Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation •http://www.archive.org/details/algebratheoreticOOedi,nrich HAMBERS'S EDUCATIONAL COURSE.— EDH, W, AND R. CHAMhERS. ALGEBRA THEOKETICAL AND PRACTICAL EDINBURGH: PTTBUSHED BY WILUAM AND ROBERT CHAMBERS. 1253 edinburgh : ;ated by w. and r. chambers. CAJORI PREFACE. present Treatise contains all the sulyects in theory and tice usually comprehended in an elementary work oji Algebra. i. Jias been a special object to explain, as clearly and concisely as j' .ssible, the principles of the science, to illustrate them fully by opriate examples, and to prescribe a sufficient nimiber of cises for solution by the student, in order to impress the principles on his memory, and enable him to acqiiire sufficient ^':'! in Algebraic Computation. preparing this new edition, the objects aimed at have been to re perfect accuracy in all the ahswers, to. supply defects, and dapt the work to the present advanced state of the science. this purpose, the questions have all been solved several s, and numerous Theorems have been given which were not imed in the former editions. MiiUiglication by Detsyjiig^d ^icients. Synthetic Divisipn^new applications oTlJndetermined l'tet6tlTli~!'!Elimination by Cross Multiplication, and a new aise on the General Solutipnof the. Higher Equations, ha\e I ihtrpclueed; together with a great number of additional rcises on the most useful kinds of Equations. In order that work might be contained within due limits, the Solution of lareterminate Equations of the second degree has been omitted, as being of little or no practical utility; and the Notes and Ai)pendix have been incorporated in the body of the work. IV PREFACE. The method adopted in this Treatise i8, to state the rule concisely and clearly ; to illustrate it by appropriate gradational examples ; to demonstrate the rule; and to prescribe a series of exercises. By this means the transition is very gradual from the more simple to the more complex parts of the subject. • jMter the student has made himself familiar with the subjects contained in this Treatise, he wiU be well prepared for entering successfully on the study of the application of Algebra to Analy- tical Trigonometry, Analytical Greomctry, and the Differential and Integral Calculus. There are some articles near the beginning of the Treatise which should be omitted in the first reading : — The discussion of insulated Negative Quantities in Addition and Subtraction ; the examples with Literal Coefficients in Addition, Subtraction, and Multiplication, which ought to be postponed till the student has studied the third case of Division ; and also those subjects marked with an asterisk in. the Contents. Jambs Petde. EvrshxmQB, May 1982. COJTTEIs^TS. FUNDAMENTAL PRINCIPLES AND RULES. Page i.lMlNARY DEFINITIONS AND PRINCIPLES, ... I VITIONS OF QUANTITIES AND SIGNS OF OPERATION, . . 2 i.KlCAL EVALUATION OF ALGEBRAIC EXPRl'SSIONS, . . 6' AtS, .8 HON, 9 -ACTION, 14 ■IPLICATION, . . 18 •IPLICATION BY DETACHED COEFFICIENTS, .... 24 i.j VISION, 27 SYNTHETIC DIVISION, 40 "•"" GREATEST COMMON MEASURE, 47 EAST COMMON MULTIPLE, 55 3RAfC FRACTIONS, . 1)9 J.UTION OF FRACTIONS INTO INFINITE SERIES, ... 78 vers OP NUMBERS, 81 ti AND COMPOSITE NUMBERS, . . . . . .83 PRODUCTS OP QUANTITIES, 86 *PRIME AND COMPOSITE QUANTITIES, ...... 87 "l.UTION, . . 89 uiiON, .95 niUATIONAL QUANTITIES, 108 ^IMAGINARY QUANTITIES, . . 1 32 SIMPLE EQUATIONS. DEFINITIONS, 136 EQUATIONS CONTAINING ONLY ONE UNKNOWN QUANTITY, . . 1 39 SOLUTION OF THESE EQUATIONS, . .. . . . . 142 EQ.UATIONS THAT MAY BE REDUCED TO SIMPLE EQUATIONS, . . 145 Ql ESTIONS PRODUCING SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY, . . . . • 147 Tl CONTENTS. NEGATIVE SOLUTIONS OF SIMPLE EQUATIONS, SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES, QUESTIONS PRODUCING THESE EQUATIONS, SIMPLE EQUATIONS CONTAINING THREE UNKNOWN QUANTITIES, ELIMINATION BY CROSS MULTIPLICATION, QUESTIONS PRODUCING THBSE EQUATIONS, .... QUADRATIC EQUATIONS. DEFINITIONS, PURE QUADRATICS, QUESTIONS PRODUCING PURE QUADRATICS, .... ADFECTED QUADRATICS, QUESTIONS PRODUCING THESE EQUATIONS, .... QUADRATIC EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES, QUESTIONS PRODUCING THESE EQUATIONS, . RATIO— PROPORTION— PROPORTIONAL EQUATIO PROGRESSIONS. RATIO, . . \ • . • • • • • PROPORTION, . ^ PROPORTIONAL EQUATIONS, EQUIDIFFERENT PROGRESSION, - EQUIRATIONAL PROGRESSION, . . , , . J14RM0NIC PROGRESSION, PROPERTIES OF NUMBERS— PERMUTATIONS- COMBINATIONS. PROPERTIES OP NUMBERS, 283 PERMUTATIONS, 287 COMBINATIONS, 289 UNDETERMINED COEFFICIENTS. EXPLANATION OF THE METHOD, . . . . . .291 RULE FOR EXPANDING FUNCTIONS, . . . . . ^ 292 DEMONSTRATION FOR THE DETELOPMENT OF AN IRKATION/T FUNCTION I or THE FORM (a + bx + 00^ r ...)*, . . . . . |29( CONTENTS BINOMIAL THEOREM. PflgC STBATION FOR POSITIVE INTEGRAL POWERS, . . . 29» > OR FINDING THE COEFFICIENTS, 301 -opjiENT OP (ic — a)"*, 302 STEATION FOR ANY POWER, INTEGRAL OR FRACTIONAL, POSITIVE ; NEGATIVE, , 304 .OPMENT OF TRINOMIALS, &C. 310 EXPONENTIAL THEOREM, jIGATION, 311 . OPMENT OP a«, 31] ICAL VALUE OF e, 312 LOGARITHMS. DEHNITIONS, 313 OEM'KAL PROPERTIES OP LOGARITHMS, 313 ; FOR THE COMPUTATION OF LOGARITHMS, . . . 316 iN LOGARITHMS, 319 SERIES. TRE DIFFERENTIAL METHOD, 7 322 INTERPOLATION OF SERIES, 32fi SUMMATION OF INFINITE SERIES, . . . . . . 32^ '^•'■r-S!ON OF SERIES, , . . 343 I'JAL SOLUTION OF HIGHER EQUATIONS. . AND THEOREMS, 346 .AHY ROOTS, . . 351 I'S CRITERION OF IMAGINARY ROOTS, . . . . 351 FORMATION OF EQUATIONS, 351 ^'S THEOREM FOR THE DISCOVERY OF IMAGINARY ROOTS, . 357 • OPMENT OF AN INTEGRAL FUNCTION OF A BINOMIAL X + J/, 359 IS OF AN INTEGRAL FDKCTION OP X FOR VERY GREAT OR VERY SMALL VALUES OF X. CONTINUITY OF INTEGRAL FUNCTIONS OF ONE VARIABLE, . '. 361 S OF THE ROOTS OF AN EQ,UATION, 365 :;i' OF EOCA/. ROOT:'- ....... 3{j'7 VJi] CONTENTS. STURM'S T/IEOREM, , . ; .a Newton's method of approximation, 37fi HORNER'S MKTHOD, CONTINUED FEACTIONS, DEFINITION, . . . -4 382 CONVERSION OP COMMON INTO CONTINUED FRACTIONS, . . 384 CONVEKGENT FRACTIONS, . 385 PERIODICAJ, CONTINUED FRACTIONS, 392 INDETERMINATE EQUATIONS.' SIMPLE INDETERMINATE EQUATIONS, . . . . . . 394 ONE EqUATFON WITH TWO UNKNOWN QUANTITIES, , . . 394 TWO EQUATIONS WITH THREE UNKNOWN QUANTITIES, . . . 402 ONE EQUATION WITH THREE UNKNOWN QUANTITIES, . . 406 EXPONENTIAL EQUATIONS. DEFINITION, 412 WHEN THE EQUATION 18 OX = b, 412 WHEN THE EQUATION IS UCX = d, uhx = cd«, OT XX z=z a^ . . 41^ COMPOUND INTEREST AND CERTAIN ANNUITIES. COMPOUND INTEREST, .* . 414 AMOUNT OP CJBSRTAIN ANNUITIES, 416 MISCELLANEOUS EXERCISES, . . . . . . .41.9 ALGEBRA. PSELBIINABT BEFISITIONS AND PEINCII'LES. Algebra is that brajicli of Mathematics which teaches tlie Liiv,i;hocl of performing caleiilations by means of lettf^rs, ■which represent nmnbers or quantities, and sjTubols 'vrhich indicate the "-" ^ations to be perforaied on tliem. Quantity is anytliing; that is capable of increase or diminu- Measurahle or mathematical quantity is that which is capable ' ng measured. i ims, lines, svafaces, time, are measurable quantities, and therefore proper objects of mathematical calculation. 4. The unit of mmsure of any quantity xh some portion of the {same kind of quantity assumed as the standard of measurement. Thus, to ascertain the length of a line, some unit of measure, ■hot, is assumed, and the number of feet in the line is called .•'?lh. Ihe numerical value of a quantity is expressed by the number aes that the assumed iinit of measure is contained in" it. lus, if a foot be assmned as the imit of measure of a lino h contains 20 feet, its namerical value is 20. Tlie same line have different numerical values according to the assumed liu.t of measure. 6. When the kind of units of which a number is composed is i'^' mentioned, the numlK-r is said to be abstract, but when tlie 1 lination is specified, it h said to be concrete. MS, 20 feet is a concrete number, tb.o kind of units being 1 ; but 20 is an abstract number, as the kind is not men- !. Since quantities may be r?prcscnted by numbers, and are .3 supposed to be so in algebra, the letters denote numbers; n the Iheor of aluebra. the letters have no particular iiumercaT vaT'ie^ bo* that they represent any numhtv ■' . : -, genet&l or abstract qn&Dtity. S iUg^bra investigates the properties of abstract i.i.iti.y ;mui (iie niles of algebraical computation: and practical j*i . l»ra is conci?riied in the solution of question?, in which tiie given quantities either have particular values, or may hare such values assigned. 9. In ft1i,'ebra, as in geometry, there are two kinds of proposi- tions — namely, theorems and problems. 10. In a theorem, it is propo*r>d to demonstrate some stated relation or property of numbers or abstract quantities. 11. In a problem, the object is to find the value of some I unknown numbers or quantities by means of given dilation- existing between them and other known numbers or quantitief DEFINITIONS OF QUANTITIES AND SIGNS OF OPEEATION. 12. Kfiptvn quantities are tliose whose niimerical values given or supposed to be known ; and unknown quantities are whose numerical values are not known. Known quantities are commonly represented by the first lr> of tijo alphabet, as a, b, c; and unknown quantities by thi letters, as x, y, «. . 13. The sign of equality, —, read equal to, denotr quantities between which it is placed are equal, ap.J expression is called an Equation. ' Thus, a = b denotes that a is equal to b. 14. The sign of additwn, ,-^f>reii6. plus, deuottT, in;u im:. tj^r.au- tities between which it is placed are to be aiidod together. ' | Tims, a -\- b means that 6 is to be added to a. If a~ 4, and \ 6 = 3, then a + 6 = 4+3 = 7. 16. The sign of subtraction, -r-, read minus, denotes that the * quantity after itf is to be subtracted from the quantity befort; h- Thus, a ~ b means that b is to be subtracted from a. If a and 6 = 5, then « — 6 = 8 — 6 = 3, ]6.-rho signs + and — are called the signs; the fon" called the positive, and the latter the negative sign ; and the • said to be contrary or opposite. 17. Quantities having positive signs, are said to be po..' and those having negative signs, are called negative. W quantity ha«? no sign prefixed, the positive is always under3t( 18. Quantities that have the same sign, are said to has- sigris ; and those that have opposite signs, are said to have signs. -~' — » ly. Tlie sign of difference, ~, read difference bclwecn, d the difference of any two quantities between which it h ];i;i . Thus, a ~ b or b ~ a, denotes the difference betwef^- - If a = 8, and 6 = 6, a ~ 6 = 8 ~ 6 = 2 ; or 6 ~ X.c, or a . 6 . c, or more comihonly by ahc. If a — Z, b = 6, i, then a*c = 3 X 5 X 4. = 15 X 4 = 60. •. The sign of division, -;-, rea^^ divided hy, denotes tliat the I ity preceding it is to be divided by the quantity following The division of two. quantities is more commonly represented hy placing the dividend as the numerator, and the divisor as the denominator of a fraction, which forms an algebraic fraction. as, a ~ b ox J, means that a is to be divided by b. If i:0, and 6 = 5, then a -t- 6 or y = 20 -4- 5, or - = 4. b 5 The following terms used in arithmetic — namely, sum, (cnd, subtrahend, remainder; multiplicand, multiplier, pro- ; and dividend, divisor, and quotient — have the same ning in algebra. The sign of majority, 'T', refid is greater than, denotes that I uantjty befoi"e it exceeds the quantity after it. 1 :is, a -ix- 6 means tiiat a exceeds b. If a = 8, and 6 = 5, then ■'• The sign of minority, ^, read is less than, denotes that the '•■' preceding it is less than the quantity following it. 4 ELEMENTS OP ALGEBRA. TIius a^h means that a is less than h. If a = 5 ana u — u, then tf^Q, 27. The sign oo or ^, denotes a qnantity greater than anv that can bfe assigned; that is, a quantitj^ indefinitely great, irifinity. 28. Tlie nvrneral coefficient of a quantity is a numher prefi: to it, Avhich shews how often the quantity is to be taken. 2 2 Thus, 4a means a taken 4 times ; -x moans - of : 3 3 means that the product ox is to be taken 5 times, ii a ~ z, and or = 3, then Bux = 5X2X3 = 5X6 — 30. ^0. The literal coefficient of a quantity is a letter by which it iS multiplied. Thus, in the quantity az^ a may be considered to be a coefficient of 2, or z a coefficient of a. Quantities having literal coefficients are commonly considered to be unknown, and the coefficients to be known, quantities. 30. The coefficient of a quantity may consist of a number together with a literal part. Thus, in .To&x, the quantity ^ab may be considered to be the coefficient of x. If a = 2, and S = 3, then 5ai = 5 X 2 X 3 = 5X6 = 30, -and oahx = 30x. When no numeral coefficient is prefixed to a quantity, xmity is understood to be its coefficient. Thus, the coefficient of x, or of ■IX, or of abx, is 1 ; that is, a — la, and ax = lax. 31. A power of a quantity is the result arising frou the multiplication of the quantity ^to itself one or more times. When the quantity is repeated twice as a factor, the product is called its square, or second power ; when it is repeated three times, the cxihe, or third power; when four times, \h& fourth powet': and so on. Instead of repeating the same quantity as a factor, a &mall figure is placed over it, to point out the.number of times tl:ut the quantity is repeated. This figure is called the exponent: and hence a small 2 is used to denote the square; 3, the cube : 4, the fourth power : 5, the fifth power ; and so on. Thus the square or second power of x is = xx, and is expressed by ,: - cube or third ... xis = .rrc.r, ... ' .,•' the fourth ... x\s, = xxxx, the fifth ... :r is = xxxxr nd generally the «th power =:a:a:x ... where a: is supposed :.< le peated n tunes, and is expressed by z". When a letter has no exponent, it is considered to be the first ■■■: simple power of the quantity, and nnity is considered to U-. its - v^xponent; thus, x (>v >■' >- i^'"-- first power of x. T)EFI>'iTIONS. 6 .. 1 . . ;^.-„:i:ily to any power, is to find thai power !iie quantity. 3. The square root of any proposed quantity is that quantii; se square, or second power, gives the proposed quantity. Tl;' ■ -. root is that quantity whose cube is the proposed quantity I so on. The nth root is that quantity whoso nth power is the ,x)sed quantity. k To extract any root of a quantity, is to ilnd that root. '">. The radical sign^ V* placed before a quantity, indicates , t some root of it is to be extracted ; and a small figure placed cv^r the sign, called the exponent of the root, shews what root i- to be extracted. Roots are also indicated by fractional exponent Thus, the square root of a is indicated either by a/« or a*; tl' , cube root, by ^a or a*; the fourth root, by >^a or oi ; and the ntl _i ^■^■>t, by ^/a or a»- '.G. A simple quantity consists of a single letter, or the product I wo, or more letters, or powers of letters, with or without a flicifent. lius, a, 5a, — a-x, ^ax^i/^ are simple quautitit^. r. A compound quantity consists of two or more simpU ntities called its terms, which are connected by the signs pli -^^I^minus. I^Hphus, a + b, da — 2x, 2a — dx^f/ -\- c;x, are compound qium- I^^Bv A simple quantity is also called a monomial ; a com- ^^Had quantity, consisting of two terms, is called a hinmnial; r^^P' of tlii-ee terms, a trinomial; of four terms, a. quadrimmial ; of more than four terms, a multinomial or poIi/nomiaL binomial, whose second term is negative, is called a,' residual ■intiiy. "'■). When any operation is to be performed on a compouxu, sntity, it is expressed by enclosing the quantity in parentheses . and inserting the sign of the operation. ^ Thus, — (a — 2ax) means that a — 2ax is to be subtracted > .'a — 2x) means t)iat 3a — 2x is to be multiplied by 5; •> ! — 2x)(a 4- h) means that 3« — 2x is to be multiplied ■i. a -\- b; (a 4- -tO -r- (a — a-) means that a + a: is to be . ided by a •— x ; (a — x)' means that a — ar is to be squared; and ^(a^ — x^) means that the cube root of a' — ^ is to 1)'. extracted. ' A vinculum is sometimes used instead of parentheses. 1 1ms, a — X . y^ means that a — a; is to be multiplied by y-. A perpeu.l'rular bar \ is sometimes used to denote that .■- comi)ound quantity is to be multiplied by another quantity. Thus, a .r' has the smwc meanine as (a — 3c 4-cZ)x'. — 2c ^" ^^'"^ '-^- ...uipie quanti. 6 ELEMENTS OF ALGEBRA. the number of first or simple powers of factors of which it is composed, ajid is indicated by the sum of its exponents. Thus, ax is of tlie second dhnension, for the sum of its exponents is two ; Za^a^ is of the fifth dimension, for the sum of its exponents is five ; and so on. A\. Quantities that are exactly the saine, except in their signs and coefficients, are called like quantities; other quantities are said to be unlike. Thus, 2a^x^ — ha'x^^ are like quantities ; and Zx^y, 4ixy% are unlike. 42. An insulated negatire quantity is one with a negative sign, which is considered to exist unconnected with a positive quantity. Thus, — a, when it is not supposed to be subtracted from any positive quantity, is called an insulated negative quantity, and, for the sake of distinction, it may be enclosed in brackets, thus [- «]• 43. Tlie sign of an insulated negative quantity is called a sign of relation or affection ; whereas in the case of an ordinary negative or subtraetive quantity, it is the sign of the operation of subtrjiction. 44. When a negative quantity is considered as insulated^ it is taken in a sense directly the opposite of that in which it would be understood if it were a positive quantity. Thus, if -I- a r(;present a number of miles in one directiouj then [— a] indicates an equal number in an opposite direction. So if -f a mean a number of pounds of property, [—a] means an equal amount of debt. If + 10 means 10 miles west or of we^inrf, [—10] means 10 miles east or of easting, or of negative westing. So if 4 a means a number of men who pay away money, [— a] means the same number who receive money. The expression that — a is a less than nothing merely means that it is an insulated negative quantity. Since aU ^gebraic quantities' represent numbers ':;, and numbers may be represented by lines, -f a may be represented by the line OA, and [— a] by the equal line OA' in the opposite direction ; and thus may lAl insu- _ ^ i ^ lated negative quantities be represented, even -ry ' ' when they denote abstract numbers. ^ O A NUMERICAL ETALUA'nOX OF ALGEBRAIC EXPRESSIOKS. 45. When numerical values are assigned to algebr^'i^ titles, the numerical value of any algebraic expression them may be found by substituting for the let equivalent numerals, and then performing on these iiUi^ibut, 'the operations indicated by the algebraic signs contained in the expression. EXAMPLES. 1. Find- the numerical value of the expression a — 2& 4- S'^ !)en a = 3, ^ = 4, a;id c = 6. a - 2b -V 3c2 = 3 - 2 X 4 -^ 3 X C" - 3 - 8 + 108 = 111 4G. Any ether nuraerical values n\i\y be assi,i;:ried to the letters fl, 6, Jind c, and the expression converted in like ma nner into its rnimerical value, which will generally be different for dilferij. i alues of the letters. 2. Find the value of c* -f 2(a — x)(a + x) — ixf, wh,.,, r:; 3, a; = 2, aud y — 5. This quantity = 3* + 2(3 - 2)(3 + 2) — 4 X 2 X 5' = 81 -f Xlx5 — 4x2x125 = 81 + 10 — 1000 = 01 - 1000- = ' 009. 8. Pind the value of (a« - x'Xa^ + -c') - ^^^^-=1^^ + 3 v when a = 9 and ar = 4. It becomes = (81 - 16X81 + IG) - |-^^ + 3V9 X 4 = 9 -f- 4 X 97 - ^-^^ + 3V36 =r 6305 - ^ -h 3 x »> =:^ -' oometrical reasoning : — 47- If equals be added to equals, the sums are equal. 48. If equals be taken from equals, the remainders are equi^h 49. If equals be multiplied by equals, the j)roduct8 are equal. 50. If equals be divideid by equals, the quotients are eqn.: ' 51. If equal quantities be involved to the same powt powers are equal. 52. If the same roots of equal quantities be extract roots are equal. ADDITION. CASE I. TO ADD LIKE QUANTITIES WITH LIKE SIGNS. r)3. Exile. To the sum of the coefficients prefix the common u, and annex the common literal part. 1. Sa 2a a 5a Smu = 11a - Uxy IGax^ — ISbf + 19 E3:AJHPLE8, 2. xy Sxij 2a.r^ - 3. - 3V + - v + 4 2 8 5 llPme I in the first example, suppose a = 2, then 3a = 3 X 2 = < !, = 2 X 2 = 4, a = 2, 5a = 5 X 2 = 10, and the sum of thf . es is 6 + 4 + 2 + 10 = 22. But the sum 22 is much more easily found from the algebra- a llo, for 11a = 11 X 2 = 22. Lu the second example, let a; = 3 and y = 2, and the values oi terms will be 6a:y = 6 X 3 X 2 = 36 xy= 3X2=6 4a;3/ = 4 X 3 X 2 = 24 Zxy = 3 X 3 X 2 = 18 1 the sum of these values is =84 ^jut this sum is very easily found from the algebraic expression -'/, Jor 14.ry = 14 X 3 X 2 = 84, And as these terms are ^ta'active, their sum is — 84. In the first example, let a mean 10 miles, then IIBR i ' 3a = 3 X 10 = 30 miles 2a = 2 X 10= 20 ... «= 10= 10 ... 5a = 5 X 10= 50 ... a the sum is the sum = 110 nules, = 11a = 11 X 10 = 110 ... 'ilto. rxprr-is(- 's Tuay aU be thus nuinprifMllv exemplified. '>y w 10 ELEMENTS OF ALGEBRA. assigning any values to the letters; but in each illustration of this kind the same numerical values of the letters nmst be retained throughout. 54. The i)rinciple of the rule may be thus explained : — If 3a and 4a are to be added, the sum is evidently 7a. If — 5a and — 3a are to be added, or, in other words, if 5a and 3n are both to be taken away from some other quantity, there is to be taken away altogether just their sum, or 8a ; but 8a to be taken away is represented by — 8a ; that is, the sum of — 5a and — 3a is = — 8a. Let 5a and 3a be insulated negative quantities, or [— 5a] and [— 3a]. Then, since these two quantities may represent lines measured towards the left from a given point (44), and these two lines together would be exactly equal to a single line 6a measured towards the left or [— Sal^ ; therefore [— 5aJ and [— 3a] taken together are equal to [— 8a]. K a represent 1 mile, and the left direction denote east, then [— 5a] = 5 miles east, and [— 3a] = 3 miles east ; and, tliere- fore, together, they are = 8 miles east. Or if a be £100, then [-5a] is. = £500 of debt, [-- 3a] is - £300 of debt, and [- 8a] is = £800 of debt ; but 800 is the sum of 500 and 300; hence [— 8a] is the sum of [— 5n] and [- Sal EXEECISJ38. 1. 2. 3. ■ Saxt/ — Gbx^ Say — Abs^ + 7 2axy — 6x' at/ — bbi^ + 9 axy — Bbx^ 15ay — libx^ -j- l-'* 5axi/ ~yib3? Say - b:i^ + 10 Uaxy - 245x* 22ay - - 24ix^ -1- 41 4. 5. iax~ Scz+ 5 Sox - 5cz+ 3 lOoX- C2+ 9 ax - \bcz + 1 2ax- 4c^-f 4 oaz — • 18az - 2az- 25az~ az — S(^+ 4by .f-f- % 2c= -f l>y 4c» + Sby c'-h Shy 25ax - 2Scz -f 22 olaz — lie'' -f 21by CASE U, TO ADD LIKE QUAJfTITIES ^TTH TOiLIKE SIGKS. 55. KuLE, Find the sum of the coefficients of the like positive quantities, and next the sum of the coefficients of the like negiitive quantities: subtract the less"^um from the greater, iv ■ _ ADDITION. 1 1 ■Terence nnncx the common literal part, prefixing the sign of EXAMPLES. 1 2. 3. 2x oxf ^^/ao^ - lOca' + 12 - Ix -2ar/ - 2^/ax^ + 5c^' - 3 \2x W 6Va,i:' + 8c?» + 16 X -9V 4Vaar'- Qcz^ - 8 - ex - V - U>Jax^ + 3C2» -24 — 4x?/* — 3js/ax^ 4. -2(z4-y)-3(a + c) + 5 5(a;+y) + 4(a + c)-8 -7(ix+y) + 8(a + c)-7 3(x+^)-X« + c)4-4 Sum = - (x + y) + 4(a + «) - 6 In the first example, the smn (Jf the positive quantities is loar, d that of the negative is — 13x, and tlxe difl?erence between 15 I 13 or 16 — 13 = 2 ; hence the sum is + 2x. In the second ample, the sum of the positive quantities is Sxy'^, and of the f/ative — 12xv^, and the difference hetween 12 and 8 is 4, the ^n of the greater being — ; hence the sum by the rule is ixy^. In the third example, the sum of the positive terms i\taining cs* is IGcx', and the sum of the negative terms is 16c«' ; the difference between 16 and 16 is 0; hence the nm is 0. In this case the quantities are said to destroy each r. e principle of the rule may l?e thus explained : — 1. \Vhe;i the quantities to be added are a positive and subtrac- Ive quantity of the same kind. Thus, the sum of 3a, — 5a, 8a, und — 2a, is evidently the same as 3a -f 8a, and — 6a, and - 2a, or 11a and — 7a; but if 11a is to be added to some ocher quantity, and 7a to be subtracted, the result is evidently The same as if their difference or 4a be added; that is, the sum of -f- 11a and — 7a is 4a. But M hen the negative quantity exceeds the positive, it is to I'C considered as an insulated negative quantity, and its meaning )(>ing the opposite of that of a positive quantity, their sum is - their difference, with the sign of the greater prefixed. Thus, 'f 8a mean property, and [— lOo] debt, their sum is [— 2a] Hi 12 ELEMENTS OP ALGEBRA. or 2a of debt. Also, since 8a and [— 2a] give Ccr, and and.[— 3a] give only 5a, the quantity [—'2a] is to be con- sidered qreater than [— 3a], just as is considered greater than ~ a (44.) But to consider this case of addition under a mwe general point of view, let a be the positive and [—6] the insuls^ '^ negative quantity ; then Avhatever quantity the former repress the latter denotes tlie same kiiid of quantity in an oppv). direction ; that is, if a be a number of miles travelled west, [— ij is a number travelled east, and tlie sum or a + [— 6] is = a — 6 miles west, when az^b, or a ~ b miles east, when h-p^a. T' sum then is found in the same manner as that of a mid — A, ^^ b is a subtractive quantity. 56. Hence it appears that, to add a negative quantity, wlv; it be a common or an insulated one, is the same as to subl an equal positive (me. 67. The process of addition, in this case is called oJgel addition, and the sum is calied the algebraic sum, in ordo distinguish them from aaithmetical addition and arithnu i sum. The object of the convention in the 44th dei&nition is to idci . the rules for the calculation of insulated negative quantities tliose for common negative quantities ; and, so for as ; concen)ed,'it appears that this object is accomplished, rule maybe proved, by suppoeing a and b to 'represem c.^ li c-. number of miles, the one towards t)ie west and the other towards the east ; then the sum, that is, the whole number of miles from the original point of departure, is — the difference between a and b. But it is evident, from this peculiar relation, that if a and b represent any other like quantities taken in directly oppo- !?ite senses or direxstions, their simi is = their difference taken in tlie sense- to which the sign of the greater refers. Thus, if a is property and [— ^] debt or negative , property, the net profK^rty is just a — b, that is, a and [—6] taken together, or a -\- l~ b'] = a — b. If the difference between a and b be d, then when a-p'b, dis property, and Avhen a .^b, d is debt or negative property = [— tfj. All the possible cases that can occur may be represaitcd by straight lines, as OA, OA', reckoned in — f^ "T \ ^i — opposite directions from a point 0. If A' O C A OA — a miles travelled west, and OA' = b miles travelled east, then the final distance from O is OC, if AC ~ OA'. For a point inoA-ing from O to A, or west, and then from A to 0, or east, will finally be in the position C, but OC = a -^ b ; that is, a -\- I— b'] ~ a — b. If OA' exceeded OA, the point C won Id lie towards the left of O, or a — 6 would be so many miles east. Or OA may denote property and OA' debt, then OC = the net property ; or, conversely, OA may denote debt, then QA' denotes liroperty, and thus OC -- the net debt. EXERCISES. 2. 3. .. {yaxy — 'ian/ -uy 3a« - 46y - 8 - 2a2 + hly + 6 Sas + (Shy — 7 - 8ac -76^ + 5 - 'laxy _. 2ar +0 - 4 4. Mx - Zcz^ 12c/ — iaxz 4- 5ma:y :hix - 505' a:c -f- 8cz^ ",ax — 4c2^ r>ax -\- Ic^ lax - acs^ 'Saf -\- axz — Swixy - 14c/ - 'l\axz — \hmxy 16c/ + 3(rax« + 177wxy X>nf + 4arz — ISmxy - I2cf — \Sax2 — Smxy -f 10c/ +0 — 1 7mxy 6. 7. 2ct:c^ - Axyz +6 3Ca + x) - 4(^r + 2) 3ax* + 4a;?/2 - 12 ~ 2(a -^ x) -^ 5(z + 2) 5ax* - 8^2/^ +8 - 8(a + x) - 3(2 -f- 2) 25ax* — a:^3 4- 9 12(a + x) — 2(3 + " ax*- + 6xyz - 11 ™ (a + ar) + 5(3 + 2) 28«.r^ — ixyz +0 4(« + ^) + (2 + 2) CASE III, — TO ADD QUANTITIES THAT ABE UNLIKE, OR PARTLY LIKE AND PARTLY UNLIKE. 58, RiTvE. Add the like quantities, by the preceding rules, and 'ter them write the unlike quantities with their proper signs and oc'fficients. EXAMPLE?-. 1. 2. 3ax 4- yz\ Sax^ — 4c^- + 8 - ii/z -j- 5ax 5m 4- ^aa^ — 15 - Sd +2yz 7 -5ax3 4- 2cz'' 14ax4-5ax -2n + 2c3'-' 4- Box^ - ■ 7 — 3?/^ 3az' — 2m -j- 3?? 27ffa- -iyz — Zd ~1 Oax-* -f 3m 4- n ELEMENTS OF ALGEBRA. EXERCISES. hey — 4a3: - 12 + 8c - 3ry -f ^az - 6 - 2cj/ as - 18 + 8c 3. Sxy — ha + 6c — 8»i + 5 + a.y 12 - 2m + 7 5m 4- 2a;^ — 24 ^xy — 5a + %c 5(a ~\- x) — 3cy -id +24 — 8cy — i.'(a 4- x) — 3(a 4- a:) + 16c_?/ 4cy - 4c? + 24 • 3(a: + y) - 4c 4- G - 14 4- 62 - 3ax - ax-5ix + y) + S 8(x+y) + 5d 6(x -f y) — 4c 4- 5s — iax -{- od SUBTKACTION. 69. liLaE, Clianjifv? the signs of the terms of the subtrahend, or conceive them to be chaugeU, and then i)roceed as :n addition. In the following examples and exercises, the quantity in the upper line is the minuend, and that in the lower line i- ^^'<^' subtrahend : — KXAMPLES. 1. 2. Sax - 2/ - 5ax - 8/ 4c^=- 8xV4-18- 5(x4-2/) 8c;2''+ 5x^/4-12 4- 8(:c4-y) 805: 4- 6/ - 4cs* - 13a:*/ 4- 6 - 13(a: 4- 2/) In the first example by the rule, — 5aa: becomes 4- 5aa:, then by adding Sax and 5ax, tire sura is = Soar. Also — 8/, by the rule, becomes -\-. Sy\ and consequently 8^'^ and — 23^^^ = 4- 6/. 3. 4. 8(rx - 6/ 4- 10 Ucx'y^ - Sz^ 4- St? - 12 8/ 4- 3ax - - 8 4- 5s- - 4 - 82^ -'r G: - 13/ 4-10 4- c ISca-y 4- Sd ~ Bz In the fourth example, 4- 5z- and — 8a* become -- 5.s' and ; . QTRAOTIOK. 15 4- fii^ ; and these being added to — Zz^, the' su'ra is 0, for they ■■ ' roy each other. > proof of the rule of subtraction depends on the principle, he difference of two qiiantities is not altered by adding to, or cting from them both, the same quantity; therefore write all the 3 of the subtraliend after both minuend and subtrahend wiih vans changed (which will be subtracting the positive from, adding the necrative terms of the subtrahend to, both), and diiference will r.ot be changed ; collect now the terms of the i-ahend, and their sum^will be = ; therefore the sum of the in the minuend will be the answer ; but the smn is evidently ,tliO subtrahend with all its signs clianged added to the minuend, which is identical with the rule. • 60. When an insulated negative quantity as [— c] is to be [subtracted from any quantity as b, apply tlie principle of the proof ■on above, and thus we have, by adding c to both minuend and abend (6 + c) — (c — c) ; but c — c = 0, therefore 6 — [— c] r c; hence, to subtract a negative quantity, whether insulate or '■- the same as to add an equal positive quantity. The process of subtraction in this case gives the difference \" r c), exceeding either of the given quantities ; the difference lis therefore called the algebraic difference, in order to distinguish it from arithmetical difference. -■. distance between two points, A and B, on a line, having their distances from a fixed point O in it, namely, OA and s - '(OB - OA) = AB. Or if OA _JL_J__JL L_ md OB = a, then AB = a - ft. When A' O A B in the position A' towards the left of 0, the distance between d B is = (OB + OA'), or the sum instead of the dilferLHce c; distances of A' and B from 0. But the rule establi-lied makes A'B also in this case the difference between OB )A'^ for OA' = [—ft]; since it is measured in an opposite •tiou from OA, consequently, « — [ — ft] = a -f ft ; that is, is the algebraic difference between OB and OA'. 1. 3a — 2ft Gax 5a — 3ft iax 2. „4/ + 3-5(c + ^) - 6/ + 2 - 7(c + z) 2tt + ft 'lax 4- lif +14- 1(<- -t- z\ 2ra + x}-18+ 3(r-fy) ~ (a + :c) + 12 + 15(x> y) M't? 4-12 3(a + x) _ 30 - 12(x + y) 16 ELEMENTS OF ALGEiiRA. 5, 6. 4.rV - ocz -1- 8m - cz 4- 2x«/ - 4cz Gar* - 3c« - 18 + 6ot _12 +7a;2- 6 + 5 2x'f + %m - x'j-Zcz- 5 + (; 7. 8. — ax-ijz + 18 - S^' + 6« ISawV - ll.ryz 4- 3a — Garys 4- 7 — 2a — 5ay? ;^- hy ■\- iaxyz — 18 — 5w laanrre' + ha — 7 ADDITION OF QUANTITIES WITH LITERAL COEFFICIENTS. 62. Rule. Collect the coefficients of the same quantity (58), enclosing them in parentheses, aryl after the sum write the common quantity. In tlie following examples, x, y, ■<:, arc considered the quan- tities, and the other letters their coefficients : — examples. 1. 2. ax -r hti^z^ — az — ihy — 6 hx 4- cy V - 2cy — tz 4- 3 dsC — ey^s?' hz -\- dy -\- Zaz (a - 6 4- + (6 4- c - e>V 2az - (43 + 2c - d)y,- 3 In the second example, the coefficient of y is — (45 4- 2o — <^ or 4- (— 46 — 2c 4- d\ wliich are both identical, or — 4^ 4-^/(59). 1. 2aa:* - Zhz^ - hx^+ cz"' Zcx^ - 2ds? EXERCISES. 2. 2aT?/-4cs2 ^ 3 - 502^ - Uxy ~ 12 15 - hcxy - 2c.£- \(2a - h 4- 3c\i:2 _ 1 1(36 - c -v^dy ; 3. 4r2*-5y 4-4z2 3cy -2c2'4- d^ — bay 4" 2ey — es^ ( (2a - 36 ~ 5c>y ) 1- (3c 4- 5ay 4- Hi 4. a.rvi; 4- 3m/ 4- ' ~ 5ny' 4- 2hxyz — \ \ 6/n — 3cxy2 - 6 ^2cz^ - (5 - 3c 4- 5a - 2<;)?/) <(a 4- 26 - Zc)xyz l4-(4 + ificients {'62). EXAMPLi^ 1. 2. -ex %'Saj Sayz - Sbxy + 10 — 6cxi/ + 11 — 4ci/z •V - (46 + 3% ■ (6a + ic)yz - (Zh — Gc)xi/ - i ! T.'ie first example, the term 2cx becomes — 2cx^ and then coefficient of x is 3a — 2c ; and that of y likewise becomes ih — ?^. or — (ih + 3c). EXERCISES. 2. Cijr — 6y* 2axyz — Zhif — 5ac ex* — df — 2ac + 5aa;y^ + 4cy'^ cy- - (h- d)y^ - Zaxyz - (36 + 4c)^2 - 87" 8. 4. ^hxy — Sc^^^ + 7 hax — 4fey + Srs - 8 + 3cu;y - %dz" — 2dy + 36x + bhy (6-' (5c - ScO-t* + 15 (.5a - 3«)a; - (96 - 2% + 3,- MULTIPLICATION. CASE I. — TO MULTIPLY SIMPLE QUANTITIES WHEIf THE MULTIPLICAND A?II> MULTIPLIER ARE DIFFERENT LETTERS. Si. Rule. Write the quantities in close succession, as in a-woid^ and in alphabetic order, prefixing the product' of the numerical coefficients to that of the quantities. When tiie signs of the two quantities are like, the sign o^ *^ ■■ product is plus ; and when the signs of the quantities are i ; that of the. product is minus. Hence like signs gwe plv nnlike sigris gwe mimis. EXAMPLHS. 1. Multiply iaxy by 3bz 4(1X1/ X Sbz — 12abxyz 2. Multiply obx^ by — iiai/J* 5bx^ X - 3a^? = - Uoha^j/^ 3 4 ,!. Mnltir ]v - (XX* by — --cy* 4 , 3 , , -4.''' ^ - ^f = -^acxY The reason of tiie rule for multiplying literal f; evident from (31); and that for the coefficients is ;, from (28). In regard to the sign of the product, it is e\ • the signs of the two factors are plus, the product For a X b means that a is to be takexri times ; !....„ ... added b times ; and, therefore, the sum is positive. Again, if — a is to be multiplied by b, this merely mean — rt is to be repeated b times, or that a is to be suLtractv i ij tunes ; that is, a, repeated b times is to be subtracted, or ab is to be siibtracted ; hence ~ a X b = — ab. ]f a is to be multiplied by — b, this merely meaii.^ be subtracted b times ; and hence the result must bo in the preceding case ; that is, — ab. If — a is to be multiplied hy — b, and — * a be c/^v > common subtractive quantity, then — a X — ' taken subti-actively is to be subtracter! b times, . intelligible idea. But if — t/ be an insulated ik,_ . ..- .. ....^. , then [— a] subtracted once gives + a (tJO); and therefore f — a] subtracted b times gives + rt^; that is, [ — aj-X ~ b ~ -\- ab. MULTTPLICA'JJ.0N. EXERCISES. ^ • ■ 3aJ by 4c(f, . . . . = \2ahcd. ~ Baxz by ohy, . . . z= — 24-abxuz. IGas^' hr - 5bcxif\ . . = - 80abcx/z'. ~ 5ax* by - Ucz\ . . == 20abcx*z\ J2c/by -5aA»xV, . = - GOaPcx^ifz*. : .. - ?.65;Mry- - 12r.'.:y, . = SGabc'x^f. CASE ir. — TO MUL-CIPLT SIMPLE QUANTITIES WHEN THE MUITIPLIEK AND MULTIPLICAND CONSIST OF POWERS OF THE 8AMS LKTTERS. 65.' Rule. Add together the exponents of each letter for its exponent in the product. 1. Multiply 3aV^* by 2ax*y 3a-z»y* X 2ax'i/ = Gc^xhf 2. Multiply 5c2/« by - 3cyV BcYz X (- 3cyV) = - 15c8j/V 3. Multiply 2a'»a;''»-'» by 3a''".r='»-« 2a"aH»"'*» X 3a2".r*"»-* = 6a»'U*'""^ rule may be proved thus ; — a^ X a' = aa X aaa (31) = aaaaa — o\ ^ Or, generally, a" X a" = aaa , a being repeated m timos ittto aaa , a boiner repeated n titnes ; that is, aiuia .a bein repeated (in -j- n) times ; but the product aaaa in which a : repeated (m + «) times, or as often as is denoted by the sum c m and n, is represented by a"*'"^ (31) ; therefore a"* X a" = a"*^". EXERCISES. 1. Multiply oa*xz' by 4a'.rV, . . , = 20a^.r"c ". 1?. ... -7(^bfzhy3abh/z\ . . = -2la*bY^' 3. ... - IGcVa;* by - 5c'a;«, . . = 80c*.rV. 4. ... 8a2"x'"y='''* by - 5a'''x'"^'^^ = - iOa'^ar'"/'-^ .;asE III. — TO MULTIPLY SIMPLE QUANTITIES WHEN THE MULTIPLICAN f AND JiCLTIPLIER CONSIST PARTLY OF DIFFERENT LETTEE8, AND PARTI/ OF POWERS OF THE SAME LETTER. Q>Q. Rule. Find the product of the quantities denoted by differert letters, by the rule in case i. (64); and that of the rowers of the same quantity, by the rule in case h. (66), writing liie Ictvers in alphabetic order. 20 ELEMENTS OP ALGEJJRA. EXAMPLES. 1. Multiply Sar^y? by 2c/ 2. Multiply 5cx*/ by — 2ct^z hac^f X (- 2cy^z). = - I0Ka. .. Multiply 2a* - ■Zax + ^x'hj 3a^ 2c?- 3a* -3ax + hx" Qa'- • Qa^x + 15aV Here 2a' is multiplied by 3a*, by the former case, which 6a* ; then ~ 3aT is multiplied by 3a*, and then Ijx^. T in .algebra to begin multiplication at the left hand, . also in addition and subtraction. It is sometimes co represent the product in this case thus, (2a=^ - 3«r -i- P^r^-^-r.^ - r.n'' ._ O--^.- -x- i w,' MULTIPLICATION. 21 S. Multiply 2ax - Sbf by Sab. 2ax — 36/ Sab ea'bx-daby 8. ... 5a'Px-2bf + Saz^hy-2afz\ Ba^^x - 2bf + Zaz^ - 2af^ - lOaWxfz^ + 4:aby'z^ - GaYz^ EXERCISES. 1. Multiply 2/ - 3x1/ + x^ by 4:X7/, = 8xy^ - 12xY + ix^i/. 2. ... 6ax^ -4aV + Ga^x hy 2a'^x', - 12aV - 8aV + 12aV. 3. ... 3ax> - ha'xy^ + 3:^^ by - Gx'y, = -18cw;y+30aV/~ ISx'y. 4. ... 2ax2 -3by* - 8x^}iy - 5abx, = - lOa'^bx^ + IBab'^xy* + 40a&x*. CASE V. — TO MULTIPLY QUANTITIES WHOSE MULTIPLICAND AND MULTIPLIER ARE BOTH COMPOUND. 68. EuLE. Multiply the multiplicand by each term of the multiplier, as in case iv., and collect the partial products by the rules of addition. The reason of the rule is evident, from the explanation given in the last case. examples. 1. 2. Multiply 20^— Sx^ by Sa^ - Sax. Multiply a + a: by a + a:. 2a2-3x2 a -\-x 3a^ — Sax a -\- x Ga* - 9aV a^ + ax — Ga^x-\-^ax^ ax +x^ 6a* — Ga'^x - 9a V + 9aa;' a^ + 2aa:+ a:^ The first partial product (ex. 1), that is, the first line of the 22 ELEMENTS OF ALGEBllA. pr<)(luct, is obtained by raultiplying the multiplicand by Ba\ aii> second line by multiplying by — Sajc; and lastly, these two p:. products ai^ added. The first term of each partial ^uAv commonly placed under the term used as multiplier. 4. Multiplv .. . . J a — X. :dv:]>\u]v a -\- x by a a ~ X a — X u^ — ax — ax ■\-x' a* -f- ax -- 2ax + 3f i ultipiy cr -f ax f .rj by o. - x. Multiply a^—ax -\- a* 4- ojci-f x' a — X 7? a^-ax + x" a +x a« 4- a^x 4- ax^ - aV - ax^ - a^ — a'x + ax"^ 4- a'x -ax^ + x^ a* + + - or a' - r* x' „'+ Jf. +x^ or a' + x' -r Multiply a^'* — 2a''.r" 4- x*" by "a" -}- 2x''. a'" — 2rt"ar" 4- a;-" o« 4 2a-'' a'" _ 2a««x'» 4- aV« -4- 2a-"a:'» — ^o"^^" 4- 2x*» o'" - Sa^a** 4- 2ar*" ^-' 8. ^Multiply a" 4 a""^x + rt"'"x^ 4- ... aa;"~^ 4- ar" by a G9. As n has no particuliir value, the in torn v' the multiplicand caimot be supplied till such a to it ; vet the manner in vThich these terms are ib i MULTirLICATIOK. 23 tenn would be a""'x^ ; the last but two vroiild ])e d^rT-'"- • -be other terms. _ a^'ar — tf'"'^* — ... — a^a:""* — oa*" — x"** J L tlie two partial products, all the terms intervening between Iirst and last destroy each other;; hence the product is a""^' —.>;""'. Uier n be an odd or an even numl)er^ and therefore whetbc-i - 1 be even or odd. But the quantity a" — dT'^x -f- a"~*a;* ~ ... ■ " has its last term positive ority when n is an even number, and ig multiplied by a + ar, it gives a""^^ + x""^^ where n + 1 is of rse an oc/r/ nmiiber. * V'hen n is an odd number, the last term is negative; thus, - a^'^x -f- a'^'^x' ~ ... — t"; which being multiplied by X, it gives a""*^^ — 3:"*^ in whicli 7i + 1 is of comrse an even .iiber. »y giviiig n particiiJar numerical values, the correspond im ■ducts will be easily found by substituting these numbers for / . he above products. vnco a and x in these examples may stand for any qnaTi- ^;5, they contain the proofs of the five following theorems, h it is important to remember, as a knowledge of th( a 1 very much facUitate the solution of many of the follovvri^i.,- /cises ; — ' iiF.OKJSM I. The square of the simi of two quantities is equal he sum of their squaix^s increased by twice their products .; >ved by example 2. EonEM II. The square of the difference of two quantities ual to the sum of their squares diminished by twice their uct. Proved by example 3. TaEOREM III. The product of the sum and difference of twv . It i ties is equal to the diSerence of their squares. Proved b _ unple 4. 50BEM IV. The sura of the squares of two quantities sod by their product, and multiplied by their diflerenee. iual to the difference of their cubes. Proved by example 6. '-' '^nRM V, The sum of the squares of two quantities d by their product, and multiplied by their sum, is lie sum of their cubes. Proved by example G. •24 elemk:nts OF \i ,.r.!ii..v. EXERCISES. 1. Multiply 2o.^ ~ 4aar + 2o!^ by 3a — 3x, 2. ... 3a* + 3x* by 2a* - 2x*, , . r- 6a« - 3. ... 2a3 -h 2a"a:+2aa^-l-2ar' by 3a - 3.t, = Ga* - 4. ... 5a* + 5a.r +6x^ by 20^ - 2ax, ~ lOo* - 1* 5. ... 3a:== 4- 3xy + 3/ by 2x2 - 2/, = 6a:* 4- 6x'^ ~ ^xf - 6. ... a* ~ 4a^a: + 6«V — 4aa^ -;- a:* by a^ — lax + --= a« - 6a*a;+15a*a^-20aV+15aV- Gax* -f 7 -• +4yby3x-2y, . . = 15ar* + 2xy - 8. ■- + a;^/ - / by X - y, . = x' - 2^^'^ -f- 9. ... 2x4-%by2x-4y, . = 4a--; 10. ... x^ — a;^ + 3'^ by a: 4- y, . = ur^ -f n. ... a^ - ax — 6.r* by a + 4ar, = a' -4- Stt'i - lOax^ ~ l 1 J . ... ^a"^ -t- lOaa: - 4.-^ by 4a — 2ar, = 24a' + 28a*x-86aa:2 + 13. ... a'"' + x*" by 2a2« - 2x^", . . = 2a*« - i 14. ... ({'" -\- d^x"" +x*»by 2a»-2x'», . = 2a^'»~ I' 15. ... «*" - a^x» + aV" ■-'•' ' - 3a" + 3x", =r 3a'» - ." iC. ... a" — a""'.r + a^'-x" - ... 4 a.r"~^— x" by 2a --)- = 2a«^^--. If the numerical exponents in any oi" the preceding exercls- multiplied by ?«, examples Avith literal exponents will be for: the products of which will be the same as. those of the for with their exponents multiplied b}" n. MULTIPLICATION BY DKTACHEB COEFFICIENTS. 70. When the powers of the letters both in the multiplicand , multiplier either increase or decrease uniformly in each term, rue multiplication may be performed with the coeflBcients alone, and The letters afterwards supplied ; for, the power of the letter in the first term of the. product will be the simi of its powers in tlie first t«rms of tlie multiplicand and multiplier, and the powers in the jMuJ.Tli-LICATION '. !ict will increase or decrease in .ir. If some of the powers must be supplied by writi . Multiply X* + 2x^ - Sx" 1+2-3 + 4+2 coe^cie^ts 1+3 + 2 -SfT Q-jyrnf' sn each 1+2-3+4+2 3 + 6-9 + 12 2 + 4 - 6 1+5+5-1+ 8 + ]^+ I .since x* (the first term of (he first term of the multiplier), V + 5.1^ + 5x* - x» . Multiply a» + 2ab^ + 36^ by a* •f the cx)eflBcient of a' in the mVtiplicai]^ riplier are both zero ; hence, by thert 1+0+2+3 1 + 0-3 product. Itiplied by of a in the 1+0+2+3 -3-0 6 -9 1+0-^1+3-6-9 8 a' - a^h" + 3a'6» - Qah* - W. coefficient of a* in the product being zero causes that term to pear. e preceding exercises can all be done in this way, and should Vrought again as shewn above. EXA3IPLES WITH LITERAL COEFFICIEKTS. i . In these examples, x, y, z, are considered as the quantities, the other letters as their coefficients. i . Multiply x' — ax + 6 by a; - c. X' — ax X — c 3C^ — ax' + bx — cx^ + OCX — be x^ - (a + c>" + (_b + ac)x — be 26 ELEMENTS OF ALGEBRA. ?. Multiply T^ — axy -t- hf by y — cy. x^ — axy + jy* X - cy - ' - {a + c)x^y + Q, 4- ac>/ ~ ic/ KXSRCISES. J . Multiply x^ — pa? -{• qx \yy x — r^ ~ X* — O 4- r)x^ + (? + P?-) '/. ... X- — aa: 4- 6 l)y .r — 1, = x' -- (a 4- l>-2 4- .'■" -^ '' ,.r y, . . o:^ — aa:^ 4- 6x — c by x" — x 4" ^ : =: X*- (1 4- a)x"' 4- (1 4- a -i- 6>' - (a 4- 6 4- c)x^ 4- ;^/' r c;- 4. IM iply 2ax — 3cy by ix 4- dy, = 2a6x2 - (3&C — 2arf)xy - 3. i_ > . ^.w.,. pound quantity be arrauged according ^" descending powers of one of the letters ; that is, ^vitli !• power first, aud the rest in order, this letter is called t; quantity. Thus, in 3x* — 2x^ 4- x"^ — 4x 4- 5, x is the leading quantity 73. A compound quantity which contains the second dimen of the leading quantity, is a quadratic. Thus, x^ — 4:3; 4- 3 is a quadratic trinomial. Since (x 4- a)(x 4- 6) = x* "4- (a 4- ft)x 4- ai, it follows thai product of two simple binomial factors is a quadratic trncT and the coefficient of the simple power of the hading 'i the sura of the second terms of the binomial, and the ; is Their product. '.Fhus — (x 4- aXx~ 6) - x'' 4- (a — b).. (x — a)(x 4- 6) ;= x^ 4- (6 — a)x ~ ub ■ (x — aXx — -6) =s X* — (a 4- b)x 4- ab Hence, when any quadratic trinomial occurs, which is the pro' of two binomials, with integers for their second terms. iViPv generally be easily decomposed into their factors. '] hition is frequently of great a^lvautage in reducing fin their simplest forms ; and may also be employed in m;;tiv , for finding the values of the imknown quantity in Quadi liquations. Decompose x^ -\- 7x -\- 10. tliis example it is easily seou that 7 = 2 + 5, and 10 = 2 X i lore, ..x' + 7x + 10 = (x + 2)(x + 5), Decompose ar* — 3x — 40. .'re — 3 = — 8 + 6, and — 40 = — 8 X : t/)r rciorc x* - 3x - (x - 8X-r + 5). 1. Dv EXERCISES. ,'^ + 10x+ 24iatot\v- simple binomial factors, = (x+ 6Xx-i 4). x^ + 9x + 20 into two factors. - (x + 5)(x -1- 4). -r X- 20, . . = (x -j- 5)(:^- - 4). j:-^ - X- 20, - (r - 5V 4 i\ X- 4- X - 2, . . = (x + i? .■ - !> x^- 2x+ 1, - (x - I /.. - : . x^ - 13x + 40, . . = (x ~ H)(.r X* - . X - 132, ^ (x-12)(x-hll; x^ ~ 3x - 40, . . =: (x - 8)(x + 5^ .•2 _ 5x - 104, = (x - 13Xx + 8) .x''+ 24x + 135, . . -- U 4- 9)(x + 15) DIVISION. JDE SIMPLE QUANXrnFS WHEN THE DIVIDEND AND DIVISOR ARE DENOTED BY DIFFERENT. LETTERS. 74. Rcx't:. Write the letters of the diA-ideud and of the divisor : the form of a fraction, makhi^ the divisor the denominator, ;L!l prefix the quotient of the coefficients with the proper sign. When the divisor and dividend have lih. signs, the sign of th^ quotient is plus; and when the ^gns are unlike^ that of t\\- quotient is luinus. The quotient of the coefficients will be either a whole numbci • ;r a vulgar fraction ; for it is not usual in algebraic division to reduce this quotient to the form of a decimal fraction. 28 ELEMENTS OF ALGEBRA. EXAMPLES. 1. Divide 12acy by 3bx. ^ Bbx hx hx When the coefficient of the dividend, divided by that of the divisor, does not give an integer for the quotient, the coefficients are treated as the terms of a vulgar fraction. Thus, if the coefficients of the divisor and dividend be 5 and 4, their quotient 5 is expressed by j. If they are 12 and 8, it is expressed by 12 3 — or -, by dividing both by 4, as in reducing fractions to their simplest form. 2. Divide 4aa:' by 126/. 3. ... - 8cj* by 12. fri '., Write the divisor and dividend in the form of a .^tion, as in the last case ; then take the difference between the >onent3 of the 'powers of the same quantity ; this will be the onent of that txuantity in the quotient, which musfc be placed ve or below the line, according as the greater exponent belongs lie dividend or divisor. ■' the same quantity occur in the divisor and dividend, it may anoelled from both. •s iu multiplying powers of the same quantity, the sum of ■(• exponents is taken, so in dividiiiff^ their difference is taken. EXAMPLES. vide '-2 by a^'»^'. ^5m-2-(2n»+l) __ ^5m-2-2m-l _. Q^m-S qj. ^'Cm-l)^ The rule may lie proverl thus : — 1. When the exponent of the quantity in the dividend exceeds that in the divisor. 76. The dividend \i just the product of the divisor and quo- tient : and therefore if a' is to he divided by a*, the quotient must evidently he a*, or — j = a* ; so — = a^ ; aiid generally — -- = a"*i for a"* . a" = a^'^. So — = a"'"", for a*"'" . «" — a^. a" a" In this last example, it is understood that mz^n. 2. mien the exponent of the quantity in the dividend ^ than that in the divisor. Before establishing the rule in this case, it will be necc previously to prove this propo.sition : — 77. The divisor and dividend may be both multiplied o? both divided by the same quantity without altering the quotient. I^or if N — the dividend, D — the divisoFf and if in, n, and d, be such quantities that N = mn, and D ~ mcL then -^=^ -r For if -. Da 'a — q, then if n be multiplied by 2, the quotient will evidently be 2q ; but if when n is doubled, d be also doubled, tlje< 2n -r- 2d will give just the same quotient q ; so 3n -r- 3c? will give the quotient q ; and, generally, mn -r- md will give for the quotient xi. 1 ^1. X • "*'* -^ TT N n the same value -q ; that is, -r— r = o', or -=r = <7. Hence -k = t ^ md ^ D ^ D d m * ■ •* ^ 4-1, ^" ^^ • A 1 '^ ^ 4.1 " Tliat is, if -J = q, then — j or -^ is = 5-. And 11 -~— q, then - is also = V. where « = — and d — ~-. m m a* 1 X o* 1 From this principle, it is evident that — .- =: -z = -- : for the quotient of 1, divided by a*, is the same as that of a* divided by a', since, if these two quantities be each divided br the snmo quantity a*, the new dividend and divisor av a* 1 , ,, a" 1 X a" 1 -— = — J ; and, generally, — r- = — — . 78. Th6 proof in the second case establishes the rule that — If DIVISION. 31 .me quantity be contained in the divisor and dividend, it may he EXERCISES. Divide r® by .r*, . . . . . . = 3^. " by x", 12aVby4.:% . . . --^ - Zz. c" ' - le^y by - 8cY, VJn'^-Jz'' by - IGaVz" - 24xyc" by 32a^y',-- 4y 7. ... Ga'"* by Sa^*", . :: 2a'». 8. ... -■a'»-> by 4«"•-^ ... 2a^ 9. ... a2'"j:"» by a'»x^'», -- -^. 10. ... ia^'^-r by Sa^'V, ":Z.^^ or ''-['' 79. It is sometimes couvenient, and it extends the meaning of Kponent, to take another view of the division of powers of the tome quantity. Instead of suUtracting the less cxiwnent from he greater, if the latter be talcen from the former, the exponent «f the quotient will be negative ; thus, a' ~ a« = ^^ = a'-" n= a"', but ^ = -V I a^ ' a« a^ ' md, similajly, it is evident that «''-^«'' = Jo = «*'-"-«-M^ut^-; = l: a* I An ilio former ease -^ is equal both to a'"^ and to -5-: and in the , a^ 11 ilatter case, -^ is equal both to a~* and to -^ ; hence a"' = -s-, ; <^ a a' 1 a'"' a'™ 1 a''' ""' -"T? 3,nd similarly, -r- = a^"'"*™ — a""*, or —r- — — : ^ rofore, generally, — = a""*. '• i it might be similarly shewn that a"* =r . 3-^ ELEMEA'TS OF ALGEBll. , 80. These negative exponents are obtained by a process di from that given in the r\ile — namely, by subtracting the v: exponent from the less ; and as no definition was formerly of quantities with negative exponents, they can have no mt further than that they ai-e merely another mode of exp; other quantities, the meaning of which is known. Thup. merely another mode of expressing — ; or a~^ is anotb for -:r or — ; so — is another mode of expressing r. or aa a^ another mode of expressing a- or att. Tliis mode ot exj quantities is sometimes found to bo convenient. It also ev pro'»v;- tliat — Any quanlittj may be taken from the divisor (/>:■''-■',■■!, or conversely, hy changing the sign of its expomnt. if ;]ie rule be applied to the case of equal exponents, Ji remai'kable mode of expressing unity is obtained. Thus, a? ■ . a^ a» 'a" Therefore a" = J. ; hence — Any quantity wii.h the exponent ? to unity. Thus, _ = -„or=:a«- ,,6 or --r-^'^tV* z= a^'*^ '-= or, a* The preceding exercises may be solved by this metho*' for the fourth, The result here found, namely, -^ , may be easily obtai divitiing the divisor and dividend by — Sc"/, for tiiey j ; written thus, - 8cy X 2/ - 8cy X ^ ' ii ad if the common factor — Sc"/ be cancelled. i lu. • V, as above j hence-^ &L The rule for this case of division mav also bo exiir in (78). DIVISION. EXAMPLES. 1. Divide 12aV by 8aV. 12aV _ 3a^ X 4aV _ 3a' "8a*x^ ~ 2x X ia*x* ~ 2x ' In this example, 4,a*x* is a factor common to both divisor and dividend, which may be expressed thus, 12aV = 4aV X 3a\ and 8a*x^ = 4aV X 2x ; and hence, expunging the common factor 4^a*x*, the quotient is found. To illustrate this method, the former examples may all be solved by it. CASE III.— TO DIVIDE SIMPLE QUANTITIES WHEN THE DIVIDEND AND DIVISOR CONSIST OF QUANTITIES DENOTED BY DIFFERENT LETTERS, AND OF POWERS OF THE SAME LETTER. 82. Etjle. Divide the quantities denoted by the different letters by the rule for case i., and the powers of the same quantity by any of the rules in case ii. EXAMPLES. 1. Divide 12c'dxh* by - Sbx~z\ -=rsb^ = - -2bz^^ ^^ canceUmg 4:rV. 27 Divide Qa^x^ by bhxy^. W -7 — 2 = TTT) ^y dividing by the common factor x. 3. Divide - IS^y^' by 9aV^«*. \%x^tFz^ 2y ~9^t^ "^ "~ ^' ^^ <^^^^x — 2ay. Here 12ax^, divided by 4ax, gives 3a;, and — Sa^xy, divided b 4aar, gives — 2ay. 2. Divide 24xy - 15aV by ISaV. 24:gy - 15aV _ 4/ _ 5aV ISaV ~ 3a2 6 * 3. ... 6aV- 8a V + lOaV by 4aV. 6aV - 8a*a:* + lOaV 3 „ o , . Sx^ __ ^__2a-.^+-^. 4. ... - 24a:y + 18aV - 15aV by - 12aV. - 24a;y + 18aV - 15aV _ 2/ Zx 5aV - 120^^0;=* ~ a^ 2 + 4 * 5. ... - 12c V - 8a;y + 7a:«s by 3ax. - 12cV - Sx^j/^ + Ix^z _ _ 4tc*x 8xy 7o^z Sax ~ a 3a 3a * 6. ... 302"* - 5a"'a;'» 4- Sx^"* by 2a"'x"'. 3a^"' - 5a"'x"' + 8x^"' _ 3a"' 5 •ix"' 2a'»x"* ~ 2a:"-' 2 "^ a"' " 7. ... Ga*"* — 8a-"'x™ + 3a"'a;*"' by 2a'^"'a:'". Qa*r^ - Sa'^"'x"' + 3a™a:*"» _ 3a^ _ 4 i ■^^'"" 2a-'"a;"' ~~ a;"* 2a'" ' DIVISION. 35 8. Divide 2:c'" - Sa;'""' + 5a;'""2 by Sx""'^. 3^m-2 "~ ~ ~3 ^ "^ 3' Here x"*, divided by ar"*"^, gives a;'""^"*"^ = x'"""''^^ = x^. EXERCISES. . 1. Divide 12aV - Sax' + So' by Sa^, . = ix^^^ ^^ 2. ... - Say + 12ay - 9ay by obf, = _8ay 12ay Oa^ 3. ... 16a'x^ - 3aY by 4a2a;*, . = 4aV - ^^f- 4x* 4. ... 8a2-24cy2''by-8ay, . = _, J_ + ^* 5. ... 12a V - 5a^xY + Sa^/'z^ by - 4ay, 6. ... 11a* - 12:cy + 13z^ by 3aV, = ll^_4xy 13^^ 32"^ aV + 3a^ 7. ... 3aV/;s2 _ Q^Y^ _j. 3^^4^2 1^^ 9a2:iV, = £_2ay _2_ 3x 3a;^« "^ 3ax 8. ... 2a:y - 3xV + 4/;s2 by 5a:y^2, _2 3 4 6^2 5/ "*" 5x2 ea^^^c^"* - 4a2'"a,'3'» by Sa'^o;™ = 2a2'»a;'» - ^^^ 3 10. ... 3aa:"* — ibx"^'^ + Scx^-^ by 2x"'-\ 11. ... 16a;"'- ix""-* 4- 6a;'""* by 4a;'""'', = 4a;^ — a:^ 4- -?- 12. ... 4a» — 3a;'" by 2a"'x"', . , = — — a;'" 2a'^ 13. ... ea*"* - 6a'"'x'rn ^ g^^n^^m . =^_^ x*" a*"' 14. ... aa;*'» - ta;"'" + cx^*" by dx^"", = — ^^ - 4- ~ ab ELEMENTS OP ALGEBRA. By applying this case, compound quantities may often be reduced to more convenient forms ; thus, being first divided by x^, which is evidently a common factor of all the terms, gives the quotient a — h -\- c; and this being again multiplied by x^, the product must be the original quantity, which is therefore =: (a — 6 + c)^"» in which the multiplication is merely represented ; thus giving a form of expression much more concise than the given one. So 2ax'^y — dbx^y -f 5cx^y = (2a — 36 + 5c)x'^y ; x^y being a common factor of all the terms. Also, ^ax^f — Bmxy"^ + Qnx^y = i^axf' — hmy -\- Qnx^)xy ; xy being a common factor. 84. Hence, to make such transformations, the terms of the given quantity must be divided by any factor that is common to them all ; and the quotient then represented as multiplied by this factor. When there is no common factor, the form of the given quantity cannot be thus altered. EXERCISES. 1. Transform 2ax — 3cx, . , . . = (2a -r- 8c)x. 2. ... 4:mxy —3nxy -{- pxy, . = (hn — 3n -\- p)xy. 3. ... 2aca^z — 5abxz, . . = (2cx — ^b)axz, 4. ... Zax^yz^ — hhx'^y'^z'^ -{■ 2cx^yz'^, = (3a - 56/ + 2cx^xhjz'. CASE V. — TO niVIDE QUANTITIES WHOSE DIVISOR AND DIVIDEND ARE BOTH COMPOUND. 85. Rule. Arrange the divisor and dividend according to the descending powers of some quantity common to them both ; that is, so that the term containing the highest power of this quantity may stand first, and the rest in numerical order. Divide the first term of the dividend by that of the divisor, the result is the first term of the quotient. Multiply the divisor by tliis term, and subtract the product from the dividend. Coiisidering tlje remainder as a new dividend, arrange its terms, and proceed as before ; and thus continue the process till there be uc '\;irkuinder, or till the exponent of the leading quantity in the remainder be less than its exponent in the first term of the divisl \ DIVISION. 37 1. Divide 4a^ 2a EXAMPLKS. Sax 4- 4^2 by 2a — 2x. - 2x)W - Sax + 4.x\2a - 2x 4:a^ — 4ax — 4aa; + 4x^ — 4:ax 4- 4:jr^ Here the dividend as well as the divisor is arranged according to the powers of a; then 4a^ divided by 2a, gives 2a for the first term of the quotient ; and after multiplying the divisor by 2a, and subtracting the product, the first term of the remainder is — iax, which being divided by 2a, gives — 2x, the second and last term of the quotient ; for there is no remainder after multi- plying the divisor by — 2x. This process of division proceeds on the obvious principle that the dividend contains the divisor as often as its parts contain the divisor. Now, the dividend contains the divisor 2a times, with a remainder — iax + 4:x^, and tliis remainder contains the divisor — 2x times, with no remainder ; therefore the dividend contains the divisor exactly (2a — 2x) times. 2. Divide 4a* - 4a;* by 2a^ - 2x^. 2a^ - 2x2)4a* - ^.x\2a' + 2:^ ■ 4a* - 4aV 4aV - 4a:* 4aV _ 4x* 9a« - 9a;« by Zc? + 3x'. 3a^ + 3a;^)9a« - 9a:«(3< 9a« + 9a^a;^ 3a:» 9a^a:' - 9a:« 9aV - 9x« ^a^ - 6a;« by 2a- - 2x^. 2a^ - 2x2)6a« - 6x«(3a* + 3aV + Sx* 6a« - Qa*:x^ Qa* 6a;« Qa^x* - 6a;« When the divisor and dividend consist of a considerable number 38 ELEMENTS OF ALGEBRA. of terms, it is convenient to write the divisor over the quotient, as in the following example : — 5. Divide 6a* - 15a^x - 8ax^ + Sx* by 2a2 - Sax + x". 6a* - 15a^x - 8ax^ + 6x* (20^ - Sax + x" 6a*- 9a3x+ 3aV \Sa^-Sax-6x^ — 6a^x- ~ 6a^x-\- 3aV 9aV - 8ax^ - Sax^ — 12aV 12aV - 5ax^ + + \8ax^ - 6x* 6x* - 2Sax^ + 12.r* 86. As the remainder contains only the first power of a, whik the first term of the divisor contains its second power, the divisior cannot be carried further ; therefore generally, when the exponeni of the leading quantity in the first term of the remainder is less than that in the first term of the divisor, the process of divisior cannot be carried further, so that in such a case it terminates with a remainder. 6. Divide 4a« - 4:r« by 2a' - 2x^. 2a^ — 2a;2)4a« - \x\2a* + 2aV + 2x* 4a« - 4a*x2 ^a*x^ — 4a:® ^a*x^ - ^a^x* 4aV - 4.r« 4aV - 4a:« 87. By division it can easily be shewn that "^^ "•" ^^ = x^-^ + x^-'a + x^-^a^ + ... + xa^'"" H — x^-^a + x^-^a^ — ... + a:a"'=^ + a""^ + a: + a The intermediate terms can easily be filled up when a particula value is given to n. The form of the quotient being now known, let it be representee by Q, and the remainder, if any, by 22, as above ; it is required t( find under what conditions there will be a remainder, and it value when there is one, without performing the ; -i'->'>'i DIVISION. 39 Writing Q for the quotient, the above expressions become — «6 "r X — a X — a , x'^ + oP- ^ , R and — = — = Q,-\ . X -\- a X -\- a Multiplying both sides of the first by x — a, and the second by a; + a, they become respectively x" ± a" = (x — d)Q + R, and x" + a" = (a: + a)Q + R. Since the above are true for all values of x, put a for x in the first, and — a for a; in the second, and they become respectively a"^ ± a^ = (a — a)Q -\- R, and (- a)" + a'' = { — a + a)Q + R. Again, since (a — a) and {— a -\- a) are each equal to 0, the first terms on the second side disappear ; hence, ^n 4. ^n _ Jl^ and (— a)" + a" = i?. Now in the first, whether n be odd or even, taking the upper sign, R = 2a" ; but taking the under sign, R = 0. Again, in the second, if n be odd, (— a)" = — a" ; but if n be even, (— a)" = + a" ; hence, taking the upper sign and n odd, R = 0; but taking n even, R = 2a" ; if now in the second the under sign be taken and n odd, R= — 2a", but if n be even, R = 0. From which the following theorems are derived : — Theorem I. re" -f- a" divided by x — a leaves a renaainder of 2a", whether n be even or odd. Theorem II. x" — a" divided by a; — a leaves no remainder, whether n be even or odd. Theorem III. a:" + a" divided by a: + a leaves no remainder if n be odd, but leaves a remainder of 2a" if n be even. Theorem IV. a;" — a" divided by ar + a leaves no remainder if n be even, but leaves a remainder of — 2a" when n is odd, I n p e I 7, exercises. Divide 4,0? — Soar + ix"^ by 2a — 2a:, . . =r 2a — 2a:. ... 8a* - 8a:* by 20" -23?, . . . =40^+ 4x-. a' + 5a"a; + 5aa;^ + a:^ by a^ + 4aa: + a:^ = a -\- x. ... x^ - 9x2 + 27a: - 27 by a: - 3, ^ ,^2 _ g^ _^ 9^ ... X* -y*hy x-y, . . = x^ + x^y -\- x/ + if. a^ — x^ by a'^ + ax + x^, . . . = a — x. x' — Bx'^y + 3x/ —y^hy X — y, = x^ — 2xy + y'. 40 ELEMENTS OF ALGEBRA. 8. Divide a" -5a*x+10aV-10aV4-5a^-'-a;'bya^-2aa;+a^ = a' — Sa-x + Zax^ — x^. 9. ... 6a:* - 96 by 3x - 6, = 2a;3 + ^x^ + 8x + 16. 10. ... 4a« - 25aV + 20az^ - 4x« by 2a3 - 5 aa^ + 2x^ = 2a^ -h hax^ — 2x^. 11. ... 6a* + 4a3a:-9aV-3ax'+ 2a;*by2a2 + 2ax-x2, = 3a2 — ox — 2a;2. 12. ... ar« - 3xy + 3a;y - y^ hy o^ - ?>x^y + 3a-/ - y^, = x^ + ^x"'y + ^x7f+y\ SYNTHETIC DIVISION. I. When the powers of the letter or letters of the divisor and dividend increase or decrease hy any regular law, the coefficients of the quotient may be obtained by observing that in performing the ordinary process of division, the coefficients of the terms of the divisor are multiplied by the coefficients of the quotient, and the products successively subtracted from the coefficients of the dividend ; by which means, when the quantities are divisible without a remainder, there is at last nothing left. If now the signs of all the terms of the divisor be changed from plus to minus, or from ininiis to plus, and the products in the corresponding columns added to the coefficients of the dividend, the sums will all be zeros ; but if the coefficient of the first term of the divisor be omitted in performing the operation, the sums will be the coefficients of the quotient, if the coefficient of the first term of the divisor be one; and the coefficient of the quotient multiplied by the coefficient of the first term of the quotient when it is not one ; in which case divide by the coefficient of the first term of the divisor, and the quotient will be the coefficient of the quotient. The powers of the letters in the quotient always follow the same law as those of the dividend and divisor ; and the exponent of the first term of the quotient is always the difference of the exponents of the Jirst terms of the dividend and divisor, which being found, the others can easily be supplied. In performing the work, the coefficients of like combinations of letters in the dividend, and the several products must be placed in the same vertical column, which will be efiected by an. raging the operation as in the following example. SYNTHETIC DIVISION. 41 1. Divide x^ - 6x' + ISx" - 20x3 + 15^2 _ e^ + 1 by x^ - 2x + 1. The coefficients of the divisor, with all the signs changed, except the first, form the first vertical column : — 1 + 2 - 1 1 — 6 + 15 — 20 + 15-6 + 1 = coefs. of the dividend. + 2- 8 + 12- 8+2 _l+4_6 + 4-l 1 — 4+6—4+1 = ... quotient. Whence the quotient is x* — ix^ -\- Gx'^ — ix -\- 1. In the above operation, since the coefficient of the first term of the divisor is 1, the coefficient of the first term of the dividend is also the coefficient of the first term of the quotient, which is there- fore 1. The second and third coefficients of the divisor are now multiplied by 1, and the products placed opposite to them and under the second and third coefficients of the dividend. The second vertical column is now added, which gives the second coefficient of the quotient = — 4 ; the coefficients of the divisor are now multiplied by — 4, and the product placed opposite to them and under the third and fourth coefficients of the dividend. The third vertical colurtm is now added, which gives the third coefficient of the quotient, and so on till all the terms of the quotient are obtained. If, after finding all the coefficients of the quotient, the sum of the coefficients or vertical columns to the right of the last coeffi- cient of the quotient be not zeros, they are the coefficients of a remainder having the same literal parts as the terms of the dividend from which they arise ; this remainder must be annexed to the qtiotient to give the true quotient. 2. Divide x« - a^ by j^ + 2x^a + 2x0^+ a\ Since, in this example the dividend wants 5 of the powers of x and a, their places must be supplied by ciphers ; thus — 1 + + + + + 0-1 = coefs. of the dividend. -2+4-4+2 -2+4-4+2 -1+2-2+1 -2 + 2 — 1 = ... quotient. Hence x^ — 2x'^a + 2xa^ — a^. Answer. 3. Divide 24x* + oix^ + SSa;^ - 8x — 7 by 6x^ + 3x - 3. The coofficiciit of the first term of the divisor being 6, the sura of the oeffieicntt, in the several columns must be divided by 6 in 42 ELEMENTS OF ALGEBRA. order to get the coefficient of the quotient. The work may be arranged thus — coefs. of the dividend. 6 24 + 54 + 33 - 8 - 7 _ 3 — 12 — 21 - 12 + 3 +12+21 + 12 6)24 + 4 2 + 24| + 1 + 5 = ... remainder. 4+7+4 = ... quotient. Hence the quotient is = 4a;^+ 7a; + 4 + a: + 5 6x^ + 3x - 3 II. The rule may be proved very simply as follows : — Let the line above the coefficients of the quotient be considered as the sign o1 equality, since the sum of the numbers in each vertical column above it is equal to the number below it ; if now all the numbers above the line, except the coefficients of the dividend, be broughl below it with their signs changed, retaining them in the sam( vertical columns, the equality will still exist ; but the numbers below the line will then be the coefficients of the product arising from multiplying the coefficients of the quotient by those of tli( divisor ; and since their product is equal to the coefficients of the dividend, the quotient, multiplied by the divisor, produces th( dividend: hence the accuracy of the process. The exercises in case iv. of Division may all be performed by this as well as by the common method, to which add th( following EXERCISES. 1. Divide I - Gx"^ + 5x« by 1 - 2a: + x^, r= 1 + 2x + 3^2 + 4x' + 5x* 2. ... x« - UOx* + 1050x3 _ 3101x2 + 3990x - 1800 bj a^ - 12x2+47ar-60, = x^ + 12a;2-43a: + 30 3. ... x^ - 6x* + 20x* - 40^3 + 50^2 - 40a: + 100 by x _2xH5x-9, =^^-Ax'^+7x+S-^^l^^^ a:**— 2x^+5a;— 9 4. ... x«-/bya:-y, = x'+x^y+xY+xY+xy^+f EXAMPLES WITH LITERAL EXPONENTS. 1. Divide a^" — x^" by a" — a;". a" — x'')a''" — x^\aP + a:" ^2„ _ ^ny.n a"x" — x"-" DIVISION. 43 2. Divide 20^" — 4a"a:" + 2r^" by a" — a:". a" — a;")2a2« - 4a"x" + 2x2"(2a'' — 2x'* 2a- 2a"x" - 2a''a:" + 1x^^ - 2a"x'' + 2a;2'» 3. . ., a:"" — 3a;^"3/" + 3a;"/" — ^^" by a;" — y". -yn-^o^n _ Sx^riyn _^ 3^^"/" — ^'"(a^" — 2x"y" + y' - 2x'-y^ + Sxy — 2x^»y'' + 2a;Y" ^yn_^3„ Any of the preceding exercises may be changed into exercises with literal exponents, by merely multiplying all the exponents in the divisor and dividend by n. The 3d example given here is the 7th of the preceding exercises, with all the exponents multi- plied by n. The quotients will also be the same as in these exercises, with the exponents multiplied by n. EXAMPLES WITH LITERAL COEFFICIENTS. 1. Divide x^ — (a + c)x^ + (h + cic)x — bchj x — c. L a^ — (a -{■ c^x^ + (i + ac)x — be (x —c \oc^ —ax + b cx^ ax^ + (6 + ac)x ax^ 4- CLCx bx — be bx — be 2. Divide x^ — (a -\- c^x'^y + (6 + ac^ocy''' — bcy^ by a: — cy. a? — (a -\- c)x^y a^ — cot^y x' — (a + c^x^y + (6 ^- ac)xy^ — bcif {x — cy V — axy + by" — as^y + (6 4- ac)^y'^ — ax'^y + acxy^ bxy^ — bcy^ bxy^ — bcy^ 44 ELEMENTS OF ALGEBRA. EXERCISES. 1. Divide a;* — (p + r)x'^ + (S" + pr^x"^ — qrx hy x — r, = x^ — px^ + qX' 2. ... x^ -(l-\- a>2 + (a + 6)x - i by X - 1, = x^ — ax -\- h, 3. ... r* - (1 + a>* + (1 + a + h)x^ - (a + 6 + c)a;2 + (6+c)a;— c by x^—x-^-l, = x^ — ax^ + 6x — c. 4. ... 2ahx^ — (36c — 2ad)xy — Zcdf by hx + c??/, = 2aa: — ocy. In the following and similar examples, the remainder is the same as the dividend, with r instead of x. Divide ax^ + h^? -\- ex ■\- d\fj x — r. ax^ + bx^ -\- ex -\- d ( x — r ax^ — aror \ ax^ + {ar + h)x -j- (a?-^ -\- hr + c) (ar + b)x^ + ex (ar + b^x^ — (ar^ + 6r)a; (ar^ -\- br -\- c)x + c? (ar^ + 6r -|- c)a; — (ar' + i?-^ + cr) ar^ -}_ 6,-2 + cr + +/, it is sometimes convenient to write it thus — ax^ — h\o(^ 4- dx +/. - c + e] EXAMPLE. . Divide ax^ + hx^ -\- cx'bj X — 1. X - - l)ax^ + 5a:2 + ex{ax'' -\- ax + « ax' - a:^ + 6 + c + a a;2 -\-cx + * + « :^-a\x + 6 -h\ quotient. + a + c + « a; — a - 6 a -\-h ^ c =i rem. (a + 6+C+ J). EXERCISES. 1. Divide ax' — 5a;^ + ex — c? by a: + 1, = a:]^— (a + b)x-\-(a-\-b+c), rem _^^, ... a;- — aa: + & hy x — ?•, I^ft = a: + (?■ — «)) rem. = ?-^ — ar I^B ... 1 — ax -{- bx^ — ex' by 1 + a-, IB = 1 _(1 +a)a: + (1 + a + 6y-(l + a + ft + c)x\ I^K rem. = (I -\- a -^ b -\- c)x*. I^H INFINITE SERIES. 89. An infinite series consists of an unlimited number of terms, vliich observe a certain law. The law of a series is a relation existing between its terms, 10 that when some of them are known, the rest may be easily '-.i'd. s, in the infinite series ax — d^x' + «'^^ — o*a:^ + &c. any . is found by multiplying the preceding by — ax^. l"). A quotient may sometimes become an infinite series, ranged by ihe ascending powers of one of the letters. 46 ELEMENTS OF ALGEBRA. EXAMPLES. 1. Divide 1 by 1 — x. 1 - x)! (1 + a: + a;2 + &c. 1-x X X - -c^ x" x^ - -x' x^ &c. It appears from this example that the quantity is equ; 1 — X to an infinite series, or — ^ = 1 + a: + a;2 + x^ + ... &c. 1 — a: After finding two or three terms of the series, the law of is generally perceptible, so that it may be extended any lengl without further division. 2. Divide 1 — ax -\- bx^ — ex' + &c. by 1 + x. 1 + -1 X + bx" — a -1 x-lx"" — a — a + 1 x" -ex' + a + & &c. &c. a + a &c. EXERCISES. 1. Divide 1 + a: by 1 — a:, . = 1 + 2a; + 2a;2 + 20;=* + & 2. ... 1 - x by 1 + a:, . = 1 - 2a; + 2a;2 - 2x' + & 3. ... Ibyl+a;, . . = ' 1 — a; + a;^ — a;' + & 4. ... 1 - ax^ + bx^ - ex^ + &c. by 1 - x^, THE GREATEST COMMON MEASURE. 91. A measure of a quantity is any quantity that is contained in it exactly, or without a remainder. A measure of a quantity is also called an aliquot part or submultiple of the quantity. Thus, 4 is a measure of 12, and 6 of 24. So 8 or 3^ is a measure of 24. Also d^ is a measure of a® ; a^x is a measure of aV ; and 2a^z^ is a measure of 4a'2^. 92. A quantity that is a measure of two or more quantities is called a common measure of these quantities. Thus, 5 is a common measure of 15 and 20 ; 6 of 12 and 18 : 3a- is a common measure of Qa^x^ and 12a^a:; and 4x^3/' is a common measure of ^a^xhf, \2x^y^^ and 20x^^*2;. 93. The common measure of the highest dimension of two or more quantities is called their greatest common measure. The greatest common measure of a^x^ and a'x^ is aV, and that of 2a*x'y and Ga^y'^z is 2a*y. 94. A simple measure of a compound quantity is a measure of each of its terms. Thus, a simple measure of 4aV — 8a^x* + 6aV is 2a^a:^ for this quantity is contained in each of the terms of the compound quantity. 95. Quantities that have a common measure are said to be commensurable; and those that have no common measure are called incommensurable. 96. Incommensurable quantities are also said to "be prime to each other, or relatively prime. 97. A quantity that has no measure except itself and unity, is said to be prime, or absolutely prime. 98. A quantity that has a measure, or which is the product of two or more factors, is called a composite quantity. Thus, ax, the product of the factors a and x, is a composite quantity. So is ic^y^ and 12a^a;^. So x^ — c?, which is the product of a; + a and a: — a, is a composite quantity. 99. The preceding terms — prime, absolutely, and relatively ; also composite, commensurable, and incommensurable — are also applied to numbers. CASE I. — ^TO FIND THE GREATEST COMMON MEASURE OF QUANTITIES WHEN THEY ARE SIMPLE. 100. Rule. The greatest common measure of the literal parts ' the quantities will contain the lowest power of each letter that common to all the given quantities, and may be easily found by spoction. 48 ELEMENTS OF ALGEBRA. If the quantities have numerical coefficients, their greatest common measure must be prefixed to that of the literal parts. EXAMPLES. 1. Pind the highest common measure of aV and a^x. Both the quantities contain a and x, and their lowest powers are a? and x ; hence a^x is their greatest common measure. 2. Find the greatest common measure of 4,aWx^y^ and Ga^x^- The greatest common measure of 4 and 6 is 2 ; therefore the greatest common measure of these quantities is 2a^x^y'^. 3. Find the greatest common measure of ia^x^z, 8aV, and 12ay^^ The greatest common measure of 4, 8, and 12, is 4 ; hence the greatest common measure is 4a^^. EXERCISES. 1 . Find the greatest common measure of x'^y*z^ and x*z^, — x^z^. 2. ... ... .... of 3ay, 6aV/,and Oay^, = Say, 3. ... ... ... of 8axYz^,12x^z\an6 24aVs2, = 4xV, 4. ... ... ... of Ga^a:/, 12aYz\ 9aVy, and2iaYz, = Say. CASE II. — WHEN THERE ARE TWO COMPOUND QUANTITIES. 101. Rule. Arrange the two quantities as a divisor and dividend ; divide that which contains the highest power of the leading quantity by the other, and then divide the divisor by the remainder ; then divide the last divisor by the last remainder ; and so on, till there be no remainder; the last divisor is the greatest common measure. It is indifferent which of the quantities is made the first dividend, when the highest power of the leading quantity is the same in both ; but in other cases, the given quantity, which con- tains the highest power of the leading quantity, must be made the dividend. If the remainder in any case does not contain the leading quantity — that is, if it is independent of that quantity — there is no common measure. If there be a simple common measure of the two given quantities, it may at first be suppressed, in order to simplify the process ; but as it will be a factor of the greatest common measure, the last divisor must be multiplied by it. If any quantity be a simple measure of one of the quantities, GREATEST COMMON MEASURE. 49 and not of the other, it must be expunged — that is, divided out (at least if it be in a divisor) — before dividing by it, as it can form no part of the common measure. If the coeflScient of the first term of any dividend be not divisible by the first term of the divisor, it may be made so by multiplying nil the terms of this quantity by any quantity that will render it divisible. EXAMPLES. 1. Find the greatest common measure of a;^ — y^ and 3^ — x^y^. The second quantity in this example contains x^ in both its terms, and the first does not ; expunging it, the second quantity becomes x'^ — y"; and dividing the first by it, ^2 _ f)x^ - y\x 3^ — xy^ or {x - y)f As the remainder is not divisible by the divisor (86), it must be made the divisor after expunging y^ ; then X — y^x^ — y\x + y X- - xy xy -y^ xy -y^ erefore x — y \& the greatest common measure of the two given quantities. This might be more expeditiously effected by (69, Theo. III.) and (69, Theo. IV.), for x^ - / = (x + y\x - y) by (Theo. IIL), and x^ — y^ = (x" -\- xy -\- y'^){x — y) by (Theo. IV.) ; therefore X — y is evidently the greatest common measure. 2. Find the greatest common measure of x^ + a^x^ and x* — aV'^. The factor x^ is common to both these quantities ; it will there- fore form a part of the common measure, or will be a factor of it. Expunging this factor, the results are a;* + a^x and x^ — a'. The former quantity still contains a factor x common to both its terms, and the latter does not ; it must therefore be expunged, which gives x^ + o,^- Hence dividing x^ — a^x ctx + or or {x 4- «)«' D 50 ELEMENTS OF ALGEBRA. X + a)3? — (V'ix — a ax ax Therefore the greatest common measure of x^ + a^ and x^ — or of X* + a^x and x^ — a^, is a; + a; and that of the given qn; titles is x\x + a) ; or more concisely, x^ — d^ = (x -\- a)(x — (69, Theo. III.), and x^-\-a^= (x'-xa + a')(x + a) (69, Theo. ^ where a; + « is evidently the common measure. 3. Find the greatest common measure of x^^ — a" and x^ — c r" — a^)x^^ — a^\x^ -f x^a^ ^13 _ ^8^5 x'a' - a" x'a' - a:^a^» :r3 or (x^ — - a')x' - a=(a;- a^y X- x^a^ — a^ or (x- - a2)a' - aO:r' - aXx x^ — xa- X - xa^ - a' or (x — d)a^ - a)x2 - aXx + a x^ — xa xa — a^ xa — c? The common measure is therefore x — a; but this might much more concisely shewn by article (87, Theo. II.) 4. Find the greatest common measure of 5a^ + 10a*x + oa and a^x + 2aV + 2ax^ + x*. 5a^ is a factor of tlje first only, and x of the second only ; hci suppressing these factors, and dividing, we have a^ + 2ax + x2)a^ + 2a'x + 2ax^ + x^a a^ + 2a2x + ox^ ax^ + x^ or (a + x)x'^ GREATEST COMMON MEASURE. 61 a -f x)a^ + 2ax + x\a + x (J? -\- ax ax + x^ ax + x^ The common measure therefore is = a + ar. 5. Find the greatest common measure of 6a* — aV — 12x* and 9a^ + Ua^x"" - Qd'x^ - %x\ Ca* - aV - 12a;*) 9a^ + 12a='x2 - Qd'x? - Sx^Sa 18a' + 24aV — 12aV - IGx^ 18a' - 3a^a;2 - 36aa;* ' 27aV — 12a2a;* + 36aa:* - 16x' or (27a^ - \2a^x + dQax" - 16x'>2 Rejecting the factor x^, and making the former divisor the (dividend, we have Qa'' - aV - 12a;* (27a* - I2a^x + SBox^ - IGx* 9 t 2a + 8x 54a* - 9aV - 108a;* 54a* - 24a='a; + 72aV ~ 32aa;' 24a*x 9 - 81aV + 32aa;^ - 108x* 216a*a; 216a*a; - 729aV + 288aar3 _ gygx* - 96aV + 288aa;* - 128x* - 633aV - 844a;* or - 211xX3a2 + 4a^) 3^2 _|_ 4a;2)27a^ — 12a2a; + ZQax" — l&x\^a - 4a; 27a^ + 36aa;"^ - \2aH - 16a;' - Vla^x - 16a;3 Hence 3a- -\- 4:X- is the measure required. In this examx)le there are three instances in which it was necessary to multiply the dividend, in order that the coefficient of the first term should be divisible by that of the divisor — once by 2 and twice by 9. 52 ELEMENTS OF ALGEBRA. The preceding rule for finding the greatest common measure quantities depends on the following principles : — 102. A quantity that measures two other quantities measures h( their sum and their difference, and also any multiples (110) of th quantities with their sum and difference. Let ilf be a quantity, whether simple or compound, th measures any two quantities A and B ; and let m and w be t quotients arising from dividing A and B by M. Then A = m3f, B = nM; therefore, adding equal quantities to equals, it follows that A+ B= mM + nM = (m + n)M; and therefore 31 is contained mA-\-Bas often as there are un: in the number, which is = m -\- n; that is, 3£ measures the su of A and B. Again, taking the differences of equal quantities, A — B = mM — nM = (in — n}M; and therefore A — B contains M exactly (m — n) times ; or 31 a measure of A — B. Also, if ;:>J. and qB are any multiples of A and B, it is evide that pA — mpM, and qB = nqM, so that pA and qB are multipl of 31; and hence 31 measures also the sum and difference of p and qB. 103. After performing the operations prescribed in the rul the last divisor is a common measure of the given quantities, ai is also the greatest possible common measure. For let A, B, be the given quantities ; let A contain B, p time with a remainder C ; and let B contain C, q times, Avith a remaind M; and let C contain M, r times exactly. This process represented below : B)A(p cr)B(q M)C(r Then, from the principles of division, A=pB+C,B = qC + M,C= vM. Since 31 is a measure of rM, it is so of its equal C, and ther fore also oi qC ov oi qC -\- 31 (102) ; and hence also of B, ai therefore ofpB or pB + C (102), and therefore also of ^. Hem 31 is a measure of A and B. It is also the greatest commc measure; for C=A~pB,M=B-qC, and any common measure of A and B, or of A and pB, must be measure of A — pB or C (102), and therefore also of q''- "• GREATEST COMMON MEASURE. 63 hence of B — qC or M. That is, every measure of A and 5 is a measure of M; but M is the greatest measure of M, therefore M is the greatest common measure of A and B. 104. The quotients that result from dividing two quantities by their greatest common measure are relatively prime. Let M be the greatest common measure of A and B, and let them contain M respectively m and n times, then m and n are relatively prime. Eor A = mM and B = nM, and M being the greatest common measure of A and B, and therefore of mM and nil, m and n can have no common factor or measure ; for if they had such a measure r, then rM would be the greatest common measure of A and B, which is contrary to hypothesis. 105. If two quantities have each a factor prime to the other quantity, and these factors be suppressed, the resulting quotients have their greatest common measure the same as the given quantities. Let m and n be respectively factors of two quantities A and B, so that m shall be prime to B, and nto A ,- also, after suppressing these factors, let the quotients be A', B'; then the greatest common measure M of the two latter quantities is that of the given quantities. Let M be contained in A' and B' respectively, p and q times, then A' = pAf, B' = qM; and therefore A = mA' =: mpM, B — nB' = nqM. But m and n, being factors of A and B, and respectively prime to B and A, are relatively prime, and so are p and q ; also m and q are so, since g is a factor of B, and also n and p for a similar reason ; hence /« and p being relatively prime to n and q, mp and nq are relatively prime (154) ; tlierefore il/is the greatest common meas^ire of A and B. 106. If a common factor of two given quantities be suppressed, the greatest common measure of the remaining quantities, multi- plied by this factor, will be equal to that of the given quantities. Let c be a common factor of A and B, and let A = cA', B = cB' ; then if M be the greatest common measure of J.', B\ and M that of A, B, M = cM'. For let A' = mM', and B' = nM', then is J. = mcM', and B = ncM' ; and since m, n, are relatively prime (104), cM' is the greatest common measure of mcM' and ncM', or of A and jB. That is, M = cM'. 107. K either of two given quantities be multiplied by a factor which is prime to the other, the greatest common measure of the product and the other given quantity is the same as that of the given quantities. Let A, B, be the given quantities, and r a quantity prime to A, 64 ELEMENTS OF ALGEBRA. the greatest common measure of A and rB is the same as that o; A and B. For let M be the greatest common measure of A and B, so tha A = mM, B = nM, and hence rB = rnM. Now, since m and n are prime relatively, and also m and r (for r ii prime to A), then m is prime to m (152) ; and hence M is tin greatest common measure of 7nM' and r«M, or of A and t-^B. Thii theorem is a particular case of that in article (105), namely, wher m in it is = 1. EXERCISES. 1. Find the greatest common measure of a;^ — a^x and x^ ... of x^ — c^x3indx'^+2cx-{-c^ = X + c ... of a^ — 5ax + ix- am a^ — a^x 4- 3aa;^ — 3x', = a — .z ... of 6a^ — Ga^x + 2ax^' — 2x and 12a^ — 15aa; + 3x^, = a — x ... of a^x* — a^^* and x^ + ^' V ... of ea* - 5a V - Gx' am 4a5 - 6aV - 2a'x^ + 3a;^ = 20,^ - 3x- ... of a* — a:* and a" — .r*^ = a — a; ... of Gx^ — 4:x* — llx^-3x Sx - 1 and 4a;* + 2a:' - ISa:^ + 3a; - 5, = 2a;' - 4a;^ + a; — 1 CASE III. — TO FIND THE GREATEST COMMON MEASURE OF THREE QUANTITIES 108. Rule. Find first the greatest common measure of two o: them, and then that of this measure and the third quantity ; tliii last measure will be the one required. Let A, B, and C, be the three quantities, M the greates common measure of A and B, and N that of M and C, then N h also that of ^, B, and C. For any measure of A and B is one also of M, therefore anj measure of ^, B, and C, must be a measure of 31 and C , likewise any measure of M being also one of A and B, anj measure of M and C is also a measure of A, B, and C. Henc< the greatest measure of M and C is also that of A, B, and C. 109. Rule. The greatest common measure of four quantitiei is found by first finding that of three of them, and then that o: LEAST COMMON MULTIPLE. 55 this measure and the fourth quantity; tliis last measure is the one required. And the greatest common measure of five, or of any number of quantities, is found in a similar manner. THE LEAST COMMON MULTIPLE OF QUANTITIES. ] ] 0. A multiple of a quantity is any quantity that contains it exactly. Thus, 6 is a multiple of 2 or of 3, and 24 of 2, 3, 4, 6, 8, 12 ; 12a"'^a:^ is a multiple of 12o, of \2ax^, of ax, &c. And 4(a — x)y'^ is a multiple of 2 (a — x), of 2?/, &c. 111. A quantity that contains two or more quantities is called H common multiple of them. Thus, 12 is a multiple of 3 and 4, or it is a common multiple of tliese numbers. So 24rta:^ is a common multiple of 12, %ax, Qx^, &c. And 8(a;^ — y^)x'^ is a common multiple of 8{x- — y"^), ix^, &c. CASE I.' — ^XO FIND THE LEAST COMMON MULTIPLE OF TWO QUANTITIES WHEN THEY ARE SIMPLE. 112. Rule I. The least common multiple of the literal parts of any two simple quantities is the product of the highest powers of each of the letters contained in them, and may be found by inspection. Rule II, The least common multiple of two quantities may also be found by dividing their product by their greatest common measure ; or by dividing one of the quantities by their greatest common measure, and then multiplying the quotient by the other; and in the same manner may the least common multiple of two numbers be found. If the quantities have numerical coefficients, their least common multiple must be prefixed to that of the literal parts. examples. 1. Find the least common multiple of 4a-a:^' and Gx*y\ Let M = the greatest common measure, and L = ... least ... multiple, then M = 2xy^ ; therefore L = ^-3 — ^ , or = 2a^ X Gx^ = Ua^xy. 56 ELEMENTS OF ALGEBRA. But in this example the least common multiple may also be found by inspection. Thus, the highest given powers of all the letters are «^ x*, and y', and the least common multiple of the literal part is therefore ci^x*y^. 2. Find the least common multiple of Gx^y^z* and 8aVs®. L = ^-^-a-xyz^ = 2ia''xYz''. 3. Find the least common multiple of \1a'(?x^ and \%o?x^. 1 9 y 1 Q L = " ^ a'cV = 2 X ISa^cV = SGa'c'x^ D EXERCISES, 1. Find the least common multiple of 12a*x^ and 18aVy*, and also of lOa-cy and 15ar^V, . = 36a*:ry and Sok-cy~'. 2. Find the least common multiple of ISx^z and 24:aVz^, and also of 25ax/ and 30aW, . . = 72aV?/V and loOaVyz^ 3. Find the least common multiple of Sac^z^ and 18cV, and ol 5aVxand8aV^^ . . . = 18ocV« and 40aVx^^ 4. Find the least common multiple of I20a'^'x*z^ and Gic^xy^z, = d60a^c^xYz'. CASE II. — TO FIND THE LEAST COMMON MULTIPLE WHEN ONE OR BOTH OF THE GIVEN QUANTITIES ARE COMPOUND. 113. EuLE. Divide the product of the two quantities by their greatest common measure ; or, divide one of the quantities by the greatest common measure, and multiply the quotient by the other. EXAMPLES. 1. Find the least common multiple of iax'^ and 3(a — ar). Here evidently M = 1, therefore L = 12axXa — x). 2. Find the least common multiple of Qa^y(a — x) and 4a(a2 - x"). M = 2a(a — x), therefore L = Say X ia^a^ — x^) = I2a^y X (a^ - a.2). 3. Find the least common multiple of 4rt(a^ — x^) and M =2{a - a:), therefore L = 2a(a + x) X 6x^(a^ — x^) = I2ax\a + xXa^ — x^) = I2ax\a* + d'x - ax^ — x*). LEAST COMMON MULTIPLE. 57 Hi. The rule for finding the least common multiple of two quantities is easily derived thus : — Let A and B be two quantities, and M their greatest common measure, and let A = mil/, and B = nM, then (104) m, n, are relatively prime, or have no common factor, and their least common multiple is mn (157); therefore the least common multiple of mM and n3f is mnM, but iM = mnM^ mM.nM A.B M M M = Z. EXERCISES. 1. Find the least common multiple of Qa^x^y and ^a\a + x), = 24aV^(a + ar). 2 ... ... ... of SaXx"" - /) and I2ax\ = 2ia^x\x^ - y-). 3. ... ... ... of 4aXa2 — X-) and 6aar* X (a* _ x% = 12a^x\a* - x*). 4. ... ... ... oi6xXa-x)sind4:xy(a-xyx (a+x), = 12xXa + xXa - x^. ■ 5. ... ... ... of 8ax(a^ -[- x^) and 12a* X (a+x)^ = 24:a*x(a+xXa^-\-x^) or 24aX«+^K«^-«^+^')- 6. Find the least common multiple of 24a*'(a^— x^) and 5a V X ir(q+^)(« - ^), |^b= 120aV(a + x)(a3 - x^) or 120aV(a + a:)(a-x)(a2 ^ax + a;2). CASE III. — TO FIND THE LEAST COMMON MULTIPLE OP THREE OR MORE QUANTITIES. 115. Rule. Find the least common multiple of two of the quantities by the preceding rule, and then the least common multiple of this multiple and the third quantity ; the result is the least common multiple of the three quantities. If there be four quantities, find the least common multiple of three of them, and then the least common multiple of this multiple and the fourth quantity ; the result is the least common multiple of the four quantities. In a similar manner, the least common multiple of any number of quantities may be found. If A, B, and C, be three quantities, and L the least common nmltiple of A and B, then the least common multiple of L and C be- iluir. of xi. i3, and C, For every common multiple of J. and 58 ELEMENTS OF ALGEBRA. jB is a multiple of L ; therefore every common multiple of A, B and C is a multiple of L and C ; also every multiple of L and ( is a multiple of A, B, and C; consequently the least commoi multiple of L and C is the least common multiple of A, B, and C. EXAMPLES. 1. Find the least common multiple of 4a^, 3a^x, and Qax^y^. For the first two the least common multiple = 12a^x ; and fo: this quantity and the third, the least common multiple = 12a^x'^y^ 2. Find the least common multiple of 8a'(a + x), ^ax^, anc 4:X^y. For the first two the least common multiple = 24a^x^(a + ar) for this and the third, the least common multiple = 24:aVy(a + x) 3. Find the least common multiple of 3(a — x), 2a*x^(a — x^ Sa\ and %x\ It is evident, by inspection, that the least common multipL = Qa*x\a — xf. 4. Find the least common multiple of Gcc^x*, Sa{x — y) UxXx"" - if), and 18a'x(x* - y'). It is evident that for the first three, the least common multiph = 12d^x*(x^ — y"^) ; and for this quantity and the fourth, the leas common multiple = 3Qc^x^(x^ — y*). EXERCISES. 1. Find the least common multiple of 12a^x-, 6a^, and 8x*y% = 24a^xy 2. ... ... ... of8xXx-y),Sa*x\saidl2axy' = 24:a*x-yXx - y) 3. ... ... ... oilQa^x'(x-y),lbr'(x-yy and 12(a;2 - if), = 60aV(x - y)Xx + y) 4. Find the least common multiple of 8(a — x), 4:x^(a^ — x^), X GaXa'' - a:^), =24a2xV_ x^'Xa'^ + ax-\- x^) or 24a V(a* + a'x - ax^-x*) ALGEBRAIC FRACTIONS. 116'. If a unit of anything be divided into any number of equal parts, one of these parts or any number of these parts is called a fraction. Thus, if one inch, as the line AB, be "T \ i T T" divided into 4 equal parts, one of these Ac a ±1 parts, as Ac, is the fourth part of an inch, and is represented tluis - ; and three of these equal parts, as Ad, is called three- I 3 I fourths of an inch, and is represented by j. In the algebraic fraction t, if b = 4:, and if 1 denote an inch, then -r means the 6 [ fourth of an inch. Likewise, if in the fraction y, a = 3 and 7 = 7 J 6 6 4 of an inch, j represents 7 of an inch. 117. The greater the number of parts into which a unit is ? divided, the less is one of these parts in the same proportion. \ Let one inch be divided into 4 equal parts, and another inch [ into 12 equal parts, then it is evident that 1 of the former parts \ 11 ' will be equal to 3 of the latter, or 7 = 3 times — ; that is, if the number of equal parts be made 3 times greater, each of the parts will be 3 times less. Thus -ism times greater than ; for if one a ma unit be divided into a equal parts, and another into ma equal parts, one of the former parts will evidently be ?« times as great as one of the latter, whatever numbers a and m may represent. 118. Algebraic fractions are represented in the same manner as numerical ones, the upper quantity being called the numerator, and the lower the denominator, which together are also called the terms of the fraction. 119. An improper algebraic fraction is one whose numerator can be divided by the denominator, with or without a remainder. ^, ax + 6 ax^ — h . n ^. ^v, Thus, , or , are improper iractions. The nume- X a + X r itnr "however, may be either numerically greater or less than the i 'tiiinator, when particular values are assigned to the letters. 60 ELEMENTS OF ALGEBRA. 120. The reciprocal of any quantity is unity divided by that quantity. Thus, - is the reciprocal of a. 121. The denominator shews the number of equal parts into which the unit is divided, and the numerator shews the number of those parts that are taken. ■ 3 Thus, in the fraction -, the 4 shews that 1 inch (fig. at 116) is divided into 4 equal parts, and the numerator 3 shews that three of these are taken ; thus the denominator indicates the denomination or name of the part, and the numerator points out the number oi those parts that are taken. In like manner, in the fraction y, I shews that a unit is divided into h parts, and a denotes that a parts of these are taken. 122. Any part of a unit, taken any number of times, is equal tc the same part of that nmnber of units. 1 3 1 Thus, J of 1 inch, taken 3 times, or j of an inch, is equal to - ol 3 inches. For 3 inches contains - of one inch 12 times, or 3 inches 12 = 12 -fourths of an inch, or = — ; and the fourth part of 12 fourths is just three-fourths, or - of one inch. So y is either the fraction -r of one unit taken a times, or it is the 6th part of a units, If a = 5, and & = 8, then - of one inch, taken 5 times, is = - oi o 8 5 inches, or = -. 8 123. K the numerator of a fraction be multiplied by anj number, the fraction is multiplied by the same number. If in the fraction "" 4' " ' "^"i-"" X.KX ^J . D, O ' ' i/i^i'iujj. - multiplied by 5, for ^ is 5 times 3 -; or 1 4 of a unit, taker 15 times 15 = 5 times T or - taken 3 times. So 20 6 5X4 6 is = 4 5 times -, or = 5 times -. o So 3a b ~ 3 times p and if m be any number, 7na b~ : in a times -r. ALGEBRAIC FRACTIONS. 61 124. If the denominator of a fraction be multiplied by any number, the fraction is divided by the same number. If in the fraction -, the 4 be multiplied by 2, the fraction - will be divided by 2, or - is the half of - ; for if 1 inch be divided into 8 equal parts, and another into 4 equal parts, any one of the latter parts is evidently equal to 2 of the former, or - = 2 times -. So - = 3 times - : - = 3 times -— . So 7- = twice -y, or = 3 times 2 6 ' 4 12 6 26 -7^, or equal m times —7. But if t = twice -, then - = twice 36 mb 4 8 4 3 1 13 3 3 3 - ; if - = 3 times -, then ^r = 3 times -. Also y = twice -7, 7- = twice :-7, and generally y = m times -^. 6 26 6 mb 125. If the numerator of a fraction be divided by any number, : the fraction is divided by the same number. ' 15 3 20 J It has been shewn (123) that — is = 5 times - ; that — = 4 7/lCf, ct 5 times -, &c. Also -7- is m times 7-, so that the numerator ma b b being divided by m, gives a, and the fraction j is an mth part of -7-. 126. If the denominator of a fraction be divided by any numberj the fraction is multiplied by the same number. It has been shewn (1 24) that - is the half of -, since 4 = one- half of 8 ; and that - is the third of -, since 6 = 3 times 2 ; also o A that y is equal 3 times -— , or -7- = one-third of v ; that -r is = 6 36 36 6 ' 26 one-half of 7- ; and generally that 7- is ra times —,. 6 6 ?nb 127. If the terms of a fraction be both multiplied, or both divided, by the same quantity, its value is unchanged. For if the numerator be multiplied by n, the fraction is also multi- plied by n (123) ; and if then the denominator be multiplied by n, the fraction is again divided by n (124), and must therefore have its . . , , ,, a na „ 3 3 X 2 6 5 20 1 original value ; thus - = ^. So - = ^^^ = - ; - = - ; and 24_2 36 ""3' 62 ELEMENTS OF ALGEBRA. CASE I. — TO REDUCE FRACTIONS TO THEIR LOWEST TERMS OR SIMPLES FORM. 128. Rule. Divide the terms of the fraction by their greates common measure. EXAJIPLES. 1. Reduce " ' to its simplest form. 6a* The greatest common measure of the terms of the fraction i 2a', and dividing them by it we have 4:a^x^ ^ 2a' _ 2^ Ga* -^ 2a' "~ 3a' 2. Reduce q~ti~ *^ ^^^ simplest form. The greatest common measure = ix^z^, and dividing the tern by it we have - 12xyz* _ _ 3/2 8a;V ~ 2 • 3. Reduce — ^- to its lowest terms. X -\- a The greatest common measure is = x -\- a; hence : X -\- a a(x — a). 4. Reduce -— ^ — r- to its lowest terms. ^a^xy* ^, ^ „ , Ga*x- Sa'x The greatest common measure = 2a''x; hence —-r, — r = —A~r- 8a^xy* 4/ X* — V* 5. Reduce — ; ^ to its simplest form. x" — xY ^ X* — 71* The greatest common measure = x" — ?/^ ; hence -^ f-j - x^ + .?/'' x' ' 129. It is evident (127) that, after dividing the terms of tl fraction by the greatest common measure, the resulting fractic will be of the same value as the given one. ALGEBRAIC FRACTIONS. 63 EXERCISES. 1. Reduce - — r, and also „ •\ r , to their simplest forms, 3a , ixy^ 2- - ^^TT^r to its simplest form, . . = jripp- to its simplest form, . c^ — 2ax + a^ x — a 5 5 to its simplest form, = -— —-,. or -{■ XT a^ — ax-^ x^ a* - a'x + Sax^ - Sx^ ^ .^ . , , ^ a^ + Sx"^ x(ia + 3a;) 4a* - 4aV + 4aa:' - X* , ., . , ,^ 6a^+4a3a:-9aV-3a..3+2x* *" ^*^ ^"^P^^^* ^"^°^' ~ Sd'- ax- 2x2' CASE II. — TO CONVERT A FRACTION INTO AN EQUIVALENT ONE, HAVING A I DENOMINATOR EQUAL TO ANY MULTIPLE OF THE DENOMINATOR OF THE { GIVEN FRACTION. 180. Rule. Multiply both terms of the fraction by such a quantity as wUl make the denominator equal to its given f multiple. The multiplier will be found by dividing the given multiple by the denominator. When the given quantity is an integer, it may be first written in a fractional form by putting unity for its denominator; the multiplier will be the given multiple itself. EXAMPLES. 1. Convert the fraction — ^ to an equivalent one, having the denominator 12a^«®. 64 ELEMENTS OF ALGEBRA. The multiplier is = / = 4a-y^ ; therefore ^ = -g^r^i = -^e- 2. Convert 2a*a:® to an equivalent fraction, having the denomi^ nator Sx^y*. 2 4 6 _ ^^1 _ ^(^*^^ X 3xy _ Ga'xY "^ ^ ~ 1 ~ IX BxY ~ ZxY ' 3. Convert the fraction — — -^ — ^ into an equivalent one, having the denominator ^a^x^- The multiplier = .^ ^ = ax; therefore - ^ ^ 3 - = .: „ . J ^ — = — „ „ „ ^"^ . 3^ ^^ 4. Keduce ^ j_ 2 *^ ^^ equivalent fraction, having 2aa;(a* — x*^ for its denominator. The multiplier = 2ax{a^ — x-), and the fraction _ (3a - 4a:) X 2ax(a'^ - a:") _ 2ax(ia^ - x'XSa-ix ) ~ (a? + x') X 2ax(a^ — x^) ~ 2aa:(a* - x*) ^' The rule for thus converting fractions depends on the principle stated in article (127). EXERCISES. Zax^ 1. Convert the fraction —j— into an equivalent one, having ^a^if for its denominator, . . . , — ^ ^ _ Qa?xY ' ' ' ' ~ Say 2. Com^ert 3a*xy^ into an equivalent fraction, having 2aV for Ga^x^i/^ its denominator, = ~-. 2a^x^ oa^x* 3. Convert --— g- into an equivalent fraction, having 14a:'yV for its denominator, = _^4^o . UxYs ALGEBRAIC FRACTIONS. 66 3a^ — AiX^ 4. Convert to an equivalent fraction, having Zx(a + x) a -\- X . ., ^ • . da'x-Ux^ tor its denomuiator, = ——p — ; — r-. ' 3j;(a + x) 5. Convert p ^ to an equivalent fraction, having 3x(a — x)^ {a — x) for its denominator, = -—7 ~. 3x(a — xy 3(a^ 4- cc') 6. Convert -^ ^ into an equivalent fraction, having (a + xYor — ar) for its denominator, = -^7 ^^7-^ 5^ — -• ^ •^^ ^ (a + a:)(a^ — a;^) III. — TO REDUCE AN INTEGRAL OR MIXED QUANTITY TO A FRACTIONAL FORM. 131. KuLE. Multiply the integral part by the denominator of the fractional part ; to the product add the numerator with its proper sign, and place the sum as a numerator, under which write the denominator of the fractional part. An integral quantity is reduced to a fractional form, by making unity its denominator. Wlien the fractional part has the sign minus before it, the signs of all the terms of its numerator must be changed (59). Thus, if the fraction be 5 ^— it becomes 5- — 5 — . a^—x^ a^ — x^ W EXAMPLES. 1. Reduce 3a + - to a fractional form. y y y Qax -\ to a fractional form. X ao? — a? Gax' 4- ax' — a? lax^ — a? Qax H = = . XXX c? — x^ c? — ax ■\- 3? ; — to a fractional form. a -{■ X £ ELEMENTS OF ALGEBRA. The fraction = —, = ; : a -\- X a -f- X 2x' a -\- X To prove the rule, let a + - be given ; then a = ~ = — (127), and h -, is = the sum of ac and b divided by c, or = . c c c And thus it appears in a similar manner that a = . EXERCISES. 1. Reduce 2 + - to a fractional form, . = . X X 2. ... 3a H to a fractional form, = -. X X 3. ... a + X + — to a fractional form, = -. a — X a — X 4. ... ^a^x to a fractional form, = , X X 5. ... xy^ ~ to a fractional form, = — . 6. ... 0? — x^ 5 — -^ to a fractional form, = r, — '■ — r,. af-\rx^ of' -^ X- a2 4.^2_5 2a2-5 7. ... a—x-\ to a fractional form, = • a-Yx a -\- X a? —a^x -\-ax^ — x^ '■ to a fractional form, a+x 2x* - 3 a -\- X IV. — TO REDUCE AN IMPROPER FRACTION TO AN INTEGRAL OR MIXED QUANTITY. 132. Rule. Divide the numerator by the denominator, and the quotient will be the integral part ; and to this annex the remainder with its proper sign as a numerator to the given denominator for the fractional part. ALGEBRAIC FRACTIONS. 67 EXAMPLES. , _ , Sax -\- or . , 1. Keduce to a mixed quantity. Sax -\- (J? c? = 3a -| . X X Here Sax is divisible by ar, and gives the quotient 3a, and the remainder = o?. 2. Reduce to a mixed quantity. 2a-x - ar^ „ x^ = 2ax . a a „ 5(a + x). to an integral quantity. 5x. a -\- X 5(a + x)x a -\- X c? — a:^ + 3 4. ... to a mixed quantity. a ■\- X The quotient is =: a — x, and the remainder = 3 ; therefore a^ - x^ + 3 a — X -\- a -\- X a -\- X r- -D J a^ + x' — 2a + X . 5. Reduce to a mixed quantity. a -{- X The quotient is = a^ — ax -\- x^ — 2, and the remainder = 3x; Sx hence the fraction = a^ — ax + x'^ — 2 -\ . I This rule is the converse of the preceding. It may be proved j thus: — Let the given fraction be , which is evidently = ax b b . ax X ^.n„\ — + - or = X + -; for— = - = x (127). a a a a \ EXERCISES. subtract — - X ~~7 2x+- y a -\- X x-\-y' " ' x^~y^ g , (a + c^y-ab ^ ac 2x__ ~ a + x* 74 ELEMENTS OF ALGEBRA. VIII.' — MULTIPLICATION OF ALGEBRAIC FRACTIONS. 137. Rule I. Multiply the numerators together, for the nume- rator of the product, and multiply the denominators together, for the denominator of the product. 138. If the numerator of any of the given fractions, and also a denominator of any of them, have a common factor, it may be cancelled, and the quotients taken as new terms. If any of the given quantities is a mixed quantity, it must first be reduced to a fractional form (131). EXAMPLES. 1. Multiply 3a ^ 5x 3a 5x loax T ^ T ~ "32"* 2. ... 3a:y 2x , 3w 3x7/ 2x St/ l^xY 7 ^ 5 ^ 11 ~ 385 • 3. ... 12a %x bx ^ 15a' 12a 8ar 96aT 32 bx '^ 15a ~ 75a.r ~ 25* But the same result may be found by cancelling the common factor 3a from 12a and 15o, and the factor x from 8x and 5a:; thus — or thus. The common factors of the terms of the given fractions are generally more easily found than that of their products, and ought therefore to be cancelled. 12a bx 8x ^ 15a 4 4 = 5^ 8 8 _ 5 "" 32 25 bx 8x 15a ~ 32 '25' ALGEBRAIC FRACTIONS. 7o ^ _. ,^. , 3a Ux , 8 4. Multiply-,— , and— . 'Sa IGx 8 _ 1 4 8 _ 32 4^^ 9 ^15^-1^9^5- 45" Here 3a and 15a have the common factor 3a, and dividing by it, the quotients are 1 and 5, which are put in place of 3a and 15a. So 16a: and 4a; have 4a; as a common factor. ). Multiply -^^ — and -5 5. a (t — 3? (a — x)x ax _ X a -\- X a -\- X ■ Here (a — x)x and a^ — 0? have the common factor a — a;, by ^celling which the quotients become a;, and a + a; ; also, ax iwd a have a as a common factor. c^ -|- x^ a^ — a;^ 6. Multiply a by a; H . a X ( J^jt^V a'-x''\_ a^-a'-x ' x^^a^-x " _-^ ^-_ a ) X )~ a ' X a ' X The rule may be proved thus : — If v is to be multiplied by 1, the product is y ; again, if it is to be multiplied by the c?th part of 1, or by -, the product must be d times less than y; that is, t-^ (124) ; and if j is to be multiplied by -z, the product must be c c • 1 times greater than before, since -, is c times -'. therefore the a a product must be ^-^ taken c times; that is, -z—^ (123). ba \ od 139. EuLE II. To multiply 4 taction by an integral quantity, either multiply the numerator or divide the denominator by it. This rule is evident from articles (123) and (126). ~, 6a , 5a 4a; 20aa; 5a 5a , 5a Thus,— X 4. = — X-= — = ^,or— X 4x = -3-. 76 ELEMENTS OF ALGEBRA. ,^ ,^. .2a .4a Multiply — and — , o o EXERCISES. 6a 5x ,8 BCa + x) , 4a; 2 a + a; a — X - X a" -, and , a — X a + x'' — i^, ^, and — , , X -y X +y X a -{- X ^lO^oc^ -i ^' ~ ^^ 2a-^^!zL^and3x + ^^-±^, a X - i^ " 15 Zx" = 6x. x\a + X a{p^ -\-y- X 2ax \a + x) 3(a2 - ax + a;^' ' + 5aV a;' ax IX. — DIVISION OF ALGEBRAIC FRACTIONS. 140. EuLE I. Invert the divisor, and proceed as in Multiplica- tion. When mixed quantities are given, first reduce them to fractional forms. EXAIMPLE8. , _^. _ Qax . 4a: 1. Divide -—- by — . o o 6ax 6 • 4a; Qax 3 5 >^l- 4x 3_9a 2 "" 10 4x2 , 9a2^^ 3a' lOx^' 4a;2 3a' ix' 10a:' 40a;» 9a2 • lOx' ~ 9?^ 3a' ■ ~ 27a^* i ALGEBRAIC FRACTIONS. 77 8(a - x ) _^ 6(0" - xQ _ 8(0-5) 3a: a-- • 3x ~ X' 5(a2 - x^) 8 3 24 X a; 5(a + x) 5x(a + x-)* a^ +x\ ^ x^ + a^ 3a p by 5x 4:a *' 3x a" + x\ ^„ x"" + a^x lla2 - Sx(na'-x'') lla^ - x^- 3x ^ Ux'' - a" 4a(14a;2 - a")' The rule may thus be proved : — Let y be divided by 1, the quotient is evidently j- ; and if it be divided by the c?th part of 1, the quotient will be d times greater than before, or — (123); again, if r be divided by c times -> or -, the quotient will be c times less than the former quotient — , or it will be -7- (124). -. ad a d ,, a c ad But y- = 7- X -, and hence ^ h- -, = r X -. be c d b c 141. Rule II. To divide a fraction by an integral quantity, either multiply the denominator or divide the numerator by it. This is evident from articles (124) and (125). ^, 12a: , 12a: 1 12x . 3 12ar , 3 Thus —- ^ 4a; = -— X y- = — - = -, or — - -i- 4a: = -. 5 5 4a: 20a: 5 5 5 , ^. , , 3a: - 5cf 1 21a: 1. Dmde-by-, . . < = — . _ 8aa: , 15a: 16a ^- - T^^-Y =T5- 78 ELEMENTS OF ALGEBRA. 3. Divide -^ by -^, = -.! 4. ... — by4ax, = -., I IQax ix i 3(a2-a;0, 2(a + x) 3(a - x)' b. ... • Dy , . . = .; X a — X 2x 32(a^ - x ^) S(a — x) _ 4(a + xX^_+_ax+ x^) : ■•• a- + x2~ ^ a -j_ a: ' ~ ^^ + x^ ' X ' ^ y ' x{2y' — b^) 2x^ a: _ 2x a^ + a;3 ^ a 4- ar' * ' " ~ a" — ax +x^' 10. ... ^^Ili;i3y^!-±^^, =1. {x — yyx—y X RESOLUTION OF FRACTIONS INTO INFINITE SERIES. 142. Any proper fraction, with a compound denominator, can be resolved, by division, into an infinite series ; for the numerator is a dividend, and the denominator is a divisor related to each other, as the remainder and divisor referred to in article (86) ; hence the quotient obtained, by dividing the first term of the numerator by that of the denominator, is not an integral quantity, but a fraction, and the process of division will never terminate, so that the quotient becomes an infinite series. The examples in Division that come under the observation in article (86), will all by this method produce an infinite series for a quotient. After a few of the terms of the quotient are found, the law of the series may be easily observed, and the succeeding terms can then be obtained without any further division. When a law is thus found, by observing it to hold in every particular instance examined, it is said to be discovered by induction, which is the source of most discoveries in science. Wlien a law is discovered by induction, its truth cannot always be implicitly relied on till it be directly demonstrated. ALGEBRAIC FRACTIONS. 79 EXAI 1. Divide c? + ^^ by a - x. a — x')a^ + x\a -\- x a^ — ax aPLES. _2^ a + ^+,.0. ax^x^ ax — x^ 2xV 1x^ a a 2x« «2 ^, &c. &c. The law of the series is evident ; for after the second term, if any term be multiplied by -, the product will be the succeeding term : this is by induction ; but it is evident from the process of the division, that this must be the law of the series. 2. Reduce -j — ; — to an infinite series. + a: 6 + x)a - x0 - (a + V)^ + (a + 5)|^ -, &c. ax " + T X^ 80 ELEMENTS OF ALGEBRA. 3, Reduce ^ , ■ to an infinite series. 6 + 1 Hence _1 3 _ 1 _ 1 3 9 1 &c. 3 + 1 3 32 ^ 3^ 3* ^ 4 Similarly, ^ = l+i, + ]5 + -l+...3=l: and generally -1^ = 1 + ^ + i, + ... EXERCISES. 1. Divide ahy a + x, = 1 1 — ^ 3- + ... to infinity. 2. ... 1 by 1 - a, . . . = 1 -\- a + a" + a^ +,&c. a ah , aP , ab^ 3. ... ahyx-b, . = ^ + •^+'7^- + "^+''^'^• 4. Resolve , into an infinite series, =:1H ;7 + -t— 5&C' a; + 1 a; a;- a;-* 5. ... ^ or -, and -, into infinite series, 3 4 — 1 2 -J- 1 = i + i + T' + - ^""-^ ^ - y^ + ^ -' '^^• a — X a a^ o? 7. ... ^^-, . . = 1 + 2ar + 2^2 + 2x' +, &c. \ — X I PRODUCTS OF NUMBERS— PRIME AND COMPOSITE NUMBERS. I. — PRODUCTS OF NUMBERS. The multiplier in the following theorems is the last number of the products. 143. Theorem I. The sura of the products of several numbers by another number, is equal to the product of their sum by that number. Or ad + hd -\- cd = {a + li -{- c)d, or = sd, if s — a + b -{- c. For a, b, and c, taken d times, are just equal to s taken d times. 144. Theorem II. The sum of the products of a number by several numbers, is equal to the product of that number by the sum of these numbers. Or da + db + dc — d(a -\- b -}- c), or = ds, ifs = a-\-b-{- c. For d taken as often as there are units in a, and b, and c, is just equal to d taken as often as there are units in s, or in the sum of «, b, and c. 145. Theorem III. The product of two numbers is the same in whatever order they are taken. Or ab = ba; that is, a taken b times, is equal to b taken a times ; or a units, repeated as often as there are units in b, are equal to b units, repeated as often as there are units in a. 1st, When b = 1, Then ab = ba, or a X I = I X a; for a units taken once is equal to 1 unit taken a times. 2d, Wlien a = b, the theorem is evident. 3d, When a is a multiple of b, or- a = nb. Then a.b = (b -{- b + ... n tini . b = bb taken n times by article (143),\ / = nb.b; and b.a = b(b-\-b-{-...n times) = bb taken n times by article (144), or = b.nb. Hence a.b = b .a, and also rib . b = b , nb, p 82 ELEMENTS OF ALGEBRA. 4th, When a contains h n times, with a remainder r, or ' a =^ nh -\- r, Then a.b = (nh + r)h — nh . 6 + r . 6 (by 143), and h .a— h{iih + r) = b .nb + b .r (by 144). Now, nb .b = b .nb (by 3d) ; and therefore, in order that a .b = b . a, this condition must exist, namely, b .r = r . b Now, b -p' r, let therefore b = n'r + r', and treating b and r exactly in the same manner as a and b, and using n' for n, it may be shewn that the equality r .b = b . r depends on the condition r .r^ = y^'.r Again, let r = n"r' + r", and treating r and r' exactly in the same manner as a and b, and using ?j" for n, it may be shewn that the equality r .r' = r' .r depends on the condition that r' . ?•" = r" . r' It is evident that the quantities n, n', n", and b, r, r', /', are = the quotients and divisors or remainders in performing on a and b, the process for finding the greatest common measure ; hence some of the remainders must ultimately become either 1 or 0. Suppose that ?•" = 1, then / . r" = r" . r' (by 1st), and therefore r . ?■' = r' .r ; and hence a .b =^ b .a. Again, if r" = 0, then the same conclusions evidently follow. 146. Theorem IV. The product of any number of factors is the same in whatever order they are taken. Let there be n factors a, 5, c, c?, . . . ; and first supiiose that the product of (?j — 1) factors is the same in whatever order thoy are taken. The products of the n factors may be divided into various groups : first, those in which a is the first factor ; second, those in which b is the first ; third, those in which c is the first ; and so on. Now, if P be the product of tlie {n — 1) factors after the first or a, then any one of the first group is = a . P, since it is assumed that the product of (n — 1) factors is the same in whatever order they are taken ; and hence all the products in the first group are equal. So if P' be the product of (n — 1) factors, excluding the second or b, any product in the second group is = b . P', and they are therefore all equal. But some product in the second group must end in a, and will there- fore be expressed by P .a; and since a.P = P .a (145) ; there- fore all the products in the first group are equal to those in the second. It may be similarly proved that all those in the second are equal to those in the third ; and so on. PRIME AND COMPOSITE NUMBERS. 83 Therefore if the proposition be true for (n — 1) factors, it is also true for n factors. But it has been proved to be true for two factors (145), therefore it is true for three ; hence therefore it is true for four ; and generally therefore it is true for any number of factors. 147. Hence, the product of a number by that of other two numbers, is the same as the continued product of the three numbers taken in any order. That is, a.bc = abc, or a multiplied by the product of be is equal to the continued product of a, b, and c. For if /> = be, a.bc = a.p=p.a = bc.a = bca, and bca is the continued pro- duct of b, c, and a, which is the same in whatever order the factors a, b, c, are taken. 148. Hence also — a product is multiplied or divided by a number, if one of its factors be multiplied or divided by that number. For it has been shewn that a.bc = ab .c; that is, if a be multi- plied by the product be, the result is equal to the product ab taken c times. The case of dividing by a number is evident from this ; for a and be being the two factors, and c a third number, be _ ab . c _ , II.— PRIME AND COMPOSITE NUMBERS. 149. Theorem I. If a number is prime to other two numbers, it is prime to their product. Let A, B, and P, be three numbers, P being prime to A and B, then P is prime to their product AB. Let B-p' P, and perform on these two numbers the process for finding their greatest common measure (101); let q, q', q", ... be the successive quotients, and R, R', R", ... the corresponding remainders ; then it is evident from that process that the following equations (103) are true. B = qP +R from which AB = qAP + AR ... [1] P = q'R -\- R' AP = q'AR + AR' ... [2] R = q"R' + R" I = q"AR' + AR" ... [3] R' = q'"R" + R'" \ J z= q"'AR" + AR'" ... [4] But since the successive remainders are continually diminishing, some one must at last be = or 1. It cannot be 0, for then B and P would have a common measure, which is impossible, for they 84 ELEMENTS OF ALGEBRA. are relatwely prime ; hence the remainder must = 1. Let the remainder that becomes = 1, be ^'", and the last equations in the two ranks above become R = q"'R" + 1, AR =z (/"AR' + A ... [5] It appears, by equation [1], that the divisibility of AB by P depends on that of AR, for qAPis divisible by P ; by equation [2], it appears that the divisibility of AR by P depends on that of AR by P, for the first member being divisible by P, the second must be so ; and if AR be so, q'AR is so, and hence also AR : again, by equation [3], it appears that if AR and AR are divisible by P, so is AR" ; and by equation [4] or [5], since R" is sup- posed = 1, it appears, if AR and AR" are divisible by P, that A must also be divisible by P. Thus, it appears that if B and P are prime, the divisibility of AB by P depends on the divisibility of A by P; and the divisibility of AB by any factor of P depends on that of A by this factor ; and, consequently, when P is prime to A and B, it is prime to their product AB. 150. Cor. 1. If a number be prime to one factor of the pro- duct of two numbers, the divisibility of the product by the first number will depend on the divisibility of its other factor by that number. For if P be prime to B, the divisibility of AB by P depends on that of .i by P by [1]. 151. Cor. 2. If a number be prime to one factor of a product, and if it exceed half the other factor, the product is not divisible by it. For if P be prime to B, the divisibility of AB by P depends on that of A by P; but the greatest divisor of A is its half; there- fore if P be greater than the half of ^, it cannot divide AB. 152. Cor. 3. If a number be prime to several numbers, it is prime to their product. Let p be prime to a, h, and c, it is prime to their product. For p is prime to ab, and if ab = d, p is prime to d and c, and there- fore it is prime to their product dc or abc. And the proposition may be similarly proved in any other case. 153. Cor. 4. A composite number can be divided only by the factors of which it is composed ; that is, either by its prime com- ponent factors, or the products of any number of them. 154. Cor. 5. If the factors of one composite number be prime to those of another, the two composite numbers are relatively prime. 155. Cor. 6. If two numbers are prime to each other, so are any powers of these numbers. This corollary is evident from Cor. 5. PHODUCTS OF NUMBERS. 85 15G, Cor. 7. If a fraction be in its lowest terms, the terms of any power of it are relatively prime. Let Y be the fraction, then a is prime to h ; hence the terms of the fraction y^ are also relatively prime (Cor. 6). 157. Theorem II. If two numbers be prime, their product is their least common multiple. Let ;j and q be prime, then \i m = pq, m is their least common multiple. If it be not, let n be their least common multiple ; then n is less than m, for pq is evidently a multiple of p and q. Since n is a multiple of p, it must be equal to the product of p by some number r ; hence n = pr, and ptr must be divisible by q, but n ^ ??i, therefore pr ^pq, therefore r ^q; and since q is prime to p, and greater than r, it is not a measure of pr or n (151). Hence n is not the least common multiple of p and q, therefore pq is so. 158. Theorem III. No root of an integer can be expressed by a vulgar fraction. Let w be any integer, and if possible let its nth root be = the a V and hence w = —; but a" and 6" are prime to each other (156) ; fraction ~, which is expressed in its simplest form. Then -^ and therefore the fraction — cannot be = a whole number ; hence 18 nth root of w cannot be = y. b PEODUCTS OF QUANTITIES — PRIME AND COMPOSITE QUANTITIES. I. — PRODUCTS OF QUANTITIES. 159. Theorem I. The simple powers of all real algebraic quan- tities represent abstract numbers. For if X be an algebraic quantity, it is admissible to involve it to any power, such as x\ Now, x'^ means that x^ is to be repeated X times (22) ; and therefore the factor x in the expression x'' = .r® X X must be an abstract number (6) ; but we never assign different values to the same symbol in the same expression ; therefore the seven factors x in x'' are abstract numbers. The same may be similarly proved of x'', when x represents a com- pound quantity. It is hence evident that the proposition is generally true. 160. Since all real algebraic quantities, at least in their simple powers, denote abstract quantities, they ought, in order to represent concrete quantities, to be multiplied by the unit of measure. 161. Theorem II. The product of any two quantities is the same in whatever order the factors are taken. Let A and B be two quantities, then A taken B times is equal to B taken A times. 1st, When A and B are simple quantities. Since A and B represent numbers, the proposition is evidently true (145). 2d, When A and B are compound quantities. In this case the quantities also representing numbers, it is evident from article (145), that the proposition is true respecting the numerical product ; but the object here is to prove that the algebraic product is the same. Since in multiplying A by B, each term of ^ is successively multiplied by each term of B ; and in multiplying i^ by ^, each term of B is successively multiplied by each term of A ; the terms of the two products must evidently be the same. 162. Theorem III. The product of any number of quantities is the same in whatever order they are taken. Having proved the proposition for two quantities, it can be PRIME A>^D COMPOSITE QUANTITIES. 87 proved for any number exactly in the same manner as for the products of numerical factors in article (146). II. — PRIME AND COMPOSITE QUANTITIES. 1G3. All the propositions formerly proved respecting prime and composite numbers are evidently true in regard to prime and composite simple quantities. For any two prime simple quantities must consist of different letters (97) ; and any two commensurable simple quantities must contain at least one common letter. 164. A quantity is said to vbe a function of one or of all the letters it contains, because its value depends on their values. 165. Wlien the letters composing any function are connected only by the signs of the four fundamental operations, or of those of involution (176) or evolution (183), it is called an algebraic function. 166. "When an algebraic function contains no fractions, it is said to be integral ; and if it contains only integral powers of the letters, it is said to be rational. 167. WTien a function is arranged according to the powers of one of the letters, these powers being integral and positive, and their coefficients integral and rational, and either simple or com- pound, it is called an entire integral and rational function of that letter ; and if the coefficients are any algebraic functions whatever, it is called an integral and rational function of that letter. Also, of two functions of the same quantity, that of the higher dimen- sion is called the higher. 168. Theorem I. If an entire integral and rational function of a quantity be divisible by another, the product of the former by any other quantity will be divisible by the latter. For the product of the first function by any term of any quantity will evidently be divisible by the second function ; and hence the sum of the partial products — that is, the whole product — is divisible by it. 169. Theorem II. A measure of an entire integral and rational function of a quantity, which is independent of that quantity, is a measure of all its coefficients ; and if the coefficient of the highest power of that quantity be unity, the coefficients have no measure. Let the function A -\- Bx -\- C^ + ••■ Mx^ have a divisor D independent of x, and let a + 6.r A "^ + ... mx"- be the quotient, then is \ A+Bx+ Cx^ -\- ... Ifx" = hya + hx-{- cx^ + ... mz") ; and hence, by the principle of undetermined coefficients (468), A = Da,B = Dh, C = Dc, ...M = Dm ; and therefore i> is a factor of each of the coefficients A,B, C, ... M. 88 ELEMENTS OF ALGEBRA. 170. Theorem III, If P and Q be two functions of x, and if the coefficients of either of them have a common measure, those of their product have the same common measure. If the coefficients of neither P nor Q have a common measure, neither have those of their product. Hence, when the coefficients of a function of x, which is the product of other two functions, have a common measure, so must the coefficients of one of the latter functions ; and when the coefficients of the product have no common measure, neither have those of either of the factors. If P and D' are two functions of'x, such that the coefficients of P have no common measure, and those of D' have one, namely, m, so that D' = mD ; then if D' be divisible by P, so is D. If P and D be two functions of x, such that the coefficients of P have no common measure, and that P is prime to P>, then P is also prime to mD, m being a simple integral quantity independent oi X. 171. Theorem IV. Let P and B be two commensurable functions of a,- P being the lower, and its coefficients not having a common measure, then if the process of (101) be performed on P and B, the last divisor is their greatest common measure ; and this divisor is a function of x. If the coefficients of P have a common measure r, so that P=rP% then the greatest measure of P' and B found, as in (101), is also the greatest common measure of P and P, provided that r and the coefficients of B have no common measure. If they have, let it be n, so that r = nr, and B = nB'. Find C the greatest common measure of P' and B', and it will be the greatest common measure of r'P' and B' ; and hence nCvnW be the greatest common measure of nr'P', and nB'— that is, of P and B. 172. Theorem V. If P, A, and B, be functions of x, and if P is prime to A and B, it is prime to their product. 173. The following theorems are also of importance : — 174. Theorem VI. The least common multiple of two integral functions that are relatively prime, is = their product. Let the functions be Ax'^ + Bx^~'^ -{■ ... = M, and a.r" + 5:r"'^ + ... = N, then MN is their least common multiple, and is of the degree m -{- n. For if not, let SxP + TxP''^ + ... = P be their least common multiple, where p = m -j- r and r ^ n. Then, since P contains J/, it will be equal to the product of M by some polynomial 11 — Wxf-^ Xr'"^ + ... ; that is, P = MR, and is also divisible by N ; and ili"and A^ being prime, P must be divisible by iV(150), although iV be of a higher dimension than P, which is impossible. INVOLUTION. 89 Again, if p = m + n, then ?• = n, and R being of the same dimension with N, and also divisible by it, it must be identical with it, or else the coefficients of R will be multiples of those of N, and exceed them in dimension. Therefore MN is the least common multiple of M and N. 175. When the division of one integral and rational function of a quantity (168) by another produces a quotient without a remainder, in which the coefficients of this quantity are not integral, the latter function is called a relative measure of the former ; and a similar measure of two or more functions is called a relative common measure. Thus, the greatest common measure of Gx^ — ix* — lla;^ — 3xr— 3a: — 1 and ix* + 2x^ — IS^r^ -{- 3x — 5 is 2x^ — ix"^ -]- x — 1, whereas 39 39 39 their greatest relative common measure is —x^—3dx^-\——x — -. 2 4 4 If in any case the leading quantity in the first term of any dividend or remainder, in the process for finding the greatest relative common measure of two functions, is divisible by that in the first term of the divisor, while the coefiicient of the former term is not divisible by that of tlie latter, the coefficient of the resulting term in the quotient will be fractional; for it is not necessary in finding this measure, in any case, to multiply the dividend by such a quantity as Avill make the coefficient of its first term divisible by that of the divisor. INVOLUTION. 176. The object of involution is to involve, or raise quantities or roots, to any assigned powers (31). Powers are said to be odd or even, according as their exponents are odd or even numbers. It appears from article 31 that any power of a quantity is equal to the continued product of tMt quantity, repeated as factor as often as there are units in\ exponent of the poAver ; but, generally, powers may be found b^ lore concise methods. CASE I. — TO INVOLVE SIMPLE QUANTITIES TO ANY ASSIGNED POWER. 177. EuLE. Multiply the exponent of the quantity by that of the power. If the quantity has a coefficient, involve it separately by actual multiplication (31), and -prefix its power to that of the literal part. 90 ELEMENTS OF ALGEBRA. Odd powers of negative quantities are negative ; in other cases the powers are positive. When the given quantity is the product of two or more factors, their powers must be found separately, and then the product of these separate powers is = the required power. EXAMPLES. 1. The square of a = {of = a}'"'' = c?. 2 0f4a^ = (4a3)2 = 42a3x2^ 16a«. 3. The cube of - a* = (- a*)» = - a>^^ = - a'-. 4 of - 3xf = (- Sxfy = - 3'xy = - 27xY, for (xfy = ix)\ff = xhf. 5. The fourth power of 5aV = (Sa^x^)* = b^a'^x^'' = 625aV^ 6. The third power of — 2axY = (— 2axyy = — 2^aVf = - 8aVf. 7. The nth power of a"* = (a"*)" = a""". The principle on which the rule is founded is that (a^y = a^ .a^ = a^''^ = a^ ; (a^y = a^ . a^ . a^ = w""^ = a}^ ; and generally (a")"* =:^ a^ .a^ .a^ ... a^ being repeated m times as a factor ; but this product is a power of a, whose exponent is = n + « + « ... n being repeated m times (65), which sum is = mn; or (jjnyn _ (jmn^ Similarly (xh/y = xhf . x^ = a;^''^ ^ = x^ ; so (x^y = x^^; and generally (x"'?/^y = x^'t/" X .r™?/" X ... x'^y^ being repeated r times as a factor; but this product (65) is = x"* . a:"* . x"* ... X 'f' .y^ .y^ ... x^ and y^ being each repeated r times ; and by the preceding proof, the product is = yrm y^ yrm — x'^^y''^. When the quantity consists of three simple factors, the proof is the same ; and this comprehends also a numerical coefficient, since it may be considered as a simple factor. The third part of the rule regarding the signs is evident from the rule for the signs in Multiplication (64). Thus, (+ x)( -\-x)= -\- x^ and (— x)(— x) = -\- x" ; also (— x){— a:)(— x) = — x^ for (— x) (— x) = x^, and x^{— x) = — x^. And when — a; is repeated as a factor an odd number of times, the product is negative ; but if an even number of times, it is positive. And + x, repeated as a factor any number of times, whether odd or even, will always give a positive product. These results may be represented generally in the following manner : — Let n be an integer even or odd, then 2n is even, and 2ra + 1 is odd, and {+xy-={{+xyY = {+x''y=+x'- ^ (_ xy^ = {(- xyY = (+ ^y = + x^" (- x)2"+^ = - x(- xy« = - x(+ x^") = - a;2»+ (+ x)2«+i = + :r(+ xy^ = + a:(+ ar"") = + a:^"^^ T INVOLUTION. 91 The last only gives a negative result, since it is an odd power of a negative quantity. These expressions may be more briefly repre- sented by taking + a (plus or minus a) to denote -\- a ov — a, as may be required. When the upper sign is taken, the first line of (yl) above will be given, and when the lower, the second line, if the exponent be 2n ; and when it is 2n + 1, the third line oi {A) above will be given when the upper sign is taken, and the fourth when the lower sign is taken. The expressions are (± xy" = {a^)T = i+^'y=+^''" (± ^yn+i = ±x(± x)^" = ±x(+ a;2") = ± x2«+i EXERCISES. ■ 1- Find the cube of 2x, . . (the power^ P=8x\ 2. square of 3a:-«/, p=dxY. 3. cube of — ia^x, P=-Q4:aV. ' 4. cube of - 8xY, P=-5l2xY. 5. fourth power of 7a^x^, . ... P= 2401a' V. 6. seventh power of — 2x7/* z\ . . . P=-128a;yV CASE II. — TO INVOLVE COMPOUND QUANTITIES TO ANY ASSIGNED POWER. 178. Rule. Involve the quantity to the proposed power by actual multiplication. EXAMPLES. 1. The square of a -f- x = (a + a;)(a -\- x) = a' + 2ax + x^. This result is found by multiplying a + xhj a -{■ x (31). 2. The square of a — x = (a — a — x) = a? — 2ax + oc^- 3. The square of ax -\- cy = (axl cyf — c^x^ + 2acxy + (?y^. 179. It appears by inspecting the coefficients of the three terms in this result, which is found by actual multiplication, that {2acf — 4a^c^ ; and hence, 180. When a trinomial properly arranged (72) is a complete square, the square of the coefficient of the middle term is equal to four times the product of the coefficients of the extremes. 4. The cube of x — y. By actually multiplying, it is found that {x—yy=:(x—y~){x—y)~ x^ — 2xy + y'^ ; and this being again multiplied hy x — y, gives (x -. yf = x'- Zxhj + 3x/ -y^=:x^-y^- Zxy{x - y). 92 ELEMENTS OF ALGEBRA. From which it appears that the cube of a binomial is = the cube of each of its terms, and three times their product into their sum, with their proper signs. The rule is evident from the definition of powers (31), as it applies to all quantities, simple or compound. EXERCISES. 1. Fin( i the square of 1 — x, . = 1 - 2x + X-. 2. square of x + 1, = :^-' + 2x + l. 3. . square of ax — cy, = aV — 2acx^ + c-y^. 4. cube of a — x, = o? - 3a-x + 3ax2 - x^ 5. square of 2x2 — 3/, = 4x* - 12x-/ + 9/. 6. fourth power of a — x, ■= a^ — 4a'x 4- Ga'x- — 4ax^ + ^''• 7. .... cube of 2x2 - 3x + 1, = 8x« - 36x5 ^ QQ^4 _ (j3_^3 _|_ 33_^2 _ 9^ ^ 2_ 8. ... sixth power of X — ?/, = x^—fSr'y + 15xy-20xy + 15xy-6x/+ /. 9. ... nth power of 2ax'", . . . . = 2"a"x'"". 181. Should the given quantity consist of two or more factors, and one or more of these factors be compound quantities, and should it be intended merely to express the power to which each factor is to be raised, the rule of the first case (177) must be applied to these factors, as if they were all simple quantities. Thus, we raise 2x'(a — x) to the second power. {2x\a - x)Y = 2V(a - x^. But when the powers and products of the factors are not merely to be represented, but actually to be expressed by a series of simple quantities or monomials (that is, to be expanded or developed), the powers to which the separate factors are to be raised may be expressed, in the first place, as above, and after- wards, the operations indicated must be actually performed ; thus, the quantity 2'-x''(a — x)^ given above is = 4:X^(^d^ — 2ax + x^) = 4:aV — 8ax^ -f- 4x«. Or the given quantity may first be actually developed, and then the result raised to the given power. Thus, the given quantity 2x^(a — x) when expanded is = 2ax^ — 2x*, which, being involved INVOLUTION. 93 to the second power by actual multiplication, gives (2ax^ — 2x'^y = iaV - Sax' + 4:x\ So {SxXa - xy(a-\-x)Y = 3V(a - x)Xa+xy = 9x«(a* - 4«'x + Gd'x^ - Aax^ + x*)(a2 + 2ax + x^) = 9a;\a« - 2a^x - a'x" + 4:0?x^ - aV - 2ax^ + .r«). Now, actually multiplying each term by Ox®, the final polynomial is obtained. If the given quantity be first expanded, it gives Sx\a- — 2ax -f- x'^)(a -\- x) = Zx^(a? — a^x — ax' + x^), and the square of this is — 9x*'(a^ — d^x — ax'^ + x^f ; and if the square of the second factor be actually found, it will be the same as the second factor above; and this being actually multiplied by 9x^ gives the final result. Powers of binomials are very easily and concisely found by means of the binomial theorem given in article (482) ; and the square of a polynomial can easily be found as follows : — Rule. To find the square of a polynomial — square the first term, and take twice its product into all the terms that come after it ; then square the second term, and take twice its product into all the terms that come after it ; in the same manner square all the other terms in succession, and take twice their product into all the terms that come after them up to the last ; thus, (a + 6 + c + «/64 = ->^(V64) = ^8 = 2, or -^64 = V(v^64) = V 4 = 2. Generally, since 7nnr = m X n X r, the root, whose exponent is mnr, may be found by extracting in succession those Avhose exponents are m, n, and r ; thus — The reader may prove these results in the same way : — EVOLUTION. 99 3 Since s/a^ or a^ means the square root of a', therefore; the square of V«^ or of a^ is just a'. So the square of V^* is ^^; the cube of >^x is x ; the wth power of ^x'^ is a;"* ; the square of flV^ is (a)\s/ xf = a^x ; the square of V — 4 is — 4 ; the square of V— «^ is — a^; and the square of V— 1 is — 1. Also the square of VC^ — ^) is « — ^« CASE in.— TO EXTRACT THE SQUARE ROOT OF A COMPOUND QUANTITY. 189, Rule. Arrange the given quantity according to the powers of some of its letters (72), then the first term must be a complete square of some quantity. Find this quantity, which is the square root of the first term, and it will be the first term of the required root ; write its square under the first term, and subtract it from the given quantity, bringing down only the second and third terms, and there will of course be no remainder under the first. Double the part of the root already found, and write it down as the first part of a divisor ; divide by it the first term of the above remainder considered as a dividend, and the quotient will be the second term of the required root, which is also to be annexed as the second term of the divisor ; multiply the divisor now formed by the last found term of the root, and subtract the product from the above remainder, and to this new remainder annex the next two terms of the given quantity for a new dividend. Proceed with the operation as before, and continue it till the given quantity is exhausted ; or if there be always a remainder, the process will of course never terminate ; the root will be an infinite series, and it can be carried out to as many terms as imiy be required. EXAMPLES. 1. Find the square root of 4a;^ + 4a^ — 8ax. Arranging the given quantity according to the powers of a, we have V {4a2 - %ax + 4x2} = 2a - 2x 4a2 4a - 2a: } — Sax + 4^^ — Sax + 4x2 100 ELEMENTS OF ALGEBRA. 9 4 2., Find the square root of —x* — a;^ + -y*. 16" 4 ,-^y +9^* Find the square root of 4a* — 12a^a: + 13aV — Gax' + x*. V {4a* - 12a3x + 13aV - 6aa;' + x*} = 20" - 3ax + x^ 4a* a'-Sax} -12a^x + 13aV -12a^x+ 9aV a^-Gax + x^ UaV-Gax' + a:* 4aV-6ax'+x* 4. Find the square root of a X^ X* x" ^ \a X \ — a 2a 8a' 16a* 2a-fj-x^ x' X*> X* ^"^ a 8a') 4a2 X* 4a2 x" x» + 8a* ' 64a« la x^ _ _x^ _ _x^> _^ _ _f^ a 4a' IGa^j" 8a* 64a« 8a* ■ ' 16a« ^ 64a« ' 256a''' 5x« x" x" 64a« 64a» 256a'» EVOLUTION. 101 In this example the root is an infinite S'^r:e^.- as 4he procese" would never terminate. EXAMPLES. Find the square roots of the following quantities : — 1. (ix* - 12a;Y' + %*)^ . . . . = 2^2 _ 3^2^ 2. (9a« + 24a V + IGx^, . . . . = 3a^ + ix*. 3. (a* - ia^x + Ga'x^ - iax^ + x% . = a" - 2ax + 3^ ..(^,.-aVH-l.y, . . . . - ^^ 3^ . 5. (9a* — 12a'5 + Ua^"^ - 20a6=' + 256*)', = 3a- - 2a& + 56". G. (49aV-24a.r^ + 25a*-30a='a: + 16a:*)J, = 5a? - 3ax + 4:x''. 7. (16a* - 40a^a; + 2oaV - BOax'^y + 64a;y + 64a%)J, = 4a^ — 5ax + 8.ry. The rule for the extraction of the square root of a compound quantity may be easily found by observing by what means the terms of the square root of d? + 2ax + x^ may be found. This quantity is the square of a + x, or « + a; is its square root. In this example, the first term a of the root is the square root of a*, the first term of the given quantity. If the square of a (a^) be now subtracted from the given quantity, the remainder is 2ax + x^, or (2a + x)x ; and if the first term of this, or 2ax, be divided by 2a, which is the double of the part of the root already found, the quotient is x, the second term of the required root ; and if this term x be added to 2a, the divisor becomes 2a + x ; and this again being multiplied by x, gives 2aa; + x^, which, taken from the above quantity, leaves no remainder. When the given square is a trinomial, a\id its root therefore a binomial, the derivation of the rule is very simple ; but it may be proved in a similar manner when the root is a trinomial or poly- nomial. Let the root be a + a: + y, then its square is a^+ 2aa; + x'^+ 2ay 4- 2xy + ?/^ ; or if a 4- a; be denoted by a single letter c, then a-\- x + y = c + y, and (a + ^ + yf = (c + yf = c^ -f 2cy + ^^ Now, 102 ELEMENTS OF ALGEBRA. nv the former case, the square root c -\- y can be found in the usual way, the operation being , ^/{c^-f 2cy+f}=c+y 2c + y } 2cy+y'' 2cy+f But c = a + X, and c^ = a' -\- 2ax -j- x- ; hence substituting their values for c and c^, the above process is changed into V{(a2 + 2ax + a:^) + 2(a + x)y + /} = (a + or) + y (a + xj = a^ + 2aa: + x^ 2{a +x) +y } 2Ca + x>+/ 2(a + a:)?/ + / so that the same rule applies when the root is a trinomial. To prove the rule when the root is a quadrinoraial, as a + x -\- y + z,\t is only necessary to assume a -\- x -\- y — c, and the root would then be = c + 2;, and the proof would be exactly similar to that of the preceding case ; and it may be similarly proved that the rule is applicable to any polynomial. 190. It is evident also that ^/{aj'x^ ± 2abxy + hhf] = ax ± by ; so that when a trinomial is a complete square, its square root is equal to those of its first and third terms connected by the sign + or — , according as the second term of the trinomial is positive or negative. The rule for extracting the square root of numbers may be easily proved by means of this rule, as it is only this rule slightly modified to adapt it to numbers. Let it be required to find the square root of 1156. The root is 34, for 342 = 1156, or (30 + 4/ ='30^ + 2 X 30 X 4 + 4^ (179), which is equal to 900 + 240 + 1^ ; the root of which is found by the preceding rule ; thus — V {900 + 240 + 16} =r 30 + 4 = 34 V 11,56, = 34 30^ =900 32 = 9 60 + 4 } 240 + 16 64 } 256 240 -1- 16 256 The second form of the process in which the ciphers are omitted, is exactly according to the arithmetical rule for extracting the square root. EVOLUTION. 103 Take now the number 64516, which is the square of 254 or (200 + 50 + 4)2 = 40000 + 24500 + 16, and V {40000 + 24500 + 16} = 200 + 50 + 4 2002 ^ 4Q0QQ 400 +50 } 24500 22500 500+4 } 2000 + 16 2000 + 16 Or thus :— V 6,45,16 = 254 2^ = 4 45 } 245 225 504 } 2016 k^ 901 fi W- The process of extracting the square root of the number 64516 or 40000 + 24500 + 1 6 is given above in two ways ; the second is the usual method. The reason of the rule for dividing the given number into periods of two figures is evident, for there are four ciphers neglected after the 4 under 6 in the second process, and two are neglected after the 225 ; and these ciphers depend on the fact, that the nominal value of figures increases 10 times by a change of one place from right to left, and therefore their squares increase 100 times. Thus, in 45, the nominal value of 4 is 4 X 10 = 40, and its square = 40^ = 1600 ; whereas 4 in the number 24 is = 4 units, and its square = 4^ = 16. CASE IV. — TO EXTRACT THE CUBE ROOT OF A COMPOUND QUANTITY. 191. Rule. Arrange the given quantity in descending order (72) ; then find the cube root of the first term, which will be the first term of the required root ; place the cube of this first term of the root under the first term of the power. Take down the next three terms for a dividend, and for the first term of the divisor, multiply by three the square of the part of the root found ; divide the first term of the dividend by this term, and the quotient is the second term of the root. To complete the divisor, annex the quantity last found to three times the root formerly found, and multiply this sum by the quantity last found ; the product annexed to the former divisor gives the complete divisor. Mul- tiply the completed divisor by the last found term of the root, 104 ELEMENTS OP ALGEBRA. 5^ a "** '^ b +3 o *« .. it a w S ^ o S-s ^^ ^•^ 0) si a !3 o =^ a 'a o ^4 |j o p, . a c2 ^ fe '^ ■5 rt^ + + CO CO + + ^ + 2 V -2 ^ + + CO CO ++ e "h •u + 1'^ + + ^'i 12 a s^^ c^ II •§a + tT %% ^^ + ++ S bo CO + s CO CO «, n: 4- ++ ^1 CO + ^ CO CO i CO + + ++ is ^J + ++ m4 + 1 "h + ++ _ *>* ci* J* ^ c H- Is ^ ^ ^0 CO CO +2 e« ^M CO + + + ++ « "I ++ ^1 1c ^H CO ^e ^^§ ft ^1 CO CO CO CO 73 n3 + + ++ + a ,<» -b C3 "^s 8A =2- 'S CO CO CO CO H .2^ + ++ ++ •»a "e"« ^^^ J Sll "h CO CO ++ '^ 9 3 + G ^-"N [> 8 CO CO J. T3 00 1 ++ n3 flj .-^ 'S + ^«^ 8 "^ CO CO M .a s ^ u o S c oi SI' EVOLUTION. 105 The principle of the rule is evident, by merely considering that the given quantity in the first example is the cube of a + x, or a + a: is its cube root ; and the rules for making up this trial and complete divisor are both obvious from the second form of the remainder. Also in the second example, after finding the first two terms of the root or a + x, the first term of the next divisor, formed according to the rule, is 3(a + x^, or 3a^ + Gax + Sx^ ; and Sarz being divided by Scr, gives z for the third term of the root ; and hence the next term of the same divisor is {B(a-\-x)-\-z}z. The rule is thus applied to a trinomial, and it may evidently be ^ applied gen,erally, or when the root is a polynomial. EXAMPLES. Find the cube root of the following quantities : — 1. (a' - Sa^x + 3ax- - x^)i, . . . . = a - x. 2. (8a« - 36a'x^ + 54aV - 27x«)*, . . = 2a2 - 3x\ 3. (x« - 6x' + 15x* - 20x' + ISz'^ - Gx-\- 1)J, =. x^ - 2x + I. 4. (27a' - Siah + 36az'' - 8z^)i, . . = 3a - 2z. =.a-.')». . . . =i-4-^-f-,^. The arithmetical rule for extracting the cube root is derived from the algebraic rule. As an example, we may extract the cube root of 14706125. Its cube root is 245 or 200 + 40 + 5. To adapt this to the second example given under the rule, let 200 = a, 40 = X, and 5 = z. ^^14706125 = 200 + 40+5 a" = 200' = 8000000 3a2=z3x2002 =120000)6706125 (3a+a;>— (3X200+40)X40 = 25600 then 3a^+3ax+x^ = 145600}5824000 3(a+xy =3X24:0^ =172800 882125 {3(a+a:)+^}2=(3X 240+5)5 = 3625 then 3(a+a:)2+3(a+x>+22 =176425} 882125 The second trial divisor 172800 is = 3 X 240- ; but this is found more easily by placing the square of the quantity last found under the complete divisor ; then adding this square, the complete divisor, and its second part into one sum, which will 106 ELEMENTS OF ALGEBRA. give the next trial divisor, that is, 1600 + 145600 + 25600. The reason of this is, that 3(a + xj = Za^ -f Qax + Sx^ = (3a^ + Sax + z^) + (3a + x')x + x^, which is precisely the former complete divisor with its second part, together with the square of the part last found. This method of finding the trial divisor is very simple compared with the common method, especially when the root extends to more than three places ; hence it is given in some treatises on arithmetic. The preceding process, freed from the unnecessary ciphers, may be thus stated — ^14,706,125 = 245 2^= 8 300 X 2^ = 1200 }6706 (30 X 2 + 4)4 = 256 1456 X 4 = 5824 300 X 24^ = 172800 }882125 (30 X 24 + 5)5 = 3625 176425 X 5 = 882125 The reason for dividing the given number into periods of three figures each, is evident from this example. It is evident, too, that 2* or 8 is the next lower cube number to 14, for 3' = 27. The nominal value of the 2, however, is 200, and its cube contains 6 ciphers, which would extend over two periods. The periods must consist of three places, for the nominal value of a figure increasing 10 times when it is changed one place to the left, its cube increases 10^ or 1000 times. CASE v.— TO EXTRACT ANY ROOT OF A COMPOUND QUANTITY. 192. KuLE. Arrange the quantity as in the two preceding cases ; then find the required root of the first term, and this will be the first term of the root. Involve this term to the power corre- sponding to the root, and place this power under the first term of the given quantity to which it will be equal ; then subtract it from the given quantity, and the remainder will be a dividend. Involve the first term of the root to a power whose exponent is one less than that of the root, and multiply this power by the exponent of the root, the product will be a constant divisor. Divide the first term of the remainder by this divisor, and the quotient will be the next term of the root. Involve the part of the root now found to the corresponding power, and subtract this power from the original quantity, and the EVOLUTION. 107 remainder will be a new dividend. Divide its first term by the constant divisor as before, and continue the same process as far as may be necessary. EXAMPLES. 1. Find the fourth root of a* — ia^x + Ga'x^ — iax^ + x*. k J/ {a* - 4a3x + SaV - ^ax" -\- x*\ = a - x ia^} - 4:a^x a — xy= a* — 4:a^x + Qa^x^ - Aax^ + x* Methods for extracting the fourth and higher roots of compound I quantities, and also of numbers, similar to those for the square and cube root, might be deduced; but they would be very complex in their application. The foundation of the rule will appear by taking a particular example. Let a+i+c + rfbe raised to the fourth power. This power may be expressed thus (494), (a + 6 + c + J)* = {(a + 6 + c) + d}* = (a + 6 + c)* + 4(a + 6 + cfd +, &c. ; and (a + b + cy = {(a + 6) + c}* = (a + by + i^a + bfc +, &c.; also (a + by = a* + Aa^b +, &c. : so that the fourth power of a-\-b+c + d, may be thus expressed, a* + 4a^6 + ... + 4a'c + ... + 4a^cf + ..., &c. Or it may be represented in these three forms : — a* + Aa'b + ..., (a + by + ia'c + ..., (a + b + c)* + ia'd + ... neglecting some of the terms which it is unnecessary to put down. After subtracting a*, the first term of the remainder is ia^b ; and this being divided by the constant divisor 4a^, gives &, the second term of the root. After subtracting (a + by = a^ + 4a^6 + ..., the first term of the remainder is 4a*c ; and this being divided by 4a^ gives c, the third term of the root. Again, after subtracting (a -\- b -{■ cy = a* -\- 4:a^b + ,.., the first term of the remainder is Aa^d ; and this being divided by 4a*, gives d, the fourth term of the root. By the same method the rule can be proved in any other case. 108 ELEMENTS OF ALGEBUA. EXERCISES. 1. Find the fourth root of 16a* - 96a'x + 216aV - 21 Sax' + Six*, =2a — 3x. 2. ... the cube root of x« — 6x^ + 15a;* - 20x^ + 15x'2 - 6a: + 1, = x" -2x-{-l. 3. ... the fifth root of 32x^ — SOx* + 80a;^ - iOx^ + lOx - 1, = 2ar - 1. 4. ... the fourth root of a*"» — 4a^'"a;" + Ga^'^x'^'' — 4a'»a;^" + X*", = a"" - X". %^'^ IRKATIONAL QUANTITIES. 193. Wheu any proposed root of a quantity cannot be exactly assigned, this root of it is called an irrational quantity, a radical, or a surd. Thus, the square root of 2 or sJ2 = 1*4142 .,., or = 1, with an interminate decimal fraction -4142 ..., which is not a periodic decimal ; or, in other words, s/2 is an interminate number, the value of which cannot be expressed by any integer or vulgar fraction (158) ; hence it is a surd. So ^2, -^3, ^/%, ^/a, ^a\ isj(a + x), are irrational quantities. ^^94. An interminate number is a non- periodic interminate decimal, either with or without a prefixed integer ; and hence it cannot be expressed by either an integer or a vulgar fraction ; a terminate number is either an integer, a vulgar fraction, or a mixed number. 195. All irrational numbers are interminate ; but interminate numbers are not all irrational ; for irrational numbers are only a particular kind of interminate numbers. 196. An irrational root of any integral quantity cannot be expressed by a fractional quantity (158). 197. Irrational quantities may be expressed either by using the radical sign, or a fractional exponent whose numerator denotes the power to which the letter is raised, and whose denominator indicates the root to be extracted. i IRRATIONAL QUANTITIES. 109 198. Rational quantities are such as have integral exponents. Thus, a, 3fla:*, 2(0^ — x% and 'j-r~i — ^5' ^*''® rational quantities. 199. A quantity is said to be rationalised when it is reduced to the form of a rational quantity. 200. Surds are said to be of the same denomination when tliey are under the same radical sign. 201. Surds are said to be similar when they consist of the same quantity under the same radical sign. CASE I. — TO REDUCE A RATIONAL QUANTITY TO THE FORM OF A SURD. |i 202. Rule. Raise the quantity to that power denoted by the ' exponent of the proposed surd, and then place this result under the corresponding radical sign or fractional exponent. EXAMPLES. 1 . Reduce 3 to the form of the square root. (3)2 = 32 = 9 ; hence 3 = V9, or = 9i. 2. ... 2 to the form of the cube root. (2)3 = 2* = 8 ; hence 2 = ^%, or = 8i 3. ... a to the form of the fourth root. (a)* = a* ; hence a = ^/a*, or = a*. a^ to the form of the square root. {a^y z= a3>'2 ^ ^6 . hence a^ = V«^ or = aS. 5. ... 2a to the form of the square root. (2a)' = ia" ; hence 2a = V^a^ or = (4a')*. — — to the form of the square root. /2a'V 4a« , 2a' 4a« (-3^) =9^'^^"^^-37=^9x^- a*" to the form of the nth root, (a"*;" == a"*" ; hence a"* = -^a™", or = (a"'")''' 110 ELEMENTS OF ALGEBRA. 8. Reduce to the form of the rth root. { I = —ir- ; hence = /Ga^ to its simplest form. I IRRATIONAL QUANTITIES. 8. Eeduce ^IQa^x* to its simplest form, ^IGa^x* = ^8a^x^^2a^x = 2ax^1o?x. 193 111 4. Simplify the fraction V 125' 192 ,64 ,3 8 8 ^125=^25^5 = 5^5- 5. Reduce /\/18aV(a^ — x^y to its simplest form. V18aV(a2 - x^J = ^QarxXa" - xy^2x(a' - x^) = Zaxiol^-x'J 8aV/ 6. Reduce V ^. ^"/^ to its simplest form. 32cy 'a:^ First simplifying the fraction, it becomes -n-, and iJ-r-r- = 4c y ic^y ts/a^x^sja _ axhja _ ax a sj^c'sly - 2^~y ~ 2^^F IGa^x* 7. Reduce ,^^, . ^ to its simplest form. ^ 81c*y' 16o,V _ >y8a^xV2ft^a: _ 2ax^ 2a^x ^81?/ ~ ^27cy^3cy ~ 3^ ^"33^" The reason of the rule is, that any root of a quantity is equal to the product of the same root of its factors (186.) EXERCISES. Reduce V18 to its simplest form, ,16 3V2. 2.. 2 ^8l ■ ■ =8^3- V8aV = 2a^xsJ2a. 8a*T3 ^27cy 2c?x 2x - 2,cY"'Zy s/a\a^ - aV) = a'^/aiy-x''). ^16/(a'-c* - x') = 2x7j4/2xyXa' - x'). 112 ELEMENTS OF ALGEBRA. CASE III. — TO TRANSFORM A FRACTION WITH A RADICAL DENOMINATOR TO AN EQUIVALENT FRACTION WITH A RATIONAL DENOMINATOR. 204. Rule I. When the quantity under the radical sign in the denominator consists of a single letter, find the least multiple of the exponent of the root that exceeds the exponent of the quantity under the radical sign ; then multiply this quantity by such a power of the simple letter as will make its exponent equal to the preceding multiple ; multiply the numerator by the same quantity placed under the radical sign, and extract the root of the denominator. EXAMPLES. 1. Rationalise the denominator of -vt-t • The least multiple of 3, that exceeds 7, is 9 ; hence multiply d} by a'^, and the numerator by -^a'^, then 2. Rationalise the denominator of 2^12 The exponent of 12 is 1, and the least multiple of 3 above 1 is 3 itself; hence multiply by 12^ or 144 under the sign -^, and 4 _ 4^144 ^ 4^/144 ^ ^144 ^ ^y8^18 ^ 2^18 ^1 2^12 "2^1728 2X12 6 6 6 3^ 4a 3. Rationalise the denominator of 3^x» The least multiple of 4 that exceeds 5 is 8 ; hence multiplying "^^^ 3<5/xV~ 3.ya;« ~ 3a^ ' 4. Rationalise the denominator of r-^-To- The least multiple of 3 above 10 is 12 ; hence 5^x''~ 5^x">x^ ~ 5^x'' ~ 5x* ' la 6. Rationalise the denominator of — r-r- IRRATIONAL QUANTITIES. 113 As the values of m and n are not known, the least known multiple of n that exceeds m is mn ; hence multiply both terms of the fraction by ^x"'^'*~\ then 2a _ 2a^/x^^^ _ 2ayx^' 3 V5' 3x 2^ a?' 1 2^x2" x^ 3V5 6 2a^ Vx" X" 3^3.m(n-2) ' 2x^^» 205. Rule H. When the quantity under the radical sign in the denominator is the product of two or more factors ; find a multi- plier for each factor as in the former case, and multiply both terms of the fraction by their product, and then find the required root of the denominator. examples. 1. Rationalise the denominator of , , . - „ ■ The multiplier is evidently a^x under the sign ^ ; hence -^aV 114 ELEMENTS OF ALGEBRA. 2a 2. Rationalise the denominator of — The multipher is a:'"^''"^)^"^'"') under the sign ^ ; and hence 2a 2a^a:'"('-'^)^"^'""') 2a>/(x'"y")'-^ 3^^,„^n 3^/x'"'-^"'- 3a;'»^" 1. Rationalise the denominator of EXERCISES. 2a 2a^l^ Sx" 3xV«F 3„^5' VaV Say' _ 3yV4aV ^2ax*' ' ~ 2a;2 ' 1 ^ yc^. 2. Reduce >^(a — xf and V(« + ^T to a common radical sign. 2 3 4 9 - and - are equal to - and - ; hence the quantities are (a — x)* O 2 DO and (a + a:)?, or >^(a — x)* and ,^(a + xf. 207. The rule is founded on this principle — that the fractional exponent of a radical may he changed into any equivalent fraction lit nifpvi'nn iho t-inliio r\-P fhn o^ti^rl K,ju^ijiiK,ii.i. uj u, luuivut, /nay ua ciiuii without altering the value of the surd For let ^a"* be the given surd then ^a™ = c, and hence a"* = «""■ = C"" ; and therefore ^a""" = c, that is, a"^ = a m m For let ^a"* be the given surd ; then ^a"* = a" ; let a" = c, then ^a™ = c, and hence a"* = c"; therefore (a^y = (c^y, or 118 ELEMENTS OF ALGEBRA. EXERCISES. 1. Reduce V2 and -^3 to similar radicals, . = 2i and 3i 2. ... a\ xi, and ?/*, to similar surds, . = a", x^^, and y^'^. 3. ... 2 V (a^ - x^) and 3^(0" + x^) to the same radical sign, = 2(a2 - xy and 8(0^ + x^. 4. ... ^(a + x)"* and ^(a — x)^ to similar surds, = (a + a:)"« and (a — 3-)"«. CASE V. — ADDITION OF IRRATIONAL QUANTITIES. 208. Rule. Reduce the radicals to their simplest form, and if the surds be similar, add their coeflacients, and prefix their sum to the radical part ; but when the surds are different, their addition can only be represented by writing them in order, and connecting them by their proper signs. EXAMPLES. 1. Add2V54and4V24. 2V54 = 2V 9V6 = 2 X 3^6 = 6^6, and 4V24 = 4V 4V6 = 4 X 2^6 = 8V6; therefore 2 V 64 + 4V24 = 6V6+8V6 =(6+8)V6 = 14V6. 2. Add 2^250 and 3^54. 2^250 = 2^125,^2 = 2 X 5^2 = 10^2, and 3^^54 = 3^27^2 =3 X 3>^2 = 9^2 ; hence the sum = 19-^2. 3. Add2v|and3vi|. 2v| = 24 = 2f =ve, and o ,32 _ 96 _ V16V6 3x4 4 4\ 7 and the sum = (1 + - )^6 = -^6. 0/ 3 » IRRATIONAL QUANTITIES. 117 4. Add 2a V— and ZhJ—^. 3u CC 0C71 ^(XOO 2a J — = 2a J — —■ = /Jxv, y f y ^ "^ and 36V x'xy Sbx" y' ~ f ^/xy. hence the sum = / 1 ^)\/xy = ■^(2ay + Zhx)i^xy. 5. Add 3V24, 2V54, and 3^aV. 3V24 = 3V4V6 = 3 X 2V6 = &^/Q, 2V54 = 2V9V6 = 2 X 3V6 = 6V6, and 3^aV=3^aV^ax2 = 3aa:^aa:2; and the sum = 12V6 + Zax^ax"^. EXERCISES. 1. Find the sum of V18 and V32, . . . = 7V2. 2. 2V54and3V294, . = 27V6. 3. ,2 ^ ,27 • • = gve. 4. ^80 and -^^270, . . = 5^10. 5. ... V27aVand V12aV, = ax(3a + 2x)ts/Sx. 6. 4a^^and3cx^^, = -.^Cia+3cy)4/x^y\ 7. Za^/x^y and Zcsfxz, = ■ 3ax/s/xy + 5c\/xz. 8. ... -Vaar, 3^a*ar, and — 2 s/cx\ I = -^A/ax -\- Sa^ax — 2x/t/cx. CASE VI. — SUBTRACTION OF IRRATIONAL QUANTITIES. 209. RtHLE. Change the sign of the subtr^^hend, and proceed as in addition of surds. 118 ELEMENTS OF ALGEBRA. EXAMPLES. 1. From 12V20 subtract 3V45. 12V20 = 12V4V5 = 12 X 2V5 = 24V5 -3V45 =-3V9V5 =-3x 3V5 =-9^5; hence the difference = (24 — 9)^5 = 15^5. 2. Subtract 4Vk from 6 Vo^- and 6 V^ = 6 V^ = ^VV30 = ^ V30 ; hence the difference = ( - — - )^J30 = — - V30. \o 5/ 15 3. From Sa'^a'x^ subtract 2x'^a'x\ and 2x2^aV = 2x2^a: V«'^'' = 2x V"""^ ; hence the difference is =(3a^—2x^')^ya^x'. 4. From 5a/J—^ subtract SbJ-x- y yZ y y^ ,x^ „ x^xy 5ax , and ^^s/ti — ^^'^''TT' — "^rr'^^J and the difference = (-^ f-)v^3/ = tC^^^ — ^ix)\/xy. 5. From 5->ya*a;2 subtract 2VaV. and 2)^1 a' 3? = 2Va*^V«^= ^c^xsjax; hence the difference = Ba^^ax^ — 2o?xslax. IRRATIONAL QUANTITIES. 119 f EXERCISES. 1. Froin5V72 subtract 2^/32, . .125 ^ax's/ax^ 2c V 4- 5V(a - xf 4V^, . . 2a^X/s/a^x, x^ 2asJ{a—x), = (3a — 5x)v'(a — x). . = 22V2, aa:(3a;2 - 2a^)V«x. -^(2cz — Saxy^/^xyz. CASE VII. — TO MULTIPLY IRRATIONAL QUANTITIES. 210. Rule. Simplify the radicals, and, if necessary, reduce them to the same denomination ; then multiply together the quantities that are under the same radical sign ; and if they have coefficients, prefix the product of these. When the quantities under the radical signs are powers of the same literal part, their product may be found by adding together their fractional exponents. The multiplication may also be performed without previously simplifying the radicals. EXAMPLES. 1 . Multiply 3 V «ar by 2/^xy. 3/s/ax X 2slxy = 6^/ax/^xy = G^ax'^y = Qxs/ay. q K 2. Multiply 2Vr and ^^/-. 3. Multiply 2aV?^ by 3xV^^^. 52 *^ %x^y ■'4=^4, 2aV— - X 3a.V 8xy 5z X 8x^y ^ iOx^yz 6ax/J SyV _ Gaxyz 3 30 _ 6ayz 40x2 Vth = ear/^TT^ = ~^V30 = j^ayz^SO 400 20 10 120 ELEMENTS OF ALGEBRA. 4. Multiply 2V(« — ar), Zax^J{a -\- ar), and bis/ax^. 2V(a — x) X 3axV(a + x) X 5\fax^ = 30ax^{axXa - x)(a + x)) = ZOax^s/{ax(a^ - x^)}. 5. Multiply 3aV:p^ by 2x^0?. 3a V^' X 2x^0? = Qax'^i^x^a^ = 6aA-JaS = Qax'^x^a^ = 6. Multiply 3aVx' and 56^x^ 13 Za^x^ X 5S^a;2 = 15a6xbS = 15a6a;t^i = 15a6a;« = 15a6^x'^ = 15a6,yxi2^x = loa&x^^«/T. 7. Multiply 3V(a — a:) and 4^(a — a;). 3V(« - x')X 4^(a - x) = 12(a - x^^a - x)i = 12(a - x)^^ = 12(a - xy = 12^(a - a:/. The reason of the first part of the rule is evident from (177 and 186) ; and the truth of the latter part of it is evident from the following example : — ^a"* . ^aF = a" . a«' m mq p np but a" = a'", and a' = a"« (207) ; and hence the product is mq np = a"^.a"^ = "y«""\/«"^ = "^a"''a"^ = "^a™^+"^, and this is mq+np ^ a "' , where the exponent is the sum of the given exponents. 1. Multiply SVax*, 2V5, and 3^^:, . . . = ISx^V^a. 2. ... 2v|and5v| = T^^' 3. ... 2V-^ and 3V— o . . . . = — ^. ^ / ^ x' ' X 4. ... 3a^(a — a:) and 5x^(a — x)^, = 15ax(a — x). 5. ... 2a^x and Sx^^/a, . . . . = ^ax^c^x^. 6. ... 5aV(a — a:), 2^ax, and 3^x^ = 30ax:;y{aV(a - x)"}. IRRATIONAL QUANTITIES. 121 7. Multiply Bs/x^ and 8^x\ . . . . = iOx^^x^ 8. ... Sa^Qc" - /) and Bb^ix"" - /), = 15a6^(x2-//. 9. ... ^(a - x) by s/a?x by 2z^aV. Zas/a^x = Za^sfax, and 2a:^aV = 2x^^aV, and ^ 3 ^ ^ = ZaXax )^ _ Za\axyi _ 3a^ oV _ 3^ oV _ 3a^ ^a^x^ _ 2x\aV)i ~ 2xXaV)S ~ 2x^^aV ~ 2x2^a''x« ~ 2x' ' ax 5. Divide 2^(a - xf by 5^(a - x/. 2^f(a-xy 2(a-x)i 2, ^^ . 2, ^, 2 „,, BA^^a — xy 5(a — xy 5^ '^ 6 ^ 5^ ^ '^ The first part of the rule is evident from (187) ; and tlie seconc part may be shewn thus — m p m? np nq/^mq ^mq mg- nj a-^a'^ = a-" - a"' = ^^, = '^{/—^ (187) = V«'"'"'^» ov a "' where the exponent is = the difference between the given expo nents — and -. n q EXERCISES. 1. Divide 3V24 by 5V6, = ^ 2. ... 2av'«'^ by SarV^^'j = -7-2 ^- - V|by3v|, = ^ 4. ... 2aV^by4V^: . . . = ^W^^ 5. ... 2V5 by 3,^4, = |^500 6. ... 5a/>^a^x hj lOx^a^x^, . . . = — y/^aV 7. ... 5^x' by 4^:r^ = ^^t" 8. ... 3aV(:c'-/)'by4T^(x^-/),= ^^^^^^(x^-y^) IRRATIONAL QUANTITIES. 1*23 CASE IX. — INVOLUTION OF IRRATIONAL QUANTITIES. 212. Rule. To involve a surd, multiply its fractional exponent by the exponent of the required power ; or multiply the exponent of the quantity under the radical sign by the exponent of the required power, and place the result under the radical sign ; then simplify the result, if possible, and prefix the same power of the coefficient. The required power is often more easily obtained if the radical is not simplified previously to applying the rule. EXAMPLES. 1. Find the square of Ba^x'. (Sa^xy = 9a%x^y = Qa^x* = da'x^/x, or (3a^a;2)2 ^ 9^2^-^* = 9a''x^x. 2. Find the cube of 2V3. (2V3)' = 8V27 = 8V9V3 = 8x3^3 = 24 V 3. 3. Find the cube of 2aV— -• (2<.V^) ^ » - 8aV-4- = 8aV— T" = — 2" V^F- 4. Find the fourth power of S^^a^ — x^). {3^(0" - x')y = 81^(a2 - xy = 81(a2 - x^') X ^(0" - x"). 5. Find the cube of 2a V ^. ~ ^l __L_^V« • • = Sayx^. 3 square root of 10a^x^^/;2S = a( 1000a VyV)i. 4 cube root of 8xV^, • = —(9Gay'z')l :. fourth root of 81^, . . .=(2x3")^. o G cube root of 8a;V(a:^ - y^f^ = 2x{xXx^ - yy}i. 214. By means of the preceding rules on the calculation of surds, the operations of addition, subtraction, multiplication, and division, and also involution and evolution, may be performed on Ciompound irrational quantities ; that is, on compound quantities, all or some of whose terms are irrational. EXAMPLES. 1. Add Sai>/x + Ax and 5^^ — Sax + 2sJ cz. The sum is = (3a + 5)^^: + (4 — 2>d)x + 2^Jcz. 2. From 8,^xy — 2ax + d^^z"^ subtract 5^y — 2^z^ + 5a:. The difference = h^z" - (2a + 5> + S^fxy - 5Vy. 126 ELEMENTS OF ALGEBRA. 3. Multiply Sa^x - 2y + 2^z by 2^y - x. Zas/x '-2y + 2^z 2sly — X Qa^xy — Ay^y-\- 4^3/V — Zaxi^x + 2xy — 2x^z As no two terms in the product are the same, addition will not alter its form. 4. Divide 0^ — xz — y ■\- z\ly by a: — sjy. X - sly\x^ - xz-y -\- z^y{x + (Vy - 2) x'' — x/s/y (Vy — 2> —y -\-Zsly 5. Find the square of a? — 2x^. a^ — 2xi a? - 2xi a« - 2a^xi — 2o?x\ + 4x a® — 4a'x* + 4a; 6. Extract the square root of 4a: — 4aa;*yJ + a^^l {4a; — 4ax^^ + a^2/?}i = 2xJ — oyS 4a; 4a;i — a^^ } — 4aa;*_y* + a%5 — 4aa:%i + a'^^S This example is performed by the rule for extracting the square root of a compound quantity (189) with the preceding rules for irrational quantities. The rule for extracting the cube root (191) may be similarly applied. EXERCISES. 1. Add 2xsly + h^a:^, Zzsjy - 5, and 8 + 2^ax^ = (2x + 3z)^y + 7^ax^ + 3, 2. Subtract 3^xy - o^z^ from 8^^:^^ - 12 + 8^y^^ = hs/xy - 12 + 13>^^*. IRRATIONAL QUANTITIES. 127 3. Multiply lays/y + 2V(a — x) by 2,s/y — ^^/{a - x\ = Qay"^ + 2(3 — 5ay)^/y(a - x) - 10 (a - ar). 4. Divide x^ - (3a + 2')x^>^z + Qaxz by x"" - 3aV«> = x' — 2a;V«. 6. Eind the square of ta-^ — 2xf^y, = a'x* — ^ax^sjy + 4x^y, 6. ... the square root of 9a^x — \2axs/xy -\- 4a;-y, = Za\fx — 2x/sjy. Having once established the existence of such quantities as mrds (196), the following important theorems may be easily proved : — 215. Theorem I. The square root of a quantity cannot be partly rational and partly a quadratic surd. Tor, if possible, let sjx — a -\- i^y ; then squaring these equals, f=- c? ■\- y -\- 2ai^y ; and hence X — a^ — y ^y= 2a ' that is, a surd is equal to a rational quantity, which is impos- sible. It may be similarly shewn that ^x = a + \fy is an impossible aquation, and generally, that any root of a quantity cannot consist of a rational and an irrational quantity. 216. Theorem II. In an equation, the members of which consist partly of rational and quadratic irrational quantities, the rational parts of each member are equal, and also the irrational parts. Let the equation be a + V-^ = ^ + \/yi and if a be not = b, let = b + z, where z may be a terminate number, but cannot be a 3urd (215) ; then Vy = 2; + ^x, w^hich is impossible (215) ; hence a = b, and therefore x = y. In a similar manner it may be shewn generally, if a + ^x = b + ^y, that a = b, and x — y. 217. Theorem III. When the denominator of a fraction is of the form c" + 3/", it may be rationalised by multiplying the terms of the fraction by a proper multiplier. 128 ELEMENTS OF ALGEBRA. The multiplier can be found thus : — In the three cases of division (87) assume a^ ■= c and x" = y; i. -2. 3. 1. 2l ^ and hence a = c", d^ = c", a^ = c" ... and x = y", /;" = y", x^ = y^ ... and these cases become 1. When n is either an even or an odd number, c — y ^li ^^2 1 tz± 1 :lzj_ _ _ = c« +c»?/» + c»y»+... +y». c" — y" 2. When n is an odd number, C+V ^^^^ 112^1 ^Z3_ 2 «;J_ — £-^ = c » - c « y« + c » 3/ '-...+ y » . 3. When n is an even number, c—y '^^ !il?i ^1132^ nil — ^ =c" — c"2/"+c"y" — ... — y". It is evident from these expressions, that if the quantity to which any of these fractions is equal be multiplied by its and assume that where the sfx and the fs/y may be, either both irrational, or one rational and the other irrational ; squaring both sides gives x^y^2slxy = a-^ sih (A) Therefore (216) a: + y = a, and 2sfxy= ^h, from which it follows, by subtracting the second from the first, that X -\- y — 2i>Jxy = a — ^Jh (B) and extracting the square root, that Vx- V^= V«-V6 (C) Multiplying the original equation by (C) gives x-y - slo? — h (D) and half the sum of (A) and (B) is x+y^a (E) (E) + (D) gives 1x ■= a -\- si 0? -A- v.-^' +^'' and (E) - (D) gives 1y = a — si <^ — b, 4 v.— ^"' Whence ^^x±Vy = ^« + ^^" " ^ + ^^ sla^ which is the general formula required. i IRRATIONAL QUANTITIES. 131 EXAMPLE. Extract the square root of 4 + 2^3. Here a = 4, V^ = 2V3, and substituting these values in the general formula gives V4: + 2V3 -^4 4+ V16 - 12 ^^ V16 - 12 = V3 - 1, .-. V4 + 2V3 = 1 + V3. [n the same manner the square root of 4 — 2>v/3 by the general formula is V4 -2V3 = V3 - 1. When a^ — b is not a square, (D) is irrational, and conse- quently the general formula in this case becomes more complex than the original quantity, and the transformation is useless ; bence the formula can only be applied so as to simplify the result ff^hen >v/«^ — 6 is a rational quantity. EXERCISES. 1. Extract the square root of5+2V6, = \/2+/v/3- 6 + 2V5, = V5 + 1. 8+ 2V7, = V7 + 1. 37 ± 2V70, = ^35 ± V2. 2j- V5, = ^V5 - 1. 41 ± 24V2, = 4V2 ± 3. IMAGINARY QUANTITIES. 218. Imaginary quantities are those whose values cannot be assigned either by terminate or interminate numbers. These quantities, which are also called impossible quantities, occur when it is required to find an even root of a negative quantity, which is unassignable in terms of any real quantities, either rational or irrational ; for any even power of any real quantity, whether positive or negative, is positive, and therefore no even root of a negative quantity can be found. Any root of a negative quantity is equal to the same root of that quantity taken positively, multiplied by the same root of- 1. For — a = a(— 1) ; therefore ^— a = ^a^— 1. When m is an odd number, J^y— 1 = — 1 (186.) When m is an even number, it will be equal to some power of 2, as 2'', or to the product of some power of 2, as 2'', by an odd number n. , Let m = 2'', then the with root of — 1 is = -y/— 1, or equal to the square root taken r times in succession. Thus, if r = 3, m = 2^ = 8, and ^— 1 = ^— I = \/{V(V— 1)}» which is ima- ginary, because V — 1 is so. Let m = 2"^, then -^— 1 = v^(^— 1) ; and as n is odd, -y -2V^^, . . . = 6V15 3. ... » + V^^ by 2 - 3 V^^, = 6 + 2V-2-9V-1+ 3V2 4. ... a — &a/ — c by a + ^V — c, . . = a'^ + 6^c 5. Find the square of a — hsl — 1, = c? — 2ahsJ — 1 — h^ 6. ... product of a — 3 V — * and c — "2,^ — d^ = ac — 2a/^ - d — 3cV — b — 6 V&c^' CASE II. — DIVISION OF QUADRATIC IMAGINARY QUANTITIES. 221. Rule. The quotient of two quadratic imaginaries is found by dividing them as if they were real, and prefixing the positive or negative sign to the quotient, according as their signs are like or unlike. For = = , = + Vr» == + ^P and -1- ^ -- ^ -^^^^__^ ^ _ ^a - V- * - V*V - 1 * examples. + V- - a + v«v- - 1 + V- —b + v&v~ ^ - v~ - a — T - V- ^ n: + v~ - a 4- V«V" — r 1. Divide 8V - 12 by 2V — 3. i^^^ = 4V^ = 4V4 = +4 X 2 :=: + 8. 2V- 3 3 IMAGINARY QUANTITIES. 135 2. Divide - 12V- 6 by 4V -3. 4V-3 3 3. ... 2 4- 3V^^ by 2 - V^^l- The multiplier that will make the denominator rational is 2 + V — 1) which may be found from the general formula (217 *), by assuming c^ = 2, and y^ = V — 1 ; or it may be found more easily by considering that (a -\- x) X {a — x) = a^ — x^. Hence 2 + 3V"^n[ ^ (2 +3V^^)(2+ yll^ _ 1 -j-8V~:il 2 - V^=-i: (2 - V^"lX2 + V^^) ~ 4 + 1 = ^(1 + 8 V"^^), or ^ + ^V~-^l- 4. Divide a + 6^-1 by c + c?V - 1- The multiplier for rationalising the denominator in this example is evidently c — rfV — 1 ; hence c+dsf-l (c+rfv-lXc-<^V-l) C' + CP EXERCISES. 1. DivideSV - 18by 2V -2, . . . = 4V9 or 12. 2. ... -15V-6by3V -3, . . . = - 5V2. 3. ... 4 _ 2 V - 1 by 2 - 3 V - 1, = ^(U + 8 V - 1). 4. ... 3-5V-3by 2 -3V-2, = ^6 - lOV - 3 + 9 V - 2 + 15 V6). 5. ... aV - & by cV - ^, . . . . = ^^/hd. ca 6. ... 2x + 3^V — 1 by 3a: — 2y V — 1, 9x2 + 4^2 EQUATIONS. Il 222. An equation is an expression stating the equality of two quantities, and generally containing at least one unknown quantity. Thus, a; — 3 = 4 is an equation which states the equality between x — S and 4 ; in which x is the unknown quantity. By the equation, it appears that if 3 be subtracted from x, the remainder is 4 ; and hence x must be 7. The equation x = 5 cannot be said to contain an unknown quantity, as the value of X is given ; and so of the equations x = 5—3, x = a, or x = b + c; in which a, b, and c, are known quantities. 223. A quantity is known when its value in numbers is given ; and when this value is not given, it is called an unknown quantity. 224. The last letters of the alphabet, x, y, z, are used to represent imknown quantities, and the known quantities are either numbers, or ar^ denoted by the first letters of the alphabet, a, b, c, ... 225. The two parts of an equation on the opposite sides of the sign of equality, are called iuembers or sides of the equation ; that on the left is called the Jii-st member or side, and the other the second member or side. Members are composed of one or more terms. 226. The solution or method of solving an equation is the process for finding the value of the unknown quantity ; and an equation is said to be solved when the unknown quantity stands alone on one side, and its value in known terms on the other side. 227. An equation consisting of literal quantities, when their numerical values are known, is called simply an equality. Thus, if the numerical values of a, b, c, and d, be known, and be such that a -\- b = c — d, the expression is an equality. If, for example, a = 2, b = 3, c = 8, d = 3, then the expression becomes 2 + 3 = 8-3 or 5 = 5. 228. A self-evident equality is called an identity, or a verified equality. Asl2 -8 = 4:Ov3a-2b - a=2a- ob + 3b. 229. If, in an equation containing only one unknown quantity, a number be substituted for the unknown ov"*--^-; . .' ^ o OF EQUATIONS. 137 equation is reduced to an identity, this number is called a value of the unknown quantity, or a root of the equation. Thus, in the equation a: - 5 = 16 - 3 + 2, if 20 be substituted for x, it becomes 20 - 5 = 16 - 3 + 2, or 15 = 15. * This result is an identity ; and hence 20 is the value of x. 230. An equation expresses some relation between the unknown and the known quantities. The value of the unknown quantity is said to satisfy or fulfil the conditions of the equation. In the equation x = 8 -5 the relation between x and the known quantities is simply that X is equal to the difference between 8 and 5 — that is, 3 ; and this value of x being put in the place of x, reduces the equation to the identity 3 = 8 — 5, or it satisfies the given equation. 231. To verify a value of the unknown quantity is to substitute this value for it in the given equation ; if the result be an identity, the value is then verified, or proved to be correct. 232. If only one quantity in any proposed question be unknown, only one relation or condition must be given. For example, if a number be required, such that being increased by 3, the sum shall be 12 ; then if x be that number, the equation ar + 3 = 12 expresses this relation or condition, and x is evidently = 9, for 9 -|- 3 = 12. But it is unnecessary to give another relation such as that the number required being increased by 6, the sum is 15, as nothing would be deduced from this further than that the number required is 9. If, however, a second condition be added, so as to give a new value to the unknown quantity different from that given by the first equation, the question becomes impossible, as the conditions would be inconsistent were a: = 9, and, at the same time, equal to some other number, while by the question only one number is sought. If, therefore, when only one unknown number is sought, a second condition be added, though it be consistent with the first, it is unnecessary, the first being sufficient for determining the required number ; and if the second condition be inconsistent with the first, the question becomes impossible. 238. Equations are divided into numerical and literal; the former contain only numbers combined with the unknown quan- 138 ELEMENTS OF ALGEBRA. tity or quantities, and the latter contain also literal quantities, whose numerical values are known, or supposed to be so. Equations are also divided into determinate and indeterminate; the former term being applied when there are as many equations as unknown quantities, and the latter when there are more unknown quantities than equations. 234. Any power of a quantity, or any product of two or more quantities, or of their powers, is called a dimension. The order of the dimension depends on the number of times that the simple power of the quantity or quantities enters into the dimension, or on the sum of the exponents in the product. When the simple power is repeated twice, it is called the second dimension ; when three times, the third dimension ; and so on : and, generally, if it be repeated n times — that is, if the sum of the exponents be equal to n, it is called the wth dimension. i Thus, if X, y, «, be unknown quantities, x', x^y, xyz, are each of 'i the third dimension ; x*, a:yV, x^yz., x'^z, are of the fifth dimension, and so on. 235. Equations containing only one unknoAvn quantity are divided into different classes, according to the highest power of the unknown quantity contained in them. An equation which contains only the first power of the unknown quantity, is called a simple equation ; one in which the highest power is the second, is called a quadratic; when the highest power is the third, the equation is called a cubic, or an equation of the third degree ; when the highest power is the fourth, an equation of the fourth degree ; when the fifth, of the fifth degree; and, in general, when the highest power of the unknown quantity has the exponent n, it is called an equation of the nth degree, and sometimes it is called an equation of n dimensions. 236. Equations containing two or more unknown quantities are similarly classified. If the highest dimension of the unknown quantities be the first, second, or third, it is called a simple, quadratic, or cubic equation ; if it contain the fourth dimension, it is called an equation of the fourth degree or dimension; and similarly for other dimensions. 237. The dimension of an equation is the same as that of any one of its terais which contains the highest dimension of the unknown quantity or quantities in the equation (40). It is understood, however, that the equation contains no fractional or negative exponents of the unknown quantity, or that it is cleared of radicals and of denominators containing unknown quantities. SIMPLE EQUATIONS. I. — EQUATIONS CONTAINING ONLY ONE UNKNOWN QUANTITY. Equations must undergo some alterations in their form, in order to prepare them for solution. These preparatory transformations depend on the following axioms : — 238. Axioms. K equal quantities be added to the two members of an equation, or subtracted from them, or if the two members of an equation be multiplied by the same quantity, or divided by it, the results will still be equal, or the equation will still subsist. From these axioms, the following rules for the process of solution are derived : — 239. EuLE I. Any term may be transposed from one side of an equation to the other, by changing its sign. EXAMPLES. Transpose the known quantities to the second member, and the unknown to the first, in the following equations : — 1. Let a; - 4 = 6. Transposing the — 4, we have a: = 6+4 = 10. I^^Let ar + 8 = 4 - 3x. I^K Transposing + 8 and — 3a:, the result is I^K a: + 3a: = 4 - 8, 1^^ or 4a: = — 4. 3. Let a; - 2a = - 4a: + 36. a; + 4x r^ 36 + 2a, or 5a: — 2a + 36. 240. The transposition of a negative quantity from one side of an equation to the other, is equivalent to adding that quantity to both sides ; but the transposition of a positive quantity is equivalent to subtracting n from both sides. Thus, let I ax — h = ex — d; by transposing ex, we have ar — ca: — 6= — d. .'.3x = 8 7x = — 6 .'.2x = 18 140 ELEMENTS OF ALGEBRA. which is the same as subtracting ex from both sides. Again, by transposing — b, the equation becomes ax — ax = b — d, which is equivalent to adding b to both sides. Or, by adding (6 — ex) to both sides, the equation becomes ax — b -\- (b — ex) = ex — d -\- (b — ex), or ax — ex = b — d; for — 6 + i destroy each other, and in like manner ex and — ex. EXERCISES. 1. Let 5a; - 3 = 2a: + 5, 2. ... 12 - 3x = G -Wx, 3. ... 5x- 6 — x=15-^2x-d, 4. ... 3x - 2a = 5e- 2x, . . .- . 5x = 2a -\- 5e. 5. ... 3a- 5x + 2c= 2x — 5e -{- 8a, .-. — 7x = 5a - 7c. 6. ... 5x4-16 -2X+ 3a = 5a — x + 8,.-.4:X = 2a- 8. 241. As a corollary from the principle of transposition, we infer that the signs of all the terms of both sides of an equation may be changed, without destroying the equality of the two members. For changing all the signs is the same as if all the terms of each member were transposed to the other side of the equation ; or the effect is the same as if both members were multiplied by-1. 242. Rule II. An equation may be cleared of fractions, by multiplying both sides successively by the denominators ; or by the least common multiple of the denominators. The last method is always the best when the denominators are not aU prime, as it gives an equation in more simple terms. EXAMPLES. 3 2 1. Clear of fractions the equation -x — -x = 11. 4 o Multiply by 4, and 3x — -x = 44. o by 3 ... 9x - 8a? = 132. SIMPLE EQUATIONS. 141 Or, more simply, multiply the equation at once by 12, and the result is 9x-8x = 132. 2. Clear of fractions the equation ^x — - = -— + -. o 4 o o Multiplying by 24, we find 16x - 18 = 15a: + 20. 3. Clear of fractions — — = — h 3. a — X a -\- X a — X Multiplying by a? — x^, we find a(a -\- x) — c{a — x) — a -\- X -{■ 3(a^ — x^). Eule II. depends on the axiom, that if both members of an equation be multiplied by the same quantity, the products are equal. EXERCISES. Clear the following equations of fractions : — 1. I - 4 = I + 6, . . . .-. 3x - 24 = 2a: + 36. 2. ^-| = | + ^, . . . .•.9x-20 = 18 4-60x. 3.i -4 3 2 "5 " 6 - " 10 ^-4, .• .10a:- -40- -12 = 18- -3a:- 120. Y' 1 - 3 = = 5 - 3 a: + 2' X -2 .-. a:+2- -3(x^- -4) = =5(z^- -4)- -3(a: -2). l- -4a: 6 3 8 _ 5 ~ 12 3a: + 10 9 ' -.36- -48a: -27= :30- -24a:- -80. r a 4 _ — X -8 = 6__ X a 4 .-. 4a:(a-|-a:) - %x{a^ - x^) = Q{a^ - x^) - 4a:(a - x). 6 3x_5a 2 ' X 4 3a: a -{- x* ... 72(a -fa:)- dxXa + x) = 20a(a + x) - 24x. a — X 3a _ c 2b X h a -\- X a'' .'. ah(a^ - x^) - da^x{a + x) = abcx — 2b^x{a -}- x). 142 ELEMENTS OF ALGEBRA. SOLUTION OF SIMPLE EQUATIONS CONTAINING ONE UNKNOWN QUANTITY. 243. EuLE. If necessary, clear the equation of fractions ; transpose to one side all the terms containing the unknown quantity, and the known quantities to the other; collect the terms on each side into one sum ; then diAdde both sides by the coefficient of the unknown quantity : the result will be the value of this quantity. EXAMPLES. 1. Given 3x — 8 = 24 — 5x, to find the value of x. Transposing, 3a; + 5x = 24 + 8. Collecting, 8x = 32. Dividing by 8, x = 4. In order to shew that this is the correct value of x, substitute 4 for X in the given equation, and it becomes 3x4-8 = 24 -5X4, or 12 - 8 = 24 - 20 ; and 4 = 4 : which being an identity, proves the value to be correct, or verifies the equation. 2. Given 2x - 10 + x — 15 = 24: —Bx -\- 2x - 33, to find the value of X. In this example, by collecting the quantities in each member, we find 3x — 25 = — a: — 9. Transposing, 3a: + a: = 25 — 9. Collecting, 4a; =16. Dividing by 4, x = 4. 3. Given ix — b = 2x — d, to find the value of x in terms of b and d, the numerical values of which are supposed to be known. Transposing 2a; and — b, 4lx — 2x = b — d. Collecting, 2x = b — d. Dividing by 2, x = ~~— or -(6 - d). SIMPLE EQUATIONS. 143 If numerical values be given to h and d, the corresponding value of X will be found. Let 6 = 5 and d=3\ then x = -(5 — 3) = - X 2 = 1. ... h = \\ and c/ = 5 ; then x — — - — = - = 3. 4. Let ax — h ■= ex — d^to find the value of x. Transposing ex and — h^ ax — ex = h — d. Collecting, {a —e^x = h — d. Dividing by a — c, a: = . a — c Suppose 5 = 13, c? = 7, a = 8, c = 5, *» 5-c?13-76„ then X = = — = - = 2. a — e 8—5 3 5. Let ^ + 3 - J = 10 - y, to find the value of a:. bo 4 Multiplying by 24, we find 4a; + 72 — 3x = 240 — Bx. Transposing, 4a: — 3a: + 6ar = 240 — 72. Collecting, 7a: = 168. Dividing by 7, x = 24. 6. Given | - - -2 5 a: 3 "~4 4- ^ -7 — , to find the value of x. Multiplying by 12 , 9 - 4a: + 8 = 15 -Sx- 9, or -ix + 3x = 15 -9 - 9-8; hence — X = 15 -26 = -11, or X = 11. 7. Let - + 6 = a - + c?, to find the value of x. Multiplying by ac we find ex f abc : = ax + acd, and abc — acd z - ax — ex, or ae(b -d) -. = (a - e)x; hence X - _ac(h-d) a—e By giving a, 6, c, and c?, numerical values, the corresponding values of x may be found. 144 ELEMENTS OP ALGEBRA. EXERCISES. 1. K 5a: - 12 = 12 - 3a:, 2. ... 15 - 2x = 6x ~ 25, . 3. ... 5x-10-2x = i0 + 4:x - 56, 4....|-2 = 6-| . . . 5....|-| + 15 = |-| + 22, - X — 3 _ X — 1 X — 5 6. ... -^ - 6 - -^ = -3~ X -\- 3 X — 3_a: — 5 vy^ 3^ + 6 ir> - 3a: _ 25 4 12 ~ 6' x-fl r + 2_ 5a: + 1 y. ... -^- + -^ - 16 J—, 10. ... 3(3x - 2) + 5(a: - 6) = 18a: - 4, - ^ + ^-^^ = ". • 12. ...^a:-^a: + 18 = J(4a: + l), . 13. 5 4' 9' X X — b de. a c ac 14. ... _A^ = _^4.1 a —b a -\-b ' a .„ ^ 26 3ca:-Z.c + 2 16. ... 3aa: = a — c a . X — - 26 • _ ab(c + d) a - -b ' , 2a + (8 - -a)hc 3ac(4a - -1) ' (0 4- dXa' -F-) 4c 17....Ilr^ + c=(^±^- ' 4a6 18. ... ^ + l = a^ + 62, T=:l. ox ax SIMPLE EQUATIONS. 145 244. Wlien the unknown quantity enters into the terms of a proportion, an equation may be formed by making the product of the extremes equal to that of the means. Thus, Let mix = a:b, then ax = bm, ... a — X : X = a : b, then ax = b{a — a:), I. — EQUATIONS THAT MAY BE SIMPLIFIED, AND WHICH CONTAIN THE UNKNOWN QUANTITY IN ALL THEIR TERMS, BUT IN ONLY TWO OP ITS POWEltS. 245. Rule. Divide each term by the highest power of the unknown quantity, which is common to them all. 9^2 1. Given Sx^ — Zx = IQx — -, to find the value of a:. 9a: ^- Dividing by a:, we find 5x— 3 = 16 — — Multiplying by 2, 10a: — 6 = 32 - 9a: 19a: = 32 + 6 = 38 x= 2 EXERCISES. 1. If 3a:2 - 8a; = 24a: - 5a:^ a: = 4. 2. ... Sx" - 15x = 2a:2 + 6a:, x= 7. 3. ... ^Ox" - Qx^ - 16a:2 = I20x^ - Ux\ . . . a: = 12. 4. ... 3ax* - lOax'^ = ^ao? + ax^ . . . . x = Si. 6 2a:=' - — + a:^ == 5a:' - 2x^, .... 23^ x^ 6 ^^ + ^- - 3 2' 7 II. — EQUATIONS IN WHICH THE UNKNOWN QUANTITY IS UNDER A RADICAL SIGN. 246. Rule. To clear an equation of a radical quantity, transpose this quantity to one side of the equation, and the rational terms to the other side; then involve both sides to a power corre- ponding to the radical sign. J 146 ELEMENTS OF ALGEBRA. If more than one term be under the radical sign, the preceding operation must be repeated. EXAMPLES. 1. Let V(^ — 1) — 1 = 2, to find the value of x. Transposing, a,/ (x — 1) = 2 -\- I = S. Squaring, x — 1 = 9 ; and a: = 9 4- 1 = 10. 2. Given V(^-5)-3 = 4-V(a;— 12), to find the value of x. Transposing, /^(x — 5) = 7 — n/(x — 12). Squaring, a; — 5 = 49 - 14 V (:« — 12) + a: - 12. Transposing, 14 V (a; - 12) = 49 — 12 + 5 = 42. Dividing by 14, >s/{x- 12) = 3. Squaring, a: — 12 =9 .-. ar= 21. 3. Let V{5 + s/(x — 4)} - 2 = ], to find the value of r. Transposing, V{5+V(^-4)}= 1 + 2 = 3. Squaring, 5 + V(^ — 4) = 9. Transposing, V (^ — 4) = 9 — 5 = 4. Squaring, ar — 4 =16 .'. X =16 + 4 = 20. 4. Given a + ar = \/{«^ + oc/^/{c^ + x^)], to find the value of a; Squaring, d^ + 2ax -\- x^ = d^ -{- x /^ (c^ + a^), or 2ax + x^ = X ^ (c" + x"^). Dividing by X, 2a + x = /^(c^ + x"^). Squaring, 40^ + 4aa; + x^ = c^ + xK ' Taking x^ from both sides, and transposing, 4aa; = c^ — 4a^ c^ - 4a2 . • . a; = — . 4a SIMPLE EQUATIONS. 147 I ^^ EXERCISES. ^(x + 3) = 4, . . . (^(x-5)=: V(x + 2)-l, . ... \/ x(a -\- x) ■= a — X, ...... V{6+ V(a;-1)} = 3, 6. ... V^ - 2 = V(a; - 8), a -\- X + V « + ^ 8. ... 1 -I- VI + ^ = vl + ^ + \/l . . X = :13. . X = :14. . . X = -. 4. X - a 3' . . X = = 10. . X = = 9. K^ a b - i- . x=- 24 " 25' J. ... — — - =1 njm^ , . X = . »/a + X — /\/a — X I -{- m 10. ... m(ax — b)i = n(cx -\- dx — f)K . x = — ^ -z — r— vvs- 1 4- a-s 1 _ :r' /a - 2V 11. (1 + xy (1 - :t')^ /a-_2\i Va + 4y (4x + l)i + 2x^ _ 4 ^''' •" (4:r + l)i - 2xi " ^' 9* QUESTIONS PRODUCING SIMPLE EQUATIONS CONTAINING ONLY ONE UNKNOWN QUANTITY. 247. The first consideration in the solution of a question in equations is, how to form the equation, by the solution of which the unknown quantity is to be found. No certain rule can be given generally applicable, and at the same time sufiBciently minute, so that the student must frequently depend on his own skill. In some questions, the enunciation of the conditions fur- nishes easily and immediately the required equation ; but in other cases it is not the conditions themselves, but other con- ditions deduced from these, that are employed in forming the equation. In the former case, the conditions are said to be explicit, in the latter implicit. As a general rule, the following may be given : — 1248. Rule. Denote the unknown quantity by some letter; 148 ELEMENTS OF ALGEBRA. upon it and the known quantities, agreeably to the conditions o1 the question, as if the object in view were to verify some parti- cular value of the unknown quantity ; and an equation will be thus formed, the solution of which will give the required value of the unknown quantity. The changing of a question into an algebraical equation will, however, be much facilitated by attending to the following principles : — 1. If the sum of two numbers be given, and one of them be represented by x, then the other will be their sum mimis x. 2. If the difierence of two numbers be given, and the less be represented by x, the greater will be x plus the difference ; and i1 the greater be represented by x, the less will be x minus the difference. 3. If the product of two numbers be given, and one of them be represented by x, the other will be the product divided by x. 4. If the quotient of two numbers be given, and one of them be represented by x, the otheB will be the quotient multiplied by x. 5. If an nth part of any quantity be taken away, -^ parts 4 will be left ; thus, if from x one-fifth of itself be taken, -~x will be o w + 1 left ; but if an nth part be added, the sura will be parts oi the quantity, which results may therefore be put down at once. EXAMPLES. 1. What number is that which, being added to its fourth part, the sum is = 10? Let X = the number required ; then the fourth part of x is expressed by -, and this added to x. must give 10 for the sum, according to the condition in the question. The relation, therefore, between the unknown and the known quantities or numbers, are expressed by this equation, x + | = 10. Multiplying by 4, ix -\- x = 40. Collecting, 5x = 40. Dividing by 5, x = 8. The number is therefore 8. In order to verify this value, H SIMPLE EQUATIONS. 149 must be tried whetlier or not it satisfies the condition of the question ; thus, putting 8 for x in the above, it becomes 8 + J = 10, or 32 + 8 = 40, or 40 = 40, an identity. Tliis result being an identity, shews that the value is correct. It is evident, by comparing the equation above with this method of verification of the number 8, that x in the one, and 8 in the other, are related in exactly the same manner with the known quantities or numbers given in the question ; so that the equation is formed exactly in the same manner as if the object were to verify any particular number. 2. What number is that from which, if one-fifth be taken, the remainder will be 8 ? Let X = the number, then, since one-fifth part is taken from it, the quantity left is -:r, which by the question -8, or 4 r = 8. Multiplying by 5, ix = 40, . ' . X = 10. It will be found on trial that the number 10 fulfils the condition of the question. As the reader must now be familiar with the various steps of transposing, collecting, &c. in the solution of an equation, it will be unnecessary to name these in future, as the steps of solution ivill be evident by inspection. What number is that which, being added to its third part, ;he sum is equal to its half added to 10 ? Let X = the number ; X 3' idded to 10 is = - + 10 ; but by the condition of the question, ;hese are equal, or l = l + '«- i\ 150 ELEMENTS OF ALGEBRA. Multiply by 6, 6x -\- 2x = Sx + 60 8a: — 3.r = 60 5a; = 60, . • . a: = 12, the number required. 4. Find two numbers such that their sum shall be = 47, and their difference = 23. Let X = the less number ; then a: + 23 = the greater, by the second condition of the question : and therefore, by the first condition, a: + a; + 23 = 47, or 2a: = 47 — 23 = 24, .'. a:=: 12, and a: + 23 = 12 + 23 = 35. The two numbers, therefore, are 12 and 35. 5. DiAdde a line 12 feet long into three parts, such that th( middle one shall be double the least, and the greatest triple tlw least. Let X = length in feet of the least part, then 2x = middle ... , and 3a: = ... ... greatest ... ; and since these three parts must together be = 12, then X -}-2x + Sx — 12 6x = 12, . • . a: = 2 the least ; hence 2a: = 4 ... middle, and 3a: = 6 ... greatest. 6. It is required to divide £470 among three persons, so tha the second may have £10 more than the first, and the third £3( more than the second. SIMPLE EQUATIONS. 151 Let X = sum to be received by the first, then a: -f 10 = second, and .r + 40 = third; therefore, by tlie condition of the question, + (x + 10) + (ar + 40) = 470, 3a; + 50 = 470 3x = 470 - 50 = 420, . • . ar = 140 = sum received by first, a: + 10 = 150 = second, a; + 40 = 180 = third. I I Proof; 470= all three. 7. The sum of two numbers is 20, and the less is to the greater as 2 to three : required the numbers. Let X = less number, then 20 -- a: = greater, and therefore ar : 20 — x = 2:3; hence (244) 3a: = 40 — 2a: 5a: = 40, .- . X = 8 = the less number, and 20 - a: = 20 - 8 = 12 = the greater. This example contains an implicit condition — namely, the pro- portion, from which the equation is deduced. 8. The difference between two numbers is 2, and their product €;xceeds the square of the less by 8 : what are the numbers ? Let X = the less, then a: + 2 = ... greater; and by the question, x(x + 2) = a:^ -f 8, or ar^ + 2x = a:^ + 8. Taking x"^ from both sides, 2a: = 8, , • . a: = 4 = the less, and a: + 2 = 6 = ... greater. A 152 ELEMENTS OF ALGEBRA. These values verify the equation, for 4(4 + 2) = 42 +8, or 16 + 8 = 16 + 8, which is an identity. 9. A cistern was found to be one-third fuU of water, and after running into it 21 gallons, it was then found to be half full : required its capacity in gallons. Let X — the number of gallons the cistern can contain ; then, by the questioUj 3 + ^^-2' or 2x + 126 = Sx, .*. ar = 126 = its content in gallons. 10. A cistern is supplied with water by two pipes; the less alone can fill it in 40 minutes, and the greater in 30 minutes : in what time will they both fill it when running together ? Let X = the number of minutes in which both can fill it ; then, since the first can fill the whole in 40 minutes, it can fill — - in 40 a minute, and therefore — in z minutes ; in the same manner it can be shewn, that — is the part filled by the second pipe in x minutes. But the sum of these two parts is the whole content of the cistern which = 1. Hence ~ + - = l. 40 ^ 30 Multiplying by 120, 3a: -f 4a: = 120 7x = 120 a: = 17- minutes, the time taken by both to fill the cistern. 11. One labourer (A) can perform a piece of work in 5, days, and another (B) can do the same in 6 days, and another (C) in 8 days : in what time can they perform the same work M'hen the three are engaged in it together ? 1 SIMPLE EQUATIONS. 153 Let X = number of days in which the three together can perform the work, then since A can do the whole in 5 days, he 1 X can do - in one day, and therefore - in a; days ; in the same 5 ' 5 manner it may be shewn, that B can do - in a: days, and C can X XXX do ^ in a: days ; therefore A, B, and C can together do - + 77 + o 5 6 8 in X days, but by the question this is equal to the whole work = 1 ; therefore - + - + -= 1. 068 Multiplying by 120, 24x + 20x + 15a: = 120, 59x = 120, 2 •*• ^ "^ ^59 ^^^^" 12. How many pounds of tea at 5 shillings and 9 shillings per pound must be mixed to make a box of 200 lbs. at 6 shillings a pound ? Let X — number of pounds at 5 shillings, then 200 - a; = ... ... 6 ... ; also 5t = the value of the former, {•nd 9(200— a:)= ... ... latter, and 1200 = ... ... mixture. But the value of the two kinds of tea must just be equal to the value of the mixture ; therefore 5a; + 9(200 — a;) = 1200, 5a; + 1800 -9a:= 1200, 4a; = 600, .'. X = 150 = number of lbs. at 5s. and 200- a; = 50 = at 9s. 13. A person at play lost one-fourth of his money, and then gained 5 shillings ; he then lost the half of what he now had, and found that he had only 7 shillings remaining : with what sum did he begin to gamble ? number of shillings he at first had ; then, after losing ■^et X 154 ELEMENTS OF ALGEBRA. one-fourth of his money, and gaining 5 shillings, he had just X -f + « = | + 5; since he now lost the half of this sum, he had remaining the half; therefore by the question. m^^^- ^' or f + 6 =U, or ¥-^ hence 3a: = 36, .-. X = 12. The equation may also be stated thus : — ^-| + 5-i(a:-| + 5) =7, 3a: or -- ■■(¥ + ^)— • Multiplying by 8, 6a: - 3a: - 20 = 16, Sx = 36, .-.a: = 12. 14 Tlie sum of two numbers is = s, and their difference what are the numbers ? Let a: - = the greater, then x-d = = the less. and x + (x-d)= - s, or 2x- d = = s, or 2x = = s-\-d, .*. X = = l(^+^=l+f, and X-d: s d s d = 2^2-^^=2-2- d: From this result is derived the following useful theorem :— 249. Theorem. Half the difference of two quantities added to half their sum = the greater, and taken from half the sxxm = less. SIMPLE EQUATIONS. 155 15. The hour and minute hands of a watch are exactly together between 8 and 9 o'clock : required the exact time at which they coincide. Let the number of minutes more than 40 be denoted by x, or X = minutes from VIII to the point of coincidence m. Then the hour-hand moves from VIII to the point of coincidence tw, during the time that the minute-hand moves from XII to the same point, or the former hand moves over x minutes, while the latter moves over 40 + r minutes ; but the latter hand moves 12 times faster than the former ; therefore ^ 40 + X = 12x, or llx = 40, 40 2 .'. a: = — - mmutes = 3™ 38' -- . Hence the required time is 43°* 38' ~ past 8 o'clock. 16. A hare is 40 of her own leaps before a greyhound, and takes 5 leaps for the greyhound's 4 ; but 3 of the greyhound's leaps are equal to 4 of the hare's : how many leaps must the grey- hound take to catch the hare ? " Let X = the number of leaps the hare takes before being caught, or from H to O, H being the hare's starting- i \ ^i point ; then a: + 40 — the number of leaps G H O of the hare from the greyhound's starting-point (G) to the place where he overtakes the hare (0). 4x But 5 : 4 = x : — , the number of leaps taken by the greyhound after starting till he overtakes the hare, for the hare takes the same time to run over HO that the greyhound takes to run over GO, and their number of leaps in the same time is as 5 to 4. Further, the whole number of leaps of the hare and greyhound in the same space, as GO, is as 4 to 3, or inversely as their length, and the number of their leaps in the space GO are respectively 4r a: + 40 and -^ ; therefore o x-f 40:^ = 4:3, 5 Zx + 120 = ~, 5 -r 15a- + 600 = 16a-, therefore x = 600 = number of hare's leaps in HO, , 4x 4 X 600 ,„^ ^ ^ , „ , . ^^ jin(l -— = = 480 = number of greyhound s leaps m GO. 156 ELEMENTS OF ALGEBRA. 17. A cistern is supplied with water by one pipe, and emptied by another; it can be filled by the former in 20 minutes, and emptied by the latter in 15 ; supposing the cistern at first to be full, in what time would it be emptied when both pipes are running? Let X = the number of minutes, then 2Q):x = 1 : ^i the portion filled in x minutes, and 15:a:=l:— , ... runout ... ; also, since the excess of the latter portion above the former is = the content of the cistern, X ^ _ 1 15 ~20~ 20a; -\ox- 300 ^^ 5.T = 300, . • . X = 60 minutes = 1 hour. 18. A person bought a number of sheep at 16 shillings a head, but found that he had not enough of money to pay for them, by 40 shillings ; had he, however, given only 15 shillings for each, he would have had 60 shillings over: how many sheep did he purchase, and how much money had he ? Let X = the number of sheep, then by the question, \Qx — 40 = 15x + 60, or a: = 100 = number of sheep, and 1600 — 40 = 1560sh. = £78 = his money. 19. A bill of £700 was paid in sovereigns, half sovereigns, and crowns, and an equal number of each was used : required this number. Let X = the number of each, then 20a; + 10a; + 5a; = 700 X 20 = 14000 shillings, or 35a; = UOOOj . • . a; = 400 = the number of each. 20. A labourer was engaged for 30 days at 15 pence a day with maintenance ; but for every day he was idle there was to be deducted 10 pence for maintenance. After his engagement ex- pired, the sura he received was 25 shillings : how many days did he work, and how many was he idle ? SIMPLE EQUATIONS. 157 Let X = number of days he worked, then SO — X = he was idle; , also 15x = wages due for his work, and 10(30 — x) = sum to be deducted ; therefore 15x — 10(30 - ar) = 25 X 12 = 300 pence, or 15x - 300 4- lOx = 300, 2ox = 600, .'. X = 24: = the days he worked, and 30 — X = 6 = ... he was idle. 250. Questions in which the given quantities are numbers, may be generalised, by assuming literal quantities for the known ones, and then the solution of the question will afford a 7-ule for the solution of all similar ones. The following question is the third of the preceding, gene- ralised : — 21. Eequired a number such that, being added to its mth part, he sum shall be equal to its nth part added to a number a. Let X = the number. then — is the expression for its mth part, and - ... ... nth part: therefore x -] = - + a; m n nultiplying by mn mnx A- nx = mx + amn mux -\- nx — mx ■= amn (mn — 771 + n)x = amn ind liPTif»ft a7nn mn — m -\- n This value of x serves as a rule for the solution of all similar problems in which particular numerical values are given for a, ?«, md n. Thus, for the 3d example, a = 10, m = 3, n = 2 ; _ 10 X 3 X 2 _ 60 _ '® "" ~ 3^2 -3 + 2 ~ y - ^'^• similar manner the 11th example may be generalised ^ can perform a piece of work w in a days, B can accom- 158 / ELEMENTS OF ALGEBRA. plish the same in h days, and C in c days : in how many days wil they finish the work when all are engaged ? Let X = the/number of days, and w = tlie whole work ; then sine rak the work w in a days, he could do a — part i; A could do — in x days : in the sam a be shew'i^ tM^t B could do -^ in x days, and that ( s : Wt ajnce the sum of these three parts i^erefore wx In the lit: therefore x = acx + ahx = abc -f ac + bc)x= abc, _ abc I * ' ab -{- ac -{- be ' ,1b = e, c = 8, 240 _ 120 _ 2 118 ~ 59 ~ 59' EXERCISES. 1. What number added to its fifth part = 24, . . = 20 2. Required a number which exceeds its fourth part by 27, = 36 3. What is that number, which being added to 7 = 5 times it fourth part ? =28 4. Reqiiired the number which exceeds its sixth part as mud as 26 exceeds its fourth part ? =24 5. The sum of two numbers is = 20, and their difference is = 8 what are these numbers ? =14 and 6 6. At an election, the number of votes given for two candidatf was = 256 ; the successful candidate had a majority of 50 votei. how many voted for each of the candidates ? . =153 and 103 SIMPLE EQUATIO 159 7. Divide a line of 24 inches into two unequal parts, so that the greater may e^j^d the less by 10 inches, = 17 and 7. 8. Divide a jffie of 42 feet in length into two par^^s, so that the one shall be plouble the other, . . . = 14 and 28. 9. Divi^ the number 30 into 3 such parts, that the second maj^e dbuble of the first, and the third triple the same, / = 5, 10, and 15. (AM )ivide a line of 60 inches into 3 such parts, that the s^e^ond 1 maj^be = 3 times the first, and the third doubleTiihe second, I . = 6, 18, an^l 36. j. 11. Divide the number 252 into three parts, such that one-third of the first, one-fourth of the second, and one-fifth of the third, shall all be equal to one another, . . = 63, 84, and 105. 12. An uncle divided £375 among three nephews. To the first he gave four-fifths of the sum given to the second, and tiie third received a fifth part of the second's share more than the second : how much did each receive? . . . = 100, 125, and 150. 13. The sum of two numbers is = 60 ; and the less is to the greater as 5 to 7 : what are these numbers ? . = 25 and 35. 14. The sum of £154 was paid in half sovereigns, half crowns, and groats (4 pence each), and an equal number of each of these coins was used : what was the number ?...== 240. 15. Two canal-boats are despatched from the same place, the first at 6 o'clock in the morning, the other at 4 in the afternoon ; the first runs 4 miles an hour, and the second 9 : in how many hours will the second boat overtake the first ? . = § hours. 16. The sum of two numbers is 10, and their product is equal to the excess of 60 above the square of the greater : required the numbers, . =4 and 6. 17. Divide the number 20 into two parts, so that 5 times the greater may exceed 6 times the less by 12, . . =12 and 8. 18. A cistern was found to be three-fo\irths full of water, but after running off 220 gallons, it was found to be one-fifth full : how many gallons could it contain ? . . , . = 400. 19. A cistern supplied by 3 pipes can be filled by the first in , 10 minutes, by the second in 12, and by the third in 15 minutes : I in what time would it be filled by the three running simultaneously ? = 4 minutes. V 20. A cistern can be filled by one pipe in 16 minutes, and expptied by another in 20 minutes : supposing it at first empty, jjvhat time would it be filled when both pipes are running ? nuirf = 80 minutes. I6(J EtiSTSlEIS'TS OF ALGEBRA. 21. Divide 48 into four such parts that the first plus 3, the second minus 3, the third multiplied by 3, and the fourth divided by 3, may be all equal to each otlier, . . = 6, 12, 3, and 27. 22. A pe^on lias a certain number of shillings in each hand, and if he tftke 8 from the left, and put them in the right hand, he would tbfra have 4 times as many in his right hand as in his left ; but atynrst he had 5 more in his right hand than in his left : how manf.had he in each at first? . . . . z= 15 and 20. 23! The hour; and niiinite hands of a watch are observed to coini;ide between 4 and 5 o'clock : how many minutes is it past 4 ? 1 =21 minutes 49 — seconds. • 24. A lauoui-er tingages to work for 3s. 6d. a day with board, but to gllow 9d. for his board each day that he is unemployed. At the end of 24 days he has to receive £3, 2s, 9d. : how many day.s has he wrought ? . . = 19. ; 25. Three workmen are employed to dig a ditch of 191 yards in i length. A can dig 27 yards in 4 days, B 35 yards in 6 days, and C 40 vaxds in 12 days : in what time could they do it if they • ;>.: simultaneously? = 12 days. ' Vo persons depart at the same time from London and ..' ' rgh, and travel till they meet ; the one goes 20 miles a day 3 other 30 : in how many days will they meet, the distance being 400 miles ? =8 days. 27. Two persons (A and B) depart from the same place to go in the same direction ; B travels at the rate of 2 miles an hour, and A 3 miles, but B has the start of A by 5 hours : in how many hours will A overtake B ? =10 hours. 28. Two persons (A and B) depart at the same time from the same place, to travel in the same direction round an island 36 miles in circumference ; A travels 3 miles an hour, and B 2| : after how many hours shall they come together ? =72 hours. 29. Divide £4400 among three persons, so that the first may have three-fifths of the second's share, and the second three-fourths of the third's share, . . . = £900, £1500, and £2000. 30. A hare has a start of 80 of its own leaps before a greyhound, and takes 3 leaps for every 2 taken by the greyhound, but one of the greyhound's leaps is equal to 2 of the hare's : how many leaps will the hare have taken before it is caught ? . . = 240.^ 31. A and B engage in trade on the same capital ; A gains qq pounds, and B loses 190, but A's money is now 8 times B's: \-^<^ how much money did they begin ? . . . = Si^j^Q, \ \ SIMPLE EQUATIONS. 161 32. Divide a number a into two parts, so that b times the greater shall exceed c times the less by d, The less = —, , the greater = -, . 6 + c ° 6 -f c 33. A person has a hours at his disposal: how far may he travel in a coach which goes b miles an hour, so as to return home in time, walking back at the rate of c miles an hour ? Also find the number of miles if a = 2, 6 = 12, c = 4, 1. = -, miles. 2. = 6 miles. 6 + c NEGATIVE SOLUTION OF SIMPLE EQUATIONS, 251. The value of an unknown quantity in an equation is sometimes found to be negative ; it will be necessary, therefore, to consider the meaning of such a solution. A negative result would at first appear to be absurd, and, in its literal sense, it is so; but a proper interpretation may be easily determined, by means of the conventional use of the negative sign (44). To illustrate this negative result, the following example may be given : — 252. Let it be required to find a number such that if a given number c be added to it, the sum shall be a given number a. If X = the required number, then X -\- c = a or X = a — c. So long as a exceeds c, the value of x is positive ; but let a be less than c, and x becomes negative ; thus, let a = 10, c = 8, then a; = 10 — 8 = 2 ; but, let a = 5, c = 9, then x = 5 — 9=— 4. In the second case, x has a negative value, which implies that, instead of adding c to x, x must be taken from c and added to a ; and hence the question, in its literal sense, is impossible with these values of a and c, or generally, when c'P' a. This sign, therefore, apprises us of the necessity of modifying the condition of the question, in order to adapt it to the case of a ^ c. For such a case, the condition, to be literally fulfilled, must be expressed thus: — Required a number such that if it be subtracted from a given number c, the remainder shall be = another given number a. 162 ELEMENTS OF ALG^x.x.a. This question will be solved for the case of a = 5, c = 9. if X = 4:, for then 9 — 4 = 5; and the general equation will be c — X = a x=:c — a = 9 — 5 = 4:. Instead, however, of making two distinct questions for the two cases of c ^ a and c ■p' a, one question and one solution will serve for both, by considering that the subtracting of a positive quantity is equivalent to adding a negative one (56), or that, in this case, the subtracting of x = 4 in this question is the same as adding X = — 4: in the preceding one, when c x' a ; therefore, with this understanding, the solution of the former question X = c — a comprehends also the solution of the latter. Thus, when a = 5, c = 9, and x = a — c = — 4, if c be added to x, by the ordinary rule of algebraic addition, the result is 9 + [- 4] = 9 - 4 c + x = 9 — 4: = 5 = a. 253. In order to explain more fully the meaning of negative solutions, let the equation ax -{- d = ex + b be given. From this equation, b-d Several cases requiring some consideration will occur according tc the relative values of the given quantities a, 6, c, d. I. Let b^^ d and a'P' c, then x is positive. n. If, while one of the quantities (b — d) or (a — c) is positive., the other be negative, then the value of x is negative, and thc; equation becomes — ax -{- d = — ex -\- b. I. "VVlien b ^d and az^ c, (b — d) is negative, and (a — cj, positive, so that ': d-b In order that the value of x may be positive, the terms in the given equation containing x are therefore in this case to be made SIMPLE EQUATIONS. 163 negative, which is the same as subtracting ax and ex ; but sub- tracting these is simply adding, by the rule of addition, — ax and — ex, or a and c multiplied by — a: ; that is, by — ^^ , or or , which is negative, because 6 — c? is so, since a — c a — c yL d. But this is exactly the value of x in the first case ; and hence the solution in that case comprehends that in this, observing the conventional definition (42), and the principle in (56). i 2. When h^ d and a^c, the result will be found to be the same as the preceding, and may be obtained in the same manner, by taking a — c for — (c — a). III. When b = d, and, at the same time, c not = a, then a — c as is evident from the original equation, which becomes ax — ex = 0, or (a — c)x = 0. Dividing by (a — c), x = 0. IV. When b is not = d, but a = c, then b~d b-d a — c the equation in this case being {a — e)x = b — d, or X X = b — d; and since no finite value of x, multiplied by 0, can give a finite product = b — d, X must = oo. V. When a = e, and b = d, from which no particular value of x can be found, or its value is indeterminate. The equation in this case is (a — c)x = b — d, or X x = ; «nd since any number, multiplied by 0,. gives 0, therefore any 164 ELEMENTS OF ALGEBRA. value whatever of x fulfils the equation, or it has no particular value. 254. The meaning, therefore, of the expression - is merely the indeterminateness of the quantity whose value it expresses ; it does not convey any sign of absurdity or impossibility in the equation. 255. An equation whose members are the same quantities expressed either in the same or in different forms, is called an identical equation. The last case of the preceding equation, namely, ax -}- b = ex -{■ d, or ax -\- b = ax -\- b, ' since a = c and b = d,\s an equation of this kind ; and so are the equations x(x - 2) = a;2 - 2x (x — a)(x — b) = x^ — (a + b)x + ab 3x(2 - 4) 4- 36 = 6x - 12a: + 36. It is evident that whatever value is assigned to the unknown quantity in an identical equation, its two members are equal ; so that the equation is satisfied by any value of this quantity, and therefore no particular value of it can be assigned. 256. The nature of a negative solution is very clearly illustrated by the following question : — Two couriers depart at the same time from two towns (A and B), distant by a miles from each other ; the former travels m miles an hour, and the latter n miles^: where shall they meet ? There are two cases of this question : — I. Wlien the couriers go in opposite directions. Let M be the point where they meet, j j j and a = AB the distance between the A MB places, X = AM, the distance that A travels, then a — a; = MB B ... ; also — = the number of hours that A travels, SIMPLE EQUATIONS. 165 but A and B travel the same number of hours, therefore m n nx = am — mx X = am m + n' an and m + n Wliatever values be given to a, m, n^ that of x will in this case be always positive^ and no diflficulty occur. II. When the couriers go in the same direction. As in the former case, let M be the point f T IT of meeting, A and B travelling in that direc- -^ B M tion, and let a=: AB, and X = AM, whence x — a = BM ; (then, as in the preceding case, X X — a Mid m n nx = mx - It appears from this value of x, that, so long as mzp'n, or A's rate of travelling exceeds B's, the value of x will be positive, and there will be no difficulty in interpreting its value. But suppose that m^n, the value of x is negative, and the equation nx = mx — am now becomes — nx = — mx — am am n — m This negative value implies an impossibility in the question 166 ELEMENTS OF ALGEBRA. understood literally, and indicates the necessity of modifying its enunciation; but, by the following considerations, this will be found unnecessary : — To subtract is the same as to add ^, by the n — in — {a — m) rules of algebraic addition (56) : and y ^ = . But — (n — m) m — n this quantity being in the present case negative, as m ^ n, the value of X is also negative, and being the numerical value of a line, its value must be measured in a direc- j j 1 j tion opposite to that of AM (44), namely, M' A B M in the direction AM' ; hence AM' m — n and M' is the point where A and B would meet. Were this negative value of x to be measured from B, it would be identical with that of x in the first case, m and n being inter- changed. For in this case the value of x will be measured from B, provided AB be added to it. But x being in this case negative, AB must be taken as negative, and = — a; then am am — am + an an an which' is exactly equivalent to the interchanging of m and n in the first value of x, in this case, and taking x negative. When the given quantities in a question are general, that is, when they are expressed by letters, interesting consequences may sometimes be deduced from the solution, by assigning particular values to the letters. Thus, from the preceding value of x in the first case, am when m = n, x m -{• n am am 7n -]- m 2m that is, each of the couriers travels over precisely half the distance, as is otherwise evident. When n = 0, X = = a, m or A has the whole distance to travel = AB. SIMPLE EQUATIONS. 167 ,,n. /v ,« X ^ When m = 0,x =\ = - = 0, -\- n n or A remains stationary ; B meets him at A, or B travels the whole distance = BA, as is otherwise evident. _ ,, - , am am In the second case, when m = n, x = = -7— = 00 ; m — n that is, A must travel an indefinitely great distance, or, in other words, he never can overtake B, as is manifest from the equality of their rates of travelling. When n = 0, X = = a, and A travels to B. rn When n = -—, x = = 2a, or A travels twice the length J m of AB. Wlien n = 2m, x = = — a, or A travels in an opposite direction from A towards M', a distance = a, and consequently B a distance = 2a. II. — SIMPLE EQUATIONS CONTAINING TVTO UNKNOWN QUANTITIES. 257. Sometimes a question requires for its solution the deter- mination of the values of two unknown quantities. In this case two conditions are stated. These conditions are necessary, and they are sufficient ; they must also be wholly independent of each other. They must likewise be consistent, for if contradictory, the solution would lead to an absurd conclusion. And, lastly, they may be either implicit or explicit. Questions of this kind may, however, sometimes be solved by using only one unknown quantity, as in some examples in the former section containing two conditions, in which one condition is used in determining the notation, and the other in obtaining an equation. (See examples 4 and 7, of art. 248.) 258. The first part of the process of solution is to eliminate one of the unknown quantities ; that is, by some combination of the two equations to derive a new one excluding one of the unknown quantities : this resulting equation, containing only the other unknown quantity, may be solved by the former method for such equations. 168 ELEMENTS OF ALGEBRA. The following is an example of dependent equations : — 3y = 2x — 4: 6l/ = 4:X— 8, the latter being derived from the former, merely by multiplying its terms by 2. The two equations, 2a: — y = 5, and 2x — 1/ = 6, are inconsistent equations^ It is evident that the same values of X and y that satisfy the former cannot also fulfil the latter ; they are therefore contradictory. 259. There are three methods of solving two equations con- taining two unknown quantities ; namely, equating, substitution^ and equalising coefficients. 1. — BY EQUATING TO SOLVE TWO EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 260. EuLE. Find a value of one of the unknown quantities in each of the equations, and equate these values ; that is, make them the members of a new equation, which will then contain only one unknown quantity, the value of which may be found as before. The simplest method in this process is to find a value of the unknown quantity which is least involved by coefficients in the given equation. EXAMPLES. 1. If i^il^Zor ""^^^^ are the values of x and y ? By finding a value ofy in each equation, we have in the first equation, ... y = S ~ x ... [3] in the second equation, ... y = x — 2 ... [4] Equating these values oiy, X -2 = % - X, or 2x = 10, .-. x = 5. By finding similarly a value of x in each of the given equations, and equating these values, which will be expressed in terms of y, the value of y may be found in the same way as that of x above. But if the value of x already found, namely 5, be substi- SIMPLE EQUATIONS. 169 tuted in either of the given equations, or in [3] or [4], the equation will then contain only y, whose value may be found. Thus, substituting 5 for x in [3], it becomes or, substituting this value of x in [4], it becomes yrzx — 2 = 5 — 2 = 3, ^hich gives the same value of y. 2. If I 3^ Z 2y = 6~ ^ } ^^^* ^^® *^® ^^"^^ ^^ X and ^ ? In the first equation, y = — - — ... [3] o In the second, y = — - — . \ +2x _ Zx-Q 3 ~ 2 2 + 4x = 9ar - 18 5ar = 20, %. .'.x= 4, lndby[3], y = L+!^| = 3. -. EXERCISES. Equating, (5x- ■• \2x- -2y= 4 ) (x= 2. 1 i .... 1^^ 3, ^_ 1^- 1 4 4~ x= 6. y= 3. a: = 12. 3+ 2- ^ J 170 ELEMENTS OF ALGEBRA. Jj 6. If = 36. y ,„ I 1^ = 24. 9~ 5- ^ C a; = 45. II. — BY SUBSTITUTION TO SOLVE TWO EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 261. Rule. Find a value of one of the unknown quantities in either of the equations ; substitute this value in the other equa- tion ; and a new equation will thus be formed, containing only one unknown quantity. The solution will be more simple if the unknown quantity whose value ig^taken be that which is least involved in the given equations. EXAMPLES. 1. If \^ 'T ~ k\ what are the values of x and y ? In the first equation y = 1 + x ... [3]. Substituting this value for y in the second equation, x + l-\-x-5 2x = 4, .'.x = 2; and by [3], y = 1 + 2 = 3. j f ^ + ^ =, 2 ) 2. If < 5 4 > what are the values of x and y ? {x-y=l} Finding a value of y in the second equation, as it is least involved in it, y = x-l ... [3], SIMPLE EQUATIONS. 171 substituting this value in the first - H — = 2, 5 4 4x + 5ar - 5 = 40 9a: = 45, .' . X = 5, and by [3], y = 5 - 1 = 4. EXERCISES. 1. If 12. 18. fa: = \ x= 16. y=\2. < x = 2 The examples and exercises under the first case may be solved by the second as additional exercises. III. — BY EQUALISING THE COEFFICIENTS TO SOLVE TWO EQUATIONS CON- TAINING TWO UNKNOWN QUANTITIES. 262. Rule I. Multiply both equations, if necessary, by such numbers as will make the coefficients of one of the unknown quantities the same in both equations ; then by finding the difference or sum of the equations, according as the two equal terms have like or unlike signs, these two terms will destroy each other ; and the resulting equation will contain only one imknown quantity. The coefficients of an unknown quantity in the two equations may be equalised by multiplying each equation by the coefficient of the unknown quantity in the other. When these two coeffi- 172 ELEMENTS OF ALGEBRA. cients, however, are not prime, find their least common multiple ; and then, by multiplication, make the two coefficients equal to this multiple. If the equations have fractional coefficients, the fractions ought to be cleared away before applying the rule. EXAMPLES. 1. If {^^ [t 2^ = !1 1 1 ^^^^^ ^^e the values of xandy? The coefficients of y being equal in these equations, it will be unnecessary to multiply them ; and* as the signs are unlike^ the equations must be added, which gives 4a; = 21 - 1 = 20, .' . X = 5. The value of y might be similarly found by equalising the coefficients of x : but it will be more readily found by substituting the value of x, as already found, in one of the given equations* Since the secopd is the less involved, it may be chosen for this purpose. Tie result is, 5-2y=-l 2y = 6, .•.y = 3. 2. If •] R^ Z 9^ — 1 I ^^^* ^^^ *^® values of x and y ? sdans idfion As X is least intolvedWts coefficients should be equalised, by multiplying the first equdfion by 2, and the second by 3, adding these, . ^ 15 and, by the first equation, x = - — ^-^ = ^ - '" = — = 5» 3 6x -iy = 14 15y -Gx = 30; 15^ -iy = 44 Uy = 44, .'.y = 4; X = 7 + 2^ 3 7 + ~ 3 SIMPLE EQUATIONS. 3. If -/ V what are the values of x and y \ Clearing the equations of fractional coefficients, the first becomes 5ar + 4y = 160 ... [3], the second ... 3ar + 5y = 135 ... [4] ; multiplying [3] by 5, 25a: + 20j/ = 800, [4] by 4, \2x + 2^y = 540, subtracting the latter from the former, 13x = 260, .-. x= 20; and by [3], ?/ = 40 - ^ = 40 - 1^ = 40 - 25 - 15. 4 4 EXEKCISES. 1 If \ 2a:- y^ 3) (x= 4. ^- -'^ ^ ".a:+ 2?/= 22; ' • • • |^= 5. .^ ( 3a: + 23/= 19) ( x = 5. - - 1 2a:- 3^= 4; • • • • t^ = 2. o (45x4- 8y = 35Q) \ x = 6. ^- "• t2l3^-13x=132i .... 1^^ ^^^ X = 144. y - 216. or X ■■ b -a' second, gives be a = d. SIMPLE EQUATIONS. 179 The values of x and t/ may be found from the first two, as in (263), which are _ <^P- hq „ _ aq-cp n. — y Z= ad — be ^ ad— be and these values being substituted in the third, give m(dp — bq) n(aq — cp) ad —he ad — be ~ ^ or m(dp — bq) + "(^^ — cp) + r(J>c — ad) = 0, and unless the values of the given quantities be such as to verify this last equation, or to make its first member = 0, no values of the unknown quantities can be found to satisfy the three equa- tions. If, however, this equation be fulfilled, the values of x and y will then satisfy the three equations ; but these values being found from the first two, the third is unnecessary. EXERCISES WITH LITERAL COEFFICIENTS. X = h-dp _ a q- cp be — ad^ ^ be — ad' _ aq -\- bp _ aq — bp ^ ~ 2ab ' ^ ~ 2ab '' be ac a + b'^ a + b nb'^ mb + y =p\ ^^ q-¥ „_ (^p- q ( x+ y =pl \ax + by = q i ax -{-by = q ) a — b P I ac(bp + d q) _ bd{ap — cq) ad + bc'' ^ ~ ad-\-be ' 180 ELEMENTS OF ALGEBRA. EXAMPLES PRODUCING SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. 1. The sum of two numbers is = 60, and their difference is =•12 : what are these numbers ? Let X = the greater, and y = ... less, then a: + y = 60, by the first condition, and X — y = 12 ... second ... Taking the sum, 2x = 72, . • . a; = 36 the greater. Taking the difference, 2i/ = 48, ^ . • . y = 24 the less. These numbers verify the conditions of the question, for a; + j^ = 36 + 24 = 60, the sum, ar - 5^ = 36 - 24 = 12, ... difference. 2. The sum of two numbers is = 44, and their ratio is that of 5 to 6 : required the numbers. Let X = the greater, andy = ... less, then, by the conditions of the question, a; + y = 44 ... [1], and a: : y = 6 : 5 ; therefore 5x = 6y ... [2]. By [1] y = U — x, substituting this value of y in [2], ' 5x = 6(44 - a:) = 264 - 6x 11a: = 264, .-.a: = 24; hence y = 44 — a: = 44 — 1 SIMPLE EQUATIONS. 181 3. The sum of two niimbers is = 16, and the sum of their reciprocals is = double the difference of their reciprocals : what are the numbers ? Let a: = the less, and y = the greater, then x+y = 16 [1], and \-l =il-l)- • [2]; multiply [2] by xy, y + x = 2y - 2x, .' . X -y 3 [3]; substituting this value of x in [1], l + y = 16 y + ^y=^ 48 4y = 48, • ••y = 12, and by [3] ^ = 1 = 4. 4. The sum of two numbers is = 10, and twice the less is to three times the greater, as the square of the less to that of the greater : required the numbers. Let X = the less, and y = the greater, then x+y=10 ... [1], and 2x:3y = x'':y^ ... [2]. By [2], 2xf = 3x'y ... [3], dividing by xy, 2y = 3x ... [4], by[l] 2x + 2y = 20, substituting for 2y, 2a: + 3x = 20, .-.a: = 4, and by [4] 3a: 12 . y = Y = Y'='- Equation [3] is a cubic equation, but it is easily reduced to a simple one by dividing by xy. 5. Divide a line of 20 inches into two parts, whose ratio shall be that of 8 to 12, or 2 : 3. 182 ELEMENTS OF ALGEBRA. Let X = the shorter part, and y = the longer ... ; then :r+y = 20 ... [1], and a::y=2:3; or 3x = 2y, by [1], 3a: + 3y = 60, taking the difference of the last two, 3y = 60 — 2y hy = 60, .-. y=12; by [1], x = 20 - y = 20 - 12 = 8. 6. To divide a line a (AB) into two parts that shall have to each other the ratio of two numbers m and n. Let AB = a the given line, X = the greater part = AC, and y = the less ... = BC, then if m ■p' w, x: y = miny or nx = my [1]; also X -\r y = a . [2], X = a — y mx = am — my ... • [3], adding [1] and [3], (m + n)x = am, am • ' ''- m + n' by [2], yz=a — x = a — am m -\- n m -\- n This question is the same in principle as the first case in (256). This example is also a simple illustration of the application of algebra to geometry. The length a of the line may be expressed in inches or in any other denomination, observing that the values of X and y are expressed in the same denomination. 7. To produce a given line AB to a point i i i i as C, so that AC shall be to CB as m to n. ~^ SIMPLE EQUATIONS. Let a = AB, the given line :c = AC, ajad y = BC, then x:y = m'.n, )r nx = my ... [1], 183 md, in this case, x — y = a ... [2], for AB is the difference between AC and BC ; then, by [2], x = y -{- a mx = my -{- am ... [3] ; subtracting [1] from [3] mx — nx = am, am by ril y = X — a = a = . "•^ ^ -" ^ 7)1 — n m — n Were m = n, x and y are each = oo, or C is at an infinite distance, or, in other words, the problem is impossible. Were m.^n,x and y are negative, and in the ordinary sense of a negative quantity the question is impossible ; but by the conven- tional meaning of a negative quantity it would then appear that the point C lies in an opposite direction, as C, so that AC would be = X, and BC = y. But X was supposed to be the greater ; and it might be con- cluded, therefore, that the equation X — y = a [would be inconsistent with this supposition. But it becomes in this case ! — x — {—y)=—x-\-y = y — x = a, and is the same as the difference between x and y considered as positive ; and this results from the principle, that to subtract a iquantity is to add it with an opposite sign. 8. The values of the properties belonging to two persons are as m to n, and the difference of their values is = a ; required the values of the respective properties. Let X = the value of the greater property, and y = ... smaller ... Then as a = the difference of the properties, the solution will lead to the same result exactly as in the last question ; therefore am , an X — — , and y = . 184 ELEMENTS OF ALGEBRA. If while X is tlie greater, m should be ^ n, then x and y are negative, and the negative value of x is less than that of y (55). Instead of x and y then representing absolute property, they represent only negative property, or debt. If an amount of property = a be added to the negative property y, the sum will be equal to the negative property x, for !/ m — n m — n where x and y are negative, because {m — n) is so, since rmLn. 9. A's age is double of B's, but 15 years ago A's age was 5 times B's : required their ages. Let X = A's age, and y = B's ... , then x = 2y ... [1], and :c - 15 = 5(y - 15) = 5y - 75 ... [2]. By [2], x = 5y- 60, subtracting [1], = 3y - 60, or y = 20, and X = 2y = 40. 10. A bill of £210 was paid in sovereigns and crowns (each = 5s.) ; and the number of the latter used was three times that of the former : how many pieces of each were necessary ? Let X = the number of sovereigns, and y = ... ... crowns, then y = Zx, and 20a: + hy = 210 X 20 = 4200, substituting the former value of y in the latter equation, 20a; + 15a: = 4200 35r = 4200, .-.a: = 120, j and y = Sx = 360. 11. There is a certain fraction, to the numerator of which if 2 7 be added, the fraction becomes = -; and if from its dcno- o rainator 2 be subtracted, it becomes equal to -: Avhat is the fraction ? SIMPLE EQUATIONS. Let X — the numerator of the fraction, and y = ... denommator ... ; then x + 2 7 y -8' and X 5 V - 2 - 6- 185 Multiply the former equation by 8y, and the latter by 6(y — 2), and they become 8x + 16 = 7y, ^nd 6x = Sy — 10. From the first member of the former, transpose 16 to the other side ; then multiply the former by 5, and the latter by 7, and they become 40x =: Zoy — 80, and 42a; = ^oy — 70. raking the difference of these, 2x = 10, .• . X = 5 ; and by the third equation, 7y = 8x + 16 = 40 + 16 = 56, .'.y= 8. 5 Hence the fraction is -. 8 By numbering the equations [1], [2], [3], &c., and using the signs +, — , X, -T-, the various operations may be more concisely expressed. The preceding steps of solution will then be repre- sented thus : — X 5 [1], [2], y-2 6 [1] X 8y, 8a: -f 16 = 7y, Sx= 7y- 16 ... [3], [2] x6(y-2), 6x= By -10 ... [4], [3] X 5, 40a: = 35^-80 ... [5], [4] X 7, 42a: = 35^-70 ... [6], [6] - [5], 2a: := 10, .• . X = 5; 186 ELEMENTS OF ALGEBRA. by [4], hy = 6a: + 10 = 30 + 10 = 40, .-. y = 8, and the fraction is = - = -. y 8 ] 2. If from the double of a certain number 6 be subtracted, th( remainder will be a number whose digits are those of the formei in an inverted order, and the sum of the digits is = 6 : requirec the number. Let X = digit in the place of tens, and y = ... ... units, then 10a- + y = the number (443) ; and therefore by the conditions of the question, 2(10x +y)- 6 = Wy + x ... [1], also x-\-y = 6 [2], by the former, 20x + 2y — 6 = lOy -\- x 19x = 8y -\-6, by [2], 8x= -8y -\- 48, adding the last two, 27a; = 54, .'. X = 2, by [2], y = 6-x = 6-2 = 4:', and therefore the number is = 10 X 2 + 4 = 24. 13. The sum of two numbers = s, and their difference = d: required the numbers. liCt X = the greater. and y = the less ; then x+y = s, and X - y = d, adding 2x = s -{- d, and x = ~(s taking the difference 2y = s - d,a.ndy = -(s d). This question has been solved in another manner in the example preceding art. (249). SIMPLE EQUATIONS. 187 EXERCISES. 1 . The sum of two numbers is = 30, and their difference = 6 : ((That are the numbers ? =12 and 18. 2. The sum of two numbers is = 60, and their ratio = that of 2 io 3 : required the numbers, . . . . =24 and 36. 3. The difference between two numbers is = 8, and twice the 5um of their reciprocals is = three times the difference of their reciprocals : required the numbers, . . . = 2 and 10. 4. The sum of two numbers is = 14, and 3 times the less is to [ times the greater, as the square of the less to the square of the greater : what are the numbers ? . . . . = 6 and 8. 5. Divide a line of 36 inches into two parts, whose ratio shall )e that of 5 to 7, = 15 and 21. 6. Find two numbers such that half the first with a third of he second shall be = 9, and a fourth part of the first with a fifth )art of the second shall be = 5, . . . . = 8 and 15. 7. Two purses together contain 300 sovereigns, and if 30 overeigns are taken out of the one purse and put into the other, here will be the same number in each : how many sovereigns are ontained in each purse ? = 180 in the one, 120 in the other. The ages of two persons are in the ratio of 3 to 4, but 10 ^ears ago the ratio of their ages was that of 2 to 3 : required heir ages, =30 and 40. 9. Seven years ago, the age of a person (A) was just 3 times hat of another (B) ; and seven years hence A's age will be just iouble that of B's : what are their ages ? . . = 49 and 21. 10. A person wished to distribute 3d. apiece to some poor >ersons, but found he had not money enough in his pocket by !d. ; he therefore gave them each 2d., and found he had 3d. emaining : required the number of poor people, and the money le had in his pocket, . . =11 poor persons, and 25d. 11. A cask, which held 60 gallons, was filled with a mixture of )randy, wine, and cider in such proportions that the cider was ; gallons more than the brandy, and the wine was as much as he cider and one-fifth of the brandy: how much was there of ach? . . Brandy = 15 ; cider = 21 ; and wine = 24. 12. A said to B, if you give me 10 guineas of your money, I ■hall then have twice as much as you will have left ; but B said o A, give me 10 of your guineas and then I shall have three imes as many as you : how many had each ? A = 22 ; B = 26. 188 ELEMENTS OF ALGEBRA. 13. A farmer wishing to purchase a number of sheep, found that if they cost him 20s. a head he would be £2 short of money ; but were they to be only 16s., he would then have £1 over: how many sheep were there, and how much money had he ? Sheep =15, ajid money = £13. 14. A person (A) departs from a certain place, and travels at the rate of 7 miles in 5 hours, and 8 hours after another person (B) sets out from the same place, and travels at the rate of 5 miles in 3 hours : how long and how many miles does the first travel before he is overtaken by the second ? =60 hours, and 70 miles. 15. There is a certain number which is equal to 7 times the digit in the place of units, and if 18 be added to it, the sum is a number whose digits are those of the given number in an inverted order : what is the number ? =35. 16. A cistern can be filled with water by two pipes running together in 6 hours, and the quantities conveyed by the pipes in the same time are as 3 to 4 : in what time would they separately fill the cistern ? =14 and 10^ hours. 17. Find that fraction which if 1 be added to its numerator its value will be -, but if 1 be added to its denominator, its value will be -r, = TT- 4 15 18. A person had two casks, the larger of which he filled with ale, and the smaller with cider. Ale being half-a-crown, and cidei 1 Is. per gallon, he paid £8, 6s. ; but had he filled the larger with cider and the smaller with ale, he would have paid £11, 5s. 6d. : how many gallons did each contain ? The larger =18, and the smaller =11 gallons, III. — SIMPLE EQUATIONS CONTAINING THREE UNKNOWN QUANTITIES. 267. By properly modifying them, the three methods given in the preceding case may be extended to this one ; and a fourth method may sometimes also be advantageously used here. I. BY EQUATING. 268. Rule. Find a value of one of the unknown quantities in each of the three equations, and equate these values, so as to form two equations, by either placing the first equal to the second, and likewise to the third ; or by putting the second equal to the first, and also to the third ; and as these equations will contain only two unknown quantities, their values may be found as in the former case. SIMPLE EQUATIONS. 189 The equations ought to be cleared of fractions, if necessary, before applying this rule. EXAMPLE. ( 2x-St/ + 5z=lB\ If < 3a; + 2y — z = S )■ what is the value of x, y, and z ? ( - a: + 5^ + 2^ = 21 j 3y the first equation, z — ^ w, ... second ... 2 = 3ar + 2y — 8 [5], ... third ... _.-5^ + 21 __ 2i [6]. liquating [4] and [5], ^^ - 2x^ + Zy ^ 3^ _^ 2y - 8, md reducing this equation, 17a: = 55 — 7^ m. Equating [5] and [6], 3a: + 2y ^-''~\ "^'^^ ind reducing 5a: = 37 — % [8]. The equations [7] and [8], resulting from the process of equating, tre two equations containing two unknown quantities, x and y, (Those values may be found by the rules of the former case. These ralues are thus found to be a: = 2, ^ = 3, lud by [5], z :=6 + 6-8 = 4 • EXERCISES. ( 2:r- y + z = 9) i.if ^ X — 2y + 3z = 14[ . . 1 . 3x + 4y - 2z = 7J ■ ^ ■1 + 1 + z It = C2'J '■■-( i* f + z 5 -J . f + 1+ 1=^« x= 3. y= 2. z= 5. (x= 24. . . • ' i^ (lOx + 5y -\-4:Z = 75) [z = 5 |+.= 3 X = X y , 4- z = 2 4^ III. BY EQUALISING COEFFICIENTS. 270. EuLE. Equalise the coefficients of one of the unknown [luantities in the three equations, as in the former case ; then take the difference between any two of them ; next that between [)ne of these equations and the remaining equation ; and there will result two equations containing two unknown quantities. EXAMPLE. Find X, y, and z, in the three equations, 3a: - 4y + z = U ... [1], -4- ^- -- 4 r2i 3+5 9-4 ... L^J, 6x - ii/ + 2z ~ 68 ... [3]. Multiplying [2] by 45, and dividing [3] by 2, they become 15a: + 9y -5z = 180 ... [4], 3a: - 2y + z = 34 ... [5]. Again, multiplying [1] and [5] by 5, in order to equalise the :ioefflcients of z, they give 15a: -2O1/ -\-6z = 70 ... [6], loo: - lOy -\-5z = 170 ... [7]. I'ho coefficients of z are now equal in [4], [6], and [7]. Taking the sum of the two former, and the difference between the two latter, there results 30a: -Ihj = 250 lOy = 100. 192 ELEMENTS OF ALGEBRA. In taking the difference between [6] and [7], it is found that x vanishes as well as z ; the cause of this elimination of x is — that its coefficient happened to be the same in [1] and [5]. Had x not disappeared, the last two equations obtained would have contained two unknown quantities. By the latter, 3/ = 10, and substituting this in the former, 30x- no = 250, .-. x= 12. Substituting now these values of x and y in one of the given equations, as, for instance in the first, 36 - 40 + ;3 = 14, .'. z=18. EXERCISES. 2x- 4y + 92: = 28 I. Jf {7x-\- Sy- 5z= 3 dx + lOy - llz = 4 -- l-i- -- 5 3 4+ 2- ^ 5+ I- l=« . <2x + 2y + 3z = 17 3y + z= 12 3x + y + 2z=lB (^ = 2. \^= 3. {.= 4. (x = 9. ^= 12. u= 10. (x = 1. h= 2. u= 4. 2 a + b- - c' 2 1.1 - + - = a X y - + - = *, ." ^ X z [ Y a -\- c As this is in general the best rule for the solution of C' , with three unknown quantities, the reader should also nt the exercises given under the two preceding cases by it. SIMPLE EQUATIONS. 193 IV. BY CONDITIONAL MULTIPLIERS.* 271. Rule. Multiply any two of the equations by some unde- termined quantities, as m and n respectively, an#from the sum of the resulting equations subtract the remaining equation ; the remainder will be an equation containing the three unknown quantities. Any two of the unknown quantities may then be made to vanish, by assuming their coefficients equal to zero ; and the value of the remaining unknown quantity may be found in terms of m and n. The quantities m and n may now have their values determined by means of the two assumed equations ; and the three unknown quantities will then be easily found. EXAMPLE. If <( Zx — 2y + hz = 11 > what are the values of x, y, and z ? 4:X-{-5y-2z= 28 j Multiplying the first equation by m, and the second by n, 2mx -f 3my — 4:mz = 10m 3nx — 2ny -\- 5nz = lln. Subtracting the third from the sum of these, (2m + 3n- 4:)x + (3m — 2n - 5)y - (im - 5n - 2)z = 10m + lln - 28 ... [1]. Assuming the coefficients of x and y equal to zero, 2m + Sn- 4: = ... [2], 3»i-2n-5=:0 ... [3]; hence — (4rw — 5n — 2)z = 10m -f llw — 28, _ 10m + l ln — 28 4m — 5n — 2 From the two assumed equations containing m and n, their values are found (by the rule of the former case) to be 23 ^ 2 hence substituting these values of m, n, in that of z, we find z = 2. * See Lacroix, Eletnens d'Algihre, seizieme edition, p. 131 ; and Gamier, J' ^irjehrf.. trnifiiMTiP! edition, p. 216. M 194 ELEMENTS OF ALGEBRA. In order to find the value of y, the coeflBicients of x and z in [I] must now be assumed equal to zero, or 2w + 3n - 4 = .. [4] ♦?« — 5n — 2 = .. [5] 10m + \ln - -28 then ^ o o K 7 and the new values of m and n may be found from these two assumed equations [4] and [5], and being substituted in that of yy give its value. The value of x is to be found in a similar manner, by assuming the coefficients of y and z in [1] equal to zero ; or it may be found from one of the given equations, by substituting in it the values found oi y and z. It is left as an exercise to find these values of x and ?/, which are respectively = 3 and 4. ELIMINATION BY CROSS MULTIPLICATION. 272, Let the three general equations of the first degree con- taining three unknown quantities be given — namely, ax + by -\- cz — d a^x -j- by]/ + CjZ = d^ a^x -\-h^ + c^= c?2. It is required to determine three conditional multipliers, /, m, n, such that if the first equation be multiplied by /, the second by ??i, and the tliird by n, and the three resulting equations added together, the coefficients of two of the unknown quantities shall each be = zero. Multiplying each of the above equations as proposed, we have the following results : — lax + Iby -\- Icz = Id ma^x + mb-yy + »«Cj2 = vidy na^ -\- nb^y + nc^z = nd^. Since three conditional multipliers have been introduced, thej may be determined so as to fulfil three conditions : let the firsi two conditions be that when the equations are added, the coeffi- cients of y and z shall each = ; then the resulting equatior will be (Ja-{- ma^ + na^x = Id -{- md^ + nd^, _ Id -{- md^ + nc?2 ~ la + moy + na^' SIMPLE EQUATIONS. 195 Again, since the coefficients of y and z are each = zero, lb + m\ + n&2 = I (Ibc -\- mbjC + nb^c = ^' " ^ 1 or i le first of these equations ting the results and transpo (bc^ — b^cyn = (b^c — bc^i by multiplying the first of these equations by c, and the second by h, and, subtracting the results and transposing, we obtain hence h^c — bc^ bci — bjC Also by multiplying the first by q, and the second by b^, and subtracting the results and transposing, we obtain hence Thus it appears that only the ratio of the multipliers has been determined, and therefore there is an indefinite number of multi- pliers that will satisfy the conditions : but to obtain the simplest forms of these multipliers, let the third condition be that each of the above ratios = unity ; then to find x, (.K - b^c)l = I n -b,< h>; ib,c. - V:) be, - by / m n ' b,c. - Vx ~ b,c- be. ~ be. - b,c Similarly and I = h^c, - h^c, \ m = b.f - bc^ V n = be, — b^c ) d(b,e^ - b^e,) + d,(b„c (A); - be,) + d,(bc. -h,c) a{b,c^ - 62C1) + a^ib^c d(a,c, - a^c;) + d,(a,c ■ - be,) + 02(601 -ac,) + d,(ac. -Key -a,e) d(a,b, - a,b,) + d,(a,b - ac,) + b,(ac, - ab,) 4- d,{ab. -a,ey -a,b) c-d (c-dy _e-f j.-df ^ a - b ^ i(a- by ~ a - b^ i^cT^^' _ (c-dy + i(a-bXe-f) 4(a - bf 216 ELEMENTS OF ALGEBRA. c -d (c-df + \{a-h)(e-f) hence x + •- = V^ —,r iv2 ~ - 2(a — 6) 4(a — by = V{(c-rfy + 4(a-&)(e-/)} 2(a - b) ^ = 2(a-6) ^^ - c ± V[(c - t^)^ + 4(a - 6X« -/)]}, which are the two values of x ; and its numerical values for any- particular values of these coefficients would be found by substi- tuting for them their values. Solving the equation by the second method, multiply by 4(a — i), 4(a - 6)V + 4(a - J)(c - d)x = 4(a - 6)(e -/) ; adding (c — dy, 4(a - &)V + 4(a - i)(c - (;)x + (c - df = (c~ dy + iia-bXe-f)', extracting the square root, 2(a-b)x + c-d= V{(c - dy + 4(« - 6)(e -/)}, or x= -^^-L-^{d-c ± V[(c - dy + 4(a - 6Xe -/)]}. EXERCISES. 1. Ifx'^ + x = a, . x= J{ ± V(l +4a)-l}. 2. ... aa:^ - bx = d — c, . x = —{6 ± V^^ + 4a(c?- c)}. 3. ... 4tt2 - 2^2 + 2ax = 18ab - 18i^ x = 2a- 3b, or - a +36. 4. ... ^ _ 3^= = a^ + i2 _ 2bx, n Tn-: - «■' ^ - ' ^ 5. . .. adx — acx'^ = bcx — bd, . d b . a: = -, or . c a 6. . .. rnqx"^ — mnx -\- pqx = np, . n p X = -, or - . 9 7. . a— si 2ax — x^ x 4 • ar = a, or a + V 2ax — x^ « - ^' In the third example, the radicals disappear, as the quj under the radical sign is a complete square. QUADRATIC EQUATIONS. 217 285. Rule II. Equations in which the exponent of one of the powers of the unknown quantity is double that of the other, may i)e solved by means of the same method as quadratics. For if a new unknown quantity be assumed equal to the lower power of that in the given equation, the square of this quantity will be equal to the higher power, and the transformed equation is therefore a quadratic. Whether the unknown quantity be simple or compound, this rule is applicable. EXAMPLES. 1. Ifx^ — 2x^=: 48, what is the value of x? Let z = x^, then x^ = z^, and the equation becomes z" -2z = 48. Completing the square, 2'^ - 2^ -f 1 = 48 + 1 = 49, 2-l = V49=±7, « =r I ± 7= 8 or - 6. Having found the value of «, that of x is easily obtained. For x^ = z, .•> X = ^z ; therefore a;=^8 = 2ora: = ,^-6 = ^6^ _ 1 = _ 1 ^6 = - ^6 ; and hence the two values of x are a; = 2, and j; = - ^&. 286. It is shewn in the theory of equations, that every equa- tion has as many roots as there are units in its degree ; and hence x has three values in the equation x^ = z, and will have three values for every value assigned to z. Therefore since z has two values here, x will have six, as it ought, since the given equation is of the sixth degree* The equation may be solved without using a new quantity, as z ; thus- completing the square, a;6 _ 2a;» + 1 =: 48 + 1 = 49 ; extracting the square root, a^ - 1 = V49 = ± 7, re' = 1 + 7 = 8 or — 6; hence x' = 8 or a; = ^^8 =2, and x^" = - 6 or a; = ^- 6 = ^-1^6 = - 1 ^6 = -\y6. Tien the unknown quantity is compound, as in the next iple, the expressions are much simplified by using a new lown quantity. 218 ' ELEMENTS OF ALGEBRA. 2. Given VC^'^ - 3x - 1) + 6 = x- - 3x - 1. Let z = a/(x^ -3x - 1), then x" — 3a: - 1 = z\ and z^ — z = G, --^ + i = e + i = ?. 1 25 5 ^ - 2 = ^T = 2' .■.z = l±l = Sov-2. In order to find the value of x, one of the values of z"^ must now be equated with x"^ — 3x — 1 ; thus, taking the first value of z or 3, x^ - 3x - 1 = z^ rz 9 ; and from this equation, by the usual process, is found a; = 5 or — 2. But by taking the other value of z, namely, — 2, the equa- tion is ai" -3x-l =z^ = (- 2f = 4:. 3 1 From which is found x = - ± -\/29. 2 2i 3 1 Hence x has four values; namely, 5, — 2, ^ + oV29, and 3 1 - — -a/29 ; and on trial it will be found that they all satisfy the given equation ; and, in fact, if the given equation be cleared of its radical terms by squaring, it will be found to be of the fourth degree, and therefore it has four roots. When the unknown quantity is compound, and is under the sign of the square root, and only the terms of it, which contain the simple unknown quantity, are equal to the corresponding integral terms, the compound integral quantity may be made equal to that under the radical sign, by adding and subtracting a proper quantity, so that the equality may still exist. Thus, 3. Let sji^x" - 3a: + 5) = 2^2 - 3a: - 15 ; adding to the second number 20, and also taking 20 from it ; that is, adding 20 — 20, V(2x2 - 3a: + 5) = 2a-2 - 3x + 5 - 20. Now, let :^ = 2ar^ — 3a: + 5, and the equation becomes z =z^ ~ 20, or «2 _ ^ ^ 20 ; and by the usual process of solution it will be found that 2 = 5 or — 4. QUADRATIC EQUATIONS. 219 Hence for z = 5, 2x^ - 3x + 5 = 25; and from this equation is found ar = 4 or — -. Also forz=- 4, 2z=^ - 3a: + 5 = 16 ; 3 1 and from this is found x = -- + tV97, 4 4 80 that the four values of a: are 4, — -, - + -^97, and - — -rs/^l. 2 4 4 4 4 4. If x^" — 2ax" = ¥, what is the value of a: ? Assume a:" = z, then a:^" = z^, and the equation becomes s?-2az = b''; and from this equation is easily found z = a± V(a2 + 62>). hence x = ^z = ^{a ± xj^a" + h"^)} ... (1). As a particular example, let n = 2, a = 4, 6 = 3, and the given equation becomes x" - 8x2 ^ 9^ from which the values of x may be found in the usual way, by assuming x^ = z; but these values are more readily found from the preceding general formula (1), which becomes X = V{4 ± V(16 + 9)} = V(4 ± 5) = V9 or V - 1 = + 3 or ± V - 1. EXERCISES. 1. If X* - 6x2 + 10 = 2, ^ X = + 2 or + V2. 2. ... a:« _ 4x' - 28 = 4, x= 2 or - ^4. 3. ... a:*» - 2x2" ^ ^^ ^ = {1 + (1 + a)i}2V 4. ... 5x2 _ 2x = V(5a-2 - 2x) + 12, X = 2 or - I or |(1 ± V46). ■■5. ... A^/{x' - 3x) = x2 - 3x, . X = 3 or or ^(3 ± V73). 6. ... 2s/ (.r2 - Br + 11) = ^2 - 3x + 8, X = 2 or 1 or -(3 ± V - 31). 220 ELEMENTS OF ALGEBRA. QUESTIONS PRODUCING QUADRATIC EQUATIONS CONTAINING ONLY ONI UNKNOWN QUANTITY. EXAMPLES. 1. Find a number such that, if 8 times the number be added tc its square, the sum shall be = 65. Let X = the number, then x^ + 8x = 65 ; and the values of x are found by the usual process of solution to be a; = — 4 + 9 = 5 or — ]3. Either of these values satisfies the equation. If, however, the negative solution — 13 be considered as the positive number 13 to be subtracted, then the enunciation of the question would require to be so far modified as was done in simple equations, by changing the condition of addition into subtraction (252) ; the cause of which change is evident, by substituting — t foro: in the equation. But the enunciation without any change will be adapted to both values, by observing that the algebraic addition of a negative quantity is the same as the arithmetical subtraction of a positive one (56). 2. rind a number such that, if 4 times its square be diminished by 6 times the number itself, the remainder shall be = 70. Let x = the number, then ix' -Gx = 70; and from this equation is found x = 5 or — -» 3. Divide the number 24 into two parts, such that their product shall be = 95. Let X = one of the parts, then 2i — x = the other, and a:(24 — x) = 95, from which is found x = 19 or 5. 4. Find a number such that, if 15 be added to its square, the sum shall be = 8 times the number. Let X = the number, then a;* + 15 = 8a:, from which is found a: = 5 or 3. 287. It appears from this example, that a qiicstion producing a quadratic equation sometimes admits of two positive solutions, QUADRATIC EQUATIONS. 221 which belong literally to the enunciation without any modification. Positive solutions are also called direct, and negative ones, indirect, 5. To divide a number (a) into two parts, such that their pro- duct shall be = b^. Let .r = one of the numbers, then a — X — the other, and xia — x) = b^. This expression for the value of x shews the limitation of the conditions ; for its value is possible, provided that ih^ does not exceed a^. When 46^ == a^, or b = -, the value of x = -, and that ofa— x=:a — - = -; so that in this case the two parts are equal, each being a half of the given number. This is also the greatest value that b can have ; and hence the product is greatest when the number is divided into two equal parts. This result is analogous to the 27th proposition of the 6th book of Euclid. 6. There are three numbers in continued proportion ; the sum of the first and second is = 10, and the third exceeds the second by 24 : what are the numbers ? Let x'= the least. then 10 — X =: ... second, and 34 — X = ... third; also (322) X : 10 - X = 10 - X : 34 - X, whence x2 - 20x + 100 = 34x - x^ .•.. = f + 1=25 or 2. When X therefore 2, 8 2, 10 - X = 8, 34 - X = 32, and and 32. TATien x = 25, 10 — x = Uien 25, - 15, and 9. , . \ person bought a number of oxen for £80, and if he had bought 4 more for the same money, the price of each would have been £\ less : how many did he purchase ? 222 ELEMENTS OF ALGEBRA. Let X = the number of oxen, 80 then — = the price paid for each, X 80 and r = the price that would have been paid for each, had X -\- 4l more been bought for the same money. ^. , , . 80 80 , Hence by the question, — = : - + 1 ; and from this equation is found by the rule a: =+ 18 - 2 r- 16 or - 20. The number of oxen bought were tuerefore 1 6. The negative value — 20 to this question suggests, as usual, th' necessity of modifying the conditions of the question, in orde that the value 20 may be literally or directly applicable. Tin necessary modification to be made will be perceived by substi tuting the negative value in the preceding equation, when it gives 80 _ 80 _ ^ - _ ^. + 4 + '' 80 80 , or 7 =1, X — 4 X which shews that the condition of buying 4 more must be change( into that of buying 4: fewer ; and hence, also, the price, instead o being £1 less, will now be £ more. The question thus modifie( has for its direct solution th value 20, and its enunciation is : — 8. A person bought a nu' iber of oxen for £80, and if he ha( bought 4 fewer for the same money, the price would have been £ more for each : how many did he purchase ? Let X = the number of oxen, 80 then — = the price of each, X 80 and ■ — — r = the price of each had 4 fewer been bought. X — 4 __ 80 80 ^ Hence 7 = 1, x — 4 X whence x= + 18 + 2 =:= 20 or — 16. The positive solution 20 is the direct solution of this quesiur and the second value, when taken posi^'vely, is the solution of tht question properly modified — t^'-t is, of tlie seventh example. Thus, the two values of the unknown quantity are the solution of two analogous questions, which are mutually convertible by i QUADRATIC EQUATIONS. 223 proper modification of those conditions wliich depend on addition and subtraction. As the transactions of buying and selling stand in a relation to each other similar to that of addition and subtraction, the former question is susceptible of another modification which adapts it to the second solution 20. The enunciation of it is this : — 9. A person sold a number of oxen for £80, and if he had sold 4 fewer for the same money, the price would have been £i more for each : how many did he sell ? The reader will find by solving this question that the values of X are 20 and — 16 ; the former of which is the solution of this question ; and the latter, or 16, is the solution of a modifica- tion of this question, which is the seventh example. The mere change of buying into selling, or the converse, does not, however, alter the solution, unless the corresponding changes be made on the other conditions. This question shews that the indirect solu- tion — 20 in example seventh, are 20 sold ; that is, the negative solution is to be taken in a sense directly opposite to that of the positive solution. 10. Several persons are bound to pay the expense of a law process, which amounts to £800 ; but three of them being insolvent, the rest have £60 each to pay additional : how many persons were concerned ? Let X = the number of persons, then = the share each had originally to pay, X 800 ^ ., and ~ = actually ... ; • •.^ = ^ + 60, Q TO whence x = - + — = 8or— 5. From the former illustrations, the reader will easily interpret the negative solution. The — 5 means five persons who are to receive equal shares of £800. 11. Given the fraction v to find a number such that, if it be added to the numerator, and then to the denominator, the former resulting fraction shall be ?« times the latter. Let X = the required number, , a + X ma . then — - I b ~ '. + x' ± ^^{iabm + (a - bf} - i(a + b). 224 ELEMENTS OF ALGEBRA. As a particular example, let a = 4:, b = 5, m = -, 11 9 then X = ± ~ = 1 or — 10, 4+13 4 532, ^^^ -5- = 2 5TI'«^5 = 2-3^^- 12, Two lights, whose intensities are as 4 to 1, are 3 yard distant from each other: find that point in the straight lir between them, which is equally enlightened by each. Let A, B, be the two lights, and C the j i | required point, also -A- C B ( let AC = a-, then BC = 3 - X, then, since the intensities of the same light at different distance are inversely as the squares of these distances, and the intensit of the light A is 4 times that of B, 1 4 (3 - xj x' This equation produces the quadratic x^ — 8ar + 12 = 0, or the simple equation = ± -. o — x X From either of which a: = 2 or 6. The value 2 refers to the point C lying between the lights, an the other value 6 belongs to a point C lying beyond B, and £ either of these points the intensities of the lights are equal. In order to generalise the question, so that the solution ma comprehend cases of any distance of the lights, and any intensitie. let AB = a, AC = x, then BC = a — x ; also let the intensities of the lights A and B at the distance 1 b denoted respectively by ??r and n^. Then, since the intensities at C are to be equal, n^ 7^ m n I or — = + , I or (a — xy x whence x = The former value refers to C, and the latter to C ; for the fennel) value is less than a, and the latter is greater. QUADRATIC EQUATIONS. 225 ,. , „ an ^ an TJie corresponding values of a — x are and or m -\- n m — n ; these values being BC and BC, observing that a — a: is eckoned from B, and that it is positive when x ^ a — that is, when ^ is between B and A ; and hence it is negative when C lies )eyond B, for which x~p^ a. EXERCISES. 1. Find a number such that its square minus 6 times the lumber itself shall be = 7, = 7 or — 1. 2. Find a number such that its square plus 8 times the lumber itself shall be = 9, . . . . = 1 or — 9. 3. Find a number such that twice its square plus 3 times the 13 lumber itself shall be = 65, . . . . = 5 or — — . 2i 4. Find a number such that its square minus 1, with f of the emainder, shall be = 5 times the number divided by 2, = 2 or — -. 5. Find a number such that, if 44 be divided by the number ainus 2, the quotient shall be = one -fourth of the number ninus 4, = 24 or — 6. 6. Divide the number 49 into two such parts that the quotient if the greater divided by the less shall be to the quotient of the ess divided by the greater as 16 to 9, . . . = 28 and 21. 7. Find two numbers whose difference is = 8, and their pro- duct = 240, . . . = 12 and 20, or - 20 and — 12. , Find two numbers whose sum is = 100, and their pro- luct =r 2059, =29 and 71. 9. Find two numbers whose sum is = ?n, and their pro- luct = /i^, = \{m ± V(/«' - 4n2)}. 10. The intensities of two lights are as 4 to 9 : at what distance rom the latter are their intensities equal, supposing their distance be 12 feet ? =7-2 and 36. 11. A certain number of persons equally concerned in a transac- ion in business lost £960; but four of ^ them becoming insol- vent, the rest had to sustain a loss each of £40 greater than le otherwise should have done : how many persons were con- erned? = 12 and — 8. 12. A store -farmer bought as many sheep as cost him £80, nd after reserving 10 for himself, he sold the remainder for ;81, and thus gained 2s. a head on them: how many did he pur- hase ? = 100 or — 80. o 226 ELEMENTS OF ALGEBRA. 13. What number is that, which, being added, first to the nume- rator and then to the denominator of the fraction two - fifths, th( former result is = 4 times the latter ? . . = 3 or — 10 14. A vintner draws a certain quantity of wine out of a full ves- sel that contains 256 gallons ; and then filling the vessel with water draws off the same quantity of liquor as before, and so on for foui draughts, after which there were only 81 gallons of pure wine left how much wine did he draw each time ? = 64, 48, 36, and 27 gallons 15. A horse-dealer pays a certain sum for a horse, which h( afterwards sells for £144, and gains exactly as much per cent, as the horse cost him : what did he pay for the horse ? = £80 16. In a parcel containing 24 coins of silver and copper, eacl silver coin is worth as many pence as there are copper coins, anc each copper coin is worth as many pence as there are silver coins and the whole parcel is worth 18 shillings : how many are then of each kind ? . . . = 6 of silver and 18 of copper 17. Find a number such that, if 9 be taken from its square, th( remainder shall exceed 100 by as much as the number itself is lest than 23, = 11 or - 12 18. Find a number such that, if 18 be added to the product of its half by its third, the sum shall be = 4 times the number, = 6 or 18 19. A horse-dealer bought a number of horses for £180, but ha( he bought 3 horses more for the /ame sum, he would have paid £i less for each : how many horses did he buy ? . = 12 or — 15 20. In attempting to arrange a number of counters in the forn of a square, it was found that there were 7 over ; and when th< side of the square was increased by one, there was a deficiency o 8 to complete the square : required the number of counters, =56 21. Two messengers (A and B) being despatched at the same time to a place 90 miles distant ; A rode one mile an hour more than B and arrived at the end of his journey an hour before him : at wlu! rate per hour did each travel ? A = 10 and 6 = 9 miles per hour 22. A grazier bought a certain number of oxen for £240 ; afte; losing 3, he sold the remainder at £8 a head more than they cos him, and thus gained £59 by the bargain: what number did li buy? = ]( QUADRATIC EQUATIONS CONTAINING TWO UNKNOAVN QUANTITIE:: 288. When there are two equations, and two unknown qu'an titles, they cannot always be solved by the preceding methods for their solution will in general depend on an equation of th fourth degree containing one unknown quantity (295). There ai\ however, several species of such equations, the solution of whii'. ultimately depends on that of a quadratic containing only oi), unknown quantity. QUADRATIC EQUATIONS. 227 A system of equations consists of two or more equations con- taining unknown quantities that have the same value ; and a system of values of unknown quantities is a set of values that Satisfy a system of equations. CASE I. — WHEN ONE OF THE EQUATIONS IS SIMPLE. 289. Rule. Find a value of one of the unknown quantities from he simple equation, and substitute this value in the other equa- ion ; the result will he a quadratic with one unknown quantity, ivhich may be solved by the preceding rules. EXAMPLES. 1. If ■] ~2 T "'•i Z 1 Qo f what are the values of x and ^? Sy the first equation, x ■= y -\-2 ... [1]. substituting this value of x in the second, we have (y + 2)2+/ = 100 ... [2], vhence y = Q or — 8 ; lence by [1], a: = 8 or — 6. There are therefore two systems of values of x and y, which are •espectively 8 and 6, or — 6 and — 8, which may be arranged hus — a; = 8, - 6 ^ = 6, - 8. Equations may frequently be solved by means of particular malytical artifices. The last example, for instance, may be solved hus — Squaring the first equation, we find x"^ — 2xy + / = 4. Subtracting this from the second, 2xy = 96, or xy = 48 ... [3]. Substituting for x in this equation its value in [1], {y + 2)y = 48, or f+2y = 48, rom which the same values of y are obtained as from [2], and lence those of x may be found from [3]. Also, if the square of the first be subtracted from twice the second, the remainder will be .r^ + 2xy -f / = 196 ; hence ;()9, Theo. I.) {x + ijf = 196 and a; 4- 1/ = ± 14 : by combining his with the first equation, the same system of values of x and y v'ill be obtained as before. 228 ELEMENTS OF ALGEBRA. 2. If •] ^2 _ ^2 Z 20 i ^^^^ ^^6 the values of x and y ? By the first equation, a: = 10 — y ... [1]. Substituting this value of x in the second, (10 - yf -f = 20, or 20y = 80, Avhence y = ^ ••- [3], and, by [1], x = 6. Only one system of values of x and y are thus obtained ; but i y = Ahe substituted in the second equation, two values of x ar obtained ; namely, x = ± 6. The system x = — 6, y = 4:, wi] satisfy the second equation only, but not the first ; and it will b found that no system except the first, x = 6, and y = 4, wil satisfy both equations. These equations admit of another simple mode of solution. B_ dividing the second by the first, the result is x-y = 2. Adding this to the first equation, 2x = 12, hence x = 6. Subtracting it from the same, ^y = 8, ... y = 4. 3. Find the values of x and y in the equations x-y= 2 ... [1], x'-xy+f = S9 ... [2]. By[l], y = x- 2 ... [3]; substituting this value in [2], a;2 _ a:(^x -2) + (_x- 2^ = 39, from which is found, x = +6 + 1 = 7 or— 5; hence by [3], 5^ = 5 or — 7. 4. Find the values of x and y in the equations \x-ly = 2 ... [1], T-f + ^ = ^ ... [2]. By[i], y =1^-^ ... [3]., substituting this value in [2], and reducing the equation to tt usual form, it becomes 147x2 - 528x = 5184, QUADRATIC EQUATIONS. 229 216 rom which may be found x = 8 or j^ ; ,nd substituting these values in [3], those of y are 390 3/ - 6 or - —. The reason of the rule is evident, for the value of one of the inknown quantities found from the simple equation, will contain inly the simple power of the other, and this value being substi- uted in the second equation, cannot produce a term containing a ligher power than its square ; the resulting equation therefore nust be a quadratic. EXEBCISES. 3 i x^ + / = 89 j 2 { ^- + y = ^\ "" \ x^ - f=l&] * 3.... I " - 2^= 2| 1 x^ -lf= Ij 3x + 2y = 22 I 5x2 _ 3^^ ^_ ^2 ^ 45 I ( 2x + y =U\ '■' \ x^ - f= ^] • (ax + 6y = 27ra 1 " i xy = n) ' ( x — 9> or - 5. \y = ^ or - 8. ' X = 5. .y = 3. (x = S or 4 \y = ^ or 1 3' r X = 4 or 76 47' \y = ^ or 403 47* (x — h or '4 ( y = 4 or -4 a a> - ahny. { y = J + -(n? - ahn)K xY + 4x3^ = 96 I J x = 4, 2, 3 ± V21. X + y = q] ' ' \y^2,4.,ZT V21. a{x' + r)- K^' - /) = 2a ] j ^ - ±V^36; ' ( (a^ _ 62)^:^ -/)=4a6/ ) „ _ , /^LZ^V 230 ELEMETJTS OF ALGEBRA. 290. An equation is said to be homogeneous when its terms contain the same dimensions of the unknown quantities. CASE II. ■ — WHEN THE EQUATIONS ARE HOMOGENEOUS, EXCEPT THE CONSTANT TERM. 291. Rule. Assume one of the unknown quantities equal to the product of the other by a new or auxiliary unknown quantity ; substitute this product instead of it in the two equations ; and the equations will now contain this auxiliary quantity and the square of the other unknown one. Then, in each of the new equations, find the value of the square of this unknown quantity, and equate these values ; the resulting equation will be a quad- ratic containing the auxiliary quantity. Eind the value of this auxiliary quantity by the ordinary rule : the values of the other two are easily found. EXAMPLES. 1. Find the values of a: and y in the equations x^+xrj=lO ... [1], 2xy-f = 3 ... [2]. Assume y = vx, in which v, the auxiliary quantity, is unknown ; then substituting this value instead of y in the given equations, they become x^ + vx^ = 10 ... [3], 2vx^-v^x^= a ... w. But by [3], x^ - -^^- ... [5], and by [4], x^- ^ . "^ ~2v-v'' hence equating these values of x^ 3 10 2U — 17* — 1 + u' or 3 + 3y = 20i; - lOy^ Hence, by the rule (284;, 13 17 3 1 ^ = ±20 + 2-0^2""5' hence also by [5], x^ = or = 4 or ---, QUADRATIC EQUATIONS. 231 OK y Q 5 md .-. X = V^orV— ^- - ± 2or IgVS; md likewise ■.y = vx = lxi±2)orlx(± ^^3) = ± 3 or + ^^3. 3 §ince the first value of u or - gave the two values of a: = ± 2, these values of v and x must be combined in finding those of y ; lamely, ± 3. The second value of v or -, from which the other values of X = ± rV^ were found, must also be combined to get o the other two values of y or + -V3. o When the equations are both quadratic, the unknown quantities have four systems of values (295). In this case the systems arranged under each other are these — 2, -2, ^V3, and- -|V3, 3, - 3, -V3, and - -^V3, + 2or + |v3, «/ = + 3 or + -V3. The reason of the rule is evident from the system of quadratics ax^ + hf ^ cxy - d ... [1], mo? -\- ny"^ -\- pxy = q ... [2]. I Substituting in these equations y = vx, they become ' ax"^ 4- bv^x^ + cvx^ = d, mx^ -\- rjvV -j- pvx^ = q, which will each afford a value of x"^ in terms of v and v^, and I these values being equated, the resulting quadratic will give two values of v, which being substituted in either of the values of x^, will give two values of it, and hence four values for x. Then the systems of values of v and x being substituted in y = vx, will give four values oi y : so that x and y will thus have four systems r of values. 232 ELEMENTS OF ALGEBRA. EXERCISES. ' "'\2x' -\-^xy -2y'' = o] ' 1 y = ± 2 or + 1^2. In this example, the value of x^ cannot be found from the second equation after substituting in it y = va: ; but after dividing it by x^, the resulting equation will be one from which the value of V may be found. ;. If j ,2 j'a;=: ±3or + ^V7. 7 x^-2f = 8 I |:r::.±4or±^V7. ^f^- -y = ^f ' ' t3,=:±2or + ^V7. 292. Quantities are said to be similarly involved in an expression, in which they may be interchanged without altering the expres- sion. Such expressions are said to be symmetrical. CASE III. — WHEN THE EQUATIONS ARE SYMMETRICAL. 293. EuLE. For the two unknown quantities substitute the sum and difference of two auxiliary unknown quantities, and the resulting equations will contain the additive quantity with its square, and only the square of the subtractive one. Eliminate the square of this quantity from the equations (262), and there will result a quadratic containing the additive quantity, whoi-e value can then be found. Hence the values of the other auxiliary quantity and the original unknown quantities can easily be found. EXAMPLE. Find the values of x and y in the equations x^ + / = 13 ... [1], 2x -xy +2y = 4: ... [2]. QUADRATIC EQUATIONS. 233 ubstituting in these equations x = u -\- v, and 1/ = u — v, they ecome (u + vf + (u - vf = 13, ad 2(m + u) - (u + v)(u — u) + 2(w — u) = 4, hich are reduced to 2u^ + 2y2 = 13 ... [3], nd 4m _ „2 _|_ y2 = 4 ... [4] . wice [4], gives 8m - ^u^ -{- 2v^ ... [5] ; ence [3] - [5], iu" - 8u =5; 3 5 1 nd from this equation is found u = + h+^ =oO^"~o» lence, by [5], u^ = - or — , > 1 > 5 .•.r^=±2or±-; ind combining the values of u with the corresponding values of v, ivhich are derived from them, we find that = « + u = |±^or-^±| = 3,2, 2or-3, md 3^ = « - u = - + - or - - + - = 2, 3, - 3 or 2, which are the four systems of values of x and y. It is evident, from the manner in which these values of x and y are obtained from those of u and v, that the first value of r must be equal to the second of y, and the second of x to the first of y. Likewise, that when u and v have each two values, J? and y will have four, and the third value of x will be equal to the fourth of y, and the fourth of x to the third of y, so that when the values of one of them are known, those of the other are also known. 234 ELEMENTS OF ALGEBRA. EXERCISES. •"§ l.If P^+ / = 25 1 0:3, = 12 \ (x= 4, 3, - 3, - 1 J 1 y = 3, 4, - 4, - 3, 2 ( X - xy ^y =1 \ 2x'' + 2/ = 20 1 J ar =r 3, 1, 1, - 3. j \y= 1,3,-3,1. 3 f ^ -^^y +f=- \ 2a;2 + 2/ = 10 l\ [x= 2, 1, - 1, - 2. f t^= 1,2,-2,-1 I ^ - ^J' + y =1 j \y= 1,2. This example gives only two systems of values. The cause ol this is, that if either a: or ^ he eliminated, the resulting equation is only a quadratic. The most general form of the equations in this case is «(^ + /) + bxy + c(a: -^ y) = d, and a'(a;^ + /) + h'xy + c'(a: + ^) = cf. Substituting in these equations u -^ v for x, and m — u for y, thej assume the forms and »i'tt^ + n'u* + />'« = ?'• Eliminating u'* from this equation by (262), the resulting equatior is a quadratic containing u, from which its value may be found The value of v is then easily found from either of these lasi two equations ; and hence those of x and y are known. This method of substituting the sum and difference of two quantities for the unknown quantities, may be employed in solving equations higher than quadratics, when there are two equations, and one o: them contains only the sum of the two unknown quantities, anc the other the sum of their cubes, fourth powers, or fifth powers. 294. The solution of equations which differ in form from thos( of the preceding cases, must be solved, if possible, by particulai methods or analytical artifices, for the discovery of which the student must exercise his own ingenuity. Many of the precedinj: exercises and examples may be solved by particular methods, ant in some cases more expeditiously than by the rule. The following examples will exhibit some of the expedients to which recourse may be had when necessary: — 1. Find the values of x and y in the equations x2+/ = 25 ... [1], xy = 12 ... [2]. QUADRATIC EQUATIONS. 235 [ultiplying [2] by 2, and adding the result to [1], we find x'+2xy-\-y^ = 49. 'he first member is a complete square ; hence, taking the square aot of both, x^y=±1 ... [3]. igain, subtracting twice [2] from [1], there remains x^ — 2xy + ?/2 = 1, {x - yy = 1, .-.x-y =±1 ... [4]; dding [4] to [3], 2x = ± 7 ± 1 = 8, - 6, 6, — 8 ; ubtracting [4] from [3], 2y = + 7 + 1 = 6, - 8, 8, - 6 ; lence x = i, — 3, 3, — 4, nd y = 3, - 4, 4, — 3. 2. Find the values of x and y in the equations x^ + / = 13 ... [1], 2xy - X -y = 7 ... [2]. \.dding [1] and [2], x"^ + 2xy -\- y^ — X — y = 20, (^ + yy -(x+y)=20; et X -{- y = z, then z' — z = 20 -, 9 1 lence z = ±- + - = 5 or — 4, >r x + y = 5 or — 4 ... [3], .♦. (x + yj = x^ 4- 2xy + / = 25 or 16; aking this from twice [1], a:^ - 2xy +/ = 1 or 10; extracting the square root, a: — y = ± 1 or + v'lO ... [4] ; lading [4] to [3], 2a: = 5 ± 1 or ± V 10 — 4 ; acting [4] from [3], 2^ = 5 + 1 or + VIO - 4; *ho Tulues are a: = 3, 2, or + \sj\0 — 2, y = 2, 3, or + |V10 - 2. 236 ELEMENTS OF ALGEBRA. In solving these equations, the first object is to obtain a quad- ratic or a simple equation containing only one unknown quantity. This unknown quantity is either one of those in the given equa- tions, or some combination of both of them, as x -\- y in the second step of the last example. In the latter case, it will be convenient generally to assume a new auxiliary unknown quantity, instead of the combination. When one of the original unknown quantities is found, the other may be easily obtained either from one of the given equations or from some other. 295. The general solution of determinate quadratic equations containing two unknown quantities, can be effected only by the method of elimination, the exposition of which is not adapted to this treatise. It may be easily proved, however, that their solu- tion depends ultimately on that of an equation of the fourth degree containing one unknown quantity. The most general form of the preceding equations is — Ax"" + Bxy + Cy"" + Dx -\- Ey + F = 0, A'x^ + Bxy + Cy + D'x ^ E'y ^ F = % which may be expressed in this form : — Ax"^ + {By + U)x + (C/ + ^y + F) = 0, A'x^ + {B'y + i)'> + {C'f -f E'y + F') =: 0. If the coefficients of o^ be equalised by multiplying the former equation by A\ and the latter by A^ and the former product be subtracted from the latter, the remainder is — \(AB - A'B)y + AU - A'U\x + (^C - A'C^f, 4- (AE' - A'E)y -f AF' - A'F = 0, from which (A'C - AC')y^ + (A'E - AE)y + (A'F - AF') ^ ~ {AB - A'B)y + {AD' - A'D) ' and if this value be substituted for x in either of the given equa- tions, the resulting equation will be evidently of the fourth degree, in terms of y. This equation would give four values of y (286) ; and hence four corresponding values of x would be obtained. QUESTIONS PRODUCING QUADRATIC EQUATIONS CONTAINING TWO UNKNOW N QUANTITIES. 1. The sum of two numbers is = 10, and their product is = 24 : required these numbers ? Let X = one of the numbers, and y = the other, then, by the question, x -{- y = 10 ... [Ij, and xy = 24: ... [2], by[l], y=10-x ... [3]; quadhatic equations. 237 ubstituting this value oiy in [2], it becomes a;(10 - x) = 24, nd from this equation, which is a quadratic, is found a:=:±l+5 = 6or4; lence by [3] ?/= 10 — 6 or 10 - 4 = 4 or 6. f X = 6, 4, The two systems of values are therefore <^ ( 7/ = 4, 6. Had X been restricted to represent the greater, and y therefore he less, there would have been but one system — namely, x = Q, ■■ 4. The solution of the above question is performed by )ase I. 2. The ratio of two numbers is that of 2 to 3, and the difference »f their squares is = 80: required ^he numbers. Let X = the greater, and y = ... less, hen x:y = S:2 ... [1], md x^ - y^ = 80 ... [2], 2 )y[l], 3y =2xoTy=~x; 4 iubstituting this value of y in [2], x^ — -x'^ = 80, rom which is found x = /s/lii = + 12 ; 2 nee y =: -X = ± S. o 3. Find two numbers such that the sum of their products by ;he respective numbers a and b may be 2s, and their product jqual to a number p. Let the numbers be x and y, then ax + by = 2s ... [1], md xy = p ... [2], L rm 2s ~ ax _ ^ t^y [i]» y - —I — - [3] ; iubstituting this value of y in [2], and reducing the result, it lecomes I ax^ — 2sx = — bp', si 1 whence x = - ± -V(s^ — abp) or -{s ± V(«^ — obp)}; 238 ELEMENTS OF ALGEBRA. substituting this value of x in [3], and reducing, we find si 1 y = -^ ± ^V(s^ - ohp) or ^{s ± slif - aph)\. 4. The equirational mean of two numbers is = 18, and th< difierenee of their square roots is = 3 : required the numbers. Let X = the less, and y = the greater, then /^xy = their equirational mean (431); and hence A/xy — 18 ... [1], also i\/y — \/ X = 3 ... [2] ; squaring these equations, they become xy -,324 ... [3], y + X —2sjxy= 9; substituting from [1] for si xy^ we find y + ^ _ 36 = 9, 3^ = 45 - a: ... [4] ; substituting for y in [3], 45j; — .x'^ = 324, 27 45 from which x = ±7^ + -^ = 36 or 9; and by [4], 3/ = 45 — ar = 9 or 36. As X is restricted to the less, the solution is the system values X = 9, and y = 36. 5. There are three numbers in equirational progression; th( sum of the first and second exceeds the third by 1, and i times the second is equal to twice the third : what are th( numbers ? Let X = the first, and y = ... second, then (428), x:y = y :^^ the third number ; hence x-\-y = ~ + loTx'^-\-xy — y"^ — x = ... [1], and 3y = -^ov Zxy - 2/ = [2] ; 3 dividing [2] by y, 3x — 2^ = 0, and ?/= -.r [3]; substituting this value of y in [1], we find x-' + ^x'-^-x'-x^O; QUADRATIC EQUATIONS. 239 leaping of fractions, and dividing by x, we obtain X = 4:; hence y = 6, and ^- = 9 ; tie numbers are therefore 4, 6, and 9. 6. There are four numbers in equidifferent progression ; the roduct of the extremes is = 28, and that of the means = 60 : what re the numbers ? Let X — y = second, X -{- y ■= third, lien 2«/ = common difference; ence x — Sy = first, nd X -^ By = fourth. )y the question, (x - SyXx + 3y) = 28 or a:^ - 9/ = 28 ... [1], nd (x - yXx + y) = 60 or x"^ — / = 60 ... [2] ; ubtracting [1] from [2], 8/ = 32, y = ^4: = ± 2, •y [2], a:^ = 60 4- / = 64, X = ± 8. Hence the numbers are x — Sy = 8~6 = 2, x— y = 6^ : + ^ = 10, and X + 3^ = 14 or 2, 6, 10, and 14. 7. There are three numbers in harmonic proportion ; the sum >f the third, with double the first, is = three times the second ; and he sum of the squares of the first and third is = 180 : required he numbers. ;hen Let X = the first, and y = ... third 2xy X +y second (438) ; md by the question, 2x + w =- — ~ ... [ll ; X +y ilso, x" +f = \%0 ... [2], 3y [1] 2x2 +/ - Zxy = ... [3], she equations [2] and [3] come under Case II. ; and hence assuming / = vx, they become x^ + v^v- = 180, and x^ =~^^~^, 2X2 _^ ^2^2 _ 3^^2 ^ Q^ or 2 + ?;2 _ 8u = ; 240 ELEMENTS OF ALGEBRA. from this quadratic (see 2d Exercise, Case II.) is found v = 2 or 1 ; 180 hence x" = ^ = 36, / = 180 - 36 = U4, and x= 6,y= 12,-^ = — = 8; the numbers are therefore 6, 8, and 12. The value of u = 1 gives 3 V 10 for the three numbers. 8. A and B depart from the same place, in the same direction, and travel at a uniform rate. A sets out 6 hours before B, and B, after travelling 52^ miles, overtakes A; but had their rate of travelling been half a mile per hour less, B would, aftei travelling only 36 miles, have overtaken A : what was their rate of travelling ? Let X = A's rate of travelling, or the number of miles lit travelled per hour, and y = B's rate. 105 Then the number of hours that A travels = 52| -^ x = ——. £iX and B ... =^^^y = -^. But had they travelled \ mile per hour less, 72 then A would have travelled = 36 4- (x — ^) = hours " IiX — 1 72 and B =36-(y-i) = — -^ ... But A travels 6 hours longer than B, lO'i IOt hence -^ = V" + 6, or 210^ - 210x - 24z^ = 0, 2x 2y and 27=n; ^ 2 - 1 ^ ^' ^^ ^^""^ ~ ■^^^'' ~ ^^""^ ^ ^ ' 72 _n dividing each of these equations by 6, they become 35?/ - 35ar - 4a:y = ... [1], 26y — 22x - ^:xy =\ ... [2] subtracting [2] from [1], \'\x — \ ^y-\Zx=-\oxy = _ ... [3] QUADRATIC EQUATIONS. 241 substituting this value of y in [1], it becomes, after reduction, 52x2 _ 14^^ ^ _ 35 . 1 7 from which quadratic is found x = 2- or -— ; 2 2b and hence by [3], 5, = 3- or — . 5 7 The system of values x = -, 1/ = -, satisfy the two conditions 7 5 of the question literally ; but the other systems x == — , y = — , Zk> 18 g'ive negative results in verifying them for the second condition ; so that in order to fulfil this condition, A and B must be supposed to travel in a direction opposite to that of their first journey, and B to depart 6 hours before A. 9. The equidifferent mean of two nmnbers exceeds their equi- :ational mean by 5, and the latter exceeds their harmonic mean 3y 4 : what are the numbers ? Let X = the less, and y = the greater number, ;hen their equidifferent mean is = ^(x -\- y\ md ... equirational ... = is/xy, md ... harmonic ... — • X +y lence ^{x + y) = s/xy + 5 ... [1], md -^^ = \^xy - 4 ... [2] ; ^ X +y ^ -^ L -J' amltiplying [1] by [2], xy = xy -\- ^/xy — 20, or /^xy = 20, and .-. xy = 400 [3]. Subtracting [2] from [1], and reducing (x + yy-18(x + y) = 4.xy = 1600 by [3]. Let X + y = z, then z"" - 18z = 1600; md hence is found 2; = +41 + 9 = 50 or — 32 ; ?rhence x -\- y = 50 ... [4], squaring, x' + 2xy -\- y^ = 2500 ; :aking 4 times [3] from this equation, 3^ - 2xy + y^ = 900 ; lence x — y = 30 ... [5], ad I I] + [5] gives 2x = 80, or x = 40; also [4] — [o] ... 2y = 20,OTy= 10. 242 ELEMENTS OF ALGEBRA. The numbers are therefore 10 and 40, and their equidiffereni equirational, and harmonic means, are respectively 25, 20, and 1( EXERCISES. 1. The diflference between two numbers is = 7, and the differenc of their squares is = 119 : required the numbers, . = 12 and I 2. The ratio of two numbers is that of 2 to 3, and the differenc of their squares is = 20 : required the numbers, = 4 and ( 3. The sum of 6 times the greater of two numbers, and 5 time the less, is = 50, and their product is = 20 : Avhat are th numbers ? , =: 5 and 4 4. The product of a certain number consisting of two places b the sum of its digits is = 160, and if it be divided by 4 times th digit in the place of units, the quotient shall be = 4 : require the number, =3' 5. If a certain number consisting of two places be di\dde by the product of its digits, the quotient is = 2, and if 27 l added to it, the digits are in an inverted order: what is tli number ? = 3( 6. There are two numbers, the sum of whose squares exceec twice their product by 4, and the difference of their squares exceec half their product by 4 : what are the numbers ? =6 and ! 7. There are two numbers whose sum is =14, and if the squai of each be divided by the other, the sum of the quotients is = 15- required the numbers, = G and i 8. The sum of four numbers in equidifferent progression = 28, and the sum of their squares is = 216 : what are tl numbers ? = 4, 6, 8, and li 9. The equirational mean of two numbers is = 12, and the sui of their square roots is = 8 : required the numbers, = 4 and 3 10. Three numbers are in equirational progression ; the thii exceeds the sum of the first and second by 2, and the third equal to twice the second : what are the numbers ? = 2, 4, and 11. There are four numbers in equidifferent progression ; tl product of the extremes is = 27, and of the means is = 35 : require the numbers, = 3, 5, 7, and 12. There are three numbers in harmonic proportion ; tl second exceeds the first by 3, and the third exceeds twice tl second by 4 : required the numbers, . . = 5, 8, and 2 13. Three numbers are in harmonic proportion ; the sura of tl extremes exceeds twice the mean by 6, and tiie pioduct of tl extremes is = 108 : required the numbers, -= 6, 9, j"-^ ^ RATIO. 243 14. Two persons (A and B) depart from the same place, in the same direction, and travel at a uniform rate. A starts 2 hours before B, and, after travelling 30 miles, B overtakes A ; but had each of them travelled half a mile more per hour, B would have travelled 42 miles before overtaking A: at what rate did they travel ? . . . . A = 2^, and B == 3 miles, per hour. 15. From two towns (C and D), distant 390 miles, two persons 'A and B), setting out at the same time, met each other, after travelling as many days as are equal to the difference of the lumber of miles they travelled per day ; and it appeared that A lad travelled 216 miles: how many miles did each travel per lay ? A =r 36, and B = 30. 16. A wine-merchant sold 7 dozen of sherry and 12 dozen of claret for £50. Of sherry he sold 3 dozen more for £10 than he did of claret for £G : what was the price of each ? = £2 and £3. RATIO. 296. As quantities in algebra are represented by letters, which denote the values of these quantities in numbers, the following theorems respecting ratios refer merely to the ratios of numbers.* DEFINITIONS. 297. The ratio of one quantity to another is the number of times that the former contains the latter. Thus, the ratio of 8 to 2 is 4 ; of 3 to 12 is :i. So the ratio of 14a to 7a is 2 ; of 6a^ to 2a is 3a. The ratio of two quantities as of m to n is denoted thus, m ~- n, or m :n, or — ; m : n is enunciated m to n, or the ratio of m to n. n If m contain n, r times, then m -=r n = r, or — = r, or m : n = r, n or TO : w = r : 1, for ?■ : 1 is = - = r. 298. The ratio of two numbers may be either a terminate or an interminate number (194). * The usual theorems respecting the ratios of geometrical magnitudes and xjuaiicities, generally, are given in the fifth and additional fifth books of the ifolume on Plane Geometry of Chambers's Educational Course. 244 ELEMENTS OF ALGEBRA. 12 8 2 8 • Thus, — - = 3, an integer ; — = - a fraction ; ov — = -6 t 9 ■ • / 5 2'^36 repeater; - = 1-285714 a circulate; or VS : 2 — — 7 '^22 = 1"118 ... which is an interminate number, for /^5 = 2-236 ... is such a number. 299. The first terra of a ratio is called the antecedent ; and th( second, the consequent. Thus, in the ratio 6:2, 6 is the antecedent, and 2 the con sequent. 300. A ratio is said to be a ratio of equality, majority, or minority according as the antecedent is equal to, or greater or less thar the consequent. 301. The antecedents of two or more ratios are called homologou terms, and so are the consequents. 302. Ratios are said to be compounded when their homologou terms are multiplied together ; and the ratio of the two product is said to be compounded of the simple ratios, and is hence called i compound ratio. The compound ratio of two simple ratios may be convenientl] expressed thus, {a:b, c:d) where a : h and c : d are the simpl( ac ratios, and (a :b,c:d) = j~z, the compounded ratio. 303. The ratios of the squares of two quantities are said to b< duplicate of the ratios of the quantities themselves ; and tliat o their cubes, ti-iplicate ; so the ratios of their square roots are saic to be subduplicate, and that of their cube roots subtriplicate of thei: ratio. 304. A multiple of a quantity is a quantity that contains i exactly. Thus, 4a is a multiple of a by 4, and 7nb is a multiple of b whicl contains it m times. 305. A submultiple of a quantity is an aliquot part or measure o that quantity. Thus, a is a submultiple of 7na.* 306. Equimultiples of quantities are multiples that contain tlien the same number of times. * The terms multiple, equimultiple, submultiple, and eqiiisubmult' in geometry applied to multiples or submultiples by an integer : but in i , mathematics, the meaning of the first two terms is extended, m as to con> • multiples by any terminate or interminate nmnbers. RATIO. 245 Thus, 4a and 46 are equimultiples of a and /> by 4 ; so ma, mb, and mc, are equimultiples of a, b, and c, by m. 307. Equisubmultiples of quantities are equal measures or equal aliquot parts of them. Thus, a and b are equisubmultiples of 4a and 46 ; so a, 6, and c, are equisubmultiples or equal measures of ma, mb, and mc. THEOREMS. 308. I. The ratio of two quantities is equal to that of their equimultiples. Let a, 6, be two quantities, and m any number, then - = — (127), where m may be either a terminate or an interminate number. Cor. The ratio of two quantities is equal to that of their sub- multiples. Thus, if a = nc, and b = nd, then y = — ,= -,^ where c and d nd d are equisubmultiples of a and b by n. 309. II. If the same quantity be added to the terms of a ratio of equality, of majority or of minority, the first will be unaltered, the second diminished, and the third increased. If a and b be any two quantities, then - = r, for each is = 1, a a + 6 or the ratio of equality - is unaltered. Let c be a third quantity ; then reducing the ratios y, , to the same denominator, they become respectively ab -\- ac ab -{- be b{b + c) 'b(V^cj' Now, when r is a ratio of majority, az^ b, and therefore ac -p' be, and the former fraction exceeds the latter, or y ^ -, . b + c Again, when r is a ratio of minority, a^b, and therefore ac ^ be, and the former fraction is less than the latter, or v -^ i • 6 b -\- c 310. III. A ratio of equality is unaltered, a ratio of majority is increased, and a ratio of minority is diminished, by subtracting the same quantity from its terms. 246 ELEMENTS OP ALGEBRA. _, a a — b , For - = =- = 1. a a — And when az^b, ac'P' be, and ab — ac, is less than ab — be and hence y ^ ^ , as is evident when they are reduced to th< b — c same denominator. Again, when a .^ b, ac ^i^ be, and db — ae is greater thai ab — be; and hence -r -y' -^ . b b — c 311. IV. Any ratio compounded with a ratio of majority, i increased ; and with a ratio of minority, is diminished. Let Y be any ratio, and -3 another, the former ratio compoundec with the latter, gives 7-,. bd rni. 4.' ^ • ^'^ The ratio 7^ is = ^-^ 6 bd When ;^ is a ratio of majority, c'p' d; therefore ac z^ ad, an( - ac ad ^ ,^ ,. ac a hence -n'^ i~i and the ratio 7-. z^ 7-. bd bd bd b When - is a ratio of minority, c^d; therefore ac^ad, an< , . ^. ac a hence the ratio 7-; -=^1 r- bd b 312. V. If a ratio of majority, and another of minority, b compounded, the compound ratio is intermediate in magnitude. Let 7- be a ratio of majority, and -. one of minority, then 5-7 ^ d bd and ■y' -.. d For 7- being compounded with the ratio of minority, is dimi b ac a nished (311), or 7-7 ^ v- bd b And - being compounded with a ratio of majority, is increasei (311), or -^z-^. 313. VI. If the homologous terms of two ratios be adde< together, the ratio of the sums is of intermediate magnitude. RATIO. 247 Let 7 and -^ be the ratios, the former being the greater, then a -\- c a - c ^ a ad jC be , ^a c . ,, IfOT y = -j—., and -, =: -—z, but r x' -^ ; hence ad -p' he. b bd d ba b d Reducing j- and r , , to the same denominator, they become ab 4- ad , ab -{- be and b(b + d) b(b + d) ' but as ad ■p' be, the numerator of the former fraction exceeds that of the latter ; and hence ~ p' -. -;. b -{- d Again reducing t— — 7 and - to the same denominator, they become ad + cd , ic + cd and (h + d)d (6 + d)d ' and since ad 'P' be, the numerator of the former fraction exceeds that of the latter, and hence , "^ , zp' -. 6 + a d 314. VII. The duplicate ratio of two quantities is equal to the ratio of these quantities compounded with itself. Let a, b, be two quantities, then a^ : 6^ is the ratio of a : b compounded with itself, or with a : b (302) ; or (a : b, a : 6) = a^ : 6^. 315. Vni. The triplicate ratio of two quantities is equal to a ratio compounded of three ratios, each of which is equal to the ratio of the quantities. Let a, h, be two quantities ; then the ratio compounded of the three identical ratios a : b, a : b, and a : b, is a^ : b^ (302), or {a : b, a : b, a : b) = a^ : ¥. 316. IX. The ratio of two quantities is equal to their sub- duplicate ratio compounded with itself. Let ^Ja : V ^ be the subduplicate ratio of two quantities a and h; then f^a:^/b, compounded with f^a: /s/b, gives a : b, or {.\/a : ^/b, /^a : A/b) = a : b. 317. X. The ratio of two quantities is equal to the ratio which is compounded of three ratios, each of which is equal to their subtriplicate ratio. 248 ELEMENTS OF ALGEBRA. Let ^a : •^h be the subtriplicate ratio of a and h ; then the three identical ratios ^a : ^/h^ ^a : ^h, and <^a : ^6, being compounded (302), give the ratio a : 6, or {^a : ^6, ^a : -v^i, -^a : ^6) == a : 6. 318. XI. The ratio of the first of any number of quantities to the last is equal to the ratio which is compounded of that of the first to the second, of tiie second to the third, of the third to the fourth, and so on to the last. For let a, 6, c, - logous in the following, and it is inferred that the femahiing ■ of the first are reciprocally proportional to the remaining ten the last. PROPORTION. 251 328. If the antecedents and consequents of two ratios are espectively equal, so are the ratios. 329. If two ratios are equal, and also their antecedents, so are heir consequents, 330. If two ratios are equal, and also their consequents, so are he antecedents. These three axioms may be more concisely stated thus : of these ;hree conditions— the equality of two ratios, of their antecedents, ind of their consequents — if any two he given, the third also jxists. 331. If one ratio exceeds another, and their consequents are jqual, the antecedent of the former exceeds that of the latter. 332. If one ratio exceeds another, and their antecedents are ;;qual, the consequent of the former is less than that of the latter. 333. If the consequents of two unequal ratios are equal, the greater ratio is that which has the greater antecedent. 334. If the antecedents of two unequal ratios are equal, the greater ratio is that which has tlie less consequent. 335. According as the first term of a proportion is greater than, equal to, or less than the second, the third is greater than, equal to, or less than the fourth. 336. If the first term of a proportion be a multiple of the second by any number, the third is the same multiple of the fourth. THEOREMS. 337. I. The terms of an analogy are proportional by inversion. Let a : b = c : d, then by inversion b : a = d : c. ■T^- a c . a ^ c y. ,^s ^ d , , For smce -.- = -3, 1 -i- t^ = 1 -r- -„ or (llO) - = - or b : a = d : c. b d b d ^ ■^ a c 338. II. The terms of an analogy are proportional by alternation. Let a:b = c: d, then by alternation a: c = b:d. For T = -p and multiplying by -, a b _ c b a _ b b'c~d'?'^^c~d' a: c = b : d. III. The terms of an analogy are proportional when taken xy regular order. 252 ELEMENTS OF ALGEBRA. Let . . . a : b — c : d ... [1] By inversion, . . b : a = d : c ... [2] ... alternation of [1], a : c = b : d ... [3] [2], b:d=a:c ... [4] .. equal ratios in [1], c : d — a : b ... [5] .. inversion of [4], d :b = c : a ... [6] .. alternation of [5], c : a = d : b ... [7] [6], d:c=b:a ... [8] There are thus eight different analogies formed by transposing the terms in regular order. 340. IV. The terms of an analogy are proportional by com- position. Let a:b = c:d, then by composition, a -\- b :b = c + d:d. c For a , c ,»a + 6 c -\- d - : therefore t + 1 = -, + 1 or — -, — = — -, — ; a b d b d that is, a -{■ b:b =^ c -\- d: d. 341. V. The terms of an analogy are proportional by addition. Let a:b = c:d, then by addition, a : a -\- b = c : c -\- d. For ^ = ^(337); and hence 1 + ^ = 1 + ^ or ^^ = ^±-^ a c ^ ^ a c a c therefore a -\- b:a = c -{- d: c, and by inversion, a : a -{- b = c : c -\- d. 342. VI. The terms of an analogy are proportional by division. Let a : b = c : d, then if a -p^ b, by division, a — b : b = c ~ d:d. a c^, „ a c a — b c — d For T = -7 ; therefore ^ — 1 = -, — 1 or — j — = — - — ; b d b d b d that is b:b = d:d. 343. CoR. If 6 z' a, then d^ c (335), and 1 - ^ = 1 - ^ o -, or 6 — a: b =^ d — c : d. b : b = c ~ d : d. * b d Therefore, generally, a 344. Vn. The terms of an analogy are proportional by con version. Let a : b = c : d, then if a 7^ t, by conversion, a : a — b = c : c — d. PROPORTION. 253 ^ b d , , b , d a-b c -d For - = - ; hence 1 — 1 or = , a c a c a c herefore a — b : a = c — d : d, nd by inversion a : a — b = c : c — d. b d b — a 345. Cor. If b ■p' a, then 1 = 1 or — -— = ' a c a d- c c ence b — a: a = d — c. c, and ly inversion, a :b — a = c : d — c. Hence, generally, a : a ~ b = c: c ~ d. 346. Vin. The terms of an analogy are proportional by mixing. Let a:b = c:d; then ii a -p- b, a + b : a — b = c -{- d-. c — d. For by addition, — rT = ' — r~j ••• LIJ > -^ a -\- c -\- d ... b d r--, )y composition and inversion, — — 7 = — —-% ... L^J ; subtracting the latter from the former, a — b c — d a -{-b~ c -\- d^ yr a — b : a + b = c — d : c -\- d-, and hence by inversion, a-^bia — b = c-\-d:c— d. 347. CoK. When b :p' a, subtract [1] from [2], and , d — c : ., whence a + b : b — a = c -\- d : d — c. c -\- d Hence, generally, a-{-b:a~b = c-{-d:c-^d. 348. IX. The product of the extremes of a proportion is equal to that of the means. Let a : b = c : d, then ad = be. For T = -7 ; therefore bd . ^ = bd . - „ or ad = be. b d b d 349. Cob. 1. The fourth term of a proportion is equal to the produpt of the secom! and third divided by the first. be For ad = be : and hence d = — . a ';■ Therefore, when three terms of a proportion are given, the fourth can be found. 254 ELEMENTS OF ALGEBRA. 350. Cor. 2. If the product of two quantities be equal to tb of the other two, these four quantities are reciprocally propo tional. Let ad = he, then —- = —-, or ^ = - • that is, a : b = c: d, ( oa ba a alternately, a: c = b: d. 351. Cor. 3. When two quantities are reciprocally propo] tional to other two, if the former two be made the extreme tern of a proportion, the latter two are means, and conversely. 352. Cor. 4. If three quantities be in continued proportion, tl product of the extremes is equal to the square of the mean. Let a : b = b : c, then ac = b . b = b"^. 353. Hence when the extremes are known, the square of tl mean proportional is easily found, and consequently the met itself. Also, when the mean proportional and one extreme a: known, the other extreme can be found, for a= — , and c = — . c a 354. Cor. 5. If the product of two quantities be equal to tl square of a third, this third is a mean proportional between then If ac = b"^, then ac = b .b, and a : b = b : c (350). 355. X. If any number of quantities be proportional, oi antecedent is to its consequent as the sum of all the anteceden to that of all the consequents. Let a, b, c, d, e, a,ndf, be proportional, then a :b = a -{- c+e\ + d+f. a b , . , 7 1 T . a I Eor - = 7- ; and since a : b = c : d, by alternation, - = - a b c ( also, a : b = e :f, therefore by alternation, - = - ; hence ti ^ / reciprocals of these equals are also equal ; that is, a be d ,e / a -\- c -\- e b -{■ d -{• f - = -,- = -, and - = y-, or = 7 ; a b a b a b a hence a -\- e -\- e : a = b -{- d -\-f: b, or (339), a:b = a + c + e:b-{-d-\-/. 356. XI. If the consequents of two analogies be the same, t: sum of the first antecedents is to their consequents, as the sum the other antecedents to their consequents. Let a:b = c:d, and e:b — f:d, then a -\- e:b — c A- f . PROPORTION. 255 For by the former proportion, r = -j^ and by the second, t = ^; a e c f a -\- e c -\- f , , ,/., t^^^n + 6 = 5 + ^' ^' -T- = -/-' °' " +''-^ ^ ' +'^=^- 357. XII, In any analogy, as one antecedent is to its conse- quent, so is the sum or difference of the antecedents to that of the ;onsequents. Let a:b = c:d, then if a z^ c, a : b = a ± c : b ± d. T^ /-oor^N c d ,,c -, , d a ± c b ± d . For (339) - = y, . ' . 1 + - = 1 + y> ov —=— = — - — ; hence a b ~ a ~ b a o ;339) a:b = a ±c:b ±d. 358. XIII. Any equimultiples of the first and second terms of m analogy are proportional to any equimultiples of the third and fourth. Let a:b = c: d, then ma :mb = nc: nd. -r^ . ci c .,__. ma nc , , . For smce t = ^? (127) — r = —u or ma:mb = nc: nd. b d mb nd 359. Cor. 1. Any equimultiples of the terms of an analogy are proportional. For when n = m, ma : mb = mc : md. 360. Cor. 2. Any equimultiples of two quantities have the same ratio as any other two equimultiples. ^ ma na . , For — ^ = —r, or ma : mb = na : nb. mb nb 361. Cor. 3. Any two quantities are proportional to any of their equimultiples. ^^ a ma , , 1 or 7- = — 7-, OT a: b = ma: mb. b mb 362. XIV. Any equimultiples of the homologous terms of a proportion are also proportional. Let a : 6 = c : c?, then ma :nb = mc: nd. For by alternation, a: c = b : d; therefore (358) ma : mc = nb : nd, and by alternation ma :nb = mc : nd. Cor. 1. When m = 1, a : nb = c : nd. Cor. 2. When n = I, ma:b = mc: d. 363. Cor. 3. According as ma is greater than, equal to, or less than nb, mc is greater than, equal to, or less than nd (335).* I'he principle in this corollary, when m and n are any integral numbers, is •operty assumed by Euclid as the definition of the proportionality of four 1 titles. 256 ELEMENTS OF ALGEBRA. The two preceding theorems, and their corollaries, are tru when equal submultiples are taken instead of equimultiples ; fo the steps of the demonstrations are true whether m or n h integral or fractional. 3G4. XV. If any number of analogies have two homologor terms in each equal to two in the following, the remaining terra are proportional by direct equality. Let a : b = c : d, and b : e = d : f; then by direct equality a:c = e:f. For (339) by the first analogy, ~ = -7, and by the second, -^ = - c a d J hence - = -, or a : c = e :/. c J If there be a third analogy, as e:g=f:h, then | = - = -, ( a'.c-=g:h. If the three analogies be arranged thus, a : b =^ c : d, b:e = d:f, e '.g = f:h, then it is evident that the terms, to which none of the rest ai equal, are a, c, g, and h, and a : c = g '. h. 365. XVI. If any number of analogies have two terms n( homologous in each, equal to two not homologous in the followin the remaining terms are proportional by indirect equality. Let a : b = c : d, and 6 : e = f:.c, then by indirect equalit; a:e=f:d. For by the first proportion (348), ad = be, and by the seconi be = ef; therefore ad = ef, and (350) a:e=f:d. If there be a third proportion, as e:g = h:f, then ef — gl therefore ad = gh, and a : g = h : d. Arranging the three analogies thus, a : b = c '. d, b :e = f: c, e :g = h:f, the terms to which none of the rest are equal, are a, d, g, and / hence a : g = h : d. 366. XVn. Katios that are compounded of equal ratios iu equal. PROPORTION. 257 Let a:b = (c:d,e:f), and let a' : b' = (c' : d% e' :/') ; also ,let c: d = c' :df, and e\f = e' :/', then is, a : b = a' :h'. For ^ = I . ^ (302), and |J- = ^ . -^ ; but since ^ = "^^ and fi fi' .1 /. '^ <*' l / 7,/ -. = - ; therefore -r = -^y, or a:b = a : b . 367. XVIII. The reciprocals of the terms of an analogy are proportional. Let a:b = c:d, then - : t = - '• 3- a b c d a c , ^ b d 1111, ^^.11 For T = -, ; therefore - = -, or --r-T = --rj; that is, - : -r b d a c a b c d a o _ 1 1 ~ c'' d' 368. XIX. If two analogies be compounded, the resulting terms are proportional. Let a:b = c:d, and e:f=g:h, then ae:bf= eg : dh. ^ a c -, e g ^. ^ a e eg ae eg For ^ = ^, and- = |; therefore^. - = ^.|, or -^=^; that is, ae:bf= eg: dh. The proposition may be proved in the same way, whatever be the number of analogies to be compounded. 369. XX. The same powers of the terms of an analogy are proportional. Let a:b = c:d, then a" : 6" = c" : c?". = c" : rf". 370. XXI. The same roots of the terms of an analogy are proportional. Let a:b = c:d, then ^a : ^b = *yc : ^d. For ^ = ^; therefore ^^ = and q be any other two numbers, it may be similarly prove' that - — =^-, = -, : and hence pc + qd d ma + n b _pa ± qb ma ± nb _ mc ± nd mc ± nd pc ± qd^ pa ± qb~ pc ± qd^ that is, ma+ nb : pa ± qb = mc ± nd : pc ± qd. PROPORTIONAL EQUATIONS. DEFINITIONS. 373. A proportional equation is an equation subsisting betwee two variable quantities, expressing merely the proportionality < the two members of the equation. Thus, if X and i/ are two variable quantities so related that an two values of x, as x' and x", are proportional to the corresponc ing values of y — namely, y' and y'' — then x' : x" = y' : ?/", whic is concisely exi)ressed thus, x oc ;/, which expression is called proportional equation ; and the sign oc is read varies as. As a particular example, let x be the rate of travelling of stage-coach, or the number of miles travelled in an hour, an 1/ the number travelled in a day; then if x', x", the rates < travelling for any two days, be 7 and 8 miles an hour respectivel; the corresponding distances travelled on these two days, or y' an y, are 168 and 192 ; and 7 : 8 = 168 : 192, or x' : x" — y' \y' The same proportion exists for any other two rates of travellinj and the corresponding number of miles travelled per day, so tlu the proportion x' : x" = y' : y" is true generally for any two rat( of travelling ; and this is expressed by the proportional equatic X cc y, which always implies tlie proportionality oi four quai titles — namely, of any two values of x, and the correspondin values of ^. PROPORTIONAL EQUATIONS. 259 Innumerable examples of proportional equations occur in ma- thematical and physical science ; quantities in the latter science, like those in the former, being represented by letters wliich express their numerical values, or the number of times that the assumed unit of measure is contained in them. 374. If two quantities are so connected that when one of theni varies, the other varies in the same proportion, the former is said to vary directly as the other. Thus, if X and y are so related that any two values of x, as x' and x'\ and the corresponding values of y — namely, y' and y" — are so related that x' : x" = y' : ;y", then x and y are said to vary directly, and this relation is expressed by x oc y. 375. If two quantities be so related that when one of them varies, the reciprocal of the other varies in the same proportion, the former is said to vary inversely as the latter. Thus, if x' \ x" = -J : —r,^ X and y are said to vary inversely, and this is expressed by a: oc -. y 876. If when one quantity varies, the product of other two varies in the same proportion, the former is said to vary as the two \?i\XQ.x jointly. Let 3/', z\ and y'\ z'\ be two systems of values of y and z corre- sponding to x' and x'\ two values of x, then if x' : x" — y' z' : y" z'\ X is said to vary as y and z jointly, or a: oc yz. Zll. If when one quantity varies, the quotient of a second divided by a third also varies in the same proportion, the former is said to vary directly as the second, and inversely as the third. Let x\ y\ z', and x'\ y'\ z'\ be two systems of values of x, y, and 2, then if x' : x" = -^ : ^, a: is said to vary directly as y, and z z inversely as z, or a; oc ^. THEOREMS. 878. I. A proportional equation may be converted into an absolute equation, by multiplying one side by some constant quantity. Let X oc ?/ be the given equation, then if wi be a constant quantity of a proper value, x = my. 260 ELEMENTS OF ALGEBRA. For if x\ y', and x", y", be any two systems of values of x and y, x' \x" = y' \y" (374), therefore (348) x" y^ — x' y" ; therefore x" = ^/'. K now x'" and y'" be another system of values of x and y^ it may be similarly proved that X = —z-y . And the same relation is proved for any other system ; hence if m x' be taken for — , then generally X = my. Thus, if, as in a former example (373), x be the rate of tra- velling per hour of a stage-coach, and y the rate per day, x' 1 then x' : v' = 1 : 24 ; therefore —- = —- = m, ^ / 24 ' and X = my is X = —y. And this absolute equation is true for any system of values of x and y. Thus, if x = 10 miles per hour, then 10 = ~y, or y = 24 X 10 = 240 miles per day. 379. Cor. Hence if any system of values of x and y be known, the constant quantity m can be found. 380. II. If one variable quantity be equal to the product oi another by a constant quantity, the former variable quantity varies directly as the latter variable quantity. Let X = my, then x cc y. For x' = my', and x" = my" ; therefore x' : x" = my' : my" = "if '. y" ', therefore x oc y. 381. III. The reciprocals of the members of a proportional equation also form a proportional equation. j Let xccy, then -cc-. ^ X y For (378) x = my: therefore - = — . - = n . -, if n = — ^ X m y y m hence (380) -oc-. 382. IV. If one quantity vary as another, any multiples oi parts of them also vary as each other. PROPORTIONAL EQUATIONS. 261 Let X cc y, then mx cc ny. For x' : x" — y' : y", therefore mx' : mx" = ny' : ny", or mx cc ny. And this result is true, whether m and n be integral or frac- tional. 383. Cor. 1. If one quantity vary as another, it will also vary as any multiple or aliquot part of the other. For when m = 1, x oc ny. 384. Cor. 2. Any multiples or aliquot parts of a variable quantity vary as each other. Let X be a variable quantity, then mx oc nx. For mx' : mx" = nx' : nx", or mx oc nx. When m = 1, x oc nx. 385. V. If both sides of a proportional equation be multiplied or divided by a constant or a variable quantity, the equation will still exist. Let X oc y, then if z be either a constant or a variable quantitj, xz oc vz, and -oc-. •^ ' z z For x' : x" = y' : y", and if z', z", be any two values of z, x'z' : x"z" = y'z' : y"z", and — ^ : -^ = - - : ^ (362) ; that is, xz oc yz, and - oc ^. z z When z is constant, z' = z" = z, and the proof is the same. 386. Cor. 1. When one quantity varies directly as another, their ratio is constant. Let X ocy, then - oc 1. For - oc - ocl. y y That is, ^ : ^ = 1 : 1 or 4 = ^- Or thus :-By (378) x = my, X therefore - = m. y o«T. Cor. 2. Any variable quantity, which is a factor of one mciaber of a proportional equation, is proportional to the other member divided by the remaining factors of the former member. 262 ELEMENTS OF ALGEBRA. Let xy oc z, then x oc-. V V Let xyz oc v, then xy oc- and x oc — . Let xyz oc — , then xy oc — and x oc . "^ 10 '' wz wyz 388. Cor. 3. Any variable quantity, which is a divisor in one member of a proportional equation, is proportional to the other part of this member divided by the other member. Let -ocz, then x oc-. X z \ z y For - oc - ; therefore (381) xoc~. X y ^ ^ z Self — oc My, - oc — (385) ; hence xoc—. X X yz uv From these two corollaries, it appears that any factor or divisor of a member of a proportional equation may be made to stand alt ne ; and that one member may be divided by the other, the quotl?nt being a constant, or 1. 389. Co.\ 4. If one quantity varies as other two jointly, either of the latter . •^ries as the first directly, and the remaining one inversely. Let X oc yz, then ^ oc - or 2 oc - (385). 390. VI. If one side of a proportional equation be unity, the other side is equal to some constant quantity. Let xy oc\, then xy = m. For x'y' : x"y" = 1 xl = m : m; and if x'y' = m, so is x:"y". and generally xy — m. The product of any particular system of values of the variables is = the value of the constant m. 391. VII. If one side of a proportional equation be equal to a constant, the other side will be any constant or unity. Let xy cc m, then xy oc n, or xy oc 1. For x'y' : x"y" = m : m = n : n = 1 : 1 ; hence xy oc n, 01 xy oc 1. 392. Cor. If the product of two quantities be constant, they vary inversely as each other. Let xy = m, then xy ocl] therefore (385) x oc -, and v ~^ PROPORTIONAL EQUATIONS. 2C3 393. VIII. If the first of three quantities varies as the second, and the second as the third, the first varies also as the third. Let X cc y, and y gc z, then x oz z. For X = my, and y = nz (378) ; therefore x = mnz, and conse- quently (380) X cc z. So if X oc y, and y cc-, x cc-. 394. Cor. I. If the first of four quantities varies as the second, and the second as tlie third, and the third as the fourth, the first will also vary as the fourth. Let X cc y, and y cc z, and z cc v, then ar oc v. For X cc z (393), and z oc v, therefore x oc v. 395. Cor. 2. If one quantity vary as a second, the second as a third, and so on for any number, each varies as each of the others. 396. IX. If the first of three quantities varies as the second, and the second as the third, the second will vary as the sum or difierence of the first and third. Let X cc y, and y cc z, then x ± z oc y. For X = my, and z = ny ; therefore x + z = (m + n)y ; and hence (380) x ± z cc y. 397. Cor. 1. If one quantity varies as another, their sum or difference will vary as the other. Let X cc y, then x + y cc y. For x = my; therefore x ± y = (m + l)y; and hence x ±ycc y. 398. Cor. 2. If one quantity varies as another, their sum and difference will vary as each other. Let X cc y, then x -^ y cc x — y. For x = my; therefore x -\- y — (7n -\- l)y, and x — y = (m — l)y. But (382) (?« + l)y cc (m — l)y ; therefore r + y oc (?« + l)y oc (m — l)y OCX — y; therefore (394) x -\- y ccx — y. 399. X. If the members of a proportional equation be respec- tively multiplied or divided by those of another, the products or quotients will form a proportional equation. Let X ccy, and z ccv, then xz cc yv. For X = my, and z = nv; therefore xz = mnyv ; and hence xz cc yv. Also, - = —J', therefore (380) - oc ^. „ z n V z V 264 ELEMENTS OF ALGEBRA. 400. Cor. 1. If the first of three quantities varies as the second, and the second as the third, the square of the second will vary as the first and third jointly. Let X Qcy, and y ccz, then xz oc y^. For X ccy, and z ocy, hence xz oc y^. 401. Cor. 2. If a factor in one of the members of a proportional equation vary also as any other quantity, the latter may be sub- stituted in place of it in the former equation. Let X oc yz, and y oc-, then a; oc — . U7/Z HZ For (399) xy oc — - ; therefore x oc — . 402. XL If one quantity vary as another, any power or root of the former will vary as the same power or root of the latter. Let X oc y, then x" oc y", whether n be integral or fractional. For X = my ; therefore ar" = m"y" = ay", if a = rn", which is constant. Hence x" oc y". 403. Cor. If one quantity vary as a second, and the second as the third, the sum or difference of the first and third will vary as the square root of their product. Let x oc y, and y oc z, then x + z oc \fxz. For (396) x + zocy, and (400) y^ cc xz-, therefore y oc ^J xz\ and hence a; + ^ oc V^'^^- 404. XII. If four quantities be always proportional, and only the first of them be constant, the fourth will vary as the product of the second and third. Let vi:x = y:z, then z oc xy. For mz = xy ; hence (380) zocxy. 405. Cor. If the first and second, or the first and third, are constant, the other two will vary as each other ; and if the first and fourth are constant, the second will vary inversely as the third. Let ni: n = x:y, then nx = my ; and hence x ocy when m and n are constant. Let m: x = n'.y, then nx — my ; and hence x ocy when m and n are constant. Let m\x = y -.n, then xy = vin, or xy ocl; hence x oc VI and n are constant. PROPORTIONAL EQUATIONS. 265 406. XIII. If three quantities be so related that when either the second or third is constant, the first varies as the other ; when both the second and third vary, tlie first will vary as their product. Let xozy when z is constant, and X ozz ... y ... , then X ccyz when both x and y vary. For if x' and x'" be two values of x and y', y" the corresponding values oiy when z is constant, then x' : x!" = y' : y". But if z be now supposed to vary, then if z' is the value of z that corresponds to the value x'" of x, and if x", z"., is another system of values of x and z^ x'" : x" = z' : z" ; and compounding the preceding proportion with the latter, x' : x" = y'z' : y"z", therefore x cc yz. This proposition is exemplified by many theorems in geometry regarding the areas of figures. For instance, if s, 6, and A, denote the area, the base, and the height of a rectangle, or parallelogram, or triangle, they have the same relation as or, y, and «, in this proposition. 407. Cor, 1. If one quantity varies as otlier two jointly, the first will vary as either of the other two when the third is constant. Let X oc yz, then when z is constant, and = m, x ac my, or x ocy; and when y is constant, and = n, x ocnz, or x oc z. The variations of y and z are supposed to be independent of each other in this theorem and corollary, as b and h are in the case of a rectangle or triangle ; but when y and z are not independent ot each other, and vary directly as each other, though the theorem then exists, this corollary cannot exist, but the following one is a consequence of the relation between y and z. 408. Cor. 2. If one quantity varies as other two jointly, and if the two latter vary directly as each other, the first will vary as the square of either of the other two. Let X oc yz, and y oc z, then x oc y"^ or x oc ^. For (401) substituting y for z, x oc /, or substituting z for y, X oc z^. This corollary is exemplified by geometrical theorems respecting the areas of similar figures. 40y. Cor. 3. If ono qr«>ntity varies as each of any number of quantities when the rest are constant, when aU are variable, it varies as their product. EQUIDIFFEEENT PROGRESSION.* 410. If the diflPerence between any term of a series of quantities that increases or decreases, and the following term, be always the same, it is called an equidifferent series or progression ; in the former ease it is an increasing, and in the latter a decreasing, series. The constant difference between any two successive terms is called the common difference. In the following, a is = the first term, d = the common difference, n = the number of terms, z = the last term, and s = the sum of the series. Thus, 1, 3, 5, 7, &c. is an equidifferent increasing series, the common difference of which is = 2. So is the series a, a -\- d, a -\- 2d, ... the common difference of which is d. The series 20, 17, 14, 11, ... is a decreasing series, whose common difference is 3. So is a -f Sd, a -{- 6d, a -\- 4:d, ... the common difference being 2d An increasing series may begin with negative terms, as — 11, — 8, — 5 — 2, 1, 4, 7, 10, ... and a decreasing one may terminate with negative terms, as 6, 4, 2, 0, — 2, — 4, — 6 ... THEOBEMS. 411. I. The last term of an equidifferent series is equ&l to the sum or difference of the first term and the product of the common difference by a number which is one less than the number of terms, according as the series is increasing or decreasing. Let a, a -\- d, a -\- 2d, a -\- 3d, ... he the series; then if the number of terms be n, the last term is evidently a + the product of c^ by w — 1 ; and hence, z = a -\- (n — l)c?. For instance, when the last term is the third, n = S, and n — 1 = 2, and z = a -\- 2d. When the last term is the fourth, n = i, and w — 1 = 3, and z = a -\- 3d. Let a, a — d, a — 2d, a — 3d, ... he a decreasing series, and it is similarly evident that z = a — (n — l)d. * By French algebraists, this series is called * progression par ^quidiff^rence,' or simply ' par difference.' Equirational series, treated of in the next section, is called by them ' progression par quotients ^geaux,' or merely ' par quotient." The term equidifferent is adopted by some English authors, and T have adopted the new term equirational, because the common terms, arithmetical and geometrical, are not appropriate. EQUIDIFFERENT PROGRESSION. 267 hese two results may be expressed thus, 2 z= a + (« - \)d ... [1], le upper sign being taken for an increasing, and the lower for a Bcreasing, series. 412. Cor. Any term of the series may be found from this )rmula. 413. II. The sum of the first and last terms of an equidifferent jries is equal to the sum of any other two terms that are equally istant from the first and last. Let the series be a -{- d, a -\- 2d, a + (n — 2)d, a + (n — l)c?; nd let the series be taken also in a reverse order, -f (n — l)d, a -{- (n — 2)d, a + 2d, a -j- d, a, hen the first term of the former, with that of the latter, is = 2a -\- (n — l)d, nd the sum of the second terms, as also of the third and fourth, s evidently equal to the same quantity ; and so on to their last erms. But the first terms of the two series are the first and last »f the first series ; their second terms are the second and last but )ne of the first series, and so on ; hence the theorem is evident. Vhen the series is decreasing, the proof is similar. 414. Cor. 1. When the series consists of four terms, the sum of he extremes is equal to that of the means. 415. Cor. 2. When the series consists of three terms, the sum )f the extremes is equal to twice the mean. 416. Cor. 3. When the number of terms is uneven, the sum of the first and last, or of any two equally distant from them, is equal to twice the middle term. For in this case the number denoting the place of the middle term is — ^ — ; and hence this term is == a -f ( — l)d = a + '—— — d, and its double is 2a + (« — l)d, which is equal 2 to a -\- z. 417. ni. The sum of the terms of an equidifferent series is equal to the sum of the first and last terms multiplied by half the number of terms. For, arrauging the series both in a direct and in a reverse order, as in the preceding theorem, the sum of the first terms of the two series bwi-mes (413) = 2a + (n-l)d=a -\-zhy [1], ■ • the sum of each succeeding pair of terms. But the 268 ELEMENTS OF ALGEBRA. number of these pairs of terms is = n ; and hence their sum is a + z, taken n times, or = n{a + z). This, however, is the sum of the two series, or it is twice the sum of the given series ; hence, s = -^{a + zy 418. When any three of the five quantities a, z, d, n, and s, are given, the other two may be found from the two equations, [1] ... z = a±(n-l)d, s = ^(a-^z) ... [2], which are then two equations containing only two unknown quan- tities, the values of which may be found by the preceding methods for the solution of two determinate equations. The equations will be simple in every case except two — namely, when a and n, or z and n, are the unknown quantities. In these two cases the equations will be quadratics, for a and n, or z and n, are in both equations, and their product is in the second. The cause of the double values of the required terms in these two cases will be easily seen from this circumstance, that in both these cases s is given, and it has the same value for two different series of terms when the progression is carried out to negative terms. For example, the series 11, 9, 7, 5, 3, 1, — 1, — 3, — 5, ... gives for the first three terms s = 27, and for nine terms it also gives s = 27 ; therefore n has two values, 3 or 9, and z has also two values, 7 or — 5. The number of cases, arranged according to the data, is ten, or the combinations of 5 quantities taken 3 and 3 (456) ; the cases are : — Given. Sought. Given. Sought. 1... a, d, n, ... 2 and s 6... d, n, z, ... a and s 2... a, d, z, ... n ... s 7... d, n, s, ... a ... z 3... a, w, 2, ... d ... s 8... n, «, s, ... a ... d 4... a, n, s, ... d ... z 9... a, c?, s, ... n ... z 5... a, 2, s, ... d ... n 10... d, 2, s, ... a ... n In all these cases, except the two last, one of the unknown quantities is directly found from one of the equations [1] or [2], and the remaining one is then directly found from the other ol these two equations. In the 9th case, n is found from [3], and then z from [2] ; and in the 10th case, n is found from [4], and then a from [2]. The last two cases are those in which the imknown quantities have two values, which may be found from [1] and [2], independently of [3] and [4]. In the foi .I'.ulai that have double signs, the upper one is to be taken in the case •^"^ -^n increasing, and the lower in that of u decreasijig, series; solutions axe contained in the following table : — EQUIDIFFERENT PROGRESSION. 269 CASES. GIVEN. SOUGHT. FORMULAS. 1. 2. d,n,z, n, z, s, a, z -(n- l)d. 2s z. n 3. d, z, s, ^±\^{{2z + dr-Ms). 4. d,n, s, ... s (n- l)d n 2 5. 6. a,d, n, a, n, s, z, a + (n- l)d. 2s a. n 7. a, d, s, -l±l^/{(2a-dy + 8ds]. 8. 9. 10. 11. 12. d, n, s, ... s (n-l)d n"^ 2 • a, n, z, a, n, s, «, z, s, n, z, s, d, z — a n-1' 2(s - an) n{n- 1)- z^ - a' (z + a)(z - a) 2s-a-z 2s-a-z ■' 2{nz - s) n{ji -ly 13. a, d, z, n, ^-a ^ ' d ' 14. a, z, s, 2s a + z 15. a, d, s, ^-^±i^^^^^-'^^'+^*^- 16. d, z, s, ^ + l±i^^^^^ + ^'-^^^ 17. a, n, z, s, P+z). i«. a, d, z, ... a + z , z"" -a' 2 ^ 2d ' a^d^n, I ... ^{2a + (n - l)d}. d,^^,., p^ - (n - 1^. 270 ELEMENTS OF ALGEBRA. EXAMPLES. 1. Given the first term of an increasing eqiiidifferent series * the common difference 2, and the number of terms 21 : to find tl last term and the sum of the series. Here a = 3, d = 2, n = 21, therefore z = a + (n - 1)J = 3 + (21 - 1)2 = 3 + 20 X 2 = 3 + 40 = 4: ■n 21 ^1 and s = -(a + z) = —(3 + 43) = ^ X 46 = 21 X 23 = 483. 2. Given the first and last terms of a decreasing equidifferei series 100 and 1 respectively, and the common difference 3 : fin the number of terms and the sum of the series. Here a = 100, z = 1, and d = 3] hence by [1], z — a a — z 100 — 1 d d 3 .-. « = 33 + 1 = 34, = 33; n S4 and s = -(a -\-z) = -^(100 + 1) = 17 X 101 = 1717 3. The first term of a decreasing equidifferent progression is 1 the common difference 2, and the sum of the series 27 : require the number of terms and the last term. This is the 9th case, and here a = 11, d = 2, and s = 2 Substituting the value of z in [1] and that of s in [2] thus, 5 = |{a + a - (« - 1)4 = ^{2a - (n - 1)^ ; .-. 27 = ^{2 X 11 - (n - 1)2} = ^(22 _ 2n 4- 2) = |(24-2«) = 1271 -n^; and from this quadratic is found n = +.3 + 6 = 9 or 3; and hence 2= 11 — 8X2 or 11— 2x2 = — 5 or 7. For n = 3, and z = 7, the series is 11, 9, 7, for which s = 2' and for n = 9, and z = — 5, the series is 11, 9, 7, 5, 3, 1, — -3,-5. 4. The first term of an equidifferent series is 1, the last 2, ar the number of terms 10 : find the common difference and the su of the series. Here a = 1, 2 = 2, w = 10, and by [1], (n — l)d = z — a, or 9c? = 2 — 1 = 1, and d = -; and by [2] 5 = J(a + 2) = ^(1 + 2) = 5 X 3 = 15 EQUIDIFFERENT PROGRESSION. 271 EXERCISES. 1. Find the sum of the natural series of numbers 1, 2, 3, 4, ... arried to 1000 terms, = 500500. 2. Eequired the last term and the sum of the series of odd umbers 1, 3, 5, 7, ... continued to 101 terms, Last term = 201, and sum = 10201. 3. Find the nth term and the sum of n terms of the natural eries of numbers 1, 2, 3, 4, ..., The nth term = n, and sum of w terms = ^n(n + 1). 4. Find the nth term and the sum of n terms of the series of dd numbers, 1, 3, 5, 7, ..., The nth term = 2n — 1, and the sum of n terms = n^. 5. Find the last term and the sum of 50 terms of the series , 4, 6, 8, ..., . . . Last term = 100, and sum = 2550. 6. The first and last terms of an equidifferent progression are and 29, and the common difference is 3 : find the number of ;erms and the sum of the series, Number of terms = 10, and sum = 155. 7. The first and last terms of a decreasing equidifferent series are 10 and 6, and the number of terms 9 : required the common difference and the sum of the series. Common difference = ^, and sum = 72. 8. The first term of a decreasing equidifferent series is 10, the number of terms 10, and the sum of the series 85 : required the last term and the common difference, Last term = 7, and common difference = ^. 9. The last term of an increasing equidifferent series is 52, the common difference 5, and the sum of the series 297 : required the first term and the number of terms, First term = 2, and number of terms = 11. 10. How many times does the hammer of a common clock strike in a week ? = 1092. 11. A carter has to gravel an avenue with 11 cart-loads of gravel, to be laid down 6 yards distant from each other ; the first load to be laid down at the end of the avenue next to a gravel pit, and at the distance of 80 yards from the pit: required the number of yards that he must travel over, supposing that he sets out from the pit, and returns to it after laying down the last load, = 2420 yards. 12. Insert four equidifferent means between the terms 5 and 7 2 4 1 3 of an equidifferent series, . . . . = 5-, 5-, 6-, and 6-. 5 5 5 5 272 ELEMENTS OF ALGEBRA. 13. Find the expression for the common difference, ; order to insert n equidifferent means between the numbers and h, Common difference = d = n + 14. Two travellers (A and B) set out from the same place the same time ; A travels at the uniform rate of 3 miles an hot but B's rate of travelling is 4 miles the first hour, S^ the secon 3 the third, and so on in the same series : in how many hours w A overtake B ? Time = 5 houi EQUIEATIONAL PROGRESSION. 419. An equirational progression is a series, of which the ten increase or decrease in a constant ratio. The ratio of any term to the preceding term is called t\ common ratio. Thus, 2, 4, 8, 16, ... is an increasing equirational series, i common ratio of which is 2 ; and 81, 27, 9, ... is a decreasing or the common ratio being |. Likewise, a, ar, ai^, ar'^, ... is a simil series, having the common ratio r ; and the series is increasing decreasing according as r is greater or less than 1 . THEOREMS. 420. I. The terms of an equirational progression are in co tinned proportion. Let a be the first term, and r the ratio, then the series is a, i ar"^, ar^, ... and ar ar^ ar^ . — = = — 5", &c. or = r ; a ar ar'- hence (319) a:ar = ar: ar"^ = ar^ : ar^^ &c. 421. IT. The last term of an equirational progression is equ to the product of the first term by that power of the commc ratio whose exponent is one less than the number of terms. For, the exponent of r in any term is evidently one less th; the number expressing the place of the term ; thus, in the thi term its exponent is 2, in the fourth it is 3, and so on, and in tJ nth. it is n — 1 ; hence if z be the last term, ^^^=— z = ar»-' ... [1]. EQUIRATIONAL PROGRESSION. 273 422. III. The product of the first and last terms is equal to ;hat of any two equally distant from the first and last. For, taking the third term, and the last but two, their product s = ar"^ X or""^ = a^/-""' ; and that of the first and last is a X ay""^ = c?r^~^. 423. Cor. When the number of terms is uneven, the product )f the first and last is equal to the sqxiare of the middle term ; md any term is a mean proportional between the preceding and ucceeding terms. 424. The sum of the terms of an equirational series is found by ubtracting the first term from the product of the last term and he common ratio, and dividing the difierence by one less than ;he common ratio. Let the sum of the series a -\- ar -{• at^ -\- ... 4" o^"~^ + ar""' = s, then ar -{■ ar^ -\- ... + ar""^ + ar""^ + ar"^ = rs, by multiplying the preceding equation by r. Subtracting the former series from the latter, rs — s ■= ar^ — a, or s(r — 1) = a(r" — 1) or = rz — a for ar" = r . ar"~* = rz. ^^ air"" — 1) rz — a Hence s = — — ^, or s = — ... [2]. r— 1 r — 1 When any three of the quantities a, z, r, and s, are given, the fourth may be found from equation [2]. There may be ten cases formed, in which three of the five quantities a, z, r, n, and s, are given, to find the other two from [1] and [2], as in the case of equidifferent series. Some of these cases, however, involve the extraction of high roots, and the use of logarithms, and higher equations than have been treated of in the preceding pages ; the formulas for the solution of these cases are all contained in the following table : — 274 ELEMENTS OF ALGEBRA. CASKS. GIVEN, SOUGHT. FORMULAS. 1. 2. 3. 4. r, n, z, r, z, s, n, z, s, r, n, s, a, Z rz-(r- l)s. a(s - ay^ = s (s - zy-K 5. 6. 7. 8. a, r, n, a, n, s, a, r, s, r, n, s, ^J z(s — zy-^ = a(s — ay-\ s — a -(hy 9. 10. 11. 12. a, n, z, a, n, s, a, z, s, n, z, s, r, ar" — sr + s — a = 0. s — a s — z (s - 2)r~ - sr»-' + s = 0. 13. 14. 15. 16. a, r, z, a, z, s, a, r, s, r, z, s, n, ^ , log. z - log. a + log. r ' ^ log. 2 -log. a log. (s - a) — log. {s - z)' log. {s(r _ 1) + a} — log. a log. r , , log. z - log-I^C^ - s) + s} + log. r 17. 18. 19. 20. a, n, z, a, r, n, a, r, z, r, n, z, s, rz — a r - 1' EQUIRATIONAL PROGRESSION. 275 EXAMPLES. 1. Find the ninth term of the series 1, 3, 9, 27, ... and the sum )f the first nine terms. Here a = 1, i- = 3, and n = 9 ; lence z = ar'^-^ = 1 X 3®"* = 3« = 6561, rz — a 3x6561-1 19683-1 19682 ^^., md s = — = :; = = — - — = 9841. r—\ 3—1 2 2 2. The first term of a decreasing equirational series is 1, the jommon ratio |-, and tiie number of terms 5 : required the last ;erm and the sum of the series. Here a = 1, r = ^, and w = 5 ; .=a.n-=ix(iy'=Qy=i, 1 j_ 1 rz - a a - rz 3 ^ 81 243 242 3 ^^^ = V-Zn^^=inV= 1 "—2— = 243^2 3 3 121 _ 40 81 ~ 81" 3. Insert three equirational means between the numbers 3 and 48. This question is the same as if the first and last term, and the number of terms of an equirational series, were given, to find the common ratio. By [1], 2 = ar"-» ; .-.r^-^rr-, and in this example a = 3, 2 = 48, and n = 5 ; z 48 hence r*"^ = - or r* = — - = 16, and r^= V 16 = 4, and consequently a 3 r = V4 = 2. The means are therefore 6, 12, and 24. EXERCISES. 1. The first term of an equirational series is 3, the common ratio 2, and the number of terms 10 : required the last term and the sum of the series, = 1536 and 3069. 2. A person walks 4 miles the first hour, 2 the second, 1 the third, and so on, in equirational progression, and continues his 276 ELEMENTS OF ALGEBRA. journey for 10 hours : how far does he travel the last hour, and what distance does he travel altogether ? 1 127 Last hour = z = — — , and entire distance = s = 7—-. 128 i-io 3. Find the sum of the series 1, 3, 9, 27, ... continued to 12 terms, Sum = 265720. 4. In order to insert seven equirational means hetween the numbers 16 and — , what must be the common ratio ? •^" Common ratio = ^, 5. The first term of an equirational series is 3, the last terra 192, and the number of terms 7 : find the common ratio and the sum of the series, . . . Common ratio = /• = 2, s = 381, 6. Insert 4 equirational means between 5 and 1 60, and find tht sum of the 6 terms, Equirational means = 10, 20, 40, and 80, and s = 315 425. In finding the sum of a decreasing equirational series carried out to an indefinite extent, or, in other words, an infinite decreasing equirational series, the last term may be considered as vanishing, or becoming = 0. Hence, since z = the sum of such a series is — a a s = -, or s =- ; r — 1 1 — r or, since in this case r is a proper fraction, if it be represented b} -, where p ^q, then a aq " — , or '' — — - — 9 EXAMPLES. 1. Find the sum of the infinite series 1, ^, -^^ ... P 1 Here a = \, and - = -, or = 1,0 = 2; q 2 _ aq q-p 2-1 = 2. 2. Find the sum of the series 1, -, -, ... carried to infinity. Here a = \, p = 2, q = ^ ; q-p 3-2 t\ EQUIRATIONAL PROGRESSION. 277 3. Find the sura of the infinite series 1, — 7, 77:, — ttt* ••• 4 lb b4: Here a = 1, r = — -, or- = — -, p = — 1, q = i ; _ aq _ 1 X 4 _ 4 q — p 4 + 1 5* EXERCISES. Ill 3 1. Find the sum of the series 1, -,-,—,... to infinity, = -. o 9 27 2 2 1, - 3, -,--,... to infinity, =-. 3. If a and x he each greater than unity, find the sum of the series a, 7, — j ... to mfimty, X ax^ ax — 1 426. Eecurring decimals — that is, repeaters and circulates — may 3e represented in the form of infinite decreasing equirational series, and their values found in the form of vulgar fractions hy umming the series ; whence the rules for converting them into vulgar fractions can be derived. EXAMPLES. 1 . Find the value of the repeater -4. 4 4 •4 = -444 ... = — 4- — - + ... 10 ^ 100 ^ Here a = — , and r = — , a 10 4 10 4 and .... = ^^-_:=--^ = _X- = -. 10 2. If n be the repeating digit, and the value of the fraction 5, *^"^ ^ = io+Too + iooo + - = -+ — + — + • 10 ^ 10^ ^ 10^ ^ '■• ' n , a To n 10 n hence .-.5 = , = ^ = _ x - = -. 10 278 ELEMENTS OF ALGEBRA. From this result the common rule for reducing a pure repeater to a vulgar fraction is derived. 3. Find the value of the circulate •24. 24 24 24 •24 = -242424 .,.= — + -^- + 100 ' 10000 ' 1000000 ' 24 1 Herea = — ,andr = — ; 24 a 100 24 100 24 hence .•.s = ^--^ = —-j- = —x— = -. 100 4. Find the value of the mixed repeater •43. •43 = 4 + -03, and -03 = 4 + j^ + jAg + ... Hence for the value of •OS. 3 a 100 _ 3 10 3 1 - r ~ J^ ~ 100 9 ~ 90' 10 and therefore . _ 4 _3^ _ 9x4 + 3 _ 10 X 4 + 3 - 4 _ 43-4 _ 39 10 "^ 90 ~ 90 ~ 90 ~ 90 "" 90' Or thus : let -43 or •43333 ... = s, then 4-333 ... = 10s, and 43^333 ... = 100s, subtracting 43 — 4 = 90s, 43-4 39 90 90 5. Find the value of the mixed circulate •435. ■435 = -4 + -085, and -035 = ^ + ^^ + ... and the value of _35_ • . . a 1000 35 ^ 100 _ 35 •035 IS s = ^— ^ = -—T = 1000 ^ "99" - 990^ 100 HARMONIC PROGRESSION. 279 incl therefore •435 = 4 10 + 35 990 4 X 99 + 35 4 X 100 + 35 - 990 990 435 - 4 431 ~ 990 ~ 990 ' -4 )rthus: let •435 or ^4353535 ... = s, ;hen 4-353535 ... = 10s, md 435-353535 ... = 1000s, mbtracting 435 - 4 = 990s, or 435 - 4 431 * ~ 990 ~ 990' EXERCISES. Find, by summing the series, the value of the pure repeater '5 ; the pure circulate -36 ; the mixed repeater •46 ; and of the mixed • • 5 4 7 13297 circulate -53241. The values are respectively =-, --, —5 and ^t^jk- HARMONIC PROGRESSION. DEFINITIONS. 427. Three quantities are said to he in harmonic progression when the first is to the third as the diiference between the first and second is to the difference between the second and third ; and the three quantities are called harmonic progressionals. Thus, if a, h, c, be in harmonic progression, then a : c ■= a — h:h — c. 428. When three quantities are in harmonic progression, the second is said to be an harmonic mean between the other two ; and the third is called a third harmonic progressional to the first and second. 429. Any series of quantities are said to be in harmonic pro- gression, or to be harmonic progressionals, when every consecutive three are in harmonic progression. Thus, a, b, c, d, e, ... are in harmonic progression if a, b, c, ' .!! 6, c, d, and c, d, e, ... be in harmonic progression. 280 ELEMENTS OF ALGEBRA. PROBLEMS J^D THEOREMS. 430. I. To find a harmonic mean between two quantities. Let a, 6, be the two quantities, and x the harmonic mean, thei (427) a:b = a — x:x — b, or a(x — b) = b(a — x) ; 2ab hence therefore x = — —7. a -{- b 431. II. To find a third harmonic progressional to any tw< quantities. Let a, b, be the quantities, and x the third harmonic progres sional, then (427) a:x = a — b:b — X, OT a(b — x) = x(a — b) ; and from this equation, x = 7. 482. ni. Hence if any two of three harmonic progressional be given, the third can be found. 433. rV. If three quantities be in harmonic progression, thei reciprocals are in equidifferent progression. / Let a, b, and c, be in harmonic progression, then -, p and - are in equidifierent progression. For a:c = a — b:b — c, OT ab — ac = ac — be, ,• .^. V ^1111 111 . .,.«. or dividmg by a 6 c, y = r > or -, y, - are in equidifferen c ' b b a c b a progression (410); in the same manner it may be proved tha if b, c, d he in harmonic progression, -^ = -7 ; an therefore, -j, -, 7- are in equidifierent progression. The same metho a c o of proof may be extended to any number of terms. Cor. By this article any number of harmonic means may b inserted between two extremes, by inserting the same number t equidifierent means between the reciprocals of the extremes, am their reciprocals will be the hannonic means required. 434. V. To insert two harmonic means between two numbers. HARMONIC PROGRESSION. 281 Let a, h, be the two numbers, and x, y, two harmonic means )etween them, hen a:y = a — x:x — y, :nd x:b = X — y :y — 6, rom which are found the two equations a(x — y) = (a — x^y, and x(y - h) = h(x — y), r ax -^ xy — 2ay = 0, and by + xy — 2bx = 0. ^ value of y being found from each of these equations, and these alues being equated, give a X — X 2bx b +x' a 2b b + 2a °'"2a- X x' > ab + ax = x = :4a6 - Sab 2bx a 4- 26' md substituting this value of x in either of the values of y, the alue of y is found, or _ Sab ^ ~ 2a + b' 435. In a similar manner, any number of harmonic means may )e inserted between two numbers, for as many quadratic equations ;o determine them could be formed. EXAMPLES. 1. Find a harmotic mean between 1 and f. Here a = 1, 6 = §, and , ..^.. 2ob 2 X 1 X § 4 3 4 ,, ^ by (430), ^=^zrb = -TTF ^3^5 = 5 '^'' ^""^^^^^ ican. The result is verified by the proportion 1:-=1 : ^ 3 5 5 3' 3 2 12 3 2 1^^3 = 3 = 5=15' °^15 = l5'°^''-' = ^ = 2- q K K Or it is verified thus by article (433) = 1 • that is 2 4 4 ' is an equidifierent mean between 1 and -. 282 ELEMENTS OF ALGEBRA. 2. Find a third harmonic progressional to 6 and 4. Here a = 6, b = i, and v r.o1^ «^ 6 X 4 24 „ by (131). •••-=2;rri = i^n = T = 8- 3. Insert two harmonic means between 1 and f . By (434), a = 1, 6 = f, and Sab 3X 1 X ^ „ 3 6 2 X - — - a -{■2b 14-2X| 7 7' Sab 3 X 1 X ^ ^ 3 3 and V = r = ~ 2 x ~ = -. ^ 2a + 6 2 + f 8 4 CO o The harmonic series therefore is 1, -, V> and ~. '7 4' 3 EXERCISES. 1. Pind a harmonic mean between 12 and 6, . 2. Find a third harmonic progressional to 3 and 2, . = ^. 3. Find two harmonic means between 84 and 5Q, = 72 and 63 4. Theorem I. If four quantities be in harnonic progression, the product of the first and second is to that of the third and fourth, as the difference between the two former to the difference between the two latter. 5. Theorem II. In any harmonic series, the product of the first two terms is to that of the last two as th) difference betweer the two former to that between the two latter. PROPERTIES OF NUMBERS. Several properties of numbers were given in a former section onneeted with the least common multiple and greatest common Lieasure of quantities ; in this section a few more properties are dded, which are interesting and useful both in a practical and heoretical point of view. THEOREMS. 436. T. If X and y represent the two digits respectively in the )lace of tens and of units, composing a number expressed by the lotation of the common or decimal system of numeration, the mmber would be expressed algebraically ; thus — \Qx +y. [f the number consisted of three digits, x, y, z, the last being that n the place of units, it would be expressed thus — lOOar + lOy + 2, or lO^x + lOy + z. For any digit has only its real value in the place of units, but when it is in the place of tens, its nominal value is 10 times greater, 10 being the base of the system ; in the place of hundreds, it is 100 times greater ; and so on ; its nominal value increasing 10 times for every place it is removed to the left. Thus, 6542 = 6000 + 500 + 40 + 2 = (6 X 1000) + (5 X 100) + (4 X 10) + 2. If a = 2, 6 = 4, c = 5, £? = 6, the number would be expressed by 1000c/ + 100c + 10& + a, or a + 106 + lO^c + Wd. 437. II. If a number be divided by 9, the remainder is the same as that resulting from dividing the sum of its digits by 9. Let the number be a + 10& + lOOc + lOOOrf -\- ... — N; then dividing this number N by 9, the remainders are a, h, c, d, ... ; hence the remainder, when N is divided by 9, is the remainder when a + 6 + c + cf + ... is divided by 9". 438. Cor. Hence when the sum of the digits of a number is divisible by 9, so is the number itself. The same property belongs to the number 3. 439. III. The remainder, on dividing a number by 6, is the same as that resulting from dividing by 6 the sum of the digit in the place of units added to 4 times the sum of its other digits. I 284 ELEMENTS OF ALGEBRA. For the number a + 10b + 100c + lOOOc? + ... being divided by 6, gives for a remainder a + 46 + 4c + 4c? + ... = a + 4(6 + c + c? + ...), when this latter quantity therefore is divisible by 6, so is the number itself. 440, IV. If two numbers be divided by a third, and the pro- duct of their remainders be also divided by the same number, the remainder arising from this last division is the same as that resulting from the division of the product of the two given numbers by the same number. Let the given numbers be JSf and N', and D the divisor, and let q and q' be the quotients, and r and / the remainders ; then N = qD -^ r, and N' = q' D -{- r^ ; hence NN' — qq'D^ + {qr' + q'r)D + rr' ; and since the first two terms of the second member are divisible by D, the remainder will arise merely from the division of ?t' by 2), which is therefore the remainder when the product NN' is divided by D. This property is useful in testing the accuracy of the product of two numbers. Any number may be taken as the divisor D ; but as the remainder, when a number is divided by 9, is so easily obtained (437), being the same as that arising from dividing the sum of the digits by 9, or, as it is called, by throwing out the nines, this number is therefore commonly adopted for proving the accuracy of multiplication. 441, V, Let r be the base of any system of numeration, and a, b, c, ... the digits of any number taken in order, beginning at the place of units, then the number is expressed by a + 6r + cr^ + (fr' + „. If r = 10, the number would be expressed in the common or decimal system. If ?• = 8, the system is the octary, and the value of a figure would increase 8 times for every place it is removed from the units' place, and the above number would be a + 86 + 8^0 + 8^d + ..., or a + 86 + 64c + 512c? + ..., but in this scale 8 would be denoted by 10, the 1 being in the place of eights, 442, VI. In any system of numeration, the difference between a number and the sum of its digits is divisible by a number one less than the base of the scale. PROPEKTIES OF NUMBERS. 285 Let r be the base, and the number = a + hr + cr^ — dr^ + ... then a-{-b + c + d-{-... s the sum of the digits, and the difference between these [uantities = t(r - 1) + c(r2 - 1) + d(r' - 1) + ..., diich is evidently divisible by r — 1. Thus, in the decimal system r = 10, and if 6432 be a number, hen 6432 -(6 + 4 + 3 + 2) = 6432 — 15 = 6417, which is livisible by 9 = r — 1. The place of units is reckoned the Jirst place. 443. Vn. If the excess of the sum of the digits in the odd )laces of a number above the sum of those in the even places be livisible by a number a unit greater than the base, the given lumber itself is divisible by the same number. Let the number be a + 6r + cr + c^r^ + ... = N, ,nd let a -\- c -\- e + ... = P, ,nd b + d+f-\-... = Q, hen iV = P - Q + &(r + 1) + c(r= - 1) + d(r^ + 1) + ••., nd the terms after P — Q in the second member are evidently livisible by (r + 1) ; and therefore if P— Q be divisible by r + 1, he number N is so also. Thus, on the decimal scale, r + 1 = 11, and if the number is 16958476, P = 21, Q = 32,' and P - Q = 21 - 32 = - 11, vhich is divisible by 11, and the number itself, therefore, is also livisible by 11. 444. Cor. Hence N — P + Q is divisible by r + 1. EXAMPLES. 1. In what system has the number 32 the same value as 14 in he decimal system ? Let X = the base of the numerical system, hen 32 = 3a; + 2 = 10 + 4 = 14, 3x = U-2 = 12, .• . X = 4. Hence the system is the quaternary. In this system 32 - 3 X 4 + 2 = 12 + 2 = 14. 2. In what system is the number 2310 equal to 45 times the base ? 286 ELEMENTS OF ALGEBRA. Let X — the base, then 2310 = 2x^ -\- Zx^ -{- x + = 4ox, or 23^ + Zx =4.^-1 = 44. From this equation is found ar = 4. 3. In the binary system r = 2, and the number of digits is two- namely, and 1 — and the greatest number consisting of 4 places i 1111, which is = lx2^ + lx22 + lx2 + l = lx8 + lX 4-1 x2 + l = 8+4 + 2 + l = 15. Hence all numbers, fror 1 to 15 inclusive, must be capable of being expressed in thi system by 1 repeated not more than 4 times. But all the possibl values of 1 in these 4 places are 1, 2, 4, or 8 ; hence all numbers from 1 to 15 inclusive, are made up of some of the combination of 1, 2, 4, and 8. Hence four weights of 1, 2, 4, and 8 ounces, will be sufficient t make up any number of ounces from 1 to 15 inclusive. This curious proposition may be easily extended. EXERCISES. 1. If the first digit in any number be divisible by 2, so also i the number. 2. Any number having 5 or in its place of imits, is divisibl by 5. 3. If a number be divided by 3, the remainder will be the sam as if the sum of its digits were divided by 3. 4. A number is divisible by 4 when the number composed of it first two digits is so ; or when the sum of the first digit and twic the second is so. 5. A number is divisible by 8 when the number composed c its first three digits is so ; or when the sum of the first digit, twic the second, and four times the third, is so. 6. If the sum of the odd digits and r times the even digits of number (r being the base) be divisible by r + 1, or r — 1, o r^ — 1, the number itself is so. 7. If the sum of the even digits of a number, expressed in th decimal system, be added to the number itself, and the sum c the odd digits be subtracted from it, the resulting number will b divisible by 11. 8. In what system is the value of the number 231 equal to tha of QQ in the decimal system ? 9. The square of every even number is divisible by 4 square of every odd number diminished by 1 is also divis PERMUTATIONS. 287 10. The difference of the squares of two odd numbers, as well iS of two even numbers, is divisible by 8. 11, The product of any two odd or two even numbers is less ;han the square of half their sum by the square of half their iifference. PEEMUTATIONS. 445. The various orders in which objects are capable of being arranged in succession are called their permutations. Thus, the letters a, 6, c, when taken in pairs, will form six [)ermutations ah, ha, ac, ca, he, cb ; and when the three are taken, they will also form six — namely, ahc, ach, bac, hca, cah, cba. THEOREMS. 446. I. The number of permutations that can be formed with n letters, taken two and two, is = n(n — 1). Let a, 6, c, ... be n letters, then a may be placed before each of the remaining letters, which are n — 1 in number, and thus n — 1 permutations are formed, in which a stands first. So h may be placed before each of the other letters ; and thus n — \ permuta- tions are formed, in which h stands first. The same may be said of all the letters which are n in number. Hence there are w — 1 permutations, repeated n times, or altogether, there are n(n — 1) permutations. 447. II. The number of permutations that can be formed with n letters, taken three and three, is n{n — 1) X (ra — 2). For any one of the permutations, taken two and two, can be placed before each of the remaining letters, which are k — 2 in number ; and thus n — 2 permutations, taken three and three, are formed. Thus, if a, h, c, d, e, ... be the n letters, then the permutation ah may be placed before each of the letters c, d, e, ... which are n — 2 in number ; and thus are formed n — 2 permuta- tions, taken three and three — namely, ahc, abd, abe, ... So the permutation ac being placed before each of the letters b, d, e, ... will form n — 2 permutations, taken three and three. The same may be proved of all the other permutations, taken two and two ; and thus all the possible permutations, taken tliree and three, will be formed. It might be said that the first permutation, taken alone — namely, ah — may not only be placed before c forming the permut>'tion ahc, but that it might also be placed after c, so as 288 ELEMENTS OF ALGEBRA. to form another permutation cab ; but this last permutation : otherwise formed — namely, by placing the permutation ca befoi h ; so that to form all the permutations, taken three and three, : is merely necessary to place each of the permutations, taken tvi and two, before each of the remaining n — 2 letters. Hence th whole number of permutations, taken three and three, will t equal to the number of them taken two and two repeated n — times, or n{n — 1) X (n — 2). 448. III. It may be similarly proved, that the number of pei mutations of n letters, taken four and four together, is n(ji — 1 (n — 2)(n — 3), by placing each of the permutations, taken thre and three, before each of the remaining letters, which are n — in number. By proceeding in this manner, the following general propositio is arrived at by induction ; namely — 449. IV. The number of permutations of n letters, taken r an r together, is = 7i(n — l)(re ~ 2) ... (ra — r + 1) ; and hence — The number of permutations of n letters, when all of them ar taken, is n{n — l)(n — 2) ... 3 X 2 X 1, or 1 X 2 X 3 X .. (n — 2)(?i — l)n. For in tliis case r = n, and k — r + l = n — n + l = l;th preceding factor isn — n-|-2=:2; the one preceding this last i = n — w + 3 = 3; and so on. 450. V. The number of permutations of n letters, taken n — and n — 1 together, is the same as when they are all taken ; o = 2 X 3 X 4 X ... (n — 2)(» — 1>. For in this case r = n — 1, and r — 1 = w — 2 ; and bene n — r -\-l = n — w + 2 = 2, n — 7- + 2 = n — n + 3 = 3: and so or EXAMPLES. 1. In how many different orders can five persons sit on a form The number of permutations of 5 objects taken altogether i = 1X2X. ..(«-!> = 1X2X3X4X5 = 120. 2. How many signals may be made with 4 flags ? Number when taken singly is 2 by 2 is «(« — 1) = 4 X 3, 3 by 3 is n{n — V)(n - 2) = 4 X 3 X 2, . 4 by 4 is n(n — ])(?i — 2)(n -3) = 4x3x2x1, . the total number is COMBINATIONS. 289 EXERCISES. 1. How many permutations of four notes each, sounded suc- cessively, can be formed with the seven musical notes of one octave? = 840. 2. In how many different ways can the seven prismatic colours be arranged ? = 5040. 3. In how many different ways can six letters be arranged when taken singly, two by two, three by three, and so on, till they are all taken together ? = 1956. COMBINATIONS. 451. The different collocations that can be formed by any number of objects, without regarding the order in which they are arranged, are called their combinations. Although several permutations may consist of the same objects, every two combinations must consist of different ones. THEOREMS. 452. I. The number of combinations of n objects, taken two and n{n — 1) For the number of permutations, two and two is = n(n — 1) ; and for each combination there are two permutations ; as, for example, the combination ah affords two permutations ab, ha ; therefore the number of combinations is | of the permutations, or n{n — 1) jthey are = ^^^ . 453. II. The number of combinations of n objects, taken three . . . n{n - 1)(« - 2) and three, is = -^- ^ — - — -. ■^ ' 1X2X3 For the number of permutations, taken three and three, is ri(n — l)(n — 2), and each combination, as abc, affords 6, or 2 X 3 permutations ; hence the number of combinations _ n(n — l)(n — 2) ~ 1X2X3' 290 ELEMENTS OF ALGEBRA. 454. in. Generally, the number of combinations of n objects, taken m and m together, is _ n{n — V){n — 2) ... (w - m + 1) "~ 1 X 2 X 3 ... m For the number of permutations, taken m and m together, is = nin — l)(n — 2) ... (n — ?n + 1), and each combination of m objects affords 1 X 2 X 3 .... (m — l)m permutations; hence the number of combinations is _ n(n — l)(n — 2) ... (n — m + 1) ~ r~X 2 X 3 X ... m * 455. lY. When all the quantities are taken together, there is evidently only one combination — as appears also from the consideration that m is then = n, and the last formula becomes n(n-l)(?i-2)....3 X 2 X 1 _ 1 X 2 X 3 ...n ~ * How many products can be formed with 5 different quan- tities ? Tlie number when taken 2 and 2 is _ <7i - 1) _ 5 X 4 _ ^Q 3 and 3 is _ nin - l)(w -2) 5X4X3 _ ~ 2X3 2X3'* 4 and 4 is _ n(n - l)(n - 2)(n -3)_5x4x3x2 _ ~ 2X3X4 ~ 2X3X4' all together, = ] Total number, . . . = 2( EXERCISES. 1. How many different tints of colour can be formed by mixing the seven prismatic colours always in the same proportion, and ii every possible way ? = 120 2. If five of eight quantities are always given to find the othei three, how many cases may be formed in reference to the data, fine how many in reference to the quantities required? =56 and 108 3. How many different notes may be rung on ten different bells supposing all the combinations to produce different notes? = 1023 UNDETERMINED COEFFICIENTS. 291 4. How many different Aveights, consisting of a whole number of pounds, can be formed by means of the six weights 1, 2, 4, 8, 16, and 32 pounds, taking them singly, and then combining them two by two, three by three, and so on, till they are all taken together? =63. 45G. It appears from this result that, as 63 is the greatest weight, or that obtained by taking all the weights together, and as there are also 63 combinations, all of which are whole numbers of pounds, these six weights, variously combined, form a series of weiglits represented by the natural series of numbers, 1, 2, 3, 4, ... up to 63 (see Art. 444, Examp. 3.) It might be shewn generally that any number of weights repre- sented by 1, 2, 2^ 2', ... 2", would afford a series denoted by the natural series 1, 2, 3, 4, 5, ... up to 2"""^ — 1, which is just the sum of the series 1 + 2 + 2^ + 2=* + ... + 2" ; for ^^g 2X2n-l r-1 2-1 METHOD OF UNDETERMINED COEFFICIENTS. 457. The method of undetermined coefficients is a method for the development of certain algebraic functions, such as rational fractions having compound denominators, or compound irrational quantities, into infinite series, arranged according to the ascending powers of one of the letters, which is considered to be variable, or to be subject to any arbitrary value. The function is assumed equal to an infinite series, such as A+Bx+Cx* + Dx^ + ..., in which the value of x is unrestricted, but the values of its coefficients are constant, though unknown or undetermined. The values of the coefficients may be determined on the following principle : — 458. Let the value of x in the equation A-\-Bx + Cx^ + Dx^ + ... = be arbitrary ; that is, let the equation be fulfilled by any value whatever of .r. An equation of this kind is said to be identical (255), to flistinguish it from an ordinary equation, which is satisfied only by (;ertain values of the unknown qiiantity. Such an equation is satisfied by its coefficients, whether the equation be a finite or an infinite series ; for the value of x being arbitrary, and those of the 292 ELEMENTS OF ALGEBRA. coefficients constant, or always the same, whatever be the value of X, if X be assumed = 0, the series is reduced to one term ; namely — ^ = 0; and as the value of A is the same for any value of x, it is always = ; and hence the series becomes Bx + Cx^ 4- Dx^ + ... = 0, or B + Cx +i)x2 + ... =0; and if a: = 0, then B = Q] and in a similar manner it may be proved that C = 0, i) = 0, &c. 459. Hence if two equal polynomials A+Bx^ Cx^ + ... =A'-{- B'x + CV + ... be identical — that is, if they merely represent the same quantity in different forms, and be therefore always equal, whatever value be assigned to x — then the coefficients of their homologous terms — that is, of the terms containing the same power of x — are equal. For this identical equation may be put under the form A -A' + {B- B')x + (C - C')x-' + (i) - D')x^ + ... = o ; and hence (458) A-A' = 0,B-B'z=0, C- C = Q,D- D' = 0, ..., and therefore A = A\B = B',C= C, D = D% ... Let F(x) denote the algebraical function to be expanded, and assume F(x) = A -\- Bx + Cx^ + Dx^ + ... which constitutes an identity ; that is, an equality between a quan- tity not developed and its development. If the function be a fraction, multiply both sides by its denominator ; or if it be an irrational quantity, involve both sides to the corresponding power ; then the coefficients of the homologous terms being equated, and the coefficients of the other terms assumed = 0, the resulting equations will determine the coefficients ; or if the first side be transposed, and the coefficients of each of the powers of x be assumed = 0, the required coefficients may be determined. 1. Find the development of EXAMPLES. a b + cx' x^ + Dx- Here the undetermined coefficients A, B, C, ... are functions, Let T-^— = A+Bx+Cx'-[- Dx^ + ... -\- cx UNDETERMINED COEmCIENTS. 293 as yet unknown, of the constants a, b, c, of the given function. To find these coefficients, multiply both members by 6 + ex, and transpose a ; tlien Ab + Bb\x + Cblx^ + ^il^' Hence (458) Ab — a Db+ Cc = 0, ... from which -\-Bc\ +Cc = 0, Bb -\- ::]-'■ Ac = 0, Cb + Be = 0, a Ac «c Be b" ac' 1^' and a _ a ac ac'' ^ ac'' ^ ''' b + ex~~b~l^'''^l^'' ~'¥^ (1 ^X' + ...) 2. Convert \/(a- — a;^) into an infinite series. If this quantity were assumed equal to the series A + Bx + Cx"^ + Dx^ +, ... it would be found in determining tlie coefficients that those of the odd powers of x, namely, B, D, F, ... are = 0; and therefore the series may be changed into one containing only even powers of x ; thus, let V(a' -x'') = A+ Bx-" + Cx* + Dx% ... squaring both sides, and transposing a^ — x^, we find A-" + 2AB -a" + 1 c^ + B^ + 2AC X* + 2AD -{-2BC x' + ... ) = 0. Hence A^ 2BC=0',... from which ; 2AB + 1 = 0'^ B- + 2AC=0; 2AD + ^-^^^--2^--^^^- 24 1 D BC A 1 . • . /J fa/' — x^) = a — r- 2a IGa* 3. To resolve 5x — 4 (^ + 1)(^ - 2) into separate or partial fractions. 294 ELEMENTS OF ALGEBRA. 5x - 4: A B Assume --; ,-r = 1 : {x + l)(x -2) x + l^a:-2' then Zx-^ = A(x-2) + B(x + 1), [1]. Since this equation is true for all values of x, let a; + 1 = Oj or X = — 1 ; then — 9 = - 3J. ; . • . ^ = 3. Next let a; — 2 = 0, or a; = 2 ; then 6 = 35; .-.B=2, 5a- -4 3,2 ^^^ ^^^^^^ (X + IXx - 2) =^ ^TI + ^~2' The same result may be obtained from equation (1) by equating the coefficients of x, and the constant terms on each side ; thus — A +B= 5, and -2A + B= -4:, from which we find by the ordinary method A = S and B = 2. 4. To resolve 7 — rrr — ■ r- into partial fractions. (x — a)(x — bXx — c) Assume -7 ^—r- -.- = 1 r + (x — a)(x — 6)(x — c) X — a x — b x — c' . • . 1 = A(x - 6)(x - c) + £(x - a)(x - c) + C(x - a)'(x — b). Let X = a ; then 1 = A(a — b)(a — c), or A = (a — 6)(a — c)* Let X = J ; 1 then l=B(b- a)(6 - c) ; or 5 = - Let X = c; then 1 = C(c — a)(c — b); or C = (a _ 6)(6 _ c)- Let X = c ; 1 (a — c)(6 — c) ' 1 1 (x — a)(x — 6)(x — c) (a — &)(« — c)(x — a) 1 1 + (a - bXb - c)(6 - x) ^ (a - c)(6 - c)(c - x) * 33-3 _ ]Q^2 I 22a: — 2 5. To resolve -r rr^ into partial fractions. UNDETERMINED COEFFICIENTS. 295 Assume 3x^ - lOx' 4- 12a: - 2 _ A ^ C D x(x - If ~ X ^ (x-lf '^ {x-lf '^ x-l' Let a: == ; then — 2 = — ^;.-.^ = 2; therefore by transposition we have 3x=> - lOx^ + 12a: - 2 _ 2 _ x^ — ia^ + 6a: _ a:^ — 4a: + 6 a:(a;-l)^ ~ x~ x(x - If ~ {x - Yf ' a;2 - 4:c + 6 B , C , D hence -^-^^ = ^--^3 + ^-— j-- + -— ^. Let X — \=zoTx = z + \, and substitute this value in both sides of the equation, and it becomes z^ ~ z z^^ z^~ z^^ z^'^ z' .-. J5 = 3, C = — 2, and Z) = 1 ; wherefore 3a:^ - lOa:^ + 12a: -2 _2 3_ x{x -Vf ~ x^ Tx^Vf (a 460. It is necessary that the form of the development of a function be known before it can be effected by this method, for sometimes the ordinary assumed series is not suitable, and the result of the process indicates this circumstance. If, for instance, x enter as a multiplier into the function, as in X 1 the expression = a: X , then when a: = 0, this function 1 — X 1 — X is also = 0, whereas the assumed series would be = A, although it ought also to be = ; hence in such a case the series ought to be Ax + Bo^ + Cx^ + ... or ar(^ ^ Bx -\- Co^ ^ ...). Again, if - enter as a factor of the given function, as in — TT = - X :; -— , then for a; = 0, the function becomes a: — 2a;'- X 1 — 2a: ^ X - = - = oo, whereas the series would also in this case be reduced to A, although it ought to be = co; hence in such a case the scries ought to be Y B -\- Cx -\- Dx^ +, ... or X 296 ELEMENTS OF ALGEBRA. 4G1. The simplest method, however, Avhen the function has any factors such as the preceding, is to reject them, and find the development of the remaining part, and then to multiply this series by the factor. 462. It may be easily proved that any rational algebraic func- tion containing one variable quantity x, which does not become either equal to zero or to infinity when x = 0, or any irrational — v function of the form (a -{- bx -\- cx^ -{- ...)*, in which - is either positive or negative, may be developed in the form A -\- Bz + Cx' + ... The series cannot contain either x^ or — as a factor, for then x"^ it would become or oo, for x = 0, while the given functions dc not become equal to or oo. The series therefore cannot contain X in all its terms, neither can x have a negative exponent in any term. m The series cannot contain a term of the form il/x", or any m fractional power of x. For if n = 2, the term Mx'^ or M^Jx" would have two values corresponding to the double value + s/x"^ ; while the rational function would have only one value. And if i1 be the above irrational function, and it be raised to the power s, then (a + 6x + cx^ + ...y = (A-\-Bx-\- Co^ + ••• + ^^^ + •••>• And it is evident that the second member when involved to the m. power s would still contain a term of the form iVir^, which would have two values : and hence the second member would have twc values at least, while the first has only one. When n is not = 2, then, whatever be its value (since it is proved in the theory of equations that the number of roots of any m quantity is equal to the exponent of the root*), the term iVic", which would exist in the second member, after being involved tc the power s, would have n values, and the second member would have at least n values, while the first member has only one. Nc fractional power of x thei'efore can exist in the development. * This is proved in the theory of binomial equations, or equations of the form x» + a» = 0. UNDETERMINED COEFFICIENTS. 297 EXERCISES. 1 1 , 1 , 1 2 , 1. Develop ^— -^, . . . ' = 3 + q"^ + 27 + •" 2. Convert 5 into a series, = l+2x + Sx^ + ix^ -\- ... 1 — 2z + X- 3. Develop V(l + ^')» • • = 1+ o^*' - s^' + TF'^" - - 4. 2 8 ' 16 + bx a' + 6'a; + cV a' \a' a'/ Va' a' "^ \a a 5. Develop ,,^,^-0:^ ' = 1 + (1 + 2^> + (^ + 2i3y + (5 + 2C>' +, ... or 1 + 3a: + 7x- + 17x3 + 41x* + ... X -\- 2> 6. Eesolve ^ —7 — — rr into partial fractions, {x - X){x + 2) 3(a- - 1) 3(a; + 2) -,— ^ mto partial fractions, = -^^^^-^ + ^^^np^y a: + l 5 4 i 7:r + 12 *** *" x — i X —3 1 (ar — a)(x — 6) (a - 6)(x - a) (a - &)(:c - b)' 3t 4- 2 10. ... -7 — —^ into partial fractions, x(x + y _ 2 __L 2 2__ ~ a: "^ (x + 17 (x + 1)2 X + 1* 463. If a polynomial, containing only one variable, and ar- ranged according to its ascending powers, be equal to another similar polynomial, and if none of the coefficients be zero, the coefficients and exponents of the corresponding terms are equal, 298 ELEMENTS OF ALGEBRA. whether the exponents be fractional or integral, positive oi negative. Let Axa + Bx^ + Cxc + ... = A'xa' + B'xb' + C'xc^ + ... ; and if the exponents be all positive, but a not = a', let a ^a' \ then dividing both sides by a;", A + Bx^-a + Cxo-a 4- ... = A'x'i'-a + B'xb'~a + CV"« + ... Now, if a: = 0, then -4 = 0; which is contrary to the hypothesis : therefore a is not ^ a' ; and in a similar manner it may be shewr that a' is not ^ a, for then would ^' = ; therefore a = a', anc if X = 0, A = A' ; and hence Bx^'a -j_ (7:rC-« 4- ... = B'x^'O' + C"xc'-a + ... ; and it may be similarly proved that b — a = b' — a, and therefon b = b', and hence also B = B'. In the same manner it may be proved that c = c' and C = C] and so on. If some of the exponents be negative, let — a be the greatest one ; then if both sides be multiplied by x% the equation woulc be reduced to the form A+Bx^ + Cxo + ... = A'xa' + B'x^' + Cx<^ + ... ; from which it may be easily shewn that the exponents anc coefficients of the corresponding terms are equal. THE BINOMIAL THEOREM. 464. The binomial theorem is a formula, by means of whicl any power or root of a binomial quantity may be found, without actual multiplication of the quantity successively into itself, oi actual extraction of the root. The demonstration of it in the case of integral powers maj be easily effected by means of the principles of combinations. Let three binomial quantities, x -^ a, x -\- b, x -\- c^hQ multiplied together, and the product will be found to ba x^ -\- a -\-b + c t" + ab + ac -\-bc abc. THE BINOMIAL THEOREM. 299 ind if this product be again multiplied by the binomial x jroduct is evidently X* -\- a'x^ -\- ab x^ + abc x + ahcd. d, the + d -\- ac + he -\-ad + hd + abd + acd 4- bed + cd 4Go. It is evident from the process of multiplication, that the irst term consists of a power of a:, whose exponent is equal to the mmber of binomial factors, and that its powers diminish by unity n each succeeding term. It is also evident, that the coefficient of ;he second term is the sum of all the second terms of the binomial ; ;hat the coefficient of the third term is the product of the second ;erms of the binomial, taken two and two ; that the coefficient of ;he fourth term is tlie product of the same quantities, taken three md three. It is therefore very likely, on the principle of induction, that the coefficient of any term whose place is denoted by n + 1, )r which has n terms before it, is the sum of the products of the second terms of the binomials, taken n and n together ; and the ast term is evidently the product of all the second terms of the ainomial. In order to prove generally the law of the coefficients, let the product of m factors be a-m _|_ ^^m-l ^ ^_^m-2 _j_ ^^^^ il/x™""*^ + Nx^'"" + ... T. Here m — n is the exponent of x in the (n + l)th term. Let this product be multiplied by a new factor x + k, and the new product is + B + kA -\-kB + kM -\-kT. By multiplying by the new factor x + k, the coefficient of the second term has been increased by /j; and tliis therefore holds true for every factor ; and hence, 466. The coefficient of the second term is the sum of the second terms of the binomials. Again, B is by hypothesis the sum of the products of the second terms of m binomials, taken two and two ; and /.:/! is the product of the second terms of the preceding m binomials by the second term k of the new binomial ; and therefore B + kA still continues to be the product of the second terms of all the binomial factors, taken two and two ; and hence, 467. The coefficient of the third term is the sum of the pro- 300 ELEMENTS OF ALGEBRA. ducts of the second terms of all the binomial factors, taken twc and two. 468. It may easily be shewn in the same manner, that tlu coefficient of the fourth terra is the sum of the i)roducts of th( second terms of all the binomial factors, taken three and three And, generally, since N is the sum of the products of the seconc terms of m binomials, taken n and n together, and as kM is th( product of the same quantities, taken (n — 1) and (n — 1) together multiplied by k, therefore N + IcM is the product of the sam( terms of (?« + 1) binomials, taken n and n together ; that is, 469. The coeflScient of any term is the sum of the products o the second terms of the binomial factors, taken n and n together where n denotes the number of the terms preceding the assumec term. 470. It is therefore manifest, that the last term is the produc of the second terms of all the binomial factors. 471. Since the law therefore of the composition of the coeffl cients, which was supposed to hold in the case of /« binomia factors, continues to be true in the case of (ni + 1) factors, i must be true generally ; for it is manifestly true for two factors and therefore for three ; and hence it is true for any number. The development of (x + «)"• will now be found by supposinj the second terms of the binomial factors all equal. When ther are m binomial factors, of which a, b, c, ... are the second terms if they be equal, then the coefficient of the second term is A = a-\-b-\-c-{-... = ma. The second term B = ab -{- ac -{- be -{-... = a- -{- ci^ +, ... a being repeated as often as there are combinations of a, b, c, .. taken two and two ; hence (452) The third term C = abc ■\- abd + bed + ... = a^ -\- a? +, ... a being repeated as often as there are combinations of a, 6, c, .. taken three and three ; hence (453) _ mjm — l)(m — 2 )^3 _ ^- 2^~3 " ' and in general the coefficient of any term, as iV, which is precede( by n terms, is ^ mjm - \Xm -2) ...jm - n + \) ^^ 1 X 2 X 3 X ... « THE BINOMIAL THEOREM. 301 472. Hence the formula is (a; + a)"* = a;"* + max"^ + 1X2 a-x" ■^ 1X2X3 Km-lX^-2)...(r.-. J:_l)^„^,_, _^_ ^^^ ^ lX2x3X...n 473. The term to which the coefficient N belongs is called he general term, because by giving to n any particular .ralues, as n = 1, n = 2, ?^ = 3, ... all the other terms may )e derived from it. The term preceding the general term, for which 71 — 1 must be taken in place of n in the latter term, is vidently m(m - l)(m - 2) ... (m - n + 2 ) .„,^ ^ 1 X 2 X 3 X ...(n- 1) and as (m — n + 1) is the exponent of the leading quantity x in this term, and n is the number denoting the place of this term, also the coefficient of the general term is found by multiplying the preceding coefficient of the nth term by (in — n -\- 1), and dividing it by n ; therefore, generally, 474. The coefficient of any term is found by multiplying the coefficient of the preceding term by the exponent of the leading quantity in that term, and dividing by the number denoting the place of that term preceding that whose coefficient is sought. It is also evident, that the first term is the leading quantity raised to the required power, and that the coefficient of the second term is the exponent of the required power ; and that the expo- nents of the leading quantity diminish by unity in each succeeding term. Also, the second term of the binomial enters the second term of the power, and its exponents increase by unity in each succeeding term, and the last term is this quantity raised to the required power. 475. When the second term of the binomial is negative, the signs of the terms of the development are alternately positive and negative, or those terms are negative which contain odd powers of the negative quantity. Por in this case the second terms of all the binomial factors would be negative, and hence the odd powers of a contained in the coefficients A, C, E, ... of the even terms; namely, (— a), (— af, (— a)\ ... being negative, the even terms of the series 302 ELEMENTS OF ALGEBRA. are negative, or the terms are alternately positive and negative that is, (x — a)"* = x^ — max^'^ A ^^; —arx'^'^ ^ ^ 1X2 m{m - l)(m - 2 ) 1X2X3 """^ +••• 476. The coefficients, taken in a reverse order, are the same as in a direct order — For the last is 1 ; the last but one is m{m — 1) ... (m — m + 2) m(jn — 1 ... 3 X 2 _ 1 X 2 X 3 X ...(m-1)' °^ lx2x3...Cm-l) ~ "" ' the last but two is m(m — 1) ... (m — m + 3) _ m(m — 1) ... 4 X 3 _ m{m — 1) 1 X 2 X 3 ... (to - 2) ~ 1X2X3... On — 2) ~ 1X2 , , , , , ,-,,,. -, , '>n(m — V)(m — 2) and the last but three would be found to be — ^:-- — — ;; — - 1X2X3 and so on. 477. The sum of the coefficients of {x + a)™ is = 2"*. Let X =1 and a = 1, and the series becomes (1 + 1)"* = 2' m(m — 1) . , w(m — 1) 2 "^ •" "*" 2 478. The sum of the coefficients of (x — a)"* is = 0. Let X =1 and a = \, then (l-l)>»^0>n:^0^1-;. + or (x + a) ". In order to demonstrate this, let (x + «)"» be put under the form x'"(l + - ) 5 and let - = ?/ ; then if the development of (1 + y)"* be found, and multiplied by x"*, the result will be the development of (x + o)™- It appears (462) that the equation (1 +y)" = i+^y + Bf + Cf + D/ + ... [1], may be assimied, its first term being 1, because when y = 0, the THE BINOMIAL THEOREM. 305 first member becomes 1, and also the second. Hence also if z be any quantity, m (1 + s)" = 1 + ^s + Bz^ + Cz^ + Z>2* +, ... in which the coefficients must evidently be the same as in the former series, for both equations are true for any values of y and 2, and therefore when y = z. Now, let (1 + yY = u, and (1 + «)" = v, then (1 + yY = m"*, and (1 + 2)" = t;*" ; also „nt _ y,n ^ ^(^ _ 2) 4. B(y^ - Z^) + €0/ - Z') + D(/-20 4- [2]. But 1 -i- y = u'^, 1 -^ z = v" ; then u^ — v" = y — z; hence l^rr^n = jzrj^^^y - ^) + ^(/ - ^') + <^(/ - ^0 + i)(/ - 2*) + ...} [3], and (87) M"* — v"* = (m — i;)(m'»~' + xC^-H + M"»"3y2 _^ _^^ .^ ^^^^-2 _|. j^m-l-J ■= {u — v)M, and M" — y" = (m — u)(m"-» + M«-2|; 4- M«-3y2 + ... + MU"-^ + y""^) = (m - t;)iV; assuming M and iV respectively for the polynomial factors. '^^'''"' N = "^?r=^" = ^ + ^(^ + z) + C( ./ + yz + ^2) + i)(/+/z+yz2+23)4. j-^-]^ And as this equation holds true for any values of y and z, it does so for y-z, for which I -\- y = 1 -\- z, ox u = v] and therefore M = T/m™"*, N = nu''-\ and by [4], TWM' zr=^+2% + 3C/ + 4i)y+ [5], mu or = M"(-4 + 2% + 3(7/ + 42)/ + ...). 306 ELEMENTS OF ALGEBRA. m Substituting for m'" and m" their values (1+ yY and 1 + y, fy6 or ^(8 -2), =2(1-^.^-^.^-^.^ -...). 484. Trinomials and polynomials of few terms may be easily developed by means of this theorem. 1. To develop any powers of a + 6 + c, assume a + 6, or 6 + c, equal to a single letter ; thus, lih -\- c = d, then (a-\.b + cy = {a + df = a" -\- 2ad -\- d" = a" + 2a(h + c) + (6 + cy = a^ + 2ah + 2ac + h" + 2hc + c=^; hence (a + 6 + c)^ = a^ + 6^ _^ c^ + 2a& + 2ac + 26c. Or if it be assumed that a -\-h = p, then (a + 6 + c)2 = (p + c)2 =/ + 2pc + c« = (a + J)« + 2(a + i)c + c^ = a^ + 2a6 + h" + 2ac + 26c + , then {a + h-\-c-\-dy = {p + dy=^p^ + ZpH + 3pcf= + d? = (a + 6 + c)3 + 3(a + 6 + c)^^/ + 3(a + 6 + c)^^ + d^. 3. So (a + 6 + c + rf)* = (a + 6 + c)* + 4(a + 6 + cfd + 6(o 4- 6 + c)2(^ -j- 4(a + 6 + c)d3 + rf*. EXPONENTIAL THEOREM. Or if a + 6 = jD and c -{- d = q, (a + b + c + dy = ip + qy=p*+ ip'q + GpY + ^p [2]. When the logarithm of the member n therefore is known, that of n -\- z will be found by this series, which will always be convergent, and will be the more so the greater n is compared with z. Let z = \, then the logarithm of n + 1 will be i(n + 1) = in + ^^^2;rTl "^ 3 • (2n + ly + 5 • (2n + 1/ + -^ - ^^'^^ which is convergent, even in the case of n = 1. 497- This series will be of the simplest form when the modulus M =\. But M= ie, and when ie = 1, e must be the base of the system (Theo. V.) It was stated in article (491) that the system whose base is e, is called the natural system, which is an appro- priate term for a system that has the simplest modulus. It was shewn in article (487) that e = 2-7182818, and that this is that value of the base which makes ^ = 1 ; and hence also y=\ = M (492). The formula for natural logarithms, therefore, is l{n + 1) = /n + ^^2n + 1 "*" 3 ' (2« + 1)' "^ 5 ' (2n + 1)'^ "^ '"^ '" ^^^' 318 ELEMENTS OF ALGEBRA. 498. For the common system of logarithms, or those of Briggs, a — 10, and M — -j—. The natural logarithm of 10 will be found by first computing that of 2 and 5 by the preceding series. If n = 1, /ra = /I = 0, and ;(l + l) = fi = 2(l + l. 1+14 + 11 + ...). The 8th term of this series is less than 1 in the 8th decimal place, or less than -00000001, and the preceding 7 terms carried to 8 decimal places, give Z2 = -69314:718, which is correct to the 8 th decimal place. Again, Ih will be found by making w = 4, for then Zra = /4 = W = 212, which is already known, and /(4 + l) = ;5 = 2fi + 2(l+l.^ + l..l, + ...). The first three terms of this series are sufficient to give a result correct to the 7th decimal place, and the first four terms give •22314355 and 14: = 212 = 1-38629436 hence 15 = 1-60943791 Also 12 = -69314718 therefore 15 + 12 = ZlO = 2-30258509 = the reciprocal of the modulus of the common system. Hence Jf = J^ = __L_ = -43429448. 499. If, then, natural logarithms be multiplied by this number, the products will be the common logarithms of the same numbers. Denoting common logarithms by L, then X2 and L5 may be found from 12 and 15 ; thus, L2 = M12 = -3010300 L5 = M15 = -6989700 500. Were it required to find the natural logarithm of a number whose common logarithm is given, it would only be necessary to divide the former by M = -j-~, or to multiply ^Y irf = ^^^ = 2-30258509; but ZlO = 4-, or Le = -^: hence i2-7182818 Le 110 = -43429448. LOGARITHMS. 319 COMMON LOGAKITHMS. 501. The logarithm of any power of 10, the base of the system of common logarithms, is the exponent of the power. Let a number m = 10", then Lm = nLlO = n X 1 = n. Thus, let m= W = 100, then ZlOO = 2, so ilOOO = 3, ilOOOO = 4, &c. 502. Hence the logarithms of numbers between 100 and 1000 are between 2 and 3, or = 2 + a fraction ; those between 1000 and 10000 are between 3 and 4, or = 3 + a fraction, and so on ; therefore, logarithms of all numbers that are not exact powers of 10, consist of an integer and a fractional part. 503. The integral part of a logarithm is called its index. The index of the logaritlun of a number is a unit less than the number of integral figures in the number. 504. The decimal part of the logarithm of a number is the same as that of the product of this number by any power of 10. For if m = n X W, Lm = Ln -\- r, for LW = rLlO = r OT the whole number r being added to Ln, gives Lm ; therefore the frac- tional part of Lm is the same as tliat of L71. So 12432 = 124-32 X 10^ ; hence Z12432 = il 24-32 + 2 ; hence if 2 be added to iyl24-32, it gives Zl 2432 ; therefore the fractional parts are the same, and only the indices difier, that of the former being 2, and of the latter 2+2 = 4. 505. The logarithm of any number less than 1 is negative. 506. Thus, let -0345 be the number, then •0345 X 100 = 3-45, and i-0345 -{- LIO^ = Z3-45, or Z-0345 + 2 = i3-45, or i-0345 = i:3-45 - 2 = 0-537819 -2= 2-537819; for Z3-45 has for its index (503). The logarithm of 3-45 is taken from the tables. The logarithm of -0345 therefore has the nega- tive index 2, the sign being, as usual, placed over it, because the fractional part -537819 is positive. If the latter part be taken from 2, the remainder is 1-402181, which is all negative; but the former mode of expressing the logarithm is commonly used, and the fact of the exponent only being negative, must be observed in calculating with logarithms. Were the number -00345, then since 320 ELEMENTS OF ALGEBRA. it multiplied by 1000, gives 3*45, the index woidd be 3, and the logarithm = 3-537819 ; and, in general, 507. The index of the logarithm of a fractional number is nega- tive, and is one greater than the number of j^refixed ciphers. The arithmetical complement of a number is found by taking its first figure from 10, and all the rest from 9, and then prefixing 1. Thus, the arithmetical complement of 573 is = 1427, which is = — 1000 + 427 = — 573 ; and hence to add 1427 is the same as to subtract 573 ; and generally, 508. To subtract a number is the same as to add its arithmetical complement. Thus, from 5648, subtract 342. 5648 5648 - 342 ar. com. 1658 5306 5306 The rule given above for finding the arithmetical complement may also be expressed thus — Subtract the given number from 1, with as many ciphers annexed as there are figures in the number, and prefix 1 to the remainder. When logarithms are to be subtracted, it is often, but not always, more convenient to add their arithmetical com- plements. The properties of common logarithms noticed in articles (502) and (504) are peculiar to this system, in consequence of the base of the system of logarithms and that of numeration being the same ; a circumstance that very much simplifies logarithmic calculations. 509. The logarithms of numbers of the natural series 1, 2, 3, 4, ... may be calculated to any extent by substituting for n in the preceding series [3] the numbers 1, 2, 3, 4, ... the results would be the logarithms of 2, 3, 4, 5, ... ; or by first finding their natural logarithms by [4], and then their common logarithms by the method in (499). It is preferable, liowever, to begin with some large number, as 10000, and to calculate the succeeding numbers to 100000, or any required extent ; for the series wHl be very convergent for these high numbers ; and besides the logarithms of the inferior numbers are easily found from these (504). For example, the fractional part of the logarithm of 124-56 is the same as that of the logarithm of 12456 ; and so on for the other inferior numbers. LOGARITHMS. 321 510. The series [3] may be expressed in a different form. Let L(n -{- V) — Ln = D, then the formula becomes ^ ^ ^^^27+"i + 3 • (2n + 1/ + 5 * (2n + ly + - ^ ' when n is 100 or upwards, the second term is = -00000004, and the first will give the value of D correct to the 7th decimal place, ^-1=^5^^ = — • and Z(100 + 1) = i^lOO + A or ilOl = 2-00432134. This logarithm, however, is not quite correct in the 8th decimal place, for if the second term of the series be added, then D = -00482138, and omitting the last figure ilOl = 2-0043214. "When n is 1200 or greater, 2n may be taken instead of 2n + 1, and then i> = ^. n Thus, for 71 == 10000, D = -000043427, and hence ilOOOl = ilOOOO -{■ D = 4-000043427. So . il0002 = ZlOOOl + D = 4-000086854, In a similar manner, the logarithms of 10003, 10004, 10005, ... may be calculated. If the logarithms were required to be correct only to 7 decimal places, then the value of Z> for n= 10000 is •0000434, and D would have the same value for several successive numbers, so that by merely adding D to ilOOOl, the sum is il0002 ; and adding D to the latter, the sum is jLl0003, and so on. When the number is still higher, as n = 99840, then the value of X> is =: -000004349, and for the value oi n = 99860, D has exactly the same value, so that for all numbers between these two, D has this constant value, even to the 9 th decimal place ; and therefore when X99840 is known, the logarithm of each of the successive numbers up to 99860 is found, by merely adding this value of D to the logarithm of the preceding number. If the logarithms be limited to a smaller number of places, the difference will be constant over a greater interval. Thus, when the logarithms are extended to only seven places, the difference i) is = 44 for all the successive numbers between 97900 and 99800. When the logarithms of prime numbers are computed, those of composite numbers are found, by adding the logarithms of their component factors (Tiieo. I.) u 322 ELEMENTS OF ALGEBRA. EXAMPLE. Given the logarithms of 2 and 7, find that of 14. i2 = 0-3010300 L7 = 0-8450980 hence X14 = 1-1461280 So the log. of 4 = L2^ = 2L2 ; L8 = L2^ = SL2. EXERCISES. 1. Given the logarithm of 2 = 0-3010300 and that of 3 = 0-4771213, find those of 4, 6, 8, 9, and 12. lA = 0-6020600, i6 = 0-7781513, L8 = 0-9030900, L9=0-9542425, X12 = 1-0791812. 2. Calculate by the formula (3) in (596) the common logarithm of 11, 13, 17, and 31, . . ill = 1-0413927, Z13= 1-1139434, il7= 1-2304489, ^31= 1-4913617. SERIES. DIFFERENTIAL METHOD. 511. The uses of the differential method are to find the succes- sive differences of the terms of a series, or any term of the series, or the sum of a finite number of its terms. The successive differences are divided into distinct orders. The differences of the^first order are the differences of the terms of the series ; those of tlie second order are the differences of the terms of the first order ; those of the thi?-d are the differences of the terms of the second order ; and so on. The differences in every case are found by subtracting each term from its succeeding term. Thus, if the series be 1, 8, 27, 64, 125, 216, ... then 7,19,37, 61, 91, ... is the Tst order of differences, 12, 18, 24, 30, 2d , and 6, 6, 6, 3d SERIES. 323 PROBLEMS. 512. I. To find the first terra of any order of difierences. Let the series be a, h, c, d, e, ... then the orders of difierences are, 1st order = b— a, c — b, d— c,e— d, ... 2d ... = c —2h +a,d—2c + &, e - 2c? + c, ... 3d ... = d - Sc +36 —a,e-3d+3c- b, ... 4th ... = e — id +Gc — 46 + a, ... The coefficients of these orders of difierences are evidently, for the 1st order 1 - 1, 1 - 1, 1 — 1, ... 2d ... <( ~ , V or 1 - 2 + 1 [1-2 + 1 \ ^ ... < V or 1 — I -1+2-1 / r 1 - 3 + 3 - 1 ) \ I or 1-4 + 6-4 + 1, I -1 +3-3 + 1/ 3d ... ^ Ur 1 - 3 + 3 - 1, 4th and so on ; being plainly the terms of the successive powers of 1 — 1, or the coefficients of the terms of the powers of a binomial X — y ; and hence the coefficients of the nth order are The last term is + 1 when n is an even number, and — 1 when it is odd. It is also evident that the last term is the coefficient of a in the first of the nth order of differences ; the last term but one is the coefficient of b ; the last but two the coefficient of c, and so on ; and these coefficients are the same in reverse order as in the direct order (476) ; therefore if d^ represent the first difference of the nth. order, and d^, d^, d^, ... those of the first, second, third, ... orders, then, when n is an even number, , , n(n- 1) n{ n - l) (n - 2) d..= a-nb + -^—c —~^ cZ+,... 324 ELEMENTS OF ALGEBRA. and when n is an odd number, , , n(rj _ 1) n(n - l)(n - 2) , , 513. It is evident from the coefficients, that when w = 1, the value of d,, has only two terms, for then n — 1 = ; when n = 2, this value has only three terms, for then n — 2 = ; and so on. EXAMPLES. 1. Find the first term of the second order of differences of the series P, 2\ 3^ 4^, ... or 1, 4, 9, 16, 25 ... Here n = 2; hence take three terms of the first value of d^, and take a = 1, 6 = 4, and c = 9 ; therefore ^^ = a_„6+<^_=i)o= 1-2x4 + ^X9 = 1-8+9 = 2. The accuracy of the result may be proved by finding the differences by subtraction ; thus, 1, 4, 9, 16, 25, ... 3, 5, 7, 9, ... 1st order, 2, 2, 2, ... 2d ... ; and hence 2 is the first difference of the second order. 2. Find the first term of the fourth order of differences of the series l^ 2\ 3^ 4^ 5^ ... or 1, 8, 27, 64, 125 ... Here n = 4 ; hence take five terms of the value of d„, and a = 1, 6 = 8, c = 27, c? = 64, e = 125, and hence d^ = ^ 4X3 ^„ 4X3X2 ^, 4X3X2X1 ,„, 1 - 4 X 8 H --— X 27 —- X 64 + -— — X 125 2 2X3 ^2X3X4 = 1 - 32 + 162 - 256 + 125 = 0, or the required difference is = 0. EXERCISES. 1. Find the first term of the second order of differences of the series 1, 3, 6, 10, 15, 21, = 1. 2. Find the first term of the third order of differences of the series 1, 6, 20, 50, 105, 196, =7. 3. Find the first term of the third order of differences of the series 1, 5, 15, 35, 70, =4. SERIES. 325 4. rind the first term of the eighth order of differences of the series 1, 3, 9, 27, 81, ... = 256. 514. II. To find any term of a series. Let d^, c/g, d,, ... represent, as above, the first terms of the diffe- rent orders of differences ; then by the preceding formula (512), b = a -\- d^, c = — a -^ 2b -\- d^, c? = a — 3& + 3c + c?3, e = -a -{- 4:b — 6c + id -{- d^J and so on to any number of terms. Hence by substituting the values of 6, c, d, ... b = a -\- d^, c=a + 2d^i- d^, d= a + Bd^ + 3d.^ + d^, e = a + 4f/i + Gc?2 + 4^3 + d^, and so on. It is evident from inspection that the coefficients of the different orders of differences in the value of any of the terms, as of e the fifth term, are the coefficients of the terms of a binomial involved to a power, whose exponent is one less than the number denoting the place of the terms ; and hence the nth term is = „ + („ _ iH + iliz^^l^^ + („-i)(.-y.-a) ^^ ^ EXAMPLES. 1. Find the 12th term of the series 2, 6, 12, 20, 30, ... 2, 6, 12, 20, 30, 4, 6, 8, 10, ... hence rfi = 4, 2, 2, 2, d, = 2, 0, 0, d, = 0, and the succeeding orders of differences are also evidently nothing ; hence the 12th term = « + («- l)t?i + ^ ^^ -d^ = 2 + 11 X 4 + ^-^"4^ X 2 = 2 + 44 + 110 = 156. 326 ELEMENTS OF ALGEBRA. 2. Find the nth term of the series 1, 3, 6, 10, 15, 21, ... 1, 3, 6, 10, 15, ... 2, 3, 4, 5, ... and d^ = 2, 1, 1, 1, d, = 1, 0,0, d, = 0, hence wth term = a + (n -1)X2+ (» " X" - ) ^ i = If n = 5, the term = 15, « = 6, ... = 21, n = 20, ... = 210. EXERCISES. 1. Eind the 15th term and the nth term of the series 1, 2^, 3*, 4', r.^^ A a -ict ilSth = 225. 2. Find the 20th terra of the series 1, 2^ 3', 4', ... or 1, 8, 27, 64, 125, = 8000. 3. Find the nth term of the series 2, 6, 12, 20, 30, ... = n(n + 1). 515. Iir. To find the sum of n terms of a series. Assume the series 0, a, a + b, a -\- b -^ c, a -\- h + c -^ d, ... whence it is evident that the (n + l)th term of this series is exactly the sum of n terms of the series a, b, c, d ... The first order of differences of the former series is also a, b, c, d, ... and the second, third, &c. orders of differences are just the first, second, third, &c. orders of differences of the series a, b, c, d ... Hence if the same values, as in art. (512), be used for d^,' d^, d^, ... the first terms of the different orders of differences of the assumed series are a, d^, d^, d^, ... and its (n + l)th term therefore is (514) „ , n(n— 1), , n(n— ])(n — 2) S = na+ ^ V , + -^ —^- d, + ... ; whence S is also the sum of n terms of the series a, b, c, d ... SERIES. 327 EXAMPLES, 1. Find the sum of n terms of the odd numbers 1, 3, 5, 7, 9, ... 1, 3, 5, 7, 9, ... 2, 2, 2, 2, ... c/i = 2, 0, 0, 0, ... c?2 = 0, and ^=„J+ S 13 Hence^^:^-^ + ^- = - + - = ^ and -^ = — = sum required. Or the sum may be found thus (530), '^Sx + 2 ^ 5 "^ 8 ■^1I"'"U"'"I7"^"' 3a; + 8 11^14^17^'"' 1 1 13 and the difference is evidently - -f- - = — - ; and taking also the 6 8 40 difference of the corresponding terms, it gives 5.11^8.14^11.17^'" 40' ^^"""^^ 631 + 04 + 1O7 + - = 6 • 40 " 240- EXERCISES. 1 9 1. Find the simi of the series 2^ rrr — ; — rr, . S = -rr. {x + 3)(a: + 5) 40 o 1 s_A (2x + 3X2x + 7)' * 35' 3 v____i____ s= 1 (4a: - SXix + 1)' 4' 534. If n terms of the series 2 be taken, and also n terms 7nx-{- a SERIES. 337 of the series 2—7 r , it is evident that the last n — r terms ra{x -\- r) + a of the former will be identical with the first n — r terms of the latter ; and therefore The difference between the whole n terms of the two series will be equal to the excess of the first r terms of the former above the last r terms of the latter. But the difference between the corresponding terms of the n terms of the two series is also equal to n terms of the series -, or rm ^2- (inx + a){7n(^x + r) + a} (jnx + a){m(x + r) + a}' This will appear evident by writing n terms of these two series, m -\-a 2m -\- a 3m -\- a " (s — 1)»* + ct rm -\- a "^ \r + l)m + a '" nm + a' ' ^^^ {? -. I i\^ ^^ + r \r + l)m + a (r + 2)m + a '** nw + a + 1 + ... L__. (n + l)m + a C?^ + r)m + a The identical terms are enclosed in braces, and the difference between these n terms of the two series is evidently the excess of the first r terms of the former above the last r terms of the latter, and the difference between the corresponding terms gives the first n terms of the series m2- "^ "(mx + a){m(x -\- r) -^ a}' When r = 1, n terms of the last series is therefore _ / 1 1 \1 n ~ \m -\- a (n -\- l)m -\- a/ m ~~ (m + a){(n + l)m + a}' and the sum of n terms for any other value of r may be similarly found. Thus, the sum of n terms of the series in the first exercise given above, or 2^ — —^. — — — ^, for which 7n = 1, a = 3, b = 5, V 338 ELEMENTS OF ALGEBRA. rm = b — a = 2, and therefore r = 2, is 1 1 ^/^-i-^-l I I )± \m + a 2m -\- a (n -\- l)m + a (n + 2)m + a/ rm _ /I 1 1 _ 1 \ 1 _ 9^2-1- 41n ~\4'^5~n + 4~ ^T"5/ ^ 2 ~ 40(n+4)(w + 5)* 535. III. To find the sum of the series whose general term is 1 Since (jnx + a){inx + b)(inx + c)" 1 1 (inx + a)(inx + 6) {inx + 6)(?«a; + c) c ~ a (inx 4- a)(;nix -\- b')(inx + c) 1 (c - a)2 (ma; -j- d)(inx + b)(j)ix + c) = 2 , . ' .. .-S. (m:c + d)(inx -\- b) (mx ■i-b)(jnx + c)* By the former problem (533), the sums of the two latter series in the second member can be found, provided that (6 — a) and (c — 6) are both multiples of m. The differences of these sums being divided by (c — a) will give the sum of the proposed series. 53(). When (b — a) and (c — 6) are different multiples of m, the sum of the proposed series may be found i by the preceding method. 537. When (b — a) and (c — 6) are the same multiple of m, the sum may be more readily found ; for then the same value of x, as x\ will make the factors in the denominator of some term of the first series equal to those in the first term of the second series, or lib — a = c — b = mr^ r terms and mx' + a = m + 6, and mx' -\- b = m 1 + c; then x' = l+r; whence the series 27 t ,. after its first (nix -j- a)(mx + b) 1 comcides with the series 2; ,^^ {mx + b)(nix + c) Hence 2rm'2i - = S' (mx + a){7n{x + r) + a}{m(x -\- 2r) -\- a} ' where S is the sum of the first r terms of the series (jnx + a){m(x + r) + a} s SERIES. 339 , and the sum of the given series is then 2rm Find the sum of the series 2 (2a; + 3X2 a- + 7X2 a; + 11) Here m = 2, a = 3, 7nr = 7 — 3 = 4 ; therefore r = 2, and 2rm = 8, also S = + hence sum = 5X9 ' 7X11' S 1/1 . 1\ 61 8V45 ^ 77/ 2m 8\45 ' 77/ 13860 EXERCISES. 1. Find the sum of the series 2 (2a; + lX2x 4- 5X2a; + 9)' 1260* x(^x + iXx + 2)' • • 4 538. IV. To find the sum of the series Avhose general term is Since (mx + a)(mx -f- b)(mx + c)(mx + d) 1 1 (inx + a)(^inx + 6) (?«a; + c)(»2x + d) __ m{c + c?)a; + cd — m{a + &)a; — ah ~ (mx + a)(?na; + b)Qnx + c)(7wa; + d) in order to make x disappear from the numerator of the second member, assume c -\- d = a -\- b, and then the difference between the series whose general terms compose the first member being divided by (cd — ah), the quotient would be the sura of the proposed series when a -^b = c -\- d. When Q) — a), (c — b), and (d — c), are the same multiples of m, the sum may be more readily found thus — 1 1 (inx + a)(nix + b)(^mx + c) (inx + b)(nix -\- c)(nix + d) _ d — a (mx + d)(inx + b)(mx + c)(mx + c?) ' 340 ELEMENTS OF ALGEBRA. and ifh — a = c — b = d — c = rm', and therefore d — a = 3rm, then the sum of the proposed series is . I =^ (mx + a){m(x + r) + a}{m(x + 2r) + a} {m(x + Br) + a} 3m ' where S is the sum of the first r terms of the series 2 I (jnx + d)(mx -\- h^^jnx + c) ' where h = a + mr, and c = a -{- 2mr. This is proved exactly as the similar formula (537) in the preceding problem. EXAMPLE. Find the sum of the series whose general term is 1 (2x + l)(2a: + B)(2x + 6)(2a; + 7) Here a=l, 6 = 3, m = 2, rm = h — a = 2-, therefore r = 1 , and Srw = 6 ; 1 ^ S I \ 1 hence b = - — - — -, and — — = ■;; X — -- = , 3.5.7' Srm 6 105 630 Or by actually subtracting the two series, ' +^- + ^- + ... 3.5.7 ' 5.7.9 ■ 7.9.11 1 1 ' rr Q 11~f"Q It iQ +J 5.7.9 ' 7.9.11 ' 9.11.13 6 6 6 _ 1 * ®" 3.5.7.9 "^ 5.7.9.11 "*" 7.9.11.13 '^ "' ^ 3.5.7' and - of this gives the sum of the proposed series o 6 3.5.7 630 EXEECI8ES. 1. Find the sum of the series 2^ (x + 2)(x + 3)(x + 4)(x + 5)' ^ = 180- , 1 8 = 1 ^x(^x + l)(x + 2Xx + 3)' 18' SERIES. 341 539. It is evident that this method may be very easily extended to any series whose general term is of the form (inx + a){in(x + r) + a}{m(x + '■^r) -\-a) ... {m(x + n — Ir) -\- a}' If the terms of the series have a constant numerator, as p instead of 1, the sum will be just p times greater. By methods somewhat similar, the sum of any series, the numerators of whose general term is of the form px + e, and denominator of the above form, may be found, when the denomi- nators contain more than two factors. 540. V. To find the sum of the series whose general term is px -\- e (vix -\- a){m(x -\- r) +a}{m(a; + 2r) + a}* Since m(^px + e) _ p {mx-\-a){m(x-\-r) + a}{m(x-\-2r)-\-a) ~ (mx + a){m(x-\-r)-\-a} p(2mr + «) — "26 (mx -\- a){w(x + r) + a}{m(^x + 2?-) + a}* If S denote the difference between the sums of the two series whose general terms form the second member of this equation, then the sum of the proposed series is proved, as in former Q propositions, to be = — . Find the sum of the series whose general term is X + 2 (2,r + lX2a: + 3)(2ar + 5)- Here p = 1, e = 2, m = 2, a = 1, mr = S — 1 = 2; hence r = 1, and p(2mr -\- a) — me = 1 ; ' ■ (2x + l)(2x + 3) 2 3 6' and 2 I =. 1 X -^ - i- (2x + l)(2a; + 3)(2ar + 5) 4 3 . 5 ~ 60' Hence the sum of the given series = -( I ^ 2\Q 60/ 1/1 1\ 1 ^_S_ 2 ^ 60 ~ 40* 342 ELEMENTS OP ALGEBRA. * EXERCISE. Find the sum of the series whose general term is 3a: - 2 S - — (2a; - l)(2a; + l)(2a: + 3)' 24 541. VI. To find the sum of a series whose general term is px -\- e (inx + a){m,(x + + «}{w*(^ + ^r) + a]{m(x + 8r) + a}* It may be shewn, as^ in the last problem, that the sum of the proposed series is the mth part of the difference between two series whose general terms are {mx + a){m{x + r) + a}{m(x + 2r) + a}' and p(Smr -\- a) — me (nix + a){7n(x + r) + a}{m(x + 2r) + a}{7n(x + 3r) + a}' EXAMPLE. Find the sum of the series whose general term is 2.r + 3 x(x-{-lXx + 2Xx-\-Sy Here p = 2, e = 3, a = 0, m = 1, rm = 1 ; hence r = 1, and p(Smr + a) — we = 3 ; ' x( + l)(a: + 2) 21X2~2' ^"^^ '^x(ix + l)(a; + 2Xx + 3) "^ 3 ^ 172:3 ^ 6' 1/1 1\ 1 and the required sum = — ( I = -. ^ m\2 6/ 3 EXERCISE. Find the sum of the series whose general term is 4a: + 6 2 • is = -. xix+lXx +2Xx-^By "-3* 343 /)x^ -\-qx + c 542. The sum of a series, such as St — ," , ' ; ,^ — ; — r, may (a: + aXx + 6)(a: + c) be found by a method somewhat similar. Thus, assume px^ -\- qx -\- c _ Ax + -B C (.-c + a)(a: + ft)(^ + c) ~ (a: + aXa: + 6) ~ (x + aX^ + ^'X^ +Ty and by reducing the fractions in the second member to one, and equating its numerator with px' -\- qx + c, the values of A, B, and C, would be found by the principle of undetermined coefficients, as their values remain constant, while those of x vary ; and if 6 — a = c — b, then by the preceding methods, the sums of the series of which the terms in the second member are the general terms, can be found, the difference between which would be the sum of the proposed series. This method may be easily extended. REVERSION OF SERIES. 543. Wlien the value of one quantity is expressed in terms of another by an ascending series, the value of the latter may also be similarly developed by the method of the reversion, or inverse method of series. 544. Let ij = ax + hx^ + cx^ + cfe* + ... [1], the coefficients a, b, c, ... being known; in order to find the development of x in terms of i/, assume the series x = Ay-]-Bf+ Cf + By* + ... [2], in which the coefficients A, B, C, ... are undetermined. Find the values of ^^ y^, y*, ... from [1], thus : — / = aV + 2abx^ + ¥ Y" + 26c r* + ... + 2acl + 2ad f - aV + Sa^fix* + 3a6V + ... / = aV + 4a^6ar* + ... / = aV + ... Substituting these values in [2], and arranging = Aa\x + Ab jx^ 4- Ac - ll -VBd^ +2Bab + Ca^ x^ -{- Ad ■j- Bb"" + 2Bac + ZCa'b + Da* x'-\- 344 ELEMENTS OF ALGEBRA. Equating with zero the coefficients of the different powers of x (459). Aa-l = 0,Ab -{- Ba? = 0,Ac-\- 2Bab + Ca^ = 0, Ad + Bb"" + 2Bac + SCa'^b + Da"- = 0, ... from which are derived . 1 ^ Ah h ^ ac-2h^ ^ ■ c?d- 5abc + 56' Hence the development of x in terms of y is by [2], 1 6 , ac - 2Z*2 «2^ _ 5„5c + 56' . ^^^ ^ = a^-^^ — ^-y ^r—y -"•^^^- 545. If the given series has a constant term prefixed ; thus, y = a -{• ax -\- bx^ -\- cx^ -\- dx* + ... [1], then assume y — a, = z, and the expression becomes z = ax -{- bx"" + cx^ + dx* + ... [2], and the value of x developed in terms of z is found in the same manner as its expansion in terms of y was found before (544), by assuming x = Az + Bz^+ Cz^ 4- Dz* + ... [3], the coefficients of which would be found to have the same value as before ; so that if z be substituted for y in [3], the result is the required development of x ; and then y — «. being substituted for z, the result is 1 b ac — 2b- x = -(y-u)- -3 (y - .y ^^ (y - «)' aV — 5abc + 56' , ,, ^.^ -7 (y - «)* - - W- 546. When the given series contains the odd powers of x, assume for x another series containing the odd powers of y. Thus, let y = ax + bx^ -\- cx^ + dx'^ + ... to develop x in terms of y, assume x = Ay-}-By'+ Cf +Dy-'+ ... Substitute in the latter series the values of ;/, ?/', ?/, ... derived from the former in the same manner as was done in (544); then, having arranged the result according to the powers of x, the SERIES. 345 coefficients of x, x^, x^, ... being each equated with zero, the values are found as formerly to be , 1 „ b ^ ac-Sb^ ^ a'd-Sabc + Ub' and hence, 1 6 , ac - 36^ a^d - 8abc + 12&* , ^ = a^--^^ --^^^ -^^ '^ -••• The following exercises will be easily solved by substituting for a, b, c, d, ... their values in the given series, and then finding the values of A, B, C, D, ... as in the following example: — ■Lety=l-^x + ^x'' + ^x' + ^-;^oc* +, ... to find the development of a: in terms of ^. By [1] HI art. (545), it appears that here « = 1, a = 1, b = -, c = - — -, d = , ... and hence the values of A, B, C, ... ■which are the same as in art. (544), are and the value of D is similarly found to be D = —-...; and hence by [4] art. (545), _ .y_-_l _ (y-1)^ , (y - 1)^ _ (y - 1)^ , ^~ 1 2"^3 4^ -^ - The given series is the value of a^ (486) when ^ = 1, or of ear = y^ where x is the Napierian log. of 7/ ; and hence the series foimd for x is just the series for the Napierian log. of i/ or /y in terms of y, and coincides with the series for L^l + x) in art. (496), when y is taken in that series for 1 + x, and consequently y — 1 for X, and / instead of L' ; for then L'e = Ze = 1, and X'(l -\-x) = ly. 346 ELEMENTS OF ALGEBRA. EXERCISES. 1. Given the series y = x — x^ -{- x^ — x* -\- ... to find the value of X in terms of ?/, . . . x = y -\- y^ -\- 1/^ -\- y* -}- . . . 2. Given y = x — ^x^ -]- ^x^ — ^x* + ... to find x in terms oi'y, . . . ^ = 3/ + ^/ + 2^/ + 2:l7l/+... 3. Given y = x- ^x^+ ^-^f- 2. 3. /.5. 6.7 "'+ - to find X in terms of y, , 1 , , 3 , , 3.5 .3.5.79, ^==^ + 2T3^ +2TT:1^ +2717677^ +2X67879^+- 4. Given x -\- ax^ + hx^ + ex* + ... = gy + hy" + Icf + k" ••• to find ?/ in terms of ar, y = - + -^. + J. + ••• GENERAL SOLUTION OF THE HIGHER EQUATIONS HAVING NUMERICAL COEFFICIENTS. 547. In order to obtain a general solution of equations of all degrees which have numerical coefficients, it will be necessary to investigate some of the properties of equations which have not been given in the previous part of the work. Every equation involving only one unknown quantity and its powers may be reduced, by collecting its terms, transposing them all to one side, and dividing by the coefficient of the highest power of X, if it is not one, to the form X" + ^x"-* + Bx""-^ +, &c., ix* + Jfx + iV = ; where the coefficients, A, B, &c., L, M, and the absolute term N are functions of the roots, and may be either positive or negative ; or some of them may be zero, and the corresponding powers of x will be wanting in the equation ; and the exponent n represents the highest power of x in the equation, and it is called an equation of the nth degree. 548. If several simple equations involving the same unknown GENERAL SOLUTION OF HIGHER EQUATIONS. 347 symbol be multiplied continually together, the product will be an equation of as many dimensions as there are simple equations employed. Thus, the product of two simple equations is a quadratic ; the continued product of three simple equations is a cubic ; that of four, a biquadratic ; and so on to any number of dimensions. For, let X be any yffriable unknown quantity, and let the given quantities a, b, c, d, ... be its several values, so that x = a, x = b, X = c, X =:■ d, &c. ; these, by transposition, become a: — a = 0, X — b = 0, X — c = 0, X — d=0, &c. If the continued product of these simple equations be taken — namely (x — a)(x — b)(x — c) (x — d), and so on — it will produce an equation (= 0) of as many dimensions as there are factors or simple equations employed in its composition. For example — Let X — a = be multiplied into x — b = The product is Multiplied into The product is Multiplied into The product is x^-a - b X +ab = 0, a qua X - c =0 x^-a - b — c x^ + aft ■\-ac + 6c X — abc = X -d = Q 0, a cubic. — a x' + ah x" - abc X + ohcd = = 0,abiquad - b + ac -abd ratic. — c + ad — acd -d + 6c - bed -\-bd + cd &c. &c. From the inspection of these equations, remembering that a, b, c, and d, are the roots of the equation, it appears that — 549. The product of two simple equations is a quadratic. 550. The continued product of three simple equations, or of one quadratic and one simple equation, is a cubic. 551. The continued product of four simple equations, or of two quadratics, or of one cubic and one simple equation, is a biquad- ratic ; and so on for higher equations. 348 ELEMENTS OF ALGEBRA. 552. The coefficient of the first term or highest power in each equation is unity. 553. The coefficient of the second term in each, is the sum of the roots with their signs changed. Thus, in the quadratic, whose roots are + a and + h,' the coefficient is — a — & ; in the cubic, whose roots are + a + 6, and + c, it is — a — 6 — c ; in the biquadratic, whose roots are -f a, + 6, + c, + it is plain that the products will all be + 560. Hence, when the signs of all the roots of the simple equa- tions are — , the signs of all the terms of the equation compounded of them will be -f- ; and conversely, when the signs of all the terms of an eqxiation are +j the signs of all its roots will be — EXERCISES. 1. Form the equation whose roots are +3, +5, and — 2. Here a: — 3 = 0, a: — 5 = 0, and x + 2 = ; hence the equation is (x - 3)(x - 5)(x + 2) = x^ -6x'' -X + 30 = 0. 2. Form the equation whose roots are + 1, + 2, — 4, and — 7, = x*+ Sx^ - 3x2 _ Q2x + 56 = 0. 3. ... ... ... — 1, - 2, + 4, and + 7, =z X* - 8x^ - 3a;2 + 62x + 56 = 0. 4. ... ... ... +6, +5, +4,-3, and — 1, . . = x^ - Ux* + 17a;'3 + ISla:^ - 258 - 360 - 0. 5. Form the equation whose roots are 2 +3V— 1,2 — SV— !» + 4, - 3, and — 1, . . = x* - 12x^ — 169a; - 156 = 0. 6. Form the equation whose roots are +7, +3, — 8, — 4, and — 2, . = x^ + 4:x* — 63x' - 202^2 + 536x + 1344 = 0. 561. Theorem. An equation cannot have more positive roots than it has changes of sign from -|- to — , or from — to +, in the terms of its first member ; nor can it have a greater number of negative roots than of permanencies, or successive repetitions of the same sign. Let the signs of a complete equation, taken in order, be ' + + - + + + -; if this equation be multiplied hy x — a, so as to introduce one more positive root, there will be at least one more change of sign ; for in multiplying by x, the signs of the product will be the given signs ; and in multiplying by — a, the signs will all be changed, and removed one place to the right, so that they will stand thus : + +-+++- - + + - + + + - + +- + ±±-4- The number of changes in the first four signs of the product is 350 ELEMENTS OF ALGEBRA. the same as in the original equation, whether the third sign be taken — or + ; and as the fifth and sixth signs are the same as before, the changes to the sixth term inclusive are the same as in the original equation. The number of changes of sign in the original and derived equations will therefore depend on the changes from the sixth term to the end. In the original equation there is evidently but one change ; but in the derived equation, if the seventh and eighth terms be both taken with the sign +, or both with the sign — , there are two changes ; and if the seventh sign be taken — , and the eighth +, there are four ; but in the original equation, after the sixth term, there is only one change of sign ; therefore by introducing one more positive root, there is at least one more change of sign. Let now the alternate signs be changed by which the positive roots will be changed into negative, and conversely, the signs of the transformed equation will then be as follows : — + + + In this series of signs there are as many changes as there were permanencies in the former, and as many permanencies as there were changes ; if now the changed equation be multiplied by X — a, or another positive root introduced, the signs will be as follows : — + + + + ±-T=F + +- + + Examining the signs of this result in the same manner as in the former, it is found that the changes cannot be less than four or more tlaan six, whereas in tlie transformed equation there were but three ; therefore by introducing a positive root into this equation, or a negative root into the original one, there is at least one more change of sign in the transformed, or one more permanency in the original equation ; hence, since the sum of the changes of sign and permanencies of sign are together equal to the number of roots, for each is equal to n, the number denoting the degree of the equation, a positive root cannot introduce more than one change, nor a negative root more than one permanency, and it has been shewn that each introduces at least one. 562. Cor. It follows, from what has been said, that every equation has as many roots as its unknown quantity has dimen- sions. Thus a quadratic has two roots, which are either botli positive, both negative ; or one positive, and one negative. A cubic has three roots, which are either all positive, all negative ; two positive, and one negative ; or one positive, and two negative ; and the like of higher equations. GEJSERAL SOLUTION OF HIGHER EQUATIONS. 351 IMAGINARY EOOT8. 563. An equation, having no surds in its coefficients, may, however, have imaginary roots, which always enter in pairs of the form a + hfs/ — 1, and a— h/>^ — 1, and are called conjugate to each other ; these being the roots of the quadratic equation x^ — 2ax + a^ 4- &^ = 0, which contains no surds. Since these roots can only enter such an equation in pairs, a complete equation, which has an odd number of changes of sign, has at least one real positive root ; and, so far as this test is con- cerned, may have any odd number not greater than the number of changes of sign ; but if the number of changes of sign be even, it has either two, four, or six, &c. real positive roots, or none at all. Again, if there be an odd number of permanencies of sign, the equation has at least one real negative root ; but if it has an even number of permanencies, it has either two, four, or six, &c. real negative roots, or none at all. DE GUA 8 CRITERION OF IMAGINARY ROOTS. 564. If a term of an equation be wanting, and the signs of the preceding and succeeding terms be the same, it will have at least two imaginary roots. For if the sign of the absent term be supposed different from the sign of the adjacent terms, these signs will give two changes indicating two positive roots ; but if the sign of the absent term be supposed the same as that of the adjacent term, there will be two more permanencies indicating two negative roots ; hence, as the roots cannot be both negative and positive, they must be imaginary. TRANSFORMATION OF EQUATIONS. 565. To transform an equation into another which shall want the second term. Rule. Divide the coefficient of the second term by the exponent of the highest power of x in the equation, and for x substitute v, with the quotient annexed, having an opposite sign from that of the second term of the given equation ; the terms of tliis equation being collected, will be the equation sought, and its roots will be greater or less than the roots of the original equation by the quotient subtracted or added ; for if x = ^ — r, then y = x -{- r; and if X = 7/ -\- r, then y — x — r. 566. To transform an equation into another whose roots shall be either multiples or parts of the roots of the given equations. Rule. For x substitute my or -^, according as the roots required are to be parts or multiples of the origirval roots, and it will at 362 ELEMENTS OF ALGEBRA. once appear that if — represent the part that the new root is to be of the former, it is only necessary to substitute y for x, and multiply the coefficient by m raised to a power denoted by its exponent in each term ; and that if the new root is to be a multiple of the original by m, it is only necessary to substitute y instead of X, and multiply the second coefficient by m, the third by irr, and so on to the absolute term, which must be multiplied by ?h", where n denotes the degree of the equation. For example, let the given equation be x*^ + 6x^ — Ix"^ + Zx — 20 = ; and first let it be required to find an equation whose roots are an mt\\ part of those of the given equation, so that x = my, which, being substituted for x in the given equation, gives mhf + QmY — Im^ + 3/ny — 20 = 0. In this equation y = — , and therefore its roots are an mth part of X, and it is also formed by the rule. Next, let y = mx, .'. x = ~; and substituting this value it becomes ^ + 6^ - 7^ + 3^ - 20 = 0, m* wr nr m which, being multiplied by m*, to clear it from fractions, gives y" + 6?«3/' - 7my + 2>mhj - 20m* = 0, an equation which may also be formed by the rule. •> 567. To form an equation whose roots shall be the reciprocals of the roots of the given equation ; that is, let ar = -, and conse- y quently y = -, so that for every value of x there will be a corresponding reciprocal value of y. Taking, again, the above equation, and substituting - for x, it becomes _l + ±_^ + ?_20 = 0. y' f f y Multiplying by ?/*, changing the signs, and inverting the terms — that is, arranging them according to the powers of y — it becomes 20y _ 3y' + 7/ - 6y - 1 = ; an equation whose roots are evidently the reciprocals of those of GENERAL SOLUTION OF HIGHER EQUATIONS. 353 the given equation, and whose coefficients are those of the original equation in an inverted order ; hence 568. EuLE. Take the coefficients of the original equation, including its absolute term, and write them for the coefficient of y in an inverted order, changing all the signs if necessary, and the roots of the transformed equation will be the reciprocals of those of the original equation. EXERCISES. 1. Change the equation x^ + 6a:^ + 9j; — 12 = 0, into another wanting the second term, . . . = 3/* — 3x — 14 = 0. 2. Change the equation x^ — 12:c^ + 15x2 + 196a: - 480 = 0, into another wanting the second term, = y^ — Zdif + lOy = 0. 3. Given the equation x* — 12a;' + ISa:^ + 196a; — 480 =0; write the equations whose roots are double, and one-half of the roots of the given equation respectively, y* _ 24/ + 60/ + 1568y - 7680 = 0. 16/ - 96/ + 60/ + 392^ - 480 = 0. = {= 4. Change the equation x^ — ar^ — 34ar — 56 = 0, into another whose roots shall be three times as great as the roots of the given equation, . . . . = / — 3/ - 306y — 1512 = 0. 5. Change the equation x^ — x^ — Six — 56 = 0, into another whose roots shall be the reciprocals of the given equation, = 56/ + 34/ + i/-l = 0. 6. What is the equation whose roots are the reciprocals of the roots of the equation x* -}- x^ — IGx^ — ix -{- 4:8 = 0, = 48/-4/-16/+y + l=:0. 569. To transform an equation into another, whose roots shall be less than those of the proposed equation, by some given quantity. EuLE. Connect the given quantity by the sign + with any letter different from that denoting the unknown quantity in the given equation, and it will form a suhi ; as y -\- e = x. Substitute this sum and its powers for the unknown quantity, and its powers in the proposed equation, and the result will be a new equation, having its roots less by e than those of the given equation. Let the equation, whose roots are to be reduced by e, be the following : — X* + ax^ -\- bx"^ -^ ex + d = 0. And let X = y + e] then if y -\- e be substituted for x in all the 354 ELEMENTS OF ALGEBRA. terms of the equation, it is evident that y in the resulting equation will be less than x in the original equation by e, for a; = y + e ; .' . y = X — e. The substitution may be arranged as under — x' = {y + ey = y'+ 4e/ + 6ey + 4.e'y + e\ ax' = a(y + e)' = ay^ + 3aey^ + Sae~y -f ae% hx" = b(y + ef = i/ + 2bey + he\ ex = c{y + e) = cy + ce, d = d = d; .-. X* + ax^ +bx'^+cx -{-d = y*+(a + ie)y^ + (6 + 3ae + 66%^^ + (c + 2he + 3ae2 + 4e*> + c? + ce + ie^ + ae^ + e* ; but the first side of this equation is equal to zero, therefore y + (a + 4ey + (& + 3ae + Ge^)/ + (c + 26e + 3ae- + 46^)7/ + c? + ce + ie^ + ae' + e* = 0. The coefficients of the resulting equation may be derived from those of the given equation, by the following very simple process: — 570. Rule. Multiply the coefficient of the first term by e, and add the product to the coefficient of the second term ; again mul- tiply this sum by e, and add the product to the coefficient of the third term ; and so on to the absolute term, which will give the absolute term of the new equation ; repeat the same operation on the various sums as was performed on the original coefficients, stopping at the last term but one, which will give the coefficient of 3/ ; again repeat the same process on the new sums to the last but two, and the sum will be the coefficient of ?/^; repeat this process continually till the last product fall under the second coefficient, and the first coefficient, with the last sum in each of the succeeding columns taken in order, will be the coefficients and absolute term of the derived equation. The process should be arranged as follows : — • 1+a -\-b + c +c?(e the multiplier. + e + ae+ e^ + 6e+ ae^+ e^ +ce-\-be^-\-ae^+e'^ a+ e 6+ ae+ e^ c+ be+ ae^+ e^ c?+ce+6e^+ae'+e'' + e ae+2e''^ + 6e+2ae'^+3e' a+2e &+2ae+3e2 c+26e+3ae^+4e' + e + ae + 3e2 a+3e 6+3ae+6e2 + e a+4e ,-. 1 + (a + 4e) + (6 + 3ae + 6e') + (c + 2be + ^ae" + 4eO -\-ld+ce + be" + ae' + e*) GENERAL SOLUTION OF HIGHER EQUATIONS. 355 are the coefficients and absolute term of the transformed equation, which has its roots less than those of the given equation by e, and they are the same as those formerly found. This is a very important process, as it contains the substance of Horner's solution of equations of the higher degrees. 571. Cor. It is evident that an equation may be found in the same manner, whose roots shall be greater than those of the given equation by e, if the sign of e be changed from + into — ; that is, by changing in the above series the signs of all the terms which contain odd powers of e. EXAMPLES. 1. By the above, find the equation whose roots are less by 1 than those of the equation a;* + 7x'^ — 5a: — 12 = 0. 1+7 — 5 — 12(1 = the multiplier. + 1 + 8 + 1 + 8 + 3 + + 3 9 - 9 + 9 +12 + 1 + 10 .•. y^ + 10^^ + 12?/ — 9 = is the equation sought. 2. Find the equation whose roots are greater than those of the equation x* - Sx^ - COa;^ + 127x + 840 = by 2. Here, applying the Cor. (571), the multiplier 2 will be — , and the work as follows : — 1_3 —69 +127 + 840(— 2 = the multiplier. - 2 +10 +118 - 490 — 5 - 2 -59 + 14 + 245 + 90 - 7 - 2 -45 + 18 + 335 - 9 - 2 -27 350 - 11 r.y* — llf — 27f + 335?/ + 350 = the required equation. 3. Diminish the roots of x* — lOx^ + 4:x^ - 20a: — 78 = 0, by a quantity greater than the greatest positive root. By the above process the roots may be diminished by 1, and those of the resulting equation again by 1, or any greater number, 356 ELEMENTS OF ALGEBRA. till the signs • of the coefficients and the absolute term be all positive, then (560) the roots will all be negative ; and hence the roots will have been diminished by a quantity greater than the greatest positive one. 572. If, in the course of the operation, the absolute term should become zero, then the quantity by which the roots have then been di- minished is a root of the equation. The operation will be as follows : — 1-10 +4 - 20 - 78(1 • + 1 -9 -5 -25 - 9 + 1 - 5 8 - 25 13 - 103 - 8 + 1 — 13 7 38 - 7 + 1 20 - 6 1-6 + 8 + 20 16 + 38 32 70 608 - 103(8 -560 + 2 + 8 + 4 80 -663 + 10 + 8 + 76 + 144 + 538 + 18 + 8 + 220 + 26 1 + 26 + 1 + 220 + 27 + + 538 247 - 663(1 + 785 + 27 + 1 + 247 + 28 + + 785 275 + 122 + 28 + 1 + 275 + 29 + 1060 + 29 + 1 + 304 + 30 1 + 30 +304 + 1060 + 122 ; .'. y + 30/ + 304/ + 1060^ + 122 = is the equation sought. GENERAL SOLUTION OF HIGHER EQUATIONS. 357 And since the roots have been diminished by 1 + 8 + 1 = 10, the greatest positive root of the equation is less than 10 ; but when the roots were diminished by 9, there remained one change of sign indicating one real positive root ; therefore tlie above equation has at least one positive root, which lies between 9 and 10. EXERCISES. 1. Find the equation whose roots are less by 1 than those of the equation x^ -^'7x^ - 4x - 12 = 0, =f + 10/ + 1% -8 = 0. 2. Find the equation whose roots are less by 2 than those of the equation x* — lOx^ + Ax^ — ox -\- 24: = 0, = y* - 2f - 32 f - lly - 34 = 0. 3. Find the equation whose roots are greater than the roots of the equation x* — 4x* — 19^^ + 46x + 120 = 0, by 4, = / - 20/ 4- 125/ - 250y + 144 = 0. 4. Find the equation whose roots are less by 3 than those of the equation x' — 12a;* + Ix^ — SOx^ + 51a: — 40 = 0, = / + 3/ - 47/ - 345/ - 831y - 697 = 0. 5. Find a number greater than the greatest positive root of the equation a:^ — 9a;^ + 26a: — 23 = 0, . . . . = 4. 6. Form the equation whose roots are greater by 2 than those of the equation 2a;^ — 5ar^ -1- 3a: — 1 = 0, = 2/ - 16/ 4- 43/ - 41^/ + 5 = 0. BTTDAN's theorem for the discovert or IMAGINARY ROOTS. 573. If the roots of an equation be reduced by any quantity ?•, and in performing the operation there be m changes of sign lost ; and if, in reducing the roots of the reciprocal equation by -, there r be n changes of sign left, then between the interval r and there are at least »i — n imaginary roots. For in reducing the roots of the equation by r, all positive roots less than r will have become negative, and there will be as many positive roots between and r as there are positive roots changed into negative — that is, as there are changes of sign lost ; whereas in reducing the reciprocal equation by -, which must be less than r any of its positive roots between - and oo, no changes should be lost ; but if there are imaginary roots, changes may be lost ; and the changes of sign left in this equation, deducted from the changes lost in the former operation, will indicate the number of imaginary roots between the assumed limits ; that is, w — n = the imaginary roots. 358 ELEMENTS OF ALGEBRA. 574. Cor. If the reducing quantity r be greater than the greatest positive root, all the roots of the depressed equation will be negative, and the changes of sign will all be lost ; whereas in reducing the reciprocal equation by -, all the changes ought to r remain if the positive roots are real ; then those changes that still remain, deducted from those originally lost, will leave the number of imaginary roots that are apparently positive. If the signs of the alternate terms be changed, the negative roots will be changed into positive, and the positive into negative ; and if the same operation be performed on the equation so changed, the number of apparently negative imaginary roots will be discovered.* Eind, by Budan's and De Gua's theorems, the number of imagi- nary roots of the equation x^ + ^x* + 2x^ — Sx^ — 2x — 2 = 0. Since this equation is of an odd degree, and its absolute term is negative, it has at least one real positive root, and as there is only one change of sign, it cannot have more than one ; therefore, changing the signs of the alternate terms, we apply Budan's theorem to ascertain the number of imaginary negative roots. The equation tlms changed becomes x^ — 3x* -{- 2x^ -{- Bx'^ — 2x -\- 2 = 0. Coefficients of the Given Equation 80 changed. Coefficients of the Reciprocal Equation. 1-3 +2 +3 +1-2+0 - 2 + 3 + 2(1 + 1 2-2+3 + 2 ± + 2-3+1(1 +3+5+2 -2 +0 +3 + 1-1-1 + 1 + 2 + 3 ±0+3 + 2+2 + 5+2+3 + 5+10 -1-1+2+3 + 1 ±0-1 + 2 + 5 + 2+4 + 10 +12 + 9 ±0-1+1 + 1 +1 + 4 + 9 + 2+6 + 19 + 1 ±0 + 1 + 6+15 + 2 + 2 + 8 1 + 2 ±0 +1 + 3 + 3 2 + 8 +15 + 19 + 12 + 3 * On the preceding theorem, together with that in (5fi4), which is De Gua's criterion, the late Professor Davies remarks, in the appendix to the Ladies' Diary of 1839 : ' I have found the criteria of De Gua and Budan quite equal to the detection of imaginary roots of every algebraical equation to which I have applied them.' GENEBAL SOLUTION OF HIGHER EQUATIONS. 359 Here, in depressing the roots of the direct equation by 1, there are 4 variations of sign lost ; for, by De Gua's criterion, since the third coefficient is zero, and the adjacent terms have the same sign, this indicates two imaginary roots ; hence that term may be taken with a + sign, and therefore there are four variations lost ; and in the reciprocal equation, when depressed, there are no variations left ; . • . 4 — = 4 is the number of imaginary negative roots. EXERCISES. 1. Find how many roots are real, and how many are imaginary in the equation x^ — lOz' + 6x + 1 = 0, The roots are all real ; 2 positive, and 3 negative. 2. Has the equation a:* — 4x^ + Sx^ — 16a: + 20 = any real roots ? None. 3. How many roots of the equation ot^ — x* — 15x^ + 38a;^ — 26a: -f- 6 = are real, and how many imaginary ? All are real. 4. How; many real and imaginary roots has the equation X* + 12x'V 29x2 _ 16a: + 2 = 0, . . .All are real. DEVELOPMENT OF AN INTEGRAL FUNCTION OF A BINOMIAL X +y. 575. Let the integral function of x be fx = Ax'' + 5x«-^ + Or"-'' + ; ... by writing x -\- y instead of x, it becomes f(x+y) = A{x 4- yr + B{x + yy' + C{x + t/)"-^ + ; ... and if we develop the powers of the binomial x -\- y, according to the decreasing powers of x, we find 1.2 ^' &c. A{x-\-yy — Ax^ + «J,.T"~^|y+n(n— l)^a;""2 +Blx+yy-^= +5x"-»+(7i-l)5x-"-2 +(rt-l)(w-2)Sx»-' + Cix+yy-^= + Cx"-2 +(n-2)Cx"-^ +(„_2)(n_3)Cx"-* + Let us now put for abridgment X = Ax" + Bx''-' + Cx"-2 + ... X' = nAx"-^ + (n - l)5x"-2 + (n - 2)Cx"-^ + ... X" = n(n - l)^x«-2 + (n - l)(n - 2)S.r»-3 + (n - 2)(n - 3) Cx^-* + ... 860 ELEMENTS OF ALGEBRA. Then the preceding result will be expressed as follows : — Xo: + 3/) = X + X'y + yt/ + y:^ + • • • Now X is the original expression, which was represented hy/c; X! is derived from X by multiplying each of the terms by the exponent of x in that term, and diminishing the exponent by unity ; X." is derived in the same way from X' — namely, by multiplying each terra of X' by the exponent of x in the term, and diminishing the exponent by unity ; in the same manner, by extending the expansion to more terms, it can be shewn that X" is derived from X'\ in the same manner as X' from X. 576. The polynomial X is called the derivative function of X; and X" is called the derivative function of X\ or the second deriva- tive function of X ; and so on. X' is also called the limiting ratio of the function X, and is a quantity of very considerable import- ance in the theory of equations, being the same as the first differential coefficient, which becomes the object of research in the differential calculus, and also identical with the coefficient of y, as found by Rule (570), if x be put for e. Since X has been designated by X^)» -^'» ^"» -^'"> ^^^^ ^^ "^'^^^ appropriately designated by f'{x), f"(x\ /'"W ; ^^d using this notation, we obtain for the development sought, As the exponent of x diminishes by unity in passing from the given polynomial to its derived function, or from any derived function to the following, a polynomial of the ?zth degree will have n derivative functions, of which the last will not contain x. It is easy to see, besides, that if the first term of the polynomial is represented by Ax''\ as above, the nth derivative function will be 1.2.3.4 ... n. A Consequently, the law of the terms which compose the develop- ment oijlpc + y) gives for the last term Ay"; which may be seen otherwise ; for it is clear that, after the substitution of x -\- y instead of x in the polynomial Ax" + Bx"~^ + Cx"~" +, ... the quantity ylx", which becomes A(x + y)", gives the term Ay% and that there is no other term in which the exponent of y can be either equal to or greater than n. 577. To apply to a practical example what has been shewn concerning the derived functions, let fQt) = x^ + 5a:* + x^ - IGx' - 20x - 16. If we wish to find what the above polynomial becomes when x is replaced by y — 1, we will calculate the successive derivatives. GENERAL SOLUTION OF HIGHER EQUATIONS. 361 The first derivative is fix) = 5x* + 20^3 + Sx'' - 32:r - 20. In forming the second derivative, and dividing it by 2, we have 4-^ = lOx' + 30x2 + 3^ _ 16.' The derivative of this last function, divided by 3, is f^- = lOx^ + 20:r + 1. The derivative of this, again, divided by 4, is 1.2.3.4-^'' + ''- Repeating the same process, and dividing by 5, we have 1.2.3.4.5 By substituting in these various functions — 1 for x, we find /(- 1) = - 0, /'(- 1) = 0, ^^ = 1, 4^!t^ = - ^' 1.2.3.4 = ^' 1.2.3.4 T5 = ^ 5 ^"^ consequently y(y-l)=/-%3+/-9. Employing the same process to find what the polynomial 2x* — 5x^ -{- 3x — 1 becomes, when i^r — 2 is substituted for x, we obtain 23/* - IG/ + 43/ - ily + 5. VALUES OF AN INTEGRAL FUNCTION OF Ol FOR VERY GREAT OR VERY SMALL VALUES OF 0'. CONTINUITY OF INTEGRAL FUNCTIONS OF ONE VARIABLE. _^78. If in the polynomial Ax"" + Bx"" + Cx^ +, &c. the exponents m, k; p, &c. being positive integral numbers forming a decreasing series,, we give to x numerical values sufficiently great, either positive or negative, the values of the polynomial will have the same sign as that of its first term Ax'" ; and a value can be given to X so great, that the value of the polynomial may be as great as we please. The polynomial given above may be written as follows : — 362 ELEMENTS OF ALGEBRA. For a very great value of x, the fraction — ^-, — — -, &c. will all TJ -1 have values very small, consequently the sum of the terms -r- . — ^-, ■A X C 1 -r . , &c. of which the number is necessarily limited, will also have a value very small ; so that, if x is made sufficiently great, the polynomial enclosed in parentheses will have a value differing very little from unity, and which will consequently be positive. The given polynomial will therefore take values of the same sign as the term Ax'". We see also that the value of the polynomial may be made as great as we wish, since the factor J.z'" increases indefinitely with x. 579. If in the polynomial Ax"" + Bx" + Cx^ +, &c. the expo- nents m, n, p, &c. being positive integral numbers, which form an increasing series, and the polynomial being composed of a limited number of terms, we give to ar a very small value either positive or negative, the polynomial will have a very small value, of the same sign as that of its first term Ax"". The given polynomial may be written as follows : — Ax^l + -jx''-"' + -jx^'"" + &c.) For a very small value of x, the quantities x"~% x^~"^, &c. of which the exponents are integral and positive, will all have values very small ; so that if we make x converge towards zero, the polynomial enclosed in parentheses will take a value differing very little from unity, and which will consequently be positive. The given polynomial will then take values of the same sign as that of its first term Ax'^'. We see also that the value of the polynomial can be made as small as we please, since the factor Ax'" decreases indefinitely with x. 580. Having given an integral function /(.r), and a particular value a of x, a quantity h can be found so small, that the difference f(a + A) —jXa) shall be less than any given quantity, however small that may be. For, if in (576) x and 3/ be replaced by a and h, and /(a) transposed to the first side, we have /(a + A) -/(a) =fXa)h +/"(«)^ +/'"^")nl73 + - Now, if a very small value be given to A, each of the terms A^ /'(a)A, /"(a)^j — -, &c. will have a very small value ; and as the number of these terms is limited, A can be conceived so small, that GENERAL SOLUTION OF HIGHER EQUATIONS. 363 the sum of all these terms may be as small as we please, which establishes the proposition. 581. Scholium. Let a and 9> be two given numbers, /B being greater than «. If we wish to determine the quantity li in such a manner that for any value a of x, comprised between « and iS, we may have /(a + li) — f(a) ^ s, indicating by £ a quantity as small as we please, it may be obtained as follows : — Putting C for a number greater than any of the coefficients which enter into the polynomials/'(a-), — -— ,"'^— — -^, &c. ; each of these coefficients, for a value of x comprised between «. and jS, will be less than C/S"*"^, if /B z' 1, and less than C, if /3 ./ 1 ; conse- quently the value of each polynomial will be less than mCli™~\ less than mC, according as /3 7^ 1, or /3 .^ 1. This being established, writing K for the number mC^'^~^, if /3 7^ 1, and for the number mC, if iS ^ 1, and developing the function j^a + A) —J{a), we have, for any value a of x, comprised between a. and li, J{a + h) -f(a) ^ Kh(\ + A + A^ + ... A-^. The second member of this inequality is equal to — ^^ — - ; consequently, if we suppose A ^ 1, this second member will be less than 7. If now — ^ s, still more will /(a + A) — /(a) ^ j. Now the first condition is satisfied when A ^ -^ . If then values be given to x, increasing from a up to ^, and such that the difference of two consecutive values may be equal to or less than the quantity „_^ -, we will obtain values of /(x) such that the difference of two consecutive values will be less than s. 582. When two numbers « and /S, being substituted for x in the first member of an equation J{x) = 0, give results with contrary signs, the equation has at least one real root comprised between u, and /3. From what has been established in the preceding article, we can give to X values, increasing from « up to /3, such that the difference between each of the corresponding values oifipc) and the following may be less than a quantity i as small as we please. There are necessarily two consecutive values of f(^x') which have contrary signs, since, by hypothesis, the two extreme values /(«) and J{(h) have contrary signs. These two consecutive values oifix), which 364 ELEMENTS OF ALGEBRA. have contrary signs, and differ from each other by a quantity less than £, are each numerically less than s. Now this conclusion holds true, however ' small i may be; and hence there exists between « and /3 at least one value of or, for which fix) is zero, ox fix) = 0. Scholium. As it is possible that the polynomial fi^x) may pass several times from positive to negative, or from negative to positive, while x varies from « to /3, the equation f(x) = may have several real roots comprised between a and /3. 583. An equation of an odd degree has at least one real root, of which the sign is contrary to that of its last term. Let x"' + Ax"^~^ + Bx'^~^ ... + i^ = 0, in which 7n, the exponent of the highest power of x, is an odd number. When the sign of the absolute term is negative, if in the first member of the equation x be made equal to zero, the sign of the result will be negative ; for it will be no other than the last term. Again, if a positive value be given to x so great that the sign of the first member will be the same as that of its first term, the sign of the result will be positive. The equation has then at least one positive root. If the last term be positive, and x be made equal to zero, the result will be positive ; but if a negative value be given to x sufiiciently great (578), the result will be negative, and the equation has in tliis case at least one negative root. 584. An equation of an even degree, of which the last term is negative, has at least two real roots — the one positive, and the other negative. For by making x — 0, the result will be negative ; and if a value be given to x sufficiently great, whether that value be positive or negative, the result will be positive, since it will have the sign of the first term (578), which, being of an even degree, will always be positive. 585. If a is a root of the equation f{x) = 0, the first member of the equation is divisible by a; — a. Let the general equation y(ar) = be of the form x™ + ^jX"-' + A^x""'^ + ... + A„,_^x -\-A,„ = 0. If a is a root of this equation, we have the equality, a"* + ^^a™-^ + A^a""-^ + ... + A„_^a + A„, = 0. From this latter equation, it is evident that Arn= — a™ — A^a""'^ — A^a""'"^ — ... — Ar„_ya. Substituting this value of A^ in the original equation, and GENERAL SOLUTION OF HIGHER EQUATIONS. 365 arranging the powers of a along with the corresponding powers of X, we obtain the following : — X'" — a"* + Ai(x"'~^ — a'""^) + A^^x"'''' — a'""^) ... + A„,_^(x -a) = 0. But X — a is a divisor of each of the binomials (x"" — a'"), a;'""^ — a™"\ &c, (87, Theo. II.) The first member of the equation f(^x) = is therefore divisible by a; — a, when a is a root of that equation. To obtain the quotient, it is only necessary to divide each of the binomials x"" — a"", x'""^ — a"'~\ &c. hy x — a, and to add after- wards the partial quotients, multiplying the second by A^, the third by A^, the fourth by A^^ &c. and the result becomes the following : — -\-A^a + a^ + A,a' + A^a + A + A,a-^-' + A^a""-* + ^. The coefficients of the terms of the quotient can be obtained, beginning at the second, by multiplying the coefficient of the preceding term by a, and adding to the product the coefficient of that term in the original equation, which is of the same rank as the term of the quotient which is to be formed. The coefficient of the first term of the quotient is the same as that of the first term of the original polynomial. LIMITS OF THE ROOTS OP AN EQUATION. 586. If a number be determined such that by putting it, or any greater number, instead of x in the first side of an equation, the result will be positive ; it is clear that it will be a superior limit of the roots. Let the equation + K=0 have the first term x"* positive, and the other terms indifferently positive or negative ; if N indicate the greatest negative coefficient, the first side will be rendered positive by satisfying the following inequality : — ^m -^ N^m-l 4_ ]^y.m-2 _|_ ___ ^ _^^ _^ JV^. the second side = iV^x"*"' + a;™"* -f ... + a: + 1) _ N{x^ - 1) 366 ELEMENTS OF ALGEBRA. The above condition is then X — i which will be verified if a: — 1 = iV, or x =: \ -{■ N, and still more if x /" 1 + iV ; hence in every equation a superior limit of the roots may be obtained by adding unity to the greatest negative coefficient of the first member of the equation. It is also evident, that since by changing the signs of the alternate terms of an equation, the positive roots become negative, and the negative roots positive ; if a negative value be given to x either equal to or greater than one +, the greatest negative coefficient of the equation so changed, an inferior limit to the roots will be obtained. 587. When the term of degree immediately inferior to that of the equation is not negative, a limit less than the preceding may be obtained. Let the equation be x"^ ... - Fx"*-" ± Gx^-''-^ ... ±K=0, the term — Fa?™"" being the first negative term, and the following terms being indifierently positive or negative. Designating by N the absolute value of the greatest negative coefficient ; the first member of the equation will be rendered positive if the following inequality be satisfied : — a;"" 7^ iVo;"''" + iVx*"""'^ ... + iV, or x'" /- ; . X — 1 K we suppose a: :7=' 1, it will be sufficient that we have a:'" z^ — ZTT"' ^'^ ^"'~\^ — ^)^ N. Also the last condition will evidently be fulfilled if we have {x — ly-^ (a: — 1) = or 7^ iV; that is, {x — l)** = or ^ N; whence a: = or 7^ 1 + ^N. A superior limit to the roots can therefore be obtained by adding unity to the root of the absolute value of the greatest negative coefficient, of which the exponent is the difference between the degree of the equation and the exponent of the first negative term. If, however, the coefficient N be less than unity, the limit 1 + xV formerly obtained will be preferable to that which has been obtained in this article. GENERAL SOLUTION OF HIGHER EQUATIONS. 367 THEORY or EQUAL ROOTS. 588. When an equation has equal roots, it may be reduced to several other equations more simple, in which each root is only- found once. To discover these equations will be the object of the present article. Let the equation be a;m ^_ J^^m-l _|. ]g^m-2 _^ ^^^ -^ Hx + K = 0. Representing by/(x) the first member of this equation, and by J^^x) its derivative function, or the polynomial. mx""-^ + (w — l)Ax'"-^ + (wi - 2)JBx"'-3 -{-...+ H. The polynomial /'(ar) is the coefficient of the first power of i/ in the result which was obtained by writing a; + ^ in place of a: in ' the polynomial /{x) (575) ; now, in representing the roots of the equation by a, b, c, ... k, we have J{x) = (x — a)(x — bXx — c) ... (x — k); then, by writing x -\- y in place of x, J{x+y) = {y + x- aXy + x - b^y -^ x - 6) ... (y -}- x - k). The second member of the last equality may be regarded as the product of m binomial factors, which have for their first term y, and of which the second terms are x — a, x — b, &c. Consequently the coefficient of the first power of y in this product is the sum of all the products of the quantities x — a, x — b, &c. taken m — 1 at a time. Besides, to form these products, it is sufficient to divide successively jf{x) by each of the factors x — a, x — b, &c. We have then Let nowX^:) = {x — a)\x — by(_x — c)« ... To apply this to the proposition which is to be established, the quotient of J{x) by x — a, must be repeated n times, for there are n factors x — a, the quotient of J{x) hy x — b must be repeated p times, the quotient of f(x) hy x — c must be repeated q times, &c. ; thus we obtain /'(ar) = n(x - ay-\x - by(x — c)' ... + jo(ar — ayix — by-\x — c)« ... + q(x - ay(x — by(x - c)'"^ ... + ... Putting D = {x- ay-\x - by-\x — cy-\ ... D divides /(jjc) and f'(x). It is the greatest common divisor of 368 ' ELEMENTS OF ALGEBRA. these two quantities, otherwise it would be necessary that one of the factors of f(x) should divide the quotient oif'(x) by Z), which is n{x — h'){x — c) ... -\- p(x — d){x — c) ... + q{x — a){x — b) ... + ... Now, each of the factors x — a, x — b, &c. divides all the parts of this sum, with the exception of one ; consequently the sum is not divisible by any of these factors. 589. Therefore, when an equation has equal roots, the first member of the equation and its derivative function have an algebraic common divisor, and this common divisor is the product of all the factors corresponding to the equal roots of the equation, each raised to a power less by unity than its power inj^x). When the equation has no equal roots, the first member and its derivative function have not any common algebraic factor ; for the exponents n, p, q, &c. being unity, the greatest common divisor of/(a;) and/'(a;) is reduced to a number. Sturm's theorem. 590. Let F = be an equation of any degree of which all the roots are unequal, and let Fj be the derivative function or limiting equation of F. We operate as if the object were to find the greatest common divisor of F and Fj, with this single difference, that we must change the signs of all the remainders before they be taken for divisors. This change of signs, which would be a matter of indifference if we only wished to find the greatest common divisor, is essential in the present case. Indicating by Qi the quotient of the division of F by F^, and by — Fg, the corresponding remainder, we will have F = y,Q - V,. Dividing Fj by V^ — that is to say, by tlie remainder of the first operation taken with a contrary sign, and indicating by Q^ the quotient, and by — F3, the corresponding remainder, we will have again F, = VA -Vr ^ By continuing thus, we will necessarily arrive at a numerical remainder — V„ since we have supposed that the equation V= has no equal roots (588), and we will have this series of relations : F = V,Q, - V„ F, =FA F, =F3Q3 F^x= F„Q„ - v„,„ ■p;..= F.,a- I - T^.. GENERAL SOLUTION OF HIGHER EQUATIONS. 369 If multipliers be introduced in the course of the operation to avoid fractional quotients, these multipliers must be positive, in order to avoid a change of sign in the remainder. The above relations being established, the consideration of the functions V, V^, V^, ... has led M. Sturm to the following theorem : — 591. Theorem. When any two numbers « and /S positive or negative, of which « is less than jS, are substituted instead of x in the series of functions V, V^, V^, ..., K_i, F„ the number of variations of the series of signs of these functions for x = (i will be at most equal to the number of variations of the series of signs of these same functions for a; = « ; and if it is less, the difference will be equal to the number of real roots of the eqiiation V = comprised between « and /3. To demonstrate this theorem, it is necessary to examine how the number of variations formed by the signs of the functions ^j V^v ^2' •••5 ^r, disposed in the order indicated, for any value of X, can alter when x passes through different degrees of magnitude. Now, there can only be a change in this series of signs by making X increase, so that one of the functions V, Fj, V^, &c., may change its sign, and consequently become zero. It presents, then, two cases to examine, according as the function which becomes zero is the first V, or one of the intermediate Fj, Fg, &c. ; for it is not the last function F, which can become zero, since F,. is a number. Case I. To examine the change which takes place in the series of signs, when x, increasing by insensible degrees, attains and passes a value which renders F equal to zero. If that value of Xj which we designate by a, be substituted in the derivative function Fj, that function will become a number positive or negative, since, by hypothesis, the equation V = has no equal roots. Representing by ^l a positive quantity so small that the equation Fi = may not have a root comprised between a — u and a -\- u; Fj will have the same sign when x = a — u, x = a and when X = a -\- u. This being established, let us designate for a moment F by J{x) and Fj by/'(a:), we will have, by observing tha.t jf{a) = 0, Xa - «) = - If (a) + ^/"(«) - r:^/'"(«) + - As there is no limit to the smallness of u, we can make u so small, that the sign of the development of J{a — u) may depend on the sign of the first term (579) ; so thaty(a — v) will have the same sign as — vf'{a), and consequently it will have a sign contrary to that of fia) ; now, /'(a) and/'(o — u) have the same X 370 ELEMENTS OF ALGEBRA. sign, therefore y(« — «) and /'(a — u) have contrary signs. Hence V and Vj^ will have contrary signs for x = a — u. By changing — u into + m in the preceding development, we have and we see here, the same as above, that J^^a + u) will have the same sign as / (a), and consequently the same sign as /'(a + ^)' Hence V and Fj have the same sign for a; = a + m. Therefore, if for x = a the sign of fXx) or of Fi is +, the sign of F will be — for x = a — u, and it will be + for x = a -\- u. If, on the contrary, the sign of Fj is — for ar = a, the sign of F will be 4- for x = a — u, and — for x = a -\- u. These results lead to the construction of the following table : — V V, V V, = a — u — -{- \ /'+■"■ For ■< a: = a + > or ■( X = a -\- u + + Consequently, when a is a root of the equation V = 0, the sign of F forms with the sign of Fj one variation before x attains the value a, and this variation is changed into a permanency after x has passed that value. As to the other functions F^, Fj, &c. each of them will have, like Fj, either for x = a — u, or for x =^ a -\- u, the same sign as they have for x = a, if none of them vanish for x = a, at the same time as F. We will now examine what will take place when one of these functions vanishes. Case II. Let F„ be the intermediate function which becomes zero for x = b. This value of x neither reduces to zero the function F„.i which precedes F„, nor the function F„^j which immediately follows it ; for if it did, the factor x — b would divide at the same time two consecutive remainders, F„_j and F„, or F„ and V„^^; consequently x — h would be a multiple factor of the polynomial F, which is impossible, since we have supposed that the equation F= has no equal roots. Besides, from the equality F„_j = VnOm — Vn+v wMch is ouc of the relations of (590), F„ being nothing for x = b, we have F„_, = — F„+i ; therefore F„_i and F„+j have contrary signs when x = b. This being established, by substituting instead of x two numbers, b — u and b -\- u, very little different from b ; the functions F„_, and F„+i will have for these two values of x the same signs as they have for x = b, since u can be taken so small that neither of the functions F„_i nor V„^^ shall change its sign while x pass^^" GENERAL SOLUTION OF HIGHER EQUATIONS. 371 the interval from b — u to b -\- u. It follows from thence, that whatever may be the sign of V„ for x = b — u, as it is placed between the signs of F„_, and F^^j, which have contrary signs, the signs of the three consecutive functions V„_■^, V„, and F„^i, when X = b — u will always form a permanency and a variation, or a variation and a permanency. It can be proved in the same manner, that whatever may be the sign of F„ for a; = 6 + w, the signs of the three consecutive functions F„_j, F„, and F„+i, when X = b + u, can only present one variation. Thus, the series of signs of all the functions Fj, ..., F^, for X = b -\- II, will contain precisely as many variations as the series of signs of these functions for x = b — u. Therefore, when any intermediate function passes through zero, the number of varia- tions in the series of signs is not changed, unless the value of x, which reduces this intermediate function to nothing, should also reduce to zero the first function F; in this case, the change of sign will cause one variation to disappear from the left of the series of signs, as was proved in the first case. It is clear that the same conclusion will subsist if several intermediate functions, not adjacent, become nothing for the value x = b. It is therefore demonstrated, that each time that the variable x, increasing by insensible degrees, attains and passes a value which renders V = Q, the series of signs of the functions F, Fj, V^, ..., F„ loses one variation formed by the signs of F and of Fj, which is replaced by a permanency; while the changes of signs of the intermediate functions l\, V^, ..., F^.j, can never either augment or diminish the number of the variations. Consequently, if we take any number a, positive or negative, and another number j3 greater than «, and if we make x increase from « up to /3, whatever number of values of x there are comprised between a and /3 which will reduce F to zero, so many will the series of signs of the functions F, F,, Fg, ..., F^, for a: = jS, contain variations less than the series of signs of these functions for x = a. The prin- ciple which has just been stated, is no other than the theorem which was to be established, expressed in other words. 592. Scholium. It may happen that one of the functions Fj, Fg, F3, ..., Fr_i, becomes nothing, either for x = x, or for x = (i. In this case, it is sufficient to consider the variations of the series of the signs of all the functions, without having regard to that which vanishes. For it has been shewn, that when the function F„ becomes nothing for x = a, if, instead of x, a quantity very little different from « be substituted, the signs of tlie three functions F„_i, V„, V„^^, will always present one variation and one permanency ; and the variation will still be found when the ^"•^rtion F, is omitted. 372 ELEMENTS OF ALGEBRA. 593. Cor. I. To know the number of real roots of an equation V = 0, it is necessary to substitute, in all the functions, two quantities, a and /S, between which all the roots may be comprised. Now, it is always possible to give to a: a value so great that each of the functions F, F„ V^, ..., F,, may have the sign of its first term (578) ; therefore, if we consider at first the signs of the first terms of the functions by supposing x negative, and afterwards the signs of these same terms by supposing x positive, the excess of the number of variations of the first series of signs above the number of variations of the second series will be precisely the number of real roots of the equation. 594. Cor. II. The theorem of M. Sturm furnishes the necessary conditions, that all the roots of an equation may be real. The equation V = being of the mth degree, it is necessary that the series of signs of the first terms of the functions F, Fj, V^, &c. should present m variations ; this requires that the number of functions should not be less than m + 1 ; now, this number cannot surpass m-\-l, since the degree of each function is inferior but one unit to that of the preceding function ; therefore it will he m -j- 1. It is necessary, besides, that the first terms of the functions, by supposing X positive, should all have the same sign. These conditions are sufficient ; for if the exponents of the first terms of the functions only diminish by unity, from each function to the following, they will be alternately even and odd ; and if the signs of these first terms only present permanencies when x is positive, they will only present variations for x negative. The number of real roots will then be equal to m. 595. Cor. III. As often as the number of auxiliary functions Fj, Fg, Fj, &c. is equal to m, the number of imaginary roots of the equation V = can be known by the simple inspection of the signs of the first terms of these functions. The equation V= has as many pairs of imaginary roots as there are variations in the series of signs of the first terms of the functions F, Fj, Fg, &c. up to the constant F„ inclusively; for if this series present n variations, it presents m — n permanencies. When x is changed into — X, the permanencies become variations, and vice versa, so tliat the number of variations is m — w ; the number of real roots is therefore m — 2n. To find the number of real and imaginary roots in an equation 3^ + ax"'-^ + hx''~^ + ... +t = Q. 596. Denote the equation by F, and its limiting equation by Fp and perform on tliese quantities the process of finding their greatest common measure ; change the signs of all the terms of the remainders before using them as divisors, and denote these divisors in order by T^, F3, F^, ..,, F„; the final remainder, with its sign changed F„, will be independent of x. GENERAL SOLUTION OF HIGHER EQUATIONS. 373 Substitute — oo in the first terms of the quantities V, Fp V^, ..., V„ ; write down in order the signs of the results, and let the number of changes in these signs from + to — , and from — to +, be denoted by m ; and when + co is similarly substituted, let the number of variations be r ; then ni — r is the number of real roots, and n — (m — r) the number of imaginary roots. Given the equation x* — 2x^ — a;^ + 8x — 12 = to find the number of its real and imaginary roots. Here V = x* - 2x^ - x^ + Sx - 12, and Fi = ix^ - 6x^ - 2x + 8 ; and performing on these quantities the process of finding their greatest common measure, the first remainder is — 5x^ + 23x — 44 ; hence V^, the next divisor is 6x^ — 23a: + 44 ; and the next remainder is found to be 248x — 1264, which, being divided by 8, and the sign changed, gives V^ — 31a: + 158; and the last remainder is a positive number, or F^ = — ; hence + 8x- 12, V = x* _ 2x' - X' Vi = 4x'- Qx^- 2x v.= hx"" - 23x + 44, Vs = - 31a: + 158, Since V^ is a number which does not alter its value by the substitution of any quantity for x, it is unnecessary to write the number, as only the signs of the results are required. When — oo and + oo are substituted in the first terms of these five quantities, the signs of the results are in order of the functions IS F, V, F„ F3, F„ for a: = — oo, + - + + -.. ,. 3 variations, for a: = + oo, + + + .. . 1 variation. In the former series of signs, the number of variations is three, and in the latter only one; hence in = 3, and r = 1, and m — r = 3 — 1 = 2 = the number of real roots. Hence the number of imaginary roots is n — (m — r) = 4 — 2 = 2. 597. To find the number of roots that lie between any two numbers. Substitute these numbers for x in all the terms of the functions F, F„ Fg, ..., F„, and write down in a row the signs of the results : and the difference between tlie numbers of variations 374 ELEMENTS OF ALGEBRA. of signs in these two rows will be the number of roots which lie between the two assumed numbers. To find the number of roots of the preceding equation that lie between 1 and 3. Substitute these numbers for x, and the signs of the results are V V V V V for X = 1, — + 4- + — ... 2 variations, for x = 3, + + + + — ... 1 variation. The difference of variations =2 — 1 = 1, or only one root lies between 1 and 3. This root will be found on trial to be 2. EXERCISES. 1. Find the number of real roots of the equation 8x'— 6x— 1 =0, The three roots are real ; there are two roots between and — 1, and one between and 1. 2. Find the number of real and imaginary roots of the equation a^ - 5x2 + 8a; - 1 = 0, There are two imaginary roots, and one real root between and 1. 3. Find the number of real roots of the equation x' — 2ar — 5 = 0, It has two imaginary roots, and one real root between 2 and 3. 4. Find the number of real roots of the equation x* — 2x^ — Ix^ + lOx + 10 = 0, It has four real roots ; two positive roots between 2 and 3 ; and two negative roots, one between and — 1, and another between — 2 and — 3. 598. In order to find the next lower integer to the root of an equation, substitute successively for the unknown quantity the numbers 0, 1, 2, 3, 4, ... or 0, — 1, — 2, — 3, ... till two successive results be found with opposite signs ; then one root or some odd number of roots lie between the two numbers producing these results ; and the smaller of these numbers is therefore the required integer. 599. If it is found that two numbers of the negative series — 1, — 2, — 3, ... produce opposite results, this indicates that at least one negative root lies between them. But if the signs of the alternate terras of the equation be changed, as the second, fourth, sixth, ... the signs of its roots are then changed, and the preceding root will therefore becomo positive. GENERAL SOLUTION OF HIGHER EQUATIONS. 375 Let the equation be a:' — 2ar — 5 = 0. Let X = 0, and the result is = — 5, ... a: = l, = - 6, ... x = 2, = - 1, ... x = 3, = + 16. As the results given by substituting 2 and 3 in the equation for x are — 1 and +16, their opposite signs imply that at least one root lies between 2 and 3 ; it is therefore = 2+7/, where y represents some fraction representing the difference between the true root and the number 2. NEWTON S METHOD OF APPROXIMATION. 600. Resuming the equation J{x) = 0, if we substitute for x, in this equation, « + y where a is a value very near to the true value of x, suppose that a differs from x by less than -1 ; by (576) we have fCx) =J{a+!/) =J{a) +fXa)y+i^f +'^-3'' + - = <>! •■• ^ = -^ -7^)^^"^4 +/"'Wo + •••> Since a differs from the true root by less than *!, y is less than a tenth, hence y^ is less than a hundredth, and y^ is less than a thousandth, &c. ; so that by neglecting, in the above value of y, all those terms which contain its powers, we will obtain its value at least to a hundredth part of a unit. And the expression is reduced to - X«) "We effect the division indicated, and stop at the second decimal place ; this value being added to o, will give a value of x differing from the true value by less than -01 ; substituting this more approximate value instead of x, and dividing out to 4 decimal places, a new correction will be found which will, in general, give a value of X true to 4 decimal places ; proceeding in the same manner, the third correction will in general be correct to 8 decimal places. 376 ELEMENTS OF ALGEBRA. EXAMPLE. Given the equation a;' — 2a; — 5 = to find a value of x. By the method explained in (598), it is found that the true value difiers from 2-1 by less than -1 ; hence a — 2-1 gives X2-1) (2-iy - 2 X 2-t- 5 _ -061 ^ . ^ ~ "" /'(2-1) ~ 3 X (2-1/ - 2 11-23 ... a + y = 2-1 - -0054 = 2-0946. Substituting this second value in the above formulary, and calcu- lating the value of ?/ to 8 decimal places, we obtain y=- — -00004851 ; hence we conclude that x — 2-09455149 nearly. Remake. In this example, the assumed value of x was too great ; hence the value of y is negative, and it was therefore subtracted from the approximate value ; but when the value of y is positive, it must be added to the approximate value. exercises. 1. Fmda root of the equation x^ — x' + 2a;- .-. X = 3 =0, 1-275682204. 2. ... x' - 15x2 + 63x — .-. X = 50 = 0, 1-028039. 3. ... ^' + 2x^ -23ar- .'. X = 70 = 0, 5-1346. 4. ... X* - 4.x' - 3x + .'. X = 27 = 0, 2-2675. Another root of the last equation lies between 3 and 4, and is = 3-6797. hoener's method. 601. This method consists in depressing the given equation tc another whose roots shall be less than those of the given equatior (570) by the highest digit in one of its roots, and then depressing the depressed equation by the highest digit in its root { and so or continually till a suflScient number of figures be obtained. After the first depression has been performed, the figure bj which it is next to be reduced is found by dividing the absolute term of the depressed equation by the coefficient immediatel) preceding it, which is the same as the method by which th( correction is found in Newton's method, for the absolute term o the depressed equation is /{a), a being the part of the root already found, and the coefficient immediately preceding it is /'(a) ; whei GENERAL SOLUTION OF HIGHER EQUATIONS. 377 this next figure is found, the equation is again depressed by this new quantity, by performing the same operation with this new figure on the coefficients of the depressed equation, as was formerly performed with the first figure on the original coefiicients ; again a new figure is found, and the operation of depression performed ; and so on continually till the root is obtained to the required degree of accuracy. The only difference in the niethod of performing the operations in Homer's method, and that already exhibited in (570), is that the absolute term is written with its sign changed ; that is, it is transposed to the second side, and then the product, which is put under it, is subtracted instead of being added, which gives — /(a) for the remainder. The first figure of each root must either be found by (582), or by substitution in Sturm's functions, which is the most certain method when all the real roots are required. EXAMPLE. 1. Pind all the real roots of the equation X* - 8x^ + 14x2 ^ 4^ _ 8 = 0. The following are Sturm's functions reduced to their simplest form : — Letx = V= x*-8x^-^Ux''+4:x-S — oo +00 -1 + 1 +2 +3 +5 6 + + + + + + V,= x'-6j^ + 7x+ 1 . — + + + _ + + l\= ^x'-nx^ 6 V,=lQx -103 . + + + + _ =F + + — + — + + + + F,= + . . . . + + + + + + + + + Variations 4 3 4 2 2 1 1 Since the variations lost between — oo and + oo is 4 — = 4, the equation has all its roots real. Again, since the variations lost from — oo to is 4 — 3 = 1, the equation has one negative root; and as the variations for — 1 is 4, and for is 3, the nega- tive root lies between and — 1. In the same manner, it is found that one variation is lost between and + 1, another between 2 and 3, and a third between 5 and 6 ; hence the positive roots are between and 1, between 2 and 3, and between 5 and 6. The equation has therefore three positive roots, and a negative one, which is also manifest (561), by a simple inspection of the signs of the given equation. In order to find the first figures of the roots that lie between and — 1, and between and + 1, we must narrow the limits by substituting in the above functions "1, -2, -3, ... ; — -1, — -2, — -3, ..., 378 ELEMENTS OF ALGEBRA. till one change is lost by the positive substitution, and one gained by the negative ; then the figure immediately preceding that which causes the loss in the positive, or the gain in the negative, will be the first figure of the root. By this means it will be found that the first figure of the positive root is -7, and that the first figure of the negative root is also -7. The same results will be obtained by substituting in the given equation till the results give opposite signs (582). The root which lies between 5 and 6 is calculated as follows : — 1* - 8 5 - 3 5 14 -15 - 1 10 9 35 *44 2-44 4 - 6 - 1 45 *U=t 9-288 8(5-236068 - 5 *13 10-6576 2 5 *2-3424 1-93880241 7 6 53-288 = d 9-784 •40359759 •39905490 12 •2 46-44 2-48 *63-072 = t 1-554747 •00454269 400954 12-2 •2 48-92 2-52 64-626747 = d 1-566321 53315 12-4 •2 *51-44 -3849 *66-193068 = < •31608 12-6 •2 51-8249 •3858 66-50915 =d -31656 12-8 •03 52-2107 -3867 66-82571 =t 12-83 •03 *52-5974 -08 12-86 •03 52-68 •08 12-89 -03 52-76 1 * 12-92 As the root ia carried out only to six decimal places, it is not GENERAL SOLUTION OF HIGHER EQUATIONS. 379 necessary to carry the real divisor for the third figure (6) to more than five decimal places ; this divisor is 66-50915 ; and this number, multiplied by -006, gives eight decimal places, and the dividend ought to be carried to seven or eight decimal places, in order that the figure in the sixth decimal place of the root may be correct. So the divisor 66-825 for the fifth figure of the root requires to be carried only to three decimal places, for the product of this number by -00006 gives eight decimal places, as it ought to do. So the divisor for the last figure (8) of the root would require to be carried only to two decimal places. The number in the vertical lines preceding the divisors requires to be carried to still fewer places, as the reader will easily perceive from the successive processes of multiplication ; and after obtaining the third decimal figure of the root (6), the numbers in the column under b do not require to be multiplied at all. The root which lies between 2 and 3 is found thus : — 4 2 2 2 0-7 7 1-4 7 2-1 7 2-83 3 2-86 3 2-89 3 2-92 + 14 - 12 2 - 8 - 6 - 4 + 4 4 8 -12 - 4 - 6-657 -10 •49 - 10-657 - 6-971 - 9-51 98 - 16-628 - -209253 - 8-53 1-47 -16-837253 - -206679 - 7-06 849 - 17-04393.2 - 1359 - 6-9751 • 858 - 17-05752 - 1359 - 6-8893 867 - 17-0.7.1,11 — 6-802.6 7 + •8(2-7320508 16 - 8 - 7-4599 •5401 50511759 3498241 3411504 86737 85356 6-795 380 ELEMENTS OF ALGEBRA. The root whose first figure is + -7 is found thus : — -8 + 14 + 4 + 8 (-763932 •7 5-11 6-223 7-1561 — 7-3 8-89 10-223 -8439 -7 — 4-62 2-989 •79211376 -e-6 4-27 13-212 5178624 -7 — 4-13 - 10104 3951341 -5*9 •14 13-201896 1227283 7 — •3084 - 28392 1184220 -6-2 _ •1684 13-173504 43063 -06 — •3048 2368 39472 -5-14 _ -4732 13-171136 3591 -06 — •3012 - 2412 2631 -5-08 — •7744 13-15872,4 960 -06 — 149 72 -5-02 — •789,3 13-1580,0 •06 — 15 - 7 -4-96 — •804 13-15,7,3 Since the fourth root is negative, we change the signs of the coefficients of the alternate terms, which changes the positive roots into negative, and the negative into positive (558, Cor.), and then calculate the same as before ; thus — 1 + -8 •7 + 14 6-09 - 4 14-063 + 8 (-7320508 7-0441 8-7 7 20-09 6-58 10-063 18-669 •9559 •89261841 9-4 7 26-67 7-07 28-732 1-021947 6328159 6171029 10-1 7 33-74 •3249 29-753947 1-031721 157130 154632 10-83 3 34-0649 •3258 30-785668 69478 2498 2473 10-86 3 34-3907 3267 30-85514,6 6952 25 10-89 3 10-92 34-7174 218 34-739,2 22 30-92467 173 30926,40 2 34-76,1 309,2,8 602. The proof of the above is very simple when all the roots are calculated, for it was proved (553) that the sum of all the roots of every equation, with their signs changed, is equal to the GENERAL SOLUTION OF HIGHER EQUATIONS. 381 coefficient of its second term : it will be found that the sum of the four roots found above fulfil this condition, when we remember that the last root is negative. It is also evident, that after all the roots of an equation except one have been found, it may be obtained by subtracting the sum of those already found from the coefficient of the second term, with its sign changed ; but it is better to calculate them all, and then apply this principle as a proof of their accuracy. 603. The student, after carefully studying the preceding ex- ample, may easily solve the following exercises, for which one root only is given in the answers, as this is suflScient for illustrating the preceding method ; but he should calculate all the roots, and prove their accuracy by (553). 2. EXERCISES. 'tltx^- 2a; - 5 = 0, . . . :c = 2-0945515. 2....x^+ 4:x^- Sx - 20 = 0, . x = 2-236068. . X* - 8x^ + 20^:^ - 15:r + -5 = 0, . x = 1-284724. . XT' -\- lOx"" - 24:x - 24:0 = 0, . ' . x = 4-8989795. 5. ...x^ + 12x2 - 18t - 216 = 0, . . x = 4-2426407. 6. ... x^ + 2a;* + 3a:^+ 4x2+ 5^; _ 54321 = 0,x = 8-414455. 7. ...x* - 59x2 4- 840 = 0, . . . x = 4:8989795. 604. When one root of an equation is found, the equation may be depressed one degree ; that is, if r be one root, and if the equa- tion be divided by x — r, there will be no remainder, and the quotient will be an equation one degree lower, the roots of which are the remaining roots of the first equation. Hence if one root of a cubic equation be r, and the equation be divided by x — r, the quotient is a quadratic, the two roots of which are the remaining roots of the cubic equation. CONTINUED FEACTIONS. 605. A continued fraction is a complex one, whose denominator is an integer with a fraction whose denominator is also an integer with a fraction, and so on. The most usual form of a continued fraction, having a prefixed integer a, is ^ a +, &c. 606. The fractions y, -, -^ ... are called component fractions. bed When the number of component fractions is limited, the continued fraction is called terminate ; and when the number is indefinite, it is said to be interminate. 607. Continued fractions may be generated in various ways. One of the most common is when the value of a quantity, wliich is not integral, is to be found by a series of approximations. Let a' be such a quantity ; and find the next lower integer to it, which denote by a : then a' — a ^ 1 ; therefore —, /'I, which a — a quantity may be denoted by h'. If h' be not integral, let b be the next lower integer to it ; then b' — b ^1] therefore — -p' 1, wliich quantity may be denoted by c'. K c' be not integral, let c be the next lower integer to it, then c' — c^l; therefore-; 7^1, which quantity may be denoted by d'. Proceeding in this manner as far as necessary, then 1 ,, / 1 or a' = a + yy, — OjOra — a — , a' — a b 1 ,1 or 6' = 6 + -i-, b' — b c and it would similarly be found that ^' = ^+Z' and that d'=d^\, CONTINUED FRACTIONS. 383 Hence by substituting successively the values of 6', c', d', ... ora' = a + l^l ^ ''+e+,... 608. Continued fractions may also be derived from common fractions, by performing upon their terms the process of finding A their greatest common measure. Thus, if -^ be a fraction, and a, b, c, d, ... the quotients obtained by this operation, and C, Z>, E, F, ... the corresponding remainders, then it is evident (103) that A=zaB-\-C,B=bC-^D,C=cD + E,... ^ A C B . D C E Hence - = « + _,_ = 6 + _,_ = c + -^, ... nBSic^Rii l-#=l-(*+#> ^=l^«=+l>... B D , C c k C D E bstituting successively the values of -^, 77,-7:, .. 609. The quantities a, b, c, ... are called partial quotients, and r* 71 7? the quantities a + -^, 6 + -^' ^ + TT ' •*• ^^® called complete quotients. The quantities a, b, c, ... here correspond with a, b, c, A Ti C ... in (607); also ^, -^, -^, ... correspond with a', &', c', ... in that article. If any remainder, as E, be = 0, then c is the complete quotient, and the continued fraction terminates with c. 384 ELEMENTS OF ALGEBRA. EXAMPLES. 1. Eeduce tt to a continued fraction. 11)25(2 „ 25 ^ 1 , — -^ +2 3)11(3 9 2)3(1 2 1)2(2 610. It is usual to represent a continued fraction by stating the quotients merely ; thus, if a' be the given fraction, a' = ^ = 2,S, 1, 2. 25 2. Keduce — to a continued fraction. As this is a proper fraction, the first quotient is ; the others are 1, 2, 12 ; hence ^12 EXERCISES. 37 1. Keduce ^^ to a continued fraction, lo a' = 2, 2, 7, or a' = 2 + J 1 2 + Y 2. ... ^ a' = 0,2, 4, 3. 3. ... ^ a' = 3, 5, 1,6. 611. Continued fractions may be easily reconverted into com- mon fractions. Thus, taking the first example, the last part CONTINUED FRACTIONS. 385 ■to I 3 2 1 1 + 2 = 2' ^^^ ^^"^® l + -~''"'^2~3' ^^^^^^"^^^ ^ + 1 j_ 1 = 3+1 = ^.-. .1^=. + ^ = . .£ = ?«. "^2 3 The other example and exercises may in the same manner be proved to be correct. . 612. Let -pr be a fraction, and let the partial quotients of the equivalent continued fraction be a, b, c, d, ... then ^ + 5+,... Approximate values to the fraction -j^, the true value of the whole continued fraction, may be found by converting into a common fraction, the continued fraction carried out to one, two, three, or any number of terms. Let the approximate fraction, found by carrying it inclusively to a, b, c, d, ... p, q, r, ... be respectively denoted by -^i, -^^ -^, T/' •" 'p'Q'R" - ^^^^ ^^ 1 (a6 + l)c + a c It is evident that — is = -^, — — jr> and this expression for -^ suggests the following theorem in reference to the formation of convergent fractions : — ■ 613. The terms of any convergent fraction are equal to the product of the corresponding terms of the preceding convergent, by the partial quotient corresponding to the former, together with the corresponding terms of the next preceding convergent. Let the theorem be true for -^ then -p;, j-, being the two pre- ceding it, it follows that R Or +P ,.,.,, 1 ., . 'S' Rs ^a R' = W+P"' ^""^ '' '' *" ^' P'"^'^ '^'^' ^ = RTT^- Q 7? 1 Since -57 differs from ^57 merely in having r -\ — instead of r, as is A a B , 1 ab + \ C A' -V'B'- = "+r b ' C 386 ELEMENTS OF ALGEBRA. evident by writing down the continued fraction carried out to hence ;r + i) + p _ (_Qr 4- P)s + Q _ Rs +Q ^'(^+.) + P' -p Hence the theorem being admitted to be true for -rrr, it is thus H S proved to be true for -^ ; that is, if it is true for one convergent, o it is true for the succeeding one. But it was before shewn to be O D E true for -4 in (612) ; hence it is true for ^r-, and therefore for -=-, ; and so on — that is, it is generally true. EXAMPLES. 71 1. Eeduce — to a continued fraction, and find the convergents. ol a' = 2, 3, 2, 4; hence ^ = ^,|,:.2+1 = ^, C_ _ 2B + A _ 2x7 + 2 _]^D_ _ 4:C + B _ 71 (7 ~ 25' + ^' "" 2 X 3 + 1 ~ y F' ~ 46" +B' ~ 3l' Hence the convergents, arranged under their respective quotients, with an auxiliary fraction - prefixed, are Quotients 2,3, 2, 4, 1 2 7 16 71 Convergents....-,-,-,-,-. The first fraction - is used merely to render the rule applicable 73 FT to the second convergent -=5;, which is here = -. The fraction X) o A 2 7? 7 -r; = - is always the first quotient. -^ or - is found by taking ^ 1 Jo o 2 the quotient over it or 3, multiplying the 2 of - by it, and adding 1 2 the 1 of -, which gives 7 ; and then the 1 of -, multiplied by 3, CONTINUED FRACTIONS. 387 and added, gives 3. So to find — , take the quotient 2 over it ; multiply 7 by it, and add 2, which gives 17; then multiply 3 by 71 . it, and add 1, which gives 7. And similarly — is found. 54 2. Eeduce — - to a continued fraction, and find the convergents. Quotients ... 0,2, 3, 5, 1, 2, 1 1 3 16 19 54 Hence the convergents - q> p g' 7' 37' 44' 125* A The first quotient being 0, ^ = y = 0. I EXERCISES. Ill 17 1103 ^ 86400 ^ .. ^r .• j« ^ Eeduce -— -, — , -^— -, and ^tttx^q, to contmued fractions, and find 40 57 oo7 JAjjAa their convergents. T. Ill \m. ^0' Quotients .2,1,3, 2, 4, I Convergents.. 1 2 3 11 25 111 • 0' r 1' 4 ' 9 ' 40 ' Quotients .0,3,2, 1, 5, 1 Convergents.. 10 12 3 17 •0'1'3' 7' 10' 57' ^r 1103 ^""^ 887' Quotients .1,4,9, 2, 1, 1, 4, Convergents.. 1 1 5 46 97 143 240 1103 ■ 0' 1' 4' 37' 78' 115' 193' 887 ' 86400 20929 ... 4, 7, 1, 3, 1, 16, 1, 1, 15, ^ 1 4 29 33 Convergents... -, -, — , --, 128 161 2704 2865 5569 86400 614. If the numerators and denominators of two consecutive convergent fractions be multiplied in a cross order, the difference of the products will be unity ; and for all of these consecutive fractions, the differences between the products are alternately positive and negative. 388 ELEMENTS OF ALGEBRA. The difference for ~ and ^,ia PQ -P'Q = D,, and for -^ and -^, observing that (613) 757 = -^, — — ^, (4 Jti It f<^r -\- Jr it is = -cm'- Q'R= QiQ'r + P')-QXQr+P)=P'Q-PQ'=r>,. Hence D^ = — D^, or the differences are equal with opposite signs. The same may be proved for the other convergents. But for -7-, and -^,, or (612) - and — -. — , this difference is ... = ah — A H 1 («& + !)=-!; Ti C CD hence for -j^. and -7^, it must be + 1 ; for -^ and ^^ it is — 1 ; and so on. Hence PQ' — P'Q is either + 1 or — 1. 615. Cor. 1. When the first of the products is that of the numerator of the first of two consecutive fractions by the denomi- nator of the following, the difference is — 1 when the former fraction is of an odd order, and -f 1 when its order is even. A ' B It has been shewn (614) that for -r, and -jr; the difference is — 1 : ^ '' A' B B C P for -^ and -p^, it is + 1 ; and so on. Hence if p; is of an even order, PQ' — RQ=l,ovD^ = 1, and since (614) D^= — D^; hence 1)2= — !• To illustrate the two last articles, take the third and fourth q 1 ft C D convergents of Ex. 2 (613), - and — , or-^ and -=r-, and CD' — CD = 111 - 112 = - 1. 616. Cor. 2. The reciprocals of these fractions possess the same property. For the difference for p- and ^ is P'Q - PQ' = - D,, and for ■jY and -^ it is Q'R — QR' = — D^, or D^, D^, have here opposite signs to those of Dj and D^ in Art. (615). 617. The convergents are in their lowest terms. p Por PQ' — P'Q= +1, being + 1 when -- is of an even order, and — 1 when of an odd order. CONTINUED FRACTIONS. 389 Now, any quantity that divides P and P' must divide 1 ; that is, P and P' are relatively prime (96). 618. The corresponding terms of any two consecutive conver- gents are relatively prime. For R= Qr -\- P, and if P and Q are relatively prime, so are Q and R. But if Q and R are not prime, let them have a common measure m ; then Qr and R being divisible by m, so must P ; but it cannot, as P and Q are supposed to be prime ; therefore R and Q must be prime. Hence, when the numerators of the first two of three consecu- tive convergents are prime, so are the numerators of the two latter. Now, B = Ab -\- 1, and any quantity that divides A and B must divide 1 ; therefore A and B must be prime. Hence B and C, C and D, ... are prime. The same property is similarly proved to belong to the denominators. 619. The difference between two consecutive convergents is equal to 1 divided by the product of their denominators ; and when the two fractions are taken in order, the sign of the difference is positive or negative, according as the first of the two fraction* is of an even or an odd order. ^^P Q PQ'-P 'Q , 1 ,„,,, p taking + or — according as the order of -p; is even or odd. 620. Cor. A convergent is greater or less than the consecutive one, according as the order of the former is even or odd. To illustrate the two last articles, take the 4th and 5th con- vergents in the first exercise in Art. (613) — namely, ^ and — r-, or 7) W rp- and -=r,- Then 25 111 1000 — 999 1 1 9 40 9 X 40 360 D'E' ' 25 D and — or yr/ i^ the greater, its order being the fourth. 621. The convergents are alternately greater and less than the complete fraction, the greater being those of an even order, and the less of an odd order ; also, any convergent approaches nearer to the complete fraction than any preceding it. For -^ =: -r-, 757, ... and if / be the complete quotient, cor- R Q,r -\- Jr responding to the partial one r, then / is equal to the whole of the 390 ELEMENTS OF ALGEBRA. continued fraction after r inclusive, or r' = r + - . ; hence s -j- &c. 7? if / be substituted for r in the preceding expression for ^^, M the result must be equal to the complete fraction; hence ... U Qr' -L. p p Ir ~ (V ' -i. P '^ '" *^®^ ^^ P' ^® °^ ^^ ^^^^ order, PQ' — P'Q = + 1 by (615), and P U {PQ'-P'QY / 7^ = D\ B V FXQV + P') PXQr'+P') a' F - Q'cav + P') ~ QXQv +p')~ Hence D' is positive, and D" negative ; therefore ^ ^ -^„ and "^jy ; that is, the convergents of an even order are greater, and those of an odd order less, than the complete fraction. Again, ^ince / is 7^ 1, for r is not ^ 1, and P' ^ Q', therefore D' -p- D'\ disregarding their signs ; that is, the fraction -rp differs less from U P -^ than -p; does, and the same is similarly shewn for any conver- gent and that which precedes it ; hence the origin of the term convergent. 622. Cor. Any convergent differs from the complete fraction by less than 1 divided by the square of the denominator of the former. For r' -p- \, and hence Q'r' + P' ■y' Q' ; and hence D" is ^ 7777' ^^ T)'2 ' *^^* ^^> Tv differs from ^ by less than -j^ ; and the same is similarly shewn for any other fractions. 623. To exemplify the two last articles, take the 2d and 3d 7 1 fi convergents of Ex. 1, Art. (613)— namely, - and — . O I _ Z7 71 P 7 (7 16 - ^^^" F = 3r'P^ = 3'C^ = T'""^ _7_71^_^ 7)'/_l^_Ii_ L ■" 3 31 "~ 93' ~ 7 31 ~ 217* CONTINUED FRACTIONS. 391 „ 7 71 . 16 71 ., 16 ._ , „ 71 , Hence - -p' — , and — ^ — . Also — differs less from — than o ol 7 ol 7 ol - does. And likewise 217 ^ ^ or — . 624. Any two consecutive convergents differ from the complete fraction by less than 1, divided by the product of their denomina- tors ; and no intermediate fraction can have either of its terms less than either of the corresponding terms of the two convergents. But -^ is intermediate between these (621), and hence it must differ from either of them by a quantity less than D. I P Q Again, let -=7 be a fraction intermediate between -p^ and -^ P I then the difference between -^, and -j-, cannot be less than D, unless /' z^ Q', since the numerator of i) is 1 ; neither can the difference between -^ and -jj be less than D, unless I' -p' P' ; Q> J- therefore I' is greater than either P' or Q!. The same may be proved of their reciprocals ; and hence / is greater than either Por Q. 625. The property of convergents proved last is of some prac- tical utility. When, for example, a ratio is expressed by large numbers, and it is required to express the ratio approximately by smaller numbers, one of the convergents will express an approximation in simpler terms than any other fraction. Let the ratio be that of the circumference of a circle to its diameter carried to 7 decimal places or "^ . The quotients are 3, 7, 15, 1, 248, &c. , 3 22 333 355 86598 „ Convergents -, y, —, -3,^^-, &c. 22 The second convergent — is that found by Archimedes, and 355 the fourth — - by Adrian Metius. No fraction can express an llo approximation so near as either of these, unless its terms exceed those of the latter. 392 ELEMENTS OF ALGEBRA. 626. When one or more of the terras of a continued fraction continually recur, it is called a periodic continued fraction. ^^'" + 1 + 1 1 b +, ... be a periodic fraction. Assume its value = X, then x — a = j \ ' + *+,... then since the part of this fraction after the first h is the same as this last fraction, x — a may be substituted for it ; hence ^-^= ^ _^^_^ > or K^ - a) + (x-df=l; .-. a;2 - (2a - 6> = 1 - a^ + ah, from which the value of a: = -(2a — V) ± ->s/ (h^ + 4). If a = 1, and 6 = 2, then x = i^2, and substituting for a and b in the given fraction, it follows that A +, ... This example aflEbrds an instance of a continued fraction, the law of whose terms is evident and simple, expressing the value of an irrational number.* As illustrations of the above example, the student may find the value of a: for any other assumed values of a and b. If the value of V2 = 1-4121356, ... or !qqqq!qq > be reduced to a continued fraction, it will be found to be the same as the preceding ; but by this method the law of the terms could not be logically inferred (142). * Lord Brounker, the inventor of Continued Fractions, has given an example of one, the law of whose terms is simple, and wMcF^Sxpresses the ratio of the area of a circle to the square of its diameter — namely, as 1 to 1 + - ^ „_ 2 +, ... the numerators of the .component fractions being tlje aqiiares of the odd numbers. i ; ; V'l t' CONTINUED FRACTIONS. 393 Let X = - 1 "+3+^1 " + 6+,... then after the first h, x may be substituted for all the terms that follow; hence a -\- Y , ab + ax + 1 -\- X And from this equation is found Where a and h may have any numerical values. Let a = 1, and 6 = 4, then x= —2 ± 2^2 ; therefore x= —2 ± 2V2 = - 1 hence 2^2 = 2+- 1 EXERCISES. ^^^ iMy*^i 1. Find the value of x = - 1 . . a+, a . 1 ., , . .. ^ = ... x^ + ax - 1 = 0, and a: = - - + xV(«^ + *)• 2 - = " + 5 + 1 The equation is hx' — (2«& — 6c)x + a'b — abc — c = 0, from which the value of x can easily be found. ^-^' -- -^^ffTi" INDETERMINATE EQUATIONS. 627. Equations are said to be indeterminate when the number of unknown quantities exceeds that of the equations. In indeterminate equations the number of values of the unknown quantities is indefinite ; but when their values are restricted by certain conditions, the number of them is generally limited. The conditions usually imposed are, that the values be integral num- bers ; or, by still further restrictions, that they be also positive, or that they be square numbers, or cube numbers, and so on. I. — SIMPLE INDETERMINATE EQUATIONS. PROBLEMS. I. Given the equation ax -\- hy = c, to find all the positive aiid integral values of x and y ; h and c being either positive or negative. Convert ^ into a continued fraction, and find the last convergent, which may be denoted by — ; then (614) an ~ bm = 1. If bm be the greater, find the value of x, — namely, — from the given equation, and multiply its numerator by w; it then becomes ~, which may be reduced by division to the form q -] -. a a Assume ~ = r, and then y = d — ar; then substitute this value for y in the preceding value of x, and its value will be of the form X = e + br. By giving r proper integral values, the positive integral values of x and y will be found. If an were the greater, then a value of y would be found, and a similar process performed. When the system of least values of x and y are known, the other systems may be found by adding the coefficients of y to the values of x, and the coefficients of x to the values of y, when the equation is ax — by = ± c. When the equation is ax -\- by = + c, and a system of values is got, such that the value of x is the greatest ; then the other values of x and y are obtained by sub- tracting from the values of x the coefficient of y, and adding to the values of y the coefficient of x. INDETERMINATE EQUATIONS. 395 That is, if m Mid n be a system of values of x and y, then the other values of x and y, for the equation ax — by =^ + c, are X ■= m -\- hr, y =z n -\- ar^ and for the equation ax -{- by = + c, they are • X = m — br, y = n -\- ar, where r may have any proper positive integral values. EXAMPLES. 1. Given 15a; — 13y = 10, to find the positive integral values of X and y. By converting — into a continued fraction, the last convergent Lo is found to be - ; hence - = —-, — =:-, and o 16 n b an - im = 15 X 6 - 13 X 7 = 90 — 91 = — 1, as bm is the greater, find a value of x ; then x = — — - ; hence 15 1.- 1 • 1. n 7(10 + ISy) 70 + 91y 10 + v multiplymg hy m = 7, -^ — j^— ^ = — ^^ = 4 + 6y+^^, therefore assume — -— ^ = r ; hence y = 15r — 10, and x = 10+13?/ 10 + 13 X 15r - 130 ,^ — 1^- = 15 =13r-8. If r were assumed = or = a negative number, the values of X and y 'would be negative ; hence r must be assumed 7=^ 0, or 1 is its least value; but its value may be assumed equal to any greater number than 1 ; the number of solutions is therefore indefinitely great. When r = 1, X = 5, y = 5, r = 2, X = 18, y = 20, r = S,x = Bl,y = 35, 628. When the least values a; = 5, ^^ = 5, are found, the other values are found ; thus (627), x = p^ lr,y = g+ ar, or a: = 6 + 13r, y = 5 -\- lor. 396 ELEMENTS OF ALGEBKA. When r = 0, 1, 2, 3, 4, 5, 6, ... then x=5, 18, 31, 44, 57, 70, 83, ... and y = 5, 20, 35, 50, 65, 80, 95, ... So tnat the successive values of x and y are found by adding respectively to their least values the coefficients of y and x; and the two series of values form equidifferent progressions, the common difference of the former being 13, and of the latter 15. 2. Given lOx + 7j/ = 763, to find the values of x and y. _ a 10 7?i 3 , ^„ ^ Here r = —,~ = ~, an ~ bm = 20 ~ 21 = 1- hence as bm is the greater, find a value of x, 763-7y_ 76 , 3^iiy The 76 may be neglected, as — -~ must be a whole number, since x and 76 are so. Hence as'm = 3, 3(3 -7y) ^ 9-21y ^ _ „„ , 9-y 10 10 ^ ^ 10 • 9 — w Let -— ^ = r, then y = 9 - lOr, 763 - 7y 763 - 63 + 7 X lOr „^ and .= — ^ = =70 + 7r. If r = — 10, then x would = 0, and if r = 1 , y would = — 1 ; the limits of r therefore are 1 and — 10 ; so that r can have only the integral values intermediate between 1 and — 10 ; namely, 0, — 1, — 2, ... to — 9 inclusive, or 10 values ; so that there are only 10 positive integral systems of values of x and y. When r= 0, x = 70, ?/ = 9, r = - 1, a: = 63, y = 19, r=-2,x=^5G,y = 29, 629. Or the values are found thus (627)ji when the system X = 70, y = d, are known, X = p — br, y = q -{• ar, or a: = 70 - 7r, y = 9 + lOr. The limits of r are evidently and 9. INDETERMINATE EQUATIONS. 397 When r= 0, 1, 2, 3, ... 9, then X = 70, 63, 56, 49, ... 7, and y = 9, 19, 29, 39, ... 99. 3. Let 3x — Ui/ = 20, to find x and y. Here - = — -, — = -, an — bm = 12 — 11 = I, and as an is 11 n 4 the greater, find a value of y, and multiply it by n = 4 y = 33: - 20 4:(Sx - 20) _ 12x - 80 _ ^ , x-3 11 ' n - n -"'"^"•""Tr* r, then X = llr + 3, 11 Bx -20 33r 4- 9 - 20 and y = —^j— = = 3r - 1. If r = 0, y = — 1- hence for positive values, r must be at least = 1. When r = 1, X = 1^, and y = 2. By making r = 2, 3, 4, 5, ... an indefinite number of values may be found. 630. Sometimes the value of x may be reduced to the form —, or -, without requiring to multiply ; and in such a case, -, or -, may be assumed = r, as in the following example : — 4. In how many different ways can £100 be paid in crowns and guineas ; and how may it be paid in this manner so as to require the smallest number of crowns ? Let X = the number of crowns, and y = guineas, then 5a; + 217/ = 2000; 2000 - 21y ,n^ , V .'. X = ^ = 400 - 4y — ^. Here the coefficient of y in ^ is 1 ; hence let ~ = r^ then y = or, and X = 400 - 21r. The least value of x will evidently be found by gi-nng r its greatest admissible value ; that is, r = 19 ; then x = 1, and ^ = 95 ; so that £100 may be paid by 95 guineas and a crown. 398 ELEMENTS OF ALGEBRA. . The least admissible value of r is evidently 1 ; for if it be less, then y would be or negative. Hence the whole number of ways required is found by giving r all the integral values from 1 to 19 inclusive ; that is, 19 different ways. 631. This rule depends on the following principles : — 1. Any integral multiple of an integral quantity is also an integral quantity. TT -n c ±hy . . ^ , . mCc + by) Hence if x = is an mteger, and so is ~ -^ . a a 2. If the sum or difference of two quantities be integral, and if one of them be an integer, so is the other. „ .r,m(c±by) , d+y. Hence it — = g + -^ be an integral quantity ; so is ± — =-^. And if this quantity be assumed = r, then y= ±ar + d; and any arbitrary integral values being assigned to r, the result will be an integral value of y. Also y + d= ± ar, and y ±dis there- fore a multiple of r by a ; hence — =^ is equal to an integer, and hence also q ± — =^ is integral, and therefore its equal is also integral; but a and m are relatively prime (618); hence c ± by is a. multiple of a (150) ; and therefore x is also an integral number. 632. The formulas given in articles (628) and (629) may be proved in this manner : — Let the equation be ax — by = + c, and let m and w be a system of values of x and y, and x', y', any system whatever, then am — bn = + c = ax' — by' ; hence a(x' — m) = b(y' — n\ and — = - = — , ^ V v^ y> y' _ n a ar where hr and ar may be both positive or both negative. Since in the preceding fraction x' and y' denote any system of values, af — /w, y' — n, will not be respectively equal to b and a, except for one system of values ; hence they must be considered equal to a fraction, whose terras are multiples of b and a — namely, br and ar, r being either positive or negative. Therefore a;' = m + br, y' = n + ar. Here a, b, m, and n, are known ; and by giving r proper integral values, those of x and y are known. INDETERMINATE EQUATIONS. 309 When r = 0, 1, 2, 3, ... r, X = m, m -{- b, m -^ 2b, m + 3b, ... m + &r, y = n, n + a, n + 2a, n + 3a, ... n + ar. The values of r may also be taken negatively. 633. Therefore the values of x and y are equidifferent series, having respectively the common differences b and a. When the equation is ax -}- by = ± c, it may be similarly proved that X = m — br, y = n -\- ar, where r may also be positive or negative. 634. The equation ax + by =+ c is always impossible in integers, when a and b have any measure that is not also a measure of c. For let mhe a measiire of a and b, but not of c, and let a — a' in, b — b'm, then a'x + b'y =■ H . And if X and y have any integral values assigned to them, the first member of this equation would be integral, and the second fractional, which is impossible. Hence, 635. In order that x and y may have integral values, a and b must be relatively prime; or, if not, their greatest common measure must measure c. 636. In order to determine when the equation ax + by = + c is possible in positive integers, let — be the convergent fraction n nearest in value to j. Then an — bm = + 1, taking + 1 when the convergent is of an odd order, and — 1 when its order is even (621). 1. For the equation ax — by = ± c. 1st, Let the convergent be of an odd order, then an — bm = 1; hence a .en — b . cm = c. Therefore one solution of ax — by = c is x — en, y =: cm', and hence its other solutions are determined by the formulas in Art. (632), which become here X = en -\- br, y = cm -{■ ar. And for positive values of r, the values of x and y are all positive. 400 ELEMENTS OF ALGEBRA. 2d, Let the convergent be of an even order ; then an — bin = — 1, hence a(— en) — h(— cm) = c. And therefore one solution of ax — bi/ = c, isx = — en, i/= — cm, and the other solutions are contained in the formulas in Art. (632), Vhich here become X = — en -\- hr, y = — cm -\- ar, and it is evidently possible to give r such positive values as will make the values of x and y positive. In the same manner it may be proved that the equation ax — by = — c always admits of positive integral values of x and y, by multiplying by c, when an — bm = — 1, and by — c when an — bm = +1. Hence 637. The equation ax — by = + c is always possible in positive integers when a, b, and c, fulfil the conditions in Art. (635), and the number of solutions is unlimited. 2. Eor the equation ax -\- by = ± e. 1st, Let the convergent be of an odd order ; then an — bm = 1, and hence an + 6(— m) = 1; therefore a . en -\- b(— cm) = c. One solution oi ax -{- by = c therefore is x = en, y = — cm, and its other solutions are (633). X = en — br, y = — cm -{- ar. 638. And if such a value can be assigned to r that en — br may be positive, and also — cm + ar, the resulting values of x and y would be positive. The limits of the values of r will shew the number of solutions. 2d, Let an — bm = — 1, then an + b(— m) = — 1, and a(— en) -\- b . cm = c. From which x = — en, y = cm, and the conclusion is similar to the preceding, observing that the values of r may be positive or negative. 639. Exactly similar conclusions are obtained for the equation ax -\- by = + c, by multiplying by — c when an + i(— 7?i) = + 1 ; and by + c when an + b{ — m) = — 1. 640. It can also be proved that the equation ax -\- by = c admits of positive integral values of x and y M^hen a and b are relatively prime, and when c 7^ {ab — (a + b)) ; but though this condition is sufficient, it is not always necessary. INDETERMINATE EQUATIONS. 401 641. When the coefficient of one of the unknown quantities is 1, the equation is easily solved, by assigning to the other any integral positive values. 642. Wlien in the equation ax + hy= +C one of the coefficients, as 6, is a submultiple of c, as c = hm, then y= ±m — j-, and when X = 0, y = ± wi ; and by means of this solution the rest may be easily obtained. 643. When the equation is ax + by = 0, one solution is evidently X = 0, y = ; the rest are x = br, y = + ar. EXERCISES. 644. In the following exercises, the required values of x and y are positive and integral : — 1. Given 7a; — 12?/ = 1 5, to find the least values of x and y, and also the number of solutions, X = ^,y = i, and the number of solutions unlimited. 2. Find the least values of x and y in the equation 20a; — 31^ = 7, a; = 5, and y = 3. 3. Given 24x — ISy = 16, to find a value of x and y, The least values are x = 5, and y = 8. 4. ... 25a; — 16y = 12, to find x and y, The least values are x = 12, and y = 18. 5. ... 8a: — 3y = 1 6, to find x and y, X = 5, 8, 11, 14, ... y = 8, 16, 24, 32, ... 6. ... 7x + lly = 47, to find x and y, Only one solution : x = 2, and y = 3. 7. ... 5ar + 7y = 19, to find x and y, Only one solution : x = 1, and y = 2. 8. ... ' 5a: + 24y = 109, to find x and y, Only one solution : a: = 17, and y = 1. 9. ... 29a: + 17y = 250, to find x and y, Only one solution : x = 1, and y = 13. 10. ... 17a; - iOy = — 8, to find x and y, The least values, x — 37, and y = 13. 402 " ELEMENTS OF ALGEBRA, 11. Given 27a; + 16y = 1600, to find x and y, There are only three solutions : -j^ Z -in' 4p' 70* Apply to this example the remark in (642). 12. A owes B £100, but A has no money but guineas, and B has only 40 crowns : how can the debt be paid, and in how many ways ? A gives B 100 guineas, and receives 20 crowns from him. This is the only solution. 13. A owes B a shilling, but A has only guineas, and B has only louis-d'ors worth 17s. each : how can A pay the debt with the smallest number of guineas ? A gives B 13 guineas, and receives from him 16 louis-d'ors. 14. A number of sheep and oxen were bought for £128, 10s. ; the sheep at 17s., and the oxen at £10 a head : how many were bought ? . . . The only solution is, 12 oxen and 10 sheep. 15. How can 21 guineas be paid in sovereigns and pistoles of ^17s. : and in how many ways can it be done ? Only in one way — namely, by 11 sovereigns and 13 pistoles. 16. Is it possible to pay £18, 19s. in guineas and sovereigns ? It is not possible. 17. Can £30, 17s. be paid in guineas and sovereigns ; and if so, in how many different ways can it be done ? The only way — with 13 sovereigns and 17 guineas. 18. A merchant wishes to mix two kinds of tea at 4s. 6d. and 5s. 4d. per lb., so as to make a mixture worth 5s. per lb. : how much must he use of each, the quantities being whole numbers ? 2 lbs. at 4s. 6d. and 3 lbs. at 5s. 4d., which is the least solution. The number of solutions is unlimited. 19. How can 78 francs be paid with pieces of 5 francs and of 3 francs, and in how many ways ? With 3 pieces of 5 francs, and 21 of 3 francs. There are five solutions. II. Given two equations containing three unknown quantities, to find positive integral values of them. 645. Let X, ^, and «, be the three quantities ; eliminate one of them, as z, and a new indeterminate equation will result, contain- ing only X and _?/; find expressions for their integral values in terms of a quantity r, as in the preceding problem ; then substitute these values of x and y in either of the given equations, and a new equation will arise, containing only r and 2:, which will be divisible by the coeflicient of z^ and will thus afford an integral value of z INDETERMINATE EQUATIONS. 403 in terms of r. Substitute this value of z in the values of x and y ; and the values of a:, y, and z, being now expressed in an integral form in terms of r, their systems of values will be easily found by assigning to r proper values. 646. Sometimes the value of z cannot be expressed in terms of r in an integral form, and then the equation in terms of z and r must be solved, as in the preceding problem ; and if s be the new quantity to be assumed, the values of z and r can be expressed in terms of s in an integral form, and then those of x and y can be expressed in the same form in terms of s ; and then by assigning to s proper values, the systems of values of x, y, and z, may be found. This exception to the general rule occurs when the equation in terms of x and y has some divisor, though the exception does not always exist when this is the case. EXAMPLES. 1. Find the integral positive values of ar, y, and 2, in the equations lOx -7y + 6z= 85, 9x + 5y — 82: = 360. Multiply the former equation by 8, and the latter by 5, and add the products, and there results the equation 125a: - 31^ = 2480. Find a value of y, and proceed in solving this equation as in the former problem, and there is found X = Sir, y = 125r - 80. Substitute these values of x and y in either of the given equations, as the second, and after collecting the terms, there results 904r - Sz = 760, or z = 113r - 95. When r= 1, 2, 3, ... a: =31, 62, 98, ... y = 45, 170, 295, ... z = 18, 131, 244, ... The number of values is unlimited. The values of x and y increase by the common differences 31 and 125, the coefficients of the above equation in terms of x and y, and that of z increases by 113. 404 ELEMENTS OF ALGEBRA. 2. rind the positive integral values of x, y, and z, in the equations &x + 7y + 4:z = 122, ll:c -\-^y -Qz = 145. Eliminate s, as in the preceding example, and there results , 80x + liy = 1312, or 40a; + B7y = 656. Solve this equation by the first problem, and there is found a; = 9 - 37r, y = % + 40r. Substitute these values of x and y in the first equation, then 42: + 58r = 12, or 2z + 29r = 6, T T and z — 14r + 3 4--. Let x = s, then r = 2s. Hence 2 = 3 + 29s, then y = 8 + 80s, and a: = 9 — 74s ; when s = 0, a: = 9, y = 8, 2 = 3. This is the only possible system of values. In this example the equation in terms of x and y is divisible by 2 ; and hence it was necessary to assume a new quantity s, in order that integral values of z might be obtained (646). EXERCISES. 1. Eind the positive integral values of a-, y, and s, and the number of solutions, in the two equations ^x— y -\-hz— 127, 2a: — 3y + 72 = 131. There are only three solutions — namely, X ^^ J-j Oj t/j y = 27, 18, 9, z = 30, 25, 20. 2. Find the positive solutions of the two equations 5x - y + 42 = 127, 7ar - 3y + 22 = 151. There are only two solutions — namely, X = 20, 25, y= 1,10, 2= 7, 3. INDETERMINATE EQUATIONS. 405 3. Find the positive solutions of the equations 4x — 3y + 6z = 157, 9x — 8i/ + lOz = 311. There are only three solutions, X = 25, 43, 61, y= 3,17,31, z = 11, 6, 1. 647. To demonstrate the rule, let the given equations be ax -\-hy -\- cz = d ... [1], a'x + b'y + c'z = d' ... [2]. In order to eliminate z, multiply the former by c', and the latter by c, and take the difference of the products ; then (ac' — a'c)x + {be' — h'cyj — c'd — cd' ... [3]. Hence if x', y\ are a system of values of x and y in this equation, their other values are expressed by (633), X — xf ■\- (he' — b'c)r, y = y' — («c' — a'cy- ; and when these values are substituted for x and y in either of the given equations, as in the first, there results a(bc' — b'c)r — b(ac' — a'cy -\- cz = e, where e represents all the other terms, which are numbers, or (a'b — ab'yr -\- cz = e ... [4]. And as c is a divisor of the first member, it must also divide the second; if the problem be possible ; hence if e = ce', z = (ab' — a'hy + e' . And thus the value of z is expressed in an integral form, in terms of '/•, as those of x and y are. When the equation [3] has a divisor, it must be suppressed, and then c is not necessarily a factor of the coefficient of r in [4] ; and hence this equation may not be divisible by c, and if not, it will be of the form cz -\- hr =■ e^ from which the values of z and r must be found in an integral form in terms of a new quantity s, by solving it for the unknown quantities z and r, by the rule of the preceding problem. When r is found in terms of s, then z, and hence also x and y, can be found in terms of s ; and by assigning to it proper values, those of x, y, and z, may be found. 406 ELEMENTS OF ALGEBRA. III. Given one equation only, and three unknown quantities, to find their positive integral values. 648. When the number of unknown quantities exceeds that of the equations by more than one, the equations are said to be moie than indeterminate ; for one or more of the unknown quantities may then have arbitrary values assigned, and the equation or equations will still be indeterminate. 649. Let the equation \)q ax -\- hy -\- cz — d •, then if d — cz = c', ax -\- by = c'. Find now from this equation the values of x and y in terms of a quantity r, as in Problem I., then the expressions I'or x and y will also contain z^ and by assigning any proper values to r and 2, those of X and y will be found. 650. Limits to the value of z may easily be found by considering that its value cannot be -/ 1 ; and as its value is greatest when those of X and y are least, and their least values cannot be ^ 1, the greatest value of z cannot exceed z = ; for when c X and y are 1, the given equation becomes a -\- b -{- cz = d. The least value of z, however, may exceed 1, and its greatest may be less than the above quantity. 651. The number of solutions will evidently be limited by the extreme values of z, and may easily be found when these values are known. Let z' and z" be the greatest and least values of z, then the number of solutions will be = z" — z' -{- 1. 652. When two of the terms of the equation containing two of the unknown quantities have opposite signs, and their coefficients fulfil the conditions in Art. (635), the equation admits of an unlimited number of solutions. Eor let the equation be ax — by -]- cz = d, or ax — by = c', then for any particular value of z, this equation admits of an infinite number of solutions (637), if its coefficients fulfil the conditions in article (635.) EXAMPLE. Let 5x -\-Gy + 20z = 187 ; then 5x -\- 6y = 187 — 20z = c', and X = z-^ = — y + ~T~^' ^^^ — ^K ~ '"' *^®" y — ^ — hr— 187 — 2O5: — 5r ; hence a: = — 187 + 2O2 + (Sr. The ,• •. o /^rnN -. ^187-5-6 176 „4 limits of z are (650) 1, and —- = -^r^ = 8-. JO 2i\j o INDETERMINATE EQUATIONS. 407 When z=l,x= - 1Q7 + 6r,i/= 167 — 5r, and the limits of r are then evidently 28 and 33 ; and it may, therefore, have 33 — 28 + 1 = 5 + 1 = 6 values. Hence, when z = 1, and r = 28, 29, 30, 31, 32, 33, x= 1, 7, 13, 19, 25, 31, y = 27, 22, 17, 12, 7, 2, the values of x increasing by 6, and those of y diminishing by 5. It is similarly shewn that when z = 2, the limits of r are 25 and 29, and the number of solutions five ; the first being x = 3, y = 22, and the last x = 27, ?/ = 2. So when z = 3, there are four solutions ; when 2 = 4, there are four ; when z = 5, there are three ; for 2; = 6, there are two ; for z = 7, two ; and for « = 8, there is only one. The reader may find these solutions as an exercise. 4. When 2 = 8, then a: = 3, j/ = 2. And 8- being the limit of z, there can be no solution for greater values of z. This appears from the values of x and y ; for when 2 = 9, a; = — 7 + 6?", y = 7 — 5r, and no positive value of r can give x and y positive integral values. 653. In the above equation x cannot have an even value, for then Gy + 20z would be equal to an odd number, while 2 is a divisor of 6 and 20. Also, since 5a: + 20z = 1B7 — 6y, and the first member is divisible by 5, the second must be so too. Hence 6y must have 2 in the units' place ; that is, y must have either 2 or 7 in its units' place. The values of z are not thus restricted, for 5 and 6 are prime to each other. It is therefore in this example more convenient to transpose z to the second member, as its values may be consecutive. The remarks in this article could all ha,ve been made a priori, or before solving the equation. It would be tedious to state all the relations that may be observed. 654. In the equation ax -\- by = d — cz = c', if a and b are relatively prime, then for any integral values of z the equation is possible in integers, either positive or negative (635) ^and if it afibrds positive integral solutions for two such values of z, it will do so for all intermediate values, so that the values of z may be consecutive. When, hoAvever, a and b have a measure m, that does not divide d and c or c', the equation is then possible only for such values of z as make c' a multiple of m ; so that in this case consecutive values cannot be assigned to z. In this case, therefore, instead of taking cz to the second member, it would be preferable to take ax, provided b and c are relatively prime. In the second of the following exercises the term Sy ought not to be transposed, as 10 and 2 are not prime to each other. 408 ELEMENTS OF ALGEBRA. EXERCISES. 1. Given 5x -\- 4:7/ -\- 3z = 24, to find the positive integral values of X, y, and z. When z = 1, X = 1, y = 4. ... z = 2,x = 2,y = 2. ... z = 5,x=l,i/=l. There are only three solutions. 2. Find the positive and integral values of x, y, and z, in the -equation 10a? -\- 3y -\- 2z = 29, When. = 1,{1 = III'} when^ = 2, {^zl] There are only four solutions. IV. To find the least integral and positive numbers that, being divided by given divisors, shall leave given remainders. 655. Let the required number be denoted by V; let x, x', x", x"\ ... be the unknown quotients ; rf, d\ d", d'", ... the given divisors ; and r, /, /', /", ... the given remainders ; then is V=dx + r = d'x' +r' = d"x" + r" = d"'x"' + r'" = ... [1]. Let the equation dx -]- r = d'x' + /, or dx — d'x' = r' —r ... [2], be solved for the positive integral values of x and x' by Problem I. ; and let x and x' be expressed in terms of a new quantity p, assumed as r was in that problem ; then the values of x and x' will be of the form X = ap -{- b, x' = a'p + h' ; and hence if c' = d'b' + / V= d'x' +r' = a'd'p + c' ... [3]. The equation d'x' -{- r' — d"x" + r" now becomes a'd'p + c' = d"x" -\- r", or a'd'p — d"x" z= r" — c' ... [4], and is to be solved for the positive and integral vahies of p and x" by Problem I., and then p and x" will be expressed in terms of some new quantity p' ; thus, p = mp' + n, x" = a"p' + b" ; and hence if c" = d"b" + r" V = d"x" + r" = a"d"p' + c" ... [5], IT^DETERMINATE EQUATIONS. 409 The equation d"x" '+ r" = d"'x"' + r'" now becomes a"d"p' + c" = d"'x"' + r"\ or o:'d"p' — d"'x"' = r'" — c" ... [6], and is to be solved as before for -p' and x'"^ which will be expressed^ in terms of some new quantity p" ; thus, •p' = m'p" + n', x'" = a"'p" + V" ; and hence V = d"'x"' + /" = a"'d"Y + C" ... [7]. If there are no more conditions than the above four to be fulfilled, assume now the least value of p" that will make V positive ; and this will be the required number. 656. Another method of solving the above problem is as follows : — Let c?i, ", therefore d"x" + r" is also -positive ; that is, the value obtained for p' renders F in [5], or a"d"p' + c", positive. It is similarly proved that the obtained value of p' affords such values to p and x" as fulfil the equation [4], and also render positive Fin [3]. . Proceeding in this manner, it appears, whatever are the number of conditions, that the value of F thus obtained is the least possible value that can fulfil the conditions of the question. EXERCISES. 1. Find the least positive integer, which, being divided by 3, 5, and 6, leaves the remainders 1, 3, and 4, . . . . = 28. 2. What number, when divided by 5, 7, and 9, leaves the remainders 1, 1, and ? =36. 3. What is the least integer that, being divided by 6, 8, and 9, leaves the remainders 3, — 3, and ? . . . . = 45. 4. Find the least integer that is divisible by the nine digits, . . = 2520. 5. Find the least numbef which, being divided by 2, 3, 5, 7, and 11, shall leave the remainders 1, 2, 3, 4, and 5, . . = 1523. 6. The number of leaves in a book is known to be between 200 and 300, and in counting them over by 6 and 6, the remainder is 4 ; by 8 and 8, the remainder is 2 ; and by 9 and 9, it is found that there is a deficiency of 2 to make up the last 9 : what is the number of leaves ? = 250. EXPONENTIAL EQUATIONS. 659. In an exponential equation containing only one unknown quantity, this quantity is an exponent in one or more of the terms. 660. I. Let the equation be a^ = h. Then x . log. a = log. b, and x = t-^ — » log. a' and as a and b are known quantities, their logarithms can be found ; and hence x can be found. EXAMPLE. Eind the value of x in the equation 25^ = 34*56. Here a = 25, b = 34*56 ; and hence log. 5 _ log. 34*56 _ 1*53857 _ ^ - log. a - log. 25 - 1-39794 " ^ ^^"^''• EXERCISES. 1. Given 75^ = 48713*8, to find x, . . . . x = 2*5. 2. Find x in the equation 100^ = 250 . . x= 1*19897. 661. II. Let the equation be ac^ = d. Assume c^ = z, then a^ = d, and hence find z in this last equation by the' preceding case ; then z is known in the equation c^ = z; and c being also known, x can be found by the first case. 662. III. If the equation were ab^ = cd^, then log. a + X . log. b = log. c -\- x . log. d, and hence x = , , ;-^^- log. b — log. d 663. rV. Let the equation be x^ = a, or x . log. X = log. a ... [1]. Find by trial two approximate values of x, and denote them by x' and x" ; and when they are substituted for x in the given equation, let the results be a' and a". Then x . log. x = log. a, x' . log. x' = log. a', x" . log. a;" = log. a" ; and since log. x, log. x% and log. x", are nearly equal, therefore the difterences x — .'■', EXPONENTIAL EQUATIONS. 413 x' — x'\ are nearly proportional to the differences log. a — log. a', log. a' — log. a", or X — x' '.sf — x" = log. a — log. a! : log. a' — log. a", nearly ; , , log. a — log. a' , , ,,- hence x — x =■ r-^ — ; — -r^ — -/x^ — a/ ). log. a — log. a The quantities in this last expression being all known but x, it can be found ; for or — x' is found by the above expression, and adding x' to it, x — x' -\- x' = x. The value of x thus found is a more approximate value than either x' or x", and it may now be assumed as a new approximate value to be again denoted by x', and the nearest of the two former approximate values (x\ x"\ or some assumed one still nearer may be taken and denoted by x" ; and the same process being performed in regard to these new values a;', x'\ a still more approximate value of x may be found. This last value may again be denoted by a;', and another near value by x", and so on. By thus repeating the same process, more and more approximate values of x can be found. EXAMPLE. Given x^ = 7-38699, to find the value of a:. It is found by trial that the value of x lies between 2 and 3. Hence if x' — 2, and a:" = 3, a;' . log. x' = 2 . log. 2 = 2 X '3010300 = -6020600 = log. a' x" . log. a:" = 3 . log. 3 = 3 X -4771213 = 1-4313639 = log. a" ; also log. 7-38699 = -8684678 = log. a ; , log. a — log. a' , , hence a; — ar = , — — — -(x' — a:") log. a' - log. a"^ ^ - -8293039 X(-l) = 0-32, and x-x'+x' = x = 0-32 +2 = 2-32. This value 2-32 is nearer to x than either 2 or 3 ; and if it now be denoted by x', and if x" be assumed = 2-4, these two new values may be used as the two former ; thus, x' . log. x' = 2-32 . log. 2-32 = -8479322 = log. a' x" . log. x" = 2-4 . log. 2-4 = -9125069 = log. a'\ and as before -8684678 = log. a ; 414 ELEMENTS OF ALGEBRA. hence x - x' - i^il^Lni^^li^ro:' - x") - -J^^SSTB X (- 0-8) = -0255, and X -x' -{-x' = x = -0255 + 2-32 = 2-3455. This last value may now be assumed = x' ; and another value nearer than 2-32 or 2*4 ; as, for instance, 2*346 might be assumed for x", and the process again repeated, in order to find a still nearer value of x. The preceding value, however, is very nearly correct, the true value carried to four decimal places being 2-3456. EXERCISES. 1. Given x^ = 42-8454, to find x, . . . x = 3-2164. 2. ... ar^ = 18-4084, to find x, . . . a: = 2-8147. COMPOimD INTEREST AND CERTAIN ANNUITIES. I. — COMPOUND INTEREST. 664. As in compound interest, interest is chargeable not only on the principal, but also on the interest as it falls due, the amount therefore becomes a principal at the beginning of each period of payment.* Let p = the principal, a = ... amount, t = ... time or number of payments, r = ... amount of £1 for one of the periods of payment, then is r = £1 -f- the interest of £1 for one of the periods of payment. 665. The interest of £1 for one period will be the rate per cent. divided by 100. Thus, if the rate per cent, be 5, and the period 5 be 1 year, the interest of £1 for 1 period is = — - = -05 ; and 4 therefore r = 1-05. So for a rate of 4 per cent, r = 1 + r^ =? 1 + -04 = 1-04. When the interest is payable half-yearly ; that is, * No definitions are here given of the terms — principal, amount, rate per cent., and present value, as those may be found in the treatise on Arithmetic in Chambers's Educational Couise. COMPOUND INTEREST — ANNUITIES. 415 when one period is half a year, and the rate is 5 per cent, per 2-5 annum, or 2*5 for one period, r = I -\- — - = •! + '025 = 1-025. If the rate is 4 per cent., and the period a quarter of a year, r = 1 4- -— - = 1 + -01 = I'Ol ; and so on for any other rate or period. Since r is the amount of £1 for one period, and since any two sums are proportional to their amounts for the same rate and time, if a' be the amount of p pounds for one period, then the following proportion is evidently true — namely, l:p = r: a' ; hence a' = pr ; that is, 666. The amount of any principal for one period is equal to its product by the amount of ,£1 for the same time. Therefore pr.r = pr^ = amount of pr pounds for one period, or pr' = ... p ... two periods ; and similarly pr^ = ... p ... three ... , or pr* = ... p ... four ... ; and generally pr*^ = ... p ... t ... . Hence a = pr'' ... ... [1]. 667. Cor. Wlienp = 1, a = r' ; and hence for t = \,2, 3, 4, ... a is = r, r'^, r^, r*, ... that is, r being the amount of <£1 for one period ; r^ is its amount for two periods ; r^ for three periods, and so on ; and generally r' is its amount of t periods. By[l], p=^^ ... [2]. Hence, by the principles of logarithms. By [1], La = Lp + tLr ... [3]. Hence Lp = La — tLr ... [4]. Lr=\{La-Lp^ ... [5], _ La — Lp ^-~~Lr • 668. It appears from these formulas, that when any three of the four quantities p, a, r, and t, are given, the fourth can be found. 669. Since the amount of a sum of money p for the time t is a = pr\ therefore p is also the present value of a sum a, which is to be paid at a future time distant from the present by t intervals, and by [2], p = 4- [6]. 416 ELEMENTS OF ALGEBRA. II.— %UtfOUNT OF CERTAIN ANNUITIES. 670. An annuity is a constant sum of money paid at regular intervals ; and it is said to be certain, when it is independent of accidental circumstances. 671. When the annuity is in arrears; that is, when it has remained unpaid beyond the period when it is due. Let a = the annuity, m = its amount, t =■ the time or the number of payments due, r = the amount of £1 for one period. Then since the last annuity is due only at the expiry of the time t, no interest is chargeable upon it, so that its amount is only a ; the annuity preceding it has been due for one period, and hence (665) its amount is ar • the annuity preceding this has been due for two periods, hence its amount is ar"^ ; the amount of the next preceding is evidently ar^ ; of the next a?-* ; and so on to the first, which has been due for (t — 1) periods, and its amount is therefore ar^~\ Hence the amount of all the annuities is = a -\- ar -i- ar^ -\- ar^ -\- ... ar'~\ But this is just an equirational series ; and hence (424) its sum is rz — a _ r.ar'"^ — a _ar* — a _ a(r* — 1) ~ r — 1 ~ r — 1 ~ r — l~ r — 1 ' a(r' - 1) ... Hence m = —■ ... [IJ ; r — 1 therefore a = \, _ ^ ... [2], and logarithmically, Lm = La-{- L(r* - 1) - X(r - 1) ... [3], La=Lm- L{r' - 1) + X(r - 1) ... [4], t = {L[a -\- mir - ly] - La] ^ Lr ... [5]; hence tLr = L{a + m(r — I)} — La ... [6], The derivation of the formulas [3] and [4] is easily effected by taking the logarithm of [1], and is left as an exercise to the student. 672. The present value of an annuity is the sum that an annuity r- -1 a{r^- -1) r - -1 a{r'- -1) r\r- -1) vr\r- -1) COMPOUND INTEREST — ANNUITIES. 417 is at present worth, supposing it to begin at the present time, and to continue till some future period. Let V = the present value of the annuity, then it is evident that, if the annuity is to continue from this date for the time ^, v must be such a sum as, if laid out at interest for that time, would just be equal to the amount of the annuity for the time t. But {Q)^^') the amount of v for that time is = vr\ and (671) the amount of the annuity for that time is = — ^ — ; and therefore whence v = ~ ~ ... [1], and a = ;, _ ^' ... [2]; and logarithmically, Lv = La-^ L(/ - 1) - tLr - L{r - I) ... [3], La = Lv - L(r' - 1) + tLr + L(r - 1) ... [4], t = {La - L\a - v(r - V)]] ^ Lr ... [5], Lr = {La - L[a - v{r - ly]} ^ t ... [6]. The student can easily derive these logarithmic formulas from [1]. In order to find [5], the value of r' must first be obtained. G73. An annuity is said to be in reversion, when it is to begin at some future time, and to continue for a limited period ; it is also called a deferred annuity. The present value of a reversionary annuity is evidently equal to the difference between the present values of two other annuities, both of which begin at the present time, and continue, the one to the beginning, the other to the expiry, of the reversion. Let t = period of the reversion, n = ... from the present time to the beginning of the reversion, then n + t = ... from the present time to the expiry of the reversion. Also, let V be the present value of the reversion, and v\ v'\ the present values of two annuities, beginning at the present time, and continuing respectively for the periods n and w + < ; then (G72) , a(r» - 1) , „ «(r"+' - 1) " — -^^ i, and v' — ^ ^ • ^\r - 1)' r«+'(^ - 1) ' 418 ELEMENTS OF ALGEBRA. a r'-l _ a(y - 1) hence ^.«(r _ 1) • r' r'^'-Xr - 1)' therefore v = ^}l^^ _ ^^ ... [1], and hence a = j^ — ; — ~ ... 121 ; and logarithmically, Lv = La + L{r' - 1-) - (n + f)Lr - L(r - 1) ... [3], La = Lv - L(/ -!) + (« + {)Lt + L(r - 1) ... [4], t = {La- L\a - r"(r - l)y]} -;- Xr ... [5], n={La + L{f - r)-Lv-tLr-L(r-V)]^Lr ... [6]. The derivation of these logarithmic formulas from [1] is left as an exercise. 674. An annuity is said to be perpetual, or it is called a perpetuity, when it is to continue for an indefinitely long period. In this case the value of an annuity may he found from [1] in (672), by considering t to be infinite ; thus — /•'(r - 1) , r - 1 (1 a(r^ - 1) ^ tM\ _ r* oo ^) and when < = go, r' is = oo, and —r — — = ; hence w = [1], T — 1 therefore o = tot • lip number of men in each side of the hoiiow square. Re*ii :■■ number of men in the hollow sqiiare, .• . . :- 80. A vintner sold 7 dozen of sherry and 12 dozen of cl . . : £50 ; he sold 3 dozen more of sherry for £10 than he did of ciar for £6.. Required the price of each, A dozen of sherry cost £2, and a dozen 31. The r. wht"-' of a carrutg^ makes 6 revolutic 422 ELEMENTS OF ALGEBRA. the liind - wheel in going 120 yards; but if the circumferci;c. each wheel be increased 1 yard, it will make only 4 revolutio; more than the hind -wheel in the same space. RecLuired t. circumference of each, The circumference of the less is i, and of the greater 5 yard 32. The sum of the squares of the extremes of four numbers i equidifFerent progression is 200, and the sum of the squares of 1 ' means is 136. What are the numbers ? " = ± 14, ± 10, +6, and + 1^. 33. The sum of the first and second of four nimibers in equi- rational progression is 15, and the simi of the third and fourth i- 60. Required the numbers, . . . = 5, 10, 20, and 4' 34. A poulterer bought 15 ducks and 12 turkeys for 5 guinea-^ . he had two ducks more for ISs. than he had of turkeys for 20s, What was the price of each ? The price of a duck was 3s., and of a turkey T 35. There are three numbers in equirational progression, tl sum of the first and second of which is 9, and the sum of the fir and third is 15. Required the numbers, . . = 3, 6, and 1:. 36. There are foxir numbers in equirational progression, the second of which is less than the fourth by 24, and the sum of the extreuies is to the sum of the means as 7 to 3. Reqxiired tl numbers, = 1, 3, 9, and 2 37. There are four numbers in equidifterent progression; tl^- sum of the squares of the "first and second is 34, and the sum of the squares of the third and fourth is 130. Required tJic numr hers, . . . . . . . .= 3, 6, 7, and 0. 38. The sum of £700 was divided among four persons, whose shares were in equirational progression ; and the difference between the greatest and least was to the difference between the means as 37 to 12. What were their respective shares ? ^ = £108, £144, £192^ and £2r.' 39. The sum of four whole numbers in equidifferent progressituj is 20, and the sum of their reciprocals is |J. Required the numbers, = 2, 4, 6, and 8. 40. The three sides of a right-angled triangle are in equidifferent progression, and the difference of the squares of the two sides^ is 63, and the difference of the sqiiares of the hypotenuse and tv> least side is 144. Find the three aides of the triangle, = 9, 12, and I 41. The equidifferent mean between two numbers exceeds il equirational by 13, and the equirational exceeds the harmonic i 12. What are the numbers ? . . . . = 104, and "" 42. There are four numbers such, that if each be multiplie( their sum, the products are 252, 504, 396, and 114. Find tlie - numbers . . — 7 ' ! ' MISCELLANEOUS EXEIlCISES. 423 io. What quantity must be added to each, of the terms of the •1 ;o ad — he ratio a : b, that it may become equal to c : a ? = — — d ' 44. If the equidiflferent mean between a and b be-twice as great .3 the equirational mean ; prov<« that a:b :: 2 •}- \/ii -.2 — ^3. 45. If the equidifferent mean t)etween a and 6 be to times the T • *v. Ml" \/'m + ,/(m-l)' ■ larraomc, then will y = — -7 rr. 46. If the equirational mean bf-.tween a and b he m times the uarmomc, then will -.- = ->-5 — r> • b m — is/(nr—\) 47. Prove that in any equirational piogression, the sum of the first and last terms is greater than the sum of any other two terms '■ quidistant from the extremes. 48. Prove that in any equirational progression consisting of an »'eu number of terms, the sumtf the odd terms is to the sura of ite even terms as 1 to the common ratio. 49. The number 2577, expressed in a particular scale of n i Ion, is 40302 ; find the root of the scale, . . . - 50. What is the root of the scale of notation in whicli a number •rhich is double of 145 will bo expressed by the same digits? =15. 51. If « be a whole number, prove that «' + 5« is divisible by G- 62. Prove that the square of every odd number, diminished by 1, is divisible by 8. 53. If from the cube of any even number there be subtracted 4 . ;me8 the number itself, the remainder wiU be divisible by 48. Required a proof. 54. What number, multiplied by 48, will give a product which s a complete fourth power ? =27. 55. Tliere are four numbers, the first three of which are in . 4ui(iitferent, and tlie last three in harmonic progression; prove .i;at the first has to the second the same ratio which the third has ■ ') the fourth. 56. Prove that an equirational mean between two quantities is mean proportional between an equidiflferent and harmonic mean netM^cen the same two quantities. 57. Form the equation whose roots are 2 + V — 3, 2 — \/ — S, L, and - 5,. . . . . = x* ^- Ux^ + 48a: - 35 = 0. 58. Form the equation whose roots are + V — 2, — V — 2, 3, and 4, . . . . = x* - Ix^ + 14x* - 14a? + 24 = 0. 69. Form the equation whose roots are 3 + V— 2, 3 — V — 2, ', and 3, - . . ~ x* - IW + 47z= - 91x + 66 = 0. 60. Find, by the method of divisors, the roots of the equai x^ - 9x2 ^ 22jr - 24 = 0, =0, ^{A + V - 7), and ^3 - V -7.; 61. Find, by the method of divisors, the roots of the equation, a:* - 4a^ - 8x 4- 32 =r 0, = 2, 4, - 1 + V - 3, and - 1- V- 3. 62. Find, bv the method of divisors, the roots of the equation, X* +x^ - 2di' - 9x + 180 = 0, = + a, - S, + 4, and - 5. 03. Find, by the method of divisors, the roots of the equation, :r* - 17x* + 101 x' - 223x=- - 18x + 360 = 0, = — 1, + 3, + 4, 4- 5, and 64, Find the least values of x and y that fulfil the conditions ol the equation 19x — 117^/' = 11, . ' . . . =56 and 9. 65, Find all tlie values of x and y that the following eqtiatioa admits of in integers, 1 3a; + 14y = 200, . . = 10 and 5. G6. Find two fractious having 5 and 7 for denominators, whose sum is equal to f^, =4 and \, 67. What number is that which, being divided by 9 and 13, shall leave for remainders 5 and 12 ? . . . . =' 77. (•■S. A root of the equation, x* + dx^ + 2x^ + 6x - 148-6 - lies between 2 and 3 : it is required to lind that root, = 2-7344 neaiiv. 69, One root of the equation, t» + 6a:* - lOx' - 112x' - 207* — 110 =: 0, lies between 4 and 5. Kequired that root, = 4-4641016. 70. ^ne root of the equation, x^ + 4a:*- 2x* + lOa? - 2x — 962 = 0, is found to lie between 3 and 4. Required the develop^' "* of it to seven decimal places, .... — 3'S54c 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recalL DEC 6 196561 . REC'D LB NOV ! .4'66-SPll fi^l4l3» HECtiiVcO AUG 7 '68 -12 M T.Tk 01 A_«rt,»-Q 'flK General Library ''( \