x) = isx^ + x2 + 1 = 0, and/(-x) = is x5-x2-l = 0. Hence n — (u + u') = 5 — (0 + 1) = 4, so that there cannot be less than four imaginary roots. But since there are five roots all told and one of them is real, the degree of x^ + ic^ + 1 = being odd, § 824, there cannot be more than four imaginary roots. Hence x» + x'- + 1 = has exactly four imaginary roots. THEORY OF EQUATIONS 449 EXERCISE LXXIII 1. One root of 2 x* - x-' + 5x- + 13 x + 5 = is 1 -2i. Solve this equation. 2. One root of 2 x* - 11 x^ + 17 x- - 10 x + 2 = is 2 + V2. Solve this equatinii. 3. Find the equation of lowest degree with rational coefficients twc of whose roots are — 5 + 2 i and — 1 + v 5. 4. Find tlie irreducible equation one of whose roots is V2 + i. 5. What conclusions regarding the roots of the following equations can be drawn by aid of Descartes's rule and § 8o2? (1) X* + 1 = 0. (2) X* - x2 - 1 = 0. (3) x< + 2 x^ + x'-; + X + 1 = 0. (4) X* - 2 x' + x^ - X + 1 = 0. (5) X' + xs + x^ - X + 1 = 0. (6) x" + x^ - x2 -1=0. (7) x5 - 4 x--^ + 3 = 0. (8) x"" - x2» + X" + x + 1 = 0. 6. Show that a complete equation all of whose roots are real has as many positive roots as variations, and as many negative roots as permanences. 7. Given that all the roots of x^ + 3 x* - 1 5 x^ - 35 x^ + 54 x + 72 = are real, state how many are positive and how many negative. 8. Prove by Descartes's rule that x^" + 1 = has no real root. What conclusions can be drawn by aid of this rule regarding the roots of x2«+i + l=0? x2«-l = 0? x2'' + i-l = 0? 9. Prove that an equation which involves only even powers of x with positive coefficients cannot have a positive or a negative root. 10. Prove that an equation which involves only odd powers of x with positive coefficients has no real root except 0. 11. Show that the equation x^ + px + q = 0, where p and q are posi- tive, has but one real root, that root being negative. 12. Show that an incomplete equation which has no zero roots must have two or more imaginary roots except when, as in x* — 3 x^ + 1 = 0, the missing terms occur singly and between terms which have contrary signs. 13. Show that in any equation /(x) = with real coefficients there must be an odd number of variations between two non-consecutive contrary signs, and an even number of variations, or none, between two non-consecutive like signs. 450 A COLLEGE ALGEBRA 14. Prove that in the product of the factors corresponding to the negative and imaginary roots of an equation with real coefficients the final term is always positive, and then show that if this product has any variations their number is even. 15. Prove that when the number of variations exceeds the number of positive roots, the excess is an even number. 16. Show that x* + x^ — x^ + x — 1 = has either one or three posi- tive roots and one negative root. 17. Show that every equation of even degree whose absolute term is negative has at least one positive and one negative root. LOCATION OF IRRATIONAL ROOTS 833 Theorem 1. //'f(a) atul f(b) liavc contrari/ signs, a root of f (^xj= lies between a and b. This may be proved as in the following example. A general statement of the proof will be given subsequently. Example. Prove that /(x) = x^ — 3 x + 1 = has a root between 1 and 2. The sign of /(I) = — 1 is minus and that of /(2) = .3 is plus. By computing the values of /(x) for x = 1.1, 1.2, 1.3, • • • successively, we find two cmsecutive tenths between I and 2, namely, 1.5 and 1.6, for which /(x) has the same signs as for x = 1 and x = 2 respectively ; for /(1. 5) =- .125 is minus, and /(l.O) = .20(i is plus. By the same meth(5d we find two consecutive hundredths between 1.5 and 1.0, namely, 1.53 and 1.54, for which/(x) has the same signs as for x = 1 and X = 2 ; for /(I. .53) = - .008423 is minus aml/(1.54) = .032204 is plus. This proce.ss may be continued indefiniiely. It determines the two never-ending sequences of numbers: (a) 1, 1.5, 1.53, 1.532, • • • (b) 2, 1.0, 1.54, 1.533, • • ., the terms of which approach the .same limiting value, §§ 192, 193. Call this limiting value c. It is a root of /(x) = 0, that is, f{c) = 0. For, by § 509, if x be made to run through either of the sequences of values (a) or (b), f(x) will approach f(c) as limit. But since/(x) is always negative as x runs tlirough the sequence (a), its limit /(c) cannot be posi- tive ; and since /(x) is always positive as x runs through the sequence (b), its limit /(c) cannot be negative. Hence /(c) is zero. THEORY OF EQUATIONS 451 Theorem 2. If neAther a nor b is a root of f (x)= 0, and an 834 odd number of the roots of i(x) = lie between a and b, f (a) and f(b) have contrary signs; but if no root or an even number of roots lie between a and b, f (a) and f (b) ha^te the same sign. Conversely, if f(a) and f (b) have contrary signs, an odd nu77iber of the roots of f (x) = lie between a and b ; but (/" f (a) and f (b) have the same sign, either no root or an even number of roots lie between a and b. Suppose that a (a) f{l>) b-ft, b-ft, b~ft,. 4>{b) (^^ In the product (2) the factor ^ (a. ) / c^ (/v) is positive. For ^ (a) and <^ (b) have the same sign, since otherwise, by § 833, between a and b there would be a root of ^ (x) = and there- fore, by (1), a root of f{x) = in addition to the roots ft^, On the other hand, each of the r factors {a — fti)/(b — fti), and so on, is negative, since each of the r roots fti, ft.2, ■ ■ ■, /S^ is greater than a and less than b. Therefore, when r is odd, /(rt)//(^) is negative, that is, f(a) and f(b) have contrary signs ; but when r is even or zero, f{(i)/f{b) is positive, that is, /(«) and f(l)) have the same sign. Conversely, when f(a) and f(b) have contrary signs, so that f(c-)/f{b) is negative, it follows from (2) that r is odd; and when f(a) and f(b) have the same sign, it follows that r is even or zero. 452 A COLLEGE ALGEBRA 835 Observe that in the proofs of the preceding theorems, §§ 833,. 834, no use has been made of the assumption that every equa- tion /(x)= has a root. Kotice also that in applying these theorems a multiple root of order r is to be counted as r simple roots. From § 834 it follows that as x varies from a to h, f(x) will change its sign as x passes through each simple root or multiple root of odd order oi f(x) = which lies between a and b, and that /(a;) will experience no other changes of sign than these. Thus, iif{x) = (x - 2) (x - 3)'-(x - 4)^, and x be made to vary from 1 to 5, the sign of /(x) will be plus between x = 1 and x = 2, minus between X = 2 and x = 4, and plus between x = 4 and x = 5. 836 Location of irrational roots. By aid of the theorem of § 833 it is usually possible to determine between what pair of con- secutive integers each of the fractional and irrational roots of a given numerical equation lies. Example. Locate the roots of /(x) = x* - 6 x^ + x2 + 12 x - 6 = 0. By Descartes's rule of signs, § 830, this equation cannot have more than three positive roots nor more tlian one negative root. To locate the positive roots we compute successively /(O), /(I), /(2), • . . until three roots are accounted for by § 83.3 or until we reach a value of X which is a superior limit of the roots, § 803. Thus, using the method of synthetic division, as in § 414, we find /(O) = - 6, /(I) = 2, /(2) = - 10, /(.3) = - 42, /(4) = - 70, /{5) = - 46, /(O) = 102. Hence, § 833, one root lies between and 1, another between 1 and 2, and the third between 5 and 6. There cannot be more than one root in any of these intervals, since there are only three positive roots all told. Making a similar search for the negative root, we have/{0)=-6, /(-!)= -10, /(-2) = 38. Hence the negative root lies between -land -2. The mere substitution of integers for x in f(x) will of course not lead to the detection of all the real roots when two or more of them lie between a pair of consecutive integers. This case will be considered in § 844 and again in § 8G4, where a method is given for determining exactly how many roots lie between any given pair of numbers. THEORY OF EQUATIONS 458 EXERCISE LXXIV Locate the real roots of each of the following equations. 1. 2 x3 - 3 x2 - 9 X + 8 = 0. 2. x« + x2 - 4 x - 2 = 0. 3. x3 - 3 x^ - 2 X + 5 = 0. 4. 2 x3 + 3 x2 - 10 X - 15 = 0. 5. x3 - 4x2 -4x + 12 = 0. 6. x* + 13x2 + 54x + 71 = 0. 7. x5 + 5 X + 19 = 0. 8. X* - 95 = 0. 9. X*- 8x3 + 14x2 + 4x -8 = 0. IQ. x* + Sx^ + x2 - 13x - 7 = 0. 11. X* - 11 x5 + 32 x2 - 4 x - 46 = 0. 12. x5 + 2x* - 16x3 _ 24x2 + 48x + 32 = 0. 13. Assuming that when x is very large numerically the sign of /(x) is that of its term of highest degree, show that (1) Every equation x" + 6iX"-i + . . . + 6„ = with real coefficients, in which n is even and b,, is negative, has at least one positive and one negative root. (2) The four roots of the equation A;2 (x — 6) (x — c) + I- (x — c) (x — a) + 7/(2 (^ _ (j) ^x _ 5) _ X (x — a) (x — 6) (x — c) = lie between — co and a, a and &, b and c, c and cc respectively, it being assumed that a, 6, c, k, I, m are real and that a3, has two roots between and 1, namely, one between \/a and 1 — 1/a and one between 1 — 1/a and 1. 15. Show that x* + (x — 1) (2 x — 1) (ax — 1) = 0, where a > 5, has roots between and 1/a, 1/a and 1 — 2/a, 1 — 2/a and 1. COMPUTATION OF IRRATIONAL ROOTS Horner's method. Positive roots. There are several methods 837 by which approximate values of the irrational roots of numer- ical equations can be computed. The most expeditious of these methods is due, in its perfected form, to an English mathematician named Horner. It may best be explained in connection with an example. 15 -59L3 21 6 18 -41 39 45 454 A COLLEGE ALGEBRA Example. Find the positive root of /(x) = 2x3 _|- x2 — I5x — 59 = 0, 1. By the method of § 836, we find that the required root lies between 3 and 4. Hence if it be expressed as a decimal number, it will have the form 3. ^yS • • • , where jS, y, S, ■ ■ ■ denote its decimal figures. 2 + 1 - 15 - 59 [3 2. Diminish the roots of f{x) =. (1) by 3. We obtain the transformed equation <^(x) = 2x3 + 19x^ + 45x - 41 = 0, (2) which has the root .^y5 ■ ■ ■ lying between 13 45 and 1. 5 Testing x = .1, .2, .3, •• • in 4>(x), we find 19 that 0(.6) is - and 0(.7) is +. Hence the root of (2) lies between .6 and .7, that is, ^3 is 6, and the root of (1) to the first decimal figure is 3.6. 2+19 +45 - 41 1^ 3. Diminish the roots of (2) by .6. We obtain i/'(x) = 2x5 + 22.6x5 + 69.96X- 6.728 = 0, (3) which has the root .0 ^5 • ■ • lying between and .1. Te.sting x = .01, .02, ■ ■ ■ m ^p (x), we find that \p (.09) is — and i/- (.1) is + . Hence the root lies between .09 and .1, that is, 7 is 9, and the root of (1) to the second decimal figure is 3.69. 4. Diminish the roots of (3) by .09, and so on. 838 The reckoning may be conducted more simply than in this example, as will be shown in the following sections. But before turning from the example, observe that the absolute terms of the first and second transformed equations, (2) and (3), namely, — 41 and — 6.728, have the same sign. This is as it should be, since - 41 = <^ (0) and - 6.728 = ^ (.6). For were ^(0) and ^(.6) to have contrary signs, the root of {.%) is +. Hence 0{x) = has a root between .2 and .3 and another between .7 and .8. By Horner's method we find that tliese roots are .25560 and .77733 approximately. Hence /(x) = has the two positive roots 1.2556 and 1.77733. 845 On locating large roots. Tn case the given equation f(x) = has a root which is greater than ten, we may employ the follow- ing method for finding the figures of its integral part. THEORY OF EQUATIONS 459 To obtain the first figure, compute the values of f{x) for X = 10, 20, • • -, or, if necessary, for x = 100, 200, • • •, and so on, applying § 833. Thus, if we found that/(400) and /(500) had contrary signs, so that the root lay between 400 and 500, the first figure would be 4. To find the remaining figures, make successive transformations of the equation, as when finding the decimal figures. Thus, in the case just cited we should diminish the roots of /(cc) = by 400 and so obtain an equation (a;) = having a root between and 100. If we found that this root lay between 70 and 80, the second figure of the root v/ould be 7. We should then diminish the roots of (f)(^x)= by 70 and so obtain an equation ij/(x)=0 having a root between and 10. If we found that this root lay between 8 and 9, we should have shown that the integral part of the root of f(x) = was 478. On solving numerical equations. If asked to find uil the real 846 roots of a given numerical equation /(a*) = 0, it is best, at least when the coefficients are rational numbers, to search first for rational roots by the method of § 802. This process will yield a depressed equation (x) = is a double root of /(x) = 0, every double root of <^ (.r) = is a triple root off(x) = 0, and so on. For, § 850, if c/) (,/•) is divisible by {x — ay, then /' (x) is divisible by (x — ay and f{x) by (x — a)'' + ^ Observe that if the quotient /(a-)/ <^(a') be F(x), the roots of F{x) = are those of f(x) = 0, each counted once. Example. Find the multiple roots, if any, of the equation /(x) = x5 - X* - 5 xs + x2 + 8 X + 4 = 0. Here f (x) = Sx* - 4x3 _ i5j;2 4. 2x + 8, and by § 465 we find the highest common factor of f{x) and /' (x) to be (x) = x^ — 3x — 2. The roots of 0(x) = may be found by § 802 and are - 1, - 1, 2. Hence /(x) = has the triple root — 1 and the double root 2, that is, its roots are - 1, — 1, — 1, 2, 2. 464 A COLLEGE ALGEBRA Observe that f{x) = (x + l)^ (x - 2)2, that /'(x) = (X + 1)2 (X - 2) (5x - 4), and that F(x) =f{x)/ so that b = |^o|> a,nd let x' denote the numerical value of x. Then b-^x + b^x'^ + • • • is numerically less than (or equal to) gx' + gx'^ -\ OT g(x' + x'^ ^ ), § 235, and therefore when x' < 1, it is less than gx' /(I — x'), § 704. But gx' / (1 — x') is less than b when x' < b/(b + g). Second, let f(x) = biX + b^x^ + b^x^ -\ , so that b=\bi\. We then have \b^x^ + b^x^ -\ 1 < |^>ia;| when \b^x + b^x^ + • • • | < |^i|, that is, when x' < b / (b + g), and so on. Thus, if /{x) = 5x + 3x2-9xS we have |3x2 - 9x*|<(5x| when x'<5/(5 + 9), that is, when x'<6/14. 466 A COLLEGE ALGEBRA 855 Theorem. Let f (x) denote a polynomial arranged in descend-^ ing poivers of x, and let a denote the numerical value of its leading coefficient and g that of its numerically greatest coeffi- cient. The leading term of f (x) will he numerically greater than the sum of the remaining terms for all values of x which are numerically greater than (a + g)/a. For let/(a;) = a^x"" + a-^pcr--^ + h «„, so that a = \a^\, and let ic' denote the numerical value of x. We have a^"" + a^x''-^ H h «„ = a;" («n + ai/x H 1- «"/.i-"). Hence |aoX"|>|aia:;''-*H |-a„| when |a„|> jaj/ic-^ |-«„/ic"|. But, § 854, |«o|>|«i/a^ -\ h «„/«"! when l/a;' {a -\- g) / a. Thus, if /(x) = 3x3 + x2 _ 7a; + 2, we have |3x3|>|x2 - 7x + 2] when x'>(3 + 7)/3, that is, wlien x'> 10/3. From this theorem it evidently follows that the number (a + g)/a is greater than the absolute or numerical value of any root of the equation f(x) = 0, whether the root be real or imaginary. 856 Theorem. If 2. is a root oft(x) = 0, the values of f (x) and f'(x) haoe contrary signs when x is slightly less than a, ayid the same sign when x is slightly greater than a. For express /(.r) and/'(a-) in powers of .r - a, § 849, and then divide the first expression by the second. When a is a simple root, so that /(a) = but/'(«) -^ 0, the result may be reduced to the form fM,=U_ „^ /'( ^)+.r(^0(^-«)/2! + --- f'{x) ^ ^ /'(a)+/"(a)(x-a) + ... The numerator and denominator of the fraction on the right are polynomials in cc — a. Hence for all values oi x — a which are small enough to meet the requirements of § 854 their signs will be those of their cominon leading term /'(a), and the fraction itself will be positive. The sign oi f{x)/f'{x) will then be the same as that of ic — a and therefore minus or plus according as x < a or x > a. But when the sign of f{x) //' {x) THEORY OF EQUATIONS 467 is minus,/(.r) and/' (a:-) have contrary signs, and. when the sign of f(x)/f' (x) is plus, /(ic) and /'(a;) have the same sign. Wlien a is a multiple root of order r we have, § 850, f(x) _ f^''\a) / r\ + terms involving (x — a) f'{x) ^ /^'H'*)/(*' — 1)! + terjns involving {x — a) from which the theorem follows by the same reasoning as when a is a simple root. RoUe's theorem. Between two consecutive roots ofi(x)—0 there is always a. root o/f'(x) = 0. For let Py and /So be the roots in question, and let c denote a number slightly greater than ySi and d a number slightly less than /5o, so that ^i < c < d < ft.. Then f'(c) has the same sign as f(c), § 856, and /(c) has the same sign as f(d), § 834 ; but f(d) has a different sign from that of /' (d), § 856. Hence /' (c) and /' (d) have contrary signs. Therefore a root of /'(a')=0 lies between c and d, that is, between fi^ and ^.,, § 833. Thus, if fix) = x2 - 3 a: + 2 = 0, then f'{x) = is 2 x - 3 = 0. The roots of /(x) = are 1 and 2, the root of f'(x) = is 3/2, and 3/2 lies between 1 and 2. Example. Prove that f{x) = x^ + x2 - 10 a; + 9 = has two roots between 1 and 2. (Compare §844, Ex.) Since /(I) = 1 and/(2) = 1, there are two roots or none between 1 and 2. If there are two roots, /' (x) = must also have a root between 1 and 2, and this root must lie between the two roots of /(x) = 0. But /'(x) = 3x'- + 2x — 10 = has a root between 1 and 2, for /' (1) = — 5 and /' (2) = fl. Solving, we find that this root is 1.5 approxi- mately. Moreover /(1. 5) = — .375 is minus. Therefore, since both /(I) and f{2) are plus, /(x) = has two roots between 1 and 2, namely, one between 1 and 1.5, and another between 1.5 and 2. Theorem. If the variable x is increasing, then, as it passes through the value a, the value o/f (x) is increasing if t' (a,) > 0, but decreasing //f'(a) < 0. /f f'(a)=: but f"(a) ^ 0, f (a) is a maximum value ofi(x) when f''(a) <. 0, a minimum value when f"(a) > 0. 468 A COLLEGE ALGEBRA For by § 849 we have f{x)-f{a)=f\a){x - a)+f"{a){x - ay/2 ! + ••.. The second member is a polynomial in x — a, and for all values of x which make x — a small enough numerically to meet the requirement of § 854, the leading term will control the sign of the entire expression and therefore that of /(a-)— /(a). We shall suppose x restricted to such values. Then 1. If f'{a) > 0, fia) {x - a), and therefore f{x) -f{a), has the same sign as (x — a). Therefore, since x — a changes from minus to plus as x passfct, through a, the same is true of f(x)—f{a), that is, /(x) is then increasing from a value less than/(«) to a value greater than /(re). 2. If f'{a) < 0, /'(«) {x - a) and (x - a) have contrary signs. Hence, reasoning as in 1, we conclude that f(x) is decreasing as x passes through a. 3. If/'(«)=0 but/"(a)^0, the sign oif(x)-f(a) is that of /"(«)(.!■ — r^y-/ 2 and therefore that off"(a); for (x — a)'^ is positive whether x a. Hence when /"{") < 0, we have/(u:-) 0, we may show that f{a) is a minlviiDii value of /(a*). It may be added that if /" {a) = but /'" (a) ^ 0, /(a) is not a maximum or minimum value of f{x) (see § 859, Ex. 2). And, in general, if all the derivatives from the hrst to the rth, but not the {r + l)th, vanish when x = a, f(a) is a maximum or minimum when r is odd, but not when r is even. Example. Is/(x) = x^ — Gx^ + 9x — 1 increasing or decreasing as x, increasing, passes through the value 2 ? Find the maxinuun and mininuim values of/(x). We find/' (x) = 3 x2 - 12 X + 9 =-- 3 (X - 1) (x - 3). Hence/' (2) = - 3 is negative. Therefore /(x) is decreasing as x passes through 2, THEORY OF EQUATIONS 469 We have /' (x) = when x = 1 and when x = 3. Moreover/" (x) = 6 x — 12, and therefore /" (1) = - 6 is negative and /" (3) =: 6 is positive. Hence /(I) = 3 is a maximum value of /(x), and /(3)= — Lis a minimum value. Variation of f (x). Let us now consider how the value of a polynomial f(x) with real coefficients varies when x varies continuously, § 214, from — co to + oc. Example 1. Discuss the variation of /(x) = x^ — 2x2 — x + 2. The roots of /(x) = are - 1, 1, 2, and /(x) = (x + 1) (x - 1) (x - 2). Hence, when x = — co, /(x) = — co • when x is between -co and — 1, /(x) is negative ; when x = — 1, /(x) >- ^J ; when x is between — 1 and 1, /(x) is positive ; when x = 1, /(x) — ; wlien x is between 1 and 2, /(x) is negative ; when x = 2, /(x) = ; when x is between 2 and cc, /(x) is positive ; when x = cc, /(x) = co. The roots of /' (x) = 3x2 - 4x - 1 = are (2 ± V7)/3, or - .2 and 1.5 approximately. When x < (2 - V?) / 3 and when x > (2 +y7) /3, /' (x) is positive, but when x is between (2 — V7)/3 and (2 -f V7)/3, /'(x) is negative. Therefore, § 858, /(x) is continually increasing as x varies from — co to (2 — V7)/3, is continually decreasing as x varies from (2 — V7)/3 to (2+V7)/3, and is again continually increasing as x varies from (2 + V7)/3 to 00. It follows from this, § 639, that /(x) has a maximum value when X = (2 — Vt) /3, and a minimum value when x = (2 + V?) /3. This is in agreement with § 858, for/" (x) = 6 x — 4 is negative when x = (2 — V7)/3, and positive when x = (2 + V7)/3. The variation of f(x) will be ex- hibited to the eye if we put y =:f{x) and then construct the graph of this equation by the method of § 389. We thus obtain the curve indicated in the accompanying figure. The points A, B, C at which the curve cuts the X-axis are the graphs of the roots - 1, 1, 2 of /(x) = 0. The por- tions of the curve above the x-axis correspond to positive values of /(x), those below to negative values. The uppermost point on the curve between A and B corresponds to the maximum value of /(x), the lowermost point between B and C to the minimum value. 859 470 A COLLEGE ALGEBRA 860 As X varies from — » to x the corresponding point on the curve moves from an infinite distance belov? the x-axis upward and to the right through A to the maximum point, then downward through B to the minimum point, then upward again through C to an infinite distance above the cc-axis. If we were gradually to increase the absolute term of /(x), the graph of y =f{x) would be shifted vertically upward and the points B and C would at first approach coincidence in a point of tangency and then dis- appear. The corresponding roots of /(j) = would at first become equal and then imaginary. Example 2. Discuss the variation of /(x) = x* — 2 x^ + 2 x — 1. The roots of /(x) = are - 1, 1, 1, 1, and/(x) = (x + 1) (x - 1)-^. Hence when x = ± oo, /(x) = co ; when x = ± 1 , f{3^) = ; when x < — 1 and when x>l, f(x) is positive; when x is between — 1 and 1, /(x) is negative. Here /' (x) = 4 x^ - 6 x2 + 2 = 2 (2 x + 1) (x - 1)2, and the roots of /'(x) = are —1/2, 1, 1. When x< — l/2,/'(x) is negative; when x is between — 1/2 and 1 and also when x > 1, /' (x) is positive. Therefore, §858, f(x) is continually decreasing as x varies from — oo to —1/2, and continually increasing as x var es from — 1/2 to 1 and from 1 to oo. Hence f(x) has a minimum value when X = — 1/2, but it has neither a maxi- mum nor a minimum value when x = 1. This is in agreement with § 858. For /" (x) = 12 x2 - 12 X = 12 X (X - 1) and /'"(x) = 24x — 12. Hence when X = — 1 /2, /" (x) is positive ; but when X = 1 , /" (X) = and /'" (x) ^ 0. The graph of y =f{x) has the form indicated in the accompanying figure. The point A where the curve merely cuts the X-axis corresponds to the root — 1 of /(x) = 0, and the point B where the curve both touches and crosses the X-axis corresponds to the triple root 1. The lowermost point of the curve corresponds to the minimum value of /(x). Its coordinates are —1/2, —27/16. As in these examples, so in general, iif(x) is of odd degree, its leading coefficient being positive, when x varies from — oo to -f- cc,f(x) increases from — oo to the first maximum value, then decreases to the first minimum value, and so on, and finally THEORY OF EQUATIONS 471 increases from the last miniuiuin value to + oo. It is possible, however, that there are no maximum or minimum values, for the equation /'(x) = 0, being of even degree, may have no real root. The graph of y =/(a-) extends from an infinite distance below the a;-axis to an infinite distance above the cc-axis. It crosses the a;-axis an odd number of times, — once at least. On the other hand, if f{x) is of even degree, f{x) begins by- decreasing from + CO to the first minimum value and ends by increasing from the last minimum value to + oo. In this case the graph of y —f{x) need not cross the a;-axis at all. If it does cross the axis, it crosses an even number of times. In most cases we can obtain a sufficiently accurate represen- tation of the graph of y =f(x) by the method of § 889, which consists in assigning a series of values to cc, computing the corresponding values of y, plotting the pairs of values of x, y thus found, and passing a " smooth " curve through all these points. Such a curve will indicate roughly where the true graph crosses the cc-axis and where its maximum and minimum points lie. But to obtain the points of crossing with exactness, we must solve the equation f{x) = ; and to obtain the actual positions of the maximum and minimum points, we must solve the equation /' (x) = 0. To every multiple root of f(x) = there corresponds a point of tangency of the graph with the X-axis. If the order of the multiple root is odd, the graph also crosses the x-axis at this point. EXERCISE LXXVII 1. Discuss the variation of /(«) = (x + 1) (a; - 2)2 = x^ - 3x2 + 4, finding its maximum and minimum values if any, and draw the graph of y=f{^). 2. Treat in a similar manner each of the following functions. (I)2x2-x + l. (2) (x + l)(x-2)(2x-l). (3) x3 - 12x + 14. (4) x3 - 5x2 + 33; + 9. (5) x3 - 3 x2 + 5. (6) (X + 1)2 (X - 2)2. (7) (x2 + X + 1) (X + 2). (8) x(x - 1) (X 4 2) (x + 3). 472 A COLLEGE ALGEBRA 3. Find the graphs of each of the foHowing fractional equations by plotting the points corresponding to x = — 1, — 1/2, 0, 1/2, 1, • • •, 4. (1),= ^^^-'^ ■ i2)y= "(^-^) ■ (X - 2) (X - 3) (X - 1) (X - 3) STURM'S THEOREM 361 Sturm functions. Let f(x) = be any equation which has no multiple roots, and let/i(a;) be the first derivative of /(cr). Divide f{x) by fi(x') and call the quotient qi, and the remainder, Avith its sign changed, /a (x). Again, divide /i (x) by /g (x), and call the quotient qo, and the remainder, with its sign changed, /a (a:). And so on, modifying the ordinary process of finding the highest common factor off(x) and/i(a;) in this respect only: tJie sl{/n of each remainder is changed, and care is taken to make no other chanc/es of sign than these.. Since /(ic) == has no multiple roots and therefore /(cc) and fi(x) have no common factor, § 851, we shall finally obtain a remainder which is a constant different from 0, § 465. Call this remainder, with its sign changed, /„,. The sequence of functions consisting of the given polynomial, its first derivative, and the several remainders in order, each with its sign changed, is called a sequence of Sturm, or a sequence of Sturm functio7is. 862 Relations among the Sturm functions. These functions are, by definition, connected by the series of identical equations f{x)^q,f(^x)-f{x), (1) M^)^'i.A{^)-M^), (2) f,(x)^q,f,{x)-f,(x), (3) fm -2^^)= 1,n - J.n - 1 (^) " fm- (m - 1) THEORY OF EQUATIONS 473 From these equations we conclude that 1. Two consecutive functions cannot vanish for the same value of X. Thus, if both/i(a;) and/2(a;) vanish when a:; = c, it follows from (2) that fz{x) also vanishes; therefore, from (3), that fi{x) vanishes; and therefore, finally, that /„, is 0. But this is contrary to hypothesis. 2. When for a certain value of x one of the intermediate functions f^ (x), fg (x), • • •, fm-i (a;) vanishes, the functions tvhich immediately precede and follow it have opposite signs. Thus, if /a (c) = 0, it follows from (2) that /i (c) = - /s (c). Sturm's theorem. Let a and b he any two real numbers neither 863 of which is a root o/f (x)= 0. The difference betiveen the number of variations of sign in the sequence r / \ * / x ^ / \ * f(a), fi(a), f2(a), ••-, f„ and that in the sequence f(b), f,(b), f,(b), ..., f„ is the number of roots of i(yi) = which lie between a and b. To fix the ideas, suppose a < b, and suppose x to vary con- tinuously from ft to ft, § 214. As X varies from a to b, the sign of the constant /„, remains unchanged, and the only changes which are possible in the signs of the remaining functions, and therefore in the number of variations of ^n in the sequence f(x),A(x),f,(x), ...,/„, are such as may occur when x passes through roots of the equations f(x) = 0, fi(x) = 0, and so on, § 835. But 1. The number of variations in the sequence is neither increased nor diminished when any function except the firsty f (x), changes its sign. 474 A COLLEGE ALGEBRA Suppose, for instance, that c is a root of /, (x) = and that /a (x) changes from plus to minus as x passes through c. Since c is a root of /g (x) = 0, it cannot be a root of either fi (x) = or /a (x) = 0, § 862. And if we take a positive number 7^ so jmall that j ig_ root of fiJ^X^^ ^^' o^ Ai^l^ ^ lies between c —Jj and c, or between g. and c + h, neither of the functions fi(x) or /^(x) will change its sign as x varies from c - A to c + h, § 835. But when x = c,f^{x) and/3(a:) have opposite signs, § 862. Suppose that /i(c) is plus; then /3(c) is minus, and we have the following scheme of the signs o^ fi{x), /^{x), f^(x) for values of x between c — h and c -\- h : fiix) h{x) fz{x) x = c — h c + h + + + a Eor all these values of x, therefore, the signs of the three functions A(x), f,(x), f,(x) present o7ie variation. The only effect of the change of the sign oif^^x), at least in this part of the sequence, is a change in the ^osiVion of a variation. And there can be no cliange in the number of variations in the part of the sequence which follows /g (ic) as x passes through c. For either no function after fz(x) will change its sign as x passes through c, or, if it does, we shall have the case of the function /j (x) over again, and the only effect will be a change in the position of another variation. 2. The sequence loses one variation for each root of i(x) = through which x passes in vanjing from a to b. For let c be a root oif(x) = 0. The functions /(a;) and/i(ir) have opposite signs for values of X which are slightly less than c, and the same sign for values of x which are slightly greater than c, § 856. THEORY OF EQUATIONS 475 In other words, the sequence has a variation between f{x) and./i(a') just before x reaches c, and this variation is lost as X passes through c. Hence the sequence of Sturm functions never gains a varia- tion as X varies from a to b. But, on the other hand, it loses a variation each time that x passes through a root oi f(x) = 0, and then only. Therefore the difference between the number of variations and in f(b), f,{b), f,{b), ••.,/„. is the number of single variations that are lost as x passes through the roots of /(x) = which lie between a and b. In other words, it is the number of these roots, as was to be demonstrated. If we apply the method of § 861 to an equatiou /(x) = which has multiple Yooi?., we obtain a sequence/ (x), /i (x), • • •,/,„(x), (1) the last term of which is the highest common factor of all the terms, § 4(55. Divide all the terms of (1) by f,„{x). We thus obtain a sequence of the form 0(x), 01 (x), • • • , 1, (2) which, as is easily shown, possesses all the properties on which Sturm's theorem depends. Hence the number of roots of (x) = 0, that is, § 852, the number of different roots of f{x) = 0, between a and 6, is the difference between the number of variations in 0(a), 0i(a), • • •, 1 and in 0(6), 1, and fx{x) is positive when x>l. Hence /i (17/13) is jiositive, and therefore /s is negative. For X = — 00 the signs of /(x), /i (x), /2(x), fz are — , +,—,—; for X = 00 they are +, +, +, — . Hence f{x) — has but one real root. THEORY OF EQUATIONS .477 The only property of the final function /„ of which any use is made in the proof in § 863 is that its sign is constant. Hence if, when computing the Sturm functions of f(x) = in order to find the number of roots between a and h, we come upon a function jf^ (a;) which has the same sign for all values of X between a and b, we need not compute the subsequent functions. For it follows from the proof in § 863 that the required number of roots will be the difference between the number of variations in f{ci), ■■•,fp(a) and in f{b), ■■ ■,fp{b). Example 3. How many real roots has/(x) = x3 + x2 + x + l = 0? Here /i (x) = Sx^ + 2 x +1, and, since 22 < 4 • 3, this is positive for all real values of x, §§ 035, 823. Hence we need not compute /o (x) and /s. The signs of /(x), /i (x) for x = - oo are -, + ; for x = co they are + , +. Hence /(x) = has one real root. EXERCISE LXXVm By aid of Sturm's theorem find the situation of the real roots of the following equations. 1. x3 - G x2 + 5 X + 13 = 0. 2. x3 - 4 x2 - 10 X + 41 = 0. 3. x3 + 5x + 2 = 0. 4. x3 + 3x2 + 8x + 8 = 0. 5. x3-x2- 15x4- 28 = 0. 6. x* - 4x3- Sx^ +18x + 20 = 0. 7. 2x4-3x2 + 3x-l = 0. 8. x* - 8x3 + 19x2 -12x + 2 = 0. 9. x-i - 12 x2 + 12 X - 3 = 0. 10. x* + 2 x3 - 6 x2 - 8 X + 9 = 0. By aid of Sturm's theorem find the number of the real roots of each of the following equations. 11. 4x3 -2x- 5 = 0. 12. x* + x3 + x2 + x + l = 0. 13. X" + 1 = 0. 14. X* - 6 x3 + x2 + 14 X - 14 = 0. 15. Let /(x) = be an equation of the ?ith degree without multiple roots. Show that the condition that all the roots of /(x) = be real is that there be n + 1 terms in its sequence of Sturm functions /(x), /i (x), • • •,/«) and that the leading terms of all these functions have the same sign. 16. By aid of the theorem in Ex. 15 prove that the condition that all the roots of the cubic x3 + px + 5 = be real and unequal is that 4j)3 -I- 27 q2 be negative. 478 A COLLEGE ALGEBRA SYMMETRIC FUNCTIONS OF THE ROOTS 865 Theorem 1. If the roots oft(x) = x" + bix"-' -\ h b„ = are /3^, /3„ ■ ■ ■ , fin, so that f (x) = (x - 13,) (x - ^,) • • • (x - (3,), Thus, suppose that 7i = 3, so that - f(x) = (x-^,)(x-^.;)(x-p,). (1) Substituting x + h for x in (1), we have f(x + A) = [(X - /30 + A] [(X -/?,)+ /O [(^ - iSa) + A]. (2) We can reduce each member of (2) to the form of a poly- nomial in h, the first member by Taylor's theorem, § 848, the second by continued multiplication, as in § 558. Since (2) is an identity, the coefficients of like powers of /* in the two polynomials thus obtained must be equal, § 284. But since f(x + h)=f(x) +f'(x)h^ , § 848, the coeffi- cient of h in the first polynomial is f'(x). In the second it is (x - (3,) {X - (3,) + (x- ^83) (X - (30 + (x- /30 (x - (3,). ^^""^^ /' (x) = (x- A) (X - ^3) + (x - A) (^ - A) X - /3i'^ X - ^, X - /3,' ^ ^ since (x — ySg) (x — (3^) =f{x)/(x — (3^, and so on. This reasoning is applicable to an equation of any degree n. In the general case there are 71 factors in the second member of (2), and when this member is reduced to the form of a polynomial in h, the coefficient of h is the sum of the products of the binomials x — ft,, x — /3.,, ■ ■ ■ , x — y8,„ taken ?i — 1 at a time, § 558. That one of these products which lacks the factor X — fti may be written f(x) / {x — /3i), and so on. THEORY OF EQUATIONS 479 Thus, if /(x) = x-^-Gx2 + llx-G = (x-l)(x-2)(x-3), we have /' (x) = Sx^ — 12 x + 11 and (X - 2) (X - 3) + (x - 3) (x - 1) + (x - 1) (x - 2) = 3x2 - 12x + n. Theorem 2. Tlte sums of like powers of all the roots of an equation f (x)= can he expressed rationallij in tervis of its coefficients. Thus, suppose that the equation is /(a-) = ic^ + b^x"" + b.x + ^3 = 0. (1) Let a, /3, y denote the roots of (1) and let s^, s^, ■ • -, s,. have the meanings si = « + /? + y, 52 = a2 + ^- + yV • ■, s, =.«'• + /S'- + y'-. We are to prove that s^, s^, ■ ■ ■ can be expressed rationally in terms of the coefficients bi, h^, b^. 1. By the preceding theorem, § 865, we have ^ ^ X — a X — ft X — y ^ Since f(x) is divisible by x — a, x — (3, and x — y, each of the fractions in (2) represents a polynomial in x which can be found by the rule of § 410. Applying this rule and then adding the results, we have f(x) I {x - a) = a-^ + (a + ^-i) a; + {a^ + b^a + b^) f{x) / (;r - /?) = a:2 + (^ 4- b,) X + {ft' + b.ft + b.:) f(x)/(x - y) = a;2 ^(y + b{)X+(y' + b,y + b,) f (x) =3x^^+(s, + 3b,)x+ (52 + b,s, + 3 ^-2) (3) But by definition, § 847, we also have f'{x)=3x'' + 2b,x + b,. (4) Equating the coefficients of like powers of x in the two expressions (3) and (4) and solving for s^, s„, we have s, + 3b, = 2b„ .-. s,=-b„ (5) S2 + b,si + 3b2 = b2, .-. s, = bl-2 b^. (6) 480 A COLLEGE ALGEBRA 2. From the values thus found for s^ and s^ we can obtain the values of s^, s^, ■ •■ successively as follows : Since a, /8, y are the roots of (1), we have a' + bia^ + ha + b^ ^ 0, (7) P^ + b,/3'+b,/3 + b, = 0, (8) y' + b,y' + b,y + b, = 0. (9) Adding these identities, we obtain S3 + ^1*2 + Ml + 3 Z>3 = 0, (10) which gives 53 rationally in terms of bi, b^, b^, s^, s.2 and there- fore, by (5), (6), rationally in terms of bi, bo, b^. Next multiply the identities (7), (8), (9) by a, ft, y respec- tively and add the results. We obtain Si + b,s, + b^s, + b,s, = 0, (11) which by aid of (5), (6), (10) gives S4 rationally in terms of bi, &2) K And in like manner, if we multiply (7), (8), (9) respectively by a^, p-, y^, by a^, [i^, y^, and so on, and after each series of multiplications add results, we obtain identities S5 + ^1*4 + ^2^3 + ^^3«2 = 0, Se + ^1^5 + l^"Ji + '''3S3 = 0, • • •, which give S5, s^, ■•• rationally in terms of b^, b.., b^. By similar reasoning the theorem may be proved for an equation /(;r) = of any degree n. Example 1. If a-, /3, 7 denote the roots oi x'^ — 2 x"^ + 4 x + 2 = 0, find Sl/a = l/a + l//3 + l/7, 2 l/a2 = l/a^ + 1/^2 + 1/^2^ S l/a3 = l/a3 + I//33 + 1/73. Applying the transformation x — \/y, and dividing the transformed equation by the coefficient of y^, we have y^ -\-2'\p- — y ■\-\ /2 = ^. For this equation, by substituting 61 = 2, 62 = — 1, &3 = 1 /2 in the formulas (5), (6), (10) above, we obtain Sj = — 2, S2 = 6, S3 = — 31/2. Therefore, §814, Sl/a = -2, Sl/a2 = 6, Sl/a3 = _3i/2. THEORY OF EQUATIONS 481 Example 2. For the equation /(i) = x* + 6ix^ + h2x'^ + 63X + 64 = 0, show that Si + 4 61 = 6i, S2 + fti'Si + 4 60 = 2 62, Ss + 61S2 + 62S1 + 4 63 = 63, Si + 61S3 + 62S2 + 63S1 + 4 64 = 0, 85 + 61S4 + 62S3 + 63S2 + 64S1 = 0, and compute Sj, S2, S3, S4 in terms of 61, 625 ^3, ^4- The preceding formulas also show that s^, s^, S3, • • ■ are infe- 867 f/ral functions of the coefficients of the equation when, as in (1), the leading coefficient is 1. Theorem 2. Every rational symmetric function of the roots 868 of an equation f(x)=0 can be expressed rationally in terms of its coefficients. Let the roots of f(x) = be or, ;8, y, • • • , v. Every rational symmetric function of a, /3, •••, v can be expressed rationally in terms of functions of the types Sor'', Sa"/??, '^crJ' fi'J-f, and so on, § 544. Hence it is only necessary to prove our theorem for functions of these several types. This was done for the type 2tt" = s^ in § 866, and we shall now show that 2«''^, and so on, can be expressed rationally in terms of functions of this type s^,. 1. The type 2a'^/ff' = a''/?' + /J^Vt' -\ . The product (a'^ + yS'' H ) (a' + /J' H ) (1) is the sum of the two symmetric groups of terms aP + n^pp + 1^ , (2) N a^P^ + I3"a'' + ---. (3) But (1) and (2) are rational functions of the coefficients, § 866. Hence (3), or Sa''/?', which may be obtained by sub- tracting (2) from (1), is also such a function. Since (1) is s^s^ and (2) is s^^^, we have the formula ^"''^^V.-^Pfr (4) When p = q, the terms of (3) are equal in pairs and (3) becomes 2 Sa^/S^. We then have 2 %aPpp = sj — s^p^ 482 A collp:ge algebra 2. The type Sa'^yS'^y'" = a'^l^y + /?'aY H • The product (a''/?' + ^''a'' -] ) (a-- + /?'- + y'- h ) (5) is the sum of the three symmetric groups of terms ^7.+ r^7^^,. + r,^7_j ^ (6) a^'^' + '- + y8"a' + '-H , (7) «'')8Y" + /3''aV +•••• (8) But we have already shown that (5), (6), and (7) are rational functions of the coefficients. Hence (8), or Sa^yS^y'", which may be obtained by subtracting the sum of (6) and (7) from (5), is also such a function. When 2^ = q, the group (8) becomes 2 Sa^'yS^y''; when p = q=zr, it becomes 6 Sa'^^S^'y''. 3. The types ^a''(3Y^", and so on. We may prove that these are rational functions of the coeffi- cients by repetitions of the process illustrated in 1 and 2. We begin by multiplying 2a''j8''y'' by a" + /3" + y" • • • . Example. Show that I,aP^iy = Sj,SqSr + 2Sj, + ,i + r - Sj, + rjSr - S^ + rSp - Sr + pS,. EXERCISE LXXIX 1. For the equation aoz^ + aiX^ + a2X + as = find S3 and 84 in terms of ao, «1, «2i <^3- 2. If a, p, 7 denote the roots of x^ + px2 + qx + r - 0, find S 1/a^, 2 1/ a^, and S afi'^ in terms of p, q, r. 3. Find the equation whose roots are the cubes of the roots of a;3_2x2 + 3x-l = 0. 4. If a, 13, 7 denote the roots of the equation x^ - x^ + 3 x + 4 = 0, find the values of the following symmetric functions of these roots by the methods of §§ 866, 867. (1) si, So, 83, 84. (2) Sa3/J2. (3) Sa3^7. (4) Sa3i327. (5) 21/ a*. (6) Sa^^/V. CUBIC AXD BIQUADRATIC EQUATIONS 483 XXX. THE GENERAL CUBIC AND BIQUADRATIC EQUATIONS Algebraic solutions. In the preceding chapter we have shown 869 that the real roots of a numerical equation can always be found exactly or approximately, and it is possible to extend the methods there employed to the complex roots of such equa- tions. As hardly need be said, these methods are not appli- cable to literal equations. To solve such an equation we must obtain expressions for its roots in terms of its coefficients. We say that an equation can be solved algebraically when its roots can be expressed in terms of its coefficients by apply- ing a finite number of times the several algebraic operations, namely, addition, subtraction, multiplication, division, involu- tion and evolution. We have already proved, § 631, that the general quadratic equation has such an algebraic solution, and we are now to prove that the like is true of the general cubic and biquadratic equations. But general equations of a degree higher than four cannot be solved algebraically. Cube roots of unity. In § Q^Q we showed that the equa- 870 tion x^ = 1 has the roots !,(-!+ i V3)/2, (- 1 - i V3)/2. Hence each of these numbers is a cube root of unity. The third will be found to be the square of the second. Hence if we represent the second by w, we may represent the third by w^. Since x^ — 1 =-0 lacks an x^ term, we have, § 805, 1 + w + w^ = 0. Similarly every number a has three cube roots, namely, the three roots of the equation x^ — a. If one of these roots be Va, the other two will be w Va and w^ Va. The general cubic. Cardan's formula. By the method of § 818 871 every cubic equation can be reduced to the form x^+px + q = 0, (1) in which the x"^ term is lacking. 484 A COLLEGE ALGEBRA In (1) put x=^y + z. (2) We obtain tf + ^,fz^Syz^ + z'^+p{y + z)+q = 0, or y^ + z^+(Zyz+p){y^z) + q = Q. (3) As the variables y and z are subject to the single condition (3), we may impose a second condition upon them. We suppose ^yz + p = 0, (4) and therefore, by (3), y^ + z"" + q = 0. (5) From (5), y^ + z'' = -q, (6) and from (4), y^z^=-p^l21. (7) Therefore, § 636, y^ and z^ are the roots of a quadratic equa- tion of the form ,,2 + ^„_^3y27 = 0. (8) Solving (8) and representing the expressions obtained for the roots by A and B respectively, we have 2/' These equations (9) give three values for y and three for s, namely, § 870, y = ^, wVCT, 0)2^, (10) z = ^, w-^B, u)2^. (11) But by (4), yz = — p/3, and the only pairs of the values of y, z in (10), (11) which satisfy this condition are y, z = -y/A, -y/B ; w -y/l, w- -^/b ; w^ "vCT, w ^/b. Substituting these pairs in (2), we obtain the three roots of (1), namely, iCi = V^ + -y/Ji, Xo = w V^ J + to^ ^B, Xs = -^B, where ., = _ | + ^fTI. B =- | - Vf^l- (12) CUBIC AND BIQUADRATIC EQUATIONS 485 Example 1 . Solve x^ - 6 x^ + 6 x - 2 = 0. By § 818 we find that the substitution x = y -\- 2 will transform the given equation to the form y^ + py + q = 0. We thus obtain 2/3 _ G ?/ - 6 = 0. The roots of this equation, found by substituting p=— 6, g = — 6in the above formulas, are 3,— 3,— 3/— a r- ^r- '^ r- V4 + V2, V4w + V2w2, V4a;2 + V2w. Hence the roots -of the given equation are 2 + v'i + V2, 2 •+ V^ t^ + V2 u)2, 2 + \/4 a.2 + 'v'i w. Example 2. Solve the equation x^ -\- o x"^ -\- Q x + b = 0. Discussion of the solution. When p and q are real, the char- 872 acter of the roots depends on tlie value of y^/4 +^j^/27 as follows : 1. i,"^ q^/4 + p^/27 > 0, one root is real, two are imaginary. For in this case A and B are real. Hence x^ is real, and x^, ^3 are conjugate imaginaries, § 870. 2. If q^/4 -f- p'/27 = 0, all the roots are real and two equal. For in this case A = B = — q /2. Hence rri=— 2V^/2, 3/ 3/ and X.2 = iCj = — (w + or) V y/2 = \q/2, since w + w'^ = — 1, by § 870. 3. i/'q"/4 + p^/27 < 0, all the roots are real and unequal. This may be proved by Sturm's theorem (see p. 477, Ex. 16). I>ut when q^ / ^ +p^ /21 < 0, A and B are complex numbers, and though the expressions for a-j, X2, Xs denote real numbers, they cannot be reduced to a real form by algebraic transforma- tions. This is therefore called the irreducible case of the culuc (see § 885). The expression q^/i +^j^/27 is called the discriminant of 873 the cubic x^ +/?x -f 5- = 0, since its vanishing is the condition that two of the roots be equal (compare § 635). 486 A COLLEGE ALGEBRA 874 The general biquadratic. Ferrari's solution. By the method of § 818 every biquadratic equation can be reduced to the form ; X* + ax^ + bx + c = 0. (1) With a view to transforming the first member of (1) into a difference of squares, we add and subtract xhi + i<^/4, where u denotes a constant whose value is to be found. We thus obtain _ X* + x^-u + u^/i - xhi — ?(x)= be not greater than eight, we may find the roots by aid of §§ 631, 871, 874. Dividing (1) by x"' and combining terms, we have But by carrying out the indicated reckoning we find that and if z = x + 1/x, and in (3) we set jc) = 1, 2, 3 • • • succes- sively, we have x' + ~ = z'- 2, x^ + \^z^-Sz, x^ x^ x' + ^, = z^-^z^ + 2,>.., (4) 488 A COLLEGE ALGEBRA 876 877 that is, we obtain for a^P + l/x" an expression of the j9th degree in z. Substituting these several expressions in (2), we obtain an equation of the mth degree in », as was to be demon- strated (compare § 645). Example 1. Solve 2x8 _ x7 _ 12 x6 + Hx^ - 14a;3 + 12x2 + x - 2 = 0, This is a reciprocal equation having the roots 1 and — 1. Removing the factor x^ 2x6 1, lOx^ + 13x3 - 10x2 - X + 2 = 0. Dividing by x^, 13 = 0. Hence by (4), 2 (z^ - 3 2) - (f - 2) - 10 z + 13 = 0, or 2z3-22_i6z + 15 = 0. Solving, z = 1, -3, or 5/2. Hence x + l/x = l, -3, or 5/2, andtherefore x=(l ±i V3)/2, (-3±V5)/2, 2, or 1/2 Example 2. Solve x^ - x^ + x* + x2 - x + 1 = 0. Every binomial equation «" + a = may be reduced to the reciprocal form by aid of the substitution x of this substitution being y" + 1 Vay, the result (compare § 646, Ex. 2). Expression of a complex number in terms of absolute value and amplitude. In the accompanying figure, P is the graph of the ^ P li / h /r 1 A a ^ complex number in § 238. + bi, constructed Tlie length of OP is Vci^ + b% the absolute value of a + bi, § 239. Repre- sent it by r. Let 6 denote the circular measure of the angle XOP, that is, the length of the arc subtended by this angle on a circle of unit radius described about as center. We call 6 the amplitude of a + bi. We call the ratio b/r the sine of 0, and write b/r = sin 0. CUBIC AND BIQUADRATIC EQUATIONS 489 We call the ratio a/r the cosine of 6, and write a fr = cos^. We thus have a — r cos 6, h — r sin 0, and therefore a -\-hi ^= r (cos 6 + i sin 6). When ^ = 0, then b = and a = r. Hence sin = 0, and cosO = 1. The circular measure of 360° is 2 tt, this being the length 878 of a circle of unit radius. Hence a point P given by r, 6 is given equally by r, ^ + 2 tt ; by ?•, ^ — 2 tt ; and, in general, by r, -{- 2 niTT, where m denotes any integer. Hence we say that the general value of the amplitude of a + ^* is ^ + 2 tnir. Theorem. The absolute value of the product of two complex 879 numbers is the product of their absolute values; and its ampli- tude is the sum of their amplitudes. For r (cos 6 -\- i sin 6) ■ r' (cos & + / sin 6') = r /•' [(cos 6 cos 6' — sin sin 6') + i (sin e cos 6' + cos sin $')'] = r r' [cos (0 + 0') + i sin {d + 6')], since it is proved in trigonometry that cos (6 + 6') = cos 6 cos 0' - sin $ sin 6', sin (6 + 6') = sin cos 0' + cos sin ^'. The construction given in § 240 is based on this theorem. Corollary 1. By repeated applications of § 879 we have 880 r (cos 6 + i sin 6) ■ r' (cos 6' + i sin 0') ■ r" (cos 0" + i sin ^") • • • = rr'/-".--[cos(^ + ^' + ^" + ...) + i sin (e + O' + 6" + ■• •)]. Corollary 2. Setting r = r' = r" = •••, and ^ = ^' = ^" = •• • 881 in § 880, we obtain the following formula, known as Demoivre's theorem : [?'(cos 6 + I sin ^)]" = ?•" (cos nO + i sin nO). 490 A COLLEGE ALGEBRA Corollary 3. For a quotient we have the formula r (cos ^ + i sin ^) i- ,^ ^,, . ■ ,^ ^„-, -^ ., ; . ■ .; = - [cos {Q - 6') + i sm {d - 6')\ r' (cos ^' + i sm ^') ?•"- ^ ■' \ jj For ^ [cos {6 - $') + i sin {6 ~ 6')']- r' (cos 6' + i sin 0') ^r(cose + isme). §879 Corollary 4. Tlie n /;th roots of a complex nvmiberare given by n/ , . ■ , V ^ + 2 A-TT v ?• (cos ^ + i sni ^j = y" I cos when k is assigned the ?i values 0, 1, 2, • rW e + 2k7r , . . ^ + 2A:7r\"|" For ?•" cos — + i sm = r [cos (^ + 2 A-tt) + i (sin ^ + 2 Att)] = r (cos O^i sin ^). §§ 881, 878 Binomial equations. The n nth. roots of r (cos 6 + i sin 6) are the n roots of the equation x" — ?'(cos ^ + t sin ^) = 0. Hence, in particular, the n roots of the equation x" — r = 0, where r is real and therefore 6 is 0, are V^(cos 2 kiT In + i sin 2 kir fn), k = 0, 1, • • •, n — 1. Thus, the roots of the equation x' — 1 = are cosO+isinO, cos2 tt/S + i sin 2 7r/3, cos4 7f/3 + isin4 tt/S, which may be proved equal to 1, (-l + zV3)/2, (-l-iV3)/2. Trigonometric solution of the irreducible case of the cubic. In the irreducible case of the cubic x^ + px -{- q = 0, § 872, 3, the expressions A and B, § 871, are conjugate imaginaries. For since in this case rf/i +2^^/27 is negative, we have ^=-2+,- ■■V-(^0— i-W-e-^ Hence the expressions for A and B in terms of absolute value and amplitude, § 877, will be of the form ^ = r (cos e + i sin 6), B ^ r(cos 6 -i sin 6), (1) CUBIC AND BIQUADRATIC EQUATIONS 491 where — q rl and cos 6 = — — — ff-MyK-0" "km (2) When J) and q are given we can find the value of Q from tliat of cos Q by aid of a table of cosines. In the formulas for the roots of x^ -^ px -\- q = ^, § 871, (12), substitute the expressions (1) for A and i?, and the expressions for w and w'-^ given in § 884. The results when simplified are o t ^ o i ^ + 27r ^ . ^ + 47r jCi = 2 /•' cos - ) a:'o = 2 r^ cos > cr, = 2 r^ cos And, ?• and Q being known by (2), these formulas enable us to compute the values of the roots by aid of a table of cosines. Example. Solve x3-x + l/3 = 0. Here Q'2/4 + -p^ /11 = — 1/108, so that we have the irreducible case. Substituting in the formulas (2) and simplifying, we find r-\/ V27, cos 6» = - V3/2, and therefore B = 150°. Hence by aid of tables of logarithms and cosines we obtain Xx = -^ cos 50° = . 7422 ; x^ = -^ cos 170° = - 1. 1371 ; V27 V27 X3 = ^ cos 290° = .3949. V27 EXERCISE LXXX Solve equations 1 — 10 by the methods of §§ 8T1 and 874. 1. x3-9x-28 = 0. 2. x3-9x2 + 9x-8 = 0. 3. x3 - 3 X - 4 = 0. 4. 4 a;3 - 7 X - 6 = 0. 5. x3 + 3x2 + 9x-l = 0. 6. 3x5-0x2 + 14x + 7 = 0. 7. X* + x2 + 6 X + 1 = 0. 8. x^ - 4 x3 + x2 + 4 X + 1 = 0. 9. x* + 12x-5 = 0. 10. x< + 8x3 + 12x2-l)x + 2 = 0. 11. Solve 3x6-2x6 + 6x*-2x» + 6x2-2x + 3 = 0, 492 A COLLEGE ALGEBRA 12. Solve 2x8 - 9x7+18x6 _ sOx^ + 32x4 - 30x^ + 18x2 - 9x + 2 = 0. 13. Solve 6 x" - x6 + 2 x5 - 7 X* - 7 x3 + 2 x2 - X + 6 = 0. 14. Find the cubic in z on which, by § 875, the solution of x^ - 1 = depends. 15. Find the condition that all the roots of x^ + 3 ax^ + 3 to -r c =-- be real. 16. Write down the trigonometric expressions for the roots of x'' - 1=0, and of x6 + 1 z= 0. 17. Solve the following irreducible cubics. (1) x3 - 3 X - 1 = 0. (2) x3 - 6x - 4 = 0. 18. In a sphere whose diameter is 3 %/3 a right prism with a square base is inscribed. If the volume of the prism is 27, what is its altitude ? 19. The volume of a certain right circular cylinder is 50 7t and its entire superficial area is 105 tt/ 2. Find the radius of its base and its altitude. 20. The altitude of a right circular cone is 6 and the radius of its base is 4. In this cone a right circular cylinder is inscribed whose volume is four ninths that of the cone. Find the altitude of the cylinder. XXXI. DETERMINANTS AND ELIMINATION DEFINITION OF DETERMINANT 886 Inversions. Odd and even permutations. When considering the permutations of a set of objects, as letters or numbers, we may fix upon some particular order of the objects as the normal order. Any given permutation is then said to have as many inversions as it presents instances in which an object is followed by one which in the normal order precedes it. And the permutation is called odd or even according as the number of its inversions is odd or even (or 0). Thus, if the objects in normal order are the numbers 1, 2, 3, 4, 5, the permutation 45^12 lias the eight inversions 43, 41, 42, 53, 51, 52, 31, 32, Hence 45312 is an even permutation. DETERMINANTS AND ELIMINATION 493 Theorem, If two of the objects in a permutation are inter- 887 changed, the number of inversions is increased or diminished by an odd number. For if two adjacent objects are interchanged, the number of inversions is increased or diminished by 1. Thus, compare ApqB (1) and AqpB (2), where A and B denote the groups of objects which precede and follow the interchanged objects p and q. Any inversions which may occur in A and B and any which may be due to the fact that A, p and q precede B are common to (1) and (2). . Hence the sole difference between (1) and (2), so far as inversions are concerned, is this : li pq is an inversion, qp is not, and (2) has one less inversion than (1) ; but if pq is not an inversion, qj) is, and (2) has one more inversion than (1). But the interchange of any two objects may be brought about by an odd number of interchanges of adjacent objects. Thus, from pabq we may derive qahp by five interchanges of adjacent letters. We first interchange p with each following letter in turn, obtaining successively aphq, abpq, abqp, and we then interchange q with each preceding letter, obtaining aqhp, qabp. There is one less step in the second part of the process than in the first part, because when it is begun q has already been shifted one place in the required direction. Had there been /x letters between p and q, there would have been /a + 1 steps in the first part of the process and /a in the second, and (/A + 1) + ^, or 2 /A + 1, is always odd. Therefore, since each interchange of adjacent objects changes the number of inversions by 1 or — 1, and the sum of an odd number of numbers each of which is 1 or — 1 is odd, our theorem is demonstrated. Thus, if in 21457368 (1) we interchange 4 and 6, we get 21657348 (2). It will be found that (1) has^we inversions and (2) eight, and 8 — 5 is odd. Of the n\ permutations of n objects taken all at a time, 888 § 763, half are odd and half are even. For from any one of 494 A COLLEGE ALGEBRA these permutations we can derive all the rest by repeated interchanges of two objects. As thus obtained, the permuta- tions will be alternately odd and even, or vice versa, § 887. Therefore, since n ! is an even number, half of the permutations are odd and half are even. 889 In what follows we shall have to do with sets of letters with subscripts, as a^, a.^, ■■-, h^, bo, ■■■, and so on. Having chosen any set of such symbols in which all the letters and subscripts are different, arrange them in some particular order and then find the sum of the number of inversions of the letters and of inversions of the subscripts. If this sum is even, it will be even when the symbols are arranged in any other order ; if odd, odd. For when any two of the symbols are interchanged the inversions of both letters and subscripts are changed by odd numbers, § 887, and therefore their sum by an even number. In particular, the number of inversions of the subscripts when the letters are in normal order and that of the letters when the subscripts are in normal order are both odd or both even. Thus, in a-^bsCi the number of inversions of the subscripts Is two; in Cia2&3 the number of inversions of the letters is tioo ; in ftsOoCi the sum of the number of inversions of the letters and that of the subscripts is four. 890 Definition of determinant. We may arrange any set of 2^, 3^ in general n'^ numbers in the form of a square array, thus : fll ^2 (h (Xo a 3 «i «2 fiz ^4 h, b. h h b. h h^ bs 64 Cl Co C3 ^2 Cs Ci d. d, d. and so on, where the letter indicates the roiv, and the subscript the column, in which any particular number occurs. We may then call the numbers the elements of the array. DETERMINANTS AND ELIMINATION 495 With the elements of such an array form all the jJroducts that can he formed by taking as factors one element and but one from each roiv and each column of the array. In each product arrange the factors so that the row marks (letters^ are in normal order, and then count the inversions of the column marks (subscripts). If their number be even (or 0), give the product the plus sigri; if odd, the minus sign. The algebraic sum, of all these plus and minus jjroducts is called the deteiininant of the array, and is represented by the array itself with bars written at either side of it. Thus, L ^ , ^ = «1&2 — «2^1 and «! a-i as 61 62 ^3 Ci C2 Cz 61 62 aih^Cz - aibzC2 + a^hiCi — aobiCz + a3biC2 — (136201. It follows from § 889 that in determining the sign of any 891 of the products above described we may arrange the factors so that the column marks are in normal order, and then give the product the plus or minus sign according as the number of inversions of the row marks is even or odd. Or we may write the factors in any order whatsoever, and then give the product the plus or minus sign according as the su7)i of the number of inversions of row marks and of column marks is even or odd. Thus, consider the product Uzb^Ci to which our first rule gave the minus sign. Writing it so that tlie subscripts are in normal order, we have Cib2as, and since the letters cba now present three inversions, the product must again, by our second rule, be given the minus sign. If the prod- uct be written hoCias, the letters present two inversions and the subscripts one, and 2 + 1 is odd. Hence again, by the third rule, we must give the product the minus sign. The number of the rows or columns in the array out of 892 which a determinant is formed is called the oi-der of the determinant. 496 A COLLEGE ALGEBRA 893 The products above described, with their proper signs, are called the terms of the determinant. 894 To expand a determinant is to write out its terms at length. 895 The diagonal of elements a^, Z»2, c^, ■■■ is called the leading diago7ial, and the product a-Jj^c^, ■ • • is called the leading term of the determinant. The leading term enclosed by bars, thus, \a^ h^ ^3 • • -I, is often used as a symbol for the determinant itself. 896 The number of the terms of a determinant of the nth oi-der is n !, and half of these terms have plus signs and half have minus signs. Por, keeping the letters in normal order, we may form n\ permutations of the n subscripts, § 763, and there is one term of the determinant for each of these n\ permutations, §890. Furthermore half of these n\ permutations are even and half are odd, § 888. Thus, for n = 3 we have 3 ! or 6 terms ; for n = 4, we have 4 ! or 24. 897 Other notations. It must be remembered that the letters and subscripts are mere marks of row and column order. Any other symbols which will serve this purpose may be substi- tuted for them. Thus, the elements of a determinant are often repre- sented by a single letter with two subscripts, as 023, the first indicating the row and the second the column. ThR symbol a^z is read " a two three," and so on. 898 A rule for expanding a determinant of the third order. To obtain tlie three positive terms, start at each element of the first row I «i «2 03 ^"^ *^^^ ^^^' ^° ^^^ ^^ possible, follow the direction i>i ^2 ^-'3 ^^ ^^^® leading diagonal, thus : ri ^2 C3 a^b^Ci, azb^Ci, a^biC^. an «12 ai3 ail a-n «23 asi az2 033 DETERMIXANTS AND ELIMIXATION 497 To obtain the three negative terms, proceed in a similar manner but follow the direction of the other diagonal, thus : — «1^3f'2, — «2^'l^3, — «3^2^1- 5 3 1 - 1 2 4 5(-l)(-l) + 3(-3)2 + 2(-l)4 -5(-3)4-3(-l)(-l)-2(-l)2 = 40. This rule does not apply to determinants of an order higher than the third. Thus, it would give but eight of the twenty-four terms of a deter- minant of the fourth order. EXERCISE LXXXI Expand the following determinants. p q r 1 X c I 1. q p s r s p 2. 1 y I 1 Z C • p -q r -q -r 3. q P - s -r s p 4. q - s r s C Find the values of the following determinants. 5. 1 2 3 3 1 2 2 3 1 6. 1-3 4 2 0-5 3-1 7 7. 8 9 2 3 1 1 4 3 6 5 tti a2 fls «! bi Cl bi ai Cl h h 63 = ffl2 b'Z Cl = - 62 a2 C2 Cl C2 C3 as 63 C3 bs as C3 «! 32 as 61 62 bs Cl Co C3 = fll &2 63 C2 C3 -as bi 63 Cl C3 Prove the following relations by expanding the determinants. \bi 62 1 10. In the expansion of the determinant | ai 62 C3 ^4 1 collect all the terms which involve as factors (1) Csdi, (2) 01^4, (3) a263di, (4) ai, (5) C3. 11. Find the signs of the following terms in the expansion of the determinant \ai 62 cs di e^]. a2biC3die5. a^boCidnes. a^biCsd^ei. 01^2036465. Cibzesaids. dsOzeibiC^. 498 A COLLEGE ALGEBRA 899 900 901 902 PROPERTIES OF DETERMINANTS Theorem 1. The value of a determinant is not chayiged if its rows are made columns, and -its columns rows, without changing their relative order. Thus, «! flo f'o a I l>i Ci h, h, h. Go bn Co ^1 Co Cs (1) ^3 h Cz (2) For every term, as aJb^Ci, in the expansion of the deter- minant marked (1) contains one element and but one from each row and column of (1). Hence it contains one element and but one from each column and row of (2) and is therefore, except perhaps for sign, a term of (2). Moreover the sign of the term in (1) is the same as its sign in (2). For if the factors of the term be arranged in the order of the letters a, b, c, the inversions of the subscripts will determine this sign in both (1) and (2), in (1) by the rule of § 890, in (2) by the rule of § 891. Conversely, every term in the expansion of (2) is a term in the expansion of (1). Hence for every theorem respecting the rows of a determi- nant there is a corresponding theorem respecting its columns. Theorem 2. If nil the elements of a. row or column of a deter- minant are 0, the value of the determinant is 0. For every term of the determinant Avill contain a factor from the row or column in question, § 890, and will therefore vanish. Theorem 3. If two rows (or columns) of a determinant be interchanged, the determinant merely changes its sign. Thus, "l Oo «3 ^1 ^2 ^3 bi h h = — b, b, b. Ci C2 Cg (1) rtj a^ «3 (2) DETERMINANTS AND ELIMINATION 499 Yov any term of (1) with factors arranged in the order of the rows of (1) may be transformed into a term of (2) with factors arranged in the order of the rows of (2) by inter- changing its first and last factors ; and conversely. But this interchange will increase or diminish the inversions of the subscripts in the term by an odd number, § 887, and therefore, since the normal order of the subscripts is 123 in both (1) and (2), it will change the sign of the term. Thus, a2b3Ci is a term of (1) and — cibsU'z is the corresponding term of (2). For in ao&sCi the subscripts present two inversions, while in Cibsa-z they present one inversion. Example. Verify the preceding theorem by expanding each of the determinants (1) and (2). Corollary. If two of the roivs (or columns) of a determinant 903 are identical, the deterniinant vanishes. For let D denote the value of the determinant. An inter- change of the identical rows must leave D unchanged ; but, by § 902, it will change D into — D. Therefore D^— D, that is, 2 D = 0, ot D = 0. a a d Thus, b b e = abf + aec + dbc — aec — abf — dbc = 0. c c f Theorem 4. If all the elements of a row (or column) are 904 multiplied by the same number, as k, the determinant is multi' plied by k. For of the elements thus multiplied by k one and but one occurs as a factor in each term of the determinant, § 890. The evaluation of a determinant may often be simplified by applying this theorem. Thus, = 480. 8 2 -3 4 1 -1 1 1 >0 5 = 2-5- 3 -4 1 = 2.5.3.4- 1 - -1 1 4 -1 3 4 -1 1 1 -1 500 A COLLEGE ALGEBRA 905 906 907 Corollary. If the corresponding elements of two columns (or roivs) are proportional, the determinant vanishes. Thus, ra a d a a d rb b e = r b b e re c f c e f ?• • = 0. §§ 903, 904 Theorem 5. If one of the columns (or roivs) has binomial elements, the determinant may be expressed as a sum of two determinants tw the manner ilhistrated below. «i + a' b, +b' Ci + c' «2 «s b, b, Co Co «1 «2 «3 a' 02 as = bi b^ bs + b' b, b. 0) Cl ^2 ^3 (2) C' Co ^3 (3) For each term of (1) is the sum of the corresponding terms of (2) and (3). Thus, (aj + a')boCz = aib^Cs + a'bnC^. Observe that any of the numbers a', b', c' may be 0. By repeated applications of this theorem any determinant with polynomial elements may be resolved into a sum of determinants with simple elements. ai + a'l 02 + a' 2 03 + a'z Example. Express the determinant hy + h\ 60 + h'n 63 + V^ as a sum of eight determinants. Ci + c\ Cn + c'2 C3 + c'3 Theorem 6. The vahie of a determinayit tvill not be changed if to the elements of any column (or row) there be added the corresjjondi^ig elements of any other column (or row) all multi- pflied by the same number, as k. Thus, by the theorems of §§ 905, 906, we have ai + ka^ a. bi + kb^ b^ Cl + kcz Co And similarly in any other case. It follows from this theorem that a determinant vanishes if one of its rows may be obtained by adding multiples of any of its remaining rows. flz 'h cio f/3 h = h, b, b. + C3 <-l (-2 C, kaa a^ a^ a-i a<2, dz kbg ^2 ba = Z»i b^ bs kc^ C2 C3 Cl C2 Cs DETERMINANTS AND ELIMINATION 501 Thus, 4 7 7 5-4 2 2 5 1 since 4, 7, 7 = 2(5, - 4, 2) + 3(- 2, 5, 1). Theorem 7. //"a determinant whose elements are rational integral functions of some variable, as x, vanishes when x = a, the determinant is divisible by x — a,. For the determinant when expanded may be reduced to the form of a polynomial in x. And since this polynomial vanishes when ic = a it is divisible by a; — a, § 415. The factors of a determinant may often be found by aid of this theorem. Example. Show that 1 1 1 a b c a- 62 C2 :{a — b){b — c) (c — a). By § 903, this determinant will vanish if a = 6, if 6 = c, or if c = a. Hence it is divisible by a — 6, b ~ c, and c — a, ^ 416, and therefore by the product (a — b){b — c) (c — a). But this product and the determinant itself are of the same degree, three, with respect to a, b, c. Hence the two will differ by a numerical factor at most. One term of the determinant is ftc^ and the corresponding term of {a — b){b — c) (c — a) is bc"^. Hence tlie numerical factor in question is 1, and the determinant is equal to (a — b) {b — c) (c — a). 908 EXERCISE LXXXII 1. Evaluate the following determinants. (1) 6 42 27 10 8 -28 36 • (2) 15 20 35 135 20 2. Prove that ffli + fcaz + Cl + kC2 + Ics 8 2 -ab ac ae 12 3 (3) bd -cd de 32 12 bf cf -ef a.2 + mQLs as ai ao as &2 + mbs ba 6i 62 63 Cl + mcs C3 c 1 C2 C3 3. By aid of the theorem of § 907 prove that the value of each of the foUowins; determinants is 0. (1) a d c d c b a b (2) 1 p 5 1 Q r 1 r s 1 s P r + s p + s p + q q + r (3) 2 1 4 3-1 2 1 2 3 - 1 - 1 -2 -2 50i A COLLEGE ALGEBRA 4. Prove that 5. Prove that 1 p p3 1 q q^ 1 r r3 iP -q){q-r){r-p){p + q + r). {b + c)2 a6 oc ab (c + a)2 6c ac be {a + 6)2 2abc{a + b + c)3. 909 MINORS. MULTIPLICATION OF DETERMINANTS Minors. In any determinant A suppress the row and column in which some particular element e lies and then form the determinant of the remaining elements without disturbing their relative positions. This new determinant is called the complementary minor of the element e in question and may be denoted by A^. Thus, in A cti 02 aa bx 62 &3 Cl C2 C3 the minor of Cj is Ac fl2 <^3 62 63 910 Theorem. In the expansion of any determinant A the stun of all the terms ivhich involve the leading element ai is aiA . Thus, in «! a^ a 3 «4 h h ^•3 ^4 Cl C-l Cz Ci di ch dz d. this sum is Oi (1) h h h C„ Cg C4 d. dz di (2) For, apart from sign, each term of A which involves a^ is formed by multiplying a^ by elements chosen from the remain- ing rows and columns of A, one from each, in other words, by multiplying a^ by a term of A„j. Moreover the sign of the A term is that of this A,,^ term, since writing rtj before the latter term will not affect the inversions of the subscripts. Thus, the term — aib^r^d^ of (1) may be formed by multiplying «! by the term — biC^d^ of (2). Conversely, the product of ai by each term of A„ is a term of A. DETERMIXANTS AXD ELIMINATIOX 503 Corollary. If e denotes the element in the ith roiv and Vth 911 column of A, the sum of all the terms of A which involve e is (-ly + '^eA,. For we can bring e to the position of leading element without disturbing the relative positions of the elements which lie outside of the row and column in which e stands, namely, by first interchanging the row in which e stands with each pre- ceding row in turn and then interchanging the column in which e stands with each preceding column. In carrying out these successive interchanges of rows and columns we merely change the sign of the determinant (i — V)-\-(k — l)ovi-\-k — 2 times, § 902. Hence, if A' denote the determinant in its final form, we shall have A' = (-l)' + ^ (- ly-^A. By § 910, the sum of all the terms in A' which involve e is eA'^. Hence in A this sum is (— l)'"'"*-eA^. For the minor of e in A is the same as its minor in A'. Thus, in the case of A = | ai 62 C3 ^4 1 the element ds, for which i = 4, fc = 3, may be brought to the leading position as follows : ax 02 as ai h 62 63 64 Cl C.2 C3 C4 di di dz di (1) di ^2 ds di «i «2 «3 ai 61 62 63 &4 Cl C2 Cs Ci (2) da di do d4 as ai 02 ai h h 62 64 C3 Cl C2 C4 (3) By interchanging the fourth row of (1) with the third, second, and first in turn, we obtain (2) before which we place the minus sign because of the three, that is, i — 1, interchanges of rows. Then by interchanging the third column of (2) with the second and first in turn, we obtain (3) before which we place the same sign as that before (2) because of the two, that is, A; — 1, interchanges of columns. The minor of ds in (1) is the same as its minor in (3). Hence the sum of all the terms of (1) which involve da is — ds • | Oi bo C4 1. Theorei6. A determinant may he expressed as the sum of the 913 products of the elements of 07ie of its 7'ows or columns by their 504 A COLLEGE ALGEBRA comijlementary minors, ivlth signs which are alternately plus and minus, or minus and plus. Thus, in the case of a determinant of the fourth order A = I «! ^2 Cg di I we have A = aiA„^ - «2A„^ + a3A„3 - ai\^. For each term in the expansion of A contains one and but one of the elements a,, a^, a^, 04. And, by §§ 910, 911, the sum of all the terms which involve a^ is a^A^ , the sum of all which involve a„ is — a^^^,^, and so on. In like manner, A = - b,\, + h\, - h\, + h\ = aiA„j — JjAjj + CiA^j — diAj^, and so on. 913 Cofactors. It is sometimes more convenient to write the preceding expressions for A in the form A = a^Ai + ^2.12 + "3-I3 + «4^4 = b^B, + boB. + bsBs + b,Bi, and so on, where Ai = A„^, A2=— A„^, and so on. We then call A I, A2, ■ ■ ■ the cofactors of ai, a^, ■ • ■. Thus, in ai a2 as 61 h h Cl cz C3 the cofactors of ai, On, as are 1^2 C3 1' |ci Car |ci Co 914 -An// sum like biAi + b2A2 + bjAg + b4A4, obtained by adding the products of the elements of one row by the cofactors of the corresponding elements of another roiv, is zero. For b^Ai + />2.l2 + ^3-43 + ^i^i denotes a determinant whose last three rows are the same as those of A =|ai b^, c^ di\ but whose first row is b^, b^, b^, b^. And since both the first and second rows of this determinant are bi, b^, b^, b^, it vanishes, § 903. And so in general. DETERMINANTS AND ELIMINATION 605 Bordering a determinant. Any determinant may be expressed 915 as a determinant of a higher order. For, by § 912, we have «! aa ^3 W h ^3 = ^1 ''2 CS 10 fti a^ b, b, Ci C2 , and so on. Evaluation of a determinant. The value of a numerical deter- minant of any order may be obtained by aid of the theorem of § 912 and the rule of § 898 916 Thus, 2 3 1 -1 2 3 2 3 2 3 1-1 4 1 1 •~ 4 10 1 1 2 — 2 1 -1 2-2 1 3 1-1 2 3 1! = — 2 1 1 2 -2 1 + 3 4 -1 1 2 ^1 -21 = — le + 87 = G9. But in most cases the value of a numerical determinant may be found with less reckoning by the following method. It. follows from §§ 904, 907, 910 that «1 «2 tts • • • (ii GiCt^ (iiao • •• bi b.^ Z>3 ••• Ci c-a Cs ••• (1) 1 <'i «i<'2 «i^3 ••• (2) 1 a^bo — a«bi a^b^ — ajby • • ■ ^^ a^c^ — aa^i a^Cs — a^Ci • ■ • 1 (3) Here (1) denotes a determinant of the 7ztli order. By multi- plying each column of (1) after the first by «! we obtain (2). From the second column of (2) we subtract the first multiplied by 02 ; from the third column we subtract the first multiplied by ftg ; and so on. The first row of the resulting determinant 917 506 A COLLEGE ALGEBRA is «!, 0, 0, • • •. Hence this determinant is equal to a^ times its minor, which is the determinant of the (n — l)th order (3). Observe that each element of (3) is obtained from the cor- responding element of the minor of a^ in (1) by multiplying that element by «i and from the result subtracting the product of the corresponding elements in the first row and the first column of (1). Thus, by two reductions of the kind just described, we have 2 2-1 3 2 13-2 2-121 3-2-2 1 _ 1 ~22 6 - 6 -10 4 2 6 -4 -1 -7 = 3 2 1 3-3 2 10 1 7 1 -15 ~3 -17 .?=--. 2- 1 - 2(- 2) = -6, 2 3-( -l)(-2 = 4, and so on. When the leading element is ^Ye begin by bringing another element into the leading position by the method explained in § 911. Example. Evaluate each of the following determinants by both of the methods just described. 1-2 3 4 1 3 - 1 2 3 1 -3 918 Multiplication of determinants. The product of two deter- minants of the same order, A and A', may be expressed in the form of a third determinant A" obtained as follows : Multiply the elements of the \th row of A hy the corresponding elements of the \.th row of A'. The sum of the products thus obtained is the element in the \th row and \th column of A". Thus, «1 ^2 \xP\ *2?3r(2) \biPi %ir(3) \\p% %2|(4) DETERMINANTS AND ELIMINATION 507 But (1) and (4) vanish, since their columns are proportional, § 905. And simplifying (2) and (3) by aid of §§ 902, 904, and adding them, we have i'i?2 h h ^Pi'z hPi. + hPi + ^ZPZ \q\ + *2?2 + *3?3 ^>1^1 + ^2^2 + ^3^3 as may be shown by resolving the third determinant into a sum of determinants with simple columns in the manner just illustrated. There will be twenty-seven such determinants, but twenty -one of them have two or three proportional columns, and therefore vanish. Each of the remaining six is equal to the determinant |ai h.^, Cz\ multiplied by a term of |pi q^, rgj, so that their sum is \a^ b^ Cz\- \pi q^ r^\. This proof may easily be generalized. The rule above given is readily extended to determinants of different orders. We have only to begin by making their orders the same by bordering the one of lower order, § 915. EXERCISE LXXXIII Evaluate the following determinants. / 2 -1 7 5 -3 2 4 7 / 1 2 3 - 1 -2 2 1 3 -2 1 1 2 _ 2 1 -1 -1 2 1 2 3 1 -1 508 A COLLEGE ALGEBRA 12 6 - 4 10 26 18 6 -30 21 12 24 40 28 9 - 2 20 14 Express each of the following products as a determinant. a b c b c a c a b a — a a b b b c c — c d d d a b a b I m n ' c d ' 8. m n I c d n I ni «1 «2 03 bi b-2 bs Ci C.2 Cs Ax A2 A3 ai fflj «3 B, B, B3 — 61 b, 63 Ci C, C3 Cl C2 C3 p r a c 6. p q a b q r b c 9. Prove that 10. Prove that a determinant reduces to its leading term when all of the elements at either side of the leading diagonal are zero. 919 ELIMINATION. LINEAR EQUATIONS Solution of a system of linear equations. We are to solve the following system of equations for x^, x^, X3 : a^Xi + a^x^ + CT3X3 = k 1 bix^ + b^x^ + 63X3 = I I. (1) c^xi + C2X2 + C3X3 = mj Let A denote |ffi b^ 03], the determinant of the coefficients of Xi, x^, X3 arranged as in (1), and, as in § 913, let A^, A^, ■ ■ • denote the cofactors of a^, a^, • • • in A. To eliminate x^ and Xg, we multiply the first equation by Ai, the second by B^, the third by Cj, and add. We obtain aiAi Xi + a.^Ai .r, + a^Ai X3 = kA^ h,lh +h,B, +h3B, +//>', ^i^'i +C2<\ +<'3<\ +m(\ But in this equation the coefficients of x., and X3 are 0, § 914 ; the coefficient of x^ is A, § 913 j and the secord member DETERMINANTS AND ELIMINATION 509 denotes a determinant found by replacing the first column of A by k, I, m. Hence the equation may be written (2) We may in like manner derive an equation which involves a-2 alone by multiplying the given equations by ylj, B^, C\ respectively and then adding them, and an equation which involves x^ alone by multiplying by .43, B^, C\ and adding. These equations are «1 0^2 «3 k a^ as b, b, b. J'l = I b, b. Ci c^ c. m ("2 fg «! CU (Iz «1 k as b, b, b. ^1 I bs Ci Co ('3 <^l m Cs (3), Hence, if A ^ 0, the required solution is »l «2 «3 ai a.j. k 1>1 b, bs Xs = b^ b^ I 2 m I is not 0, it follows from (2), (3), (4) that the given equations (1) have no finite solution (compare § 394). 920 510 A COLLEGE ALGEBRA If A and all the determinants \k b^ ^3], |ai I Cg], |ai Jg ^1 vanish, the equations (2), (3), (4) impose no restriction on the values of x-^, x^, x^. In this case the given equations (1) are not independent. This follows, by § 394, from the manner in which (2), (3), (4) were derived from (1), unless all the minors A^, An, ••• vanish. And if all the minors vanish, it may readily be shown that the three equations (1) differ only by constant factors, so that every solution of one of them is a solution of the other two. These results are readily generalized for a system of n equa- tions in 71 unknown letters. 921 Homogeneous linear equations. When k = I = rn = 0, the equations (1) of § 919 reduce to a system of homogeneous equations in x^, x^, Xs, namely, ajXi + a^x^ + «33:"3 = Q 1 b,})pi + hx, + hx,=il, (1) CiXi + C2X2 + CsXs = e) J . and the equations (2), (3), (4) of § 919 become Aa-i = 0, Ax^ = 0, Axs = 0. (2) Evidently the equations (1) have the solution Xj = a'g = Xg = 0, and it follows from (2) that this is the only solution unless A = 0. But if A = 0, the equations (1) are satisfied by Xi = ?vli, Xo = rAo, X3 = rAz, (3) where r may denote any constant whatsoever. For, substituting these values in (1) and simplifying, we have a^Ai + a^A^ + «3'43 = 0, b^A^ + /^2.42 + ^3^4 3 = 0> CiAi + C2A2 + ^3^43 = 0, and these are true identities, the first one because A = 0, the other two by § 914. The same thing may be proved as follows : If we solve the second and third of the equations (1) for Xi and X.2 in terms of x^, we obtain x^/ A^ = x^/Ao = x^/ A^, or, if DETERMINANTS AND ELBIINATION 511 r denote the value of these equal ratios, x^ = rA^, x^ = rA^, Xs = rA-i. And as just shown, if A = 0, these values will also satisfy the first of the equations (1). From this second proof it follows that when A = Xi : Xz'. Xs = Ai_ : A^: A 3 = Bi : B2 : Bs = Ci : C\ : C\, that is, the minors of corresponding elements in the rows of A are proportional. It is assumed that these minors are not 0. From the system of three non-homogeneous equations in x, y 922 a^x + a.^ij + ag = 1 M + % + ^>3 = 0l, (1') CiX + C.JJ + (-3 = J we may derive the homogeneous system (1) of § 921 by sub- stituting X = Xi/xg, y = x^/x^ and clearing of fractions. Hence A = is the condition that the equations (1') have a common solution. EXERCISE LXXXIV Solve the following systems of equations by determinants. r2x + 3?/-5z = 3, r2x-H4?/-32 = 3, '' 1. -! X - 2?/ + z = 0, ^ 2. <: 3x- 8y + 6z = 1, l3x + 7/ + 3z = 7. ^ l8x-2?/-0z = 4. r2x-4y + 3z + 4« = -3, ( az + by + cz = a, I ' 3x-2y-h6z + 5« = -l, I I 5x + 8?/ + f)z + 3« = 9, [a^x + ¥y + c^z = d^ nn <, -.0 Lx - lOy - 3z - t t = 2. Show that the following systems of equations are consistent, and solve them for the ratios x-.y-.z. c X + 2 y - z = 0, fciix + biy + (kcii + Ibi) z =. 0, 5. \sx-y + Az = 0, 6. J aox + h.y + (ka2 + lb..) z = 0, L 4 X + 2/ + 3 z = 0. [asx + bsy + {has + ^^3) z = 0. 7 . For what values of X are the following equations consistent ? 4 X + 3 y + z = Xx, 3x — iy + 1 z = \y, x +1 y -6z = \z. 512 A COLLEGE ALGEBRA RESULTANTS 923 Resultants. By the resultant of two equations f{x) = and (^ (x) = is meant that integral function of the coefficients of f(x) and (f) (x) whose vanishing is the necessary and sufficient condition that/(x) = and (f)(x)= have a common root. Thus, the resultant of a^x,^ + aiX. + a^ = (1) and x — b = (2) is aob"^ + aib + a^ ; for when aolfi -f a^b + ao = 0, the equations (1) and (2) have the common root b. 924 The resultant of any two equations /(a-) = 0, <^ (x) = may be obtained by eliminating x by the following method due to Sylvester. To fix the ideas, let f{x) = Qox'^ -f a^x"^ + a^x + a 3 = 0, (1) (x) = b,p:' + hx + b, = 0. (2) Multiply (1) by x and 1, and (2) by x^, x, and 1 successively. We obtain 4,3,2, n a^jx* -\- a^x^^ -\- a (x), .T<^ (jc), cf) (x). Hence, if fx-i, 1x2, h-s, H-i^ H-b denote the cofactors of the elements of the fifth column of D, we have, § ^^^' D = (fJi.X + fJi.)f(x) + (/X3.f'^ + f,,X + ix,) (x) = 0. It follows from this identity that each factor a- — /8 of f(x) must be a factor of (fx^x^ + fiiX + /xg) ^(a:*), and therefore, since f(x) is of the third degree and /xsX^ + fXiX + /x^ is of only the second degree, that at least one factor x — ^ oif(x) must be a factor of f^{x), in other words, that one of the roots of f(x) = must be a root of <^(x)= 0, § 795. It is here assumed that the minors fxi, fx^, •■-, fx^ are not all zero. If the minors of all the elements of D are 0, it can be proved that/'(.r) = and <^(a') = have more than one common root. If Xi, A2, • • •, A5 denote the cofactors of the elements of any 925 row of D, it follows from § 921 that when D = a-* : x^ : ic^ : a- : 1 = Xi : X2 : A3 : A4 : A5, whence x = Ai/Ao = A2/A3 = • • • = A4/A5. Therefore when f(x) = and (f> (a-) = have a common root, the value of this root is Ai/Ao. In the general case when the degrees of f(x) = and 926 (f) (x) = are m and u respectively, the resultant D will be a determinant of the (w + ^>)th order whose first n rows consist of the coefficients of f(x) and zeros and whose remaining m rows consist of the coefficients of ^(x) and zeros, arranged as in § 924, (3). Hence in the terms of D the coefficients of f(x) enter in the degree of (x, y)= is R (y) = 0, and one of the roots of R(y)= is )8, the corresponding value or values of x can* always be obtained by finding the highest common factor of f(x, ft) and <^ (x, /3). Usually this highest common factor will be of the first degree, when but one value of x will correspond to y = /3. But it may be of a higher degree, and then more than one value of x will correspond to y = /3. 928 Properties of the resultant. Suppose a pair of equations to be given of the form /(x) = a;'" + . • . + «,„ = 0, ) denote the resultant of f(x) = and (/, (x) = 0, § 923, we have The product n (a- — {3^) may be written K-A)('^i-A)---K-)8„), («.-A)(«2-/3.)---(a.-^„), K-/3i)K-A)---K-A0- But since <^ (x) = (x — (3i) (x — /S-.) ■■■(x — y8„), the product of the factors in the first row is (tri), in the second row (a2), and so on. Hence n(cr, -/?,) = <^W-<^(«.)-- -^^K)- Again, since f(^x) = (x — a^) (x — a,,) ■ ■ -(x — a-„,), the product of the factors in the first column is (— l)"'/(/8i), in the second column (— l)"'/(/32), and so on. Hence n (a, - A.) = (- lyvm -/m ■ ■ -/m- When the given equations have the form f{x) = a.x"' + • • • + a,„ = 0, (x) = b^x" + ... + &„ = 0, that is, when the leading coefficients are not 1, the product of the factors in the first row is {a{) /b^, and so on; and the product of the factors in the first column is (— l)"'/(/8i)/ao) and so on. Hence, in this case, to make 11 (a^- — yS^) an integral 516 A COLLEGE ALGEBRA function of the coefficients of f(x) — and (f>(x)= we must multiply it by ao^'". We then have i?(/, <^) = <^™n(a,-;8,) = a'o (a-i) • (0-2) • • • <^ (or,,.) 929 In the resultant of a pair of equations f (x)= 0, ^(x)= the coefficients of f (x) = e7iter in the decree of <^(x) = 0, and vice versa. For the product (a^) (^ (a^) • •• («„) contains m factors, each involving the coefficients of (a;) = to the first degree ; and the product /(/3i) -/(ySo) ■ • • /(/8„) contains n factors, each involving the coefficients oif(x) = to the first degree. We thus have another proof that the determinant D described in §§ 924, 926 is the resultant of f(x) = and cf>(x) = 0, that is, that D^ R(f, ). 930 The su 7)1 of the subscripts of the coefficients of f(x) = and cf) (x) = 171 each term of R (f , ) is van. For, by § 812, if we multiply each coefficient of f(x) and <^ (x) by the power of r indicated by its subscript, we obtain two equations /i (x) = aoX"* + miCC™- ^ + r^aox'"- ^ -\ \- r"'a„ = 0, <^i (x) — boX" + j'bix" ~ ^ + 7'%„x" '- + • • • + r''b,^ = 0, whose roots are r times the roots oi f(x) = and ^(.t) = 0. Each term of R (f, i) will be equal to the corresponding term of R (f (f>) multiplied by a power of r whose exponent is the sum of the subscripts of the coefficients of /(a-) and (x) which occur in the term. Hence our theorem is demonstrated if we can show that in every term this exponent is m?i. But since there are m?i factors in the product n (a, — yS^.), we have R (./;, <^0 = a'l/j'i; n (/-a, - 7'/3,) = 7""" ■ R (f <^). detp:rminants and elimination 517 Discriminants. The discriminant oif(x) — a^"^ -\ 1- «„ = 931 is that integral function of the coefficients oi fix) whose vanish- ing is the necessary and sufficient condition that/(a:-) = have a multiple root (compare §§ 635, 873). If D denote the discriminant of /(*") = 0, then For, by § 851, f{x) = has a finite multiple root when and only when f{x) — and /' (cc) = have a finite root in common. By § 928, the condition that/(.r) = and /' {x) = have a root in common is R(^f,f')= 0. But Uq is a factor of R (/,/'), as may be shown by expressing R (/, /') in the determinant form of §924. Hence R(f,f')=0 when «« = 0. But the root which in this case is common to f(x) = and/' (x) = is oo, §81G. Tlierefore, since oo is not a multiple root of /(a;)= unless both a^ and a^ vanish, we have D — R(f,f')/aQ. Thus, for aox2 + «iX + a2 = 0, Z> = The diserimijiant of f (x) = is equal to the product of the 932 squares of the differences of every t%vo of the roots of f (x) = multiplied by a certain power of the leading coefficient ao. Thus, if /(x) = ao (X - ft) (X - ft) (x - ft), (1) then, § 865, f'{x) = a^ [(x - ft) (x - ft) + (x - ft) (x - ^i) + (x-ft)(x-ft)]. (2) By § 928, B (/, /') = alf (^i)/' (ft)/' (ft). (3) But from (2), /'(/3i) = «o(/3i - ft) (/3i - ft), and so on. (4) Substituting (4) in {?,) and simplifying, we have R if, n = - aS (^1 - ^2)- (ft - ft)- (ft - ft)2, whence B = - a^ifi^ - ft)--2 (ft - ft)2 (ft - ft)2. (5) On the number of solutions of a pair of equations in two unknown 933 letters. Observe that if in the equation f(x, y)= we make the substitutions x = Xi/x^, y = x^/x^, and clear of fractions, we transform f{x, y) = into a homogeneous equation of the flo dl ^2 2ao ai -^ ao = - (a^ - 4 aoai). 2ao ai 518 A COLLEGE ALGEBRA same degree in Wi, a-g, x^. Thus, x^ + xy + y-{-l = becomes *t/i ~|~ *)C-^pCt^^^ "Y" OC^'iCo ~j~ OCq — U. Observe also that a homogeneous equation of the wth degree in iCg, Xg which is not divisible by Xg determines n finite values of the ratio Xo /x^. Thus, from a-| — 3 x^x^ + 2x1 — Owe obtain x^/x^ = 1 or 2. 934 Let f(x, y) = and <^ (x, y)=0 denote two equations whose degrees are ?n and ?i respectively. If they involve the terms a;"* and x", then by substituting x — Xi/xs, y = x^lxz, clearing of fractions, and collecting terms, we can reduce them to the form F(a-i, a-2, x^ = ao^cr + ai^^ " ^ H h «,„ = 0, (1) $(a:i, a-2, x^ = b^xl + h^xl'^ -[ \- b,^ =0, (2) where each of the coefficients ciq, a^, ■■■, bo, b^, ••• denotes a homogeneous fmiction of Xo, x^ of the degree indicated by its subscript. Hence R, the resultant of (1) and (2) with respect to Xi, is a homogeneous function of X2, x^ of the degree ?»n, § 930. By § 928 the necessary and sufficient condition that (1) and (2) be satisfied by the same value of x^ is that ^ = 0. (3) If R is not divisible by X3, then R = is satisfied by ??m finite values of x^/x^ or y, § 933. If /? denote any one of these values, the equations f(x, (3)= 0, (x, /3)= have a common root, and if this root be a, then x = a, y = fi is a solution oif(x, y) = 0, (x, y) = 0, all different or some of them equal. Hence f(x, y) — 0, (x, y) = 0. Hence /(a-, y) - 0, <^ {x, y) = also have mn solutions. In the preceding discussion it is assumed that R does not vanish identically. If R does thus vanish, f(x, y) and <^(x, y) have a common factor and therefore f{x, ?/) = 0, ^ {x, y) = have infinitely many solutions. We therefore have the theorem : If i(x, y) and 4>(x, y) are of the degrees m and n respec- tively and have no common factor, the equations f(x, y)=0, <^ {x, y)—0 have mn solutions. EXERCISE LXXXV 1. By the method of §§ 924, 925 show that the equations 6 x2+5 x-6 = and 2x3 + x2 — 9x — 9 = i^ave a common root and find this root. 2. Form the resultant of aoX- + aiX + a2 = and hoz- + 6iX + 60 = 0. 3. Find the resultant of ax^ + hxr + ex + fZ = and x^ = 1. 4. By the method of § 931 find the discriminants of the equations (1) x3 + px + g = 0. (2) ax3 + &x2 + c = 0. 5. By aid of § 931 show that x^ + x^ - 8 x - 12 - has a double root and find this root. 6. Solve the following pair of equations by the method of § 927. x2 - 3 x?/ + 2 2/2 - 16 X - 28 y = 0, x2 — xy — 2 2/2 _ 5 X — 5 2/ = 0. 520 A COLLEGE ALGEBRA XXXII. CONVERGENCE OF INFINITE SERIES DEFINITION OF CONVERGENCE 935 Infinite series. If i/^, v. 2, • • •, ic^, • • ■ denotes any given never- ending sequence of numbers, § 187, the expression "1 + ?'2 H \- iin -\ is called an infinite series (compare § 704). For Vi + n. + ■■■ we may write 2»„, read " sum of n„ to infinity." The series 2«„ is called real when all its terms Vi, «„, • • • are real, jjositlve when all its terms are positive. In what follows we shall confine ourselves to real series. A series is often given by means of aformula for its nth term m„. Thus, if M„ = Vn/{n + 1), the series is Vl/2 + V2/3 + V3/4 -\ . Sometimes such a formula is indicated by writing the first three or four terms of the series. Thus, in 1/2 + 1 • 3/2 • 4 + 1 • 3 • 5/2 • 4 • 6 H we have u„ = 1 ■ 3 • ■ • (2 u - l)/2 • 4 • • • 2 ji. 936 Convergence and divergence. Let S„ denote the sum of the first 71 terms of the series u^ + v., + • • •, so that Sj = Ui, So = Ui + u^, and in general S„ = ir^ -\- v„ -\- ■ ■ . -}- y^. As n increases, *S',^ will take successively the values ? and grj, ^2, • • • denote the sums of its first two terms, its next four terms, and so on, the series fifi + g'2 H will have the same sum as Wi + t<2 + ■ • • • For if Un denote the last term in the group g,n, we have gi + 92 + ■■■ + 9m = Ul + Mo + ... + «„, and the two members of this equation approach the same limit as m and therefore n is indefinitely increased. In the same manner it may be shown that a divergent positive series remains divergent when its terms are grouped. 941 We may therefore introduce parentheses at will in a con- vergent series. It is also allowable to remove them unless, as in the following example, the resulting series is divergent. The convergent series 1/2 + 1/4 + 1/8 + • • •, § 936, may be written (11 _ 1) _f (1| _ 1) + (11 _ 1) -I- . . .. But here it is not allowable to remove parentheses since IJ - 1 + 1^ — 1 + 1^ — 1 + • • • is divergent. 942 It is sometimes possible to find the sum of a series by the removal of parentheses. Thus, the sum of 1/1 • 2 + 1/2 • 3 + 1/3 • 4 + ■ • • is 1. For Sn = — + — + • • ■ + 1-2 2-3 ?i(?i + l) 1223 n n+1 n+1 Hence -S = liin S„ = lim (l ") = 1. V n + 1 / Example. Find the sum of the series whose nth term m„ is 1 /?i (n + 2). 943 Remainder after n terms. If the series «i + t'2 H (1) is convergent, that portion of the series which follows the nth. term, namely, ??„ + , + ?^, + ._, + ••• (2), will also be convergent, § 938. Let R„ denote the sum of (2). It is called the remainder after n terms of (1). Evidently lim /.'„ = 0. CONVERGENCE OF INFINITE SERIES 523 POSITIVE SERIES Theorem 1. A positive series n^ + «2 + • • • *« convergent if, as 944 n increases, S^ remains always less than some finite number c. For since the series is positive, .S",^ continually increases as n increases. But it remains less than c. Hence, § 192, it approaches a limit. Therefore, § 937, the series is convergent. Theorem 2. Let Ui + Uj H (1) denote a given positive series, 945 a7id let a^ -j- c''2 + ■ ■ ■ (2) denote a positive series known to he convergent. The series '(1) is convergent in any of the cases: 1. When each term of (1) is less than the corresponding term of {2). 2. When the ratio of each term of (1) to the correspc^nding ■term of (2) is less than some finite number/ a. ) 3. When in (1) tlie ratio of each term ^ the immediately preceding term is less than the corresponding ratio in (2). 1. For let 5„ denote the sum of the first n terms of "i + "2 H J and let A denote the sum of the series a^ + ag -I • If Ui < «!, i<2 < ^2) • ■ ■) we shall always have S^ < A. Hence «i + «2 + • • ■ is convergent, § 944. 2. For if — < c, — < c, ••■, then u-^ < ca^ «« < caz, ••■. Ol «2 Therefore, since cai-{-ca2-] is convergent, § 939, the series III + ^2 + • • • is convergent, by 1. 3. For if Ui Oi «2 «2 «4 04 Us as en W2 ^ «i — < — > rt2 «1 «3 «2 W4 «3 a4 a3 It follows from these inequalities that each of the ratios Uo/a^, Us/ds, • • • is less than the finite number Wi/^i. There- fore «i + M2 + • • • is convergent, by 2. 524 A COLLEGE ALGEBRA It follows from § 938 that the same conclusions can be drawn if any one of the relations 1, 2, 3 holds good for all but a finite number of the terms of the series (1) and (2). Example. Prove that 1 + 1/2 + 1/2 • 3 + 1/2 • 3 • 4 + • ■ . (1) is convergent by comparing it witli the convergent geometric series 1 + 1/2 + 1/2 • 2 + 1/2 . 2 • 2 + . . . (2) by each of the methods 1, 2, 3. First. Each term of (1) after the second is less than the corresponding term of (2). Hence (1) is convergent, by 1. Second. The ratios of the terms of (1) to the corresponding terms of (2), namely, 1, 1, 2/3, 2 • 2/3 • 4, • ■ • , are finite. Hence (1) is convergent, by 2. Third. The ratios of the terms of (1) to the immediately preceding terms, namely, 1/2, 1/3, 1/4, • • • , are less than the corresponding ratios in (2), namely, 1/2, 1/2, 1/2, • • •. Hence (1) is convergent, by 3. 946 Theorem 3. Let Ui + Uo -| (1) denote a given positive series, and let bj + bj + • • • (2) denote a positive series known to he divergent. The series (1) is divergent in any of the cases: 1. Wheyi each term of (1) is greater than the corresponding term of (2). 2. When the ratio of each term of (1) to the corresponding term of (2) is greater than some positive number c. 3. When in (1) the ratio of each term to the immediately preceding term is greater than the corresponding ratio in (2). The proof of this theorem, which is similar to that given in § 945, is left to the student. 947 Test series. The practical usefulness of the preceding tests, §§ 945, 946, evidently depends on our possessing test series known to be convergent or divergent. The most important of these test series is the geometric series a + ar -\- ar^ -\- ■ ■ -, which has been shown, § 704, to be convergent when r < 1, and which is obviously divergent when r>l. Another very serviceable test series is the following. 948 The series 1 + 1/2p + 1/3p + • ■ • + I/up + • • • , when p > 1, divergent when p < 1. CONVERGENCE OF INFINITE SERIES 525 1. p > 1. Combining the two terms beginning with 1/2^, the four terms beginning with 1/4^, the eight terms beginning with 1/8^, and so on, we obtain the equivalent series, § 940, ^ 1, and therefore 1/2^"' < 1, the geometric series (3) is convergent. Hence (1) is convergent, § 945, 1. 2. J) — 1. Combining the two terms ending with 1/4, the four terms ending with 1/8, the eight terms ending with 1/16, and so on, we obtain -^(^0-G-^^O (4) Evidently each term of (4) after the second is greater than the corresponding term of the series that is, greater than the corresponding term of 12 4 111 1+2 + 4 + 8 + -, orl + 2 + 2 + 2 + -. («) But (6) is divergent. Therefore (4) is divergent, § 946, 1. 3. p <1. In this case the series 1 + 1/2" + 1/3^ -] is divergent since its terms are greater than the corresponding terms of the series 1 + 1/2 + 1/3 + • • •, which has just been proved to be divergent, § 946, 1. 526 A COLLEGE ALGEBRA 949 Applications of the preceding theorems. The following exam- ples will serve to illustrate the usefulness of the theorems of §§ 945, 946. Example 1. Show that 1/1 • 2 + 1/2 • 3 + 1/3 • 4 + ■ • • is convergent. It is convergent because its terms after the first are less than the corresponding terms of the convergent series 1/2^ + I/32 + I/42 + • . • , §945, 1. Example 2. Show that 1 + 1/3 + 1/5 + 1/7 + • • • is divergent. The ratios of the terms of this series to the corresponding terms of the divergent series 1 + 1/2 + 1/3 + 1/4 + • • • , namely, 1, 2/3, 3/5, 4/7, • • • , 71/(2 n — 1), are all greater than 1/2. Hence 1 + 1/3 + 1/5 + • • • is divergent, § 946, 2. Example 3. Is the series In which m„ = {2n + l)/{n^ + n) convergent or is it divergent ? Here ^^^^_2n + l_ n 2 + 1/n ^1 2 + 1/n v? + n n^ 1 + 1/n^ n^ 1 + 1/n^' Hence the ratio of m„ to 1/n^ is (2 + l/n)/(l + l/n^), an expression which is finite for all values of n, and which approaches the finite limit 2 as n increases. But 1/n^ is the nth term of the convergent series 1 + 1/2- + 1/3- + • ■ • . Therefore the given series is convergent, § 945, 2. 950 By the method employed in Ex. 3, it may be proved that if 7i„ has the form n,^ = f(7i)/(n), where f(n) and <^(??) denote integral functions of 71, the series is convergent when the degree of <^ (n) exceeds that of /(?i) by more than 1 ; otherwise, that it is divergent. Example 1. Show that the following series are convergent. 1 1 > a(a + b) (a + b){a + 2b)'^ {a + 2b){a + 3b)'^"'' Example 2. Show that the following series are divergent. V2 V3 V4 1-1-2 V2 1 + 3 V3 I+4V4 CONVERGENCE OF INFINITE SERIES 527 Example 3. Write out the first four terms of the series in which m„ has each of the following values and determine which of these series are convergent and which divergent. 2n-l .^, V^ ,„, n2_(^_l)2 (1) Un = (2) Un = (3) Un — ^^ -• ^ ' (n + 1) (n + 2) n2 + 1 n^ + (n + 1)3 Theorem 4. The positive series u^ + u^ + ■ • • is convergent if 951 the ratio of each of its terms to the immediately preceding term is less than some number r which itself is less than 1. For in «i + n^ + 1/3 + • • • (1) the ratio of eacli term to the immediately preceding term is less than the corresponding ratio in the geometric series Vi + Uir + Uir"^ + • • • (2), since in (1) the ratio in question is always less than r, while in (2) it is equal to 7\ But (2) is convergent since r < 1. Therefore (1) is convergent, by § 945, 3. If the ratios above mentioned are equal to 1 or greater than 1, the series is divergent ; for in this case lim «„ ^ 0. Corollary. If as n increases the ratio u„ + i/Uj, apjjroaches a 953 definite limit X, the series is convergent ichen A < 1, divergent when A > 1. 1. For if X < 1, take any number r such that X < r < 1. Then, since lim (?/,„ + j/^^,) = X, after a certain value of n we shall always have Vn + i/Un — X < r — X, § 189, and there- fore u„ + i/i(„ < r. Hence the series is convergent, §§ 938, 951. 2. If X > 1, after a certain value of 71 we shall always have ^^< + i/"n > 1- Hence the series is divergent, § 951. When «„ + i/«„ > 1 and lim (m„+i/?<„) = 1, the series is divergent; but when «„ + i/'/„ < 1 and lim (w„_^i /?/„) = 1, no conclusion can be drawn from the theorem of § 951. Example 1. Show that - + ——- + "^" ^ -f • ■ ■ is convergent. 5 5 • 10 5 • 10 • 15 The ?xth term of this series is 3 • 5 • 7 • • • (2 n + l)/5 ■ 10 • 15 • • • 5 n, and the ratio of this term to the term which precedes it is (2 n + l)/5 n. But since (2 n + l)/5 n = 2/5 + 1/5 n, lim (2 n + l)/5 n = 2/5, which is < 1. Hence the series is convergent. 528 A COLLEGE ALGEBRA Example 2. When is ■ 1 1 f- • • • convergent, X being positive? 1 + x 1 + 2x^ 1 + 3x3 M„ + i _ I + nx" X" + l/n Here l + (>i + l)x" + i x» + i(l + l/n) + l/ft M 1 and therefore lim -^^^ — - • Un X Hence the series is convergent when 1/x < 1, that is, when x >1. 1 1 • 3 1 • 3 • 5 Example 3. Show that - -\ '- ^ '^^ 1- • • • is convergent. 1 1-4 1-4-7 ^ Example 4. Show that - + - 1 1 is convergent. 2 2 5 2 5 2 X x^ x^ Example 5. When is - H 1 f- • • • convergent, x being positive ? Example 6. When is 1 h 1 convergent, x being positive? 1+^ 1+^'^ 1 + -^' 953 Series in which lim (Un + i/Un) = 1. In a series of this kind the ratio ii„ + i/Un can be reduced, to the form Wn + i/»„ = l/(l + «-„M where lim (a„/n) — 0. We proceed to show that if, as n increases, cr„ ultimately becomes and remains greater than some number which is itself greater than 1, the series is con- vergent ; but that if a„ ultimately becomes and remains less than 1, the series is divergent. 1. For suppose that after a certain value of n, which we may call k, we have cTb > 1 + a, where a is positive. Then ^!^ = ;; < , when n%k. Un 1 + a„/n 1 + (1 + a)/n But we may reduce this inequality to the form Un + i< — [nun-(n + l)u„ + i], when n%k. (1) In (l) set n = k, k + I, ■ ■ ■ , k + I — 1 successively, and add the resulting inequalities. We obtain Uk + i + UK- + 2 + ■■■ + ui- + i< ~[kuk - (k +l)u^.+ ,]. (2) It follows from (2) that as I increases the sum of the first I terms of the positive series Uk + i + Uk + 2-^ remains always less than the finite CONVERGENCE OF INFINITE SERIES 629 number ku^/a, which proves that this series is convergent, § 944. There- fore the complete series vi + M2 + • • • is convergent, § 938. 2. Suppose that when n > A; we have «» < 1. Then -^^^ = > , when n>k. w„ 1 + a„/n 1 + 1/n But 1/(1 + l/ri) is the ratio of the corresponding terms of the divergent series 1 + 1/2 + 1/3 + • • ■ ; for l/(n + 1) ^ 1/n = 1/(1 + 1/n). Hence the given series Mi + mj + • • • is divergent, § 946, 3. If a„ remains greater than 1 but approaches 1 as limit, the preceding test will not determine whether the series is conver- gent or divergent. But in this case «"„ can be reduced to the form a„ = 1 + (3,,/n, where lim ;8„/?i = ; and if j3„ remains less than some finite number b, the series is divergent. For since ^„ < 6, we have Un + l _ 1 _ 1 1 M„ " 1 + a„/n ~ 1 + 1/n + /3„/n2 l + l/n + b/n^' But 1/(1 + 1/n + b/n'^) in turn is greater than the ratio of the corre- sponding terms of the divergent series 1/(1 — 6) + 1/(2 — 6) + 1/(3 — 6) H . 1 1 n-b 1 1 For {n + l)-bn-b (n - 6) + 1 l + l/(n-6) l + 1/n + b/n^ 111 1 6 62 + -, + -^ + n — b n 1 — b/n n n^ n^ Therefore the given series iti + W2 + • • • is divergent, § 946, 3. It follows from the preceding discussion that a series in 954 which Un + i/i(n can be reduced to the form v,^^Ju,^ = (iiP + ««''-! H )/(nP + a'nP-'^ -\ ) is convergent when a' — a > 1, divergent when a' — a<_ 1. For dividing the denominator of this fraction by its numerator, we have u„^i _ np + anP-^ + ■■ ■ _ 1 Un ~ ni> + a'nP - 1 H ~ I + (of - a)/n + ^JrC^' where /3„ is finite. Example. Prove that the " hypergeometric series" a-/3 a(a + l)^(^ + l) a(a + 1) (a + 2)/3(^ + 1) (^ + 2) ■^1-7 1-27(7 + 1) l-2-37(7 + l)(7 + 2) is convergent when 7 — a — /3 > 0, divergent when 7 — a — )3 ^ 0. 630 A COLLEGE ALGEBRA EXERCISE LXXXVI Determine whether the following series are convergent or divergent. 1 1 1 2+1 "^22 + 1 23+1 3. -^ + ^ + ^H 2-3 3-4 4-5 5. 1 1 1 V3 ^ ^ 7. 2 2-4 2-4.6 4 4-7 4 • 7 • 10 8. 2 2-3 2-3.4 4 4-5 4.5C 9. 1 1 • 3 1 • 3 • 5 2 ' 2-4 2-4-6 a^ a^ + 1 a'2 + 2 2 • 4 • 6 • • • 2 /I 4 • 7 • 10 • • • (3 n + 1) 2-; (n+l) + ••• + 4 ■ 5 • 6 • • • (ji + 3) 1.3..5...(2n-l) 2-4-6---2?i Write out the first four terms of the series in which ?t„ has the follow- ing values and determine whether these series are convergent or divergent. 3/— n + l ^, Vn 10. u„ = ■ 11. u„ 12. Un = V?l- + 1 — ?l = 'Vn- + 1 + ?i Determine wliether the series in which itn + \/Un has the following lues are convergent or divergent. u„ + i 2n , ^ u„ + 1 _ 3 ?i3 — 2 ?i2 ~ 3 n^ + 9i2 + 1 ■ 13. 14. Un 2 n + 3 For what positive values of x are the following series convergent? ,. . 3 3-6 , 3-6-0 , 15. 1 + -X + a;2 + x3 + .... 5 5-8 5-8.11 16. 1 + X 1 + X- 1 + X^ 1 + X'* a(a + 1) a(a + l)(a + 2) 1-2 1 •2-3 is divergent 17. Show that - + when a is positive. 18. If for all values of n we have Vm^ < r, where r is positive and less than 1, show that Wi + uo + • • ■ is convergent by comparing it with the convergent series r + r'- + r'' + • • • . CONVERGENCE OF INFINITE SERIES 531 SERIES WHICH HAVE BOTH POSITIVE AND NEGATIVE TERMS General test of convergence. By definition, § 937, an infinite series of any kind Wj + ^2 + • • is convergent if 5„ approaches a finite limit as n is indefinitely increased. But, §§ 195, 197, S'„ will approach a limit if the sequence of values through which it runs as n increases, namely, S-^, S^, S3, •■•, possesses the property that for every given positive number 8, however small, a corresponding term S^. can be found which differs numerically from every subsequent term S^.^^ by less than 8. If this condition is not satisfied, .S^„ will not approach a limit, § 198. Since .s:^. = k^ + . . . + w^,^ and ,S\.+^ = «i -I f- Wt + i'k+i-\ + %+p, we have S^+^ - 5^. = ? — -1{ h • • • to A; terms, i. e. > — • A: or - . k + \ k + 2 k + k 2k 2k 2 k 2 Hence k cannot be so chosen that u^+i + • • • + wt + t is less than every assignable number, and the series is divergent (compare § 948, 2). 532 A COLLEGE ALGEBRA 956 Corollary 1 . A series ivhich has both positive and negative terms is convergent if the correspo?iding positive series is convergent. For let «i + ifo + ■ ■ ■ (1) be the given series, and let u\ + u'2 H (2) be the same series with the signs of all its negative terms changed. Then |%+i + «/i-+2 H + %+p| < «'i + i + «'i- + 2 -\ h w'i+p- Hence, if by taking k great enough we can make «',+ ! + ■•• + «', + ;.< 8, the same will be true of l^ + i H h i is convergent. Thus, t/1 + iV22 + ■i-V32 + ... is convergent since 1 + 1/22 + 1/32 ^. . . . is convergent. 958 Corollary 2. A series whose terms are alternatehj positive and negative is convergent if each term is numericalhj less than the term which precedes it, and if the limit of the nth term is 0. For let the series be ^i — 02 + "s — • • •, where «i, a.,, • • • are positive. Using the notation of § 955, we here have I%+i + '«i + 2 H h ^'i+,.| = |«i+i - «A- + 2 H +(- l)''~'oi.+,.|. We can write a^. + i — a^.^^_ -\ f- (— 1)^"'%.^^ (1) in the form {n, ^^ - a,^^) + {a,^^ - a^^,) + ■ ■ ■ (2) and in the form r/^_^, — (rti_^2 ~ '^i + a) ~ ' ' "• (2) Since a^.^, > a^.^g > '''a + s > •••> each of the expressions in parentheses in (2) and (3) is positive. Hence it follows from (2) that (1) is positive, and from (3) that (1) is algebraically less than %.+ !, and therefore from (2) and (3) combined that (1) is numerically less than %+*• But since lim a„ = 0, we can choose h so that a^. + i < 8. Therefore «! — a^ + a^ — • • • is convergent, § 955. CONVERGENCE OF INFINITE SERIES 533 Absolute and conditional convergence. A conyergent real series 959 is said, to be absolutely converfjent if it continues to be convergent when the signs of all its negative terms, if any, are changed ; conditionallij convergent if it becomes divergent when these signs are changed. Thus, 1 — 1/2 + 1/4 — 1/8 + • • • is absolutely convergent since the series 1 + 1/2 + V-4 + 1/8 + • • ■ is convergent. But 1 — 1/2 4- 1/3 - 1/1 + • ■ • , which is convergent by § 958, is only conditionally convergent since 1 + 1/2 + 1/3 + 1/4 + ■ • • is divergent. Theorem. In an absolutely convergent series the positive terms 960 hy themselves form a convergent series, and in like manner the negative terms by themselves. And if the sums of these two series be P ayid — N respectively, the sum, of the entire series is P — N. But in a conditionally convergent series both the series of posi- tive terms and the series of negative terms are divergent. Por let Vi + ?<2 + • • • be a convergent series which has an infinite number of positive and negative terms. Of the first n terms of this series suppose that^j are positive and q negative. Then if S,^ denote the sum of all ?t terms, Pp the sum of the p positive terms, and — N^ the sum of the q negative terms, we shall have 6'„ = P^ — N^. When n is indefinitely increased both p and q will increase indefinitely, and since S,^ will approach the finite limit S, one of the following cases must present itself, namely, either (1) both Pp and N^ will approach finite limits which we may call P and N, or (2) both P^ and N^ will approach infinity. In the first case lim .§„ = lim (P^^ — N^ = lim P^ — lim N^, § 203, that is, S = P — N. The series is absolutely convergent. In fact, after the change of the signs of the negative terms the sum of the series is P + A^. In the second case the series is conditionally convergent. Por if S'„ denote the sum of the first n terms of the series obtained by changing the signs of the negative terms, we have lim 5'„ = lim {Pp + N^) = c^. 534 A COLLEGE ALGEBRA 961 Corollary. The terms of a conditionally conx'erge7it series may be so arranged that the sum of the series will take any real value that may he assigned. For, as just shown, in a conditionally convergent series the positive terms by themselves and the negative terms by themselves each constitute a divergent series the limit of whose nth term is 0. Hence, for example, if we assign some positive number c, and then, without changing the relative order of the positive terms or that of the negative terms among themselves, form .§„ by first adding positive terms until the sum is greater than c, then negative terms until the sum is less than c, and so on indefi- nitely, the limit of this S„, as n is indefinitely increased, will be c. Hence the commutative law of addition does not hold good for a conditionally convergent series. EXERCISE LXXXVII 1. Determine whether the following series are convergent or divergent. n^ 3 3 • 5 3 ■ 5 • 7 .'>. • 5 • 7 • 9 ^''3 3-63-6-9 3-6-9.12 2. For what real values of x are the following series convergent and for what values are they divergent ? (i)-J- + -J- + -^L_ + ... + '- + .... ^^l_x l + 2x l-3x l + (-l)«nx (2)-^ + ^l- + -^^ + ---+ ^"'"' +■••. ^ ^ l + x2 1 + 2x* 1 + 3x6 l + nx'-'" 3. If ii\ -{- Ui -V v-z + ■ • • is absolutely convergent, and Oi, a2, as, • • • denote any sequence of numbers all of which are numerically less than some finite number c, prove by the method of § 956 that the series aiMi + aiU2 + azUz + • • • is also convergent. 4. If .S denotes the sum of a series of the kind described in § 958, show that the sums ai, ai — 02, ai — a^ + a^, ■■ ■ are alternately greater and less than S. convergp:nce of infinite series 535 CONVERGENCE OF POWER SERIES Power series. This name is given to any series which has 962 the form «o + «i^ + <^2^^ + • • • + «„»;" + • • • (!)> where a; is a variable but a^, a^, • ■■ are constants. The values of x and Uq, Ui, • ■ • may be real or imaginary. By § 957, the series (1) is convergent if the positive series lao| + |«ix| + [a^x^l + • • • + |«„a;"| + • • • (2) is convergent. When (2) is convergent we say that (1) is absolutely/ convergent (compare § 959). Whether (1) is convergent or divergent will depend upon the value of x. Hence the importance of the following theorems. Theorem 1 . If tvhen x = b evenj ter'm of slq + ^iX + • • • w 963 numerically less than some finite positive number c, when x|<[b| the series is absolutely convergent. For since [ a,p" | < c for every n, have \a,^x" \ = | a,p'" | • y < A for every n. Hence each term of |ao| + l^i^;] + |a2*^ the corresponding term of c + c + • • • (1) is less than + •••(2). But (2), being a geometric series, converges when \x/h | < 1, that is, when \x\ < \b\. And when (2) converges, so does (1), § 945, 1. Thus, l + 2x + x2 + 2x3 + --- converges when [x] < 1. Corollary 1 . If Sio -\- aiX + ■ • • is converge7it when x = b, it is 964 absolutely convergent when |x| < |b|. This follows immediately from § 963. For since ao + «i^ -\ is convergent when x = b, all its terms have finite values when X = b. Corollarj^ 2. If slq -}- ajX + • • • is divergent when x = b, it is 965 also divergent when [x|>|bl. 536 A COLLEGE ALGEBRA Por were a^ + a^x ^ to converge for a value of x which is numerically greater than h, it would also converge for x = b, §964. 966 Limits of convergence. It follows from § § 964, 965 that if we assign to a class A^ all positive values of x for which fffl + ajx H converges and to a class A^ all for which it diverges, each number in .4i will be less than every number in A2. Hence, § 159, there is either a greatest number in A^ or a least in A^. Call this number X. It represents the limit of convergence of a^ -\- a^x + ■ ■ ■ , the series being absolutely con- vergent when |ji'l < A, divergent when |x-| > A. Thus, in both x + xV2 + x^/Z + •••(!) and x + x'V^^ + x^/Z^ + • • • (2) the limit of convergence \ is 1. Observe that (1) diverges and (2) con- verges when X = X = 1. It is possible to construct a series in which X = ; for example, the series x + 2 ! x^ + 3 ! x^ + . . •. What we have called the limit of convergence is more frequently called the radius of the circle of convergence. For if we picture complex num- bers by points in a plane in the manner described in § 238 and draw a circle whose center is at the origin and whose radius is X, the series ao + aix + • • • will converge for all values of x whose graphs lie within the circle, and it will diverge for all values of x whose graphs lie without the circle, § 239. 967 Theorem 2. If in ao + ajX + • ■ ■ the ratio \ a„/a„ + j [ approaches a definite limit fi, tlien p, is the limit of convergence. For, by § 952, the series |r/o] + |rti./'l -\ converges when lim "^''', — < 1, that is, when l.rl < lim -^ • Similarly |«„| + |aix-| H diverges when \x\ > lim M^ • Example L Find the limit of convergence of the series 5 5 10 5- 10- •• 5ft bmce = = — , wo have fi = lim — = -. a„ + i 2n + 3 2 + S/n 2 + 3/n 2 CONVERGEXCE OF IXFIXITE SERIES 537 Example 2. Find the limits of convergence of the series ■ • • + 23x-3 + S-'^x-s + 2 x-i + 1 + x/3 + xV32 + x^fi^ + • . • . Here 1 + x/3 + x-/3- + • • ■ Is a geometric power series In x which converges when |x|<3, for a„/a„ + i = 3. On the other hand, 2x-i + 22x-2 + osx-"^ -|- ... is a geometric power series in x-i or 1/x which converges when |x-i| 2. Hence the given series converges when 2 <|x|<3. Example 3. For what real values of x will the series x/(l + x) + 2 xV{l + x)2 + 3xV(l + x)3 + . . . converge ? This is a power series in x/(l + x) which converges when |x/(l + x) |< 1, for 11m a„/rt„ + i = 11m n/{n + 1) = 1. But |x/(l + x)|< 1 for all positive values of x and for negative values which are greater than — 1 /2. Hence the series converges when x > — 1/2. The binomial, exponential, and logarithmic series. We proceed 968 to apply the preceding theorem to three especially important power series. 1. The exponential series, § 990, namely, i + f + f; + - + n^ •■ 1 i ! nl is convergent for all finite values of x. ■r^ , a„ 1 1 i or here = = = n + 1. a„+i n\ (ri + 1)! Hence 11m — '— = 11m {n + 1) = oo, that is, ^ = oo. O-n + l 2. The logarithmic series, § 992, namely, J, 6 ^ ^ n is convergent when |a'j < 1, divergent when |xj > 1. ^ , dn 1 1 n + 1 if or here = ; = ■ a„ + 1 n n + 1 n Hence lim — ~ = - lim = - lim ( \ -\- -\ = — \^ that is. u = 1. (1,1 + 1 n \ 11/ ' ' The series converges when x = 1, § 958, diverges when x = — 1, § 948. 538 A COLLEGE ALGEBRA 3. The binomial series, namely, where m is not a positive integer, is convergent when \x\< 1, divergent when |a;| > 1. For here a,, _m {m — \) • • ■ {m — n + \) m {m — 1) • • • (m — n) _ n + 1 a„ + i ~ l-2---n l-2---(n + l) " m - n TT 1- ^n ,• J^ -*■ ^ ,. 1 + l/n 1 ^, ^ • 1 Hence lim = lira — =-i- -- — lim = — 1, tkat is, /u = L a„ + 1 m — 11 1 — m/n When X = 1 the series converges if m> — 1, diverges if ?n < — 1 (see § 1001, Ex. 2). When X = — 1 the series converges if m > 0, diverges if m < 0. For when x = — 1, by setting m = — a we may reduce the series to the form a(a + l) a(a + l)(a + 2) 1-2 1-2-3 Evidently from a certain term on all the terms are of the same sign, so that the test of § 954 is applicable, § 956. But here Un^ ^a + n -I ^n + (a - \) _ u„ n n Hence, § 954, the series converges if — (a — 1) > 1, that is, if — a > 0, or, since — a = m, it converges if m > 0. But it diverges if — (a — 1) < 1, that is, if m < 0. EXERCISE LXXXVIII Determine the limits of convergence of the following series. 1. 1 + mx + ??i2xV2! + ?n3xV3 ! + •••. 2. 2(2x)2 + 3{2x)'' + 2(2x)* + 3(2x)5 + -... m (m — 2) , m (m ■— 2) (m. — 4) „ 3. mx + —^^ X'- + — ^ x3 + . . . . For what real values of x will the following series converge ? 3x 1/ 3x \2 1/ 3x \3 x + 4 2Vx + 4/ 3\x + 4/ X2 + 1 \X2 + 1/ Vx2 + 1/ ■(3x)-3+ (3x)-2 + (3x)-i + l + 2x + (2x)2 + (2x)8 + .-.. OPERATIONS AVITH INFINITE SERIES 539 XXXIII. OPERATIONS WITH INFINITE SERIES SOME PRELIMINARY THEOREMS When a given power series a^ + r/^.r + • • • is convergent, its 969 sum is a definite function of x which we may represent by f{x), writing f(x) = «o + a^x + • • •. In what follows when we write /(a") = a^ -\- a^x + • • •, we assume that a^ + a^x + • • • has a limit of convergence X whicl "is greater than 0, and suppose that \x\< A. Theorem 1 . Give7i that ^ (x) = aiX + aaX^ + . . . , and that when 970 X has the j^ositive value b every term of <^(x) is numerically less than some finite positive number c. If any positive number 8 be assigned, hoivever small, then l(x)| < 8, ivheJiever |x| < b8/(c + 8). For, as was shown in the proof in § 963, when [a:;|< J, \4.{x)\ «i = ^u ''2 = ^2> • • •• This theorem is called the theorem of undetermined coeffi- cients. It asserts that a given function of x cannot be expressed in more than one way as a power series in x (compare § 421). OPERATIONS WITH INFINITE SERIES 541 OPERATIONS WITH POWER SERIES Since many functions of x can be defined by means of power series only, it is important to establish rules for reckoning with such series. These depend upon the following theorems, §§ 974, 976, which we shall demonstrate for infinite series in general. Theorem 1. If the series xi^ + u-z + • ■ ■ end Vj + Vj + • • • 974 converge and have the sums S atid T respectively, the series (ui + Vi) + (u2 + Vo) + • ■ • converges and has the surn S + T. For, § 203, lim [{u^ + v,) + (uo + v^) + • • • + (»„ + O] = lim {ui + iin^ h 11,;) + lim (;'i + v^ -] 1- -?;„) = S+T. The like is true of the series obtained by adding the corre- 975 sponding terms of any J7?iite number of infinite series. Hence the rule for adding any finite number of functions defined by power series in x is to add the corresponding terms of these series, that is, the terms which involve like powers of x. Thus, if /(x) = 1 + X + x2 + • . . and

+ ■ • • (1), U. = u^P + u<|> + • • • (2), • • .. 542 A COLLEGE ALGEBRA Again, let U/, U2', • • • denote the sums of the series obtained by replacing the terms of (1), (2), ■■■by their absolute values, so that and so on. If the series Uj' + Ug' + • • • is co7ivergent, the several series obtained by adding the corresponding terms of (1), (2), •••, namely, the series u<}> + u^f) ^ , u^p + u(|) -i , and so 07u are convergent, and if their sums be denoted by Vi, Vj, • • •, we shall have Ui + U2 + U3 + • • • = Vi + V2 + V3 + • • • . For let us represent the remainders after n terms in the series (1), (2), • • • by R<1\ R^f^, • • -, § 943, so that C/i = ?<(p + ?/p H + M^p + R^l\ U^ = « + w<|> H h z<'2) + R(i\ U^ = u(\^ + tf<^ + • • • + u^';^ + R^^\ Each of the cohxmn series u^\^ + «*-p -| , «/p + w^f) H , and so on, is convergent since each of its terms is numerically less than the corresponding term of the convergent series Ui + U2+^--, §945, 1. Let the sums of these series be denoted by l\, l\, ■■-, V„, R„. If we add the corresponding terms of these n + 1 column series, we obtain the original series Ui -{- U^ -\ . Therefore, since n is finite, we have, § 975, C^i + C^2 + • • • + f^t + • • • = Fi + F2 + • • • + F,. + i?,.. To prove our theorem, therefore, we have only to show that when n is indefinitely increased lim R„ = 0. But if the remainder after k terms in 7?^^ + 7?^;' + • • • be denoted by 6<*>, we have R„ = R^l^ + R^-^ -\ h R^t? + ^'^T- Let 8 denote any positive number, it matters not how small. Since each term of R^]^ + A";-;' H (a) is numerically less than OPERATIONS AVITH INFINITE SERIES 543 the corresponding term of Ui + U^' + ■ ■ -(h), the remainder after k terms in (a) is numerically less than the corresponding remainder in (b). But since (b) is convergent we can so choose k that the latter remainder will be less than 8/2. Hence we can so choose k that ivhatever the value o/n may be, we shall have S'^l^ < 8/2 numerically. But again, since each of the row series u^\' + u^^^ + • • • > w'^f + ?//^' 4- • • •, is convergent, as n increases each of the k remainders R^]}, 7i^,f , • • • R'^^^ will ultimately become and remain numerically less than 8/2 k, and therefore the sum of these remainders, namely, R'^]^ + iZ^,f + • • • + R'-'i^ will become and remain less than (8/2 k)k, or 8/2. Therefore, as n increases, /?„ = R^l^ + R^-J -\ 1- R^l'> + S^^;^ will ultimately become and remain numerically less than 8/2 + 8/2, or 8. Hence liui R,^ = 0, § 200 ; and therefore U,+ U^+U, + ---= V, + V, + F3 + . . ., as was to be demonstrated. A series f/i + f/2 + • • • each of whose terms is itself an infinite series is called a doubly infinite series. Thus, consider the series x/{\ + x)- xV(l + x)2 + xy{\ + x)3 (1) which converges for all real values of x which are greater than — 1/2 (also for imaginary values of x whose real parts are greater than — 1/2). Is it possible to transform (1) into a power series in x, that is, into a series which will converge for any value of x except ? When |x||. But since (1) is absolutely convergent, Hm (|tt„ + i] + • • • + |Mn + p|) = 0« Therefore lim |S;„ - S„l = 0, that is, lim S;„ = lim S„. 2. We may break the series up into any number (finite or infinite) of series the terms of each of which occur in the same order as in the original series. For we can recover the original series from every such set of series by applying one of the theorems of §§ 974, 976. Thus, if we form one series out of the terms of Mi + M2 + Ws + • • • which have odd indices, and another out of those which have even indices, we have, by § 974, Wl + W2 + Ma + M4 + • • • = (Ml + M3 + U5 + • • •) + ("2 + W4 + "6 + • • •)• OPERATIONS WITH INFINITE SERIES 545 Or again, arrange the terms of Mi + U2 + ?«3 + • • • as follows : Ml In this scheme there are an infinite number of u<2 + 7(3 columns, each forming an infinite series. M4 + Ms + We The sum of Ui + W2 + M3 + • ■ • is equal to the sum of W7 4- "8 + Mg + Uio the terms of the scheme added by rows, § 940. And the sum by rows is equal to the sum by columns, § 976. Hence Mi + M2 + M3 H = ("i + W2 + "4 H ) + (ws + U5 H ) -\ . And similarly in every case. 3. Every possible rearrangement of the terms of iti + «2 + M3 ^ \-^in-i may be had by combining 1 and 2. Products of power series. If the functions/(a3) and (x) are defined, when | •?- 1 < A, by the power series/(a;) = ao + ^1^ H (1)> (f, (a-) = ^-0 + ^'i^' + • • • (2), their product f(x) ■ <^ (x) will be defined, when |a'| < A, by a power series derived from (1) and (2) by the ordinary rules of multiplication (compare § 314). Thus, /{x)=ao + aix + Uox"^ + aax" -\ (1) (X) = 60 + hx + 6ox2 + 63x3 + • • • (2) /(x) • <^ (X) = ao6o + aibo I x + «2^o x^ + a-zK x^ + ■ • • (3) ao6o + aiho x + a^ho x2 + azh» + ao6i + ai6i + a^hi + aob-i + ai62 + ao63 For when |x]{x) are replaced by their absolute values. We may therefore add the corresponding terms of /(x) 60 = aobo + aib(yc H , /(x) 6iX = + ao&iX + • • • , and so on. The result is the series (3). Example. Express (1 + x + x2 + • • •) (1 + 2 x + 3 x^ + • • •) as a power Transformations of power series. Suppose that the power series rr,, + aii/ + a„y/'^ + ■■•(!) converges when jy] < X. Suppose also that y may be expressed in terms of x by the power series ]/ = b^ -\- b^x -\ (2), where [6o| < A. 546 A COLLEGE ALGEBRA By repeatedly multiplying (2) by itself, we obtain expres- sions for y, if', i/, • • • in the form of power series in x which converge when (2) converges. If we substitute these expres- sions in the terms a^y, a^^i/'^, ■ • • of (1), we obtain a series of the form a^ + a^ (b^ + h^x + ---) + a^ {!>% + 2 hjj.x + ...) + ... (3), and this, when the terms which involve like powers of x are collected, becomes a power series in x of the form K + «A + •••) + («i^'i + 2 aAh + • • Oa- + • • • (4). This final series will converge and have the same sum as (1) for all values of x such that |&„| + |Jia-| + • • ■ < X. For in this case the condition of § 976 is satisfied by the doubly infinite series (3), the series f/j' + Uo,' + • ■ • being |ao| + |«i| (l^ol + \^i^\ + •••)+■■■' which by hypothesis con- verges when l^ol + I ^1-^1 + • • • < A. 980 Quotients of power series. A fraction whose numerator and denominator are power series, as («o + a,x + -. ■)/{]>, + b,x + ■■■), where ^o "^ 0, may be transformed into a power series which will converge for all values of x for which a^ + ciiX + • • • con- verges and \hix\ + \h^x'^\ -f • • • < \b^^\. For let y = biz + hoz"- + • • • . (1) 1 111 Then bo + 6ix + 62x2 + . . . 60 + 2/ 60 1 + y/bo (2) 60 since, by hypothesis and § 232, |?/|<|6ix| + \biX-\ + ■ • • <]6o|- In (2) replace y by its value (1) and then apply § 979. We shall thus transform (2) into a power series in x which converges when |5lX| + |?^2X2|+-.-<|6o|. Multiply this power series by a^ + ctiX + 02x2 + ■••,§ 978. The result will be a power series in x which will converge and be equal to the given fraction for all values of x for which ao + aiX + • • • converges and|6ix| + |62-c21+ •v<|&o|- 1 + 3 + 7 + ■ • • 1 + 1 + 1 + --- 2 + 6 + --- 2 + 2 + • . • OPERATIONS WITH INFINITE SERIES 547 The quotient series may be obtained to any required term by the process of cancelling leading terms described in § 406, or by the method of undetermined coefficients, § 408. Example. Expand (1 + 2 x + 2^2 + . . .)/(i ^ x + x^ + ■ • ■) to four terms. Using detached coefficients, we have 1 + 2 + 4 + 8 + ■■• 1 1 + 1 + 1 + 1 + ... 1 + 1 + 1 + 1 + ■ • • |l + l + 2 + 4 + ... Hence the quotient series is 1 + a; + 2 x2 + 4 x3 + . . . . It converges when |x| < 1/2. 4 + ... When instead of being infinite series the numerator and 981 denominator are polynomials in x, so that the fraction has the form (^?,) -\- a^x -{-■■• + ci',n^'")/(^o + biX + ■ • ■ -\- b,flf), the quo- tient series will converge for all values of x which are numer- ically less than the numerically smallest root of the equation h^ + h^x + • ■ ■ -\- h^x" = 0. This will be evident from the first of the following examples. The second of these examples illustrates the form of the quotient series when b^ = 0, and the third illustrates the method of expressing the fraction as a power series in 1/x. Example 1. Find the limit of convergence of the series which is the expansion of (3x + 8)/(x2 + .5x + 6). By the method of partial fractions, § 537, we find 3x + 8 3x + 8 21 and x2 + 5 X + 6 (x + 2) (X + 3) X + 2 x + _2_ xT2 1 + x/2 1 But-^ = — ^- — = ^1 + -") '=1-- + -" when|xl<2, -x/2 \ 2/ 2 22 II' X + 3 3 1 + x/3 = Vl+^^ '^l_^ + ?! when|xl<; 3 \ 3 / 3 32 33 ' ' ^^ r 3x + 8 4 llx 31x2 V , ,^0 Therefore = when x < 2. X2 + 5X + 6 3 18 108 ' ' 548 A COLLEGE ALGEBRA Example 2. Expand (1 - x)/(x- + 4 x^) in increasing powers of x. We have ] ~f ^ = l-l::!^ = 1(1 - 5x + 20x2 - 80x3 + • ■ •) x2 + 4x3 xn + 4x x^^ ' = x-2 - 5x-i + 20 - 80x + • • ■. Example 3. Expand (2x2 + x - 3)/(x3 + 2x + 4) in powers of 1/x. 2 x2 + X - 3 _ 1 2 + 1/x - 3/x2 _ 1 /q 3 5 \_2_3 5 x3 + 2x + 4~x'l + 2/x + 4/x3 ~x\ x ^")~x x^~x^"'' 982 Reversion of series. From the equation j/ = a^x + a^x^ + • • •> defining y in terms of x, it is possible to derive another of the form X = b^y + b^ij^ + • • •> defining x in terms of y. The pro- cess is called the reversion of the given series a^x + a2X^ + • • •• It will be observed that this series lacks the constant term a^, and the understanding is that a-^ ^ 0. It can be proved that if aiX -f Uox'^ + •■• has a limit of convergence greater than 0, the like is true of the reverted series h^y + b^y'^ -\ . Example. Revert the series y = x + 2x'^ + Zx^ -\- ■ ■ • . Assume x = feiy + h-zy- + h-nr + ■■■. (1) Computing y'^, y^, ■ ■ ■ from the given equation by the method of § 978, and substituting the resulting series in (1), we have x3 + . . . (2) M + 2 6i x2 + 3 6i + h + 46, + h Equating coefficients, 1 = 6i, = 2 6i + 60, = 3 61 + 4 62 +63, • • • whence 61 = 1, 60 = — 2, 63 = 5, • • • . Therefore x = y - 27/^ + [>y^ + ■ ■ ■. (3) By the same method, from an equation of the form y = aQ-\- a^x -f a^x' + • • ■ , or // — a^J = (tiX -\- a„x' + • • ■ , we can derive another of the form x = b^ (y — r?„) + b„ (y — a^'^ + • • • . And from an equation of the form y = a^x' + a.^x^ + • ■ • we can derive two others of the form x — b^y^ + b^y + b.^y^- + • • •. Expansion of algebraic functions. An algebraic equation of the form f{x, y) = which lacks a constant term is satis- fied when cc = and y = 0. Hence, if we suppose f{x., y)= OPERATIONS WITH INFINITE SERIES 549 solved for y in terms of ar, one or more of the solutions must be expressions in x which vanish when x vanishes. It can be proved that these expressions may be expanded in series in increasing powers of x which have limits of convergence greater than 0. In ordinary cases these series may be obtained to any required term by the method illustrated in the follow- ing examples. Example 1. The equation 7/2 + ?/ — 2x = lacks a constant term. Find the expansion for the value of y which vanishes when x = 0. When X = 0, the equation 2/2 + y - 2 X = (1) becomes 7/2 -f ?/ = 0. Since one and but one of the roots of this equation is 0, one and but one of the solutions of (1) for y in terms of x vanishes when X vanishes. Suppose that when this solution is expanded in a series of increasing powers of x its first term is ax'^, so that 2/ = ax'^ + -.-. (2) Substituting (2) in (1), we have a2a;2M + [- cfx'^ + 2 X = 0. (3) Since by hypothesis (3) is an identity, the sums of the coefficients of its terms of like degree must be 0. Hence there must be at least two terms of lowest degree ; and since \j. is positive, these must be the terms ax'^ and — 2 x. Therefore /^ = 1 and a — 2 = 0, or o = 2. We therefore assume that 2/ = 2 X + 6x2 + cx^ + • • • . ^2') Substituting (2') in (1), we obtain (4 + 6)x2 + (4 & + c)x3 + • • . = 0. Hence 4 + 6 = 0, 46 + c = 0, ■••, and therefore 6 = — 4, c = 16, • • • . Therefore the required solution is y = 2 x — 4 x2 + 16 x^ + • • • . Example 2. Find the expansions of the values of y in terms of x which satisfy the equation 2/3 — x?/ + x2 = and vanish when x = 0. When X = 0, the equation 2/3 _ a;2/ + x2 = (1) becomes y"^ = 0, all three of whose roots are 0. Hence we may expect to find three expansions of the kind required. Let ax^ denote the leading term in one of these expansions, so that 2/ = ax/^ + ■ • • . (2) Substituting (2) in (1), we have aH^>^ + • • ■ - ax*^ + 1 + f- a;^ = 0. (3) 550 A COLLEGE ALGEBRA By the reasoning of Ex. 1, at least two of the exponents 3yu, yit + 1, and 2 must be equal and less than any other exponent of x in (3). Setting 3ja = ^ + 1, we find ij. = 1/2. This is an admissible value of fj., since when /j. = 1/2, both 3/i and fi + 1 are less than 2. Setting ;u + 1 = 2, we find /x = 1. This also is an admissible value of /n, since when m = 1, both /x + 1 and 2 are less than 3^. Setting 3^ = 2, we find /j. — 2/3. But this is not an admissible value of M, since when /x = 2/3, 3 /a and 2 are greater than /u + 1. Hence jit must have one of the values 1 or 1/2. When ,Lt = 1, (3) becomes o?x^ + • • • — ax- + • • • + a;^ = 0, from which it follows that — a + 1 = 0, or a = 1. When At = V-i (3) becomes a^x' + • . . - ax? H + x'-^ = 0, from which it follows that a^ — a = 0, or, since a ?i 0, that a = ± 1. We therefore assume that the required solutions are of the form y = X + 6x'- + cx^ H ^ y = x^ + 6x + cx^ + . • • , y = — x^ + 6x + ex? H . And substituting these expressions for y in (1) and determining the coefiicients as in Ex. 1, we obtain , o , o ■? , 1 X 3x? 1 X 3x? 7/ = X + X- + 3x3 + •.., 2/ = x^ ------+..., 2/ = - x= -- + —- + ••• . 2 o 2 o In this method it is assumed that if the leading term of one of the required expansions is ax'' , the expansion will be in 1 powers of x"^. In exceptional cases this is not true and the method fails. But the following method is general. Having found the leading term ax'* of an expansion as in the examples, set y = x** (a + v) in the given equation. It becomes an equa- tion in V and x. From this equation find the leading term of the expansion of v in powers of x, and so on.* Thus, in Ex. 2, setting ?/ = x^ (1 + ?j) in 2/3 - xy + x2 = (1) and simplifying, we have 1)3 + 3 u2 + 2 V + x^ = ; (2) whence r= — X-/2H , and therefore 2/=x^(l— xV2H ) = x'— x/2H , To find the next term, set u = x^(— 1/2 + v') in (2), and so on. *For a fuller discussion of the iiiothods of thi.s section and the use in con- nection with them of Newton's parallcloijvain see Chrystal's Alr/ebra, II, pp. 349-371 ; also Frost's Curve Tracing and Johnson's Curve Tracing OPERATIONS WITH INFINITE SERIES 551 Taylor's theorem. If f{x) = a^ + a^x + a^x^ + ••• when 984 ja;| namely, /' (x) = ai-\-2aoX + S a^x^ H , /" (x) = 2 (12 + 3 -2 a^x + •••, and so on. If in the preceding identity we replace x by a and h by 985 X — a, where |a| + |a5 — a] < X, we obtain the expansion oi f{x) in powers of x — a, namely, f(x)=f(a)+f'(a)(x-a)+...+r(a)^^-^ + .... From this last expansion and § 971 it follows that if 986 /(x) — a^-\- a^x + • • • when |a-| < A, and if [a] < X, then \\mf(^x)=f(a). EXERCISE LXXXIX 1. Show that (1 + X + x2 + . . ■)2 = 1 + 2 a; + 3 x2 + 4 z' + . . • . 2. Show that (1 + a; + x^ + • • •)3 = 1 + 3 x + 6 x^ + 10 x^ + . . .. 3. Show that (1 + x2 + x^ + • • ■)/{! + z + x- + ---) = l-x-\- x^- ■ -. 4. Assuming that (1 — x + 2 x^)^ = 1 + aiX + 02x2 + • • • , find ai, Os, as, Ui by squaring the given equation and applying § 973, 552 A COLLEGE ALGEBRA 5. By a similar metliod find the first four terras of tlie expansions of (l)(8-3x)i (2) (l + x-x2)l 6. Expand eacli of the following fractions in ascending powers of x to the fourth term by the method of the example in § 980. 2 + x-3x2+ 5a;3 x + 5 x^ - x^ ^' l + 2x + 3x2 ■ ^ ^ l-x + x-i-xs" 7. Expand each of the following fractions in ascending powers of x to the fourth term by the method of undetermined coefl&cients. 3x2 + x3 » + 5x* ^ ' l + a; + x2 ^ x3 + 2x* + 3x5 8. Expand each of the following fractions to the fifth term by the method of the first example in § 981 and indicate the limits of con- vergence of the expansions. 9X-22 5X + 6 ^ ^ (x2-4)(x-3) ^ ^ (2x + 3)(x + l)2 9. Expand each of the following fractions to the fourth term in descending powers of x. For what values of x will the first of these expansions converge ? 2 x2 + X - 15 x* + x3 + x2 + X + 1 10. Revert each of the following series to the fourth term. (1) ?/ = x + x2 + x3 + x* + ..-. (2) 2/ = x-|- + |--|- + --.. 11. From 2/ = 1 + X + xV2 + xV3 + • • • derive to the fourth terra a series for x in powers of y — 1. 12. From y = x^ + 3 x^ derive to the fourth term a series for x in powers of y^. 13. By the method of § 983 find the first three terms of the expansions of the values of y in terms of x which satisfy the following equations and vanish when x = 0. (1) x2 + 2/2 + 2, _3x = 0. (2) x8 + 2/3 - xy = 0. 14. By aid of the theorem of § 976 show that X 2x2 3x3 X . X2 . X8 , 1 - X 1 - X2 1 - X3 (1 - X)2 (1 - X2)2 (1 - X^f THE BINOMIAL SERIES 553 XXXiy. THE BINOMIAL, EXPONENTIAL, AND LOGARITHMIC SERIES (ffi The binomial series. 7)1 (m When VI is a positive integer, 98? 1 + T-- a- + 1-2 n,. + !!Ll!!L !)(»'- 2) l'J'3 •' + (1) is a finite or terminating series and its sum is (Iji,^)'". When m is not a positive integer, (1) is an infinite series, ' but one vrhich converges, that is, has a sum, when \x\< 1, § 968. We proceed to demonstrate that if m has any rational value whatsoever, this sum is (1 + or)"'. The series (1) is a function of both x and in, but since we are now concerned mainly with its relation to m we shall represent it by (»«). For convenience let w,. denote the coefficient of x^ in (1), so that v}y = m (rn — !)■ ■ • (in — r + 1)/'" !• Then if m and n denote any two numbers, we have (^ (m) = 1 + ni^x + vux" + m^x^ + ■•■■> ^ (/;) z= 1 + n^x + tux' + Ji^x^ H , <^ (m + n) = 1 + (m + n)iX + (??i + n).yx'^ H . We can prove that (ft (w) • ^(n) = (^(w + 71). For when \x\< 1, so that (2) and (3) converge, we have (-0 (3) (4) (5) But in §§ 773, 774 it is shown that 77?j 4- nj = (m + ?/)i, m„ + 7n^ni + w., = (^" + n)n, ■ ■ ■, VI,. + w^_i"i + • • • + Wi7j,._i + 71,. = (m 4- w),.. Hence <^ (//;)• <^ (w) = 1 4- (wi + ")i-^ + i^"- + ")2-''''' + 0^ + ^Os'^"" H = (m + n). (6) <^ (w) • <^ (n) = 1 + ?«-i X ■{- Tn^ X- + 7^3 + Wi + ??ii?ii 4- wij^^-i + n. 4- miWo + W3 554 A COLLEGE ALGEBRA By repeated applications of (G), we have (m)-cf)(n) ■(m + ??)-<^(;/)= <^(y«, + ?i 4- ;y), and so on, for any finite number of factors of the form (??«), (n),{p),(q),--: We are now prepared to prove the binomial theorem for all rational values of the exponent ?«, namely : 988 Theorem. //* m he any rational number whatsoever, the sum of the series . s . 1" m(m — 1) ., m(m — l)(m — 2) „ , (m)=l + jX+ \ ,, ' yi'+ ^ ^^;^ '-x'+--; when [x| < 1, is (1 + x)'". Notice first of all that when m = the series reduces to 1, and that when ni = 1 the series reduces to 1 + cr. Hence <^(0)= 1 and <^(1)= 1 + ir. (1) 1. Let VI be a positive integer. Then <^ (ot) = <^ (1 + 1 + • • • to m terms) = ^(1) • <^(1)- •• to m factors = [«^(l)]"' = (l+:r)"', by(l), (2) which proves the theorem for a positive integral exponent. 2. Let m be any positive rational fraction 7) Z^-. Then [<^ {p/q)y = 4> {p/q) ■ ^ (p/'i) • • • to ? factors = i> ivl'i + pIi H to ? terms) = <^(/0 = (l+x)^ by (2). Therefore <^ {plq) = (1 + a-)". (3) For it follows from the equation [<^ (/?/!?)]' = (^ + ^Y ^"^ § 986 that the values which ^{^pl'i) takes for all values of x such that la*! < 1 must be the corresponding values of one and the same qXh. root of (1 + a-) p. THE BINOMIAL SERIES 555 Moreover this root must be the principal qth. root, namely, p (1 + xy ; for this is the only one of the ^■th roots of (1 + xy which has the same value as ^ {p/q) when x = 0. 3. Let m be any negative rational number — s. Since <^(- s) • <^(s) = <^(- 5 + s)= <^(0) = 1, by (1) we have <^ (- «) = l/<^ (s) =^ 1/(1 + x)' by (3) = {l+x)'% (4) which proves the theorem for any rational exponent. It is not difficult to extend the theorem to irrational values of the exponent. Example. Expand (1 + 2 x + 3 a;^)' in ascending powers of x. We have (1 + 2 x + 3 x^)^ = [1 + (2 x + 3 x2)]3 = 1 + 1 • (2 X + 3x2) + ilzill (2 X + 3x2)2 \(_ 2W_5\ + '^ ^[l ^\ 2x + 3x2)3 + ... _ 2 X 5 x2 68 x3 ''"T'''~9 81 ' The expansion converges when 2|x| + 31x2| - 1-2 1 m(m — 1) „ . (a + x)™ = a™ + ma™-^x + — \ ^ a'"-''x''4- For r x m{m — l)x'^ ~\ (1) = a'" + ma"-ia; + '^^'^' ^ ^^ a^-^x^ + • • •, (2) where (1) and therefore (2) converge if \x/a\ < 1, or [a:[ < [a|. 990 556 A COLLEGE ALGEBRA The exponential series. We have already shown that the series 1 + x/l+xy2l + 0:73! + • •• + x"/7il + ••• (1) converges for all finite values of x, § 968. Let e denote its sum when a- = 1, so that e = l + l + l/2! + l/3! + --- = 2.71828---. We are to prove that the sum of (1) for any real value of x is e'. For letf{x) denote the sum of (1), so that f(x) = 1 + :r/l + x'/2 ! + xys ! + ••■+ x^/n ! + •.., /(y)= 1 + y/1 + f/2l + ,//3! + ... _^ yV«! +••.. Then by the rule for multiplying infinite series, § 978, /(^) -Ai/) = 1 + (^ + y) + (iJ + ^-y + f-J) \3l^ 211^ 121^ 3lJ^"' From this result it follows that /(^) -/(y) •/(«) =/(^ + y) ■f{^)=f{x + y-\-z), and so on. Hence, observing that/(0)= 1 and /(I) = e, we may prove successively, precisely as in § 988, that 1. When a; is a positive integer m, 2. When cc is a positive fraction ji/q, 3. When a; is a negative rational — s, /(-.)=l//(.)=l/e« = e-. THE LOGARITHMIC SERIES 557 Therefore when x is rational we have f{x) = e% that is, e- = 1 + X + a;V2! + ^yS! + • • • + a;"/^! + • • •• (2) Moreover (2) is also true for irrational values of the expo- nent X. For if h denote any given irrational number and x be made to approach h as limi-t through a sequence of rational values, for all these rational values of x we have fix) — e" and therefore ^'^^'^fix) = l^n e^. x=b^ ^ ^ x=b But ^i"V(^) =/(^)' § 986, and lif^ e^ = e\ § 728. There- fore /(6)~i e^ that is, 1 + ^ + by 2 !'+••• = e\ The second member of (2) is also a convergent series, that is, has a sum, when x is imaginary. Hence (2) may be used to define e-' for imagi- nary values of the exponent x. Thus, by definition, e' = 1 + i + iV2 ! + iV3 ! + ••• + i"/nl + ■■■. Series for a*. Let a denote any positive number and x any 991 real number. Since a = e^^Se", § 732, we have a^ = e^'^-^", § 730. Therefore, substituting x log, a for x in the series, § 990, (2), we have a* = 1 -f a: log.a + x2(log,a)y2! + ■■■ + x"{\og,ay/nl -f • • •. It can be proved, as in § 968, 1, that this series converges for all finite values of x. The logarithmic series. If in the series just obtained for a^ 992 we replace a by 1 -f ic and x by y, we have (l+xy = l + \og,(l + x)-i/ + l\og,(l + x)Yif/2l + .-..(l) But by the binomial theorem, § 988, when |a;| < 1, By carrying out the indicated multiplications and collecting terms we can transform (2) into a power series in y. The coefficient of y in this series will be ^ + "i:^'^ 1-2.3 +---'0^^ 2 + 3 • 558 A COLLEGE ALGEBRA Equate this to the coefficient of y in (1). We obtain, if ' log,(l + ic) = a;-a;72 + a;73-a;V4 + -.-. (3) This series is called the lorjarlthmlc series. In § 968 we proved that it converges when |a;| < 1. In the proof just given we have assumed that the series (2) remains equal to (1 + x)^ after it has been transformed into a power series in y. But this follows from §976 when la;|, §§ 726-729, 73L 993 Computation of natural logarithms. The logarithms of num- bers to tlie base e are caWed their 7iafurallof/arith7)is. A table of natural logarithms may be computed as follows : We have log, {l + x) = x- xy2 + x^S - x'/4 + • • ., (1) and therefore log, (l-x) = -x- xy2 - x^S - x*/4: . (2) Subtract (2) from (1). Since 1 + iC log,(l +a;)-log,(l - x)^\^g^YZr^' 1 + 1 + a; / x^ x^ \ weobtain loge ^^T^ = 2 (^x + 3 + ^ + ■■). (3) THE LOGARITHMIC SERIES 559 I A_ X n + 1 1 In (3) set = and therefore x = • We obtain n+l_^f 1 1 1 \ ^^' n *'\27i + l ^3(2 w + 1/^ 5(271 + 1)^"^"7' or log^(/i + 1) = ^°S^^^ + ^ {jlhl + 3 (2 n + 1)3 + 5 (2 7.^ 1)^ + • • •)• ^^^ Setting ?i = 1 in (4), log,2 = 2(1/3 + 1/3* + 1/5 . 35 + • • •) - .6931 • ■ •. Setting ?t = 2 in (4), log, 3 = log, 2 + 2 (1/5 + 1/3 • 5» + • • •) = 1-0986 • • •, and so on to any integral value of 71. Modulus. By § 755, log„n = logg7i/log,a. Hence the loga- 994 rithnis of numbers to any base a may be obtained by multi- plying their natural logarithms by l/log^a. We call l/\og^a or its equivalent, § 756, log^e, the modulus of the system of logarithms to the base a. In particular, the modulus of the system of common logarithms is logigg = .43429 • • •. EXERCISE XC 1. Compute loge4 and logeS each to the fourth decimal figure. 2. Show that e-i = 2/3 ! + 4/5 ! + 6/7 ! + • • •. 3. Show by multiplication that / , X x2 x3 \ / , X x2 x3 \ (' + I + 2! + 37-^---)('-i + 2--3!+---)='- 4. Show that (e'^ + e- ^^)/1 = 1 - xV2 ! + xV-4 '• - xV6 ! + •••. 5. Show that (e'^ - e- '^)/2 i = x - xV3 ! + xV5 ! - xV7 ! + • • • 6. Show that the (r + l)th term in the expansion of (1 — x)-" by the . 1 ,, . n(n-fl)---(n + r-l) ^ bmomial theorem is — ^ ^ z^. r\ 7. Find the term in the expansion of (8 + x)^ which involves x*. 8. Find the term in the expansion of (1 — x^)~^ which involves x^. 560 A COLLEGE ALGEBRA 9. For what values of x will the expansion of (9-4a;2)i and of (12 + z + X-)' in ascending powers of x converge? 10. Expand (1 — x + 2 x2)Mn powers of x to the term involving x*. 11. Find the first three terms of the expansion of (8 + 3 x)^ (9 — 2 x)~ ' in powers of x. For what values of x will the expansion converge ? 12. Find the limiting values of the following expressions. (1) lim ^'-^ - ^'\ (2) lim (1 + x^)^ - (1 - x'^)^ ^"° 3x ''^"°(H-3x)^-(l + 4x)i" 13. Prove that '""("!+ - V= e^. 14. Prove that li«i (e^ + x)' = e^. 15. Expand logc(l + x + x^) in powers of x to the term involving x*. For what values of x will this expansion converge ? /in — n\^ l/?/i — n\8 16. Show that loge — = 71 n 2 17. Show that loge ^^ = 1 + — + — + .. .. n2 - 1 n2 2 ?i* 3 n6 18. Show that n + 1 2(n + l)2 3(n + l)3 n 2n^ 3n^ XXXV. RECURRING SERIES 995 Recurring series, A series a^ + a^x + a^x^ -\ in which every r + 1 consecutive coefficients are connected by an identity of the form «n + i'lOn - 1 + i^2«„-2 "I h i^/'n - r = 0, where ^1, p2, • • •, j^r ^^6 constant for all values of n, is called a recurring series of the xth order, and the identity is called its scale of relation. Thus, in 1 + 3x + 5x2 + 7x3 + . • • + (2n + l)x» + • • ■ (|) every three consecutive coefficients are connected by the formula a„-2a„_i + a„_2 = 0; (2) for 5-2-3 + 1 = 0, 7-2.5 + 3 = 0, 2n + 1- 2(2n -1) + 2n - 3 = 0. RECURRING SERIES 561 Hence (1) is a recurring series of the second order whose scale of relation is (2). A geometric series is a recurring series of the first order. Every power series which is the expansion of a proper frac- tion whose denominator is of the rth degree is a recurring series of the Ah order. Thus, if i^ — ^ = ao+ aix + a^z^ + h a„x» + • • ■ , (1) we have 2 + x = ao + ai I x + ao + 2ao| +2ai X2 + ---+ an + 2a„_i + 3n„_2 + 3rto Therefore «o = 2, ai + 2 ao = 1, a™ + 2 ai + 3 ao = 0, and, in general, o„ + 2a„_i + 3a„_2 = 0. (2) Hence (1) is a recurring series of the second order whose scale of rela- tion is (2). If a few terms at the beginning of a power series be given, it is always possible to find a scale of relation which these terms will satisfy. By aid of this scale the series may be continued, as a recurring series, to any required term. Example. Given 1 -i- 4 x -|- 7 x2 -i- lOx^ -|- 13 x* -|- • • • . Find the scale of relation which these terms satisfy, and then find two additional terms. As 1 + 4 X -1- 7 x2 -f- • • • is not a geometric series, we begin by testing the scale of the second order, a„ + pan-i + qan-2 = 0- If all the given terms are to satisfy this scale, we must have 1 +ip + q=:0, 10 + 'lp+4q = 0, 13 + 10^ + 7^=0. Solving the first two of these equations, p = — 2, g = 1. And these values satisfy the third equation, since 13 — 10 • 2 + 7 = 0. Hence the required scale is a„ — 2 a„_i + a„_2 = 0. We may therefore find the required coefficients as, a^ as follows : as - 2 • 13 + 10 = 0, .-. as = 16 ; as - 2 • 16 + 13 = 0, .-. ae = 19. If the given terms had been 1 + 4 x + 7 x^ + 10 x^ + 14 x*, they would not have satisfied a scale of the form a„ + pa»-i + qa„-2 = 0. But we might then have assigned a sixth term ax^ arbitrarily and have found a scale of the third order, a„ +pa„_i + ga„_2 + ra,i-3 = 0, which the six terms would satisfy, namely, by solving for p, q, r the equations 10 + 7p + 4g + r = 0, 14 +102) + 7g + 4r = 0, a +14j) +105 +7r = 0. 562 A COLLEGE ALGEBRA Fiom the example it will be seen that ordinarily when 2 r terms are given the series may be continued in one way as a recurring series of the rth order, and that when 2 r + 1 terms are given it may be continued in infinitely many ways as a recurring series of the {r + l)th order. 998 The generating function of a recurring power series. Every recurring power series of the rth order is the expansion of a proper fraction whose denominator is of the rth degree (com- pare § 996). This fraction is called the generating function of the series. It is the sum of the series when the series is convergent. Thus, let ao + aix + a-infl + h a„x" + • ■ • (1) be a recurring series of the second order ^'hose scale of relation is «« + i'ctn-i + ga„-2 = 0. (2) Set (S„=ao+ aix+ aoX^H + a„_ix"-i .•.px) ■■■: in which the number of the factors is supposed to be infinite. Such a product is said to be convergeyit or divergent accord- ing as (1 + fli) (1 + "2) •••(! + a„) approaches or does not approach a finite limit as n is indefinitely increased. 1001 Theorem. If all the nuvibers a^ are positive, the infinite product n (1 + a^) is convergent or divergent according as the infinite series Sa,. is convergent or divergent. First, suppose that 2a^ is convergent and has the sum S. Then since 1 + a; < e^ when x is positive, § 990, we have (1 + a-^) (1 + r/o) • • • (1 + a,,) < e"! ■ e"^ ■ ■ ■ e"-, that is, < e"i + "2 + - •• + «'.< e"?. Hence, as n increases, (1 + a^) (1 + a„) • ■ -(1 + «„) increases but remains less than the finite number e^. It therefore approaches a limit, § 192, that is, 11(1 + 0^) is convergent. Second, suppose that 1a^ is divergent. In this case lim (a^ + ^2 + • • • + "„) = ^• But (1 + «i) (1 + «2) ••■(! + fl„) > 1 + (fli + oo + • • • + a„). Therefore lim (1 + Oj) (1 + Un) ■ ■ ■ (1 + o,,) = 00, that is, n (1 + a,.) is divergent. Thus, n (1 + 1/n'') is convergent when j) > 1, divergent when p < 1. Example 1. If 2ar is a divergent positive series whose terms are all less than 1, show that n (1 — a^) = 0. Since a^ < 1 and 1 — af. < 1, we have 1 — a,. < 1/(1 + a,) numerically. Hence (1 - a{) (1 - aa) • • • (1 - a,) < 1/(1 + a^) (1 + a.) ■ ■ • (1 + a„). But lim (1 + ai) (1 + a.) • • ■ (1 + a„) = 00. Therefore n (1 - a,) = lim (1 - ai) (1 - a^) ■ ■ ■ (1 - a„) = 0. INFINITE PRODUCTS 565 Example 2. Show that when x = 1 the binomial series, § 987, con- verges if jn + 1 > 0, but diverges if m + 1 < 0. When X = 1, the binomial series becomes l + m + '^^i^+...+ - -'- -■-' +•••. (1) In this series we have + • , m{m-l)---(m-n + l) , ' 1.2. .-n ' m - - n + 1 _ m + 1 (2) Hence, if m + 1 < 0, we have | w„ + 1/«„ | > 1, and (1) is divergent, § 951. But if m + 1 > 0, after a certain term the series will be of the kind described in § 958 and will therefore converge. For if r denote the first integer greater than ?h + 1, it follows from (2) that when n>r, u„ + i/u„ is negative and numerically less than 1. Hence the terms of (1) after Ur are alternately positive and negative and decrease numerically. Therefore (1) converges if lim w„ = 0. „ in m — 1 m — n + 1 But u„ + i = -.^ 12 n =<-'K'-^)('-'^>-(-'^)- and it follows from Ex. 1 that as n increases the product on the right approaches as limit. EXERCISE XCII , ^^ ^ 3 5 9 17 .5 10 17 26 1. Show that - — -.-... and - • — — — . • • are convergent. 2 4 8 16 4 9 16 25 2. For what positive values of x are the following infinite products convergent ? <^> "(' "0-s)K'-f,)('^i:)o-i-;)-- <'> "0-i-:)=o-i)(>+f)('-s>- 3. Show that u^„'^(« + ^)(« + 2)---(« + n)^ ^ ^^ ^ according as ^^ b{b + l)(b + 2)...{b+n) o > or a «3) • • • ^^^ positive integers. The numbers a^, a.^, •■• are called the first, second, • ■■partial quotients of the continued fraction. According as the number of these quotients is finite or infi- nite, the fraction is called terminating or nonterminating. 1003 Terminating fractions. Evidently every terminating simple continued fraction has a positive rational value, for it may be reduced to a simple fraction.. 1 1 ^ 4 .30 1 1 1 13 Thus, 2 + -^^ = 2 + - = ~; 2 + 3 + 4 = 35- Conversely, every positive rational number may be converted into a terminating simple continued fraction. This will be evident from the following example. Example. Convert 67/29 into a continued fraction. Applying the method for finding the greatest common divisor of two integers to 67 and 29, we have 29)67(2 = ai 58 9) 29 (3 = aa 27 2) 9 (4 = as 8 1)2(2 = 04 2 ^ 2+ '^ =2+ ' . 29 29 29/9 (1) !=-h-^- (2) ^-i (3) CONTIXUED FRACTIONS 567 ' Substituting (2) in (1), and (3) in the result, we have 67 „ 1 „ 1 1 1 ■ — = 2 + - =2 + - - -,as required. 29 3 + 1 3 + 4 + 2 ^ 4+1 2 Since 29/67 = 1 h- 67/29, we also have 29_ 1 1 1 1 67 "2 + 3 + 4 + 2* Convergents. The fractio^^ .-> Oi+— ? «i + — — ' •••, 1004 1 a^ «2 + <^z are called the first, second, third, • • • convergents of the fraction 11 oi + — — «2 + '^'3 H Q When Oj is 0, the first convergent is written — • Theorem 1. Each odd convergent is less and each even con- 1005 vergent is greater than every subsequent convergent. This follows from the fact that a fraction decreases when its denominator increases. Thus, 1. aiai -\ , since — > — a^ a2-\ a2 02 + • • • 1 1 .11 3. ai ^ (H-\ » 1 „ , 1 «3 as + --- 02-1 02 H Os Os + • • • by 2; and so on. Reduction of convergents. On reducing the first, second, third, 1006 • • • convergents of aj H — to the form of simple . . , . «2 + *s + • • • tractions, we obtain «i (i\(^2 + 1 ayi„^nj^r_a^j\-_a^ 1 ftj ''a^s + 1 ' 568 A COLLEGE ALGEBRA Let 2>i, i?2) V^i ■ ■ ■ denote the numerators, and g-j, g-j, g'g, • • • the denominators of the convergents as thus reduced, so that IH = «i, IH = «i«2 + Ij i^s = aia2«3 + «! + «3, • • • (2) ^1 = 1, q2 = a^, 3-3 = a2«3 + 1, •••• (3) Since «!, a.^, a^, ■ ■ • are positive integers, it follows from (2), (3) that as n increases p^ and q^ continually increase, and that they approach oo if the given fraction does not terminate. By examining (2) and (3) it will be found that 2h = «3i>2 + Pi and q^ = a^q^ + q^. (4) This is an illustration of the following theorem. 1007 Theorem 2. The numerator and denominator of any conver- gent are connected with those of the two preceding convergents by the formulas Pn = a^Pn-l + Pn-2, qn = a„q„ _ i + q„_2- For suppose that these formulas have been proved to hold good for the A;th convergent, so that Vk = (^kPk - 1 + Pk-2, qk = %?i- -i + qk-i, (1) Pk ^ (^kPk-l +Pk-2 ^2) qk (ikqk-i + qk-2 The (k + l)th convergent may be derived from the kth. by merely replacing a^ by a^. + l/a,^_^.^, § 1004. Therefore, since Pk-v Pk-if qk-ij qk-2 do not involve a^., it follows from (2) that Pk + i _ (ak + 'i~/ak + -i)Pk-i +Pk-2 qk + 1 («x- + i/«i- + 1) qk -i + qk-2 ^ ak+i{fi'kPk-x + Pk-2)+Pk-x ^ ak + ^Pk + Pk-i by(l)^ that is, pk+i = ^h-^iVk+Pk-v qk+i =-- "k+iqk + qk-v We have thus proved that if the formulas^,, = a„p„_i +jo„_j> q^ = a„q„_i + q„_2 hold good for any particular convergent, CONTINUED FRACTIONS 569 tliey hold good for the next convergent also. But we have already shown that they hold good for the third convergent. Hence they hold good for the fourth, hence for the fifth, and so on to every convergent after the thh-d (compare § 791). Example. Compute the convereeiits of 3 + - ^ ^ ^ 2+3+4+5 Since 3 = 3/1 and 3 + 1/2 = 7/2, we have pi = 3, Hence P3 = 3 • 7 + 3 = 24, p4 = 4 • 24 + 7 = 103, ps = 5 • P2 = 7, gi = l, 103 + 24 = 539, and 53 = 3 • 2 + 1 = 7, 54 = 4 • 7 + 2 =: 30, 55 = 5 • • 30+ 7 = 157. ^, , , 3 7 24 103 539 Therefore the convergeuts are -, -, — , , 1 2 7 30 157 Theorem 3. The numerators and denominators of every two 1008 consecutive converyents are connected hy the formula Puqa-l-Pa-iqn=(-l)°- The formula holds good when n — 2. For, by § 1006, we have /»2?i — P\<1'2. = («i«2 + 1) — «'i«2 = 1 = (— 1)^- Moreover we can prove that if the formula holds good when n = k, it also holds good when 71 — k -{- 1. For 2Jk +i1k- JMk + 1 = ((h + i2h + ]h - 1) 'Ik - Pk («i- + iqk + «-i?«-2-i\-29'„-i) = (- !)""'«„• The theorem of § 1005 may be derived from these formulas. 1011 Theorem 4. The nth convergent of a no7iterminating simple continued fraction approaches a definite limit as n is indefinitely increased. For, by § 1005, the odd convergents pi/qi, Pz/q^, • • • form a never-ending increasing sequence, every term of which is less than the finite number p^/q^. Hence, § 192, a variable which runs through this sequence will increase toward some number A as limit. Similarly a variable which runs through the sequence of even convergents p^/q^, Pi/qa • • • will decrease toward some number /a as limit. But/A = X,since/x-A= lim f^^-^^i^!!^"] =, lim (-1)"" ^^^ "'^ " L'^ '» ^2 m - 1 J "'= - q-Z ,„q-2 ,n - 1 Therefore a variable which runs through the complete sequence of convergents Pi/qi,P2/qi, Ih/qz, ••• will approach X as limit. 1012 By the value of a nonterminating simple continued fraction is meant the number li^ii (pjq^). It follows from § 1003 that this number is always irrational. The value of a terminating fraction is that of its last con- vergent, § 1004. COXTIXUED FRACTIOXS 571 In the statements of the following theorems, §§ 1013, 1014, the understanding is that when the fraction terminates, " con- vergent'"' means any convergent except the last one. Corollary 1. The value of a simple continued fraction lies 1013 hetwmn the values of every two consecutive convergents. Corollary 2. The difference betiveen the value of a continued 1014 fraction and that of its nth co7iverr/ent is numerically less than l/^n^ln + i ^'^'^ greater than a„ + ,/qnq„^2- For let X denote the value of the fraction, and to fix the ideas sujipose that n is odd. We then hav( Hence and 3 X- X In In + 2 'In + l Pn^J'n + l P", • ^ 1 , 1005, 1013 § 1010, 1 § 1010, 2 ?„ 'lu + r 9n 'Jn'In + l '/« 'In + 2 'In 1n'In + 2 Evidently l/q^qn + i < ^/'I% ^^^^ ^J leaking use of the rela- tion «7„_^2 = ^'« + 2'Z«+i + 'Inj § 1007, it may readily be shown that •^» + 2/7n'7n + 2 = !/'/«('?« + I7n + i)- Hcncc the difference between X and 2Jn/'Li is less than 1/y^ and greater than l/'/„(q„ + q„ + i)- Corollary 3. JSach convergent is a closer approximation to the 1015 value of the fraction than is any preceding convergent. For, by § 1014, if X denote the value of the fraction, the dif- ference between X dcnfi p ,J q ,^ is numerically less than '^/q„qn + \, while the difference between A. and p„_ ,/2'„_i is numerically greater than o„ + ,/./„ _ ^q„ + 1 ; and l/^?,//,, + 1 < «« + i/q„ _ ,7,, + 1, since q„-i< <'n + iqn^ § 1006. Corollary 4. The convergent Pn/cin ^^ "' <^loser approximation 1016 to the value of the fraction than Is any other ratloiial fraction whose denominator does not exceed q„. For if a/b is a closer approximation to the value of the fraction than Pn/qn i^? ^^ must also, § 1015, be a closer 572 A COLLEGE ALGEBRA approximation than 7>„_i/j„_i is, and. must therefore, §1013, lie between ])„lqn ^^^ Pn - il'L, - 1- To fix the ideas, suppose that n is even. We then have qn-i ^ qn Hence qn that is, 1 ^ "V«-i-^7>.-. qnq. - 1 "q,. - 1 or i>> q„{'iqn-i-k\-\)- But a'7„_i — ^>Pn-i is positive, since a/h > ^>„ _ i/q„_ ^ Hence b > q„; that is, if a/b is a closer approximation to the value of the continued fraction than p,Jq„ is, its denominator b must be fj renter than //„. 1017 Recurring fractions. A nonterminating continued fraction in which a single partial quotient or a group of consecutive partial quotients continually recurs is called a recurrlmj frac- tion. And such a fraction is called pure or mixed according as it begins or does not begin with these recurring partial quotients. The value of a recurring fraction may be found as follows. Example 1. Find the value of2 + - =2 + - - - 3 + --. 3 + 3 + 3 + --. This is a pure recurring fraction with ihe period 2 -\ Hence, if x denote its value, we have '"' x:.2 + l 1, .•.x=I^+^, .•.3x^-Gx-2 = 0, .■.x = ^±^. 3 + X 3x + l 3 Example 2. Find the value of 4 + - 5 + 2 + 3 + . • . This is a mixed recurring fraction with the period 2 + 1/3. Hence, if X denote the value of the recurring part 2 + ^ , and y tlie value of the entire fraction, we have, by Ex. 1, 1 1 21 X + 4 21 (3 + vT5)/3 + 4 75 + 21 VTs y = 4 5 + x 5x + l 5(3 + Vl6)/3 + l 18 + 5 VTs CONTINUED FRACTIONS 573 In general, if x denote the value of a pure recurring fra.ction with the period. «i + • • • — ' we have, § 1007, + ('k x = a, + ... 1 l^P^+JP^, + a^ + x q„x+qf,_^ and therefore qf.x^ + (5-^, _ 1 — 2h) *' ~ 2h _ 1 = 0. Since the absolute term — p<. _i of this quadratic is negative, it has one and but one positive root, and this root is the value of the fraction. Again, if 1/ denote the value of the mixed recurring fraction i J_ J_ + a, + a, +1 H + a,._^i. H we find the value x of the recurring part as above, and then have, § 1007, 1 1 ]->,.x + J)r-1 « = «, + ••• — -= — - — — • + a^ + x q,:x + qr-i On converting irrational numbers into continued fractions. 1018 Every positive irrational number is the value of a definite nonterminating simple continued fraction which may be obtained to any required partial quotient by the following process. If h denote the number in question, first find a^, the greatest integer less than h. Then b = Oi + l/^i, where h^ is some irra- tional number greater than 1. Next find (u, the greatest integer less than b^. Then b^ = a^ + l/b^, where b^ is some irrational number greater than 1. Continuing thus, we have & = oi + — = rtj H — = ... = «j H — ^^1 "o + b^ 02 + «3 H It can be proved that when 5 is a quadratic surd the con- tinued fraction thus obtained is a recurring fraction. 574 A COLLEGE ALGEBRA Example. Convert Vu into a continued fraction. The greatest integer less tlian VTl is 3, and, § 60.3, Vll = 3 + (Vrr-3) = 3 + — J =3 + -— J (1) _Vll + 3 (Vll + 3)/2 The greatest integer less than (Vu + 3)/2 is 3, and ^ ^ 2 ( Vn + 3) Vu + 3 The greatest integer less than vTl + 3 is G, and Vri + 3 = G + (VIl -3) = 6 + ^ =6 + ? (3) Vll + 3 (VTi + 3)/2 The last fraction in (3) is the same as the last in (1). Hence the steps from (3) on will be (2), (3) repeated indefinitely ; that is, the partial quo- tients 3 and G will recur. Hence, sub.stituting (2) in (1), and (3) in the result, and so on, we have Vll = 3 + - 3 + G + ... 1019 A given irrational number can be expressed in onli/ one tray as a simple continued fraction. This follows from the fact that two nonterminating simple continued fractions cannot be equal unless their corresponding partial quotients are equal. For if a + or = c + 7, where a and c denote positive integers and a and 7 denote positive numbers which are less than 1, then a = c, since otherwise it would follow from a — c = y — a that an integer, not 0, is numerically less than 1. Hence, if aiH — = Ci -I — , where ai, a2, as, •• •, "2 + "3 H C2 + C3 H ... ^ 11 Ci, C2, Cs, • • • denote positive integers, we have Oi = Ci, .-. — — a2 + «3 + • • • 11 11 = — — , .-. tto H — = ("2 H — , .-. 02 = C2, and so on. C2 + Cs H ' 03 H Cs + ■ ■ • 1020 If we compute the continued fraction to which a given irrational number b is equal as far as the nth partial quotient, we can find its nth convergent p„/q„, and this rational fraction p„/q„ will express b approximately with an error less than 1/ql, § 1014. Moreover p„/qr. will be a closer approximation to b than is any other rational fraction whose denominator does not exceed q„, § 1016. CONTINUED FRACTIONS 575 Thus, the first four convergents of vll = 34-- - _ are 3 + G + 3 + ... 3 10 68 100 ^ 100 /— .^^ , ^, 1 - , — , — — , and expresses Vll with an error less than 1 3 19 GO GO 60'^ Solution of indeterminate equations of the first degree. Given 1021 any equation of the form ax + hij = c, where a, b, c denote integers of which a and h have no common factor, § 672, If we convert a/b into a continued fraction, the last convergent of this fraction will be a/b itself, and if the convergent next to the last be p/q, we have aq — bjy = ± 1, § 1008. This fact makes it possible always to find a jjair of integral values of x and // which satisfy ax + bi/ = c. The method is illustrated in the following example. Example. Find an integral solution of 205 x + 93 y = 7. Asin §1003, Ex., wefind — =2 + i 111. 93 4 + 1 + 8 + 2 2 11 97 205 The convergents, found as in § 1007, Ex., are ", -, — , — , ^• ^ ' ^ ' ' 1 4 5 44 93 Hence 205 • 44 - 93 ■ 97 = - 1, or, multiplying by - 7, 205 (- 44 • 7) + 93 (97 • 7) = 7. Therefore x = - 308, y = 679 is a solution of 205 x + 93 y = 7. The general solution is x = - 308 + 93 1, y = 679 - 205 1, § 674. Similarly we may show that 205 x — 03 y = 7 has the solution x = — 308, y = - 679. EXERCISE XCIII Compute the convergents of the following : 1.3 + 1 1 '. 2. 1 ^- 1 ± 1. 4 + 1 + 5 1 + 1+3 + 10 + 12 Convert each of the following into a continued fraction. For each of the last three compute the fourth convergent and estimate the error made in taking this convergent as the value of the fraction. 3. '± 4. 1^. 5. 11?. 6. 3.54. 12 56 513 7. .1457. 8. ?^. . , 9. ^. 10. '^. 177 > ^ 972 31827 576 A COLLEGE ALGEBRA Convert each of the following into a recurring continued fraction and compute the fifth convergents and the corresponding errors for the first four of them. 11. Vl7. 12. V2G. 13. V(3. 14. VsS. 15. Vl05. 16. I/V23. 17. VlO. 18. VtT. 19. 3 Vs. 20. (Vl0-2)/2. 21. (V2 + l)/( V2 - 1), Find the values of the following recurring fractions. 22. i 1 i . 23. i 1 i . 24. 34-^ i i • 1 + 2+3 + --- 2 + 1 + 3 + --- 4 + 5 + 2+ •• 25. ^ + 1 ^ i . 26. 1 1 i ^ i . 3 + 4 + 5 + ... 2 + 7 + 1 + 2 + 1 + --. 27. Show that Va^ + 1 29. Show that ill _ - {abc + a-b + c) + V(abc + a + 6 + c)^ + 4 a + b + c + ---~ 2 (a6 + 1) 30. Convert the positive root of x- + x - 1 = into a continaed fraction. 31. Show that ^-^ = ^ + -i L + .-. + lnil:'. Qn 3i qiq2 (Mz qn-iqn 11 111 32. Show that - — = 1 . a-i + as -] 91^/2 5293 qsqi 33. What rational fraction having a denominator less than 1000 will most nearly express the ratio of the diagonal of a square to its side ? Estimate the error made in taking this fraction as the value of the ratio. 34. Find the simplest fraction which will express tt = 3.14150265- •• ■with an error which is less than .000001. 35. Compute the sixth convergent of e = 2.71828 • • ■ and estimate the error made in taking it as the value of e. 36. Find an integral solution of 127 x -2Uy = 6. 37. Find an integral solution of 235 x + 412 y - 10. 38. Find the general integral soluticJh of 517 x - 323 j/ = 81. PROPERTIES OE CONTINUOUS EUNCTIONS 577 XXXVIII. PROPERTIES OF CONTINUOUS FUNCTIONS FUNCTIONS OF A SINGLE VARIABLE Functions. If the variable y depends on the variable x in 1022 such a manner that to each value of x there corresponds a definite value or set of values of y, we call y a. function of x. In what follows when we say that ?/ is a function of x and write y =f(x), we shall mean that it is a one-valued function; in other words, that to each value of x there corresponds but one value of //. And f{a) will denote the value of y which corresponds to the value a of x. Evidently ?/ is a function of x if it be equal to an algebraic expression in X, as when y = x^ + 1. But a relation which defines y as a function of X may be one which cannot be expressed by an equation. Thus, y is a function of x if y is 1 for all rational values of x and — 1 for all other values of x. But this relation between y and x cannot be expressed by an equation. We call y a function of x even when there are exceptional values of x foi which the given relation between y and x fails to determine y, § 1024. •Sometimes y is defined as a function of x only for a certain class of values of x or only for values of x which lie between certain limits. Thus, the equation ?/ = x + 2x2 + 3x'' + • ■ •, by itself considered, deter- mines y for those values only of x which are numerically less than 1. Continuity of a function. Let f{x) denote a given function 1023 of X. We say that f{x) is continuous at a, that is, when x = a, if /(«) has a definite finite value, and if ^^"^ f{x)=f{a). In the contrary case we say that /(a?) is discontinuoiis at a. Here and in what follows the notation ^^"^ f(x) = f(a) means that f(x) will approach /(a) as limit whenever x approaches a as limit, that is, no matter what the sequence of values may be through which x runs in approaching a as limit. In the case of a function y defined by a given equation 1024 y=f(x) it may happen that the expression /(x) assumes an 578 A COLLEGE ALGEBRA indeterminate form when x = a, §§ 513-518. The equation y =f(^x) by itself considered then fails to define ?/ when x = a. But if ^^^^ f(x) has a definite finite value, we assign this as the value of /(«), § 519, which makes f(x) continuous at a. If li^ /(a;) = 00, we assign to /(«) the value oo, § 515; f(x) is then discontinuous at a. Finally, if ^^^^ /(^) is indetermi- nate, we have no reason for assigning any single value to f{o-). Evidently we can assign none for which ^i^^ f(^) =f{^)- In this case also f(x) is discontimwus at a. 1. Thus, every rational function/(a;) is continuous except perhaps when the denominator of some fraction occurring in/(x) vanishes. For example, consider the function /(x) = (x — l)/(x^ — 1). This function is continuous except when x"^ — \ = 0, that is, when a; = 1 or — 1. For if a is not 1 or — 1, /(a) = (a — l)/(a'- — 1) has a definite finite value and ''i'^^/(x) ==/(a), § 509. When X = 1, the expression (x — Vj/{x^ — 1) assumes the indeterminate form 0/0. But l™/(a;) = ^"^^ [(^ - l)/(x-^ - 1)] = ^'i" [l/(a; + 1)] = 1/2, and by assigning to /(I) the value 1/2 we make /(x) continuous when x = \. When X = — 1, /(x) is discontinuous ; for ^l"^ /(x) = oo. 2. Consider the following function : /(x)=4^=ii^==i^^. 2^+1 1 + 1/2^ 1/2 ^- + 1 Here/(0) has the indeterminate form oo/oo, § 517. But if we write /(x) in the second form and tlien make x approach through positive values, we have lim 2-^ = oo, and therefore lim/(x) = 1. If we write f(x) in the third form and then make x approach through _ 1 negative values, we have lim 2 * = co, and therefore lim/(x) = .3. Finally, if we make x approach through values which are alternately positive and negative, /(x) will not approach any limit. Hence /(x) is discontinuous at 0. No value can be assigned to/(0) for which I'm /(x)=/(0). -' 1025 From the definition of continuity in § 1023 it immediately follows, § 189, that V 1026 PROPERTIES OF CONTINUOUS FUNCTIONS 579 The sufficient and necessary condition that f (x) he coritimwiis at a is that f (a) have a definite finite value, and that for every positive number S which can be assigned it shall be possible to find a corresponding positive miniber e such that |f (x) — f (a)| < 8 whenever [x — a[ < e. Thus in the neighborhood of a value of x, as a, at which f{x) is continuous, very small changes in the value of x are accompanied by very small changes in the value of f{x), and the change in the value of x can be taken small enough to make the corresponding change in the value oi f{x) as small as we please. This is not true of f{x) in an interval con- taining a value of x at which f(^x) is discontinuous. See the examples in § 1024. Theorem 1. If both of the functions f (x) and ^(x) are con- tinuous at a, the same is true of f(x)±<^(x) and f(x)-^(x)j also of f (x)/<^ (x) unless ^ (a) = 0. Ifi{^) is continuous at a, the same is true of vf (x). This follows immediately from the definition of continuity at a, § 1023, and the theorems of §§ 203-205, according to which lim \f{x) + ^(.r)] = \\m.f(x) + lim ^{x), and so on. Real functions. In what follows x will denote a real variable, 1027 that is, one which takes real values only, and f(x) will denote a real function of x, that is, one which has real values when x is real. Number intervals. The practice of picturing real numbers by 1028 points on a straight line, §§ 134, 209, suggests the following convenient nomenclature. Let us call the assemblage of all real numbers between a and b, a and b themselves included, the number interval a, b, and represent it by the symbol (a, b). Moreover, it being understood that a „ — a,,) = lim (b — «)/2" = 0. Thenf(xo)=0. PROPERTIES OF CONTINUOUS FUNCTIONS 581 For since/(£c) is continuous at x„, lim/(a„) = lim/(6„) =/(xo). But since /(«„) is always positive, its limit /(xq) cannot be negative, and since /(^„) is always negative, its limit /(x^) cannot be positive. Therefore /(x^) is 0. Thus, if f{x) = 1 - xV2 ! + a;V4 ! - x^/6 ! + •••, it may readily be shown that /(I) is positive and /(2) negative. Hence this f(x) will vanish for some value of x between 1 and 2. Simpler illustrations of the theorem will be found in §§ 833, 836. Maximum and minimum values. Superior and inferior limits. 1031 Consider the following infinite assemblages of numbers : 2,H, li,li, ...(A), 2, 2i,2|,2|, ...(B). In (A) there is a greatest number, namely 2, but no least number ; and in (B) there is a least number, namely 2, but no greatest number. On the other hand, while there is no least number in (A), among the numbers which are less than those in (A) there is a greatest, namely 1. Similarly among the numbers which are greater than those in (B) there is a least, namely 3. The like is true of all infinite assemblages of finite numbers, that is, of numbers which lie between two given finite numbers a and b. In other words, Theorem 3. Let aj, a^, •••, a„, ••. (A) denote any infiniie 1032 assemblage of finite numbers. Then 1. Either among the different numbers in (A) there is a greatest or among the numbers greater than those in (A) there is a least. 2. Either among the different numbers in (A) there is a least or amo7ig the mimbers less than those i^i (A) there is a greatest. To prove 1 assign all numbers greater than those in (A) to a class 7?2) and all other real numbers, including those in (A), to a class Ri. Since each number in 7?i will then be less than every number in i?.,- there will be, § 159, either a greatest num- ber in i?i or a least in R2, — which means either a greatest 582 A COLLEGE ALGEBRA among the different numbers in (A) or a least among the num- bers which are greater than those in (A). By similar reasoning 2 may be proved. 1033 If among the different numbers of an assemblage there is a greatest, we call that number the maximum number of the assemblage ; if a least, its minimum number. The superior limit of an assemblage is the maximum number, if there be one. If not, it is the least number which is greater than every number in the assemblage. The inferior limit of an assemblage is the minimum number^ if there be one. If not, it is the greatest number which is lesa than every number in the assemblage. An assemblage like 1, 2, 3, 4, • • • which contains numbers greater than every assignable number is said to have the supe- rior limit 00. Similarly an assemblage like —1, —2, —3, — 4, • • • is said to have the inferior limit — co . Evidently, if an assemblage has a finite superior limit X, either A. is its maximum number or we can find in the assem- blage numbers which differ from X as little as we please. 1034 By the " values of /(a-) in (a, h) " we shall mean those which correspond to values of x in (a, h). And if this assemblage has a maximum or a minimum value, we shall call it the abso- lute maxim/um or minimum, value of f{x) in (a, li). The maxi- mum and minimum values defined in § 639 may or may not be the absolute maximum and minimum values. 1035 Theorem 4. If f (x) is continuous thronghout the interval (a, b), it has an absolute maximum and an absolute minimum value in (a, b). For since the values of /(a-) in (a, b) are finite, § 1023, they have finite superior and inferior limits. Call these limits A and /A respectively. We are to demonstrate that in (a, b) there is a number Xq such that f{x^ — k, and a number x^ such that f{x{) = fi. PROPERTIES OF CONTINUOUS FUNCTIONS 583 As the proofs of these two theorems are essentially the same, we shall give only the first of them. Divide (a, b) into any number of equal intervals, say into two such intervals. Evidently A. will be the superior limit of the values of f(x) in at least one of these half intervals. For convenience call this half interval («!, b^). Deal with the interval («i, ^i) as we have just dealt with (a, b), and so on indefinitely. We thus obtain a never-ending sequence of intervals within intervals, (a,b), {a„b,), (a,,b,), ■■., («„, ^.„), .••, in each of which A is the superior limit of the values of/(x). As 71 is indefinitely increased, «„ and i„ approach the same number as limit (see § 1030). If we call this limit x^,, then /(a"o) = X. For if not, since both /(x^) and X denote constants, their difference must be some constant, as a, different from 0, so that X-f(^,)=a. (1) Since f(x) is continuous at x^, we can make the interval (a„, b„) so small that for every value of x in (a„, 6„) we have, § 1025, \f(x)-f{x,)\„) and (2) so that, § 1033, X -f(x) < a/2. (3) But it will then follow from (2) and (3) that X -/(xo) < a. (4) Therefore, since (4) contradicts (1), (1) is false ; that is, \ —f(xo) = 0, or A =f(Xo), as was to be proved. 584 A COLLEGE ALGEBRA 1036 Corollary. T/f (x) is continuous throughout the interval (a, b), it will have in (a, b) every value intermediate to its maximum and miyiimxim values in this interval. For let c denote the value in question and consider the function f(x) — c, which is continuous in (a, b), § 1026. If/(.ro) and/(a-i) denote the absolute maximum and mini- mum values of /(.r) in (a, b), /(r„) — c is + and /(xj) — c is — . Hence, § 1030, between x^ and a-i there is a number, call it x^, such that/(j:-.,) — c = 0, or/(a-2) = c, as was to be proved. 1037 Oscillation of a function. By the oscillation of f(x) in (a, b) is meant the difference between the superior and inferior limits of the values of /(.r) in (a, b). 1038 Theorem 5. Let f(x) he continuous tliroughout (a, b). If any positive number a be assigned, hoicever small, it is possible to divide (a, b) into a finite number of equal intervals in each of which the oscillation off(x) is less than a. For divide (a, b) into any number of equal intervals, say into two such intervals, each of these in turn into two equal intervals, and so on. The process must ultimately yield inter- vals in each of which the oscillation of /(.«) is less than a. For if not, there must be in {a, b) at least one half interval in which the oscillation of f{x) is not less than a ; in this, in turn, a half interval in which the oscillation of f(x) is not less than a ; and so on without end. Let this never-ending sequence of intervals within inter- vals be i<',h), (a„b,), (a,,b,), ..., (a„,b„), .■■, and, as in § 1030, let lim a„ = lim b„ = .r„. Since f(x) is continuous throughout (a, h), it has an abso- lute maximum and an absolute minimum value in each of the intervals (a, b), (a^, bi), •••, (a„, b„), •••, § 1035. Let /(«-„) denote the absolute maximum and f(^„) the absolute minimum value of f(x) in (a„, b„). PROPERTIES OF CONTINUOUS FUNCTIONS 585 Then, by hypothesis, /K)-/(A.)>«. and therefore lim f{a^) — lim /(/?„) > a. But this is impossible. For since a„ and /?„ are in (a„, &„), and lim a„ = lim h„ = x^, we have lim «„ = lim ^„ = x^. Therefore, since /(a-) is continuous at x^, lim/(a„) = lim/(^„); that is, lim/(a„) — lim/(/3„) = 0, .'. not > a. FUNCTIONS OF TWO INDEPENDENT VARIABLES Functions of two variables. We say that the variable u is a 1039 function of the variables x and y when to each pair of values of X and ?/ there corresponds a definite value or set of values of u. We shall confine ourselves to the case in which to each pair of values of x, y there corresponds a sijigle value of u. The notation u —f(x, y) will mean that w is a function of x and y, and f(a, h) will mean the value which u has when x = a and y — h. Thus, M is a function of x and y if u =/(x, y) = x"^ - 2y -\-\. Here, when X = 1, ?/ = 2, we have u =/(!, 2) = 1-4 + 1= — 2. The note at the end of § 1022 applies, mutatis mutandis, here also. Continuity of such a function. Let f(x, y) denote a given 1040 function of x and y. AVe say that /(a;, y) is continuous at a, b, that is, when x = a and y = b, if f(a, b) has a definite finite value and \if(x, y) will always approach /(rt, b) as limit when X and y are made to approach a and b respectively as limits. In the contrary case we say that f{x, y) is discontiruwns at a, b, that is, when x = a and y = b. From this definition and § 189, it immediately follows that The sufficient and necessary conditio7i that f (x, y) be continu- 1041 ous at a, b is that f (a, b) have a definite finite value, and that for every j^ositive number S which can be assigned it shall be possible to find a corresponding positive number e such that |f (x, y) — f (a, b)| < 8 whenever |x — al< e and [y — b| < c 586 A COLLEGE ALGEBRA Theorem 1. If both of the functions f (x, y) and ^ (x, y) are continuous at a, b, tlte same Is true of i(x, y)± (x, y) and f (X, y) • <^(x, y), also ofi(x, y)/(x, y), u?iless <^(a, b)=0. Ifi(x, y) Is co7itlmious at a, b, the same is true of Vf (x, y). This follows immediately from § 1040 and §§ 203-205. Number regions. In what follows it is to be xmderstood that X and 7/ denote real variables, and /(.«, y) a real function of these variables (compare § 1027). As is shown in § 382, pairs of values of x and ?/ may be pictured by points in a plane. Evidently, if employing this method we draw the lines wh. :h are the graphs of the equa- tions X = a, X = b, y = c, 7/ = d, i 384, the rectangle bounded by these lines will contain the graphs cf all pairs of values of X, y such that a (x, y) and i/^ (x, y) denote real polynomials in x, y, and therefore have, § 232, \f{z)\ = ^<^i^x,yy + ^{x,yyi^\ By § 855, we can find a positive number, as c, such that the roots of /(,?)— 0, if there be any, are all of them numerically less than c ; and if c' = c/^2, evidently |,t|, or (a-^ + t/^)^, is less than c for all values of x, y such that — c' (x, y)'^ + i(/(x, ?/)^]^ is a continuous function of x and y, § 1042. It therefore has a minimum value in this region, § 1045, say when x = Xq, y = y^. If z, = x,+ iy„ then 1/(^0)1 = [<^(^o, l/oY + ^(xo, yo)l' = 0. For since |/(«o)| is the minimum value of |/(s;)], we cannot make \f(z)\ < \f(zo) \. Therefore \f(zo) \ = 0, since otherwise, §1047, we could so choose z thai \f(z)\<\f(zo)\. Hence |/(«)|, and therefore f(z), vanishes when z = z^; that is, Zg is a root of the equation /(s) = 0. INDEX Numbers refer to pages Abscissa, 138 Addition of integral expressions, 93 of numbers, 10, 19, 35, 50,71, 72 of radicals, 274 of rational expressions, 217 of series, 541 Amplitude of complex number, 488 Angle, circular measure of, 488 Annuities, 391 Approximations, 48, 55, 453 Assemblage, infinite, 3 Associative law of addition, 11, 22, 35, 54, 74, 521 of multiplication, 14, 23, 35, 54, 74 Asymptote, 335 Base of power, 39 of system of logarithms, 377 Binomial theorem, 256, 283, 554 Binomials, products of, 102, 253 Biquadratics, 112, 48G Cardan's formula for cubic, 483 Chance, 409 Clearing of fractions, 118, 231 Coefficient, 86 Coefficients, detached, 99 undetermined, method of, 152 undetermined, theorem of, 172, 540 Cologarithms, 386 Combinations, 393 Commensurable, 37 Commutative law of addition, 11, 22, 35, 54, 74, 534, 544 of multiplication, 14, 23, 35, 54, 74 Completing the square, 187, 300 Condition, necessary, sufficient, 93 Constants, 79 Continuity of functions, 577, 585 of real system, 46 Convergence of infinite series, 520 absolute and conditional, 533 limits of, 536 tests of, 523, 531 Convergents of a continued frac- tion, 567 Converse, 92 Coordinates, 138 Correspondence, one-to-one, 1 Cosine, 489 Counting, 9 Cube root. See Roots Cubics, 112, 483 irreducible case of, 485, 490 Cyclo-symmetry, 248 Degree of equation, 111 of polynomial, 87 of product, 98 Density of rational system, 34 of real system, 46 Derivatives, 460 Descartes's rule of signs, 447 591 592 A COLLEGE ALGEBRA Determinant, 494 bordering a, 505 cofactors of, 504 diagonals of, 496 elements of, 494 evaluation of, 505 minors of, 502 order of, 495 products of, 506 properties of, 498 terms of, 496 Differences, method of, 364 Discriminant, 517 of cubic, 485 of quadratic, 304 Distributive law, 14, 23, 35, 54, 74 Divergence of infinite series, 520 Divisibility, exact, 28, 155, 161 Division of integral expressions, 107 of numbers, 27, 35, 54, 73, 489 of radicals, 287 of rational expressions, 219 of series, 546 Division, synthetic, 166 Division transformation, 155 by aid of undetermined coeffi- cients, 160, 163 Elimination, 131, 143, 317 by determinants, 508, 514 Ellipse, 334 Equality, 3, 8, 32, 34, 45, 72 algebraic and numerical, 18. '^5 rules of, 13, 15, 24, 36, 54, 57 Equations, binomial, 313, 490 biquadratic, 112, 486 complete, 426, 448 conditional, 110 cubic, 112, 483 depressed, 427 equivalent, 117, 131 exponential, 390 Equations, fractional, 111, 231, 300 identical, 89 inconsistent, 133, 146 indeterminate, 342, 575 integral, 111 interdependent, 133, 145 irrational. 111, 288, 313 irreducible, 445 linear, 112, 139 literal. 111 logarithmic, 390 numerical. 111, 429, 459 quadratic, 112, 298 rational. 111 reciprocal, 311, 438, 487 roots of. See Roots simple, 11:2, 118 simultaneous simple, 127, 143, 508 simultaneous, of higher degree, 135, 317, 514 simultaneous symmetric, 326 solution of, 112, 1.8, 483 solution of, by factorization, 194, 309, 318 transformation of, 114, 129,436 Errors of approximation, 55 Evolution, 39, 56, 76, 83, 260, 276. 490' Expectation, value of, 411 Exponents, integral, 39 irrational, 376 laws of, 57, 279, 376 rational, 279 Expre.s.sions, algebraic, 85 finite and infinite, 85 integral and fractional, 85 rational and irrational, 86 Factor, 14, 176 highest common, 196 INDEX 593 Factor, irreducible, 211 prime, 177, 208, 212 rationalizing, 285 Factorial n, 395 Factorization, 178, 249 Ferrari's solution of biquadratic, 486 Fractions, 32, 213 continued, 566 improper, 213 irreducible, 37 partial, 236 proper, 213 reciprocal, 219 recurring, continued, 572 Functions, 88, 571, 585 defined by power series, 539 expansion of, 371, 548, 551 integral, 85 rational, 86 symmetric, 245 Fundamental theorem of algebra, 427, 588 Graphs of e»i'nations, 139, 333 of numbers, 27, 38, 66 of variation of functions, 469 Groups of things, 1 equivalence of, 1 finite and infinite, 3 Homogeneity, 87, 99 Horner's method, 453 Hyperbola, 335 Identities, 89 Imaginaries, conjugate, 295 Incommensurable, 65 Indeterminateness of rational func- tions, cases of, 223 Induction, mathematical, 424 Inequalities, solution of, 340 Inequality. See Equality Infinitesimal, 63 Infinity as limit, 224, 229 Interest, compound, 390 Interpolation, 371 Inversions, 492 Involution, 39, 56, 76, 82, 105, 276, 489 Lagrange's formula of interpola- tion, 373 Length, 26, 37, 66 Limit of variable, 58 Limits, superior and inferior, of assemblages, 582 Logarithms, 39,^377 characteristic of, 381 common, 379 mantissa of, 381 modulus of, 559 natural, 558 table of, 384 Maximum, 307, 467, 582 Mean, arithmetical, 355 geometrical, 359 harmonical, 362 Measure, 26, 37, 65 Minimum, 307, 467, 582 Multinomial theorem, 408 Multiple, lowest common, 205 Multiplication of integral expres- sions, 97 of numbers, 14,20,35,52,72,489 of radicals, 275 of rational expressions, 218 of series, 545 Number, cardinal, 2, 10 complex, 71 fractional, 33 imaginary, 70 integral, 18 irrational, 46 594 A COLLEGE ALGEBRA Number, natural, 6 negative, 18 positive, 18 rational, 34 real, 45 Number intervals, 579 regions, 586 Numbers, theory of, 211 Odds, 410 Ordinal, 7, 33, 45 Ordinate, 138 Origin, 137 Oscillation of a function. 584 Parabola, 333 Parentheses, rule of, 95 Part (of group), 3 Permanences of sign, 446 Permutations, 393 odd and even, 492 Polynomials in x, 87 products of, 103 Powers. See Exponents and Invo- lution perfect, 260 Power series, 535 convergence of, 535 products of, 545 quotients of, 546 reversion of, 548 transformation of, 545 Probability, 409 Products, continued, 252 infinite, 564 Progressions, arithmetical, 354 arithmetical, of higher order, 364 geometrical, 357 harmonical, 362 Proportion, 347 continued, 350 Quadratics, 112, 298, 304 simultaneous, 317 Radical expressions, simple, 277 Radicals, 271 index of, 271 similar, 273 simple, 271 Radicand, 271 Ratio, 69, 347 Rationalization, 285 Remainder theorem, 169 Resultants, 512 properties of, 514 Rolle's theorem, 467 Roots of equations, 112, 426 extraneous, 116 imaginary, 444, 448 infinite, 229, 306, 439 irrational, 453 location of, 452, 458, 475 multiple, 428, 463 number of, 427 rational, 429 superior and inferior limits of, 430, 441, 466 symmetric functions of, 305, 434, 478 Roots of integral functions, 260 cube, 266 square, 262 Roots of numbers, 39, 56, 76 cube, 268,. 483 principal, 271. square, 265, 292, 295, 296 trigonometric expression of, 490 Scale, complete, 6 natural, 17 Sequence of numbers, 58 regular, 00 INDEX o% Series, alternating, 632 binomial, 538, 553 doubly infinite, 543 exponential, 537, 556 geometric, 360 hypergeometric, 529 infinite, 520 logarithmic, 537, 557 recurring, 560 Sign, rules of, 95 Simultaneous, 127. See Equa- tions Sine, 488 Solutions of systems of equations, 128 infinite, 230, 318 integral, 342 number of, 517 Square root. See Roots Sturm's theorem, 472 Substitution, principle of, 128 Subtraction of integral expressions, 93 of numbers, 16, 19, 35, 51, 72 of radicals, 274 Subtraction of rational expres- sions, 217 of series, 541 Surds,^ 291 Symmetry, absolute, 245 cyclic, 248 Taylor's theorem, 461, 551 Term, 86 absolute, 426 Transformation of equations, 114, 129, 436 Transposition of terms, 115 Value, absolute or numerical, 18, 75, 488 Variable, 58, 79 continuous, 69 Variation, 351 of integral functions, 308, 469 Variations of sign, 446 Zero, 17 as limit, 63 operations with, 19, 25, 31 RETURN CIRCULATION DEPARTMENT TO— ^> 202 Main Library LOAN PERIOD 1 - HOME USE 2 : 3 4 5 ( b ALL BOOKS MAY BE RECALLED AFTER 7 DAYS Renewals and Recharges may be made 4 days prior to the due date. Books may be Renewed by calling 642-3405. DUE AS STAMPED BELOW Jl}L24196 f AUTO. 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