$B 3Db M36 .:»;,;»;; i'^Pfwjiiilfirc UHl/lKSnVQp CAilfOiNIA f^W^^^^THW UK* / ^"^^ 9 /' "^-r u - • (-^^e ni L 7 n Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/elementaryalgebrOOsmitrich ELEMENTARY ALGEBRA jS&^^ ELEMEiNTARY ALGEBRA^^^ ,, BY // . / CHARLES SMITIJ, M.A. f' Author of "A TREATWE'ffN Algebra," *rAN Elkmentab* TR^Iiraa ON Conic Sections," Eia y^ *^ REVISED AND ADAPTED TO AMmiCAN SCHOOLS BY IRVING STRINGHAM, Ph.D. Professor of Mathematics and Dean of the College Faculties IN the University of California COMPLETE EDITION THE MACMILLAN COMPANY LONDON: MACMILLAN & CO., Ltd. 1898 All rights reserved iDXJCATioi lob; Copyright, 1894, By MACMILLAN AND CO. Set up and electrotyped June, 1894. Reprinted September, 1894; February, August, 1895; January, September, 1897; Jan- uary, August, 1898. ( cUutau (B'^A Norton otj ^rras : J. S. Gushing & Co. — Berwick & Smith. Norwood, Mass., U.S.A. Add to Iiib. GIFT EDUC. UBRARY PREFACE TO THE AMERICAN EDITION. The transition from the traditional algebra of many of our secondary schools to the reconstructed algebra of the best American colleges is more abrupt than is necessary or creditable. This lack of articulation between the work of the schools and the colleges emphasizes the need of a fuller and more thorough course in elementary algebra than is furnished by the text-books now most commonly used. It is with the hope of supplying this new demand that an American edition of Charles Smith's Elementary Algebra is published ; a work whose excel- lencies, as represented in former editions, have been recognized by able critics on both sides of the Atlantic. In the rearrangement of the work and in its adaptation to American schools many changes have been made, too many to be noted in a short preface, and a considerable amount of new subject-matter has been introduced. The following are innovations of some importance : Chapter I., consisting of a series of introductory lessons, is wholly new, and Chapter XIII. is partly new and partly transferred from Chapter XXVIII. of the second edition. Horner's synthetic division is made prominent in the chapter on division, an early introduc- tion to quadratic equations finds its appropriate place in the chapter on factoring, the binomial theorem for posi- v 929 VI PREFACE. tive integral exponents is demonstrated by elementary methods in the chapter on powers and roots, the chapter on surds has been enlarged by a short discussion of com- plex numbers, the chapter on exponentials and logarithms has been re-written, and Chapters XXVII, XXVIII, XXX, XXXI, XXXII, XXXVII, and XXXVIII are wholly new. Some new collections of examples have been introduced, and several of the older lists have been extended. As thus reconstructed, the book constitutes a rounded course in what may be called the newer elementary alge- bra, and includes all the subjects prescribed for admission to American Colleges and Schools of Technology. It will prove especially helpful to students preparing for such colleges as are using Mr. Smith's Treatise on Algebra for advanced work. The most important sources that have been drawn upon in the preparation of the new chapters are Professor Chrystal's Text-Book of Algebra and Mr. Smith's own larger book, the Treatise, I am indebted to my colleagues of the mathematical department of the University of California for valuable suggestions, but especially to Professor Haskell and Dr. Hengstler for contributions to subject-matter and for reading many of the proof-sheets. Special thanks are due to Mr. Smith, for allowing free scope for this revision. IRVING STRINGHAM. University of California, February, 1895. CONTENTS. CUAPTEB PAGE I. Introductory Lessons 1 II. The Language of Algebra. Definitions 17 Positive and Negative Quantities 27 III. Addition 32 Subtraction 37 Brackets 43 IV. Multiplication 46 Product of Multinomial Expressions 56 Continued Products 67 V. Division 71 Division by Multinomial Expressions 75 Horner's Synthetic Division 78 Miscellaneous Examples I 88 VI. Simple Equations 92 VII. Problems 101 VIII. Simultaneous Equations of the First Degree 109 Elimination by Addition and Subtraction ^ . 110 Other Methods of Elimination 114 Equations with Three Unknown Quantities 119 IX. Problems 122 Miscellaneous Examples II 128 X. Factors 132 Factors of ax'^ + hx-\- c 142 Factors by Rearrangement of Terms 147 Quadratic Equations 151 vU VIU CONTENTS. CIIAPTKB PAGE XI. Highest Common Factors 156 Xll. Lowest Common Multiples 108 Xill. Miscellaneous Theorems and Examples 174 Mathematical Induction 174 Factor Theorem. Remainder Theorem 179 XIV. Fractions 18(5 Reduction to a Common Denominator 193 Addition of Fractions » 194 Multiplication of Fractions -. 201 Division of Fractions 202 Reciprocal. Infinity 203 Two Important Theorems 208 XV. Equations with Fractions 210 Simultaneous Equations 221 Miscellaneous Examples III 229 XVI. Quadratic Equations 234 General Properties 239 Irrational Equations 245 Relations between the Coefficients and the Roots. . . 249 Special Forms 251 Equations with Given Roots 253 XVII. E(iuations of Higher Degree than the Second 200 Will. Simultaneous Equations of the Second Degree 200 XIX. Problems 279 Miscellaneous Examples IV 285 Miscellaneous Equations 289 XX. Involution 297 The Binomial Theorem 302 Evolution. 309 Square Root 310 CONTENTS. IX CHAPTEK XXI. XXII. XXIII. XXIV. XXV. XXVI. XXVII. XXVIII. XXIX. XXX. XXXI. XXXII. XXXIII. XXXIV. XXXV. XXXVI. PAGB Fractional and Negative Indices 319 Surds 328 Complex Quantities 337 Ratio 344 Proportion 349 Variation 357 Miscellaneous Examples V 363 Arithmetical Progression 367 Geometrical Progression 380 Harmonical Progression 393 Other Simple Series 398 Miscellaneous Examples VI 403 Inequalities 409 Limits 417 Indeterminate Forms. . 421 Exponentials and Logarithms 426 Exponentiation . . 427 Logarithmic Operation 430 Natural Logarithms 436 Convergency and Divergency of Series 447 Indeterminate Coefficients 465 Application to Integral Functions 469 Application to Partial Fractions 472 Application to Expansion of Functions 476 Application to Summation of Series 480 Permutations and Combinations 485 The Binomial Theorem for an Integral Index 406 The Binomial Theorem : any Index 507 Exponential and Logarithmic Series 514 Logarithmic Computation 521 CONTENTS. CHAPTER PAGE XXXVII. Continued Fractions 530 XXXVIII. Determinants 648 XXXIX. Scales of Notation 572 Miscellaneous Examples VII 578 •^ ALGEBRA. CHAPTER I. Introductory Lessons. 1. Signs and Symbols in Algebra, In algebra a letter is frequently used to denote a number. When so used in a connected series of operations, a given problem for example, the letter must be regarded as standing for the same number in all the operations of the series. Several distinct letters may be used to denote several distinct numbers in the same problem. In these Introductory Lessons the signs -}-, — , x, -J-, and =, have the usual significations given to them in arithmetic. Their formal definitions are given in Arts. 12, 13, 14, 15, and 21, of the next chapter. When letters are used to represent numbers, the sign of multiplication is frequently omitted. Thus ah means a X 6. 2. The Method of Algebra. The statement of an equality, made by placing the sign = between two numbers, or two sets of numbers, indicating that the two numbers, or two sets of numbers, are equal to one another, is called an equation. 2 INTRODUCTORY LESSONS. Thus 2x = 5 and x-\-2 = a-\-3 are equations. Problems stated in the form of equations are said to be stated algebraically. In this introductory chapter some of the uses of alge- bra will be explained by stating and solving, with the help of equations, certain classes of problems already familiar to the learner who has studied arithmetic. In applying the method of algebra, the first step is to translate the ordinary language in which the problem is expressed into a concise symbolical statement, and the form used for this purpose is the equation. The solution of this equation then solves the problem itself. Algebra has no one universal rule that will put every problem expressed in ordinary language into the algebraic form, but rather it devises ways and means for solving the various classes of equations that are produced by dif- ferent practical problems. Our first concern is to state the given problem in the form of an equation, and the way to do this must be sought in the language of the problem itself. A few simple examples will best explain how. Ex. 1. If $ 10 were added to twice the money I have, the result would be $ 150. How much money have 1 ? The specifications of this problem may be expressed more con- cisely thus : Twice the money I have^ added to $ 10, equals $ 160, or again, using the signs -f and = in the ordhiary arithmetical sense, Twice the money I have + f$ 10 = $ 150. The problem is here stated in the form of an equation, which as yet, however, is not purely algebraic. INTRODUCTORY LESSONS. B This analysis will not be in any way disturbed by the replacement of 10 and 150 by any other numbers we choose to insert, and we may analyze with equal facility the following problem, in which a and b stand for any numbers, although we impose, for the present, the con- dition that b shall be greater than a. Ex. 2. If a dollars were added to twice the money I have, the result would be b dollars. How much money have I ? Stated in the briefer form, this problem says ; Twice the number of dollars I have +^a = $b. Ex. 3. The sum of two numbers is- 50, and their difEerence is 20. What are the numbers ? The first condition of this problem asserts that The larger number + the smaller number = 50 ; and since we obviously get the larger number by adding their dif- ference to the smaller, the second condition asserts that 20 4- the smaller number = the larger number. Hence, replacing ' the larger number ' in the former of these two statements by its equal '20 -|- the smaller number,'' we obtain 20 + the smaller number + the smaller number = 50, or, in briefer form, 20 -f twice the smaller number = 50. From this statement the smaller number is at once seen to be 15. Observe that here, as in Ex. 1, the numbers 20 and 50 may be replaced by letters representing any other num- bers, without in any way affecting the analysis of the problem, or the form of its solution. These examples point to the fact (they do not prove it) that no matter what the relation of the quantities involved in a determinate problem may be, it can be expressed in the form of an equation. 4 INTRODUCTORY LESSONS. But although the abbreviated statements thus far pro- posed are all more concise than the corresponding state- ments in ordinary language, they can be reduced to still more convenient form. Thus, in Ex. 1, if we briefly indicate 'the number of dollars I have' by some symbol or letter, as $ a;, the equation Twice the number of dollars I have -f $ 10 = $ 150 becomes 2xfa; + f 10 = ^ 150, and is now more strictly algebraic. Any symbol, for example ($?), may here be used to stand for Hhe num- ber of dollars I have, but it is the universal custom to use a letter for this purpose, and obviously it may be any letter. It must be noticed that, in the algebraic statement of a problem, x is an abstract number, which may turn out to be 5, or 11, or |, or in fact any number, and that, therefore, x dollars means x times one dollar, or a; x (1 dollar). Similarly, a; inches means a; x (1 inch), etc. In other words, x dollars is a product, the first factor of which is the abstract number x, the second being the concrete unit one dollar, in terms of which all the quan- tities involved in the problem must be expressed. The unit being then well understood, we drop it out of the equation, which in the case of Ex. 1 then becomes 2 a; -f- 10 = 150, a pure algebraic equation. The purely algebraical state- ment of Ex. 2 is, similarly, 2a;-f a = &, and that of Ex. 3 is 20 -f- 2 a; = 50, in wliich x stands for ' smaller number.' INTRODUCTORY LESSONS. 6 It is important that the beginner clearly understand what these algebraic equations mean. They can be easily translated backwards into the ordinary language in which the problems were originally expressed. An equation is thus simply a sentence in algebraic language expressing the conditions of the problem. And besides its conciseness of statement, giving at a glance the rela- tions between the quantities involved in it, the theory of algebra will eventually show that out of it may be evolved, by simple processes, the value of the impor- tant quantity x, the unknown quantity, which, when found, answers the question propounded by the problem. State the following problems algebraically : Ex. 4. What number is that whose haK added to 73 will pro- duce 85 ? Ex. 6. What number is that whose nth part added to a will produce b ? Ex. 6. A boy had a apples. He gave a certain number of them to his companions, and then had b times as many as he had given away. How many did he give away ? Let the learner now read the axioms of Art. 93. 3. Problems involving the Four Fundamental Processes of Arithmetic. Ex. 1. If the sum of x and 2 x and 3 a; be 42, what number does X stand for ? The algebraic statement of the problem is a: + 2x + 3a; = 42; and this is the same as 6 a; = 42 ; for, obviously, any number added to twice itself and three times itself is six times the nuuiber. Hence, since 6 times x is 42, x = 1. 6 INTRODUCTORY LESSONS. Ex. 2. If 5 a; be subtracted from the sum of 3 sc, 4 at, and OJ + 5, the result is 35. What number does x stand for ? The algebraic statement of the problem is 3a; + 4ic + a! + 5- 5x = 35. In order to find what x stands for, we first combine the four terms containing x into one term, which, by the rules of addition, subtraction, and multiplication in arithmetic, obviously is 3 a;; hence 3 x + 5 = 35. From both sides of this equation subtract 5 ; then 3x = 30, whence x = 10. Ex. 3. The sum of two numbers is 50, and their difference is 20. What are the numbers ? This is the third problem of Art. 2. It was there stated alge- braically ; we now solve it in the purely algebraic form. Let x represent the smaller number ; then 20 + x is the larger, and, by the conditions of the problem, 20 + a; 4- a; = 50. We now subtract 20 from both sides of the equation, and obtain 2 a; = 30 ; whence x = 15, the smaller number, and 20 + ic = 35, the larger number. Ex. 4. The larger of two numbers is 3 times the smaller, and their difference is 92. What are the numbers ? Let X represent the smaller number ; then 3 a; is the larger, and by the second condition of the problem, 3x-a; = 92. But three times any number less the number itself is obviouslj twice the number. Hence 2a; = 92, and X — 46. INTRODUCTORY LESSONS. 7 Ex. 5. A, B, and C together have ^75 ; A has twice as much as B, and B has $5 less than C, How much has each ? Let X represent B's share ; then 2 x is A's share, and x + 5 is C's share, and the algebraic statement of the problem is, x + 2x + a;+5 = 75. From this equation are obtained, by the processes already explained in the preceding examples, x=\l\, 2x = 35, x+5 = 22i. Ex. 6. An indicated sum, as (w + ^ 7i — 1), is frequently placed in a parenthesis in order to indicate that an operation is to be performed upon the number which the sum represents. Thus, if n = 6, n+ |n- 1 =8, and 2 X (n + hn - l) = 2 X S = IG. This result will again be the same if each term of the sum be multiplied by 2 before the substitution of 6 for n is made. Thus, whatever number n may represent, we shall find it to be always true that 2 X (n + i w - 1) = 2 X n + 2 X i 71 - 2 X 1, and Ave verify this result iu the present instance by put- tiug G in place of n, thus obtaining 2x6 + 2x^x6-2 = 16. It is of extreme importance to observe that we have here, not sl proof oi a principle or formula, but a verifica- tion of it in a particular instance. Ex. 7. If n stand for 9, what numbers are represented by 2 «, 5 ?^, n + 3, /i - 5, 2 » + 3, 3 n - 2, 3(» + 1), 3(n - 4), 3 n + 2 >i - 4 «, \{n - 3), 3(^ « - 1)? 8 INTRODUCTORY LESSONS. Ex, 8. John is 4 years older than William, and 4 times John's age is 5 times William's. What are their respective ages ? Let X represent John's age ; then a; — 4 is William's age, and the algebraic statement of the problem is b(x — 4) = 4x. Here the parentheses enclosing x — 4 indicate that both x and 4 are to be multiplied by 5. Performing this multiplication, we obtain 6x-20 = 4x; and from this equation, by adding 20 and subtracting 4x, perform- ing the operations on both sides of the equation, we find X = 20, John's age, and then x - 4 = 16, William's age. examples" I. 1. Divide 15 into two parts whose difference is 7. 2. The sum of two numbers is a, and their difference is '). What are the numbers ? 3. A and B together own 140 acres of land, and B owns 3 times as much as A. How many acres has each ? 4. A man having $02 spent a part of it, and then had 3 times as much as he had spent. How much did he spend ? 6. A farmer sold 15 bushels of barley at a certain price, and 25 bushels of wheat at twice the price of the barley, and he received $14 more for the wheat than for the barley. What were the prices per bushel of the wheat and of the barley ? 6. Twenty-four cohis, consisting of dimes and half-dollars, amount to $0. How many coins of each kind are there ? 7. A boy gave one-third of all the apples he had and one-third of an apple more to his sister, and then had one apple left. How many apples did he give to his sister ? INTRODUCTORY LESSONS. d 8. What number is that whose double exceeds its half by 27 ? 9. Verify the following equations by putting some number (any number) in place of n : (1) A«+W:=fw, (2) 3x(w+l)+l = 3» + 4, (3) 3 X (w - 4) + 2 = 3 w - 10, (4) 5m-2(w- l)=5?i-2« + 2 = 3« + 2, (5) i(n-l)+2 = ^»i-i + 2 = ln-{- I, (6) i n - 2(2 - i w) = ^ ?i- 4 + n = f w - 4. ' 10. A can do a piece of work in 10 days, A and B together in days. In how many days can B do it alone ? 11. The age of a father is now 7 times that of his son, but in 3 years it will be only 5 times that of his son. How old are each ? 12. At what time between 2 and 3 o'clock are the hour and minute hands of a watch together ? 4. Problems involving the Rule of Three. Ex. 1. What number has to 12 the same ratio as 57 to 9? Let X represent the number. Then if the ratio of 57 to 9 be denoted by 57 : 0, the conditions of the problem require that X : 12 = 57 : 9 ; or this statement may be made in the equivalent fractional form ^^57 12 9' Multiplying both sides of this equation by 12, we obtain 9 10 INTRODUCTORY LESSOi^S. Ex. 2. How long will it take to fill a cistern of 165 gallons by a pipe that fills one of 120 gallons in 8 minutes ? If X represent the number of minutes it takes to fill the larger cistern, then by the rule of three (proportion) ^^165 8 120' and by multiplication of both sides of this equation by 8, „ 8 X 165 120 = 11. Ex. 3. A train goes at uniform speed one-third of a mile in 20 seconds. What is its speed per hour ? Let X represent the number of miles per hour which denotes the speed of the train. Then, by the rule of three, or proportion, the algebraic statement of the problem is 3600 20' since there are 3600 seconds in an hour. From this equation, by the proper multiplications and reductions, we find x to be „ 3600 = 60. 3x20 Ex. 4. If it require 15 men to build a house in 92 days, how many men must be employed in order to build it in 60 days ? This problem may be stated in the form of a proportion, or as follows. It requires 15 x 92 days' work for tlic buildinii; of the house ; hence, if x stand for the number of men that can build il, in 60 days, X X 60 = 15 X 92. Therefore 15x_92^23. 60 INTRODUCTORY LESSONS. 11 Ex. 5. If it require a men to build a house in h days, how many men must be employed in order to build it in c days ? It requires a x h days' work for the building of the house ; hence, if x stand for the number of men that can build it in c days, • XX c = a xh. Therefore a xh EXAMPLES II. 1. Divide $20 into two parts whose ratio to one another is the same as the ratio of 3 to 7. 2. If 5 tons of hay cost $ 84 , how many tons can be bought for .1^270? 3. If the speed of a train be 45 miles an hour, how long will it take it to go 360 miles ? 4. How long will it take to fill a cistern containing a gallons by a pipe that fills a cistern containing h gallons in c minutes ? 5. A train goes at uniform speed a miles in h minutes. What is its speed per hour ? 6. It requires a men to build a house in h days. How long will it take c men to build it ? 7. In Ex. 6 put 15 for «, 75 for &, and 20 for c, and work out the consequent numerical result. 8. A's age is now twice B's age ; but in five years their ages will be in the ratio of 7 to 4. What is the age of each ? 9. If the wages of 6 men for a days be $00, what will be the wages of 13 men for the same time and at the same rate ? 10. If 7 men earn $105 in 6 days, how many men, working for the same rate of wages, will earn $ 157.50 in 7 days ? 12 INTRODUCTORY LESSONS. 11. If 10 men can reap in 3 days a field whose len^h is 1200 feet, and breadth 800 feet, what is the breadth of a field whose length is 1000 feet which 12 men can reap in 4 days ? 5. Problems in Percentage and Interest. Ex. 1. Of a herd of cows, 280 are Jerseys, and these are 35% of the entire herd. How many cows are there in the herd ? Let X stand for the number of cows in the herd. Then y^^ x x is the number of Jerseys. Hence, the algebraic statement of the problem is tVff X X = 280, and the value of oj, derived from this equation, is „ 100 X 280 35 = 800. Ex. 2. A town lost 7% of its population, and then had G045 inhabitants. What was its population before the loss ? Let X be the original, unknown number of inhabitants. Then yj^^ X « was the loss, and x — ^^^ x x was the number of inhabi- tants remaining after the loss. Hence X - -f^j X X = 0045. Now X less seven hundredths of x is ninety-three hundredths of x ; therefore yVo x x = 6045, , 100 X 6045 r.rr,r, and X = — = 6500. Ex. 3. The amount due on a note for $800 at 5% simple in- terest was $ 000. For what length of time was interest reckoned ? The interest for one year is yfij^ x 800 dollars, or .$40 ; hence the interest for x years is 40 x x dollars, and the amount due after X years is 800 4- 40 x x dollars, and this, by the conditions of the problem, is $000. The equation expressing the conditions there- fore is 800 + 40x = 900. INTRODUCTORY LESSONS. 13 Subtract 800, and divide by 40, performing these operations on both sides of the equation ; then 900-800_„. 40 ' Thus the length of time for which interest was reckoned is 2i years. Ex.4. What principal amounts to $1456 in 2 years at 6% simple interest ? If X be the principal, j^^ X a; is the interest for one year, twice this is the interest for two years, and x + (2 x j^^ x x) is the ataount. Hence the parenthesis being here used to indicate that the entire product within it is to be added to x. (See Exs. 6, 7, 8, Art. 3.) But j^o ^^ ^ added to .r is ]-^g of x. Therefore {Uxx = 1456, and ^^100x1456^^3^^ 112 The required principal is $1300. Ex. 6. The amount due, after 3| years, on a note for $ 1500 bearing simple interest, was $1788.75. What was tlie rate of interest ? Let X stand f on, the rate of interest ; then 1500 x x is the interest for one year, 3^ x 1500 x Xy or 5250 x x, is the interest for 3| years, and 1500 + (5250 x x) is the amount due. Hence 1500 + (5250 xx) = 1788.75. Subtract 1500, and divide by 5250, performing these operations on both sides of the equation ; then 1788.75 - 1500 ^.. X = = .055. 5250 Thvis the rate of interest was 5| per cent, 14 INTRODUCTORY LESSONS. EXAMPLES III. 1. What number increased by ^ of 25 % of itself equals 315 ? 2. A city gained 13 % in population, and then had 80,456 inhabi- tants. What was its population before the gain ? 3. Tlie population of a city increased from 31,000 to 33,046 in ten years. What was the average per cent of increase per annum ? 4. The annual rent of a house is $240, and this is 8% of its value. What is its value ? 5. What sum bearing interest at 5^% will yield an annual income of $1500? 6. In what time will any sum of money double itself at 8% simple interest ? 7. The amount due, after a years, on a note for b dollars bear- ing simple intert'.st, was c dollars. What was the rate of interest ? 6. Problems involving the Use of Negative Numbers. In the vsecoiid problem of Art. 2, whose statement in the algebraic form was found to be and whose solution is therefore x = \{b-a)y we said that h must be greater than a. Now while this restriction is necessary in arithmetic, the method of algebra obtains a solution for all values of a and h. In accordance with the enlarged algebraic- view, the problem nndor consideration may be stated as follows: Ex. 1. I have no money ; but balance my accounts by .setting off debits ajjalnst credits, then double the thus ascertained value INTRODUCTORY LESSONS. 15 of my possessions, and add a dollars to it, and the result will be h dollars. Can I 'pay my debts with the money due me, and how much money, if any, shall I have after all accounts are settled ? If the excess of credits over debits be x dollars, then by the con- ditions of the problem 2 x + a dollars is equal to h dollars, or, in the purely algebraic form, '2x-[- a = h. From this equation we obtain, as previously, x = \{h-a). Thus, when all accounts are settled, I shall have \ (b — a) dollars. Let us see what will happen under the three different suppositions that may here be made. (i.) If the credits exceed the debits (in amount), I shall be able to pay my creditors with the money due me and have some- thing left. This something is | (6 — a) dollars, and obviously b must be greater than a. (ii.) If the credits be just equal to the debits, they cancel each other in the final settlement of accounts, and I shall have just nothing left. This nothing is again i (& — a) dollars, and in order that I (b — a) may be zero, b must be equal to a. (iii.) But if the debits exceed the credits, I shall fail to cancel all my indebtedness with the money due me, and shall have some debts remaining and no money to pay them with. I no longer have possessions, but debts. The amount of my indebtedness is exactly I (a — 6), and the condition for this state of affairs is that a shall be greater than b. Now algebra takes account of this condition by the simple device of writing the minus sign before the positive number V (a — 6), and calling the result negative ; thus x = \{b-a) = -l(a-b), a negative number, representing indebtedness, if a be greater than b. The number i (6 — a) is therefore the true answer to the ques- tion propounded by the problem, under all circumstances. Consider a copcret-e illustration : 16 INTRODUCTORY LESSONS. Ex. 2. Restate the problem with 55 in place of a, and 5 in place of b. The solution then is X = 1 (5 - 55) = - 1(55 - 5) = -25 ; and the interpretation of this negative result is that the debits exceed the credits by $ 25. Thus, if credits = $ 100, then debits = f 125 ; or if credits = $ 5000, then debits = ^ 5025, etc. We may verify the conclusion here reached by performing the operations indicated in the problem upon — 25. We first diiuble — 25 and obtain — 50, then to this negative number — 50 we add the positive number 55 and obtain 5. These are some of the ordi- nary processes of algebra whose meaning will be more fully ex- plained in the next two chapters. Ex. 3. A steam launch, whose speed in still water is aleet per minute, goes up an estuary a distance of b feet in c minutes. Show that the tide is running at the rate of (6 -r- c) — a feet per minute, — inward if 6 -^ c be greater than a, outward* if 6 -r- c be less than a. 7. The foregoing problems have brought to view some of the purposes of algebra and have shown how, by its larger methods, some of the inconvenient limitations to which arithmetic is necessarily subject may be removed. Other like problems, to be considered in subsequent chapters, will still further attest the superiority of these methods. But since many of them cannot be success- fully attempted until tlie fundamental operations of algebra are learned, our next task must be a study of the fundamental operations of algebra. LANGUAGE OF ALGEBRA. DEFINITIONS. 17 CHAPTER II. The Language of Algebra. Definitions. 8. Algebra is that branch of mathematics which treats of the relations of numbers as expressed in equations and in what are known as algebraic forms, or expressions. These algebraic equations and expressions constitute the language of algebra. 9. The Operations of Algebra* are conveniently grouped in four main divisions, each of which comprises two con- trasted kinds of operation^ called respectively a direct process and an inverse process. These eight processes, thus classified, are : Direct Processes : Inverse Processes: I. 1. Addition, 2. Subtraction. II. 3. Multiplication, 4. Division. III. 5. Involution, 6. Evolution. IV. 7. Exponentiation, 8. Logarithmic Operation. The direct process having been defined, its inverse is described as that operation which annuls, or undoes, the direct process. * Another point of view enables us to regard all the operations of algebra as forms of the addition process. But the investigations neces- sary to establish this conclusion are not appropriate in a merely ele- mentary treatise. 18 LANGUAGE OF ALGEBRA. DEFINITIONS. A full explanation of the character of these processes, as fundamental parts of algebraic science, forms a neces- sary introduction to the problems that algebra under- takes to solve. They will be discussed in the order here indicated, the first four in Chapters III., IV., and V., the others somewhat later, as occasion for their use arises. 10. The Symbols of Algebra. For the convenient rep- resentation of these operations an algebraic symbolism has been devised; and, as has been exemplified in the problems of the Introductory Lessons, the signs used in aritlimetic are adopted as a part of it. But for the adequate representation of all its operations, algebra requires an enlarged symbolism of which arithmetic needs only a part. The symbols of elementary algebra are of five kinds : (i.) Symbols of Quantity, usually letters of the alpha- bet, employed to represent numbers. (ii.) Symbols of Operation, usually called the signs of algebra, employed to indicate processes to be applied to the symbols of number. ( + , — , ~, X, -f-, ", -y/, log.) (iii.) Symbols of Relation, by means of which compari- sons of numbers are expressed. ( = , =, >, <.) (iv.) Abbreviations of frequently recurring words or phrases. (.*., •.-, •••.) (v.) Symbols of Aggregation. (Parentheses, braces, square brackets, the vinculum.) SYMBOLS OF QUANTITY. 11. In arithmetic, numbers are represented by figures^ each of which has one, and only one, meaning. LANGtJAGE OF ALGEBRA. DEFINITIONS. 19 In algebra, numbers are represented either by figures or by the letters of the alphabet. In an arithmetical product, such as 3 x 5, it is obvious that the factors 8 and 5 may be interchanged, and this law of commutation, as it is called, is easily proved to obtain for any two numbers. But the statement of the law in the form 3x5 = 5x3 is the state- ment of a particular case only. Now algebra states it in the form a xb = b X a, in which a and b represent any two numbers what- ever. The gain in conciseness and generality of statement by the use of letters is thus apparent. The word quantity is used in algebra as synonymous with number. SYMBOLS OF OPERATION. 12. The sign -f, which is read 'plus/ is placed before a number, or a number represented by a letter, to indicate that it is to be added to what has gone before. Thus 6 -f- 3 means that 3 is to be added to 6 ; 6 + 3 + 2 means that 3 is to be added to 6 and then 2 added to the result : so also a-\- b means that the number which is represented by b is to be added to the number which is represented by a ; or, expressed more briefly, it means that b is to be added to a. 13. The sign — , which is read 'minus,' is placed before a number to indicate that it is to be subtracted from what has gone before. Thus 6 — 3 means that 3 is to be subtracted from 6, a — b means that b is to be subtracted from a, and a — b -\- c means that b is to be subtracted from a and then c added to the result. It should be noticed that in a series of additions and subtractions the order of the operations is from left to right. The meaning of the sign ^ is explained in Art. 46. 20 LANGUAGE OF ALGEBRA. DEFINITTOKS. 14. The sign of multiplication is x, which is read * multiplied by' or ^into.' Thus 6x3 means that 6 is to be multiplied by 3, a x ft means that a is to be multiplied by 6, and a x h x c means that a is to be multiplied by h and then the result multiplied by c. The sign of multiplication is generally omitted between two letters, or between a number and a letter, and the letters are simply placed side by side. Sometimes the x is replaced by a point. Thus ah ov a 'h means the same as a x 6, and 2 ahc or 2 a • ft • c the same as 2 x a x 6 x c. 15. The sign of division is -t-, which is read 'divided by' or 'by.' Thus 6 -=- 3 means that 6 is to be divided by 3, a -f- ft means that a is to be divided by 6, and a ^ ft -r- c means that a is to be divided by 6 and then the result divided by c ; also a ^h x c means that a is to be divided by ft and then the result multiplied by c. The operation of division is often indicated by placing the dividend over the divisor with a line between them, or by separating the dividend from the divisor by an oblique line called the solidus ; thus either - or a/b is frequently used instead of a-^h. It should be noticed that in a series of multiplications and divisions the order of the operations is from left to right. 16. When two or more numbers are multiplied together, the result is called the continued product, or simply the product ; and each number is called a factor of the product. LANGUAGE OF ALGEBRA. DEFINITIONS. 21 17. When the factors of a product are considered as divided into two sets, each is called the coefficient, that is the co-factor, of the other. Thus in 3 ahx, 3 is the coefficient of ahx ; also 3 a is the coeffi- cient of 6x, and 3 ah is the coefficient of x. When one of the factors of a prodnct is a number expressed in figures, it is called the numerical coefficient of the other part of the product. EXAMPLES IV. Calculate the values of 1. 7 4- 6 + 4. 3. 11 + 7 - 12 - 6. 5. 5 - 3 x 4. • 2. 5-3 + 4. 4. 7 X 6 X 4. 6. llx7-l-J-(;. If a = 1, 6 = 2, c = 3, and d— 4, find the numerical values of 7. c - 6. 12. 13a-6/; + 7c-5d 8. d-a. 13. 18 6-3c-4fZ + 9a. 9. 7a -36. 14. 20a6-3cd 10. 106 -6c. 16. 4da-2 6c. 11. 5a -26 + 6c-4(?. 16. ahc -^ hcd -{- cda -{■ dah. . If a = 6, 6 = 2, c = 5, and d = 0, find the values of 17. 3ac + 2 6c + ca. 19. a x c h- 6. 21. 2c -=- a -f- 6. 18. 7a indicates that the number which precedes the sign is greater than that which follows it. Thus rt > 6 means that a is greater than b. The sign < indicates that the number which precedes the sign is less than that which follows it. Thus a%^ are likp. terms ; but althougli 3 a~hx^ and 7 a%'^x^ contain the same letters, they are not like terms, for all the letters are not raised to the same power. 25. A monomial expression is one which contains only one term, and a multinomial expression is one which con- tains more than one term. Thus ^ah'^cx is a monomial expression, and a -{- h is a multino- mial expression. An expression whicli consists of two terms is often called a binomial expression, and an expression which con- sists of three terms is called a trinomial expression. Monomial and multinomial ex])ressions are sometimes called respectively simple and compound expressions. LANGUAGE OF ALGEBRA. DEFINITIONS. 25 SYMBOLS OF AGGREGATION. 26. The parentheses ( ), the square brackets [ ], and the braces \ ], are used to indicate that all operations denoted by signs within the enclosure are to be consum- mated before any operation without it is performed. Thus {a + b)c means that b is to be added to a and that the result is to be multiplied by c ; again, (a + by means that b is to be added to a, and that the cube of the result is to be formed ; also (a + 2&)(c — 3cZ) means that 2 6 is to be added to a and that 3 d is to be taken from c, and that the first of these results is then to be multiplied by the second. A line called a vinculum is often drawn over the expres- sion which is to be treated as a whole : thus a — b — c is equivalent to a — (b — c), and Va + & is equivalent to When no vinculum or bracket is used, a radical sign refers only to the number or letter which immediately follows it : thus ^2 a means that the square root of 2 is to be nniltiplied by a, whereas \/2 a means the square root of 2 « ; also -yj/a + x means that x is to be added to the square root of «, but Va + x means the square root of the sum of a and x. The line between the numerator and denominator of a fraction acts as a vinculum, for "^ is the same as Ma + b). Note. — It should be carefully noticed that every term of an algebraical expression must be added or subtracted as a whole, as if it were enclosed in brackets. Thus, in the expression cf -j- 6c — fZ -^ e + /, b must be multiplied by c before addition, and d must be divided by e before subtraction, just as if the expression were written a + {be) — {d h- e) + /. 26 LANGUAGE OF ALGEBRA. DEFINITIONS. Ex. Find the values of (i.) 2a-bG + cd-6b -^ a 4- — , (ii.) (a + 6)3(2 6 - 3c)2, a (iii.) a +6« + c*, (iv.) ^{Ta'* + (6 + c)8+d3}, when w=i, 6 = 3, c = 1, d = 0. The results are : (i.) 2a-bc-\-cd-6b-i-a + — a = 2 X 4 - 3 X 1 + 1 X - 6 X 3 -- 4 + ^-^ 4 = 8-3 + 0-1 + 1 = 1. (ii.) (a + 6)3(2 6-3c)-^=(4 + 3)8(2 x 3 - 3 x l)-^ = 7^ x 3^ = 343 X 9 = 3087. (iii.) a» + 6*' + c« = 43 + 31 + 1* = 64 + 3 + 1 = 68. (iv. ) ^{7a3 + (6 + c)3 + d^} = ^{7 x 4^ + 4^ + O^} = ^{7 X 64 + 64 + 0} = ^(448 + 64) = ■;/512 = 8. EXAMPLES V. 1. Writedownthe valuesof2^33, 43, 4^ ^64, ^64, ^16, 4/125, ^625, and ^32. If a = 2, 6 = 3, c = 4, and 4, that is -f- 4 will stand for f 4 that he possesses, or that is owing to him ; so also — 4 will stand for whatever decreases his property by f 4, that is — 4 will stand for ^ 4 that he owes. If, on the other hand, we are calculating tlie amount of a man's debts, + 4 will stand for whatever increases his debts, that is -t- 4 will now stand for a debt of $ 4 ; so also — 4 will now stand for whatever decreases his debts, that is — 4 will stand for $ 4 that he has, or that is owing to him. If we are considering the amount of a man's gains, -h 4 will stand for what increases his total gain, that is + 4 will stand for a gam of 4 ; so also — 4 will stand for what decreases liis total gain, that is — 4 will stand for a loss of 4. If, however, we are calculating the amount of a man's losses, + 4 will stand for a loss of 4, and — 4 will stand for a gain of 4. Again, if the magnitude to be increased or diminished is the distance from any particular point, measured in LANGUAGE OF ALGEBRA. DEFINITIONS. 29 any particuiar direction, + 4 will stand for a distance of 4 units in that direction, and — 4 will stand for a distance of 4 units in the opposite direction. 29. From the above examples it will be seen that the signs + and — will serve to distinguish between magni- tudes of opposite kinds. Thus whatever -f 4 may rep- resent, — 4 will represent an equal magnitude, but of the opposite kind. The signs -f- and — are therefore used in algebra with two entirely different meanings. In addition to their original meanings as signs of the opera- tions of addition and subtraction respectively, they are also used as marks of distinction between magnitudes of opposite kinds. 30. A quantity to which the sign -f is prefixed is called a positive quantity, and a quantity to which the sign — is prefixed is called a negative quantity. The signs -1- and — are called respectively the positive and negative signs. 31. The signs -|- and — are often called signs of affection when they are used to indicate a quality of the quantities before whose symbols they are placed. The sign -f-, as a sign of affection, is frequently omitted; and when neither the -f- nor the — sign is prefixed to a term, the + sign is to be understood. 32. The range of positive and negative algebraic numbers is obviously double that of the numbers which belong to arithmetic. Thus, the series of integers in arithmetic is +1 +2 4-3 4-4 ..., 80 LANGUAGE OF ALGEBRA. DEFINITIONS. while the corresponding algebraic series is ... _4 -3 -2 -1 +1 +2 +3 +4 ... In comparing the terms of this double algebraic series, we adopt the convention that they are here arranged in ths order of magnitude, so that -3<-2<-l<0 6. [See Art. 47.] 33. The magnitude of a quantity considered inde- pendently of its quality, or of its sign, is called its absolute magnitude, or its absolute value. Thus a rise of 4 feet and a fall of 4 feet are equal in absolute magnitude ; so also + 4 and — 4 are equal in absolute magnitude, whatever the unit may be. 34. Note. — Although there are many signs used in algebra, the name sign is often used to denote the two signs -f- and — ex- clusively. When the sign of a quantity is spoken of, it means the + or — sign which is prefixed to it ; and when we are directed to change the signs of an expression, it means that we are to change the -f or — before every term into — and -f , respectively. EXAMPLES VI. Calculate the values of 1. 5-3-4. 2. 6x3-19. 3. 21-3-9. If a = 1, 6 = 2, and c = 3, find the values of 4. a - 2 b. 6. be -11 a. 8. 6 -f- a - c. 6. ab — c. 7. ab — be. 9. a -e- 6 — c. LANGUAGE OF ALGEBRA. DEFINITIONS. 31 10. The reading of a thermometer was + 40°, and its njercury column then fell 53°. What was its reading after the change ? 11. If a = 2, b = S, and c = — 10, which is the gi-eater, a x ft or c ? 12. Can the result of subtracting one positive number from another ever be greater than the minuend ? 32 ADDITION. SUBTKACTION. BRACKETS. CHAPTER III. Addition. Subtraction. Brackets, addition. 35. The process of finding the result when two or more quantities are taken together is called addition, and the result is called the sum. Since a i)ositive quantity produces an increase, and a negative quantity produces a decrease, to add a positive quantity we must add its absolute magnitude, and to add a negative quantity we must subtract its absolute mag- nitude. Thus, when we add -f 4 to + 6, we get + 6 + 4 = -f- 10 ; and when we add — 4 to + 10, we get + 10 — 4 = + 6. So also, when we add + 6 to a, we get a-\-b', and when we add —b to a, we get a — b. Hence a-f-(+6) = a + 6 and a-{-(—b) = a — b. We therefore have the following rule for the addition of any term : To add any term, affix it to the expression to which it is to be added, with its sign unchanged. Ex. A boy i)layed two games ; in the first game lie won 6 points, and in the second lie won — 4 points (that is he lost 4). How many did he win altogether ? The total gain in the two games together is what is meant in algebra by the sum of the gains. To obtain the total gain, we must add — 4 to 6, and this opera- tion is indicated by 6 + (- 4), which by the above is 6 - 4 = 2. ADDITION. SUBTRACTION. BRACKETS. 33 Wlieii numerical values are given to a and to 6, the numerical values of a + 6 and a — b can be found ; but until we know what numbers a and b represent, we can- not take any further step, and the process is considered to be algebraically complete. 36. It should be noticed that when b is greater than a, the arithmetical operation denoted by a — 6 is im- possible ; for we cannot take any number from a smaller number. Thus, if a = 3 and b = 5, a — b will be 3 — 5, and we cannot take 5 from 3. But to subtract 5 is the same as to subtract 3 and 2 in succession, so that 3 — 5 = 3 — 3 — 2 = — 2 : we then consider that — 2 is 2 which is to be subtracted from some other algebraical expression, or that — 2 is two units of the kind opposite to that repre- sented by 2. And if — 2 is a final result, the latter is the only view that can be taken. In some particular cases the quantities may be such that a negative result is without meaning ; for instance, if we have to find the population of a town from certain given conditions ; in this case the occurrence of a nega- tive result would show that the given conditions could not be satisfied, but so also in this case would the occur- rence of a fractional result. EXAMPLES VII. Find the sum of 1. 4 and - 3. 5. 3 and - 11. 9. 2 a and - 3 6. 2. 3 and - 2. 6. - 3 and - 9. 10. - 3 a and— 2 6. 3. 6 and - 3. 7. G, - 2, and 7. 11. 5«,-66, and -2c. 4. 7 and - 8. c 8. - 3, - 2, and 6. 12. -3 a, -4 6, and 7 c. 34 ADDITION. SUBTRACTION. BRACKETS. 37. It follows at once from the nature of addition that the sum of two or more algebraical quantities, whether positive or negative, is the same in whatever order the quantities may be taken. This is kr.ov/n as the Law of Commutation in addition. For example, to find how much a man is worth, we can take into account the different items of property ;deT)ts being con- sidered as negative) in any order we please. It also follows that to add any algebraical expression as a wJiole gives the same result as to arfd its terms sepa- rately. This is known as the Law of Association in addition. But to add any term, we have only to write it down, with its sign unchanged, after the expression to which it is to be added. We have therefore the following Eule. To add two or more algebraical expressions, write down all the terms in succession ivith their signs unchanged. For example, the sum oi a -{-h and c — d!isa4-& + c— d; also the sum of a — 6 + c and — d + e— /isa — 6 + c — d + e— /. 38. If some of the terms which are to be added are 'like' terms, we can and must collect all such terms together before the process of addition is considered to be complete. Of this we have three cases, as follows : I. The sum of ' like ' terms which have the same sign is a 'like' term which has the same sign, and whose coefficient is the arithmetical sum of their numerical coefficients. For example, to add 2 a and 6 a in succession gives the same result, whatever o may be, as to add 7 a ; that is, + 2 a + 5 a = + 7 a. ADDITION. SUBTRACTION. BRACKETS. 35 Also to subtract 2 ab and 5 ab in succession gives the same result as to subtract 7 ab ; that is, — 2 a6 — 5 a& = — 7 ab. II. The sum of two 'like' terms whose signs are different is a ' like ' term whose coefficient is the arith- metical difference of their numerical coefficients and whose sign is that of the greater. For example, +5« — 3a=+2rt + 3a — 3a = +2a; also +'Sab — ^ab =-\- Hab — oab — 2ab = — 2ab. III. If there are several -like' terms some of which are positive and some negative, the positive terms can be collected into one sum by I., as also the negative terms : the final sum is then obtained by II. Thus any number of 'like' terms can be reduced to one term. Ex. 1. Add'2 a + 3 6 and a -5b. The sum is 2 a + 3 6 + a — 5 6 ; or, since the terras can be taken in any order, the sum is 2a-|-a + 36 — 56 = 3a — 2 6, by col- lecting the like terms. Ex. 2. Find the sum of 6a'^ -6ab + 4&2, 2b^- ab- a^, and 5 a6 - 9 62 _ 4 a^. The sum is 6 a2 - 6 a6 + 4 62 + 2 62 - «6 - a2 + 5 a6 - 9 62 - 4 a2 = 6 a2 - a2 - 4 a2 - 6 a6 - a6 + 5 a6 + 4 6- + 2 62 - 9 6-. The terms 6 a2, — a'^, and - 4 a2 can be combined mentally ; and we have a'^. Similarly we have — 2 a6 and — 3 62. Thus the required sum is a2 _ 2 ^6 — 3 62. 39. It is best for beginners to place the different sets of 'like' terms in vertical columns; so that the last example would be put down in the following form, the ^6 ADDITIOlsr. SUBTRACTION. BRACKETS. -h sign being put before the terms 2b^ and 5ab, which have no signs : and then the sets of ' like ' terms can be combined mentally. -4a2 + 5a6-962 a2 - 2 rt6 - 3 62 EXAMPLES VIII. Simplify the following by combining ' like ' terms : 1. 2a + 6 + 3c+26 + 3a + 2c + 5a + c. 2. 6a-3b + 2G-4a-Sc-h2b. 3. 5a2-3a + 6-4a2 + 6a + 3. 4. 7 a8 - 4 a + 9 - 3 a2 + 2 a + 7 - 3 a8 - 16. 6. 5 a3 - 4 n'^b + Sab^-5a^-i a% - 3 bK 6. 3 a2 4. 6 a?) - 4 62 _ 2 rt2 _ 4 «?, + 3 ?>2 _ «2 _ 2 a& + 62. Add together 7. a 4- 6 and a — b. 9. .] « + ^ 6 and — \a-\-\b 8. 1x- y and 2 a; + y. 10. ^ a + f 6 and 5 6 - | a. 11. a2 _ flj and a^ + a. 12. a + a2 _}. 4 „» and 2 «» _ ^-i _ 4 q. 13. w2 + WW 4- ^2 and m2 — ?wn — n\ 14. 3p2 4. ^pq _ g2 and bq^-Apq- 3;)2. 16. 3 a2 - 2 a6 + 62 and a2 _ 2 ah - I hK 16. 2 a + 6 - 3 c, 2 6 + c - 3 a, and 2 c + a - 3 ?>• 17. 4 a — 3 6 — c, 4 6 — 3 c — «, and 4 c — 3 a — 6. 18. 4 a2 _ 3 ah + 62, 4 a6 - 3 62 + a2, and 4 62 - 3 a2 + ah. 19. a-\b -^ \c,h - \c-\- \a, and c - J a + i 6. 20. 4x-2y + l, -3x + 2-?/, anda; + 3y-3. 21. -4a - 6 + 2, 2 + 8a-56, and -a-46-2. ADDITlOK. SUBTRACTION. BRACKETS. S7 22. x3 - 2 x2y - 2 xy'^, xhj - 3 xy^ - y^, and 3x?/2 - 2 y^ _ ^^3, 23. a^ + 4 &3 - 5 g3 + 3 ahc, ft^ + 4c'3 _ 5 a^ _^ 6 a6c, and ci + 4 a3 _ 5 53 _ 9 a6c. 24. 5 a^ _ 2 a2;, 4. 9 ^52 _^ 17 53^ _ 2 «=? 4. 5 ^2^ _ 4 05^2 _ 12 53^ 63 _ 4 a62 - 5 a'^ft _ ^js^ and 2a^h -"la^ -Qh^ - aV^. 25. I a^ - I a^ft + 3 53^ ^^3 _. 2 ^^,-2 _ .| ^,3^ and \ar-h -I ab^ - i ft^. 26. 3x3 -4a; + 5, 2x^-6x+ 7, 6^3 -2x2 -2a;, and 3 + 8x-4x3. 27. x3 - 3 ax2 + 5 a2x - a3, 2 x3 + 4 ax2 - % a'^x, 6 rtx2 - 3 a2x + «», and — 2 x3 4- 4 a% — 5 a^. 28. 3 x2 + y2 _ 3 ^2; - 2:2, 2 x?/ - 3 ^2 ^ 3 y^^ and - 4 x2 - 2 X?/ + ^2 _^ ^2. 29. Show that, if x = & + 2c-3a, ?/ = c + 2a-3 6, and = a + 26 — 3c; then will x + ?/ + 2 = 0, 30. Show that, if rt = 5x — 3y — 2^;, h = 5y — Sz — 2x, and c = 5^ — 3x — 2y; then will a + b -\- g = 0. SUBTRACTION. 40. Since subtraction is the inverse of addition [Art. 9], an addition and a subtraction of the same quantity pro- duces no effect. Hence a is the same as a-^b — b. Now if we take + b from a -\-b — b, what is left is a — b. So that if we take away + b from a, the remainder is a — b; that is a — (4-6) = a — 6. Again, if we take away — b from a + ?> ~ 6, what is left is a-\-b. So that if we take away — b from a, the remainder is a + 6 ; that is, a — ( — b) = a-\-b. 38 ADDITION. SUBTRACTION. BRACKETS. Thus to give a numerical example, + 10-(4-4) = 4- 10-4 = 6, and -f 6- (-4) = 4- 6 + 4 = 10. We therefore have the following rule for the sub- traction of any term : to subtract any term, affix it to the expression from which it is to he subtracted^ bat with its sign changed. e: XAMPLES IX. Subtract 1. 3 from - 4. 3. - 6 from 4. 5. - - h from a. 2. - 4 from 3. 4. a from — h. 6. - - a from - b. Show that : 7. -4-(+3) = - -7. 10. - -h- -(+a): = -b-a. 8. 3-(-4)=7. 11. - -h- -(-a): = - b i- a. 9. 4 -(-6)= 10. 12. - -(- .6)-(- a)=b-\- a. 41. Since subtraction and addition are inverse opera- tions, and since we know that to add any algebraical expression as a whole gives the same result as to add its terms separately, it follows that to subtract an alge- braical expression as a whole is the same as to subtract the terms in succession. (Law of Association.) We have therefore the following Eule. To subtract any algebraical expression from any other, write doivn its terms in succession ivith their sigiis changed, after that other. Thus, if 2 rt + 6 — 4 c be subtracted from 3 a — 4 6 + c, the result is Sa-4b + c-2a-b-\-ic=^a-2a-4b-b-\-c-\-ic = a-Gb + 5c. ADDITION. SUBTRACTION. BRACKETS. 39 The law of commutation for subtraction is the same as that for addition, and has been enunciated in Art. 37 ; but in applying it, the signs, as well as the letters to which they are prefixed, must be interchanged. Thus, -f a — 6 = — 6 + a is an immediate consequence of + tt-f(-5) = + (-6) + a. 42. The expression which is to be subtracted is often placed under that from which it is to be taken, ' like ' terms being for convenience placed under one another; and the signs of the lower line are changed mentally before combining ' like ' terms. Thus tlie example considered in Art. 41 would be written as follows : 3a -46 + c 2a+ &-4c a — 66 + 5c The terms of the result being obtained by combining mentally 3 a and — 2 «, — 4 6 and — 6, and c and + 4 c. As another example, if we have to subtract 3 a3 - 4 d^h + 2 a6- - 6^ from 4 a^ -|- a^i) _ ab'^, the process is written as follows : 4 a^ + a% - ab^ 3a3-4a26 + 2a62_58 a3 + 5 a26 _ 3 ab^- + b^ The terms of the result being obtained by combining 4 a^ and 3a3, a'^b and +4a% - ab- and -2ab^ and + b'\ 43. We have hitherto supposed that the letters used to represent quantities were restricted to positive values ; 40 ADDITION. SUBTRACTION. BRACKETS. it would, liowever, be very inconvenient to retain this restriction. In what follows therefore it must always be understood, unless the contrary is expressly stated, that each letter may have any value positive or negative. Since any letter may stand for either a positive or a negative quantity, a term preceded by tlie sign + is not necessarily a positive quantity in reality; such terms are, however, still to be called positive terms, because they are so in appearance ; and the terms preceded by the sign — are similarly called negative terms. 44. We must now carefully examine whether terms can be added and subtracted without knowing whether the letters really represent positive or negative quantities. Now we have seen in Arts. 35 and 40, that when h is really positive, + (+&) = + &, +(_&) = _6, _(+6) = _6, and — ( — 6) = -I- & ; and we have to see whether the same laws hold good although h may really be negative. If h be really negative and equal to — c, where c is posi- tive, then +6 = H-( — c) = — c, and — Z> = — ( — c) = + c, since c is positive. Hence, putting — c for -f h, and + c for — h, the laws expressed above will be true, although h is negative, provided + (-c) = -c, +( + c) = + c, -(-c)= + c, and — ( -f c) = — c, are tnie for all positive values of c, and this we know is the case. ADDITIOK. StJBTRACTlON. BRACKETS. 41 Hence terms are added or subtracted in iwecisely the same ivay whether the letters really stand for positive or for negative quantities. 45. Zero, whose symbol is 0, may now be defined as a difference, in the form a — a = 0. When it appears in any algebraic expression, as a term in the form 0, or as a pair of terms in tlie form a — a, or as a series of terms reducible to either of these forms, it may obviously be erased without in any way affecting the numerical value of the whole expression. 46. Def. The algebraical difference between any two quantities a and b is the result obtained by subtracting the second from the first. For example, the algebraical difference of 5 and 4 is 5 — 4 = 1, and the algebraical difference of 4 and 5is4 — 5=— 1. The algebraical difference between two quantities de- pends upon the order in which they are given, and may therefore not be the same as the arithmetical difference, which is the result obtained by subtracting the less from the greater. The symbol a '^ 6 is used to denote the arithmetical difference of a and b. 47. Def. One quantity a is said to be greater than another quantity b, when the algebraical difference, a — 6, is positive. From this definition it is easy to see that in the series 1, 2, 3, 4, etc., each number is greater than the one before it ; _^and that, 42 ADDITION. SUBTRACTION. BRACKETS. in the series —1,-2, — 3, — 4, etc., each number is less than the one before it. Tiius 7, 5, 1, 0, — 5, — 7 are in descending order of magnitude. [See Art. 32.] EXAMPLES X. Subtract 1. a — b from a + b. ^. ^x^\y from ^x- ^y. 2. 2a- 36 from 3a -26. 5. Sx- ix"^ trom 4x - Sx^. 3. 2x-\-y from 2x- y. 6. x^ - 2 from 1 - 2 a;2. 7. 4a-26 +3cfrom2c4- 46--3a. 8. 4 a2 - 2 rt6 + 3 62 from 2b'^-\-iab- Sa^ 9. 3x2-4a; + 2froma;2 + 6ic- 7. 10. 3x3 - 2 ic2 + 5 from 3 x^ - 2 x + 5. 11. a — lb — Ic from 6 — Ic — ^a. 12. ^x^-lxy-^'^y^ from I y^-^xy-hl x\ Find the difference between 13. a + 2 6 and a - 2 6. 14. 3 a - 7 6 and 7 a - 3 6. 15. a2 + a6 + 62 and a2 - a6 + 6^. 16. x'^ -Zxy and 3 x2 - 4 xy. 17. 3x3 + 5 x22/ + 4 xy^ and ix^y + Q xy^ + 7 y». 18. 2 x8 - 7 x22/ + 9 xy2 _ 4 ?/3 and 4 x* - x2i/ + xy2 - 4 y«. 19. What must be added to 2 a — 3 6 in order that the sum may be 4 a - 6 6 ? 20. "What must be added to a2 + 62 in order that the sum may be 2 a2 - a6 ? 21. What must be added to 6 a6 — 2 6c + 3 ca in order that the sum may be 7 a6 + 2 ca ? 22. What must be added to a2 + 3 62 + 2 c^ that the sum may be62-3a2? ADDITION. SUBTRACTION. BRACKETS. 48 23. Subtract from 3 a - 4 6 the sum of2a+7 6, -4a -6b, and 6 a — db. 24. Subtract from 3 x2 - 2 x + 7 the sum of x"^ - x -^ 9, 2x^ + 7 X — 6, and 3 x^ — 4 x — 5. 25. Subtract the sum of a^ - 4 a5 - 3 6^ a6 - 4 62 _ 3 ^2^ and 52 _ 4 <^-2 _ 3 ^5 from 2 a-2 + 2 b'^ -\-2ab. BRACKETS. 48. To indicate that any algebraical expression is to be added as a ivhole, it is put between brackets (paren- theses, braces) with the -j- sign prefixed. But, as we have seen, to add any expression, we have only to write down the terms in succession, with their signs unchanged. Hence, when a bracket is preceded by a + sign, the brackets may be omitted. Thus -f-(2a — 6 + c) = 2rt — 6-l-c. Conversely, any number of terms in an expression may be enclosed in brackets with the sign -f- placed before the bracket. For example, we may write a-26 + c + 2d-3e+/ in the form a + (-26-fc) + (2d-3e4-/), or a_26 + (c + 2cZ)-l-(-3e-f/). When the sign of the Jii'st term within the brackets is +, it is generally omitted for shortness, as in the preced- ing example. 49. To indicate that any algebraical expression is to be subtracted as a whole, it is put between brackets, and 44 ADDITION. SUBTRACTION. BRACKETS. the — sign prefixed. But, as we have seen, to subtract any expression, we have only to write down the terms in succession with all their signs changed. Hence, when a bracket is preceded by the — sign, the brackets may be omitted, provided that the signs of all the terms within the brackets are changed. Thus -(2a-b-\-c) = -2a + b-c. Conversely, any number of terms in an expression may be enclosed in brackets with the sign — placed before the bracket, provided that we change the signs of all the terms which are placed in the brackets. Thus a-2b + c-\-2d-3e-{-f may be written in the form a-(2b-c)-(-2d-\-3e-f). 50. Sometimes one enclosure is put within another: in this case the different pairs of brackets must be of different shapes to prevent confusion. Thus a — [& + f c — (d + e) j ] ; which means that we are to add to b the whole quantity within the braces \ j, and then subtract the result from a; and to find the quantity within the braces {}, y^e must add d and e, and then subtract the sum from c. When there are several pairs of brackets, they may be removed one at a time by the rules of Arts. 48 and 49 ; and it is best for beginners to remove at every stage the innermost bracket. Thus a -\b + \c - {d + e)\'\ = a - [b + \c - d -e\^ ^ a— [b-\-c — d — e'] = a — b — c-\-d-\-e. ADDITION. SUBTRACTION. BRACKETS. 45 EXAMPLES XI. Simplify the following expressions by removing the brackets and collecting like terms : 1. (a-\-b)-(a-b). 6. a -[a - {a -(2a - a)}]. 2. a-b-{a-hb). 7. 1 -[2 - {3 -(4 - 5)}]. 3. a -(6 + c) + (6 -c-a). 8. a + &-[a-6+{a + &-(a-6)}]. 4. 3x-(?/-2x) + (24-2/-5x). 9. 5 -[4 + {5 -(4 + 5^^)}]. 5. x-{y -{z ~ X)}. 10. x-\:y -{z-{x-y - z)]']. 11. 3x-{2y + 52 - 3x + 2/}. 12. {2x -(5y- 3^ + 7)} -[4 + {X - (3y + 20 4- 5)}]. 13. [2a-{36 4-(4c-36 + 2a)}]. 14. x-(y ~z)-\-{2z- Zy - 5x}. 15. a - 2 6 - {3 a - (6 - c) - 5 c}. 16. «-[3 6 + {3c-(<2-6)+ a}-2a]. 17. 3rt-[2 6 — {4c-12a- (4 ft - 8c)} - (6 6 - 12c)]. 18. {2x-(3y-72;) + (3x-2y + 90)} - {(y - 50)-(3x - y - 22) + 82}. 19. a2 _ (o 055 _ 4 62)_('2 a2 _ 3 056 4. 662) -{562_(3rt6-7a2-62)}. ). (wi2 - n2) - {3 m;i - (5 7*2 _ w2)| + [n2 - {3 w/i - (5 iiiP- - 6 ?i2) j 4- 8 mn\. 46 MULTIPLICATION. CHAPTER IV. Multiplication. 51. In arithmetic, multiplication is first defined to be the taking one number as many times as there are units in another. Thus, to multiply 5 by 4, we take as many fives as there are units in four. As soon, however, as fractional numbers are considered, it is found necessary to modify somewhat the meaning of multiplication, for by the original definition we can only multiply by whole numbers. The following is therefore taken as the defi- nition of multiplication : Def. To mnltiply 07ie number by a second ive do to the first what is done to unity to obtain the second. Thus 4 is 1 + 1 + 1 + 1; .-. 5 X 4is 5 + 5 + 5 + 5. Again, to multiply 4 by J, we must do to -f- what is done to unity to obtain J; that is, we must divide ^ into four equal parts and take three of those parts. Each of the parts into which 4 is to be divided will be -, and 5x3 ^ X ^ by taking three of these we get -• So also (_5)x4=-5+(-5) + (-5) + (-5) = _5_5_5_5 =^ -20. MULTIPLTCATION. 47 With the above definition multiplication by a negative quantity presents no difficulty. For example, to multiply 4 by —5. Since to subtract 5 by one subtraction is the same as to subtract five units successively, -5 = -l-l-l-l-l; ... 4 x(- 5) = -4 -4-4-4-4 = -20. Again, to multiply — 5 by — 4. Since -4=:-l-l-l-l; ... (_5)x(-4) = -(-5)-(-5)-(-5)-(-5) = + 5 + 5 + 5 + 5 [Art. 40.] = + 20. We can proceed in a similar manner for any other numbers, whether integral or fractional, positive or nega- tive. Hence we have the following laws : ax 6 ==+ a6 . . . (i.) (— a)x b =—ab . . . (ii.) ax(—b) = — ab . . . (iii.) {-a)x{-b) = -\-ab . . . (iv.) The rule by which we determine the signs of the prod- ucts is called the Law of Signs; this law is sometimes enunciated briefly as follows: like signs give +, and unlike signs — . 52. The factors of a product may be taken, in any order. It is proved in arithmetic that when one number, whether integral or fractional, is multiplied. by a second, 48 MULTIPLICATION. the result is the same as when the second is multiplied by the first. The proof is as follows : First, when the numbers are integers, a and b suppose, write down a series of rows of dots, putting a dots in each row, and take b rows, writing the dots under one another as in the following arrangement : a in a row b rows. Then, counting by rows, the whole number of the dots is a repeated b times ; that k, a x b. Also, counting by columns, the whole number of the dots is b repeated a times ; that is, b x a. Hence, when a and b are integers, a xb = b X a. Next, when the numbers are fractional, for example, ^ and f , we prove as in Art. 51 that ^ x f = ^ — j- And, 5x3 3x5 by the above proof for integers, ' '^ = ; hence y X :f — 4 X y. Hence ab = ba for all positive values of a and b ; and being true for any positive values of a and b, it must be true for all values, whether positive or negative; for, from the preceding article, the absolute value of th^ MULTIPLICATION. 49 product is independent of the signs, and the sign of the product is independent of the order of the factors. Hence for all values of a and h we have ab = ba (i.) This is the Law of Commutation in multiplication. If in the above arrangement of dots we put c in the place of each of the dots, the whole number of c's will be ab ; also the number of c's in the first row will be a, and this is repeated b times. Hence, when a and b are integers, c repeated ab times gives the same result as c repeated a times and this repeated b times. So that to multiply by any two whole numbers in succession gives the same result as to multiply at once by their product, and the proposition can, as before, be then proved to be true without restriction to whole numbers or to positive values. Thus, for all values of a, b, and c, we have a xb X c = a X (be) (ii.) This is the Law of Association in multiplication. From (i.) and (ii.) it follows that the factors of a product may be taken in any order without altering the result, however many factors there may be. 53. Since the factors of a product may be taken in any order, we are able to simplify many products. For example : 3ax4«=:3x4xaxa = 12a2, (-3a)x(-4&) = +3ax46= + 3x4xax6 = 12a&, (a6)2 = abxah = axaxhxh = a'^b^, {^ay= y/'2a Xy^a=^ Xy/2 X ax a = ^a^, 50 MULTIPLICATION. Although the order of the factors in a product is indifferent, a factor expressed in figures is always put first, and the letters are usually arranged in alphabetical order. 54. By definition, a^ = aa, a^ = aaa, a* = aaaa, etc. Hence a^ x a^ = aa x aaa = a^ = a^+* ; also a^ X a^ = aaa x aaaaa = a^ = a^+^ ; and a X a* = a X aaaa = a^ = a^+^. In the above examples we see that the index of the product of two powers of the same letter is equal to the sum of the indices of the factors. We can prove in the following manner that the above is true whenever the. indices are positive integers : Since, by definition, a"* = aaaa • • • to m factors, and a" = aaaa • • • to n factors ; .-. a"* X a" = {aaa • • • to m factors) x {aaa ... to 71 factors) = aaa • • . to m -f n factors = a'"^**, by definition. Thus, for any positive values of m and n a"" X a" = a'"+". This result is called the Index Law. 55. Product of Monomial Expressions. The results arrived at in the preceding articles will enable us to find the product of any monomial expressions. These results are : (i.) The sign of the product of two quantities is MULTIPLICATION. 61 -f when the factors are both positive or both negative ; and the sign of the product is — when one factor is positive and the other negative. (ii.) The factors of a product may be taken in any order. (iii.) The index of the product of any two powers of the same quantity is the sum of the indices of the factors. Ex.1. Multiply 3a262by 6o268. 3 a-b'^ X rt2ft3 = 3 X 6 X a2 X a2 X 6-2 X fts, from (ii.) = 18 rt2 + -212 + 3^18 a^b^, from (iii.) Ex. 2. Multiply -Sa^bhy - 5 ab^. ( - 3 a%) X ( - 5 a6°) = + 3 a^ft x 5 ab^ from (i.) = 3 X 5 X a^ X a X 6 X 6^, from (ii.) = 15 a2 + ifti + 5 = 15 a^b^ from (iii.) Ex. 3. Multiply 2 a'^b^c* by - 5 a^b. The whole work of finding the product of two monomial expres- sions can be performed mentally and the result written down at once. First put down the sign of the result ; then multiply the numerical coefficients to find the numerical coefficient of the product ; then take each letter which occurs to a power whose index is the sum of the indices of that letter in the factors. Thus the product of 2 a263c4 and - 5 a^b is - 10 a^b^c^. Ex. 4. Find the cube of 2 a%. The cube is 2 a^ft x 2 d^b x 2 a^^ = 8 a^b^. 10. 2 a by - 4 6. 11. 36 by -4a. 12. a2 by - a. 13. - 6 a^b by 4 ab. 14. -2ab^hy-7a^b'^. 15. - 3 a^-bd^ by C ab'^c^. 16. - 3 a&2c by 2 a&3c-2. 17. -2a:y*by -Sa^V- 18. 2 ax2?/30 by - 5 a^xi/*z^. 52 MULTIPLICATION. EXAMPLES XII. Multiply 1. 3a by 6a. 2. 6a2by7a. 3. 2a3by5a2. 4. ab by a%^ 5. 3a2&by2a62. • 6. 4a63by7a*62. 7. 3 a2&c3 by 6 a6. 8. 5 ab''^c'^ by 3 ab^. 9. 2 a^ftSc by a6c. 19. 6 a%'^cH^y^z by - 12 a63c*a;22^3;^^ 20. Find the values of (- a)% (- a)^, and {-ay. 21. Find the values of (- ic'^)^, (- a;2)3^ and (- x'^y. 22. Find the values of (- aby\ (- aby, and (- a6)*. 23. Find the values of {a%^y^, {a%'^y, and {a^^y. 24. Find the values of (- 2 a%^y, (-2 a%^y, and -(2 a^M)^. 26. Find the squares of 3 a&V, - 2 a^&c*, and - 4 a^ftSc^. 26. Find the cubes of a'^, —a*, ab, and — a^fe. 27. Find the cubes of 2a&2, _ 3a«6^ - 4a6^ and - 1 a%^(^. 28. Find the values of (- a)2 x (- 6)3, ( - 2 a)^ x {a'^y, (- a62)3 X (- a26)3, and (a262)3 x (-rt?>3)4. If a = 2, 6 = — 3, c= — l,d = 0, find the numerical values of 29. 6a6. 35. 3 a^&V + 5 ftV. 80. 4a6c. 36. 2 a&2 + 3 ?>c2 + 4 cd^. 31. 8a262c2. 37. (a + 6) (c + d). 32. a6 + 6c + ca. 38. (a - 6)2(c - d)2. 88. a8 + 6» + c8 + f7'. 39. (a^ + 6c) (6"^ + c -f c H )M= aM-\- hM-\- cM-] = Ma-\-Mb-^Mc-\- ••- [Art. 52.] = {x -{- y + z + •")a + {x -{- y -\- z -\- •••)& -\-{x + y-\-z+-")c-\-"' = ax -\- ay -{- az -\- • • • -\- bx -\- by -\- bz -{- • • • -{- ex -{- cy -{- cz -\ 1---- Hence (a + & + c-f •••)(.'» + ?/ + 2:4- •••) = ax -^ ay -{- az -\ 1- bx-\-by -\-bz-\ [- ex -j- cy -\- cz -\ Thus the product of any two multinomial expressions is the sum of the products obtained by multiplying every term of the multiplicand by every term of the multiplier. For example, {a -\- b) {c -\- d) — ac -\- be -\- ad + bd. Also (2a-f-r)?>)(3a4-26)=2ax3a-fr)&x3a + 2ax2&+56x26 =:(\a'-\-ir)ab-^\((h^\Ob''=Cur-\-nab-\-\Ob\ MULTIPLICATIOK. 57 Again, to find {a — b) x (c — d); we must first write this in tlie form \ a -h {—b) \ \c -\- (— d) \, and we then have for the product ac-\-(-b)c-\-a{-d)-\-{-b){-d), which by Art. 51 is equal to ac — be — ad -f bd. Note. In the rule given above for the multipUcation of two algebraical expressions it must be borne in mind that the terms include the prefixed signs. 60. The following are important examples : (i.) (a 4- by = (a -{-b)(a -\-b)= aa-{-ba -\- ab -^bb; .-. (a-\-by = a'-^2ab-\-b\ Hence, the square of the sum of any tivo quantities is equal to the sum of their squares plus twice their product. (ii.) (a-by = (a - b) {a-b)=aa + {- b)a + a{-b) -^(-b){—b) = d' — ab — ab + 6'- ; .-. (a-by = a^-2ab-\-b'. Hence, the square of the difference of any tivo quantities is equal to the sum of their squares minus twice their product. (iii.) (a ^b){a — b) = aa + ba + a( — 5) -f 6(— 6) = a^ -\- ab — ab — b^; ... (a-\-b){a-b) = a^-b\ Hence, the product of the sum and difference of any two quantities is equal to the difference of their squares. 58 MULTIPLICATION. 61. It is usual to exhibit the process of multiplication in the following form : a^ - 2 a6 + 62 a'b^-[-2ab^-b* a* -4,a^b^-{-Aab^-b' or by a rectangular arrangement as follows : a^-\-2ab-b^ -2ab a^^2a%- a^b^ -2a^b-4:a'b^ + 2ab^ + a'b^-{-2ab^-b* a* _4a262 + 4a63-6* The multiplier is placed under the multiplicand, arranged horizontally, or to the left arranged vertically. The successive terms of the multiplicand, namely a^, + 2 ab, and — b% are multiplied by a^, the first term on the left, or at the top, of the multiplier; and the products a\ + 2 a% and — arb^, which are thus obtained, are put in a horizontal row. The terms of the multiplicand are then multiplied by — 2 ab, the second term of the multi- plier, and the products thus obtained are put in another horizontal row, the terms being so placed that 'like' terms are under one another. The terms of the multi- plicand are then multiplied by b^, the last term of the multiplier, and the products thus obtained are put in a third horizontal row, 'like' terms being again placed under one another. The final result is then obtained by MULTIPLICATION. 59 adding the rows of partial products ; and this final sum can be readily written down, since the different sets of * like ' terms are in vertical columns. 62. The following are additional examples of multipli- cations arranged as in the preceding article : ci + 6 a + b a a -b -b a-{-b a-b a^-ab -ab 4-6' a^-^ab -ab- -b' a^^2ab-h b' a +-^2b a' -2ab + b' a + 6 + c a' + 6^ + + -b' a -^2b a^-\-^2ba -V26a-262 a 4-c ci" + ab-{- ac H- ab -f ac be 6c + c2 262 a2 + 2a6 + 2«c4-62 + 26c + c2 63. While these arrangements are helpful to the be- ginner, the practised worker in algebra writes out the product of two simple multinomials without going through this somewhat long and formal process. Let the results in some of the examples of the following list be written out in this way. Thus, in order to write down the product of « + 1 and x^ -\-x-\-l, observe that there can be only four kinds of terms ; namely, terms in a^,a^, X, and a term independent of x. The only partial product containing x^ is x xx^; two partial products, XXX and 1 x a^, contain x^ ; two, a; x 1 and 1 x a;, con- tain X ; and one, 1 x 1, is independent of x. Hence the product is {x + l)(x' + x + l)=:x^-\-2x'-h2x-{-l. 60 MULTIPLICATION. EXAMPLES XIV. Multiply 1. x + 2yhy x-2i/. 11. 2y + 5bhy Sy - ib. 2. a - 36 by a + 3 ?>. 12. 3 w^ - 1 by 3m'- + 1. 3. 2x-\-Syhy '3x-2y. 13. 2 w^ + 5 n^ by 2 m"^ - 5 tA 4. 5 a + 4 & by a — &. 14. a + | 6 by a — ^ 6. 5. X + 7 by X + 6. 15. 2 a + i & by 3 a + I b. 6. X - 7 by X - 6. 16. \ a - i 6 by i a - [ b. 7. X + 7 by X - 6. 17. x^ + x + 1 by x - 1. 8. a + 9 by a - 5. 18. x^ - x + 1 by x + 1. 9. 2 X - 4 by 2 X + 6. 19. a"- + ab -\- b'^hy a - b. 10. 3x - 7 by 2x - 1. 20. a'- - a6 + &^ by a -|- 6. 21. 4 a^ + 6 a& + 9 &2 by 2 a - 3 b. 22. IGp^ + 20 pq + 25 (?2 by 4p - 5 q. 23. x3 - 3ax2 + 2 a^x by x + 3a. 24. a3 _ 4 0^2^ _,. 0,52 by a2 + 4 a&. 26. x3 - 3x2 + 2x + 1 by x2 + 3x + 2. 26. x3 -f x2 - 2 X -f 1 by x2 - X + 2. 27. x2 + xj/ 4- y^ by x2 — x?/ + ?/2^ 28. a* + a262 + ^4 by a* - a262 ^ 54. 29. 2 x8 - 3 x2?/ 4- 2x?/2 -f ?/ by a;'2 + 3x?/ + 2y^. 30. x3 - 4 x2?/ + 6 x.v2 - 3 y3 by 3 x2 - 4 xy + 5y2. 64. If in any expression consisting of several terms which contain different powers of the same letter, the term which contains the highest power of that letter be put first on the left, the term which contains the next highest power be put next, and so on ; the terms, if any, MULTIPLICATION. 61 which do not contain the letter being put hist ; then the whole expression is said to be arranged according to descending poivers of that letter. Thus, the expression a^ + a% + aU^ + h^, is arranged according to descending powers of the letter a. In like manner we say that the above expression is arranged according to ascending powers of the letter h. 65. Although it is not necessary to arrange the terms either of the multiplicand or of the multiplier in any particular order, it will be found convenient to arrange both expressions either according to descending or accord- ing to ascending powers of the same letter : some trouble in the arrangement of the different sets of like terms of the product in vertical columns will thus be avoided. Hence, before beginning to find the product of two ex- pressions, it is often desirable to rearrange the terms. 66. A term which is the product of n letters is said to be of n dimensions, or of the nth degree. Numerical factors are not to be counted in reckoning the number of dimensions. Thus ahc is of three dimensions, or of the third degree ; and 5 a^h'c, that is 5 aabhc, is of five dimen- sions, or of the fifth degree. 67. The degree of an expression is the degree of that term which is of highest dimensions. In estimating the degree of a term, or of an expression, we sometimes take into account only a particular letter, or particular letters : thus we say that ax^ -j- 6a; + c is of the second degree in x, or is a quadratic expression in x ; also that ax~y -f bxy^ is of the second degree in x, and of the third degree in x and y. 62 MULTIPLICATION. When all the terms of an expression are of the same dimensions, the expression is said to be homogeneous. Thus a^ -\-3a^b — 5 b^ is a homogeneous expression, every term being of three dimensions ; and ax^ + bcxy + dy'^ is a homogeneous expression in x and y. EXAMPLES XV. Arrange the following expressions according to descending powers of a : 1. a^-{-b^-2ab. 3. a^ -\- b"^ -^ a% -{- ab'^. 2. 2-4a2 + 5«_6a3. 4^ ba'^ - ^ - 6a^ -2a. 5. a^ -{- b^ -^ c^ - S abc. 6. a^ + b^ + c3 -f a'^b + ab'^ + a^c + ac^ + &% + ;,c2. 7. Wliat are the degrees of the above expressions, and which are homogeneous expressions ? Bracket together the different powers of x in each of the fol- lowing expressions : 8. x^ -\- ax + bx + ab. 9. x^ — ax — bx — ab. 10. x^ + aa;2 + bx^ + cx^ + bcx + cax + rt6x •+ a6c. 11. ax^ + 6^2 + cx2 -I- ftcx + cao; + abx + a6c. 12. x'^(ij - z)-h y'^(z - x) + 2;2(x - y). 13. Simplify (6 + c - a)x + (c + a - &)« + (a + 6 - c)x. 14. Simplify (& - c)x+ (c - a)x + (a - b)x. 15. Simplify {(?> - a)x + (c - (Z)?/} + {(a + b)x + (c + d)?/}. 68. Product of Homogeneous Expressions. The product of any two homogeneous expressions must be homogeneous; for the different terms of the product are obtained by multiplying any term of the multiplicand by any term MULTIPLICATION. 63 of the multiplier, and the number of dimensions in the product of any two monomial quantities is clearly the sum of the number of dimensions in the separate quan- tities ; hence if all the terms of the multiplicand are of the same degree, as also all the terms of the multi- plier, it follows that all the terms of the product are of the same degree. When two expressions which are to be multiplied are homogeneous, that fact should in all cases be noticed by a student ; and if the product obtained is not homo- geneous, it is at once seen that there is an error. 69. We now return to the three important cases of multiplication considered in Art. 60, namely, (a + by = (a-br = (a + 6) (a ■ Def. A general result expressed by means of symbols is called a formula. Since the laws from which the above formulae were deduced were proved to be true for all algebraical quan- tities whatever, we may substitute for a and for b any other algebraical quantities, or algebraical expressions, and the results. Avill still hold good. We give some examples of results obtained by substi- tution in the above formulae. In the first place, let us put — 6 in the place of b in (i.) ; we then have Ja+(-6)p = a^4-2a(-6) + (-&)^ a? + 2ab + b^ . ■ ■ (i) a'-2ab + b^ . . ■ • (ii.) -b)=^a^-b' . . . (iii.) 64 MULTIPLICATION. that is, (a - 6)2 = a2 _ 2 a6 4- h\ Thus (ii.) is readily seen to be inchided in (i.) Now put 6 + c in the place of h iu (i.) ; we then have Ja4-(6 + c)S-^ = a2 + 2a(5-f c) + (6 + c)^ .-. \a-\-h + cY = rt- + 2ah + 2ac + li- + 26c + cl Thus Ja + 6+cP=a=^4-6-+c2+2a6+2ac+26c . (iv.) Now put — c for c in (iv.), and we have Ja + 6 + ( -c) P= a^ + 6^ + (- c)2 + 2a6 + 2a(- c) +26(-c); ... (a + 6 - c)2 = a2 + 62 -I- c^ + 2a6 - 2ac - 26c. Again put 6 + c in the place of 6 in (iii.) ; we then have Sa+(6+c)i>-(6 + c)J = «'-(^ + c)l The following are additional examples of products which can be written down at once : (a2 + &2)(a2 - &2) ::^ (^2)2 _ (^2)2 = a* - 6*, (a - & + c) (a + 6 - c) = {a - (6 - c)}{a + (6 - «)} = a2 - (6 - c)2 = a2 _ (^,2 _ 2 &c + c2) = a2 _ ^,2 + 2 6c - (l», (a2 + a6 + 62) (^2 - a6 + 62) = (a~2 +~&2 ^. ^j^) (a^T^^ _ ^h) = (a2 + 62)2 _ ^262 ^ a* + 2 a^h'^ + 6* - a262 = a* + 02^2 ^ 54. 70. Square of Any Multinomial Expression. We have found in the preceding article, and also in Art. 62, the square of the sum of three algebraical quantities ; and we can by the same methods obtain the square of the sum of more than three quantities. The square of the sum of any number of algebraical quantities may also be found in the following manner. MULTIPLICATION. 66 Suppose we wisli to find We know that the product of any two algebraical expressions is equal to the sum of the partial products obtained by multiplying every term of the multiplicand by every term of the multiplier. If we multiply the term a of the multiplicand by the term a of the multi- plier, we obtain the term a^ of the product : we similarly obtain the terms 6^, c^, etc. We can multiply any term, say b, of the multiplicand by any different term, say d, of the multiplier ; and we thus obtain the term bd of the product. But we also obtain the term bd of the product by multiplying the term d of the multiplicand by the term b of the multiplier, and we can obtain the term bd in no other way, so that every such term as bd, in which the letters are different, occurs twice in the product. The required product is therefore the sum of the squares of all the quantities a, b, c, d, etc., together with twice the product of every pair. Thus, the square of the sum of any number of algebraical quantities is equal to the sum of their squares together with twice the product of every pair. Ex. 1. Find the square of a + 6 + c. The squares of the sepa- rate terms are a!^, b^, c^. The products of the different pairs of terms are ab, ac, and be. Hence (a + 6 + cf = a^ + ?>2 + c^ + 2 a& + 2 ac + 2 6c. Ex. 2. Find the square of a + 2 6 — 3 c. The required square = a2 + (2 &)2 + (- 3 c)2 + 2 a(2 6) + 2 «(- 3 c) + 2(2 6)(- 3 c) = a^ + 4 62 + 9 c2 + 4 a6 - 6 ac - 12 be. 66 MULTIPLICATION. Ex. 3. Find (a - b -\- c - dy. (a-6 + c-(Z)2 = a2+(-6)2+c2 + (-(^)2-f2a(-6)+2ac 4-2«(-d)+2(-6)c + 2(-6)(-cZ)+2c(-d) = a2 4- &2 + c2 + d2 - 2 a& + 2 ac - 2 atZ - 2 6c H- 2 6c? - 2 cd. After some practice the intermediate steps can be omitted and the final result written down at once. T^o ensure taking twice the product of everij pair, it is best to take twice the product of each term and of every term which follows it. EXAMPLES XVI. Write down the squares of the following expressions : 1. 2a-\-b. 5. a^-bab. 9. a - b - c. 2. 4a + 36. 6. 2a2-3a6. 10. 2a + 26-c. 3. 3a -6. 7. -Sxy-^2y^. 11. 4a + 26-Sc. 4. 5a -66. 8. 4x2-72/2. 12. 2a -66 -3c. 13. a:2 + X + 1. 19. 3 a + 2 6 - 4 c + d. 14. X2-X + 1. 20. 5a + 6-4c-3d. 16. x'^-xy-\- 2/2. 21. x3 + x2 + X + 1. 16. X* + x22/2 + 2/4. 22. x3 - x2 4- X - 1. 17. a - 6 - c + d. 23. 2 x8 - x22/ + xy^ - 3 y^ 18. 2a-26-3c + 3d. 24. 2 x^ ^ x22/ + 2 X2/2 - 2/3. Multiply 25. X — y-^zhyx — y — z. 26. x-22/ + 42!byx-22/-4 2;. 27. 3x- 2/- 6^; by 3x + 2/ - 6«- 28. -x+22/-3;«;byx + 22/-3 2;. 29. x2 + X2/ + y^ by x^ - xy -{- y\ 80. 3x2-x2/ + 22/2by 3x2 + x2/ + 2 2/2. MtTLTIPLICATION. 67 31. x^-x + lhy x^-x-7. 32. 2 x2 - 3 a; + 7 by 2 a;2 + 3x + 7. 33. a + 6 + c + d by a + 6-c — (?. 34. 2o-36 + 2c-4dby2a-35-2c + 4d 35. 2a + 36 + c-2dby2a-36-c-2d. 36. a-36-4c + dby a + 36-4c-d. 71. Continued Products. The continued product of several algebraical expressions is obtained by finding the product of any two of the expressions, and then multiplying this product by the third expression. Ex. 1. Find the continued product {x + a)(x-{- h)(x + c). The process is written as under : x-ir a x-\-h x^-\- ax bx + ab x^+ia-hb)x + ab « + c x^ + ia + b)x^' '' + abx cx2 + c(a + b)x + abc a^ + (a + 6 + c)ic- + (ab -\- ac -^ bc)x + abc li a = b = c, we have (x -\- a)^ = x^ + S ax- + 3 a-x -\- a^. The above result is arranged in a way which is frequently re- quired, namely according to powei-s of x, and all the terms which contain the same powers of x are collected together. Ex. 2. Find the continued product of x^ -{-a^^x-i- a, and x — a. The factors can be taken in any order ; hence the required prod- uct = (x - a) (x + a) (x2 + a2) = (x^ - a^) (x^ -f a2) = x* - a*. 68 MULTIPLICATION. Ex. 3. Find (x - aY{x + a)2. Since factors can be taken in any order, (x — a)\x -\- ay ={x - a){x -\-d)(x — a) (x + a) ={(x-a){x-\-a)f = (x2 - a2)2 = x4 - 2 a2x2 + a*. 72. Powers of a Binomial. We shall in a succeeding chapter show how to write down any power of a binomial expression, by a formula which is called the Binomial Theorem. The square and the cube of a binomial should however be learnt at this stage. These are given by (a±by = a'±2ab + b^ and (a ± by = a^ ±3a'b + Sab' ± b^ EXAMPLES XVII. Multiply 1. 3 a^ + a6 - 62 by a2 _ 2 «& - 3 b'\ 2. x^-xy-8y^ by 3x2 -\- xy - y\ 3. 5 x3 - 4 x2y + 3 xy2 by x2 -f 4 xy + 6 y'^. 4. x3 - 7 x2?/ + 3 xy2 by X* + 7 xhj - 3 xhj"^, 5. 3x3-7x2 + 6x-3by 2x3 + 7x2-5x + 4. 6. a3 + 3 a2 - 7 a + 6 by 3 a3 - «2 _|. 5 a - 4. 7. |X2 _ I icy + ^2 by ^ X2 - ^ xy - 2/2. 8. ix2-fxy+12?/2by 12x2-f fx2/-i2/2. 9. a2 + 62 ^. c2 _ 6c - ca - a6 by a + 6 + c. 10. 4 a2 + 9 62 4- ^i _ 3 ?,c - 2 ca - 6 a6 by 2 a + 3 6 + c. 11. a2 + 62 _|. c2 -I- /^c 4- crt - ah by « + 6 - c. 12. 9 a2 + 62 4- 9 c2 + 3 6c - 9 ca + 3 a6 by 3 a - 6 + 3 c. MULTIPLICATION. 69 IS. x^ -\-xy + y^ by y"^ - xy -\- x^. 14. a2 _ 2 aft + 4 6-2 by 4 62 + 2 a6 + a^. 15. ax + a2a;2 + a^ofi by a2a;2 -ax+1. 16. a;2^ 1^ a;.}. 1^ anda; - 1. 17. «2 ^ ajs^ ^ _|_ a;, and a — ic. 18. a;2 + 4 2/2, X + 2 ?/, and x-2y. 19. 9 a;2 + 25 ?/2, 3 a: + 5 ?/, and 3 a; - 6 1/. 20. X* + ?/*, a;2 -\- y^^ x + y, and ic — ?/. 21. (a2 + ?>-2)2, (rt + 6)2^ and (a - 6)2. 22. (a:2 + X + 1)2 and (^2 - a; + 1)2. 23. a;2 - a^y -f y^, x"^ + xy -\- y% and x^ -x-y^ + y*. 24. a* - a262 ^ ^,4^ «2 _|_ ^fft + 62, and a^ - ab + 62. 25. a;2 — ax + a2, a;2 _|_ ^^-j. ^ (^i^ ^ ^ x, and a — x. 26. x + y + z^ — X + y + z, X — y + z, and x 4- ?/ — sr. Find the following cubes : 27. (2 a + 3 6)3. 29. (3 a - 2 6)3. 31. (a - b +cy. 28. (2a -3 6)3. 30. (a + 6 + c)3. 32. (a - 6 - c)3. 33. Show that (i.) (2x + 1)2 + (x-iy = 4x^+(x + 1)2 + 1. (ii.) (2x + 1)2 + (x + 2)2 =(x - 2)2 + 4x(x + 3)+ 1. (iii.) (x2 + X + 1)2 + (x2 - X + 1)2 = 2 (x* + 3x2 + 1). 34. Show that a3 - 63 = (a - 6)(a2 + a6 +62) = (rt - 6)3 + 3a6 (a - 6) = (a + 6)3 _ 3 a6(rt + 6) - 2 63. 35. Show that (x + 2/)(x + 0)+(?/ + z)(y + x) + (;? + x)(s -\-y)-(x-\-y-}- zy = yz + zx -\- xy. 70 MULTIPLICATION. 86. Show that (6 - c)^-\-(ic - ay -{-(a - by - 3(& - c)(c - a)(a -b) = 0. 37. Simplify (x-^y + zy-(-x + y-{-zy^-\-{x-y + zy-(x-hy- zy. 38. Simplify (x-\-y-\-z) (x^ + 2/2 + 2!2) - yz(y -\- z)- zx^z -\-x)- xy{x + y). 39. Show that (6 + c){c + a){a +6)+ ahc = {a-\- h -\- c){hc + ca + ab^. 40. Show that, if x=2y-^6z; then will x^=Hy^-{-\2bz^+30xyz. 41. Show that, if x=b + c — 2a, y = a-\-a — 2b, and z=a+b — 2c; then x2 + ?/2 + ^2 + 2 2/^ + 2 ^a; + 2 a;y = 0. 42. Show that a34.&Hc3+3(6 + c)(c+a)(a+&) = (a + 6 + c)3. 43. Show that a2(6+c)2+&'^(c+a)2+c2(a+6)H2a6c(a + 6+c) = 2(6c + c'a + a6)2. 44. Show that (a + 6)3 + 3 c(a + 6)2 + 3 c^(a + 6) + c^ = (6 + c)8 + 3 a(6 + c)2 + 3 a2(ft + c) + a\ 45. Show that {(a + 6)a;2 - (a2 + 62)iB + a^ + 63} {(a - 6)a;2 - (a^ - b'^)x + a' -68} = (a2 - 62)x4 - 2(a3 - 63)x8 + 3(a* - 6*)x2 -2{a^- b^)x + «« - 66. DIVISION. 71 CHAPTER V. Division. 73. In multiplication we have two factors given, and we have to find their product. In division we have the product and one factor given, and we have to determine the other factor, so that to divide a by 6 is to find a quantity c such that h x c = a. 74. Since division is the inverse operation to that of multiplication [Art. 9], and successive multiplications can be performed in any order [Art. 52], it follows that successive divisions can be performed in any order. Thus a^6-^c = a-7-c-=-6. It also follows, from Art. 52, that to divide by any quantities in succession gives the same result as to divide at once by their product. Thus a-f-6-4-c = a-=- {he). 75. Not only may a succession of divisions be per- formed in any order, but a succession of divisions and multiplications may be performed in any order. For example, « x 6 -^ c = a -f- c x 6. For a = a-i-c X C', .'. axb = a-T-cxcxb = a-i-c xb X c; [by Art. 52.] 72 DIVISION. therefore, dividing each by c, we have axb-i-c=:a-i-cxb. Hence we get the same result if we divide the product of two quantities by a third, as if we divide one of the quanti- ties by the third and then multiply by the other. 76. The operation of division is often indicated by- placing the dividend over the divisor with a line between them, or by separating the dividend from the divisor by an oblique line called the solidus : thus j- or a/6 means a-i-b. When a-^b is written in the fractional form -, a is b called the numerator, and b the denominator. By putting a = linax6-r-c = (x-j-cx6, we have lx6-J-c = l-^cx6, that is 6^c = -x6 = 6x-, c c so that to divide by any quantity c gives the same result as to multiply by — c 77. Since a^ xa^ = a*, and a^ x a^= d^^, we have con- versely a* -f- a'^ = a' = a*"^, and a^^ -i-a^ = a^^^ = a^ and similarly in other cases. Hence, if one power of a letter be divided by another power of the same letter, the index of the quotient is equal to the difference of the indices of the dividend and the divisor. DIVISION. 73 78. We have proved, in Art. 51, that ax(— ?>) = — a6; .-. (— a6)-^(— 6) = a, and ( — aft)-j- a = — 5. We have also proved that (—a) X {— b) = + ab, and {-{-a) x (+6) = 4-a6; .-. {-{- ab) -T- {— a) = — b, and (f a6)^(+ a) = + &. Hence, if the signs of the dividend and divisor are alike, the sign of the quotient is + ; and if the signs of the dividend and divisor are unlike, the sign of the quotient is — ; we therefore have the same Law of Signs in division as in multiplication. 79. Division of Monomial Expressions. The results arrived at in the preceding articles will enable us to divide any one monomial expression by another. These results are: (i.) The sign of the quotient is + when the signs of the dividend and divisor are alike, and the sign of the quotient is — when the signs of the dividend and divisor are different. (Law of signs.) (ii.) The operations of multiplication and division may be performed in any order. (Laws of commutation and association.) (iii.) When one power of any letter is divided by any smaller power of the same letter, the index of the quotient is equal to the difference of the indices of the dividend and the divisor. (The index law for division.) Ex. 1. Divide 18 a^b^ by 6 a^b^. 18 a*b^ -f- 6 a263 = IS ^ 6 x a^ ^ a'^ x b^ -i- b% from (ii.) = 3 a*-265-3, from (iii.) = 3a262. 74 Divisioi^. Ex.2. Divide ISa^fte by - Saft^. 16a366H-(-6a65) = -15--5xa8H-ax6«-^65, from (i.) and (ii.) = -3a3-i66-5^ from (iii.) = -3a26. Ex. 3. Divide - 6 a'b^c* by - 7 a^b^C*. The whole work of dividing one monomial expression by another can be performed mentally and the result written down at once. First put down the sign of the result ; then divide the numerical coefficient of the dividend by that of the divisor ; then take each letter which occurs in the dividend to a power whose index is the difference of the indices of that letter in the dividend and the divisor. Thus (- 5 a'^b^c^) -^ ( -7 a^b^c^) = + f a*b% since c* ^ c* = 1. If, in the above example, we had used the rule for finding the index of the quotient of c* -=- c*, we should have been led to the at present meaningless result c^ : but c* -^ c* is obviously 1. 80. Division of a Multinomial Expression by a Monomial. When a multinomial expression is divided by a monomial, the quotient is equal to the sum of the quotients obtained by dividing its separate terms by that monomial. Thus (a + 6H ) ^a;= a-^ a; + &-^a; + ••• To prove this, multiply by x ; then (a-{-b-\ )-^x XX = a-\-b + '•-', also {a-i-x-{-b-i-x-] — ) xx=:a-i-xxx-\-b-i-xxx-\ — = a4-&+--- [Art. 57.] Hence {a-\-b-\ )-^ a; = a-J-a;^-&-^a; + ••• Ex. 1. Divide a^ + 2a^ by a. The result is a» -^ a + 2 a-^ -r- a = a'^ + 2 a. DIVISION. 75 Ex. 2. Divide a^'^ — 3 ax by ax. The result is aH^ -=- ax + ( - 3 ax) -=- aa; = ax - 3. Ex. 3. Divide 12 x^ - 5 ax^ - 2 a^x by 3 x. The result is 12x8-T-3x + ( -5ax2) -- 3x +(- 2a2x) -^3x = 4x2 - fax-fa^. EXAMPLES XVIII. Divide 1. 10 a by - 5 a. 10. 25 a'6V by - 6 d^hi^. 2. - 10 6 by 2 6 11. - 27 a%^^<^ by - 6 d?h^. 3. - 2 X by - 3x. 12. abed by - ah. 4. a2 by - a. 13. a263c4(f by - 2 a62c#. 5. 8 a6 by - 2 ft. 14. 8 a^x^y?^ by 6 ax^?/*. 6. - 4 a6 by 3 a. 15. - 3 a^h^cx'y^ by - 2 abcx^y^. 7. 12 a2&3 by - a&2. 16. 3 x2 - 5 ax by x. 8. -6 x2y3 by - 4 xy. 17. 6 ?/* - 6 y^ by - y2. 9. -3x*y3by2xy2. 18. 4a6 - Sa^ + 2a* by a*. 19. 12 a3 + 9 a* - 6 a^ by - 3 a2. 20. 15 a*66 _ 7 aW + 9 a264 by - 3 a^b^. 21. 5(x - 1)2 - 3 a(x - 1) by (x - 1). 22. 12 a5(x - a)* - 6 a*(x - a)3 + 3 a3(x - a)2 by a^(x - ay. 81. Division by a Multinomial Expression. We have now to consider the most general case of division, namely, the division of one multinomial expression by another. The object of division is to find by what algebraical expression the divisor must be multiplied to produce the dividend, 76 DIVISION. Consider, for example, the division of Sa^ + Sa'b-^Aab^ + b^hy 2a + 6. The dividend and the divisor must first be both arranged according to descending (or ascending) powers of some common letter, and the terms of the quotient will be found separately in the same order. In the present case the arrangement is according to descending powers of the letter a. Now the term of highest degree in a in the dividend, namely, 8 a^, must be the product of the terms of highest degree in a in the divisor and the quotient ; hence the Jlrst term of the quotient must be 8 a^ ^ 2 a = 4 al Now multiply the divisor by 4 a^ and subtract from the dividend: we then have the re- mainder 4 a^6 -f- 4 ab^ + b^. This remainder must be the product of the divisor by the other terms of the quotient ; and hence the first term of the remainder, namely 4 a^b, must be the product of the first term of the divisor and the next term of the quotient. Hence the second term of the quotient is 4 a^6 -^ 2 a = 2 ab. Now multiply the divisor by 2ab and subtract the product from the re- mainder : we then get the second remainder 2 ab^ + b^ The third term of the quotient is similarly 2 ab^ -i-2a = b^. Multiply the divisor by b^ and subtract the product from 2 ab^ + b^, and there is no remainder. Since there is no remainder after the last subtraction, the dividend must be equal to the sum of the different quantities which have been subtracted from it; but we have subtracted in succession the divisor multiplied by 4:a% by 2ab, and by 6^; we have therefore subtracted altogether the product of the divisor and 4. a^ -\- 2 ab -\- b^. Hence tlio required quotient is 4 a'^ -f- 2 ab -f- b'K DIVISION. 77 The process is written in the following form 2a + 6 4:a^b + 2ab^ 2ab' + b^ 2ab^-}-b^ 4a2-|-2a& + 62 From the above example it will be seen that in order to divide one multinomial expression by another we proceed as follows : (1) Arrange both dividend and divisor according to ascending or both according to descending powers of some common letter. (2) Divide the first term of the dividend by the first term of the divisor : this will give the first term of the quotient. (3) Multiply the divisor by the first term of the quotient found above, and subtract the ^yroduct from the dividend. (4) Now treat the remainder as a new dividend, and go on repeating the process. 82. The following are additional examples : Ex. 1. Divide 3 x^ - 4 a:2 4. 25c - 1 by 1 - x. The order of the terms in the divisor must first be changed to - x + 1. 3x3-4a;2+2x-l -x + 1 3aj8-3x2 -3x2 + x- -1 _ x^+2x-\ - x^-\-x x-1 78 DIVISION. Ex. 2. Divide a* - a% + 2 a^"^ - ab^ + b* by a2 + b^. a* - a^b -^ 2 a^b"^ - ab^ -^ 6» a* + a262 a2+62 i'^ -ab-\-b'^ -a^b-\- a%'^ - - ab^ + 64 -a% - -ab^ + a'^b-^ + 6* + a^b^ + M Ex. 3. Divide a* + a^b^ 4. 54 by ^s _ ^^^ ^ 52. ^4 + a252 ^ 54 I g-^ - a6 + &^ a* - a^ft + ^252 a2 4- aft + 62 + a36 + 64 + a^6 - a2ft2 _^ cf63 + a262 - - ab^ + 64 4- a262 - - a68 + 6* In the last example the terms of the dividend were placed apart in order that ' like ' terms might be placed under one another without altering the order of the terms according to descending powers of a. The subtractions can, however, be easily performed without putting < like ' terms under one another ; but the arrange- ment of the terms according to descending {or ascending) powers of the chosen letter should never be departed from. 83. Horner's Synthetic Division. In the foregoing process of division there is a redundancy of work wliich may be avoided by a different arrangement of the successive steps. Fixing upon the idea of division as the process that undoes multiplication [Art. 9], let us first multiply together the divisor and quotient of Ex. 3 of the last article, obtaining the dividend as their product and arranging the work in rectangular form as follows: DIVISION. D a^-\-ah +6^ =Q 7d — ah a^ -\- a^h -\- a'h'' =Pi — a^b — a-b^ — aW = P2 For brevity's sake the several miiltinoitiials of the diagram will be referred to by means of the attached letters A Q, Pi, P2, Psy P- In order to retrace the steps here indicated and recover the quotient Q from P and D (the dividend and divisor), regarded as known or given, we have merely to subtract from P the sum of the two partial products pa and p^ and divide the result by a^ ; for P was obtained by adding together p^, Ps, and a^Q. Or better, instead of subtracting P'2-{-Ps from P, we may change all the signs in p2 and ^3, add together —P2, — jPsj and P, and then divide the result by a^, as before. Now we can obtain the succes- sive terms of —^2 and —p^ by multiplying the negatives of the last two terms of D by the successive terms of Q as fast as found; and tliese are found, by addition, as fast as the columns in the above arrangement, proceed- ing from left to right, are completed by the successive multiplications here indicated. In order to exhibit the process in a convenient form, we modify the arrangement in the multiplication process by interchanging P and Q, suppressing jh, and changing all the signs in the lines in which po and jh stand. Thus, a^ a'+ +a'b'^-\- +b' . . . = P -{-ab ^a'b + a'b' + ab^ =-P2 __62 -a'b^-ab^-b* . . . =-^3 a^^ab-^b'-: 80 DIVISION. The successive steps of the work are as follows : (1) The dividend and the divisor, with all its signs changed except that of the first term, are written down as indicated, arranged according to descending powers of a, the zero coefficients of a^ and a being retained. (2) The first term a^ of the quotient is found as the quotient of a'* by a^. (3) The oblique column ~ _ 2u2 is got by multi- plying 4- ab and — b^, the last two terms of the divisor (with signs changed), by a% the first term of the quotient. (4) The second term ab of the quotient is the sum of the terms in the second column, divided by a\ (5) The terms -f a^b\ — ab^ of the second oblique column are the pr®ducts of ab, the second term of the quotient, by + ab and — b\ (6) The third term b^ of the quotient is the sum of the terms in the third column, divided by a^. (7) The terms ab^, — b* of the third oblique column are the products of b% the third term of the quotient, by -h ab and — b^. (8) The terms of the last two columns destroy one another. In this process, since the quantities sought are merely the coefficients of the different powers of a in the quotient, these powers of a may be omitted until the coefficients are found, and be then inserted in their proper places in the quotient. The work then appears in the following abbreviated form ; 1 + b -62 DIVISION. l + O + ft' + O +6* + 6 + 6^ + 6' -b'-b'-b* 1 + 64.62. Quotient : a^ + ab -{- b^. 81 The method here described is known as Horner's syn thetic division. 84. Other examples of the method : Ex. 1. Divide x* - 2 x2 + 8 a; - 3 by x2 + 2 x - 1. 1 -2 + 1 1+0-2+8-3 -2+4-6 +1-2+3 l_2 + 3; Quotient : x2 - 2 x + 3. Ex. 2. Divide 4xS - 6 x* + 4 x^ - 11 x^ + 1 by 2 x^ - 3x - 1. 2x2 f-3x + 1 4x5-6x* + 4x3-llx2 + 0x+l + 6x* + + 9x2-3x + 2x3+0 +3x-l Quotient: 2 x^ + 0x2 + 3x - 1 ; Observe that the terms of the quotient are here obtained by dividing the sums of the terms in the respective columns by 2 x^. 85. If the division be exact, zeros will appear at the right of the quotient, in the diagram of the synthetic method, and their number will be one less than the number of terms in the divisor. For, when the dividend, divisor, and quotient are made complete by inserting all the zero-terms (powers of the leading letter with zero-coefficients), the number of terms 82 DIVISION. in each is one greater than the highest exponent of its leading letter, while the highest exponent of the dividend is the sum of the highest exponents of the divisor and the quotient. Hence the number of terms in the dividend is one less than the number of terms in the divisor and quotient combined. But the last row of our diagram has the same number of terms as the dividend, and therefore Number of zeros = (number of terms in dividend) — (number of terms in quotient) = (number of terms in divisor) — 1. Should the zeros not make their appearance, as re- quired in exact division, the algebraic expression which takes their place is called a remainder. [See Art. 89.] It is well to separate this remainder from the quotient by a distinctive mark, a line, or a semicolon. 86. Note. The foregoing exposition of the division process in its two principal forms (Arts. 81-84) has employed only particular examples for the purpose, and is, therefore, not in the form of strict algebraic proof. But in many such cases a formal proof is not deemed necessary. A general principle can sometimes be recognized through its application to a particular example. 87. The following formulae are very important, and should be remembered ; they can easily be verified. (a^ -f 2 ax + a^) -^ (« + a) = a; -f a, (pc^ — 2ax-\- a-) -i-(x — a) = x — a, {x^ — a^) -i- {x — a) = X -{- ay (x^ — a^) -^(x — a) = x^ + ax-]- a^, (ar^ -f a^) ■^(x-\-a) = x^ — ax + a\ {x* — a*) -i- (x — a) = x^ -{- ax^ -{- a^x + a\ («'' — a*) -i-{x-\-a)=x^^ax^-\- a^x — ct^. DIVISION. 8S EXAMPLES XIX. Divide 1. x2 - 5x + 6 by ic - 2. 11. x'^ - ^y* hy x - ^y^. 2. a;2^5x-24by x + 8. 12. ^x^ - j\a^ hy ^x^ - \aK 3. a;2-23x+132bya;-ll. 13. 3x2 + 8a; + 4 by 6a: + 12. 4. x2 - 4x - 77 by X + 7. 14. x^ - ^'^ x - 1 by 4x + 3. 5. 3x2-4x-4by x-2. 15. a^ - b^ hy a - b. 6. 3x2-llx+10by3x-5. 16. Sa^ + 27 6^ by 2a + 36. 7. a2 _ fc'2 by a - 6. 17. a^a^ _|_ ^,3^3 by ax + by. 8. x2-9y2byx-32/. 18. 8 0^^x6-1252/6 by 2 a2x2- 5 y2. 9. x2 - 16 2/2 by X + 4 y. 19. x^ - 8 x - 3 by 3 - x. 10. 9x2 -642/* by 3x4-82/2. 20. 2x3 - 5x2 + 4 by 2 - x. 21. 2x*-9x3 + 10x2- 5x + 6by 3-x. 22. X* + 2 x2 - X + 2 by 1 -^ X + x2. 88. We now give examples which require greater care in the arrangement of the terms. Ex. 1. Divide 2a2-2 62_3c2_56c-5ca-3a6 by a-26-3c. Where, as in this example, more than two letters are involved, we must not merely arrange the terms according to powers of «, but b also is given the precedence over c. The terms are there- fore arranged as under. 2a^-Sab - 6ac-2b'^ - 6bc-3c^ \a -2b -Sc 2a^ -4ab — 6ac 2a + 6 + c ab-\- ab ac - -262_66c-3c2 - 2 62 _ 3 6c + ac ac - 2 6c - 3 c2 -2 6c-3c2 84 DIVISION. Ex. 2. Divide a^ -\- b^ + c^ - S abc hy a -\- b -[■ c. First arrange the dividend according to powers of a, and give b precedence over c throughout. a»-SabG + b^ + c^ a8 + a'^b + a^c a+ b -h c a'^ - ab -ac-\-b^-bc + €^ - a%- - a%- -ab^ - 3 abc f 63 + c3 — abc + a62- 2 a6c + 63 + c8 a6c - ac2 + a62_ a6c + «c2 4- 68 + c8 4- 63 + 62c — abc + ac"^ - 6^0 + c* a6c - 62c - 6c2 ac2 + 6c2 + c8 The work of this example is much shorter by the synthetic method. Arrange the terms according to descending powers of a, and retain only their coefficients and the term not containing a. Thus, 111 -36c +(63 + (.8) -(6 + c)| -(6 + c) + (6 + c)2 _(6 + c)(62 + c3-6c) l-(6 + c) + (62-f c2-6c); Hence the quotient is a2 - (6 + c)a + 6"^ + c2 - 6c = a2 4. 52 4. ^2 _ ftc - ca - ab. Ex. 3. Divide a2(6 - c) + 62(c - a) + c2(a - 6) by a - 6. Arrange the terms according to descending powers of «, and retain only their coefficients and the term not containing a. Thus, 1 1 (5 _ c) - (62 - c2> + (62c - 6c2) + 6 1 4- (ft _c)6+(6c2-62c) (6-c) + (c2-6c); Hence the quotient is (6 — 2(c - a) + c2(a - 6) . 40. a\b - c) + 64(c - a) + c4(a - 6) by a\b - c) + 62(c _ a) + c2(a - b). MISCELLANEOUS EXAMPLES I. A. 1. Find the value of (y - zY +(0 - xy +(«— yy when X = — 1, ?/ = 0, and 2; = 1. 2. Add together 3 a2 - 2 ca - 2 aft, 2 62 -|- 3 6c + 3 6a, and c2 — 2 ac — 2 6c. 3. Show that 08 -f 68 + c8 - 3 a6c - a(a2 - 6c) - 6(62 _ ^a) - c(c2 - ab) = 0. 4. Multiply 2x2-Ja; + ^by^x + 3, and i ^2 + ^ a;?/ + 2/2 by 1^ a;2 _ j^ a;^ _l_ ^2. 5. Simplify (a; + l)(a; + 2)(x + 3)-(a; - 1)(« - 2)(a; - 3). 6. Divide 2 - 12x5 + lOx^ by 1 - 2x + x2. B. 1. Find the value of c2 + 6(c + «), and of V2 6c — a, when a = 11, 6 = 3, and c = 6. 2. Subtract 6 a* - 3 a36 + 4 a262 from 5 6* - 3 63a + 4 62a2. 3. Simplify x - (1 - 1 - x), 3 x - 7 - 4x, and 6 — 2a — {c — a— (6 — a + c)}. 4. Multiply 9 x2 - 1 by x2 + I, and a + 6 + c by a + 6 - c. 6. Show that 2(a2 + 62 + c2 - 6c - ca - a6) = (6 - c)2 + (c - a)2 + (a - 6)2. 6. Divide x8 - 3 x2 + 3 x + y8 _ 1 i,y x + ?/ - 1. C. 1. Find the numerical value of Va2 + 2 6c J V62 + eg ^ Vc^ + a 6 c when a = 4, 6 = 3, and c = — 2. MISCELLANEOUS EXAMPLES I. 89 2. Add together Sa^b-6 ab"^ + 7 ft* and 2 a^ - ^ a^b -\-5ab^- 1 b^. 3. Multiply ax^ — X + a hy ax^ + X -{- a, and find the square of 2a + b-Sc. 4. Simplify (x-\-yy-(ix-\-y)(x - y)- {x(2y-x)-y(2x-y)]. 5. The product of two algebraical expressions is x^ -\-x^y-\- x^y^ - x^y^ + y^, and one of them is x^ -{■ xy -^ y'^ ; what is the other ? 6. Prove (i.) a(b - c)+ 6(c- a) + c(a - 6)=0, (ii.) a2(6-c) + 62(e_a) + c2(a-6) + (6-c)(c-a)(a-6) = 0. D. 1. If a = l, 6=2, and c=-3, find the value of 6a + 36+4 c, and of a^ + &3 + c^ — 3 abc. 2. Simplify S{x - 2(y - z)} -I4y + {2y -(z - x)}]. 3. What must be added to (a + 6 4- c)^ that the sum may be (a-b-cy? 4. Multiply a* + 6^ by a — 6, and divide the result by a + 6. 5. Divide x^ + z^hy x -\- z, and from the result write down the quotient when (x + yy -\- z^ is divided hy x + y -\- z. 6. Prove (i.) (a; + y)(x -y) + (y+ z)(y -z) + \ bx -}- ay = 1. 28. ax-^ by = a^ - b'\ , bx + ay = a^ - ft^. 35. - + - = 1, X y 29. 62a;_a2y = 0, ^+«:^1. bx ^- ay = a ■\- b. x y ' 30. x + y = rt + 6, 36. (a + 6)a:+(a + c)y = a + 6, ax-by = ft2 _ «2. (a + c) X 4- (a + 6) y = a + c. 31. bx-ay = b% 37. (a + 6)a; -(a -6)?/ = 3a6, ax— by = a'. (a + ft) 2/ — (a — ft) x = aft. 108. Simultaneous Equations witli Three Unknown Quantities. Three simultaneous equations of the first degree, con- taining the three unknown quantities x, y, and 2, can be solved in the following manner : Multiply the first and second of the given equations by such quantities that, in the resulting equations, the coefiicients of one of the unknown quantities, z suppose, may be equal ; then by addition or subtraction we elimi- nate z. Then take the first and third, or the second and third of the given equations, and eliminate 2 in a similar man- ner. We thus obtain two simultaneous equations con- taining the two unknown quantities x and y ; and these can be solved as in Art. 104. 120 SIMULTANEOUS EQUATIONS. Ex. 1. Solve the equations 2a; + 42/4- 2; = 7, 6x — 4:y + 4 z = 9. Multiply the first equation by 2 ; then 4x-{-Sy + 2z=U. Subtract the second equation ; then x + 6y = 6 (i.) Now multiply the first equation by 4 ; then Sx+16y -\-4z = 2S. Subtract the third equation ; then Sx + 20y = 19 (ii.) From (i.) and (ii.) we obtain x = S, y = a. Substitute these values in the first of the given equations ; then 6 + 2 + 2 = 7; therefore z= -1. Thus X = S, y = ^, and z =: — 1 are the values required. Ex. 2. Solve the equations + =6, y z X 1 + 1-1 = 3. z X y 1+1-1=1. X y z Add the first two equations ; then 2 1 - = 8, whence z = — z 4 Add the first and third equations ; then 2 1 - = 6, whence w = -• y 3 Add the second and third equations ; then - = 4, whence x = — X 2 SIMULTANEOUS EQUATIONS. 121 EXAMPLES XXV. 1. y + = 14, 10. x + y -\- z = 1, x + y = 24. 2 + 4^ '^-^' 2. y + .=:2«, ^4.§1_| = 0. ^+x=2fe, 3 4 2 «' + 2/ = 2^. 11^ ax-{-by=l, 3. X + y + 2 = 1, 6y + C0 = 1, 2a; + 3?/+ 2 = 4, cz + ax=l. 4x + 9y-\-z = 16. 12. cy + 62; = 6c, 4. 5 rK + 3 y + 7 2 = 2, a^; + ca; = ca, 2x-4y-\-9z = 7, bx -\- ay = ab. Sx-\-2y + 6z = S. 5. 2x-y + ^ = 4, 13. x+y + ^ = 3, 6x+2/+32 = 5, 1 2x-3y+4^ = 20. 2x-\-Sy + - = 2, 6. x + 2y-30=36, Sx-2y + ^=n. 2x-\-4ky -1 z = 9, z 3 jc - 2^ - 5 = 8. 7. a; + 22/ + 3^ = 4, *' a^ft^ z ~ ' 2a; + 3y + 42r = 6, ^.§J_. c_o 3a; + 4y + 6^ = 8. a^ b z~ * 8. 3a; + 2y + 52; = 21, ^ _ ^1 _j. 3^ ^ 11^ 2x-Sy-\-4z = n, a b z x + ^y + lz = 20. 15, a:-ay + a% = a3, 9. 10 a; - 2 y + 4 ;2 = 5, x-by^b'^z = b% 3 X + 5 1/ - 3 = 7, X - cy + 0^2 = c3. x + 3w-2^ = 2. 122 PROBLEMS. CHAPTER IX. Peoblems. 109. We shall now give examples of problems which involve more than one unknown quantity, and in which the relations between the known and unknown quantities are expressed algebraically by means of equations of the first degree. Many of the problems given in Chapter VII. really con- tain two unknown quantities, but the given relations are in those cases of so simple a nature that it is easy to find an equation giving one of the unknown quantities in terms of the known quantities, and when one of the unknown quantities is found, the other is immediately determined. Ex. 1. Find two numbers such that the greater exceeds twice the less by three, and that twice the greater exceeds the less by 27. Let X and y be the numbers, of which x is the greater. Then we have by the conditions of the problem X - 2 y = 3, and 2 X - y = 27. Multiply the first equation by 2 ; then 2x-4y = 6. Now subtract the members of this last equation from the corre- sponding members of the second equation, and we have 8y = 21, ory = 7. PROBLEMS. 123 Then from the first equation Thus the numbers are 17 and 7. Ex. 2. A number of two digits is equal to seven times the sum of its digits, and the digit in the ten's place is greater by four than the digit in the unit's place. What is the number ? Let X be the digit in the ten's place, and y be the digit in the unit's place. Then the number is equal to 10 x + y, for the x represents so many tens; also the sum of the digits is x -\- y. Hence we have lOx + y = 7(x-\- y), that is 10x-\-y = 7x + 7y; .♦. Sx = 6y^ or x = 2y. We have also x = y + 4. Hence 2y = y -\- i, or ?/ = 4, and therefore x = S. Thus the required number is 84. Ex. 3. Find the fraction which is equal to ^ when its numerator is increased by unity, and is equal to ^ when its denominator is increased by unity. Let X = the numerator of the fraction, and y = the denominator. Then we have by the conditions of the problem x+l_l^ ^„^ X _1 y 2 y+l 3 Multiply the first equation by y ; then Multiply the second equation by ?/ + 1 ; then x = Ky + i). 124 PROBLEMS. Subtract the corresponding members of the last two equations ; and we have from which we find that y = S. Then, since y is 8, x = ^ -1 = 4-1 = 3. 2 Thus the fraction is f . Ex. 4. A man and a boy can do in 15 days a piece of work which would be done in 2 days by 7 men and 9 boys. How long would it take one man to do it ? Let X = the number of days in which one man would do the whole ; and let y = the number of days in which one boy would do the whole. Then a man does -th of the whole in a day ; and a boy does X -th of the whole in a day. y Now by the conditions of the problem a man and a boy together do ^igth of the whole in a day. Hence we have - + - = — X y 15 We have also, since 7 men and 9 boys do half the work in a day, 1 + 5 = 1. X y 2 Multiply the first equation by 9 and subtract the second ; then 97^9^1 X X 15 2' 2 1 11 that is - = — , or - = ^ ; X 10 X 20 .-. a; = 20. Thus one man would do the work in 20 days. PROBLEMS. 125 EXAMPLES XXVI. 1. A and B have $50 between them, but if A were to lose half his money, and B | of his, they would then have only $ 20. How much has each ? 2. A number of two digits has its digits reversed by the addi- tion of 9. Show that the digits differ by unity. 3. A man bought 8 cows and 50 sheep for $ 1125. He sold the cows at a profit of 20 per cent, and the sheep at a profit of 10 per cent, and received in all $1287.50. What was the cost of each cow and of each sheep ? 4. Twenty-eight tons of goods are to be carried in carts and wagons, and it is found that this will require 15 carts and 12 wagons, or else 24 carts and 8 wagons. How much can each cart and each wagon carry. 5. A and B can perform a certain task in 30 days, working together. After 12 days, however, B is called off, and A finished it by himself 24 days after. How long would each take to do the work alone ? ' 6. If the numerator of a certain fraction be increased by 1 and its denominator diminished by 1, its value will be 1. If the nu- merator be increased by the denominator and the denominator be diminished by the numerator, its value will be 4. Find the fraction. 7. Find the fraction such that if you quadruple the numerator and add 3 to the denominator the fraction is doubled, but if you add 2 to the numerator and quadruple the denominator the frac- tion is halved. 8. The first edition of a book had 600 pages, and was divided into two parts. In the second edition one quarter of the second part was omitted and 30 pages added to the first. The change made the two parts of the same length. What were thejr in t-be first editioij ? 126 PROBLEMS. 9. If A were to receive $ 10 from B, he would then have twice as much as B would have left ; but if B were to receive $ 10 from A, B would have three times as much as A would have left. How much has each ? 10. A farmer sold 30 bushels of wheat and 50 bushels of barley for $93.75. He also sold at the same prices 50 bushels of wheat and 30 bushels of barley for $96.2S. What was the price of the wheat per bushel ? 11. A rectangle is of the same area as another which is 6 yards longer and 4 yards narrower ; it is also of the same area as a third, which is 8 yards longer and 5 yards narrower. What is its area ? 12. A and B can together do a piece of work in 1 5 days. After working together for 6 days, A went away, and B finished it by himself 24 days after. In what time would A alone do the whole ? 13. An income of $ 120 a year is derived from a sum of money invested partly in 3i per cent stock and partly in 4 per cent stock. If the stock be sold out when the 3 1 per cents are at 108 and the 4 per cents at 120, the capital realized is $ 3672. How much stock of each kind was there ? 14. A number of two digits is equal to seven times the sum of its digits ; show that one digit must be twice the other. 16. Find all the numbers of two digits, each of which is equal to four times the sum of the digits. 16. $ 1000 is divided between A, B, C, and D. B gets half as much as A ; the excess of C's share over D's share is equal to one-third of A's share, and if B's share were increased by $ 100, he would have as much as both C and I). Find how nuich each gets. 17. A number has two digits, of which the second is double the first ; and, if the digits be reversed, the new number exceeds the original number by 36 ; find the number. PROBLEMS. 127 18. A certain number consists of two digits, and another num- ber is formed from it by reading it backwards. If the sum of the two numbers is 99 and the difference is 45, tind the digits. 19. In a certain proper fraction the difference between the numerator and the denominator is 12, and if eacli be increased by 5, the fraction becomes equal to f . Find it. 20. The wages of 10 men and 8 boys for a day amount to $ 31 ; and four men receive $ 5.50 more than six boys. How much does each boy receive ? 21. A farmer has two farms, for each of which he pays a rent of $ 7.50 an acre, and his total rent is $ 3375. If the rent of one farm were reduced by $ 1.25 an acre, and that of the other by $2.50 an acre, his rent would be $2500. What is the acreage of each of the farms ? 22. A man has one pound's worth of silver in half-crowns, shillings, and sixpences j and he has in all 20 coins. If he changed the sixpences for pennies, and the shillings for sixpences, he would have 73 coins. How many coins of each kind has he ? 23. The price of a passenger's ticket on a French railway is proportional to the distance he travels ; he is allowed 25 kilo- grammes of luggage free, but on every kilogramme beyond this amount he is charged a sum proportional to the distance he goes. If a journey of 200 miles with 50 kilos, of luggage cost 25 francs, and a journey of 150 miles with 35 kilos, cost 1G| francs, what will a journey of 100 miles with 100 kilos, cost ? 24. A has twice as many dimes as dollars ; B, who has 60 cents more than A, has twice as many dollars as dimes ; together they have one more dime than they have dollars. How much has each ? 25. Of the candidates in a certain examination, one-quarter fail. The number of marks required for passing is less by 2 than the average marks obtained by all the candidates, less by 11 than the average marks of those who pass, and equal to double the average marks of those who fail. How many marks are required for passing ? 128 MISCELLANEOUS EXAMPLES II. MISCELLANEOUS EXAMPLES II. A. 1. Find the value of ab-\-2bc-Scd a» + 6-^ - c« b-\-c-\-d a2 + 6'^-c2' when a = 1, 6 = 2, c = — 3, and d = 0. 2. Subtract 2 a - 3 (a - b - a) from 2 6 - 3 (6 - a- b). 3. Show that (a:+3)(i/+3)-3(x+l)(2/+l)+3(x-l)(2/-l)-(x-3)(2^-3)=0, and that (a;+2)(?/+2)-4(x+l)(y+l) + 6xi/-4(x-l)(y-l) + (x-2)(y-2)=0. 4. Divide a;5 - 5x* + 7 x^ - x^ - 4x + 2 by x^ - 3x« + 3x - 1. 5. Solve the equations : ,. s X x±l x±2 x±_S_. ^'^2346 (ii.) ^ + 1 = 5, 2x+|-17=0. (iii.) ax — by = d^ — 6^ &x — a?/ = a' - 62 . = 62-a2/ 6. A number of marbles were divided among three boys so that the first boy had 10 less than half of the whole, the second had 10 more than a quarter of what was left, and the third had 20. How many marbles were there ? B. 1. Find the numerical value of (a - 6)2 + (6 - c)2 +(c - a)2 +(6 - c){c - a){a - 6), when a = 1, 6 = 2, and c = ^. 2. From the sum of (2 a - 6)2 and (a - 2 6)2 lake the square of 2(a-6). MISCELLANEOUS EXAMPLES II. 129 3. Show that n^ = n(n- !)(» - 2)+ 3n(n - 1)+ n, and n4 = w(w-l)(ri-2)(n-3)+6n(w-l)(n-2) + 7»t(n-l) + n. 4. Find the algebraical expression which, when divided by x2 — 2 X + 1, gives a quotient x^ + 2 x -f 1 and a remainder x + 1. 5. Solve (i.) 3(x + 3)2 + 5(x -f 5)2 = 8(x + 8)2. (ii.) ^(x + y)-K^-y)=-S -) X y X y 6 6. A man paid a bill of £100 with sovereigns and crowns, using in all 130 coins. How many coiris of each sort did he use ? C. 1. Simplify a -[a + b - {a -\- b -\- c - a + b -{- c -h d}]. 2. Find the sura, the difference, and the product of a -\- b and a_b 2 2 3. Show that ab - 3(a - 1)(& - 1) + 3(a - 2)(6 - 2) -(a - 3)(6 - 3) = 0, and that ab -4(a-l)(6-l)+6(a-2)(6-2) - 4(a - 3)(& - 3) + (a - 4)(6 - 4)= 0. 4. Divide a^+b^-\-<^—S abc by a + ft 4- c, and from the result write down, without division, the quotient when 8 x^ + 8 y^ + 2^ _ 12 xyz is divided by 2 x + 2 1/ + 0. 6. Solve the equations : (i.) x = K5y + 2), y = K^-l)• (ii.) ax + by = a^, bx -\- ay = b^. 6. A and B each shoot 30 arrows at a target. B makes twice as many hits as A, and A makes three times as many misses as B, Find the number of hits and misses of each. I 130 MISCELLANEOUS EXAMPLES II. D. 1. Find the value of abc{ah + hc-\- cd -\- da) ^ (a + &) (a + 6. Find two numbers such that twice their difference is greater by unity than the smaller number, and is less by two than the larger number. F. 1. Find the value of 3(x+ y -\-zY -{t^ + 2/^4- z^), when a: = 3, ?/ = — 5, and z = 1. 2. Find the continued product of a* + x*, a- -{- x^, a + x, and a — X. 3. Show that (1 -a)(l - &)+ a(l - 6)+ ft = 1, and that (1 - «)(1 -&)(1 -c) + a(l - 6)(1 - c)+ 6(1 - c) + c = 1. 4. Divide 4y^ - 9y^ + 6y - Ihy 2y^ -{- 3y - 1. Divide also a^ -\- 2b^ - Zc^ + be -{- 2 ac + S ab hy a -^ b - c. 5. Solve the equations : (i.) (a + x)(6 + a;) = (c + a:)((Z + x). (ii.) Sx-7y = 4,1x-9y = 2. 6. If 12 lbs. of tea and 3 lbs. of coffee cost $8.76, and 12 lbs. of coffee and 3 lbs. of tea cost $6.24, what is the price per lb. of tea and of coffee ? 132 FACTORS. CHAPTER X. Factors. 110. Definitions. An algebraical expression which does not contain any letter in the denominator of any term is called an integral expression. Thus a -\-h and \ a^ — ^h'^ are integral expressions. An expression is said to be integral with respect to any particular letter when that letter does not occur in the denominator of any term. Thus — H is integral with respect to x. a a-\-h An expression is said to be rational when none of its terms contain square or other roots. 111. In the present chapter we shall show how factors of rational and integral algebraical expressions can be found in certain simple cases. In arithmetic we mean by the factors of a number its integral divisors only ; and similarly, by the /actors of an algebraical expression, we mean the rational and integral expressions which exactly divide it. 112. Monomial Pactors. When any letter, or group of letters, is common to all the terms of an expression, each term, and therefore the whole expression, is divisible by that letter, or group of letters. FACTORS. 133 Thus ab-{- ac = a{b + c), a^b + ab^ = ab(a-\- 5), and 3 aoi^y + 6 ax^^ = 3 ax^y (x + 2 y^) . Monomial factors, if there are any, can be seen by inspection, and the whole expression can be at once writ- ten as the product of the monomial factor and its co-factor, as in the above examples. In what follows therefore we need only consider expressions which have been freed from monomial factors. EXAMPLES XXVII. Find the factors of the following expressions : 1. a;2 ^ X. 7. x^ -5x^y + 20x^y^. 2. a- - ab. 8. a' - a»x + a^x^. 3. ab - be. 9. 6ax* -b a%2 + 20 a^x^. 4. 2ax + I x\ 10. 3 aH'^y'^ - \ a^xhf. 5. 4 a;3 - 3 x^. 11. 11 a%d^ - \ ab'^t^. 6. a^ - 3 a%. 12. p^-q^r^ - 1 p^q\ 113. Factors found by comparing with Known Identities. Sometimes an algebraical expression is of the same form as some known result of multiplication : in this case its factors can be written down. We proceed to aj^ply this principle in the case of the most important forms of algebraical expressions. 114. We know that a^^2ah + Jr = {a + hy, and a" -2ah + h~ = (a - by. 134 FACTORS. Hence, when a trinomial expression consists of the stLm of the squares of any two quantities plus twice the product of the quantities, it is equal to the square of their sum. And when a trinomial expression consists of the sum of the squares of any two quantities minus tivice their product, it is equal to the square of their difference. Hence it is easy to recognize when a trinomial is a complete square ; for two of the terms must be squares, and the remaining term must be twice the product of the quantities whose squares are the other terms. The two terms which are squares must both have the same sign. Thus a2 _j. 6 a6 + 9 5^, that is a^ + (3 lyi -j. 2 a (3 6), consists of the squares of a and 3& together with twice the product of a and 3 ft, and hence a2 + 6a6 + 9Z>-2= (a + 36)2. Again, 4 a5 _ ^2 _ 4 62 = _ (a2 + 4 62 _ 4 a6) = - {d^ + (2 6)2 - 2 a (2 6)} = -(a-2 6)2. As other examples, we have 16a* + 8a2 _|. i =(4a2)2 + 2 (4a2) + i = (4a2 + 1)2, jc^ - 4 ^2^2 -j- 4 icy* = X (x2 - 4 xy"^ + 4 ?/*)} = X {x2 - 2 X (2 2/2) + (2 y2)2j = a; (X - 2 2/2)2, and (a + 6)2 - 2c (a + 6)+ d^ = {(a + 6)- c}2. Note. — We may consider that a2 — 2 a6 + 62 is equal to (p _ a)2 instead of (a - 6)2, for by the Law of Signs (6 - a)2, that is {-(a - 6)}2, is equal to (a - 6)2. In fact, when we find that an expression is equal to tlie product of two factors, we may equally well consider it is as the product of the same two factors with all the signs changed. FACTORS. 135 EXAMPLES XXVIII. Find the factors of the following expressions : 1. 4a;2 4.4x4-1. 9. 3 a^ -f 6a6 + 362. 2. 9x2 - 6x + 1. 10. 5 a* - 10 a^h + 5 62. 3. l-8x2 + 16x*. 11. a3 - 6 a2^, + 9 a62. 4. 4 a2 _ 12 a6 + 9 62. 12. 3 a^ - 30 a* 6-^ + 75 a'^h 5. 9 a* + 24 a262 + 16 6*. 13. 4 x2t/2 _ x* - 4 y*. 6. x2 + xy + i 2^-^. 14. 8x2-4x4-4. 7. 4 a2x2 + 4 ahxy + 62y2. 15. 4 xy^ - 4x2y2 + x^y. 8. 25 aix2 - 30 a262xy + 9 ¥y^. 16. y^y- + a^y + i a;y3. 17. (a + 6)2 + 4c(a + 6)+4c2. 18. (X2 + 2/2)2 _ 2 (X2 + y'O ^2 + 24. 19. 4 x22/2 4- 4 (a + 6) xy + (a + 6)2. 20. 9(a + 6)2-6c(a + 6)+c2. 115. From the formula a2-62 = (a + 6)(a_6), we see that the difference of the squares of any two quan- tities is equal to the product of the sum and the difference of the quantities. Thus a2 _ 4 62^ or a^ - (2 6)2, is equal to (a + 2 6) (a - 2 6). Also 9x3 - 4x?/2, that is x{(3x)2 - (2 y^}, is equal to x(3x + 2?/)(3x-2?/). As other examples, we have : 8 axy - 18 a^x^y^ = 2 axy (4 - 9 a^xY-) = 2 axy {22 - (3 axy)2} = 2 axy (2 - 3 axy) (2 + 3 axy) , a* - 6* = (a2 + 62)(a2 - 62)= (a2 + 62)(a + 6)(« - 6), and 792 _ 712 = (79 + 71) (79 _ 71) ^ 159 x 8 = 1200. 136 FACTORS. We may deal with the squares of multinomial expres- sions in precisely the same way as with the squares of monomial expressions. Thus (a + by - c^ = {(a + 6) + c} {(a + 6) - c} = (a + 64-c)(a + 6-c), a' + a^h' + &* = (a2 + 6^)2 _ ^^h)'^ ^ (^2 4. 52 4. ^5) (^2 + 52 _ ah), and (2a- & + 2c)2-(a4-46 + c)2 = {(2 a - 6 -}- 2 c) + (a + 4 6 + c)}{(2 a - 6 + 2 c)- (a + 4 6 + c)} = (3a + 3Z) + 3c)(a-5& + c). EXAMPLES XXIX. Find the factors of the following expressions : 19. x^-yK 20. x^ - 16 ?/. 21. 81 a* -16 ft*. 22. 625 a > -256 a;*. 23. x^y^ — a'&*. 24. a*?)t-81c*#. 25. 81a;*i/*-l. 26. 16rt46»ct-l. 27. 16-81a;V. 28. x^-y\ 29. a8 _ ft8c8. 30. «io - a^. 31. (a + ^)2-c2. 32. (rt + 6)'2-4c*. 33. 4(a; + 2/)2_l. 34. 9(a;-y)2_4, 35. (a; + y)2_(a;_y)2. 86. (2 a + 6)2 - (2 6 + a)^ 1. a2_9. 2. 16 - 62. 3. 25 aP- - 62. 4. X2-92/2. 6. 16x2-92/2. 6. 64 a2 _ 49 62. 7. 4a2-8162. 8. a;2-9?/t. 9. 36 a* - 49 62. 10. 4a262-9c2. 11. 9 a^x2 - 49 62?/2. 12. 49a-'6V-36a;2 13. 4x2/2- 9 x'J. 14. 8a62-18a3. 15. 3a!i-108a3. 16. 7 rt'i - 28 a«. 17. 8x^y--^2x2/8. 18. 7a6c2-7a868. FACTORS. 137 37. x'^-(x-yY. 41. (a-2 + fe2)2 _ 4 «2ft2. 38. a^-{2h- a)2. 42. (a + 6 + c)2 _ (a - 6 - c)2. 39. 4 (a +6)2- (a -6)2. 43. (3a +6- 2c)2-(«+36-c)2. 40. 9 (ic + 2^)2 - 4 (x - 2/)2. 44. (« - 2 & + 3c)2 -(« _ c)*. 46. (3x2 + a;-2)2_(x2-a:-2)2. 116. From the formulae Q? -\- o? = {x -{- a) {x^ — ax -\- a?), and a?-a^ = {x-a)(x^ + ax-\-ar), [Art. 87.] we see that the sum of the cubes of any two quantities is divisible by the sum of the quantities ; and that the dif- ference of the cubes of any two quantities is divisible by the difference of the quantities. Thus 8 aS _|. 27 h\ that is (2 a)^ + (3 5)3, is divisible by 2 a + 3 6, and the quotient is (2 a)2 - (2 a) (3 b) + (3 6)2, that is 4a2-6«6 4-962. Also 27 a^ - 8, that is (3a:)8 - 28, is equal to (3 x - 2)((3 a;)2 + (2.3 a;) + 22} = (3 a; - 2) (9x2 + 6 a; + 4). As other examples, a%^ _ 1 c8 = (a6 - 1 c) (a262 + i a6c + ^ c2), and 06 - 66 = (a3 + 53) (^3 _ 53) = (a + 6)(a2 - a6 + 62)(a - 6)(a2 + ah + 62). The cubes of multinomial expressions may be dealt with in precisely the same way. Thus (a + 6)8 - c3 = {{a + 6) - c]{{a + 6)2 + (a + 6) c + c2}, and {x-2yY-{y-2xY ^(^r^ - y - 2x){(ix - 2yy +(x -2y)(y -2x) + (y -2x)^} = (3x-32/)(3x2-3a;y + 3y2). 138 FACTORS. EXAMPLES XXX. Find factors of each of the following expressions ; 1. a8-868. 11. da*b + 2iab*. 2. 8a»4-ft». 12. 40 a^ftc - 5 6*c*. 3. 8a3-125a:3. 13. a^ - 64. 4. «3_ 1250^6. 14. 64a6-72966. 6. 4a8 + 32 63. 15. xi-^ - ae^s. 6. 27 x^ - I- 2/3. 16. (x + 2 y)3 - y8. 7. x^y^ + 2V a^&^- 17. (a; + 2 y)8 + (2/ + 2 a;) -. 8. 8 aSfts 4. a;3. 18. (2 y-xy -(2x- yy. 9. 2 xy^ - \ X*. 19. {x-Syy-{y-S xy. 10. 9 a46'^ - I ah^. 20. (2 ?/ - xY +{2x- yy. 117. By multiplication wo have (a; + a) (aj + 6) = aj2 + (a + 6) a; -f- a6. Hence conversely, if an expression of the form cc^ -\- px -\- q he the product of the two factors x-^a and x-{-b, the given expression must be the same as ar^ 4- (a H- 6) a; + a6 ; we must therefore have p = a-i-b and q = ah. Hence a and b are such that their sum is }), and their product q. For example, to find the factors of x^ -\-Tx-{- 12. The factors will be X + « and x + 6, where ah = 12 and a + 6 = 7. Hence we must find two numbers whose product is 12 and whose sum is 7. Pairs of numbers whose product is 12 are 12 and 1, 6 and 2, and 4 and 3 ; and the sum of the last pair is 7. Hence x2 -I- 7 X 4- 12 = (x + 4) (a; + 3). Again, to find the factors of x^ — 7 x + 10, we have to find two numbers whose product is 10, and whose sum is — 7. Since the product is + 10, the two numbers must both be positive or both negative ; and since the sum is — 7, they must both be negative. FACTORS. 139 The pairs of negative numbers wliose product is 10 are — 10 and — 1, and — 5 and — 2 ; and the sum of the last pair is — 7. Hence a;2 _ 7x + 10 ={x - 5)(ic - 2). Again, to find the factors of x^ + Sx — 18, we have to find two numbers whose product is — 18, and whose sum is 3. Since the product is — 18, one of the numbers is positive and the other negative. The pairs of numbers whose product is — 18 are — 18 and 1,-9 and 2,-6 and 3,-3 and 6,-2 and 9, and - 1 and 18 ; and of these pairs the sum of 6 and — 3 is 3. Hence x^ + S X - IS ={x + 6)(x - 3). EXAMPLES XXXI. Find the factors of each of the following expressions : 1. x^ + 4x-\-S. 13. x2 + 6x-14. 2. a;2 - 4a; + 3. 14. x^-x- 132. 8. a;2_6a; + 8. 16. x2 + 18x+72. 4. x2-8a; + 15. 16. x'^-bx-Si. 6. x2 - 11 X + 18. 17. x2 - 25 X + 150. 6. x2 + 9x+20. 18. ic2+6x-150. 7. x2+2x-3. 19. x2+llx-180.. 8. x'^ + Ax-b. 20. x'^-x-166. 9. a;2 + x-6. 21. x2- 31x4-240. 10. x2 - X - 6. 22. x2 - 17 X - 200. 11. x2+2x-35. 23. x2-34x + 288. 12. x2 - 3x - 10. 24. x2 - 35x - 200. 118. By multiplication we have {ax + b) {ex -\-d) = acT? -f- {ad -\-hc)x-\- bd. Hence conversely, if an expression of the form px^ -{- qx + r be the product of the two factors ax -^ b and ex 4- df the given expression must be the same as 140 FACTORS. acx^ -\- (ad-\- bc)x -\- bd; we must therefore have ac=p, bd = r, and ad-\-bc = q. We can in simple cases find by trial the values of a, b, c, d which satisfy these relations. For example, to find the factors of 3 x^ — 16 x + 5. The 3 x^ can only be given by the multiplication of 3 x and x. The 6 could only be given by the multiplication of 5 and 1, or — 6 and — 1 ; and, since the middle term is negative, the latter pair must be taken. We have now only a choice between (3 x — 6) (x — 1) and (3x — l)(x — 6), and it is at once seen that the latter must be taken in order that the coefficient of x in the product may be — 16. Thus the factors required are 3 x — 1 and x — 5. Again, to find the factors of Sx^ + 32x — 21, the 5x2 can only be given by the multiplication of 6 x and x. The end term — 21 can be the product of —21 and 1,-7 and 3, —3 and 7, or —1 and 21. The possible pairs of factors, so far as the two end terms are concerned, are therefore (5x =F 21) (x ± 1), (6x T 7)(x ± 3), (5x =F 3)(x ± 7), and (6xT l)(ic ± 21). It will be found that of these pairs the one which will give 32 x for the middle term is (5x-3)(x + 7). 119. The factors of x^ — 5xy -{-4:7/ can be found in the same way as the factors of x^ — 5x -{-A. For we must find two quantities whose product is 4y^, and whose sum is — 5 .V : these are — 4 y and — ?/. Hence a^ — 5xy + 4:y^ = {x - 4:y) (x - 1/). So also the factors of 3 a*^ — 16 ay + 6 y^ can be found in the same way as the factors of 3 ar^ — 16 .r -f- 5 ; and the factors of either can be written down as soon as the fac- tors of the other are known. For example, if we know that 5 x^ + 32 x - 21 = (5 x - 3) (x + 7), we must have 6 x'^ + 32 xy - 21 y'^ = (5 x - 3 y) (x + 7 y). FACTORS. 141 EXAMPLES XXXII. Find the factors of each of the following expressions : 1. 3x2-10x4-3. 20. 7x2 + 123x-54. 2. 3x2-17x+10. 21. 24x2-30x-75. 3. 2x2+ 11 x+ 12. 22. x2-f 4x?/ + 3y2. 4. 2x2 + 3x-2. 23. x2-6xy + 8y2. 6. 3x2 + 7x-6. 24. x2 - llxy + 18y2. 6. 4x2 + X- 3. 25. x2 + 5xy-14y2. 7. 6x2-38x + 21. 26. x2 - 25 xy + 150 y2. 8. 3x2 + llx-20. 27. x2-35xy-200?/2. 9. 7x2-33x-54. 28. 3x2 - 17xy + 10y2. 10. 5x2-38x + 48. 29. 7x2 - 33xy - 54!/2 11. 7x2 + 75x-108. 30. 24 x2 - 70 xy - 75 y2. 12. 9x2 + 130x-75. 31. x*- 13x2 + 36. 13. 4 x2 + 21 X - 18. • 32. x* - 25x2y2 + 144 f. 14. 4x2 + 4x-15. 33. 36 x* - 97 x2y2 + 36 y4. 15. 6x2+55x-50. 34. x3-3x2-18x. 16. 10 x2 + 3 X - 1. 35. y?y - x2y2 _ 2 xy^. 17. 132x2 + X - 1. 36. 15 x^y - 4x3y2 _ 4,xhf^. 18. 4 x2 - 5 X + 1. 37. 75 xy^ - 130 x'^y^ - 9 x^y^. 19. 12x2 + 50x-50. 120. It is clear that the method of finding by trial the factors of an expression of the form px^ -\-qx-\- r, where p, q, r are known numbers, would be very tedious if there were many pairs of numbers whose product was p, and man}^ pairs whose product was r, for there would then be very many pairs of factors which would agree with the given expression so far as the end terms were concerned, and out of these the single pair which would ^ive the correct middle term would have to be 142 FACTORS. sought. It would, for example, be almost impossible to find the factors of 2,310 x" - 2419 x - 9009 in this way. Again, we not unfrequently meet with such an expres- sion as a;^ + 6 .T + 7 which cannot be written as the prod- uct of two factors altogether rational, and in such a case it would be impracticable to try to guess the factors. We therefore need some method of finding the fac- tors of a quadratic expression which is applicable to all cases. This method we proceed to investigate. 121. We first note that since ar^ + 2 aa; -|- a^ is a perfect square, namely (x -\- a)^, it follows that in order to com- plete the square of which y? and 2 ax are the first two terms, we must add the square of a ; that is, we must add the square of half of the coefficient of x. For example, x^ + 6 ic is made a perfect square by the addition of (l)'-^ ; and the square is x'^ + 6x + 9, which is {x + 3)2. So also, x^ -f 5 X is made a perfect square by the addition of (1)2 ; and the square is aj2 ^ 5^ -f- ( Y), which is (x + |)2. And x2 — 7 X is made a perfect square by the addition of (— l)\ that is of Y ; and the square is x- — 7 x + Y» which is (x — f )2. Whatever a may be, x2 4- ax is made a perfect square by the addition of ( - j , for 122. We will now find the factors of ax^ -\-hx-\-chy a method which is applicable in all cases. The problem before us is to find two factors which are to be rational and integral with respect to x, and are there- fore of the first decree in Xj but which are not necessarily FACTORS. 143 rational and integral with respect to arithmetical num- bers or any other letters which may be involved. We first apply the method to some examples. Ex. 1. Find the factors of a;2 + 6 a; + 8. As we have seen in the last article, x^ -\- 6x is made a perfect square by the addition of 3^ ; hence, adding and subtracting S^, we have x^-\-6x + S = x'^ + 6x-\-9-9-\-S = (x4-3)2-l. The last form of the expression is the difference of two squares^ and the factors are therefore x 4- 3 + 1 and x -\- S — 1, that is a; + 4 and x + 2. Hence the factors of x^ -\- 6 x -{- S a.Te x -\- 4 and x + 2. Ex. 2. Find the factors of x2 - 3x - 28. a;2 _ 3 X is made a perfect square by the addition of (— |)2, that is of f ; hence, adding and subtracting |, we have x2 _ 3x - 28 = x2 - 3x + f - I - 28 = (X-3)2_i|l Now the factors of the last expression are x — | + ^^ and Thus the required factors are x + 4 and x — 7. Ex. 3. To find the factors of x2 + 6x + 7. x2 + 6 X is made a perfect square by the addition of 3^ ; hence we write x2 + 6x + 7 = x2 + 6x + 9-9+7 = (x + 3)2-2 = (a^ + 3)2-(V2)2. Now the factors of the last expression, and therefore of »;2 -j- 6x + 7, are X + 3 + V^ ^^^ x + 3 - y2. 144 FACTORS. Ex. 4. To find the factors of 3 x2 - 10 x + 3. 3x2 _ lOx + 3 =: 3 (x2 - YaJ + 1). Now x2— yx is made a perfect square by the addition of (- 1)2, that is of V- ; hence a;2 _ .n) a; + 1 = a;2 - i/x + -^^ - -%^- + 1 The factors of the last expression are x — f + | and x — | — |, that is X — I and x — 3 ; hence 3x2 - lOx + 3 = 3 (x - ^)(x - 3). From the above examples it will be seen that the method of finding the factors of a quadratic expression consists in changing the expression into an equivalent one which is the difference of two squares. The factors of ax^ -i-bx -^c are therefore found in the following manner : aa^ + 6.'c + c = a ^ar^ + - ic -h - Y \ a aj Now a^-] — X is made a perfect square, namely (x-\ — - j , by the addition of f — - ) = -— -• And, by adding and sub- ,2 \2aJ Aa? tracting to the expression within brackets, we have 4a^ -«I(-fJ-(5-:)! FACTORS. 146 The factors of the expression in square brackets are, by Art. 115, 2a \\4a2 aj 2 a \\Aa^ aj Hence aoc^ -^bx-\-c ="l-^.W(£-.-:)}l-^-C--3i- This formula solves the most important case of factor- ing in elementary algebra, and in form represents a class of cases, also important, in which the problem is: to resolve an algebraic expression of higher degree in x, or in X and y, into a product of factors of the first degree in the same letters. In this problem the coefficients of X and of y in the factors may, and frequently do, involve irrational quantities (surds). The principle of factoring, thus enunciated, as the stu- dent will have occasion to observe in the sequel, underlies all elementary methods for solving algebraic equations. 123. Instead of working out every example from the beginning, we might use the formula found in the last article, and we should only have to substitute for a, b, and c their values in the particular case we are consid- ering. Ex. 1. Find the factors of x^ -h 7 x + 12. Here a = 1, 6 = 7, c = 12. Hence x^ + 7x-\-12 = (^ + I + V(-V - 12)} {a^ + 1 - V(¥ - 12)} = (x + 4)(a; + 3). 146 FACTORS. Ex. 2. Find the factors of 5 a:2 -f 32 xy - 21 y'^. Here a = 5, h = Z2y, c = -2ly\ Hence ± = ^_2y^iey 2 a 10 6 ' // 256 ?/2 4- 1051/2) 19w -Vi ¥5 — '-f=-Er' Rencebx^+32xy-2ly^=5lx + ^-\-^\ fx + ^-^X I 5 5Jl 5 5.' = 5{x + Uj)(x-^-l\ = (x+7 2/)(5x-32/). 124. If in the formula of Art. 122 we put 2 for a and 4 for b, we obtain the factors of the expression ar^-f 2a;4-4 j these factors are x-\- 1 + -^(1 —4) and x + 1 — ^(1 — 4); that is, a;4-l+V(— 3) and a; + l— V( — 3). Now all squares, whether of positive or of negative quantities, are positive. Hence it is impossible to find any real number whose square is —3. Such an expression as V— 3 is, however, often used in algebra; and the meaning to be given to V— 3 is simply that expressed by V^ X V^^ = - 3. The square root of a negative quantity is said to be imaginary ; and a; + 1 -{- V— 3 and a; + 1 — V — 3 are said to be imaginary factors of ar* + 2 a; + 4. By referring to the formula of Art. 122, it will be seen that the factors of ax^ -\-bx-\-c are imaginary whenever b^ c — • IS negative, 4ct=* a FACTORS. 147 125. Pactors found by Eearrangement and Grouping of Terms. The factors of many expressions can be found by a suitable rearrangement and grouping of the terms. No general rule can be given, but in many cases it will be sufficient to arrange the expression according to powers of one of the contained letters, and the factors will then often become obvious, particularly if the given expression only contains one power of that letter. Ex. 1. Find the rational factors oi x^ — Zx^ + x — ^. a;3 _ 3a;2 + a; - 3 = a:2(a; _ 3) + (x - 3) = (a;2 + l)(x - 3). Ex. 2. Find the factors of ax + by ->r hx -{■ ay. Arrange in powers of x ; then we have X (^a -\- h)-\- by -\- ay = X {a + h)-\- y {a -\- h) = {x -{- y){a -{■ b). Ex. 3. Find the rational factors of aofi — x + a — 1. Arrange in powers of a ; then we have o(a^+l) — (x+1), and it is now obvious that a; + 1 is a factor. Ex. 4. Find the factors of the first degree of a\b - c)+ 62 (c - a)+ c2 (a - b). Arrange in powers of a ; then we have a2(6 - c)- a (62 _ c2)+ be (6 - c). It is now obvious that 6 — c is a factor, and the given expression = (6- c){a^-a{b + c)+ bc}=(b - c){a - b){a - c). Ex. 5. Find the factors of the first degree of ^2 _ 3 52 _ c2 _ 2 aft + 4 6c. This is a quadratic expression in a, or in 6, or in c. We therefore proceed as in Art. 122. The given expression = a2 - 2 a6 - 3 62 - c2 + 4 6c. 148 FACTORS. The terms which contain a, namely a^ — 2 ab, will be made a complete square by the addition of 6'-. We therefore write the expression in the form a2 - 2 a6+ 62 _ 4 62 _ c2 -f 4 6c = (a - &)2 - 4 62 - c2 + 4 6c = (a-6)2-(2 6-c)2={(a-6)-(2 6-G)}{(a-6) + (2 6-c)} = (a - 3 6 + c) (a + 6 - c). 126. The expression a^ -{- b^ -\- c^ — S abc is of frequent occurrence and its factors should be known. It is easy to verify that = (a + 6 + c) (a^ + 52 4. c2 _ 6c - ca - a6) . The expression a;^ + a;^ -f 1 is also of frequent occur- rence. Writing it in the form (x^ -\-iy — x^^ we see that a;2 -|- 1 -j- a; and a^ -f 1 — a; are factors ;* and we do not pro- ceed any further, for both ar^ + a; -f 1 and a^ — a; -f 1 have imaginary factors of the lirst degree. The following miscellaneous examples will afford suit- able practice in the methods above described. EXAMPLES XXXIII. Express in rational factors : 1. 16x*-625?/. 7. (a + 6)2 - (c - d)2. 2. 81 a*- 16 6*. 8. (a 4- 6)2 - 4 (c - d)2. 3. x^-Sl X. 9. (a + 6)2 - (a -H c)2. 4. 1 + 27 a8. 10. (a; + y)* - (x - y)*. 6. 27 + 8x8. 11. (rt + 6 + c)2-4ca. 6. 27x* + 8a;. 12. (a + 6 - 3c)2 - 9c«. 18. (a2 + a6 + 62)2 _ (^2 _ 52)2. 14. (a2 + a6 + 62)2 _ („2 _ ab -\- 62)2. 16. (x8 + 3x)2-(3x2 + i)2. FACTORS. 149 16. (x2 + 6xy + 2/2)2 _ (a;2 - xy -\- 2/2)2. 17. (a + h + c + d)^-(a-b + c-d)K 18. (3 a + 2 6 + c)2 - (a + 2 6 + 3 c)2. 19. (2x-{-Syy-h{Zx-\-2yy. 26. 72 (x2 - 1)- 17x. 20. (a + 36)3 -(3a + 6)3. 27. tc*-5a;2 + 4. 21. x2 - f a:!/ - i 2/^. 28. a;* - 13 x'^y^ + 36 2/*. 22. x'^ + (a + ^\xy-{-y^. 29. a6 (x^ + 2/2)+ a;2/0a? + 62). 23. x3 + J/x2 + x. 30. a6(x2-2/2)+x2/(a2-62). 24. 6x2 -5xy- 62/2. 31. (a + 6)2 - 5c (a + 6)+ 6c2. 25. x^-|x82/-«V- 32. (x + 2/)2-72(x+2/)+lO02. 33. (a + 6)2-8(a+6)(c + d)+15(c + d)2. 34. x(x + 2)-2/(2/ + 2). 35. X (x + 4) -2/ (2/ + 4). 46. a262 _ ^2 - 62 + 1. 36. x3 - 5 x2 + X - 5. 47. ac + 6fZ -ad- be. 37. x3 + x2 - 4x - 4. 48. ac^ + 6(f2 _ ad^ - bc^. 38. 2x3-3x2-2x + 3. 49. x:^ - y'^ -\- xz - yz. 39. 5x3-x2-5x + l. 60. a^-b^-{a-by. 40. ax3 + x + a + l. 51. a* ^ a'^b'^ - b'^c^ - c^. 41. ax3 + 6x + a + 6. 52. a2 _ 52 4. 5c - ca. 42. x8 + 6x2 _ cfi^ _ a^b. 53. a^ - a - c^ + c. 43. 6x3 + ax2 + 6x + a. 54. 1 + 6x - (a2 + ab) x\ -^^ _ <3l^ 44. ax2 + 62/2 + (a + 6) X2/. 55. 1 - a6x3 + (6 - a^) x2. ^>^ ^ -^ 45. a262 + a2 + 62 + 1. 56. a2c2 + acd + abc + 6(2. 57. a2x + a6x -\- ac ■\- b^y + aby + 6c. 58. a2 - 62 + c2 - d2 - 2 (ac - bd). 59. 4 a252 _ (^2 + 62 - c2)2. 60. (a2 - 62 + c2 - d2)2 _ (2 ac - 2 bdy. 61. X* - 23 x2 + 1. 63. X* - 11 x22/2 + y^. 62. X* - 7 x22/2 + 2/*. 64. x*- 3x2 + 1. 65. (x2 + 4x)2-2(x2 + 4x)-15. 66. (x2 + 7x + 6)(x2 + 7x + 12)-280. 150 FACTORS. 127. Zero Pactors. Since zero was defined [Art. 45] as a difference, in the form a — a, a zero factor in any prod- uct may be replaced by a — a. If, then, F have a finite value, the product X F= (a - a) F= aF- ciF is itself zero, and it is obvious that a product will not be zero unless one of its factors is zero. Hence, exclu- sive of infinite factors,* In order that a product may he zero, it is necessary and sufficient that one of its factors be zero. 128. Quadratic Equations. In particular, the theorem of Art. 127 says : the product (x — 2)(x — 4) is zero, if ic — 2 is zero, or if a; — 4 is zero ; and, in order that this product may be zero, one of these factors must be zero. The roots of the equation (a;-2)(a;-4) = are therefore 2 and 4. [Definition of a root. Art. 91.] Again, the product (a; _ 3) (a;- 4) (a; ~ 5) vanishes if a; — 3 = 0, or if a; — 4 = 0, or if a; — 5 = 0, * The combination of zero and infinite factors in a product introduces what are known as indeterminate forms. Thus, for the value x = a, the expression (x^ — a^)xl/{z — a) assumes the form 0x1/0, or X 00, where 1/0 is said to be infinite and is represented by the symbol oo. But this product is not necessarily zero, when x = a; for, suppress- ing the common factor z — a from x^ — a^, and x — a, we have (x2 - a2) X 1/ (x - a)= ?^^^= X + a, X — a and X -h a = 2 a when x = a. [See Treatise on Algebra, Art. 217.] FACTORS- 151 and in no other case ; therefore 3, 4, and 5 are the roots of the equation {x - 3) (a; - 4) {x - 5) = 0. Applied to the general quadratic expression ax^ -\-hx +c, this principle of vanishing products asserts that ax'^ + hx-\-Cy or its equivalent [Art. 122] 1 ^2a \\4La' a) S \ 2a \\Aa a)) becomes zero, if and only if one of the factors here writ- ten is zero, and that therefore the values of x that satisfy the equation a7? -f- 6a; 4- c = are derived, one of them from the equation 2 a \ \4 a^ a J the other from the equation Thus the problem of solving any equation of the second degree is reduced to that of solving two equations of the first degree. This subject will be taken up again in the chapter on quadratic equations. The general principle, of which the above examples are particular cases, asserts that the equation {x — a){x- h) {x - c) = is equivalent to the system of alternative equations oj — a = 0, or a; — 6 = 0, or x — c = 0, etc. Note. — Observe the distiuction we make between alternative equations and simultaneous equations [Art. 101]. 152 FACTORS. 129. From the examples considered in the last article it will be apparent that the solution of an equation of any degree can he written down at once provided the equation is given in the form of a product of factors of the first degree equated to zero. The following are examples of such equations : Ex.1. Solve (a; -l)(a;+ 1)=0. The equation is satisfied if a; — 1 = 0, or if a; + 1 = 0, and in no otlier case. Hence we must have a; - 1 = 0, or x + 1 = ; that is, oj = 1, or aj = — 1. Tims the roots of the equation are 1 and — 1. Ex.2. Solve a;(a;+ l)(x+ 2)= 0. The equation is satisfied if a; = 0, or if x + 1 = 0, or if a; + 2 = 0, and in no other case. Hence we must have a; = 0, ora; + l = 0, orx + 2 = 0; that is, x = 0, or X = — 1, or x = — 2. Thus the roots of the equation are 0,-1, and — 2. Ex. 3. Solve X (2 X - 1) (2 X + 3) = 0. The equation is satisfied if x = 0, or if 2 x — 1 = 0, or if 2 X + 3 = 0, and in no other case. Hence we must have x = 0, or2x-l = 0, or2x + 3 = 0; that is, X = 0, or X = I, or x = — f . Thus the roots of the equation are 0, J, — |. 130. Since all tlie terms of any equation can be trans- posed to one side, an equation can always be written with FACTORS. 163 all its terms on one side of the sign of equality, and zero on the other side. It therefore follows from the last article that the prob- lem of solving an equation of any degree is the same as the problem of finding the factors of an expression of the same degree. Hence the process of solving any equation which in- volves only one unknown quantity is as follows : First lorite the equation with all its terms on one side of the sign of equality, and zero on the other side ; then resolve the whole expression into factors {of the first degree in the unknoivn quantity), and the values obtained by equating each of these factors separately to zero will be the required roots. In the following examples the resolution into factors can be performed by inspection. Ex.1. Solve the equation a;^ - 3 a; = 0. Since x^-'6x = x{x-Z), we have a;(x-3)=0. Whence a: = 0, ora;-3=0; the roots required are therefore and 3. Ex.2. Solve the equation a;2 _ g = o. Since a;2-9 = (a;-3)(a; + 3), we have (a;_3)(iB + 3) = 0. Whence cc-3 = 0, ora;-f3 = 0; the roots required are therefore 3 and — 3. Ex.3. Solve the equation a;^ — 2 = 0. Smce x2_2=(x-v2)(a; + V2), we have (x-V2)(x+V2)=0. Hence x-^=0,ovx-\-y/2 = 0. Thus V2 and - ^2 are the required roots of the equ 164 FACTORS. Ex. 4. Solve the equation x^ -4x = 0. Since a^ - ^x = x(x^ - ^) = x^x - 2){x -\- 2), we have x(x-2)(x + 2) = 0. Hence x = 0, or x - 2 = 0, or a; + 2 = 0. Thus 0, 2, and - 2 are the roots of the equation. Ex.5. Solve 9x^ = 4x. Transposing, we have 9 x3 - 4 a; = ; that is, x(9x''-4) = 0, or x(3x-2)(3a; + 2) = 0. Hence x = 0, or 3x - 2 = 0, or 3x + 2 = ; that is, x = 0, ora; = f, orx = - i- Ex.6. Solve x2 + 6 = 5x. Transposing, we have a;2_ 5x4-6=0; that is, (a;-2)(x-3)=0. Hence x - 2 = 0, or x - 3 = 0. Thus 2 and 3 are the required roots. EXAMPLES XXXIV. Solve the equations 1. (x-l)(a; + 2) = 0. 7. x (x - 2)(x + 3) = 0. 2. (x-3)(x-4)=0. 8. x(x + 2)(x-4)=0. 8. (x + l)(x + 2) = 0. 9. x(x-3)(x+4)=0. 4. (2x + l)(2x-l)=0. 10. x(2x-l)(3x + 4) = 0. 6. (3x-l)(3x + l) = 0. 11. x(5x-2)(6x-7) = 0. 6. (2x-5)(x-4) = 0. 12. x(x-3)(3x4-7)=0. 13. (x-l)(x-2)(x-3)(x-6) = 0. 14. (x-2)(x-l)(x+l)(a; + 2) = 0. 16. (3x-l)(4x+l)(5x-2)(2x + 7) = 0. 16. (2x-3)(3x-4)(4x-6)(6x + 6) = 0. FACTORS. 155 17. x'^~x = 0. 25. Sx^ = 5x. 18. x2-2x = 0. 26. 5x2=r-6x. 19. a;2 + 3x = 0. 27. x^ = 6a;. 20. 2 x2 - 3 X = 0. 28. ax^ = ?)x. 21. 2x2-5x=:0. 29. 6x2 = x2 4-5. 22. 3x2 + x = 0. 30. 5(x2 + 5)=3(x2 + 25). 23. x2 = 5x. 31. 5(x2 + 4) = 4(x2 + 9). 24. 2x2 = x. 32. 2(x2 + 7)=7(x2-|-2). 33. 5(x2 4-3)-(x-5)(x4-6)=76. 34. 7(x2-l)-(x + 3)(x-3) = 56. 36. 3x2 + (5x + 2)2 = 20x + 32. 43. x2 - 5x - 84 = 0. 36. 17 + 3x = I (a^ + 3)2 - 28. 44. x2 - x - 156 = 0. 37. x2-5x + 6 = 0. 45. x2 + 5x- 150 = 0. 38. x2-7x+12 = 0. 46. x2 + 2x = 3. 39. x2-12x + 20 = 0. 47. x2 + 4x = 45. 40. x2 - 9x + 20 = 0. 48. x2 - 3 X = 10. 41. x2 - 11 X + 28 = 0. 49. x3 + 11 x2 - 180 X = 0. 42. x2 - 25x + 160 = 0. 60. x^ - x2 = 132 x. 156 HIGHEST COMMON FACTORS. CHAPTER XL Highest Common Factors. 131. A Oommon Factor of two or more algebraic expres- sions is an expression which will exactly divide each of them. Thus a is a common factor of ab and ac. The Highest Oommon Factor of two or more algebraic expressions is the expression of highest dimensions which will exactly divide each of them. Thus a^ is the highest common factor of a% and aH. Instead of Highest Common Factor it is usual to write H. C. F. We proceed to show how to find the H. C. F. of given expressions. 132. H. 0. F. of Monomial Expressions. The highest com- mon factor of two or more monomial expressions can be seen by inspection. Take, for example, aHy^c and a%H*, The first expression is clearly divisible by a, or by a^^ or by a', but by no higher power of a ; and the second expression is divisi- ble by a, or by a^, but by no higher power. Hence a^ will divide both expressions, and it is the highest power of a which will divide them both. Also h^, but no higher power of 6, will divide both expressions ; and c, but no power of c above the first, will divide both expressions. HIGHEST COMMON FACTORS. 157 Hence the H.C.F. of a^b'^c and a^ft^c* is a^b^c. Again, to find the H. C. F. of ab^cM^ and a^c^d^. The highest power of a which divides both expressions is a ; 6 will not divide both expressions ; the highest power of c which divides both is c^ ; and the highest power of d which divides both is (P. Hence the H. C. F. is a<^d^. Also a%c*y a%HH^ and a*bc^d^ are all divisible by a^, by 6, and by c* ; and therefore the H. C. F. of the three expressions is a^bc*. From the above examples it will be seen that the H.C.F. of two or more monomial expressions is obtained by taking each letter which is common to all the expres- sions to the loioest power in ivhich it occurs. If the expressions have numerical coefficients, the G. C. M. of these can be found by arithmetic, and pre- fixed as a coefficient to the algebraical H. C. F. EXAMPLES XXXV. Find the H. C. F. of 1. a362anda263. 7. 9a^b^y^ a.ndSxY' ,2. a6c2 and «26c3. 8. f aSft^cS and 2 68c'5. 3. 9 ab^ and 6 a^b. 9. 42 axy^z^ and 77 bY- 4. 4 x^y and 10 xy^. 10. a6^, a^bc^ and abcK 5. 24 a353^* and CO a264a^. u. Sx^yz^, 16 xy^z'^^&nd 10 x'^y^z'^. 6. a'^b^a^ and 3 b^x. 12. ab^(^x\ a*bc^ofi, and a^b'^cx^ 133. H. 0. F. of Multinomial Expressions whose factors are known. When the factors of two or more multinomial expressions are known, their H. C. F. can be at once written down. The H. C. F. will be the product obtained by taking each factor which is common to all the expres- sions to the lowest power in which it occurs. 168 HIGHEST COMMON FACTORS. Consider, for example, the expressions {x — af{x -\- by and (x — ay(x + by. It is clear that both expressions are divisible by (x — ay, but by no higher power of (x — a). Also both expressions are divisible by (x + by, but by no higher power of x + 6. Hence the H. C. F. is (x - ay{x + bj^ Again, a'^b (a - b)'^ (a + bf and ab^ {a - b)'^ (a + by are both divisible by «, by b, by (a — by^ and by {a + by. Hence the H. C. F. is aft (a - 6)2 (a-f by. In the following examples the factors can be seen by inspection, and hence the H. C. F. can be written down. Ex. 1. Find the H. C. F. of a^b'^ - a%^ and a^b^ + a^bK Since a^"^ - a^b* = cfib"^ (a^ - 62) ^ 0,%"^ (a - 6)(a + 6), and a*63 ^ a%^ = a%^ (a + 6), we see that the H. C. F. is a262 (a + 6). Ex. 2. Find the H. C. F. of «3 + .3 a^b + 2 a62 and a* + 4 aSft _|. 3 ^1252. a8 4- 3 a26 + 2 a62 = « (a2 + 3 (z6 + 2 62) = a (a + 6) (a + 2 6) , and a* + 4 a36 + 3 a^b"^ = a^ (a2 + 4a6 + 3 62) = a2 (a + 6)(a + 36); hence the H. C. F. is a (a + 6). Ex. 3. Find the H. C.F. of 3a8 + 2 a2 - a and 5a* + 3 a^ - 2 0,"^. 3a«4-2a2-a = a(3a2 + 2a-l)=a(3a-l)(a+l), and 5a* + 3a8-2a2 = (352(5^2_|_3a_2)=a2(5a-2)(a+ 1); hence the H. C.F. is a (a + 1). EXAMPLES XXXVI. Find the H. C. F. of 1. x2 (x - ay and x8 (x - a)». 2. a^ - 62 and (a + by. 3. 3a6 (a + 6)2 and 2 (a - 6) (a 4- 6)». 4. 6*c6 (6 + c)2 and 6«c« (6 + c)*. 6. 0868 + ab'^ and a^ - a26*. 7. a2x8 + 2 a8x2 and a^x* - 4 a*x^. 6. a26 + 3 68 and a« - 9 a26*. 8. a''x2 - 4 a*x* and a«x2 - 16 a2x«. HIGHEST COMMON FACTORS. 159 9. x2 + 3x + 2 and x2 + 6x + 8. 10. x3 4- 3 x2y + 2 xy'^ and x* + 6 x^y + 8 x?y'^. 11. 3x2-4x+ 1 and4x2-5x+ 1. 12. 3 a2 - 4 aft + ft2 and 4 a* - 6 a^h + anjK 134. Although we cannot, in general, find the factors of a multinomial expression of higher degree than the second, we can always find the highest common factor of any two expressions by the following process, which will be seen to be analogous to that used in arithmetic to find the greatest common measure of two numbers. Kule. Arrange the two expressions in descending powers of some common letter, and divide the expression which is of the highest degree in the common letter by the other : if both expressions are of the same degree, it is immaterial which is used as the divisor. Take the remainder, if any, after the first division for a new divisor, and the former divisor as dividend; and continue the process until there is no remainder. The last divisor will be the H. C. F. of the two given expressions. For example, to find the H. C. F. of x^ + x^ - 2 and x^ + 2 x^ - 3, the process is as follows : First dividend, Second dividend, X3 4-2 X8 + x2-3 x2-2 X3 + X2 - 2 1 X8 + X2 - 2 X2-1 X8-X X +1 First divisor Second divisor Third dividend, X2 X -2 -1 X2-1 X2-X X -1 X +1 Third divisor -1 -1 Thus the H. C. F. is x - 1. 160 HIGHEST COMMON FACTORS. The work can be arranged more concisely as follo-vrs : 1 x3 + 2 x2 - 3 a;3+ a:2-2 X8+X2 X^-X -2 X2- 1 X^^X X2 -2 -1 X -1 X - 1 X -1 Note. — It is only multinomial factors which are to be found by the above rule ; the H. C. F. of any monomial factors of the given expressions must be found by inspection. For example, to find the H. C. F. of a^x* + a^x^ - 2 a^x and ahx^ + 2 aftx* - 3 ahx\ Since aV + a^x^ -2.a:^x = aH (x^ + x^ - 2), and abx^ + 2 a6x* - 3 ahx^ = abx^ (ac* + 2 x2 - 3), it is clear that the H. C. F. of the monomial factors is ax. We have already found that the II. C. F. of x^ + x^ - 2 and x^ + 2 x^ - 3 is X — 1. Hence the whole H. C. F. is ax (x — 1). 135. The rule given in Art. 134 for finding the H. C. F. of two expressions which have no monomial factors, may be proved as follows. Let A and B stand for the two expressions, which are supposed to be arranged according to descending powers of some common letter, and let A be of not higher dimen- sions than B in the common letter. Let B be divided by A, and let the quotient be Q, and R the remainder, so that the first stage of the process of finding the H. C. F. of A and B is as under, A)B(Q AQ HIGHEST COMMON FACTORS. 161 Then we have B = Aq-\-R (i.), and R = B-AQ (ii.). Now an expression is exactly divisible by any other if each of its terms is so divisible; hence, from (i.), B is divisible by any common factor of ^4 and R. Thus any common factor of A and R is also a common factor of A and B. Again, any factor of both B and A will be a factor of B — AQ\ that is, from (ii.), will be a factor of R. Thus any common factor of A and B is also a common factor of A and R. It follows therefore that the common factors of A and B are exactly the same as the common factors of A and R. Hence the H. C. F. of A and B is the H. C. F. of A arid R. If now we divide A by R, and the remainder is S, the H. C. F. of S and R will similarly be the same as the H. C. F. of A and R, and therefore will be the H. C. F. required. And so on ; so that the H. C. F. of any divisor and the correspoyiding dividend is the H. C. F. required. But if at any stage there is no remainder, the divisor must be a factor of the corresponding dividend, and that divisor is clearly the H. C. F. of itself and the corre- sponding dividend. It must therefore be the H. C. F. required. It should be remarked that by the nature of division the remainders are successively of lower and lower dimen- sions ; and hence, unless the division leaves no remainder at some stage, we must at last come to a remainder which does not contain the common letter, in which case the given expressions have no H. C. F. containing that letter. We have already remarked that the process we are 162 HIGHEST COMMON FACTORS. considering is only to be used to find the H. C. F. of the multinomial factors of the given expressions ; and, bear- ing this in mind, it is clear that any of the expressions which occur may be divided or multiplied by any mono- mial expression without destroying the validity of the process ; for the multinomial factors will not be altered by such division or multiplication. Ex. 1. Find the H. C. F. of x3 + 4ic2 _ 8 X -I- 24 and x*-x^+Sx-S. Neither expression has any monomial factors ; we therefore proceed as in Art. 134, performing the work by the synthetic method. x^ a:*-x3+0 +8x-8 -4x-^ -4x3 4-20x2 -f 8x + 8x2 -40x -24 -24X + 120 x-5; 28x2-56x4-112 The remainder 28 x2 - 56 x -f 112 = 28 Cx2 - 2 x -|- 4) ; and since the numerical factor 28 is clearly not a factor of the given expres- sions, we reject it, and continue the process with x2 — 2 x -}- 4 as the new divisor : X2 a:3 4.4a;2_ 8x4-24 -f 2x -H2x2 4-12x -4 - 4x-24 x-f 6; Thus the H. C. F. is x2 - 2x + 4. Ex. 2. Find the H.C.F. of x8-4a2x+15a« and x X8 X4 + 4- a-^^-\-0 4-25a« -|-4a2x -|-4a2a;2 -16a2 -15a«x Now 5 a2x2 x; 6a2x2- 16a8x4-26a* 15 a^x 4- 25 a« = 6 a2(x2 - 3 ax 4- 5 a^). HIGHEST COMMON FACTORS. 163 We reject the monomial factor 5 a^, and continue the process with x^ — 3 are + 5 a- as the new divisor : X2 ic3 + _ 4 «% + 15 a3 -\-Sax -\-Sax'^+9a^x -5a2 -5a%-15a3 a; + 3a; Hence x- — S ax -\- 3 a^ is the required H. C. F. Ex. 3. Find the H. C. F. of x^ - y^ and x'^ - if x} - y"^ x' — xP-y^ x^ - y^ Before using the remainder as a divisor, we reject the monomial factor y^ ; we then have y? — y'^. x^ — y^ x^ 4- xy^ x^y'^ xy^ xy^ - y^ We reject the monomial factor ?/* of the remainder. a;2 -xy x-y xi^- x-^y xy-y^ xy-y^ Hence x — y is the required H. C. F. Ex. 4. Find the H. C. F. of 2x2-5a; + 2 and x^-\-ia^^-ix-lQ. To avoid the inconvenience of fractions, we multiply a:3 + 4x2-4a;- 16 by 2. We have shown in the preceding article that we may do this, as no additional common factors can be thereby introduced. It will be seen that we multiply a second time in the course of the work. The following arrangement will be found convenient : 164 HIGHEST COMMON FACTORS. First dividend x^ + 4iX^ — 4 a; Multiply by X 13 16 2 First divisor. 2x3 + 8a:2- 8x-32 2a;3-5a;2+ 2x 2a;2-5a; + 2 2a;2-4a; 13x2 -10a; -32 Multiply by . . 2 -a; + 2 -a; + 2 26a;2-20x-64 26a;2-65a; + 26 2 a; Divide by ... . 45 |45x-90 Second divisor ... a; — 2 Thus a; - 2 is the H. C. F. 136. The H. C. F. of more than two expressions is sometimes required, when the factors cannot be deter- mined by inspection. Now it is clear that any factor which is common to each of three or more expressions, must be a factor of the H. C. F. of any two of them. Hence we first find the H. C. F. of two of the given expressions, and then find the H. C. F. of this result and of the third expression, and so on. Ex. Find the H.C.F. of a;8 + a;2 - a; - 1, a;3 + 3a;2 3, and a?-\-x^-2. The H. C. F. of the first two expressions is a;2 - 1. The H. C.F. of x2 — 1 and the third of the given expressions is a; — 1. Hence a; — 1 is the H. C. F. required. 137. If the numerical coefficients of the first terms of the given expressions are large, it is often desirable to arrange the terms of both expressions in the reverse order before applying the rule for finding their H. C. F. HIGHEST COMMON FACTORS. 165 For example, in order to find the H, C. F. of 7 X* + 2 x3 - x2 + 8 X + 1 and 9 X* - 2 x3 + 3 x2 -f 6 X + 1 it is best to arrange the expressions in the form 1 + Sx-x^-{-2x^ + 7a:iandl + 6a: + 3x2 -2x3 + 9x*. 138. The highest common factor of two expressions is sometimes, but very inappropriately, called their greatest common measure [G. C. M.]. The inappropriateness of the term greatest common measure can be seen by an example. If a^ is a factor of two expressions, so also is a, and a- is of higher dimen- sions than a ; but, as a may represent any number what- ever, a^ is not necessarily greater than a ; in fact, if a is positive and less than unity, a^ is less than a. It should also be noticed that if we give particular numerical values to the letters involved in any two ex- pressions, and in their H. C. F., the numerical value of the H. C. F. is by no means necessarily the G. C. M. of the values of the expressions. This may not be the case even when the given expressions are integral for the particular values chosen. For example, the H. C. F. of 2x^-f- 15a;-f 13 and. 6 a?^ + 17 aj + 11 will be found to be a; + 1 ; but if we suppose X to be ^, the numerical values of both expres- sions, and therefore their G. CM., will be 21, whereas the numerical value of the H. C. F. will be f . EXAMPLES XXXVn. FindtheH.C.F. of 1. x2 - 5 X + 4 and x^ - 5 x2 + 4. 2. x^ — 5 xy + 4 ?/2 and x*^ — 5 x^y + 4 xy^. 3. 2x2 - 5x + 2 and 4x3 + 12 x2 - x - 3. 166 HIGHEST COMMON FACTORS. 4. 2 x2 - 5 ary + 2 «/2 and 4x3 + 12 x'^y - xy^ - S yK 5. x* + 3x- - 10 and x* - 3x2 + 2. 6. x6 + 3x*y - 10 x2y2 and x* - 3 x2y + 2 y^. 7. 2a2-5a + 2 and 2a3-3a2-8rt+ 12. 8. 2 62 _ 5 5 + 2 and 12 63 - 8 &2 _ 3 5 4- 2. 9. x2 - 3 X + 2 and x3 - 3 X + 2. 10. x*y* - 3 x2y2 -f 2 and x'^y"' - 3 x2?/2 + 2. 11. x3 - 3a2x - 2 a3 and x^ _ ax2 - 4 a^ 12. 2 a3 + 3 a2ft _ ^3 and 4 a^ + aft^ - bK 13. a^ + ?,3 and a* + a262 + 6K 14. 8 a3 + 1 and 16 a* + 4a2 + i. 15. x3 + 2x2y - X2/2 _ 2 y3 and x^ - 2x2?/ - x?/2 -f 22/^. 16. 2x3 + 5x2 + X - 8 and 3x3 - 4x2 + 9x - 8. 17. 3x3 + x2 + X - 2 and 2 x3 - x2 - X - 3. 18. x3 - 4 X + 15 and x* + x2 + 25. 19. 3 x2 _ 38x + 119 and x8 - 19x2 + 119x - 245. 20. 3x8 - 3x2y + xy2 _ y8 and 4x2y - 5xy2 + yS, 21. 12x2 _ i^xy + 3^/2 and 6x3- 6x2y + 2x?/2 -2y^ 22. 2 x2 - 14 X + 20 and 4x (x2 + 6) - 25 (x + l)(x - 1). 23. 16 X* + 4x2 + 1 and 8xt - 16x3 + x - 2. 24. a2 - 4x2 -f 12 X - 9 and a2 ^ 2 a - 4x2 + 8x - 3. 26. 21 - 7 X + 3x2 - x3 and 35 + 19 x2 + 2x*. 26. 2x* +9a;3+ 14x + 3 and 2 +9x+ 14x3 + 3x*. 27. x« + 3 x2 - X - 3 and x* + 4x3 - 12x - 9. 28. 3x* + 5x3 - 7 x2 + 2x + 2 and 2 x* -|- 3 x3 - 2 x2 + 12 x + 5. 29. 2/3 — 2 y2 4- 3 y _ 6 and y* — y^ — y'^ — 2 y. HIGHEST COMMON FACTORS. 167 30. 2/4 -Syz^ + 20 z^ and 5y^ -Sy^z + 6izK 31. 2x^- 11x2 + 11 a: +4 and 2x*-3a:3 + 7 x- - 12a; -4. 32. 2a:4 + 4x3 + 3x2-2a;-2 and 3x^ + 6x^+ 7ic2 + 2x + 2. 33. X* - a;3 + 2 x2 - X - 1 and 2 X* - 2 x2 + X - 1. 34. 2x* - 6x3 + 3x2 - 3x + 1 and x' - 3x6 + cc^ - 4x2 + 12x - 4. 35. x3 + (w - 3) x2 - m (2 m + 3) X + 6 m2 and x^ + (5 ?w - 3) x2 + 3 w (2 m - 5) X - 18 m^. 36. mn (x2 + y2) .^^^^ ^„j2 ^ .,^2) aj^tl mw (x^ + y^)-\-xy {ni^y + n2x). 168 LOWEST COMMON MULTIPLES. CHAPTER XII. Lowest Common Multiples. 139. A Oommon Multiple of two or more algebraical expressions is an expression which is exactly divisible by each of them. The Lowest Oommon Multiple of two or more algebraical expressions is the expression of lowest dimensions which is exactly divisible by each of them. Instead of lowest common multiple it is usual to write L. C. M. We proceed to show how to find the L. C. M. of given expressions. 140. We first consider monomial expressions, the factors of which can be seen by inspection. Take, for example, a^b^c and a^b^c*. The highest power of a which occurs in either expression is a' ; hence any common multi- ple of the given expressions must contain a^ as a factor. Any common multiple must also contain b^ as a factor, and it must also contain c* as a factor. Any common multiple must therefore contain a^H* as a factor; and hence the common multiple of lowest dimensions must be a^b^e*. Again, to find the L. CM. of a^bc*, a%^c^d and a^bcH^ the highest power of a in the given expressions is a* ; hence any com- mon nniltiple of them must contain a* as a factor. Any common multiple must also contain ?>», r^^ and d^ as factors. Any common multiple must therefore contain a*6Vd* as a factor ; hence the L. C. M. required is a'^i^bH^. LOWEST COMMON MULTIPLES. 169 From the above examples it will be seen that the L. C. M. of two or more simple expressions is obtained by taking every letter which occurs in the different expres- sions to the highest power which it has in any one of them. If the expressions have numerical coefiBcients, the L. C. M. of these can be found by arithmetic, and pre- fixed as a coefficient to the algebraical L. C. M. EXAMPLES XXXVIII. Find the L. C. M. of 1. a^b^ and a^bK 7. Oa^ft^V and Sa^syj, 2. a6c2 and a22,c3. 8. ^a^b^c^ and 2 b^c^. 3. 9 a63 and 6 a26. 9. 42 axy'^z^ and 77 bY- 4. 4 x^y and 10 xy^. 10. afe^, a^bc^ and abd^. 5. 24: a^b^x* and QOa^b^sfi. 11. S x'h/z^, 15 xy^z'^, and 10 x^^z^. 6. a^b^ and 3 b^x. 12. ab^c^, a*bc'h^, and a^b^cx^. 141. When the factors of two or more multinomial expressions are known, their L. C. M. can be at once written down. The L. C. M. will be the product obtained by taking every factor which occurs in the different expressions to the highest power which it has in any one of them. Consider, for example, the expressions (x — a){x — by (x — c)* and (x — ay (x — b)(x — c). It is clear that any common multiple must contain (x — a)* as a factor ; it must also contain (x — 6)^ and (x — cy as factors. Any common multiple must therefore contain (x — ay {x —by (x — cy as a factor ; and hence the common multiple of lowest dimensions must be (x — ay {x — 6)^ (x — cy. 170 LOWEST COMMON MULTIPLES. Ex. 1. Find the L. C. M. of a*b^ - a%^ and a^h^ + a%^. Since a^h"^ - a'^b* = a'^b^ (a - 6) (a + &), and a*b^-\-a^b^ = a^b^(ia + b), we see that the L. C. M. is a%^ (a — 6) (a + 6). Ex. 2. Find the L.C.M. of a^i-Sa%-\-2ab^&nd a^-\-ia%+Sa^b^. a3 + 3 a^b + 2 a6-2 = a(a + 6)(a + 2 6), and a* + 4a36 + 3 a'^b'^ = a'^ia + &)(a + 3 &) ; hence the L. C. M. is a'^{a + 6) (a + 2 6) (a + 3 6) . We leave the L. C. M. in the above form, as it is generally' most convenient to have it expressed in factors. Ex. 3. Find the L. C. M. of x2 + 2a; + 1, a;2 _ 2 ac - 3, and x2 4-2a;+l=(x + 1)2, x^ - 2x - S =(x + l){x - 3), and x2 + 4ic + 3=(x+ l)(a; + 3); hence the L. C. M. is (x + l)2(x - 3)(x + 3). EXAMPLES XXXIX. Find the L. C. M. of 1. (a - x)(a-2x) 2ind (a-2x)(a-Sx). 2. ax^(a -x)ia-2x) and a^x(a - 2 x) (a - 3 x). 3. a2 _ 52 and (a + by. 4. 6 ab{a + 6)2 and 4 a2(a2 _ 52). 6. a;2 ^ 3a; + 2 and x2 + 5x + 4. 6. x2 ^ 3 a;y 4. 2 y2 and ^2 + 5 x?/ + 4 y*. 7. a;2 _ 4a: + 3 and ac2 - 6x + 6. 8. 3x2 - 4x2/2 + ?/< and 6x2 - 6x?/2 4- y4. 9. (a + fe)2, (a - ?))2, and a* - h\ 10. (x + 2 2/)2, (X - 2 2/)2, and x2 - 4 y2. 11. x2 + 7x + 12, x2 + 6x + 8, and x2 + 6x + 6. 12. %^-lxy-\- \2y\ x^ -Qxy -\-Sy% and x^ - bxy -\-fiy^. LOWEST COMMON MULTIPLES. 171 142. When the factors of the expressions whose L. C. M. is required cannot be seen by inspection, we must use the rule for finding the H. C. F. given in the preceding chapter. Thus, to find the L. C. M. of ar^ + x^ - 2 and a^ + 2 x^ - 3. The H. C.F. of the given expressions is x — 1, and we find by division that x3 + a;2 —2 =(x - l)(x2 + 2x + 2), and x3 + 2a:2-3=(ie-l)(x2 + 3a-, + 3). Then, since x^ + 2 x + 2 and x^ + 3 x + 3 have no common factors, the L. C. M. required is (x - l)(x2 -f 2x + 2)(x2 + 3x + 3). 143. Let A and B stand for any two algebraical expressions, and let H stand for their H. C. F., and L for their L. C. M. Let a and b be the quotients when A and B respec- tively are divided by IT; so that A = Hxa, and B — Hxh. Since H is the highest common factor of A and B, a and h can have no common factors. Hence the L. C. M. of A and B must be jH" x a x 6. Thus L = H'a'h. . . . • • (i-)- It follows from (i.) that ^ rr Hh . B . . (ii), and also that Lxff=^HaxHb^AxB. . . (iii.). 172 LOWEST COMMON MUIiTlPLES. From (ii.) we see that the L. C. M. of any two algebrai- cal expressions is found by dividing one of the expressions by their H. C. F., and multiplying the quotient by the other expression. From (iii.) we see that the product of any ttvo eoppres- sions is equal to the product of their H. C. F. and L. C. M. 144. To find the L. C. M. of more than two expres- sions, whose factors cannot be determined by inspection, we first find the L. C. M. of two of the given expressions, and find the L. C. M. of this result and of the third expression, and so on. EXAMPLES XL. Find the L. C. M. of 1. 3a26c, 27 a^b^c^, and 6a6%. 3. 4a^ 6a% and Sab^. 2. a4&2, 68c, and aV. 4. 2a^, 6ab% and ia^b'^. 6. x\x - y)% y\x + y)2, and xyix"^ - y"^). 6. «» -H a%, a(a-b), and a^ - b'^. 7. xy^ — y^, xhf^ + xy^, and x^y — y. 8. 2axy{x-y), S €ix:\x^ - y^), and 4y^(x-{-y)^. 9. 6(x2-9), 9(x-|-3), 15(a;-4), and 10(a:2 - x - 12). 10. 4 a + 4 6, Qa^ - 24 62, and a^-Sab + 2 b\ 11. 4 a68 + 4 bH, a-^b^ - b^d^, and 8 a^bd - 8 abd^. 12. x^ - lOx + 24, a;2 - 8a; + 12, and x'^-Qx-^ 8. 13. a;2 _ 9a. _ 10, x^-7x- 30, and x* + 4 x + 3. 14. 2x2-8, 3x2-9x + 6, and 6x2 + 18x+ 12. 16. x2 - 3x + 2, 2x2 - X - 6, and 3x2 -2x - 1. 16. 6x2 + x- 2, 21x2+17x4-2, and 14x2 - 6x - 1. 17. 3x2 - lOxy + 32/2, 3^2 _ 4xy + y2, and x2 - 4xy + 3y2. 18. x« + 2x2_33j and 2x» + 6x2-3x. LOWEST COMMON MULTIPLES. 173 19. x^y'^ - 9 ?/2, x'^y -yx-6y, and x^ + x^-Qx. 20. x^ - a^, x^ + a^, and x* + a'^x'^ + aK 21. x^ - 1, x^ + x2 + X + 1, and x^ - x^ + x - 1. 22. 9 x^ - X - 2 and Sx^ - lOx^ - 7 x - 4. 23. x3 - ax2 - a^x + a^ and x^ + ax^ - a^x - c^. 24. x3 + x2-4x-4 and x^ + 6 x^ + 11 x + 6. 25- x* - x3 + 8x - 8 and x-5 + 4x2 - 8x + 24. 26. o3 + 6a2 H- 11a + 6 and a^ + lOa^ + 29a + 20. 174 MISCELLANEOUS THEOREMS. CHAPTER XIII. Miscellaneous Theorems and Examples. 145. Mathematical Induction. A method of proof, ordi- narily called mathematical induction, is frequently em- ployed in the demonstration of mathematical propositions. The method is best explained through its application to simple examples. Ex. 1. The following equations are evidently true : 1 + 3= 4 = 22, 14.3 + 5= 9 = 32, 1 + 3 + 5 + 7 = 16 = 42, 1 + 3 + 5 + 7 + 9=^25 = 52; and they at once suggest that perhaps the sum of the first n odd numbers is equal to n^. Let us assume provisionally that it is. This hypothesis algebraically expressed is l + 3 + 5 + ... + 2w-l = n2, in which n is an integer and equal to the number of terms in the sum. To both sides of this equation add 2 n + 1. The result of the addition is 1 + 3 + 5 + ... + 2n-l + 2n + l = n2 + 2n + l = (n + 1)'^ and shows that if the sum of the first n odd numbers is n2, then the sum of the first n + 1 odd numbers is {n + 1)2. lint the formula above written is obviously true when n = 1 and when ?i = 2 ; hence it is true when n = 3, And being true when MISCELLANEOUS THEOREMS. 176 n = 3, it is tme when n = 4, and so on indefinitely. It is there- fore true when n is any integer whatever. Thus the proposition that the sum of the first n odd integers is equal to nr is proved. Ex. 2. Let it be proposed to find the sum of the first n natural numbers. We begin by writing down the following equations, which are evidently true : 1 + 2 = 3 =1-2(2+1), 1+2 + 3 = 6 =i.3(3+l), 1 + 2 + 3 + 4 = 10 = ^4(4 + 1), 1 + 2 + 3 + 4+5 = 15 = i. 5 (5 + 1). They at once suggest that perhaps the sum of the first n integers is I n (n + 1). We accordingly assume, provisionally, that 1 + 2 + 3 + 4 + h w = i 71 (w + 1). Adding n + 1 to both sides of this hypothetical equation, we obtain 1 + 2 + 3 + .. . + n + n + l = |w2+in + ?i + l = Kw + !)(« + 2), a result which shows that if |n(n + l) is the sum of the first n integers, then i (w + l)(w + 2) is the sum of the first w + 1 integers. But the formula is obviously true when » = 1 ; hence it is true when n = 2. And being true when n = 2, it is true when n = 3, and so on indefinitely. It is therefore true, whatever integer n may represent. These examples may suffice to explain a method of proof that will be made use of, as occasion requires, in subsequent pages of this book. If its logical rigor be not at first entirely evident to the student, a careful study of its applications in a few examples that require its use should remove the doubt. Ex. 3. Prove that the sum of the first n even integers is n(w+l). 176 MISCELLANEOUS THEOREMS. Ex. 4. Prove that the sum of the squares of the first n integers is^n(n+ l)(2/i+l). Ex. 5. Prove that 1.2 + 2.3 + 3.4+... + n(w+l)=in(7i+ l)(w4-2): 146. The term induction does not correctly describe the above method of proof. In the natural sciences the true method of induction is used for the purpose of inferring the universality of natural laws from particular mani- festations of their truth in natural phenomena. It does not seek to attain absolute certainty, but achieves its object as soon as it establishes a high degree of proba- bility. The so-called method of mathematical induction, how- ever, is as logically rigorous as any other process in mathe- matics. It is more correctly described as reasoning by progressive transformations, in which the uniqueness of an algebraic form persists, — monomorphic transforma- tions, or monomorphosis. 147. Theorem. TJie expression «" — a" is divisible by X — a for all iJOi^itive integral values of n. It is known that x — ayX^ — a^, and x^ — a^ are aU divisi- ble by a; — a. We have a;" — a" = a;" — ax""-^ -f aa;"~' — a** = a;"~^ (x — a)-^a (a;""^ — a**"*). Now, if x — a divides a;""^ — a'*~\ it will also divide x^-^ {x — a) + a (a;*-^ — a"-^), that is, it will divide a** — a". (The if of thig sentence is important.) MISCELLANEOUS THEOREMS. 177 Hence, if x — a divides a;"~^ — a"~^, it will also divide ic" — a". But we know that x — a divides ar^ — a^ ; it will there- fore also divide a;* — a*. And, since x — a divides x^ — a*, it will also divide a^ — a^. And so on indefinitely. Hence a?" — a" is divisible by x — a, luhen n is any posi- tive integer [Art. 145]. Since a;" -}- a" = af — a" + 2 a", it follows that when x^ -f a" is divided by a; — a the remainder is 2 a", so that a;" 4- a** is never divisible by a; — a. If we change a into —a, x—a becomes a;— (—a) = a; -fa-, also af — a'' becomes a;" — ( — a)" ; and a?" — ( — a)" is a^ -f- a" or a;" — a" according as n is odd or even. Hence, when n is odd, a;** -f a** is divisible by a; -f a, and when n is even, x"" — a" is divisible by a; -f- a. Thus, when n is a positive integer, x — a divides a;" — a** always, x — a .... af* 4- a* never, a; -f a .... af" — a" when n is even, and a; 4- a .... a;" + a" ... . odd. We have in the above shown that the four cases are all included in the first : we leave it as an exercise for the student to prove each case separately. The above results may be written so as to show the quotients: thus M 178 MISCELLANEOUS THEOREMS. X — a ^'^^'^ = x^-' - x^-^a + x^-^a" ± a" 1 x-\-a the upper or lower signs being taken on each side of the second formula according as n is odd or even. EXAMPLES XLI. 1. Find the factors of x^ — x^ — x -\- 1. 2. Find the factors of x^ — x* — ic^ + 1. 3. Find the factors of 1 — x — x* -{- x^. 4. Find the factors of 1 - x^ - x^ -\- x^^. 5. Find the factors of x^ - y^ - 2 ax - 2by -\- a^ - h\ 6. Find the factors of x"^ — y'^ —Zx — y -\-2. 7. Find the factors of ar^ - 2 xy - 8 «/2 _ 2 x + 20 y - 8. 8. Find the factors of a^ -ly^ ■\- c"^ - p -2ac -2 bf. 9. Write down the quotient in each of the following divisions : (i.) (x» - r) - (aJ - y), (ii.) («' + y^) - (« + y), (iii.) (x^-y»)^(x-\- y). 10. Show that, if n is any positive integer, 7" — 1 is divisible by 6 ; show also that SS'^^+i + 1 is divisible by 36. 11. Show without actual division that (3 x2 - 2 X + 1)3 - (2 x2 + 3 X - 6)8 is divisible by x^ — 6 x + 6. 12. Write down the result of dividing (2 a + 36)8 + (3 a +2 6)8 by 5 a + 6 6. 13. Write down the result of dividing (2 a + 4 6 - 4 c)8 + (a - 6 + 7 c)8 by a + 6 + c. 14. Show that (1 — x)^ is a factor of 1 — x — x* + x*. // MISCELLANEOUS THEOREMS. 179 15. Show that (1 — x)- is a factor of 1 — x — ^ + a;"+i, n being any positive integer. 16. Show that (x — 1)^ is a factor of wx"+i — (n 4- l)ic" + 1, and also of x*^ — nx + n — I, n being any positive integer. 148. Factor Theorem. If any rational and integral ex- pression which contains x vanish when f is put for Xj then will x—fhe a factor of the expression. Let the expression, arranged according to powers of X, be aotf" -h fea?"-^ + cx""-^ -\ Then, by supposition. Hence oaf + hx""-' + ca;"-^ + . . • = ««" + 6a;"-i + cx""-^ H (a/" + &/"'^ + cf-^ -\ ) = a{^ -/«) + h {^-^ -f-^) +c (af»-2 -/"--) + — But, by the last article, a;" -/«, a;«-i -f-\ a;— ^ -/""^ etc., are all divisible by a; — / Hence also ax" + bx""'^ + cx""'^ + •••is divisible by x—f 149. Kemainder Theorem. If any expression which con- tains X be divided by x—f the remainder is equal to the result obtained by putting f in the place of x in the expression. Divide the expression oaf 4- bx""'^ -f cxf*'^ + • • • by a; — /, continuing the process until the remainder, if there be any remainder, does not contain a;; and let Q be the quotient, and R the remainder. Then, by the nature of division, ax^ 4- bx^-^ + cx""-"^ + ••• = Q(a; -/) + R, and the two members of this equation are identical. \ 180 MISCELLANEOUS THEOREMS. Now since R does not contain x, no change will be made in R by changing the value of »; put, then, «=/, and we have the required result, ar H- &r ' + cf-' + ... = Q(/-/) + E = R. This is known as the remainder theorem. It includes the factor theorem of Art. 148 as a particular case, or corollary; for, a/" + 5/**"^ + c/*-^ + . • • = when i2 = 0, and if i? = 0, the expression aa?" + 6a;"~' + ca;"~^+ ••• is exactly divisible by x —f. Ex. 1. Find the remainder when ic* — 4 a;*2 + 2 x + 1 is divided by a; - 3. The remainder is 3^ - 4 • 32 -}- 2 . 3 + 1 = - 2. Ex. 2. Show that a; - 2 is a factor of a;* - 3 x^ + 2 a; - 8. The condition is 24-3- 22 + 2. 2-8 = 0; and this condition is satisfied. Ex. 3. Determine whether a^ - 3 a; + 2 and a;^ - 13 a; + 12 have a common factor. The factors of x^ — 3 a; + 2 Are obviously x — 1 and x — 2 ; and of these x — 1 does, and x — 1 does not, satisfy the above condition of being a factor of x^ — 13 x 4- 12. Ex. 4. Show that any rational and integral expression in x is divisible by x — 1 if the sttbi of the coefficients of the different powers of x is zero. For if the sum of the Coefficients is zero, the expression will vanish when we put x = 1, ttnd therefore x — 1 is a factor. 160. Symmetry. An expression is said to be symmetri- cal with respect to any two letters when it is unaltered by an interchange of the two letters. Thus a-\-h and a^ -f W are symmetrical with respect to a and h, for neither expression would be altered by changing a into h and 6 into a. MISCELLAI^EOUS THEOREMS. 181 Also a-\-b -\-c and a^ + 6^ + c^ — Sabc are symmetrical with respect to any two of the three letters a, b, c. The only expression of the first degree which is symmetrical with respect to the three letters a, b, c is pa -\- pb -\- pc, where p is some numerical coefficient. These examples suggest the following definition : Def. An expression is (completely) symmetrical with respect to a specified set of letters when it is symmetri- cal with respect to every pair of them. Ex. The following expressions are symmetrical with respect to the letters contained in them : a^ + b^ + c^-bc-ca- ab, (6-c)4+(c-a)*+(a-6)*, bed + cda + dab + abc. 151. A partial symmetry may exist in expressions containing more than two letters. Thus, although ab^ -{-bc^ -\- CO? is not completely symmetrical, but in fact changes its form when any two of its letters are interchanged, yet it remains unaltered when a is changed to 6, b to c, and c to a. Such a series of changes is called a cyclic displacement, and the symmetry, repre- sented in such an expression as ab^ -\-b(^ -\- ca^, is called cyclo-symmetry. Def. An expression is cyclo-symmetrical with respect to the letters a, b, c, d, • • • Z if it remains unaltered when a is changed to b, b to c, c to d, etc., and I to a. Ex. The following expressions are cyclo-symmetrical with respect to the letters contained in them : a2(6 - c) + b-\c - a) + c2(a - 6), (6-c)3 + (c-a)3+(«-6)3, (6 _ c - l)(c - a - l)(a - & - 1). 182 MISCELLANEOUS THEOREMS. 152. The arrangement of the terms in a symmetrical expression is of some importance. Consider, for example, the arrangement of the expression bG-{- ca-\- ah. The term which does not contain the letter a is put first, and the other terms can be obtained in succession by chang- ing a into h, b into c, and c into a. In the expression a\b — c) + b^{c — a)-\-c^{a — b) the same arrangement is observed ; for by a cyclic displacement of the letters in a^{b — c) we obtain b'\c — a), and another cyclic change will give c^ (a — 6). By reason of this law of the derivation of terms from one typical term, symmetrical expressions are fre- quently not written out in full, but indicated by placing the Greek letter ^ before the typical term. Thus, when two letters are involved, ^a'b = a^b -\- b\ or, if three letters are involved, Sa^ft = a'b -f a^c + Wc + b^a + c^a + (?b. Similarly, when three letters are involved, Sa^ -f- %bc = a^ -^ y^ + c^ ■\-bc + ca + ab. 153. We now proceed to consider some examples which will illustrate the theorems proved in the pre- ceding articles. Ex. 1. Show that, if n be a positive integer, jc" — y" is divisible hy x-y. If at = y, then ic« — y« = ic* — a;" = 0. Hence the proposition is an immediate consequence of the factor theorem, which is itself a corollary of the remainder theorem. MISCELLANEOUS THEOREMS. 183 Ex. 2. Show that {h - cy +{c- ay + (a - by is divisible by (6 — c){c — a) (a — b). If we put 6 = c in the expression ib-cy+(c-ay+(a-by, the result is (c — ay + (a — cy, which is zero. Hence (b — c) is a factor, and we can prove in a similar manner that c — a and a — b are factors. Ex. 3. Show that (a -{- b -\- cy -(b -h c - ay -(c + a -by-(a -\- b - cy = 24 abc. If we put a = in the expression (a + & + c)3 - (6 + c - ay ~(c + a - by -(a -]- b - cy (i.), it is easy to see that the result will vanish for all values of b and c. Hence « is a factor of (i.). So also b and c are factors of (i.). Now (i.) is an expression of the thi7'd degree ; it can therefore only have thi'ee factors. Hence it is either equal to abc, or is equal to abc multiplied by some number. Thus (« + & + c)3-(6 + c - a)3 -(c+ « - by- (a -{- b - cy =Labc (il), where L is some number which is always the same whatever a, b, and c may be. We can find the value of L by giving particular values to a, 6, and c. Thus, let a = & = c = 1. Then (ii.) becomes 38 _ 13 _ 18 _ 13 ^ i. .-. i = 24. Ex. 4. Find the factors of rt3(6_c)+ 63(c_a)+c3(«-6). If we put ^ = c in the expression a3 (6 - c) + b^{c - a) + c^ (a - 6), the result is (^(c — a)-\- c^ {a — c)^ which is clearly zero. 184 MISCELLANEOUS THEOREMS. Hence b — c is a, factor of the given expression ; and we can prove in a similar manner that c — a and a — h are also factors. Now the given expression is of the fourth degi'ee ; hence, besides the three factors we have found, there must be one other factor of the first degree, and as this factor must be symmetrical in a, &, c, it must be a + ft + c. Hence the given expression must be equal to L{h- c)(c- a){a- 6)(a4- & + c), where Z is a number. We can find L by giving particular values to a, h, and c ; or, by comparing the coefficients of a^, we have at once b -c = -L{b-c). Hence L =— I. Thus Sft3(c- a) = -(b -c)(c- a)(a-b){a-{- b + c). 154. The following is an important identity : aa+63_|_c3_3a6c = (a + b + c) {a^ + b^ + €^-bc- ca- ab). It should be noticed that ^24- b'^c'-bc- ca-ab=i\ (6-c)2+ {c-ay-\- {a-bYl Since a-\-b-{-c is a factor of a^ -\-b^ + c^ — Sabc^ it follows that a-^ + &3 H- c^ - 3a&c = if a + 6 + c = 0. Hence a^ + 6^ H- c^ = 3 a&c for all values of a, b, and c, provided only that a -f 6 -h c = 0. That is, the sum of the cubes of any three quantities is equal to three times their product, provided that the sum of the three quantities is zero. For example, the letters involved being a, ft, c, 5 (ft - c)8 = 3 (ft - c)(c -a)(a- ft), S (ft + c - 2 a)8 = 3 (ft + c - 2 a)(c + rt - 2 ft)(a + ft - 2c) ; also, since (x - a) (ft - c) + (x - ft) (c - a) + (x - c)(a - ft) = 0, we have 2 (X - a)8 (ft - c)8 = 3 (X - a)(x - ft)(x - c)(ft-c)(c -rt)(a - ftV 4. MISCELLANEOUS THEOREMS. ISS EXAMPLES XLII. Let the student employ the 2 notation for the purpose of abbre- viating such of the following expressions as admit of its use. 1. Show that a2 (5 _ c) + 52 (c- a)+ d^ (a - 6) J^ = hc(h — c)-\- ca {c — a)+ ah {a — h) = — {h — c){c — a){a —b). 2. Show that a* (^ _ c) + 6* (c - a) + c* (a - 6) ^ = - (b - c)(c - a)(a - b) (a^ + b'^ + c^ ^ be -\- ca -\- ab). 3. Show that (6 - c)5 + (c - ay + (a - 6)° = 5 (6 — c)(c — a){a — b){a'^ + 6^ 4- c'^ - be — ca — ab). 4. Show that (a + & + cy -(6 + c)* -(c 4- ay - (a + &)* + a* + 6^ + c* = 12 a6c (a + 6 + c). 5. Show that a(b - cy -\- b (c - ay -\- c (a - by = (b- c)(c - a)(a - b){a -\-b-\-c). 6. Show that b-2c^ (b - c) -{- c'^a^ (c - a) -{- a^b^ (a - b) = — (6 — c)(^c — a)(a — b) (be + ca -\- ab) . "7. Show that «» (62 _ c2) + fts (02 _ a2) + c^ (a^ - 62) = — (6 — c) (c — a) (a — 6) (be -^ ca -\- ab). 8. Show that a* (62 _ c^) 4- 6* (c2 - a^) + c* (a2 _ 52) ^-(6 + c)(c + a)(a-f 6)(6 - c)(e- a)(a- b). 9. Find the factors of be (62 - c^) + ca (c2 - ^2) _|_ „;, (^2 _ 52). 10. Find the factors of a (b + c - a)^ + b(c+a-by^-\-c(n + b-c)^ + (6 + c - a) (c + a - 6) (a + 6 - c). 11. Find the factors of a2 (6 -f c- a) + 62 (c+a - 6) + c2 (a+6-c) - (b + e - a)(c -\- a - b) (a + b - c). 12. Find the factors of a (6 + c) (62 -f c2 - a2) + 6 (c + a)(c2 + a2 _ ^,2)4. c (« + 6)(a2 + 52 _ ^2). 13. Find the factors of (y - 0)3 -{-(z- x)^ -\-(x - y)^ - k(y - z)(z - x)(x - y), 14. Find the factors of 06(52 _ c2)3 _^ jF/(c2 _ ^2)3 4. cf'(a'^ - b^)\ 186 FEACTIONS. CHAPTER XIV. Fractions. 155. To obtain the arithmetical fraction |, we must divide the unit into 7 equal parts and take 5 of those parts. So also to obtain the fraction -, where a and h h are positive integers, we must divide the unit into h equal parts and take a of those parts. 156. The numerator and the denominator of a frac- tion, defined as in Art. 155, must both be positive inte- gers: we cannot, for example, with that definition, have 3 such a fraction as -^ j for t« say that the unit is to be divided into — | equal parts, and that | of such parts are to be taken, is without meaning. We must therefore suppose the letters in - to be re- ft stricted to positive integral values, or we must alter the definition of - ; and as we cannot restrict the values of h the letters, we must entirely dispense with the fractional form, or make some modification in its meaning. Now, with the definition of Art. 155, to multiply the fraction - by 6, we must take each of the a parts 6 h FRACTIONS. 187 times ; we thus get ab parts, the parts being such that each b of them make up a unit, and therefore the whole ab parts will make up a units. Thus ^Xb = a (i.). Dividing each side by b, we have Now we may define the fraction - as that quantity which when multiplied by b becomes a; for, as we have just seen, this new detinition agrees with that of Art. 155, whenever the definition of Art. 155 has meaning ; and by taking this new definition we do away with the necessity of ascribing only positive integral values to the letters. We may similarly define the fraction - as the quotient obtained by dividing a by b. Hence, instead of the definition of Art. 155, which is inapplicable to an algebraical fraction, we have either of the following equivalent definitions. Def. I. The algebraical fraction -, where a and b are b supposed to have any values whatever , is that quantity which, when multiplied by b, becomes equal to a. Def. II. TJie algebraical fraction - is the quotient b obtained by dividing a by b. The fractional form — has already been used with the 188 FRACTIONS. meaning a-i-b, and henceforth the notation a/b will also be frequently employed to denote a fraction. 157. We now proceed to consider the properties of algebraical fractions ; and we shall find that algebraical fractions are added, subtracted, multiplied, divided, and simplified, precisely in the same way as arithmetical fractions. It will be assumed throughout the discussion that the quantities involved are all finite and different from zero. 158. The value of a fraction is not altered by multiply- ing its numerator and denominator by the same quantity. We have to prove that a _am b bm^ for all values of a, b, and m. Leta;=-. Then a; x 6 = - X 6. b b But - X 6 = a, by definition. b .'. xb = a\ and therefore xbm = am. Divide by bm, and we have x = am -i- bm ; . , , . a am that IS, - = ~ — b bm Thus the value of a fraction is not altered by multiplying its numerator and denominator by the same quantity. 159. Since, by the last article, the value of a fraction is not altered by multiplying both the numerator and the denominator by the same quantity, it follows conversely FRACTIONS. 189 that the value of a fraction is not altered by dividing both the numerator and the denominator by the same quantity. Hence a fraction may be simplified by the rejection of any factor which is common to its numerator and denom- inator. For example, the fraction a^y/oc^y takes the sim- pler form a^/x^, when the factor y, which is common to its numerator and denominator, is rejected. When the numerator and denominator of a fraction have no common factors, the fraction is said to be in its lowest terms. 160. Reduction of Fractions to their Lowest Terms, 'i'o reduce a fraction to its lowest terms, we must divide; its numerator and denominator by their H. C. F. ; for we thus obtain an equivalent fraction whose numerator and denominator have no common factors. Ex. 1. Reduce ^ to its lowest terms. 6 cfixy The H. C. F. of the numerator and denominator is 3 axy ; and Zaxhi _ 3 ax^y h- 3 axy _ x_ 6 d^-xy 6 d^xy -f- 3 axy 2 a Ex. 2. Reduce ^^-^ to its lowest terms. The H. C. F. of the numerator and denominator is a'^b^ ; and Ex. 3. Reduce ^ a^^^^lf'^ to its lowest terms. 3 ab^x^y'^ The H. C. F. of the numerator and denominator is ab^xy^ ; and 2 a^b'^xy^ ^ 2 aV)^xy* ^ ab'^xy'^ ^ 2a^ 3 ab^x^y^ ~ 3 abH^y^ -f- ab'^xy'^ 3 bx'^ ' 190 FRACTIONS. 161. Instead of reducing a fraction to its lowest terms by dividing the numerator and denominator by their H. C. F., we may divide by any common factor, and repeat the process until the fraction is reduced to its lowest terms. Thus g^fe^c^ ^ a^hH^ = «!£!=«!. a%^di^ ¥cJ^ be* be The above process may be written down more compactly as follows : a2 a%*c* ~ be be EXAMPLES XLIII. Reduce the following fractions to their simplest forms : - Orb - 1 5 a%'^e'x^ ab'^' ' 25a2&*c3a;8* 2 ^ g 125a6W[^ xV ' imd^b'^d^d 3 2 gg&c^ g 3 a^x^yz^ 4a*6'c ' bab^xy'^z ^ 6 w'We* jQ 6 a%*c^xy^ 4a^65c** ' lb*cHy^ ' - x^y'''z^ -J 14 ab^c^xy'^sfi x^y'^z'^ 21 a^b'^cx^y^z l2^V£i^ 12 3 a^bc^x^y'^z 162. When the numerator and denominator of a frac- tion are multinomial expressions whose factors can be seen by inspection, write the numerator and denominator as the product of factors of the lowest possible dimen- sions; the factors which are common to the numerator FRACTIONS. 191 and denominator will then be obvious, and can be removed. a?' — ax Ex. 1. Simplify ^2 _ ^^2 * a^ — ax _ a(a -x) _ a a^ - x2 (a _ x) (a + x) a + x Ex. 2. Simplify ^* Ex. 3. Simplify x*- 1 x^-x^ _ x^ (x^ - 1) _ x^ X*-l (x2-l)(x2-f 1) ~x2 + l" a;2 - 7 a: + 10 x2_5x+ 6 7a;+ 10 ^ (a;-5)(a;-2; Ex. 4. Simplify x2_5x + 6 (x-3)(x — 2) x-S x'-^ — ax x(x — a) cfi — x^ {a — x){a + x) Now X — a = — (a — x) ; hence, dividing the numerator and denominator by a — x, we have tlie equivalent fraction, — X a + X If we divide the numerator and denominator by x — a, we have X -(« + x)' By the Law of Signs in Division ~ ^ and are both ^ a + X - (a + x) equal to — . and this latter is the form in which the result is a + X' usually left. Note. — It should be remarked that the value of a fraction is not altered by changing the signs of all the terms in the numerator and also of all the terms in the denominator ; for this is equivalent to multiplying both numerator and denominator by — 1. 192 FRACTIONS. EXAMPLES XLIV. Simplify 1. 2 a?) a2 + a& 9. 2. X2 - X-^?/'^ 10. 3. a2 - ah a2 + a& 11. 4. a;2+ ax x2-a2 12. 5. (x-iy X2-] 13. 6. X2 - x2?/2 (X + X?/)2 14. 7. x2 + 2x x2-4 15. 8. X* + X2 x4-l 16. 25. x2- 9x + 20 x2 + 6 X - 65 26. l-92/2 + 20y4 l+6y2_ 562/4 27. X2-8X2/ + 7y2 x2 -Zxy -28y2 oa 1 _ 8 a262 + 7 a*&* 4x- -16 X2- -16 2x3 -4x* X2- -4x* X — 2 4- X2 a - 3 9- a2 a2x2 -X* X*- -a* X*- -a2 rt2- ax2 X6- a2x3 X*- -a* a2x2?/2 _ x'.y4 29. 80. 31. 82. " 1 - 3 a2&2 _ 28 a4&* (a" - 68)(a* - 6*) 163. When the factors of the numerator and denom- inator of a fraction cannot be found by inspection, their H. C. F. can be found by the rule given in Chapter XI. ; and the fraction will be reduced to its simplest form by dividing the numerator and denominator by their H. C. F. 17. oajG — LOW x2 - 9 a2 18 3x2 -12 ax 48a2-3x2 19 a2 - 2 rtx 4- 3^2 a2 _ a;2 20. a» + 2 a262 + fti a* - 6* 21. x2-l x8-l 22. X4-1 x«-l 9)(6-c)- 1 1 1 (c-a)(c-&) {_(a-c)}{-(6-c)} (a-c)(ft-c) Hence we have to simplify ^-1.1 (a — 6)(a — c) (a — 6)(6 — c) (« — c)(& — c) The L. C. M. is (a - 6) (a - c)(6 - c) ; and 1 h-c (a — 6) (a — c) (a — 6) (a — c) (6 — c)' -1 ^ -(«-c) (a-6)(6-c) (a_6)(6_c)(a-c)' and 1 - «-^ {a-c)(h-c) (a-b){a-c)ib-c) Hence we have b — c — (a — c)+a — b _ b—c—a+c-\-a—h _q ia-b){a-c)ib-c)~ {a-b){a-c){b-c)~ {a-b){a-c)i_b-c)~ EXAMPLES XLVI. Reduce to their lowest common denominator : 2 4 7 „ 3 4 1. 3a;' 6x' 30x * 2a;-2' 3a;-3 2. -i-, -^, J-. 7. ^ ^ 2aa;' 6 6a;' 8ca; Q a & c a be ca ab 4 be ca ab *» • — » -rt — «»• a b e 6. -i-, ^-^. 10. 6a; + 12' 8a; + 16 1 3 6 a;+ 1' 2 a; 4- 2' a;'^-l 6 2 4 6a; -5' 3a; + 3' a;^ - 1 a X a^ a;2 X + l' 2 a; -t- 2 ' x - a' a - x' x^ - ««' a^ - «« FRACTIONS. „ 2a h 3a2 5 62 ' a-b' 2b -2a 4( a^-62)' 6(6^ - a2) 12 ^ 2^ (^ 3x2 + 1)« • X + 1' (x+ ly' 13 ^ 1 1 • (x-a)(x-6)' (6 -x)(c- X)' (x-c)(x-a) 14. 1 (a-6)(a-c) 1 1 (& -c)C6- a)' (c-a)(c-6) Reduce to one term : 15. a-l+-^. a + 1 o^ 6a- 5ft 4a- 76 ^ 3 2 1-2 16. a + x + ^ — a —X 21 a -3ft , 3a-ft 4 5 17. x + 22/+-i4- x-2y 22. 2^-y 3^ + ^ 3 4 18. ^7' + "-'- a^6 ab^ 23. 5^-2y 3^-2* -g a-66 a-36 5 3 «, X X-4X-5 ^'4 3 ' 6 26. 2^-^^ 3 . a; + 2y ^ 4 3 X - 2 y 6 Simplify : a - 6 6 - a 30 ^ ^ -"" 2 - X 4 - x2 27. ^ + " . X — a a — X 31. ^ + ^ . ^^- 3 + X x-2 - 9 28. * 1 "* x2 - a2 ' a2 - x2 32. 1 1 . *^' a;-3 x-2 «n 1 2 1000 > 100 > 10 > 1. 204 FRACTIONS. If then a be very small, 1/a is very large, and if a be very large, 1/a is very small. When a becomes zero, the fraction a/b assumes the form 0/b and, by the proposition- of Art. 127, is itself zero. But when b is zero, the fraction assumes the form a/0 and, regarded as a quotient, has no meaning ; for division by zero is an impossible operation. It becomes merely symbolic. Yet it. frequently makes its appearance in algebraic operations, and since it may be regarded as having arisen by virtue of a decrease of the denominator from a finite quantity to zero, that is, by an increase of the fraction itself beyond measurable limits, it is called an infinite quantity, or infinity, and for convenience the special sym- bol QO is used to represent it. But this nomenclature must be regarded as conven- tional, and the symbol oo must be used in algebraic operations with extreme caution. A complete study of its legitimate use requires more extended discussion than is at present appropriate. [See Treatise on Algebra, Art. 217.] EXAMPLES XLVII. Reduce the following fractions to their simplest forms : 2 a ^ 3 c a ^ c bd 1. — X — 5. - X - X — 3c 4a b d ac 6a2^ 362 ai b-i ^ c2 2. • X • 6. — X — X — • 6 6c 10 ca be ca ab a _^c^ b ^b c 3. 7. - X - -. — c ' d a c a 4. 2 a* . 3a6 &2 ^ 62 c2 ; . 8. — X — — • be c* 62 c2 a« FRACTIONS. 205 9 2a 2& 2^ jj Saxy^ . 6ay^ ' be ca ab ' 6b^ ' 10 b-^x 10 2x^ X ^ X ^- 12 ^^'^^ - 5a^^ ' 3^0 50X 2ic?/' ■ SfeVa ' Sab'^c^' 13. By what must '-^ be multiplied, that the product may be — ? 5 x^^y 5 X 14. By what must — ^ be divided, that the quotient may be — ? Express in their simplest forms : 16. ^-y X ^ + y . 21. ?^=i X ^^ X ^::i^. xy — y'^ X — 2 X — S X— 4: 16. ^^ -r .^ ^ ab - b"^ 22 x+1 ^ a;+2 ^^ a;+3 2a. x-y .. x'^-^xy. a + b ^ a^ - a^b x-^ + 2x,. a:-^-9 '^ x'^-y' .. a:2_4y2' x3+3x2 . x + 4 a-f 4?) rt6 + «2 (x+2)2 (x+3)2 (a;+l)2 ^^ .. -r^^ ^ x^-Sx 23 a;2-3a;+2 ^^ a;2-7a;+12 " " x^-4 ' x^-6x+6 x2-5x+4* 18. --^ X '-^^- 24. ^'-^ X ^'-^^ . x + y x2+3x-10 x2-3ic-4 jg ^ -r » ^ ^ x + 3 gg a^ -g^ ^ x + 2 a x2 4- 4 a; ic2 — 4 a2 x — a ab + 4 &2 2g q^-x^ _^ (« - x)2 a2 4- 5 a6 a'* + 5 a^ft' " a^ + a^ ' a2 _ 3^2 * 2,y q + X X q^ - a;2 _ a* - X* 30. 31. 32. (a - x)2 a2 + a;2 (a + x)3 (q+&)2 (a2 -^ 62)2 ^ (a_+_6)8 (a -6)3 ^ a*-6* "02 + 62' (a - 6)2 - c2 62 - (c - a) 2 (a - c)2 - 62 ^ c2 - (o - 6)2* x'^-(y + zy _ y2-(x-hzy oc:'-(y-zy ' y^-(x-zy x^ + y^ X ^~y ^ x^ - x2y2 + y4 X'5-l/« X + ?/'x* + X2^2 _|_ yA 206 FRACTIONS. 169. We will now give some examples of more com- plex fractional expressions. a d ad b c be Ex. 1. Simplify ^ /^. 6/ d a / £_ bl d~ Ex.2. Simplify ^1+1^^+1^ {^jf.\\ I l^A. i\- <^+ ^ / b -\- a _ a-\-b a _a \b 1/ \a I b I a b 6 + a b . Ex.3. Simplify f «L±^ _ ^JZJi'Wf ^^^^ + ^^=^V \a — X a + x// \a — X a + xj a-\-x _ a — x _ (g + x)(a -\- x)-(a — x) (a - x) _ 4 ax a — x a + x~ {a-x){a + x) ~ a^ - x:^' and ^"^^ + ^~^ _ (« + a;) (g + g) + (g - a;) (a - a;) _ 2a^-[-2x^ a- x a + x~ (g - «) (g + «) ~ a^ - x;^ ' Hence the given fraction is equal to 4ax /2g2 + 2a:2 4ax ^, a^ - x"^ 2ax / 2g2 + 2a;2 _ iax a^ '/ g2-a;2 a'- - x^ 2g2 + 2 x2 g2 4- a;2 Ex. 4. Simplify x-\-2 x+2-^±i x + 2 ^ (a;+2)x a; a; a;(a;2-|-a;- 1) ^_J^_±2x_ a;(x2 + a;- l)-(a;2 + 2x) x- + a; — 1 a;(a;'^ + x - 1) _ x'^ + x - 1_ FRACTIONS. 207 b b — c ■r. - o- vr 1 + aft I + be Ex. 5. Simplify — — — • 1 (a-fe)(&-c) (1 +ab){l + bc) a-b b- c ^ (g - 6)(1 + 5c)-i-(& - c)(l + g6) 1 + a6 1 + &c (1 + ab) (1 + 6c) __ a — b -\- abc — b'^c + b — c + ab^ — abc (1 + «&)(! + 6c) _ a-c + ab'^ - cb^ (l + a6)(l + 6c)' (a-b)(b-c) ^ (1 + a6)(l + 6c)-(a-&)(6-c) (1 + a6) (1 + 5c) (1 + a6) (1 + ^c) _ 1 + a6 + &c + gft^c - q& 4- 52 ^. q^ _ ^^ (l + a6)(l + 6c) _ 1 + qc + 62 ^ fltfigg (H-a6)(l + 6c) * Hence the given fraction is equal to a- c + ab'^ - cb^ . 1 + «c + 6^ _|. ^^2^ (l4-a6)(H-6c) ■ (l + a6)(l + 6c) ^ a-c + ab^-cb'^ ^ (a- c)(l + b^) ^ a-c 1 + ac + 62 + ac62 (1 + ac) (1 + 62) 1 + ac EXAMPLES XLVIII. Simplify 1. f x-\-2aa-\-2x ^^f 3 1_1 la — 2x X — 2a/ \2a— x a — x) "ll-fx X Jll+x X /* 8. f^-af£+ y.Uf^ + ^-U^--^V \y xl\y^ x-V ■ Vl/ xJU^ y.i) 4.^ + i_i i_^ i_? X y X y 208 FRACTIONS. . y 2 1 + X 1 + x'-' 5. y X a; y 1 -\-x^ 1 4- a;« 1+ x'* 1 + x* ?+l? + 2 ? + ^-2 ,, 1 1 y x w x 11. + X x-\-y x-y x-\- 3.-1 a-\-h h ^ 7. ^t^^ ,^ . 12. xs ?? a + ft 1 + a a+ 26 1 X + 1 13. x + ^ + ^^TT 1 + " + ' 3 x + 24a;4-5 ^4 1 2x4-3 5x + 6 ^2 a^^ + 1 2x + 3 _ 3x+4 ' ^ ^ _|_ 3x + 4 4x+ 6 X — 170. The following theorems are of importance. Theorem I. If the fractions a/h, c/d^ e/f etc., he all equal to one another, then will each he equal to the fraction pa + qc + re-\ — ph + qd + rf-^-'" Let each of the equal fractions be equal to x. Then, since a/h = x, c/d = x, e/f= x, etc., .•. a=hXf c = dx, e=fx, etc.; ,'. pa =phXy qc = qdXj re = rfx, etc, FRACTIONS. 209 Hence, by addition, pa -\- qc -\- re -{- ••• =pbx -\- qdx -}- rfx -\- ••• = (pb-{-qd-{-rf-\-'-')x; pa -{- qc -\- re -\- " ' _ _ ^ _ As a particular case, each of the equal fractions a/b, c/d, e/f, etc., is equal to" + ^ + ^"*'" • b+d+f+ Theorem II. If the denominators of the fractix)ns a/h, c/d, e/f etc., be all positive, then will the fraction n I c I e I • • • — i- — — — — — be greater than the least, and less than the & + («+/+••• greatest, of the fractions a/b, c/d, e/f, etc. Let a/b be the greatest of the fractions, and let a/b = X. Then c/d < x, e/f< x, etc. Hence, as b, d,f,"' are all positive, we have a=bx, c < dx, e ) f ^'^ + ?/•' ^' - y' ] X ^ . /a;+y g-y] ^ ^la;2-?/2 X2 + 1/2J x2-a;y + y2 \a;-y aj+yi ■ ^ 2a6 ya3 + 63 • ^2 - a6 + 62* - (-:-!M-MM->"r-{E-:)(^^)(-:-:-)- 47. I ^ + 2 \^J.x±y__x-y^)^ l3a;2-14a;y+15y2 3a;2-2xy-5y2i lx-3y x+3y/ 48 ^ — ^ I c — « , a— b , (b — c)(c — a) (a — b) \ a-\- X b + x c + X (a + x) (6 + x) (c 4- x) Ha + 6/ o + 6i \\a~bl a-b ) 50. {f«_^y+3f«^y+3«^ + i| l\a + 6/ \a-hbj a+b i 51. 16 21 16 21 x-17 x-9 x + 4 x + 7 11 7 11 , 7 x-5 x-7 x+2 x+4 X + 7X+8 X4-9 x+10 x+1 x+2 x+3 x+4 214 FKAGTIONS. ' (a + by-{c + d)2"^ (6 + cy - (a + dy'^ {c-\-ay -{b-\-dy' 55. Simplify -^^ + axy a{x - a){y - a) y{x - a){y - a) 56. Simplify __«Jl^ + « + y + «_+^ . x{x-y){:x-z) y(y-z){y -X) z{z-x){z-y) 57. Find the value of ^~^^ when a; = "^ 6 -2x a + 6 58. Find the value of a + 6~2c a-\-b-2d when rt + 6 = ^ ^^ c + d 59. If ^ = -, prove b d ^ a - 2 6 c - 2 d ^ a^ + ft"^ c'^ + d^ ,jj X _ jt^ ^ c2 ,.^ . j)«-^ + gab + rft ^ ^ pc^-\-qcd -\- rd^ ^ '^ (a + 6)2 (c + d)2 ^ '^ ia-^ + mtt6+n62 ZcHwicd + nd* 60. If --^— = -1— = -?— ; then will « + y + « = 0. a — b b — c c — a 61. If -l±g-= ^ + ^ =^±iL; then will each fraction = -^. ay-{-bz az + bx ax-\-by a-\-b 62. Prove that, if a, 6, c be unequal, and ~ ^ = ^~ ^ = ^~ , X y z then will « + y + 2; = and ax + &2/ + c;? = 0. 63. Prove that, if «±4 = -Ail^ = -f+«-, a - 6 2(6 - c) 3(c - a) then 8a + 96 + 6c = 0. FRACTIONS. 215 64. Prove that, if ^^ ~ ^^ = ^^~^^ , then each is equal to Mn^. 6 - c c—a a — b 65. Prove that, if a b c 2y-\-2z-Sx 2z-h2z -Sy 2x-\- 2y -Sz then ^ = y = ^ a + 2b + 2c 6 + 2c + 2a c4-2a + 26 66. Prove that, if X — a _ y — b _ z — c p ~~ q ~ r ' and x-{-y-^z = a + b-{-c, then a; = a, y = b, and z = c. a 6 c prove that — ^'^^ = ^ = — = abc, b + c — a c-i-a — b a + b — c and therefore x= bc{b -\- c — a), y — ca{c-^ a — b), » = ab (a -\- b — c). 216 EQUATIONS WITH FRACTIONS. CHAPTER XV. Equations with Fractions. 171. In the present chapter we shall give examples of equations which contain fractional expressions. x-1 3a;-l 43 -5a; Ex. 1. Solve 5 8 6 We may multiply both aides of the equation by 120, the L. C. M. of the denominators of the fractions, without destroying the equality ; we thus get rid of fractions, and h^ve 24(a; - 1) - 15 (3a; - 1) = 20 (43 - 5a;) ; . .. 24a; - 24 - 45a; + 15 = 860 - 100 a;. Transposing, we have 24a;- 45a; + 100a; = 800 + 24 - 15; .-. 79 a; = 869. Hence a; = — =11. 79 Ex. 2. Solve — ^ + a;-3 a; + 3 a; + 5 The L. C. M. of the denominators is (x - 3)(a; + 3)(x + 6). Multiplying both sides of the equation by the L. C. M.,* we get rid of fractions, and have (a; + 3)(a; + 5)+ 2 (a; - 3)(a; + 5) = 3 (a; - 3)(a; + 3); .-. a;2 + 8a;+15 + 2a;2 + 4x-30 = 3x2-27. ♦ See Art. 181. EQUATIONS WITH FRACTIONS. 217 By transposition, we have x2-|-2x2-3a:2 + 8x + 4a; = -27 -15+30; .-. 12x = -12, or x = — l. Ex.3. Solve — 1- x + l x + 7 x + 3 x + 5 In this case it is best not to multiply at once by the L. C. M. of the denominators ; the work is simplified by proceeding as under. Wehave -^- W + -^ ^=^5 x+1 x-^S x + 1 x+5 a; + 3-(a;+l) a;H-5-(a; + 7) _ •• (x+l)(a; + 3) (a;4.5)(a; + 7) thatis ? =0; (x+l)(x + 3) (x + 5)(x+7) .-. (x+l)(x + 3) = (x+5)(x+7), that is x*2 + 4x + 3 = x^ + 12 x + 35 ; .-. 4x-12x=:35-3; .-. -8x = 32, .-. x = 32/(-8) = -4. Ex. 4. Solve the equation x-l|X+5_x+l|X+3 x+1 x+7 x+3 x+5 Since^-1-1 2 a. + 5_j 2 x + 1 j, x+1 x+lx + 7 X + 7X + 3 2 x + 3' and '= + ^ = 1 \, X+5 x+5 wehave 1 2_ ^ _ _J_^ ^ __2 ^ 2 x+1 x+7 x+3 x+5 2,2 2,2 • • . , ' . « — . rt 1 . ,.' x+1 which is the same equation as that in Ex. 3. 218 EQUATIONS WITH FRACTIONS. Ex . 6. Solve the equation 4a;+5 x + 6 _2a x + l a;-f 4 X ^4-5 a + 2 ;2-10 x + 3 + «. By division, we have ix + 6_ x+l .4+ 1 , x + l x + 5 X + 4: - 1 4- 1 X 4-4 2a; + 5. .9,+ 1 , and-^ -10 = x- 3 ic + 2 x + 2 x + S x + 3 Hence x+l x + 4 a; + 2 1.1 1.1 4 + -l-+l + ^— = 2 + -^^ (x-S ~] + x x+l x + 4: a; + 2V x + 3J that is x+l x+4 x+2 x+3 1 1 ^1 1_. *'x+l x + 2 x + 3 x + 4' Cx+ 2)-(x + 1) ^ ( X + 4)-(x + 3) (x + l)(x + 2) (x + 4)(x + 3) ' 1 1 (x + l)(x + 2)~(x + 4)(x + 3)' .♦. (x + l)(x + 2) = (x + 4)(x + 3), that is x2 + 3x + 2 = x2 + 7 X + 12 ; .-. x2 + 3x-x2-7x= 12-2; .-. -4x = 10; "^- 4" 2 EXAMPLES li. 3 a; X — 2 _x 3 ^~^ ' e 6 ~6 4 * 2. "*- ^ -2 *"^ = ^"-*^ . 4. ^zi2_3^i:i=x-6^:=^. 11 6 10 3 6 6 x-2 x-3 x-7 6 4 10 3x-l -2*- 1 2- EQUATIONS WITH FRACTIONS. 219 5. l-3a; 3x + l_ 2 2 2 l-3a; 6. 3 -4a; 5 -8a; 1 - a; 6 12 1 + a; 7. 3a;-l ^x-\-l o 3-6a; 4 \-3 ^ 8 8. 2a; 3a;-l_3 1-4 a; 5 a; + 1 10 9. 1 , 1 _ 2 4x + 6 6x + 4 2x + 3 10. 1 2 _ 2 . 3x + 9 5x+l x + 3 11. 1 3 4 4X+12 4x+l 2x+6 12. 3 5 _ 7 2x + 3 4x + 6 6x + 8 13. 4x X _3 x+l x-2 14. 3x X _o - x+6 x+5 15. 6x X _5 x-7 x-6 16. 2x 4x ^2=0. x+3 x+7 17. 1 , 2 _ 3 x+4 x+6 x+6 18. 3 2 1 x+1 x+2 x+3 19. 5 13 1 1 20. ?J 5^__J_^o. 21. 2x + 2 4x + 3 4x 2x- 5. 3x-7 2x-7 3x-5 Bx-2 8x-5 _3x+7 4x+8 x-9 3x-7 _2x-5 6x-4 3-2x x-6 2x-7 4-x x+1 x-l x-3 8 x + 3 X x + 2 x-2 8 23. 25. X — 2 x + 2 x+1 27. ^Izl_^dL2+10^o. x+3 X— 4 X oo 3x _^x-l_^_9_^Q x + 2 x-2 x+1 29. 3?— ^ + 2?-ii = 5. x + 1 x-l 30. 5^^^ -2^^ = 3. 31. x + 2 x + 3 8x 1 1 x^ — 1 X — 4 x + 4 12x _^ 2 2 x3-9 x + 3 x-3 33. ^^+ 1 ' 34. — ^ + x+1 1 -X 1 2 2x + 4 2x+2 x + 6 x2-9 x + 3 3-x 220 EQUATIONS WITH FRACTIONS. OK 2x-\-l 8 2x-l 2a;-l 4x2-1 l + 2x 36. Sx + b 5 _8 + 3a; 3x-l 1-9x2 l + 3x* 37. 1 3 _ 5 x_5 2x-6 (x-3)(x-5) 38. 1 3 _ 7x+l 2x-4 x-5 (x-2)(x-6) 39. 2-3x 6 -6a: 10 + a^ 40. « 1 7 1-0. 41. 42. 3x-l 3-7x d-\-x 1111 x+5 x+6 x+6 x+8 _l 1_ ^ ^ 1 x + 7 x+1 x + 1 x + s' 43.-A_ + _l = 1 + 1 44. x + 2 x+10 x + 4 x + 8 1 ■ L_ = __i L_. x-2 x-6 x-4 x-8 «. -^+^=-L-.+ 1 X-6 x+2 x-4 x+1 46 Jg . a;-9 _ x+1 , x-8 *x — 2 x-7 X — 1 x-6 47 2x + l . 2x + 9 ^ 2x + 3 ^ 2x + 7 48. x+1 X + 6 x + 2 X + 4 2x-3' 2x-4 2x-7 2x-8 2x-4 2x-5 2x-8 2x-0 x + 7 , x + 9 ^ x + 6 ^ x+ 10 x + 6 x + 7 x + 4'x + 8 EQUATIONS WITH FRACTIONS. 221 gQ 16 a; - 13 4 a; - 43 ^ 32 a; - 80 20 a; - 24 4x-3 8a;-9 8x-7 4x-5 ' 61 a;-7 a;-8 ^ 2x-7 2a;-ll X — 5 X — 6 2x — 5 2x — 9 B2. ^-^ = ^—-1 x-9 x-4 x+2 x+3 63 1^ _ 21 ^ 16 _ 21 "x-lT x-9 x + 4 x + 7 X— b X — a 65 ^ + « x-&_ 2(a + 6) X — a X + 6 X -« ax , 6x , , 56. 1 = a + 0- a + X & + X g- q + c & + c _ g + ft + 2c x + 26 x + 2rt x + a + 6 58 a; — 6 _ X — a _ 2(a — b) X — a X — b X — a — b 172. Simultaneous Equations. Any pair of equations of the first degree in two unknown quantities can be reduced to the form ax -\-hy -\- c = 0, a'x + b'y 4- c' = 0, and any formula that completely solves them solves also every pair of equations of this class. We shall now derive such a formula, applying for this purpose the method of elimination by undetermined multipliers. [Art. 106.] To solve the simultaneous equations ax + 6y + c = 0, a'x + b'y + c' = 0. 222 EQUATIONS WITH FRACTIONS. Add the first equation to k times the second, thus : (a + ka')x + (& + kh')y + c + A:c' = 0, . . . (i.) and make 6 + A:6' = 0, for the purpose of eliminating y. Then k = -^. . h' and, substituting this value of k in equation (i.), we obtain («-!•«') X + c - - . c' = h' whence a; =[ - c + -^ • c'W( a - - • a'Y Again, for the purpose of eliminating x, make a + ka' = 0. Then fc = --^, a' and, substituting this value of k in equation (i.), we obtain {'-V") a y + c-^.c' = 0; whence y = (-c + — - d\/ Ih - —•h'Y When simplified, these fractional expressions for the values of X and y have the form x = (bc' -h'c)/{ah' -a'h), y =(ca' - c'a)/(ab' - a'b). Observe that either one of them may be derived from the other by interchanging a with b and a' with b'. Thus the principle of symmetry here finds an application to elimination ; and it fre- quently happens that by taking advantage of this principle, one-half or two-thirds of the work of elimination by the ordinary process ia avoided. EQUATIONS WITH FRACTIONS. 223 173. A convenient device in elimination, sometimes called the rule of cross-multiplication, is suggested by the cyclo-symmetry observable in the equations a; = (6c'-6'c)/(a6'-a'6), when they are written in the form a; _ 1 y _ 1 6c' — 6'c a6' — a'6' ca' — c'a a6' — a'6 or more concisely, in the form a? ^ y ^ 1 . be' — b'c ca' — c'a ab' — a'b For the purpose of making the symmetry of the expressions more apparent, c is changed to cz and c' to c'z, in the two equations to be operated upon, which then become ax -{- by -{- cz — 0, a'x 4- b'y + c'z = ; and the process of elimination (in any of its forms) gives, not now the values of x and y, but of the fractions x/z and y/z, in the form already written above, namely, X _ y z be' — b'c ca' — c'a ab' — a'b^ in which z has now taken the place of 1 in the third numerator. The rule of elimination by cross-multiplication may now be stated as follows : Write three fractions with x, y, and z for their respective numerators. Under x write the difference of cross-products be' — b'c, formed from 224 EQUATIONS WITH FRACTIONS. the coefficients of the terms that do not contain x, and under y and z respectively write the similar differences ca' — c'a and ab' — a'6, observing the cyclic order in the displacement of the letters. By this simple rule the values of x/z and y/z are written down by inspection. These cross-product differences are sometimes exhibited in the form of square matrices, which are called determi- nants, thus : b c c a a ^ 1 b' c' J c' a' y a' b' and when so written they suggest at once the rule of cross-multiplication. For example, J " =bc'-b'c, b' c' and the other two similar identities are derived from this one by a cyclic displacement of the letters. The method here used is applicable to the solution of simultaneous equations with any number of unknown quantities. [See Treatise on Algebra^ Arts. 145, 432.] Ex. 1. Write, by the rule of cross-multiplication, the values of x/z and y/z which satisfy the simultaneous equations 2x-\-1y-\-^z = (i. Solution 20-14 4-12 21 10 or x_ y _z , 6 -8 11' A, y 11 z EQUATIONS WITH FRACTIONS. 225 Ex. 2. Write, by the rule of cross- multiplication, the values of x/z and y/z which satisfy the simultaneous equations 3a;-5y + 20 = O, Solution : a; __ y _ z 20-14 4 + 12 21 + lo' or ? = X = A; 6 16 31' " z Z\ z 31* Ex. 3. Determine the relations that must exist between the coefficients of ax^ + 6x + c and a'x^ + 6'x + c', in order that the equations ax2 -f 6x + c = 0, a'x2 + 6'x + c' = 0, may have a common root. By a common root is meant a value of x which satisfies both equations. Hence, if we assign to x, for the moment, the particular value in question, the two equations become simultaneous in x and x^. Then, by the rule of cross- multiplication, we have x2 X 1 hcf — b'c ca' — c'a ah' — a'b and, placing the product of the first and third of these fractions equal to the square of the second, we obtain (6c' - b'c) (ab' - a'b) (ca' - c'a)-^' whence, after dividing by x^ and inverting, (6c' - 6'c)(a6' - a'6) = (ca' - c'a)2, which is the relation sought. 226 EQUATIONS WITH FRACTIONS. 174. When a set of equations can be conveniently written in terms of a series of equal fractions, Theorem I. of page 208 provides still another useful method of elimination. Ex. 1. Solve the simultaneous equations a b c ' By Theorem I., page 208, we have -y -z+ z + x-\-x-\-y _ 2x _j^ — a -\- h + c — a 4- 6 4- c .-. X = \k {— a -\- h -^ c) ; and by a cyclic displacement of the letters y -lk{a-h^-c), «= |A;(a + 6 — c). Ex. 2. Solve the simultaneous equations a h c By Theorem I., page 208, each of these fractions is equal to a -\- h + c .-. y/x-\-y-\- z = K« + & + c), and the original equations become a h c ' Putting ^(a + ft + c) in place of k in the results of Ex. 1, we have a; = ^(a4-6 + c)(-a + 6 + c), y = J(a + 6 + c)(a-6 + c), « = i(a + 6 + c) (a + 6 - c). EQUATIONS WITH FRACTIONS. 227 Ex. 3. Solve the simultaneous equations y + Sz _ z + x _ Sx + 5y _ ^ 4 ~ 5 ~ 2 By Theorem I., page 208, we have 6y-\-lbz—16z — l5x — Sx—5y _ ^ 20-75-2 .-. - 18a; = -57, x = f | ; y-\-Hz-Sz-Sx-\-Sx-\-5y _^ 4-15 + 2 ... 62/ = -9, y = -i', 5y + 15g + 3g + 3a;-3a;-5y _ ^ 20 + 15-2 EXAMPLES LI. Solve the following simultaneous equations, obtaining one set of values of the unknown quantities in each case. 1. a;-2/ = l-l, ?_^=2_2 a b a b a'^ b^ o a; y _ 3 y x _ 1 _ a + 6 a — b ab a -\- b a— b ab a b b a «+ 1 y+ 1 1 -2/ a;+l 6. — ^ ^^=1, x + 2^ = a + 6. X — a y — b 6. ? + ?^ = l, ^ + 1 = 1. * a b c a' b' c' » a b a' b^ , 7. _ + - =r C, — + - = C'. X y X y 228 EQUATIONS WITH FRACTIOKS. 8. a; + l _ 3y + 1 _ x-{-y X Zy xy 9. ? + * = !, » + £ = !, £ + ? = !. a; y y^z ' z^ x 10. 1 + 1 = 7, 1 + 1 = 9, i + l = 12. y z z X X y 11. ^ = f = ^z:.(x + y + 2;)2. a b c 12 '^y"^ = g-3a; _ x+y _ , 2 3 6 * 13. -^4-y + ^ ^ a:-y + ^ ^ x + y-g ^ ^^^ ,^ a ,6 c j^ cy + &3; _ az + ex _ bx + ay — i- * Z m n 16. Find two sets of values for x, y, z, that satisfy the equations X t-l- a b c x-\- y -\- z 16. Find three sets of values for x, y, z^ that satisfy the equa- tions ?=f=?=^y.. a 6 c 17. Find a value of c that will make the equations 2ic2_a; + c = 0, 4x2+ 4a: + 1 = have a common root, and then determine the common root. 18. Determine two values of a, either of which will make the equations x2 + 2 X - 3= 0, ax2 - 11 X + 10 = have a common root, and find the common root in each case. ♦ See Ex. 3, of Art. 170. MISCELLANEOUS EXAMPLES III. 229 MISCELLANEOUS EXAMPLES III. A. 1. What must be added to x^ — Sx'h/ + 3 xy^ that the sum may be Sx^y-Sxy'^-\-y^? 2. Show that x(y-\-zy-{-y(z-{- xy -^ z (x -\- yy - 4 xyz = {y+z){z + x) (x -f ?/) . 3. Find the factors of (i.) x^ — y\ (ii.) a- — 5'^ + 2 a — 2 6, and (iii.) x^-Sx-2S. 4. Find the H. C. F. of x^-i-dx- 20 and 6x* + 9a^ - 64. Simplify 5x6 + 9x4- 64 a; 5. Solve (i-) x-l 2 ,.-^ = .-.. (ii.) 3a;- 4y -^2 = 5x-Qy -2 = -.7x : + 2i^ + 4. (iii.) a 1 b _a+b x+b x + a X 6. If A were to give $ 10 to B, he would then have three times as much as B ; but if B were to give ^ 5 to A, A would have four times as much as B. How much has each ? B. 1. Find the value of a (a + 6) (a + &+ c) - « (a - 6) (a — 6 - c), when a = 5, 6 = 3, and c = — 6; also when a = — 3, b = — 2, and c = 4. 2. Show that a'^ + 3(a - 2 6)2 = 3(a - 6)2 + (a - 3 6)2, and that 3. Find the factors of (i.) x^ - 4 a;2 + 4 x, (ii.) 2 x2 - 5x + 2, and (iii.) a;3 - a;2 + X - 1. 230 MISCELLANEOUS EXAMPLES III. 1 1 A 4. Simplify 1 , and show that ^ x-S x + 3 x2-9' \ y -X )\ y - X } \ y -X I 5. Solve (i.) -l- + -l- = -l-, (ii.)^ ' • ^ ^ x-\ x-2 a;- 3' ?.|^ = « hah 6. Find the fraction which becomes \ when 2 is added to its numerator, and \ when 2 is taken from its denominator. C. 1. Prove that (a;- l)2(y2 + i)_(a;2 +!)(?/_ 1)2 = 2(a; - y){xy - 1), and that (X + y)(a; + 2;) -a;2 = (y + 2;) (2/ + x) - 2/2 = (2; + x){z + t/) - 2^2. 2. Prove that the product of any two numbers is equal to a quarter of the difference of the square of their sum and the square of their difference. 3. Find the factors of (i.) 1 + 18 a; - 63 a;2, (ii.) Zxhj-2Ay^, and (iii.) (a;3 + 3x)2 -(3x^ + 1)2. 4. Simplify ^:-";^ + ^; . x^ — 46 X — 21 and (x-3)(x-2) (x-l)(x-3) (x-l)(a;-2) 6. Solve ,. V x — 3 X — 1 X + 1 X — 3 _ r. 00-^ r'^~Q 8~-"' (ii.) ax-\-hy = c, a^x + h'^y = c2. 6. A and B have $ 100 between them ; but if A were to lose one- half of his money, and B were to lose one-third of his, they would then have only $55 between them. How mucli has each ? MISCELLANEOUS EXAMPLES III. 281 D. 1. Find the value of ^/a^ + 63 + c^ - (a - 6 - c)2, when a = 3, 6 = — 3, and c = 4. 2. Show that (x-5)(a:-4)-3(x-2)(x-l)+3(x+l)(x-h2)-(x+4)(x+5)=0, and that 1 + n + ^n(n + 1)+ in(w + l)(n + 2)=l(n + 1)(/H- 2)(n + 3). 3. Find the factors of (i.) x^ + x^y -6xy^, (ii. ) x^ + ax'^ - a^x - a\ and (iii.) xV _ x'i - y^ + 1. 4. Simplify (i.) ^Lzi^ + a + x a-^-x^ a-x X — 4 X — 5 X — 6 6. Solve (i.)^-2C6-x) = ^-^-^2. (ii.) 3y + - + 6 = 0, y + - = 8. X X 6. A hare is pursued by a greyhound, and is 60 of her own leaps before him. The hare takes 3 leaps in the time that the greyhound takes 2 ; but the greyhound goes as far in 3 leaps as the hare does in 7. In how many leaps will the greyhound catch the hare ? E. 1. Show that {m + w)(m + ?i — 1)= m(m — 1)+ 2mu + n(n - 1), and that (m + n) (m + n - 1) (w + w - 2) = m(in - 1) (»» - 2) + 3wi(m-l)«4- Zmn{n- 1)4- m(h - l)(n-2). 2. Divide x* — y* by x — y, and from the result write clown the quotient when (x + yY - 16 2* is divided by x + y - 2 2. 232 MISCELLANEOUS EXAMPLES ill. 3. Find the factors of 6 a;2 + 24 a; - 6, ahd of a^ -2 abc - ab^ - ac\ 4. Find the L. C. M. of 8 a^ - 18 ah\ 8 a^ + 8 a^ft - 6 ah\ and 4 a2 _ 8 a& 4- 3 hK 6. Show that (i.)l + -^+ ^^ ^' X — a {x — a)(x-b} {x — a)(x- 6)' (ii.)l + _^+ ft^ ic — a {x — a)(x —"7)) (x — a) (» — 6) (x — c) ^ ^ (x — a) (a; — 6) (x — c) 6. A debtor is just able to pay his creditors 25 cents on the dollar ; but if his assets had been five times as great, and his debts two-thirds of what they really were, he would have had a balance in his favor of $ 700. How much did he owe ? F. 1. Find the value of {a-^b- c)}2 + {6 -(c - «)}2 + {c-(a- b)Y when a=— 1, 6 = —3, and c = — 6. 2. Divide x^ -2 a*x* + a^ by x^ + ax"^ + a'^x + a^ 3. Find the L. C. M. of x2 - 1, (x - 1)2, (x + 1)2, x^ - 1, and x3+l. 1 1 3x 4. Add together » » and : and show 2x + y 2x-y y^-ix'^ that (X - a) (y - a) ( x-b){y-b) (x-c)(y-c) _ (a - 6)(a - c) "^ (b - c)(6 - o) "^ (c - a)(c - 6) ~ * 6. Solve (i.) —^ + ^ ^ xfa X4-6 x + a + 6 (ii.) 5x + 2^ + 32 = 131 Sx-\-7y-z= 2 ^• x-2y + z= sJ MISCELLANEOUS EXAMPLES III. 233 6. When the arable land of a farm was let at $7.50, and the pasture at $10 an acre, the total rent of a farm was $2750. When the rent of the pasture was reduced by $ 1.25 an acre, and that of the arable land by $2.50 an acre, the whole rent was $ 1937.50. What was the total acreage of the farm ? G. 1. Solve the simultaneous equations X — a y — h _-. c a + c — 6 x + a — o, y — a _a — c a + c a — h a + c 2. Solve the simultaneous equations a b G 8. Solve the simultaneous equations ^/y + c/z __ c/z + a/x _ a/x + b/y _ <» ■ ^ . c i a b c X y z ' 4. A man rides one -half of the distance from A to B at the rate of a miles an hour, and the other half at the rate of b miles an hour. If he had travelled at the uniform rate of c miles an hour, he would have ridden the whole distance in the same time. Prove that cab 234 QUADKATIC EQUATIONS. CHAPTER XVI. Quadratic Equations. 175. Let it be required to state the following problem in the form of an equation. If the speed of a certain train were increased by 4 miles per hour, the time it now requires to make a trip of 120 miles would be decreased by one hour. What is the train's speed in miles per hour ? Let X = the speed of the train in miles per hour. Then 120/a;=the number of hours the train now requires for the trip of 120 miles ; 120/(a;-|-4) = the number of hours it would require for the same trip, were its speed increased as indicated, and by the conditions of the problem this is one hour less than 120/a; hours ; 120^ 120 ^ X a; -f 4 Multiplying both sides of this equation bya;(a; + 4), we have 120 a; + 480 = 120x + JB^ ^ 4^.^ whence ar* + 4 a; = 480. The problem of finding the speed of the train is thus reduced to that of solving an equation which contains both the first and second powers of its unknown quantity. QUADRATIC EQUATIONS. 235 Some equations of this class were considered in Arts. 128, 129. We now proceed to a more complete analysis of them. 176. A quadratic equation is an equation which contains the unknown quantity to the second, but to no higher power. Thus oj- = 4, 3a^ + 4 ic = 7, and aocF -\-bx-{-c = are quadratic equations. When all the terms are transposed to one side, a quad- ratic equation must be of the form aoc^ + bx -{- c = 0, where a, b, c are supposed to represent known quantities. It was shown in Art. 122 how the factors of the expression on the left side of this equation can be found by changing the expression into an equivalent one which is the difference of two squares. We first apply the method to some examples ; but the student will profit by reading Art. 122 again before considering them. Ex. 1. Solve x^ + 12 X + 35 = 0. Since x^ -{- 12x is made a perfect square by the addition of the square of half the coefficient of x, that is the square of 6, we add and subtract G^, The equation then becomes x:^ + 12 X + 62 - 62 + 35 = 0, that is (x + 6)2 - 1 = ; .-. {(x+ 6)-M}{(x + 6)- 1} = 0, that is (x + 7) (x -f 5) = 0. Hence x + 7 = 0, or x + 5 = 0. Thus X = — 7, or X = — 5. Ex. 2. Solve 3 x2 = 10 X - 3. By transposition we have 3x2- 10x + 3 = 0. 236 QUADRATIC EQUATIONS. We must first divide by 3 in order to make the coefficient of x^ unity ; the equation then becomes aj2_-i/a;+l =0. Half the coefficient of oj is — f , and therefore x"^ — ^f x will be made a perfect square by the addition of (— |)2 ; and adding and subtracting (— f)^, that is ^., the equation becomes a;2 _ Y X + -2g5 _ 2^5 +1=0; .•• (x--|)2--V- = o, that is (a;- 5)2 _ (1)2 = 0; ... |:«-5 + 4||x-f-|} = 0; ... (a:-i)(x-3)=0. Hence a; — i = 0, or x — 3 = 0. Thus X — J , or ic = 3. Note. — In many cases the factors can be -written down at once, as in Arts. 117 and 118, without completing the square. The student should always see if this can be done : he will thus save himself a great deal of unnecessary work, Ex. 3. Solve 4 X - a;2 = 2. By transposition, we have 4 x - x2 - 2 = 0. Change all the signs so as to make the coefficient of «* unity ; then we have jk2- 4x4-2 = 0. Add and subtract 4, the square of half the coefficient of x. Then x2-4x + 4-4 + 2=0; ... (X -2)2-2 = 0; that is (x - 2)2 - ( ^2)2 = ; .-. (x-2-f V2)(«-2-V2)=0. Hence x - 2 + v^2 = 0, or x - 2 - ^2 = 0. Thus X = 2 - v'2, or X = 2 + V^- QUADKATIC EQUATIONS. 237 EXAMPLES LII. Solve the following equations : ^1. a:2_4x = 45. 22. 17x2 + 8 = 70 a;. 2. x2-6x = 91. 23. 51x2 + 19x = W. ''• (x - 3)2 = a; + 3. 24. 110x2 -21x + 1=0. 4. (x - 4)2 = re _ 2. 25. 21x2-13x = 20. 5. (x-l)(x-2) =20. 26. 21x2 + 23x = 20. 6. (x + l)(x + 3)=2(x + 3). 27. 6x2 + 6 = 18x. 7. ix-\-3 = x(x + 2). 28. 6x2 = 5x + 1. 8. 4x-S=x(2-x). 4^9. 9x2 -63x + 68 = 0. 9. (x+l)3=(x-l)3 + 26. 30. 16x2 + 3= 16x. 10. (x-l)3 = (a: + l)3-56. 31. 29x2-41x-138 = 0. 11. 9x2 4-6x + l=0. 32. 29x2 + llx = 138. 12. 9x2+ 16 = 24x. 33. x2-16 = 216-10x. 13. x2+150 = 53x. ^^ pr84. (x + 2)2 = 4(x - 1)2. -^14. x2 = 2x + 99. 35. (x + 6)2 = 16(x - 6)2. 15. 3x2 + 3 = 10x. 36. (x + 8)2 = 9x2. 16. 3x2+llx = 20. 37. (X - 7)2 = 49(x + 2)2. 17. 4x2+ 21 x= 18. , 38. x2 - 3 ax + 2 a2 = 0. -4«- 6x2 + 55x = 50. 39. x2 + 3 a2 = 4 ax. 19. 24x2 = 30x+75. 40. x2 + 2 a& = 62 + 2i,rtx. 20. x2-200 = 35x. 41. 4x2 + 4ax = 62_rt2. 21. 19x2- 39x + 2 = 0. 42. x2 + 2 ax = 62 + 2 ab. 177. To solve the equation ax^ 4- 6a; + c = 0. Divide by a, the coefficient of x-, the equation then becomes 238 QUADRATIC EQUATIONS. Now add and subtract the square of half the coefficient of X, that is the square of - — Then we have Z a a \2ciJ \^«/ « The first three terms are now a perfect square, namely, Hence we have lx-\ i — i — -„ 1=0, V 2aJ \4.a^ a) ""'■ (-.^)'-{x-^)r=»- Hence [Art. 115] We therefore have 2 a 2 a or else a, + 1. + ^^111^ = 0. 2a 2a Thatis . = -^ + vfci^=-*±y(^^l4ac) 2a 2a 2a - _ b v(feg - 4 ac) _ - 6 - V(6' - 4 ac) or X — — 7? — — '~7z — pj * 2a 2a 2a It follows from the above that a quadratic equation has two and only two roots. QUADRATIC EQCJATIONS. 239 178. Instead of working out every example from the beginning, we may use the general formula found in Art. 177, and substitute for a, h, and c their values in the particular equation we are considering. Thus, the roots of the equation ax^ + 6x + c = being 2a 2a we find the roots of 3 a;'^ — 10 a; + 3 = by putting 3 for a, — 10 for &, and 3 for c in the above formula. Hence the roots of 3 x-^ — 10 x + 3 = are - 10 VCIOO - 36) 6 6 that is ^^- ± f ; and hence the roots are 3 and \. 179. General Properties. In discussing the quadratic equation ax^ -\-hx-{-c = 0, it is assumed, unless a contra- dictory hypothesis be made, that the coefficients a, 6, c are finite and different from zero ; and the general prop- erties of this equation are enunciated under this limi- tation. In Art. 177 we found that the equation ax^ + 6a; + c = had two roots, namely, 2a 2a 2a 2a (i.) If b^ — Aac be positive, ■y/{b^ — 4ac) is the square root of a positive quantity and has a numerical value, either rational in the form of an exact square root, or irrational in the form of an indicated square root. The quantity ^^(6^ — 4ac) is here said to be real The roots of the quadratic equation are then also real, and they are obviously different from one another. Also, unless 240 QUADRATIC EQUATIONS. b^ — Aac be positive, V(^^ — 4ac) has no numerical value and is not real. Hence : In order that the roots of aa^ + bx + c =0 may be real, it is necessary and sufficient that 6^ > 4 ac. From the formula of Art. 177 it is clear that the roots of the equation aa^ -\- bx -\- c = are irrational unless 6^ — 4 ac is a perfect square. It is also clear that if either of the roots of a quad- ratic equation be irrational, they are both irrational. Let- the student here call to mind the meaning of the expression irrational quantity. It is a quantity whose exact value cannot be expressed either as an integer, or as a fraction whose numerator and denominator are integers. In algebra it is usually expressed symbolically, as in the case under consideration by means of the sign V- [See Art. 20.] (ii.) If 6^ — 4ac = 0, both roots reduce to — 6/2 a, and are thus equal to one another. In this case we do not say that the equation has only one root, but that it has two equal roots. Also, the roots are unequal unless b^ — 4:ac = 0. Hence : In order that the roots of aa^ -}-bx-{-c = may be equal to one another, it isnecessary and sufficient that b"^ — 4:ac = 0. When 6^^ — 4 ac = 0, the expression ax^ -\-bx-^c is a perfect square so far as x is concerned; for ax^-{-bx-\- c. that is, aly? -\--x-\-^\ will then be aficH-— -], 7,2 , \ a ^y V 2 a/ since 4a^ a (iii.) When &^ — 4ac is negative, y(6* — 4ac) is the indicated square root of a negative quantity, and has no QUADRATIC EQUATIONS. > - " Z41 numerical value. It is therefore excluded from the category of so-called real quantities, and is called an imaginary quantity. [Art. 228.] Hence : In order that the roots of ax^ -{-bx-\- c = may be imagi- nary, it is necessary and sufficient that 6^ < 4 ac. It should be noticed that if either of the roots of a quadratic equation be imaginary, they are both imaginary. 180. The above criteria of the character of the roots of a quadratic equation are more concisely stated thus : (1) If 6^ — 4ac be positive, the roots are real and unequal. - ' (2) If 6^ — 4ac = 0, the roots are equal. (3) If 6^ — 4ac be negative, the roots are imaginary* and unequal. (4) If 6- — 4ac be a perfect square, the roots are rational and unequal. (5) If 5^ — 4 ac be not a perfect square, the roots are irrational and unequal. 181. By multiplying both sides of an equation by the same quantity, we do not destroy the equality, and therefore the new equation is satisfied by all the values which satisfy the original equation; if, however, we multiply by any integral expression which contains the unknown quantity, the new equation will be satisfied not only by the values which satisfy the original equation, * Complex quantity is a mor e gen eral and more accurate name for expressions of the form m-\- v — n. The term imaginary is then ap- plied only to expressions of the form V — n. [See Chapter XXII.] 242 QUADRATIC EQUATIONS. but also by any value which makes the expression by which we multiply vanish. For example, the equation x^ = 9 is satisfied by the values X = 3, or a; = —3. If we multiply both sides of the equation by x — 2, we have the new equation x%x - 2) = 9(x - 2) ; and this new equation is satisfied not only by x = 3, and by X = — S, but also by x = 2. Thus additional roots are introduced whenever both sides of an integral equation are multiplied by any inte- gral expression which contains the unknown quantity. When an equation contains fractions in whose de- nominators the unknown quantity occurs, the equation may, however, be multiplied by the L. C. M. of the denominators without introducing any additional roots ; for we cannot divide both sides of the resulting equation by any one of the factors of the L. C. M. without re-intio- ducing fractions. The student must, however, be careful to multiply by the lowest common multiple of the denominators, for otherwise the resulting equations will have roots which are not roots of the original equation. We may, for example, multiply both sides of the equation 2x 10 ^ 7 x-l x2-l x + 1 by a;2 — 1, the L. C. M. of the denominators of the fractions ; we thus obtain 2a;(.x+l)-10 = 7(x-l), and this new equation is not satisfied by either of the values obtained by equating to zero the factor by which we have multi- QUADRATIC EQUATIONS. 243 plied, and hence no additional roots have been introduced by the multiplication. If, however, we multiply the original equation by (x— l)(a;2— 1), we shall have 2x(x + l)(x - 1)- 10(x - 1)= 7(x - 1)2, and one of the roots of this equation, namely x = 1, is not a root of the original equation. Note. — The student should always remember that when both sides of an equation are divided by any factor which contains the unknown quantity^ the resulting equation will not give all the roots of the original equation ; to find the remaining roots we must equate to zero the factor by which we have divided. For example, if we divide both sides of the equation (x2-4)(x + 2) = (x2-4)(2x + l) by x2 — 4, we obtain the equation x+2 = 2x-l, from which we obtain x = 3. But x = 3 is not the only solution of the original equation ; to obtain the other solutions we must equate x^ — 4 to zero. EXAMPLES LIU. Solve the following equations : 1. . + i=| ^e. X 2 X 6 3. ? + ?^^. 8. 2 X 3 —4. x+ 5 4 _3 4 x+ 5 2 6 " x + 1 x + 10 3x-3 5. _£_ + ^ZlI^13. 10. x + 2^ x + 2. x+1 -29. 10* 1 x-2 2 x + 2 _3 5* ) x + 4 1 4-x _5 "3* ^.^ 4 x-f 1" ::3. 5 10 2 244 QUADRATIC EQUATIONS. 11 1 I 1 I 1 =0 17 2a;-l 13^ 3a; + 5 "2 3 + x 2 + 3a; * *2a; + l H 3x-5 12. 1+1+^=0. 18. 2x^ + 3-30. 3 x + S x + 11 2ic-3 3a:-2 8x-12 13. 3^^-2^±i=6. 19.-^^ l_ + _L. = o. x+l x-1 x-2x + 32-a; 14. _l-+-^=^i-. 20. ^ 6_ J[9_^0 x-1 x-S x+3 x-3 x+2 3-x 15 4x + 3 _ 6a;-13 7x-46 ^i a;-lx+l _ 6x 9 18 5x + 3' 'x + 1 x-1 x2-l' -« 2x-3 x-5 , 5x-16 „„ 1 , 1 _ 9 •••O' : — = ~T7: — r 12 x-1 x2-3x x2 + 4x 2x2 1 1 13 23. x^-Sx 9-x2 16x 24. ^ +-^+ ^^ =0. 3x-6 4-x2 72(x+2) ^ 26. ^-^ = ^1 27.1+ 1 xa_i l_x 8 x+1 X x + 4 x+l x + 2 1 3_ _ J 1 gg X x+l _ x-2 x-1 aj2_4 2-x 3x + 6' 'x+l x+2 x-1 x „o X a;- 3, X x + 3_2 x-3 X x + 3 X 3 80 a^ + 3 ■ x-6 _ x + 4 x-5 *x+l x-4 x+2 x-3 gjX-3 x-4_x-6 x-7 X — 4 X — 6 X — 7 X — 8 32. x2 + 2a2 = 3ax. 35. a(x2 + 1) = x(o2 + 1). 33. 9x2 - 6 ax = a2 - 62. 36. x2 - 2(a - 6)x + 62 = 2 ab. 84. a(x2 - 1) + ic(a2 - 1) = 0. 37. x2 + 2(6 - c)x + c2 = 2 6c. 38. (6-c)x2 + (c-a)x + (a-6) = 0. QUADRATIC EQUATIONS. 245 39. (a-{-b)x^ + cx-a-b-^c=0. 41. bcx^-\-(b^ + c^)x + 6c = 0. 40. abx^-(ia^ + b^)x + ab = 0. 42. (a^ - b^) (^x^ - 1) = ^ abx. 43. (62 _ a2)(x2 + 1) = 2(a2 + 62)^;. 44. a X 49. a; , « _ 6 , a^ a ic a 6 45. a +x a — X = 4. 60. ic+a+6 X a 6 46. X — a x — b a b 51. 1 111 X — a — b X a b 47. a ^ 6 X — a X— b 6 a 52. 1 + 1 -1 + 1 . x-\-a x-\-b c-^a c+6 48. a2 1 62 X— b X — a = a + b. 53. X X __ c c x-\-a x-^b c+a c+6 g^ x + g ■ x + & _ a; — g . x — b X — a x — b x-\- a x + b 55. ^ +-^= "^"^ + ^ + ^ x+a x+6 x+«— c X+6+C 182. Irrational Equations. An irrational equation is one in which square or other roots of expressions con- taining the unknown quantity occur. In order to rationalize an equation it is first written with one of the irrational terms standing by itself on one side of the sign of equality : both sides are then raised to the lowest power necessary to rationalize the isolated term ; and the process is repeated as often as may be necessary. Ex. 1. Solve the equation y/(x^ — 9)+ x = 9. By transposition, we have V(x2 - 9) = 9 - X. 246 QUADKATIC EQUATIONS. Square both sides ; then x2-9=(9-x)2j .-. 18x = 90; .-. x = S. Ex. 2. Solve the equation V(2 X + 8)- 2 V(x + 5) = 2. Squaring both sides, we have 2 « + 8 + 4 (a; + 5) - 4 v(2 X + 8) V(a; + 5) = 4, that is . eic + 24-4v'(2a: + 8)V(a^ + 5) =0; .-. 3x4- 12 = 2V(2x + 8)v'(a;+5). Square both sides of the last equation, and we have 9x2 + 72x + 144 = 4(2a; + 8)(x + 5); .-. a:2 - 16 = 0. Hence a; = 4, or x = — 4. Note. — If we put 4 for x in the original equation, we should have vi6-2>/9 = 2, thatis4-6 = 2, which is not true. Again, if we put — 4 for x, we should have V0-2V1=2, which is not true. Thus neither of the values we found for x appears to satisfy the equation. The failure is due to our having supposed that ^(2 a; + 8) and ^(x + 5) were necessarily positive, whereas every algebraical quantity has two square roots, one positive and the other negative, and the square-root symbol does not necessarily represent one only to the exclusion of the other. Bearing this in mind we shall find that a; = 4 does satisfy the equation, if the ambiguous signs be inserted ; for the condition is ±y/\Q- 2(i:V9) = 2, that is ±4 -(±6)= 2, which is true when the lower signs are taken. QUADRATIC EQUATIONS. 247. If an irrational equation is to be solved on the supposition that the radicals it contains are to he necessarily positive {or necessarily negative)^ the equation will he accompanied by a definite statement to that effect. Ex. 3. Solve the equation y/i2x + 7) + y/{Sx - 18) + V(7a; + 1) = 0. Since the sum of three positive numbers cannot be zero, this is not a possible equation unless one or more of the square roots involved in it be negative. By transposition, we have Square both members ; then 2 a; + 7 + 2 V(2 X + 7) ^(Sx - 18) + 3x - 18 = 7 a; + 1. By transposition and division by 2, we have V(2 X + 7) V(3x - 18) = X + 6. Square both members ; then (2a; + 7)(3x-18) =(rc + 6)2; .-. 5x2 - 27 X- 102 = 0. Hence x = 9, or ic = — -U. By applying the test for a solution [Art. 91] it will be found that a; =3 9 is a solution when -^(7 x + 1) is taken negatively, and that X = — 18/5 is a solution when y/(Sx — 18) is taken negatively, the two remaining square roots being taken positively in each case. The process of rationalizing equations introduces factors that contain the unknown quantity. Hence, in solving irrational equa- tions it is important to apply the test for a solution. [See Art. 181.] Ex. 4. Solve the equation + V(^'-9)-(x-9)=0, wherein + y/(x^ — 9) indicates that only the positive square root is to be taken. Multiplying by + ^(x"^ - 9) + (x — 9), we have x2-9-x2 + 18x- 81 =0; whence 18 x = 90, and x = 5. 248 QUADRATIC EQUATIONS. Btk 5 is not a root of and it is a root of the equation + V(a^'-9) + (x-9)=0, by whose left member the original equation was multiplied. It may be shown that there is no finite number that will satisfy the equation + V(^'-9)-(^-9)=0. EXAMPLES LIV. Solve the following equations, taking account of both positive and negative square roo.ts of the quantities under the radical sign. Apply the test for a solution in each case. 1. ^(x-S)=S. 14. y/(0-^4x)=2x-S. 2. ^(x-7)=4. 15. 3a;=5+V(30x-71). "" 3. ^(Sx+l)=i. 16. 2x-^y/x=-S. ^4. 5V(a; + 7)=4V(3a;-2). 17. a; + 3 + VC^k + 3)= 6. 6. 3V(x + 3)zz2V(3x + 6). 18. 2a; + 1 = V(6 x + 3). ^. V(a5 + 5)=2V(3a; + 4). -^9.. 7a-=V(3a;-ll)+33. '7. 3V(a;-4-7)=5v(3x-2). ^^ ^{x -{- 10) + ^(x ■]- l)=l. 8. y/(x-{-2) = x. 21^ Va; + V(5a;+1)=5. 9. y/(x-\-20)=x. 22. y/x-i-y/{6x-\-l)=2. 10. x-{-^(x + l)=6. A2S. V(2a; + 9)-V(aJ + 4) = a.rtf 11. x-^ix + 2) = 4. 24. V(7a;4-l)-V(3a:+10) = l. 12. a;-2 + 3V(«-2)=0. 25. ^{2x-{-U)- y/i2x-6)=2. 13. x-5 + 2V(aJ-5)=0. 26. y/(4x -{- I)- ^(x -\- S)=2. 27. V(8a;+5)-2V(2x+l)=l. 28. V(3-' + 4) + V(a^+20)-2V(x+ll) = 0. 29. V(4x+l)-V(a; + 3) = V(»:-2). QtJADRATIC EQUATIONS. 249 30. V(2« + 4) + v(3x + 7) = V(12:c + 9). 31. ^(Gx + 1)+V2{1 - x) = ^0 X + Q). 32. 2V(2x + 3)-V(5x + l) = V2(a;-l). 33. V(x - l)(a; - 2) + V(x- 3)(x - 4) = V2- 35. v/(a^+l) + V»^= ^ V(l+«) 36. VCS + a;)^^/^^^-^- 37. 38. V(a - a;) + V(^ -x) = V(« + 6 - 2 X). 39. V(««+^^) + \/(^^ + «^)=«-^- 40. V(« + »') + V(^ + a;) = V(« + ft + 2x). >*-41. V(«-3;) + V(^-^) = \/(2« + 26). 183. Eelations between the roots and the coefficients of a quadratic equation. We have seen [Art. 122] that x-+px-{-q can always be expressed in the form (x — a) (x — (3). We have also seen [Art. 128] that if a^ + px + q = (x — a) (x — P), then a and fi are the roots of the equation a^+px-i- g = 0. But, if a;2 -{-px -{'q = {x — a){x — P) = x^ — (a-\-/3)x + ajS, we have a + ^ = — p and a/3 = ^. 250 QUADRATIC EQUATIONS. Hence in the equation 7? +pa; + g = 0, the sum of the roots is —p and the product of the roots is q. The equation aa^-f 6a;+c=0 becomes, on dividing by a, a a Hence, from the above, if a and ^ be the roots of aoi? + 6a; -|- c = 0, h c we have a-^ 3 = and a^ = - • a a The above relations between the roots of a quadratic equation and the coefficients of the different powers of the unknown quantity are of great importance. Analogous relations hold good for equations of the third and of higher degrees. [See Treatise on Algebra, Art. 129.] These relations may also be deduced as follows : We have found in Art. 177 that the roots of ax^ -\-bx-\-c = b ■y/(b''-4.ac) are — o " ± o ' 2a 2a Call these roots a and fi respectively ; then « = __+VV___^^ and R = — - ^• '^ 2a 2a QUADRATIC EQUATIONS. 261 By addition and by multiplication we then have a Again, the quotient of a-\- /3hj aft is ajS fc p" c' and observing that (a-\-fiy-2a/3=a' + p' and Wl3j afi~a'^/i'' we find 2 ^2 b^ 2c b^-2ac and 11 W 2a W-2ac 184. Special Forms. If one or more of its coefficients vanish, the quadratic equation and the formula for its roots become simplified. (i.) If c = 0, the equation reduces to X {ax -\- b) = 0, the roots of which are and — b/a. (ii.) If c = and also 6 = 0, the equation becomes aa^ = 0, both roots of which are zero. (iii.) If b = 0, the equation reduces to ax^ -J- c = 0, 252 QUADRATIC EQUATIONS. the roots of which are -\--\J —cja and —-\/—c/a. The roots of a quadratic equation are therefore equal and opposite when the coefficient of x is zero. (iv.) If a, b, and c are all zero, the equation is clearly satisfied for all values of x. (v.) The case in which either a = 0, or a = and 6 = (c not zero), is best studied through the general expressions for the roots of ax^ + 6a; + c = 0, but in the changed form 2c 2c -b--y/(b'-4.acy _6+V(6'-4ac)* That these expressions are respectively equal to -6 + V(6'-4ac) -6-V(6'-4ac) 2a ' 2a follows from the identity J - 6 + V(^' -4.aG)\{-b - V(^' - 4ac) i = 4ac. If now a = 0, one root becomes — c/b, and the other 2c/0. And if a = and 6 = 0, both roots become 2c/0. A fraction, such as 2 c/0, whose numerator is a finite quantity and whose denominator is zero, was called, in Art. 168, an infinite quantity, or infinity, and was denoted by the symbol oo. Thus 2 c/0 = oo. In accordance with this definition and with the results here obtained, when it becomes necessary to interpret the equations 0.aj2-f 6a; + c=0, QUADRATIC EQUATIOTSIS. 253 we say that the former has one infinite root, the latter two infinite roots. Such equations present themselves in investigations in analytic geometry. Ex. 1. Solve the equation VoxTl + Vx + 4 = 3. Squaring both members, we have (a+l)a; + 5 + 2\/(ax+ l)(a; + 4)=9; then transposing the rational terms to the right side and squaring again, we obtain 4(ax + 1) (a; + 4) = 16 - 8(a + l)x + (« + 1)^x2, whence, by the proper reductions and transpositions, (a-l)2x2_i2(2a + l)x = 0. Hence, one root is x = 0, and for this one both square roots must be taken positively. A second root is 12(2 g+l) and it requires Vx + 4 to be taken negatively. If a = 1, the coefficient of x"^ is zero, and the second root is infinite. Ex. 2. Solve the equation y/(iax) + y/{x + l) = y/b, and determine for what value of b one root is zero, and for what value of a one root is infinite. 185. Equations with given Eoots. Although we cannot in all cases find the roots of a given equation, it is very easy to solve the converse problem, namely, the problem of finding an equation which has given roots. For example, to find the equation whose roots are 4 and 5. We are to find an equation which is satisfied when x = 4, or 254 QUADRATIC EQUATIONS. when X = 6; that is, when ic — 4 = 0, or when x — 5 = ; and in no other case. The equation required must be (iC-4)(x-5)=0; for this is an equation which is a true statement when a; — 4 = 0, or when x — 6 = 0, and in no other case. If we get rid of the brackets by multiplying out, the equation becomes ic2_9a; + 20 = 0* Again, to find the equation whose roots are 2, 3, and — 4. We have to find an equation which is satisfied when x — 2 = 0, or when x — 3 = 0, or when x + 4 = 0, and in no other case. The equation must therefore be (a;-2)(x-3)(x + 4)=0, that is x^-x'^-Ux-\-2i = 0. Similarly, the equation whose roots are 0,-1, and — i is x(x+l)(x+^)=0, that is «»+ 1x2+ ia; = o. Ex. 1. Find the equation whose roots are the squares of the roots of the equation x"^ + 5 x — 7 = 0. Let a, /3 be the roots of the given equation ; then a", ^^ will be the roots of the required equation. Hence the required equation is (a;_a2)(x-^) = 0, that is x2 - (a2 + /32)x 4- a2/32 = (i.) * The equation a;2 _ <) a. -|- 20 = is certainly an equation with the proposed roots; but to prove that it is the only equation with the pro- posed and with no other roots we must assume that every equation has a root. If the equation x^-{-7 x^ — 2 = 0, for exam[)le, had no roots, then (x — 4) (x — 5) (x** 4- 7 x2 — 2) = would be an equation with the pro- posed roots and with no others. Tlie proof of the proposition that every equation has a root is too difficult for an elementary book. QUADRATIC EQUATIONS. 266 We have therefore to find a'^ + p^ and a^^. Now, by Art. 183, we have a+/9=-5, and aj8 = — 7. Hence a2 + ^2 ^^(a + /3)2 _ 2a/3 =(- 5)2 - 2(- 7) = 25 + 14 = 39 ; also a2/32 = 49. Substituting in (i.), we have the required equation, namely ic2_39aj + 49 = o. Note. — We might have obtained the required equation by finding the roots of the given equation, and squaring them ; but it is best to use the relations found in Art. 183. Ex. 2. If a, /3 are the roots of the equation ax^ + 6x + c = 0, find the equation whose roots are - and ^• The required equation is 0. 6^_2c a , _ cC^ a ... a2 + /32 ^62 ^ 0/3 ac Hence the required equation is \ac J or, multiplying by ac, acx^ - (62 _ 2 ac)x + ac = 0. (^-^)Kf)-' that is .2-(«^^).H-l = Now a . ^_a2+^2^ /3 a a^ ' and, by Art. 183, we have a^ = - and a2 -f /32 : 256 QUADRATIC EQUATIONS. Ex. 3. Show that no numerical value of x will make x^— 4x4- 5 zero ; and find its least value. Since x^ - 4 x + 5 = (x - 2)2 + l, and (x — 2)'^ is always positive,. 1 + (x — 2)2 is always greater than 1, except when x — 2 = 0, and then it is equal to 1. Thus x2 — 4x + 5 can never be zero, and its least value is 1. EXAMPLES LV. Write down the equations whose roots are 1. 2 and - 2. 8. and - 4. 2. 3 and - 4. 9. 6, - 3 and 0. 3.-3 and - 2. 10. V2 and - ^2. ^^4. \ and -\. 11. V^ and - ^5. 6. \ and — \. 12. 0, ^3 and - ^3. 6. - i and - \. 13. 2 - ^3 and 2 + V3. 7. and 3. 14. 5 - ^7 and 5 -f ^7. 15. a — y/h and a + ^h. 16. Write down the product of the roots of each of the follow- ing equations : (i.) x2 + 4x+l=0, (iii.) 3x2 + 5x-7 = 0, (ii.) x2 + 7 X - 2 = 0, (iv.) 5x2 - x - 1 = 0, (v.) 9 ax2 + 3 6x + 4 c = 0. 17. Write down the sum of the roots of each of the following equations : (i.) x2 - 4 a2 = 0, (iii.) x2 - 5x + 3 = 0, (ii.) x2 + 3x - 5 = 0, (iv.) 2x2 - x - 7 = 0, (v.) 6ax2 + 76x + 8 = 0. 18. Show that, if a, /9 be the roots of ax2 + 6x -f c = 0, then 1^1= V(&''-4gc) a^/3 c QUADRATIC EQUATIONS. 267 19. Show that, if a, /3 be the roots of the equation aa;2+ 6x+ c=0, then (^ (iii.) ^+^^ = ^i^4|%c 20. Find the sum of the squares of the roots of the equation / x2 + 4 X + 2 = 0. ^ 21. Find the sum of the squares of the roots of the equation a;2 + 4pa; + p2 _ q. ^ 22. Show that the sum of the squares of the roots of the equa- tion -^ -\- ax -\- h — ^ is the same as the sum of the squares of the roots of the equation 0?-- t U. x2 + 3 ax + 6 + 4 a2 = 0. 23. For what value of a are the roots of the equation M 3x2 + 4x + a = / equal to one another ? 24. For what values of a are the roots of the equation 4x2+(l + a)x+ 1 =0 equal to one another ? 25. Find the value of a in order that One of the roots of 100 x2 + 60 X + a = may be double the other. 26. Show that one of the roots of x2 + px + g = is double the other, if 9^ = 21)2. 27. Show that, if a, /3 be the roots of 2x2 - 5x + 3 = 0, the equations whose roots are - and " is6x2— 13x + 6 = 0. ^ a 28. Show that, if a, /3 be the roots of 2x2 - 15x + 4 = 0, the equations whose roots are - and ^ is 8 x2 — 209 x + 8 = 0. o 258 QUADRATIC EQUATIONS. 29. Form the equation whose roots are a + /3 and - 4- - , where a ^ a, /3 are the roots of x^ — 11 a; + 22 = 0. 1/ 30. If a, /3 be the roots of x^ -\- 7 x -\- 9 = 0, find the equation whose roots are ?Lz_£ and ° "^ ^ . a /3 31. Show that the roots of x^ - (p2 _ 2 g)x + g2 _ q are the V^ squares of the roots oi x^ -{- px + q = 0. 32. Find the condition that the roots of a'^x'^ + 6% + c^ = may be the squares of the roots of ax'^ -{- bx -\- c = 0. 33. Show that, if a, j8 are the roots of px"^ -\- qx -\- r = 0, the equa^ tion whose roots are a + /3 and "^ is pqx^ + {pr + q'^)x-^ qr = 0. a + )8 34. Show that, if a, /3 are the roots of ax^ -{- bx -\- c = 0, the equation whose roots are a/3 and — is acx^ — (a^ + c^)x-^ac = 0; a^ 1 show also that the equation whose roots are a + iS and — —^ is aboc^ -f (a2 + 62)^ + a6 = 0. 35. Show that, if the difference between the roots oix^-^ax-{-b=0 be equal to the difference between the roots of x'^ -^ px -\- q = 0, then a2-46=j32_4g. V 36. Show that, if a, j3 be the roots of x^ -\- px -\' q = 0, the equa- tion whose roots are (a + /S)^ and (a — /3)2 is x2 _ 2(p2 _ 2 (2')ic + p2(p2 _ 4 g) = 0. 37. If a, /3 be the roots of x^ -{■ px -\- q = 0, show that the equation whose roots are a 4- - and /3 + - is P a qx^ + p(l + (?)a; + (1 + qy= 0. 38. Show that, if o, /3 be the roots of ax^ -\- bx -^ c = 0, the roots of the equation (2 b^ + ac)x^ + 3 abx + a2 = are ^ and ^ o-f-2i8 /3H- 2a QtTADUATlC EQUATIONS. 259 39. Find the relation that must exist between k and the coeffi- cients of the equation ox^ + 6x + c = 0, in order that one root may be k times the other. 40. Determine the relation that must exist between a and 6, in order that one of the roots of the equation ^{ax) + y/(x + a) = ^{x + h) may be zero, the value of a that will make one of its roots infinite, the value of a that will make both of its roots infinite, and the relation between a and h that will make its two roots equal. 260 EQUATIONS OF HIGHER DEGREE. CHAPTER XVII. Equations of Higher Degree than the Second. 186. It is beyond the range of this book to show how to solve equations of higher degree than the second, when the equations are in their most general forms; we give however some easy examples of such equations. Ex. 1. Solve a^ _ 6 a;2 + 8 = 0. Here the equation contains only two powers of the unknown quantity, one of which is the square of the other ; we therefore proceed as in Art. 176. We have a;* -6 a:2 + 9-9 + 8 = 0, that is (a;2 _ 3)2 - 1 = 0. Hence {(x2 - 3) - 1} {(x^ _ 3) + 1} = ; . •. x2 — 4 = 0, giving a; = ± 2, or x2 — 2 = 0, giving x = ± y/2. Thus there are four roots, ±2, ±y/'2,. Ex. 2. Solve (a:2 + x)2 + 4(a;2 + x) - 12 = 0. In this, as in the former example, the unknown quantity occurs in two terms, one of which is the square of the other. In all such cases we can bring all the terms over to the left side of the sign of equality, and then resolve the expression on the left into two fac- tors as in the preceding chapter. In the present instance the factors can be seen by inspection. For ^2 + 4^_12=(^ + 6)(^-2); therefore the equation may be written {(x2 + a;)+6}{(a;2 + x)-2} = 0. EQUATIONS OF HIGHER DEGREE. 261 Hence x^ -{- x -\- Q = 0, or x^ + x - 2 = 0. The roots of x^ + x + 6 = are - J^ ± '^ - The roots of x^ + x - 2 = are 1 and - 2. Hence the given equation has the four roots 1, -2, _J±^V^=^. Ex.3. Solve _^ + ^±2 = 37. x + 2 x2 6 Here one of the terms in which x occurs is the reciprocal of the Other. If we put y = , we find y from the quadratic equation x + 2 y 6 or, multiplying by 6 y, 6«/2_37 2/ + 6 = 0, that is (6y -l)(y -6)=0. = 6. X — 2 = 0, the roots of which In the second case, we have x2 - 6 X - 12 = 0, the roots of which are 3 ± ^y^2l. Hence the given equation has the four roots, \ I, -1, 3±V21. Ex. 4. Solve 4 x2 - 6 X + 3^(2 x2 - 3 x + 7) =r 30. Equations of this form, in which the ratio of the coefficients of x^ and x in the terms under the radical is equal to the ratio of the coefficients of x^ and x in the terms outside the radical, are not uncommon. Hence y = ■h ory = 6. Thus x2 x + 2 .1 6' or - X X2 + 2 In the first case, we have 6 X2 - X - -2 = are f and - h 262 EQUATIONS OF HIGHER DEGREE. To solve the equation, put V(2x2-3x+7)=2/, then 2a;2-3x + 7 =1/^ 2a;2_3x = ?/2- 7, and 4a;2_6ic = 2^2_ 14. Hence, from the given equation, we have 2y2_ 14 + 3?/ = 30; that is 2y^ + Sy -U = 0, that is (y -4)(2y +n)=0; hence y = 4, or y = —^^ ; . •. ?/2 _ iG, or ?/2 = ifi. Since y^ ^^x"^ — Zx-\- 1 ,^e. have I. 2ic2-3x+ 7 = 16, that is 2 ic2 _ 3 X - = 0, that is (2 « + 3) (x - 3) = ; .': « = 3, or X = — |. II. 2x2-3x4-7= ifi, that is 2 x2 - 3 X - 'V' = 0, the roots of which are - ± a/1^. 4 4 187. It has been shown [Art. 148] that if a rational and integral expression contiiiniiig x vanishes when any particnlar value a is given to x, then a; — a is a factor of that expression. For example, x^ — 7 x + 6 vanishes when 2 is put for x ; there- fore, by the above theorem, x — 2 is a factor, and by division we find that a;3 _ 7 a; ^. 6 = (X - 2) (x2 + 2 X - 3) . Again, x' — 4x2 + 2x+ 1 vanishes when x = 1 ; therefore « — 1 is a factor, and by division we find that a;8 _ 4x2 -j- 2x + l = (a; - l)(x2 - 3x - 1). EQUATIONS OF HIGHEK DEGREE. 263 188. It follows from the theorem enunciated in the last article that when one root of an equation is known, the degree of the equation can be lowered. By the application of this principle, the roots of the important equation a:^ — 1 = can be found. Ex. 1. Solve the equation x^ — 1 = 0. Since x^ - 1 =(x - l)(x^ + x + I), we have (x - 1) (x^ + x 4- 1) = 0. Hence x = 1 ; or else x^ + x + 1 = 0, the roots of which ai*e — ^ i v/(— |). Hence there are three roots of the equation xi^ = 1, one being real and the other two imaginary. Thus tliere are three quantities whose cubes are equal to 1 ; that is, there are three cube roots of 1, which are 1- -i + V(-f) and -i-V(-l). [See Treatise on Algebra, Art. 139.] Ex. 2. One root of the equation x^— 7x-\-G = is 2; find the other roots. Since x^ — 7x4-6 vanishes when x = 2, x — 2 must be a factor, and by division we find that x3 _ 7 X + 6 = (X - 2)(x2 + 2 X - 3). Hence (x - 2) (x2 -f 2 x - 3) = 0. Hence the other roots of the equation are those given by x2 + 2 X - 3 = 0, that is (x + 3)(x-l)=:0. Thus the cubic equation x^ — 7x + 6 = has the three roots 1, 2, and - 3. 264 EQUATIONS OF HIGHER DEGREE. Ex. 3. For what values of x will and jc*-8x3+ 10x2- 14x + 3 both vanish ? If both expressions vanish for the same value of x, say x = a, they will both have x — a as a factor. Now we can find the com- mon factor of any two expressions by the ordinary process of finding their H. C. F. In the present case, the H. C. F. will be found to be x2 - 5 X + 3 ; and since x^ — 5 x + 3 is a factor of both expressions, and is their highest common factor, both expressions will vanish for the values of x given by x2 - 5 X + 3 = 0, and for no other values. Thus the values required are the roots of a;2 _ 5 X + 3 = 0, and these roots are 5±V13 . 2 EXAMPLES LVI. Find the roots of the foljowing equations : 1. X* -5x2 + 4 = 0. 2. x4 _ 2 x2 - 8 = 0. 3. x4- 10x2 + 9 = 0. 4. a;4 - 7 x2 - 18 = 0. 6. 9 ^x2 6. x2+l^f = 20. X2 7. ^^+i=«^+i 8. X2 - 1 _ X2 :a2 - 1 9. x* + 1 _ a* + 1 0. 10. (x2-x)2-8(x2-x) + 12=0. 11. (x2+x)2-22(x2+x) + 40=0. 12. (x2 + x)(x2 + x+l)=42. 18. -^i- + ^-+I = 2. X x2 ~ ^ ' a* «« + 1 ' « 2 14. g | a;«+l _6. EQUATIONS OF HIGHER DEGREE. 265 .. x^ ^x-1 17 x-1 x^ 4 ^ 16. (x2 + x+l)(x2 + a;-f)+l=0. 17. ix^-^x+l){x^-i-x+2) = 12. ^ ^2,^11^ 42 18. (.^1)V4(. + 1).12. X2 + X 21. 2x2-4 x-Vx2-2ic-3=9. 19 a;2+2 x2+4x4-l _5 . • x^+ix+l x2+2 ~2 22. x2+Vx2+3x+7=23-3x. 23. 2x2 + 6x = l-V(a;2 + 3x + l). 24. x2 + v(4a:2 + 24x)=24-6x. 25. 2(2 X - 3) (X - 4) - V(2 x2 - 11 x + 15) = 60. 26. x2 + (x - 2)(x - 3) + \/2 x2 - 5 X 4- 6 = 6. 27. (x + 5) (x - 2) - 36 = \/(x + 4) (x - 1) . Solve the following cubic equations having given one of the roots of each : 28. x3-2x+l =0, [x= 1]. 29. x3-5x + 4 = 0, [x = 1]. 30. x3 - 49x + 120 = 0, [x = 3]. 31. x3-3x2-7x + 21 = 0, [x = 3]. 32. x3 - 2 x2 - 7 X - 4 = 0, [x = - 1]. 33. 5x8 - 15 x2 + 3 X + 14 = 0, [x = 2]. 34. For what values of x will x^ + 2 x2 + 9 and x^ — 4 x + 15 both vanish ? 35. The equations 2x3 + 21 x'^ + 34x- 105 = 0, 2 x3 - x2 - 76 X - 105 = have one root in common ; find it. Solve the following equations : 36. x3 - 21 X + 20 = 0. 39. x^ - 3x2 - 60x - 100 = 0. 37. xH2x2-9x-18=0. 40. x3-19x + 30 = 0. 38. x3-x2-22x+40=0. 41. 2x3 + 7x2 - 5x - 4 = 0. 42. 4x3 + 4x2 -9x- 9 = 0. 266 SIMULTANEOUS EQUATIONS. CHAPTER XVIII. Simultaneous Equations of the Second Degree. 189. We now proceed to consider simultaneous equa- tions, one at least of which is of the second degree. We first take the case of two equations which contain two unknown quantities, one equation being of the first degree and the other of the second degree. Example. Solve the equations x+2y = 5, x2-f22/2 = 9. From the first equation we have x = 5-2y. Substituting this value of x in the second equation, we have {6-2yy + 2y^ = 9; .-. 6|/2_20y + 16 =0; .-. 3y2_i0y + 8 = 0; that is (3 2/ - 4) («/ - 2) = 0. Hence y = |, or y = 2. If y = faj = 6-2y = 5-1 = 1; and if y = 2, a; = 6 - 2 ?/ = 5 - 4 = 1. Thus we have x = |, y = | ; or x = 1, y = 2. [The result should not be written in the ambiguous form x = |, orl; y = |, or 2.] SIMULTANEOUS EQUATIONS. 267 From the above example it will be seen that two equations, one of which is of the first degree and the other of the second degree, can be solved in the following manner : From the equation of the first degree find the value of one of the unknown quantities in terms of the other unknown quantity and the knoivn quantities, and substitute this value in the equation of the second degree : one of the unknown quantities is thus eliminated, and a quadratic eqxiation is obtained, the roots of which are the values of the unknown quantity which is retained. Ex. 1. Solve the equations ^x + Ay = 5, 2x2 - xy + y2 _ 22. From the first equation, x = -^ — ^- Hence, substituting in 3 the second, we have which reduces to 53 ?/2- 95!/ -148 = 0, that is (y+l)(53?/-148) = 0. Hence y = — 1, or y = 53 And,if y = 'A x = '-^^=-'^. ^ 53 ' 3 53 Ex. 2. Solve xy -\-x = 2b, 2xy-3y = 2S. Multiply the first equation by 2, and subtract the second ; then 2x + 3?/ = 50-28 = 22. 268 SIMULTANEOUS EQUATIONS. Hence y^ 22-2x . o and substituting in the first equation, we have {^^)- = - .-. 2x2 -25aj + 75 = 0, whence we have x = 5 or x = y-, the corresponding values of y being 4 and |, respectively. Ex. 3. Solve 2x2 -3x- 4?/ = 47, 3x2 + 4x + 2?/ = 89. Multiply the second equation by 2 ; then 6x2 + 8x + 4y = 178. Adding the last equation to the first, we have 8x2 + 5x = 226, whence x = 5 or x = — Y . The values of y are then found by sub- stituting in either of the given equations. EXAMPLES LVII. Solve the following equations : 1. x4-2/ = 6, 6. x + y = 16, x2 — y2 — 24. 4 X — 4 ?/ = xy. 2. x-y = 10, 7. 2x-i/ = 5, x2 + ?/2 = 58. X + 3 y = 2xy. 3. 3x+3?/ = 10, 8. 3x-31 = 6y, xr/ = l. x2 + 6xy + 25 = y2. 4. 2x + 3y = 0, 9. 3x + 2?/ = 5, 4x2 + 9xy + 9i/2 = 72. x2-4xy + 5y2 = 2. 6. 2x-5!/ = 0, 10. 2x-7i/ = 25, {c2_3 2/2=i3, 5a;2 + 4xj^ + 3j/2 = 23. 11. x + y=2. 2 + 3 = 6. X y 12. x + 2y = 7, X y 13. x + y = b. 1^1 = ^-. X y Q 14. X— y = I, y x~6 15. 2x-y = 4, X y SIMULTANEOUS EQUATIONS. 269 16. 7x-3y + 7 = 0, X y 17. icy + X = 15, xy — x = S. 18. a;2/ + 2x=5, 2xy-?/ = 3. 19. x2-y = 29, 2c2 + a: + y = 49. 20- x2 + 3a;-22/ = 4, 2x2-5x + 3i/ + 2 = 0. 190. It should be remarked that the methods we have thus far explained do not enable us to solve any two equations which are both of the second degree ; for the elimination of one of the unknown quantities will fre- quently lead to an equation of higher degree than the second, from which the remaining unknown quantity would have to be found ; and we have not yet learned how to solve an equation of higher degree than the second, except in very special cases. For example, if we have the equations x^ -\- x + y = S and a;2 4- y2 — 5, We have from the first equation y = 3 — x — x^ ; and, by substituting this value of y in the second equation, we get a;2 + (3 _ X - a;2)2 = 5, that is X* 4- 2 a;3 - 4 0:2 - 6 X + 4 = ; and this equation of the fourth degree cannot be solved by any methods which are within the range of this book. 270 SIMULTANEOUS EQUATIONS. 191. We can always solve two equations of the second degree when all the terms which contain the unknown quantities are of the second degree. The method will be seen from the following example. Ex. 1. Solve the equations xy + 4 2/2 = 8. Divide the members of the first equation by the corresponding members of the second equation ; we then have x'^-\-Sxy ^ 28 ^ 7^ xy + 'iy'^ 8 2 Hence 2 (a;2 + 3 xy) = 7 (a;y + 4 y^) ; .-. 2x^-xy-2Sy^ = 0, that is (2x + 7y)(x-4y) = 0. Hence x = 4 y, or a; = - ^ y. I. If X = 4 ?/, we have from the second equation 42/2 + 4?/2 = 8; .-. y = ±l. And therefore x = 4 y = ± 4. II. If X = — I ?/, we have from the second equation, -iy^ + iy^ = S; .-. 2/ = ±4. And therefore x = - | y = =F 14. Thus there are four sets of values, namely, x=4, ?/ = l;x = — 4, y = -l; x = 14, y=— 4; and x = — 14, 2/ = 4. We have in the above written down the factors of 2x2-xy-28y2 by inspection : when this cannot be done, the factors can be found by the method of Art, 1^2, SIMULTANEOUS EQUATIONS. 271 Ex. 2. Solve x^-Sxy = 0, 5x2 -\-3y^ = 48. The first equation is x(x — Sy)=0. Hence x = 0, or x = Sy. If 05 = 0, the second equation gives 3 y^ = 48 ; . •. y = ± 4. If x = Sy, the second equation gives iby'^+3y'^=iS ; . *. y = ± 1, and then x = ±d. 192. Any pair of equations which are homogeneous, so far as the terms which contain the unknown quantities are concerned, can be solved by the method adopted in the last article. Sometimes, however, the equations can be solved by special methods. For example, to solve x^+y'i = 74, 2xy = 70. By addition, we have x2_,.2xy + 2/2 = i44, that is (« + y)2 - 122 = ; .-. (a; + y-12)(x + y+12)=0. Hence x-^ y=V2 (i.), x + y = -12 (ii.). Again, from the given equations, we have by subtraction, x^-2xy-{-y^ = i, that is (X - ?/)2 - 22 = ; .-. (x-y-2)(x-y + 2)=:0. Hence x — y = 2 (lli-)> x-y = -2 (iv.). From (i.) and (iii.), we have x = 7, y = 5. From (i.) and (iv.), we have x = b, y =7. From (ii.) and (iii.), we have x =—6, y = —7. And, from ^ii.) and (iv.), we have a; = — 7, y = — 5, 272 SIMULTANEOUS EQUATIONS. Thus there are four pairs of values, two of which are given by X = ± 7, y = ± 5. and the other two by x = ± 5, y = ± 7, it being understood that in both cases the upper signs are to be taken together, and the lower signs are to be taken together. Again, taking the equations considered in Art. 191, namely, a;2 + 3 xy = 28, xy + 4 2/2 ^ 8. We have by addition x^ + 4 x?/ + 4 y2 = 36, thatis (x + 2?/)2-62 = 0; .-. x + 22/ = 6 (1.), or x + 2i/=-6 (ii.). We can now complete the solution as in Art. 189, by taking (i.) and (ii.) with either of the given equations. EXAMPLES LVIII. 1. x2-2.xy = 0, 9. x2-3y2 = i3, 4x2+ 9?/2 = 225. 3x2-2/2 = 71. 2. 2 x2 - 3xy = 0, 10. 3xy + x2 = 10, 1/2 + 5xy = 34. 5 xy - 2 x2 = 2. 3. x2+3xi/ = 45, 11. x2 + 3a;2/ = 54, 1/2 _ X2/ = 4. xy + 4 y2 = 115. 4. x2-x2/ = 03, 12. ^.(^a. ^ y) = 40, y2 + X?/=22. y(^^_y)^0. 6. x2 + x?/ = 24, 13. x2-5x2/ + 2y2 = 3, 2y2 + 3xy = 32. 2x2 + y2 = 6. 6. x2 - 3 xy = 10, 14. x2 - 2 xy + 5 = 0, 4 y2 - xy = - 1. (a; - y)2 - 4 = 0. 7. x2 + xy-2y2 = -44, 15. x2 + 5y2 = 84, xy + 3 y2 = 80. 3x2 + 17 xy - y2 = - 84. 8. x2 + 3xy = 7, 16- a;"^ - 7a;y - 9y2 = 9, j/2 -f xy = G, »2 + 6xy + 11 y^ = 6, SIMULTANEOUS EQUATIONS. 2T3 17. 2/2 - xy = 6. 23. 3 2 3 18. x2 + 3 xy = 40, iy^-{-xy = 9. X y 19. x^^xy + y^ = 7, Gx^-^xy -^y'^ = 6. 24. 2 5_5 X y 6 20. 21. 2x2-2x2/ + 3?/-^ = 18, 3x2-21/2 = 19. X + ?/ + 1 = 0, 22. x y V2 X + 2/ + 3 = 0, 1 + 1-1 = 0. X y 6 25. 26. (x+l)(2/4-l)=10, X2/=:3. (x + 3)(2/ + l)=4, xy + 1 = 0. 193. The following examples will show how to deal with some other cases of simultaneous equations. Ex.1. Solve x-y = 2 (L), 3(^-y^ = S8Q (ii.). From (i.), we have x = y -{- 2. Substitute this value in (ii.), and we have (y + 2)3 -y^ = 386, that is 6 2/2 + 12 y + 8 - 386 = ; . •. 2/2 4. 2 2/ - 63 = 0. Hence y = 1^ or y = — 9. Ify = 7, x = 2/ + 2 = 9. If 2/=- 9, x=y+2z=-7 Thus X = 9, 2/ = 7, or X = - 7, y=-9. Ex. 2. Solve X - 3 2/ = 2, x2 -9 2/2 = 8. 274 SIMULTANEOUS EQUATIONS. Divide the members of the second equation by the correspond- ing members of the first ; then and from cc + 3 y = 4 and x — 3 y = 2, we have x = S, y = ^. Ex.3. Solve i + - = 7 (i.), X y ^ + -\=25 (11.). X^ ?/2 We may consider - and - as the unknown quantities, as in Art. X y 107, Ex. 3. Square both members of (i. ) ; then i + i- + i = 49 . a;2 xy y^ . . . . . (ill.). we have A = 49 -25 = 24 . xy . . . . . (iv.). From (ii.) and (iv.), we have l--^ + l = 25-24 = l, x2 xy y'^ \x y) ±1 (v.). that is y . 1_1 ' ' X y The values of - and - are at once found from (i.) and (v.). X y Ex. 4. Solve x2 - xy + y2 _ 61 (i.)^ «* + x2y2 + y4 = 1281 (11.). Divide the members of (ii.) by the corresponding members of (1.) ; we then have »2 + xy +2/2 = 21 (ill.). SIMULTANEOUS EQUATIONS. 275 From (i.) and (iii.) we have, by subtraction, 2xy = - 40, or acy = — 20 (iv.). From (i.) and (iv.), we have 06^ - 2 x?/ + ?/2 = 81 ; .'. x-y = +9 (v.), or x-y = — Q (vi.). From (i.) and (iv.), we have also x2 + 2 xy + y2 = 1 . .'. x + y = \ (vii.)i or x-\-y = —\ (viii.). Then combining either of the equations (v.) and (vi.) with either of the equations (vii.) and (viii.) we get the four pairs of values aj = ±5, y=T4j ic = ±4, 2/ = =f5- Ex.5. Solve x^~2xy = Zy (i.), 2a;2-9|/2 = 92^ (ii.). Multiply both sides of (i.) by 3 ; then 3a;2-6a;y = 9y (iii.). Hence 2x'^ -Qy"^ = ^x^ -Qxy; .: aj2_6a;y + 9y2 = o, that is (x - 3 y)2 = ; . •• x = Sy. Substitute in (ii.) ; then 2(3y)2-92/2 = 9y, or 9 y2 _ 9 y = ; .-. y(y -1) = 0. Hence y = 0, or y = 1. liy = 0, X = 3 y = 0. lty = l, x = Sy = S. Thus X = 0, y = ; or else x = 3, y = 1. 276 SIMULTAKEOtTS EQUATIONS. Ex. 6. Solve X - y = 2, x^-y^ = 242. Put « = «+!, then y = z — \. Hence x^-y^={z + \y-{z-\y = 10 ^* + 20 z^ + 2. Hence lO;?* + 20 ^2 + 2 = 242 ; .'. 04 + 2^2 = 24; .-. (22_4)(5;2 + 6)=0. Hence 2 = ± 2, or z = ± V— 6. If ;s = ± 2, 0! = 3 or - 1, and y = 1 or - 3. It z=± V^^, x = i ± V^ and y = -l± V^^. Ex. 7. Solve xy + xz = 27 (i.), yz-{-yx = S2 (ii.), zx + zy = S5 (iii')' From (i.) and (ii.), we have by addition 2 xy -\- xz -{■ yz = 69 (iv-)* From (iv.) and (iii.), we have by subtraction 2 icy = 24 ; .'. xy = 12 (v.). Hence, from (i.), xz = 16 (vi.). And, from (ii.), yz = 20 (vii.). From (v.) and (vi.), we have by multiplication x'^yz = ISO (viii.). From (vii.) and (viii.), we have by division a;2 = 180/20 = 9 ; .-. x = ±3. Hence, from (v.), y = 12/(± 3) = ± 4. And, from (vi. ) , z = 16/(±S) = ± 6. Thus ic = ± 3, y = ± 4, z=±b. All the upper signs being taken together, and all the lower signs being taken together. SIMULTANEOUS EQUATIONS. 277 EXAMPLES LIX. 1. X - y = 3, IZ. x + y = l, x^-y^ = 279. x^y^ + lSxy + 12 = 0. 2. X - y = 2, 14. X + 2/ = 5, x^-y^ = 98. 4xy = 12 - x2y2. 3. X + y = 7, 16. x2 - xy + 2/ = - 6, x3 + y3 = 91. yi-xy-X = 12. 4. x + y = l, 16. x2 + a;y-y = 9, x3 4-2/^ = 61. ^24.3;^ _a;=_3. 6. X2 + X2/ + 1/2=13, j^ ^^1 X* + x2|/2 + y* = 91. 6. x2-xy + y2=,9^ X* + x2i/2 + y4 ^ 243. 7. xy(x + y) = 240, 18 x^-\-y^ = 280. 8. xy(x-y)=l2, x^-y^ = 63. 19. 2x2-xy + y2=:2y, 9. 1_1 = 2, 2x2 + 4xy = 5y. X 1/ X2 w2 x + - = = 3, X 4 "3* 18 ^-^ _7^ "4* 20. X3 + 1 =9; + -, = 20. 3.2 _^ a; ^6; 21. 2x3 + 5y3 = 115, 10. ? + i = l, 3x3 + 7w3 = 186. XV 2 3 1 _ . 22. x2y + X2/2 = 180, ^■^^"^y2-^- X2y2:=400. 3 1 23. x* + x2y2 -j- y* = 91, ^^" ^~^"^' (x2-xy+y2)(a._y)2 = 28. 1_J:__^1^3. 24. x2+3xy+y2_,.4(a;+y)=i3, x2 xy 2/2 * 3x2-x2/+32/2-|_2(x+?/)=9. 12. -^-;^-3, 25. X2/ + ? = |, x2 4|/2 2/3 L_l + J_ = 9. =»!, + 3^ = 5. x* «y 4y2 a: 6 278 SlMULTANEOtrs EQUATIONS. 26. ^ + J- = 20, 28. aCx-a)=b(y-b), X xy Z X 5 xy = ax-^ by. XV 4- 29. X = y = a — b. y X 27. ax = by, (x — a){y — b)= ab. 30. yz = 4, zx = 9, xy = 16. 31. x'^yz = 6, y^zx = 12, ^j^^jy = ig. 32. x(a; + y + «) = 8, y(x + y4-^)=12, 2(x + ?/ + 2;)= 5. 33. ic(y + 2!)=6, y(2 + a;)=12, ;2(cc + ?/)= 10. 34. (x + y)(x + ;2;)=2, (y + z)(y + x)= S, (^z -\- x) (z + y) = 8, 36. 1/2; = a^, zx = 62, ic^/ = c^. 36. ILiLi = ^±i= = ^±J^^2a:y;r. a 6 c 37. iy-\-z)(x-\-y-\-z)=b-{-c, (z + x)(x + y + z)=c -\- a, (x-\- y){x-l V + z) = a + b. PKOBLEMS. 279 CHAPTER XIX. Problems. 194. We now give examples of problems in which the relations between the known and unknown quantities are expressed algebraically by means of quadratic equations. In the solution of problems it often happens that by solving the equations which are the algebraical state- ments of the relations between the magnitudes of the known and unknown quantities, we obtain results which do not all satisfy the conditions of the problem. The reason of this is that the roots of the equation are the numbers, whethev positive or negative, integral or fractional^ which satisfy it; but in the problem itself there may be restrictions, expressed or implied, on the numbers, and these restrictions cannot be retained in the equation. For example, in a problem which refers to a number of men, it is clear that this number must be integral, but this condition cannot be expressed in the equations. Thus there are three steps in the solution of a prob- lem. We first find the equations which are the algebrai- cal expressions of the relations between the magnitudos of the known and unknown quantities ; we then find the values of the unknown quantities which satisfy these equations ; and finally we examine whether any or all of the values we have found violate any conditions 280 PROBLEMS. which are expressed or implied in the problem, but which are not contained in the equations. The following are examples of problems which lead to quadratic equations. Ex. 1. How many children are there in a family, when eleven times the number is greater by five than twice tlie square of the number ? Let X be the number of children ; tlien we have 11 ic = 2x^4- 6; .-. 2a;2_iia;_|_5==o, that is (2 a;- l)(x -5)=0. Hence ic = 5, or sc = \. Thus there are 5 children, the value \ Ixiing inadmissible. Ex. 2. Eleven times the number of yards in the length of a rod is greater by five than twice the square of the number of yards. How long is the rod ? This leads to the same equation as before, only in this case we cannot reject the fractional result. Thus the rod may be five yards long, or it may be haif a yard long. Ex. 3. A number of two digits is equal to twice the product of the digits, and the digit in the ten's place is less by 3 than the digit m the unit's place. What is the number ? Let X be the digit in the ten's place ; then ic + 3 will be the digit in the unit's place. The nuqjber is therefore equal to 10 x + (x + 3). Hence, by the question, 10aj + (a; + 3)=2a;(x + 3) ; .♦. 2x2 _ sa: _3 = o, that is (a;-3)(2a;4- 1)=0. Hence a; = 3, or x = — ^. PROBLEMS. 281 Now the digits of a number must be positive integers ; hence the second value is inadmissible. Therefore the digits are 3 and 6, and the required number is 36. Ex. 4. The square of the number of dollars a man possesses is greater by 1000 than thirty times the number. How much is the man worth ? Let X be the number of dollars the man is worth ; then, by the conditions of the problem a;2 = 30x + 1000; .-. x2 - 30 a; - 1000 = 0, that is {x - 50) {x + 20) = 0. Hence a; = 50, or a; = — 20. Both of these values are admissible, provided a debt is considered as a negative possession. Hence the man may have $ 50, or he may owe $ 20. Ex. 5. The sum of a certain number and its square root is 42 : what is the number ? Let X be the number ; then x-{- Va; = 42, that is y/x = ^2 — x. Square both sides ; then, after transposition, we have x2 - 85 X 4- 1764 = 0, the roots of which are 36 and 49. The value 49 will not however satisfy the condition of the problem, if by a square root of a number is meant only the arith- metical square root. Ex. 6. The sum of the ages of a father and his son is 100 years ; also one-tenth of the product of their ages, in years, exceeds the father's age by 180. How old are they ? Let the father be x years old ; then the son will be 100 — x years old. Hence by the conditions of the problem, 282 PROBLEMS. j\ X (100 - x) = X + 180 ; ... x2-90x+ 1800 = 0, that is (x - 60) (x - 30) = 0. Hence x = 60, or x = 30. The second value is inadmissible, although it is a positive integer, for it would make the son older than his father. Hence the father must be 60, and the son 40 years old. EXAMPLES LX. 1. Find two numbers one of which is three times the other and whose product is 243. 2. Find two numbers whose sum is 18 and whose product is 77. 3. Find two numbers whose difference is 20, and the sum of whose squares is 650. 4. Divide 25 into two parts whose product is 156. 5. Divide 80 into two parts the sum of the squares of which is 3208. 6. A certain number is subtracted from 36, and the same number is also subtracted from 30 ; and the product of the re- mainders is 891. What is the number ? 7. A rectangular court is ten yards longer than it is broad ; its area is 1131 square yards. What is its length and breadth ? 8. The product of the sum and difference of a number and its reciprocal is 3| : find the number. 9. The number of tennis balls which can be bought for a pound is equal to the number of shillings in the cost of 125 of them. How many can be bought for a pound ? 10. The number of eggs which can be bought for 25 cents is equal to twice the number of cents which 8 eggs cost. How many eggs can be bought for 25 cents ? PROBLEMS. 283 11. A cask contains a certain number of gallons of water, and another cask contains half as many gallons of wine ; six gallons are drawn from each, and what is drawn from the one cask is put into the other. If the mixture in each cask be now of the same strength, find the amount of water and wine. 12. The cost of an entertainment was $ 20, which was to have been divided equally among the party, but four of them leave without paying, and the rest have each to pay 25 cents extra in con- sequence. Of how many persons did the party consist ? 13. A man buys a certain number of articles for $ 5, and makes $3.82 by selling all but two at 4 cents each more than they cost. How many did he buy ? 14. A man bought a certain number of railway-shares for $9375; he sold all but 15 of them for $10,450, gaining $20 per share on their cost price; how many shares did he buy ? 15. A crew can row a certain course up stream in 8^ minutes, and if there were no stream they could row it in 7 minutes less than it takes them to drift down the stream ; how long would they take to row down with the stream ? 16. A boat's crew can row 8 miles an hour in still water. What is the speed of a river's current if it take them 2 hours and 40 minutes to row 8 miles up and 8 miles down ? 17. Two trains run without stopping over the same 36 miles of rail. One of them travels 15 miles an hour faster than the other, and accomplishes the distance in 12 minutes less. Find the speed of the two trains. 18. A person having 7 miles to walk increases his speed one mile an hour after the first mile, and is half an hour less on the road than he would have been had he not altered his rate. How long did he take ? 19. A and B together can do a piece of work in a certain time. If they each did one half of the work separately, A would have to work one day less, and B two days more than before. Find the time in which A and B together do the work. 284 PROBLEMS. 20. The price of photographs is raised 50 cents per dozen ; and, in consequence, six less than before are sold for $ 5. What was the original price ? 21. What are eggs a dozen when two more for 30 cents would lower the price 2 cents a dozen ? 22. A woman spends 75 cents in eggs ; if she had bought a dozen less for the same money, they would have cost her 3| cents a dozen more. How many did she buy ? 23. The price of one kind of sugar per pound is 2 cents more than that of a second kind, and 10 pounds less of the first kind can be got for $2.40 than of the second. Find the price of each per pound. 24. One-half of the number of cents which a dozen apples cost is greater by 2 than twice the number of apples which can be bought for 30 cents. How many can be bought for $ 2.50 ? 25. Divide | 3620 among A, B and C so that B shall receive $ 20 less than A, and C as many times B's share as there are dollars in A's share. 26. Find two fractions whose sum is |, and whose difference is equal to their product. 27. Two men start at the same time to meet each other from towns which are 25 miles apart. One takes 18 minutes longer than the other to walk a mile, and they meet in 6 hours. How fast does each walk ? 28. The men in a regiment can be arranged in a column twice as deep as it is broad. If the number be diminished by 206, the men can be arranged in a hollow square three deep, having the same number of men in each outer side of the square as there were in the depth of the column. How many men were there at first in the regiment ? 29. The area of a certain rectangle is equal to the area of a square whose side is six inches shorter than one of the sides of the rectangle. If the breadth of the rectangle be increased by one inch MISCELLANEOUS EXAMPLES IV. 285 and its length diminished by two inches, its area would be unal- tered. Find the lengths of its sides. 3D. The diagonal and the longer side of a rectangle are together five times the shorter side, and the longer side exceeds the shorter by 35 yards. What is the area of the rectangle ? 31. If the greatest side of a rectangle be dimini.^ied by 3 yards and the less by 1 yard its area would be lialved ; and if the greater be increased by 9 yards and the less diminished by 2 yards its area would be unaltwed. Find the sides. 32. Two trains A and B leave P for Q at the same time as two trains C and D leave Q for P. A passes C 120 miles from P, and D 140 miles from 1*. B passes C 126 miles from Q, and D half way between P and Q : fuid the distaace from P to Q. •MISCELLANEOUS EXAMPLES IV: A. 1. Simplify 2x- [3x - 9?/ - {2 x - 3y -(x + 5?/)}]. 2. Multiply a2+25&-^+4c'H5rt&-2«c-}-10 6cby a-56-|-2c. 3. Dividex34-(4a&-62)x-(a-2 6)(rt2+3 62) hyx-a-\-2b. 4. Find the factors of (i.) (2x + y - zy -{x -{-2y -\- 4 zf. (ii.) xV - x2 - ?/'^ + 1, and (iii.) x-y^z^ - x?z -y^z + 1. ar.3 + 4x2-8x + 24 5. Simplify (i.) S-Sx + x^-x* (ii.) ^ZLi — __2 ^-^ (x-2)(x-3) (ic-3)(x-l) (x-l)(x-2) 6. Solve the equations : X — a X — h x — c (ii.) 4x2 -25x- 21 = 0. (iii.) x + 2/ = i + i = ^. X y 2 7. Show that x^ — 5x + 7 can never be less than f. 8. The difference of the cubes of two consecutive integer num- bers is 919 : find the numbers. 286 MISCELLANEOUS EXAMPLES IV. B. 1. Simplify (x + 3)3 - 3(a; + 2)8 + S(x + ly - a^. 2. Show that (X - a)^-\-(y - 6)2 + (a2 -\- b^ - l)(a;2 + y^ - y) = (xa -\-by - 1)2 + (bx - ayy\ 3. Divide x^ - 2 a^x^ + a^ ^y y-^ _ 2 ax + a2. 4. Find the L. C. M. of 8x3 + 27, 16x^ + 36x2 + 81, and 6x2 -5x- 6. 6. Simplify 1 x + 3 (i-) X— 1 X+1 X2+1 ai^ {(« + h){a + 6 + c)+ c2}((a + &)2 - c^ {(a + 6)3 - c3}(a + 6 + c) 6. Solve the equations : (i.) (6-x)(l + 2x)+3x(x + 5) = (x+l)2_a;. (ii.) 5x2 + 7x = 160. (iii.) x2 + x?/ = 10, y^-xy = S. 7. Show that the sura of the squares of the roots of the equation x2-5x + 2 = is 21. 8. At a conceit $ 6 was received for reserved seats, and the same sum for unreserved seats. A reserved seat cost 60 cents more than an unreserved seat, but 6 more tickets for unreserved than for reserved seats were sold. How many tickets were sold altogether? C. 1. Simplify 12 a - 3(6 - 2 (a - 3 6) - 2 a}. 2. Multiply a3 ^ 2 a26 - a62 + 2 63 by a^ - 2 a-b -ab'^-2 6'. 3. Divide 24 x* - 10x3y - 8 x2y2 + lo xy^ - 4 y* by 2 y^^xy-ijc-, 4. Find the factors of 9 x2 + 9 x + 2, and of 4(a6 - cd)^ - {a^ + b^ - c^ - ^2)2. MISCELLANEOUS EXAMPLES IV. 287 x + 5. Simplify I + xy ^ l + xy and show that ^^:^ + "-^=^+ ^^^+ ^^ " '^^' " ^^^^ " ^^^ ^ 0. a b c abc 6. Solve the equations : ^■^3 5 ~ 6 * (ii.) X + y = 2 a, a;2 + 2/2 = 2 a2. 7. Find the least value of x^ + 6 x + 12, and the greatest value of 6x-x2-4. 8. A and B have 45 coins between them, which are all dollars and dimes, A has four times as many dollars as dimes, and B has just as many dollars as dimes ; also A has $ 9.50 more than B. How much money has each ? D. 1. Show that (6 + c)2 _ a2 _^ (c 4- a)2 _ 52 + (a 4- 6)2 - c^ = (a + 6 + c)2. 2. Arrange (1 + x)* + 2(1 — x + x2) according to ascending powers of X. 3. Show that the difference between the squares of any two con- secutive numbers is one more than double the smaller number. 4. Show that if 1 + 1 + 1 = 0, a^ -\- h'^ + c'^ = (a + h -\- c)K abc 5. Find the L. C. M. of x2 - 5x - 14, x2 - 4x - 21, and x» - 3x2 - 25x - 21. For what value of x will all three expressions vanish ? 6. Solve the equations X y X y X a b X — a — b (iii.) X + y = 3, 3x2 - 11 y-' = 1. 288 MISCELLANEOUS EXAMPLES IV. 7. If xi, X2 are the roots of axP' + 6aj + c = 0, prove that a?i I Xz _ h^ — '2, ac X'i xi ac 8. Out of a cask containing 60 gallons of alcohol a certain quantity is drawn off and replaced by water. Of the mixture a second quantity, 14 gallons more than the first, is drawn off and replaced by water. The cask then contains as much water as alcohol. How much was drawn off the first time ? E. 1. Find the numerical value of 2 13 4/2 ^ 2. Show that (a + by - (a^ _ &2)2 ^ 4 ^^(^ + i^y^^ a„d that 2(a - 6)(a - c)+ 2(6 - c)(& - a)+ 2(c - a){c - b) = (& - c)2 + (c - ay + (a - b)'^. 3. Divide (a + 2 5 - 3 c + dt)2 - (2 a + & + 3 c - J)^ by a + &. 4. FindtheL.C.M. of 6a;2- 6ax - Ga^ and 4x3 - 2ax2 - Oa^. 5. Simplify (1^ + ^) ^1^ + -^ - ^^^. \x y/y^-x^ xy-y^ x^ -{- y^ 6. Solve the equations : (1.) H — ^. X — 6 X — a (ii.) (X - 1) (X - 2) + (x - 2) (X - 3) + (x - 3) (x - 1) = 11. y 10 X 3 7. For what values of x are ? + - and - + ^ equal to one an- a x a c other ? 8. A tricyclist rode 180 miles at a uniform rate. If he had ridden 3 miles an hour slower than he did, it would have taken him 3 hours longer. How many miles an hour did he ride ? MISCELLANEOUS EQUATIONS. 289 F. 1. Add together 3 «» _ 4 aSft + 2 ah\ 3 a^ft - 4 a62 + 2 6^, 3 a52 _ 4 ^3 _,_ 2 a3, and 3 6^ _ 4 a^ + 2 a^h. 2. Show that (a^ + 62 + c^) (^2 + «/2 + z^) - {ax -\- by + czy = (ay - bxy + (Jbz - ci/)2 + (ex - a;?)^. 3. Divide 2 + 11 x + 11 ic2 + x^ - x* by x2 + 3x + 2. 4. Find the H.C.F. of x3-x2-2x+2 and x*-3x3+2x2+x-l. What value of x will make both expressions vanish ? 5. Simplify (i.) /^ + - — i - + (ii.) x(x-2) x2-5x + 6 x(3-x) 1 1 1 1 1-x 6. Solve the equations : x + 3_6-x^^_l. ^ ^ 5 10 10 /ii>) x+J/_^^^^^2, ax + 6y = a2 + 62. a-\-h a—b x+i x_+2^2X + 3. ^ x-1 x-2 x-3 7. Find the difference of the squares of the roots of the equation x2 - 7 X + 9 = 0. 8. What number exceeds its square root by 156 ? MISCELLANEOUS EQUATIONS. 1. (X - 1) (X - 4) + (X - 3) (X - 5) = 2(x - 3)2. 2. l(x-2)-l(x-4:)-\-{(x-5)=0. 3_ 1 1 1 1 X — 3 X — 5 X — 2 X — 4 4. 15x+17y = 79, 17x+15y = 81. T 290 MISCELLANEOUS EQUATIONS. 6. A man drives to a certain place at the rate of 8 miles an hour. Returning by a road 3 miles longer at the rate of 9 miles an hour, he takes 7|- minutes longer than in going. How long is each road ? 6. l(x-2)-^-=-^+5?-=-i=0. 2^ -^9 6 7. 3(a; + l)(x - 3) + (a; + 7)(3a; - 13) = (2a; - 5)(3x - 3). 8 2a; + 9 Sx~2 ^ bx+U x + 2 x-\-S ~ x-\- 4: ' 9. ?4.^ = 82, ? + ^ = 83. 7 8 ' 8 7 10. A pound of tea and 6 pounds of sugar cost 00 cents, but if the tea were to rise 20 per cent and the sugar 25 per cent in price, they would cost $ 1.10. What was the price of the tea ? 11. K6aJ + 3)+A(7a; + 6)+TV(9aJ + 2)=2x+l. ^n X , X a b abba 13 a; + 3 . x — G _ x-\-4: x — 5 ' x+1 x-i~ x-\-2 x-S 14. x-\-y = 2, (a + b)x -\-(b-a)y = 2 b. 15. A poultry keeper obtained 100 eggs more in May than in April, and the daily average in May was 3 more than in April. How many did he have in May ? 16. 3a; - 3[4a; - 2(2a; - 6)]= 9 - 2[3a; - 5{x - 5)]. 17. ^_6^2, 4 + 12^3. X y X y 18. 15x2 + 34x + 15 = 0. 19. a?-f2 a^-3 _^ * » — 4 « — 6 MISCELLANEOUS EQUATIONS. 291 20. A father's age is three times that of his youngest son, and in 7 years he will be twice as old as his oldest son ; also the oldest son is 5 years older than the youngest. What are their present ages ? 21. K^ + l)(x + 3)= i(x + 5)(x + 2)+ l(x -l)(x- 4). 22. {b + c)(x- rt)-(c + a)ix- b) = {a-\- 6)(x-c). 23. 11x-l9y = i, 21x-29y =24,. 24 2a;-3 3x + 8^^ x-S 2x+l 25. Two passengers, who have altogether 360 lb. of luggage, are charged 80 cts. for excess above the weight allowed free ; if one passenger had taken the same weight of luggage, he would have been charged $ 1.30. What weight is allowed free ? 26 •^^~ ^ + ^^+ 1 _ 9a;+ 1 _ I -x 5 6 8 3' 27. ^L±y = 2x-y-2,^y^=y-x-\-5. 28. (a + b)x + (a - b)y = a^ - b'^, (a - byx + (a + b)y = a'^ + b^. S-x S-x 30. A father's age is equal to the united ages of his five children, and five years ago his age was double their united ages. How old is the father ? 31 2a; a; + 3 _ 3a;+19 ' x— I x-{- I x + 5 33. . l(x-2)=^(l-y), 26x-\-3y + 4=0. 117 1 x2-l 1-x 8 x+1 34. V^ + v6x+l=2. 292 MISCELLANEOUS EQUATIONS. 35. Two cyclists started at the same time, and one rode from B to A and the other from A to B along the same road. They reached their destinations 3 hours and 1 hour 20 minutes, re- spectively, after they passed each other. When did they meet ? 36. 37. 40. A certain sum of money was divided between A, B, and C. A had one-ninth of the shares of B and C together, B had one- third of the shares of C and A together, and C had 6 dollars more than A and B together. How much had each ? 6x 6x- ^- 9 1-25X-2 1 + lOa; l + 5x 32/ 2_ x~ 5y-? + 9 —I- ix -7 , 2a;+l_ipi Sx -4 ' x-1 ■^i- V4 x + 4 _ Va; + 8 = --Vx-4. -i(-^-!)-('-T)=^-*^^- 42. x+_3 x-S 43. X + 2 = ^{4: -\- xVS - x}. 44. 3x-2y = 12, 9 x"^ - 4 y^ = b76. 45. A man about to invest in the 2| per cents observed that, if the price of the stock had been 7^ less, he would have got f per cent more interest for his money. What was the price of the stock ? 46 4a;8 + 4a;-^4-8x-|- 1 ^ 2a;^ + 2a; -H . 2a;2-|-2a;-f-3 x-\-l 47. 3J,(5x-h6)-,\(ll2/-6)=ll, 5^5(55y-12) = ^-37. 48. 1^+ J^ = 6.. X x — i 49. y/x -{- Vx + i = -=— y/X MISCELLANEOUS EQUATIONS. 293 50. A fast train left Cambridge for London at the same time as a slow train left London for Cambridge. They arrived at their destinations three-quarters of an hour and one hour and twenty minutes respectively after they passed one another. Find how long each took for the journey. ' x-i sU-l 3J 10(x- 1) 52. 6x-\-2y -1 = 53. y/(x'^-a-) = 2 6x-\-2y -l=Sx-y-\-U=x-\-l9y + e. 1 54. 3 a;2 _ 2 a;y + 4 ?/2 = 36, 4 x^ _ y2 - 7. 55. Find three consecutive integers such that the sum of the cubes of the greatest and least exceeds twice the cube of the middle number by 42. ^-12 na x-h 3 2 o; - 1 9.', - 57. ^J + ^Zl2^_2__^ ^ 2 3x-3x-2 58. ax-by = a^ -b'^-2 ah, bx -\- ay = 2ab + a'^ - h"^. 59. x^ -\- xy -{■ y"- = 9, x* + xf-y'^ + y^ = 243. 60. A man bought a certain number of articles, of which he sold half at 5 per cent profit, one-third at 10 per cent profit, and the rest at 30 cents each. His total profit was at the rate of If per cent. What was the cost of each article ? 61. (a + 6)(a; + a-6) + (a- 6)(x - a - 6)+ 2a(a; - 2 c)=0. 62 3x-2 2x-3 ^2 2x-5 3a;-2 3 63. 2 x2 -SVCx^ -f 2 X + 14) 4- 4 X - 49 = 0. 64. 2x2 4- 3a;y = 8, y^ - 2xy = 20. 294 MISCELLANEOUS EQUATIONS. 65. A man bought a certain number of sheep for $ 290. Having lost five of them, he sold one quarter of the remainder for $ 63, making a profit of 5 per cent on the sale of these. How many did he buy ? aa 1 X + 6 3 5a; -1 l-25x-^ 1 + 5x 67. (a + b)x + by = ax +(a-{- h)y = a^ + h^. 68. ^{x-a)^-y/{x-h) = —-^ + ^ V(x-a) y/(X-h) 69. x2 + ?/2 = xy + 7, x2 - y'^ =zxy — 1. 70. A number of articles were bought for $25, and sold again at an advance of 12.] cents in the price of each ; and at the advanced price .$25 was received for the sale of all but 10 of the articles. How many were there ? 71. iiSx+l)-^(2x-S)=M^x-l)-iax-S). 72. - — £ — - — -. c x~ a c 73. y/(x-hl) + V2lc = y/(i6x-{-l). 74. a;2 - a;y 4- 3 y = 11, y^ - xy -Sx -]- I = 0. 75. The value of 185 coins consisting of dollars, dimes, and half-dimes amounts to $30.50. If there were twice as many dimes, halt: as many half-dimes, and three times as many dollars, the total value of the coins would be $ 72. How many coins are there of each kind ? 76 ^^+ ^ _ 1 -5a; _ x -\- 1 _ 2 -9a; 77 a; + l . a;g+1 ^29 • a;2 -I- 1 "^ a; -f 1 10 78. (x — a)(y — h)= ah, bx = ay. 79. x2 = 2 1/ -f 24, ?/ = 2 a; +'24. MISCELLANEOUS EQUATIONS. 295 80. 1120 square feet of paper will just cover the four walls of a room which is 8 feet longer than it is wide ; but, if the room were 4 feet higher, the same quantity of paper would just cover the two smaller and one of the larger walls. What are the dimensions of the room ? 81. i(x-2)-^ + 5^ = 0. 82. {a-xy+{h-xy={a^h-2xf. 83. (a;-6)-^+(y-6)2 = 2(xy-40), x = y + \. 84. 4 xy = 96 - x'^y^, x + y = 6. 85. A merchant gained as many eagles on the sale of a certain quantity of coal as there were half-dollare in the cost price of a ton, or half-dimes in the retail price of a cwt. How many tons did he sell ? 2x-3 2x-4 2x-6 2x 2x-4 2x- 5 2x- 7 2x-8 87. 2x(a-6-2x) + (a-l)(6-l)=0. 88. y/{Qx-\- 1) + V2(1 -x)=V7x+ 6. 89. xy(x + y) = 30, x^-\-y^ = - 91. 90. A rectangular enclosure is half an acre in area, and its perimeter is 198 yards. Find the length of its sides. 91 x + 3 3x + 7 ^ 2x+l '2x+l 2x-2 x-2 92. x + 2?/ + 3;2 = 4, x + 3y4-2 = 4 2, x + 2^ + 3 = 4y. 93. ^(a-x) + ^(x-6) = ^(a-6). 94. (x-3)2 + (y-3)2 = 34, xy-3(x+?/)=6. 95. Two trains start at the same instant, the one from B to A , the other from A to B, the distance between A and B being 100 miles. The trains meet in 1 hour 15 minutes, and one train gets to its destination 1 hour 20 minutes before the other. Find the rates of the trains. 296 MISCELLANEOUS EQUATIONS. gg ax - by ^ (2a + &)x-(a + 2&)y _ ^2 _ 52 3 7 97. 13x2+12 = 80x. 98. (a; + l)(a: + 2)(a; + 4)(a; + 5)=4. 99. a;2(y + 1) + 2,2 (^^ ^. i) = IO9, xy = 12. 100. A starts to walk from P to Q at 10 a.m., B starts from Q to walk to P at 10.24 a.m. They meet 6 miles from Q. B stops 1 hour at P, and A stops 2 hrs. 54 min. at Q, and returning they meet midway between P and Q at 6.54 p.m. Find the distance from P to Q. 101. ?/2 + z'^ = z^ + x'^ = x^-\-y^ = axyz. 102. ax(y -\- z) = hy{z -\-x) = cz(x + y) = xyz. 103. x'^ -{■ 2yz = y"^ -\- 2 zx = z"^ ^- 2xy = 12. 104. Resolve x^ -\- y^ -\- z^ — Z xyz into the three factors x-\- y + z, x-\- wy -\- (jfiz., x + (o^y -\- wz, where w=— ^ + ^V'— 3, and determine three alternative relations between a, 6, and c, one of which must be satisfied in order that the equations JCS 4. y3 + ^3 _ 3 xyz - 0, ax -\- by -{- cz = 0, a b c may be simultaneous in sc, y, and z. POWERS AND HOOTS. 297 CHAPTER XX. Powers and Roots. 195. The process by which the powers of quantities are obtained is called involution ; and the inverse process, by which the roots of quantities are obtained, is called CYolution. [Art. 9.] INVOLUTION. 196. When m and n are any positive integers, we have by definition a*" = aaaa. . . to m factors, and a" = aaaa... to n factors ; .*. a'^xa''^ (aaaa... to m factors) x (aaaa... ton factors) = aaaa... to m + w factors = a"'+**, by definition. Hence when m and n are any positive integers, a"'-^a'' = a'"+'*. Thus, the index of the product of any tivo powers of the same quantity is the sum of the indices of the factors. This result is called the Index Law. From the Index Law we have a*" X a" X a^ = «'*+" X a'' = a"*+''+^, and so on, however many factors there may be. 298 POWERS AND ROOTS. Hence a"* x a** x a^ • • • = a'""*""+*+'". Thus, the index of the product of any number of powers of the same quantity is the sum of the indices of the factors. 197. We have a X a X a X a... to m factors a"" /a" = a X a X a X a... to 71 factors Now, if m is greater than n, the n factors of the denom- inator can be cancelled with n of the factors of the numerator : we then have m — n factors left in the numerator. Thus, when m is greater than n, If, however, n is greater than m, the m factors of the numerator can be cancelled with m of the factors of the denominator : we then have n — m factors left in the denominator. Thus, when m is less than n, 198. To find (a"*)" when m and n are positive integers. By definition (a*")" = a™ X a*" X a"* X ••• to n factors _.^m+m+m+... to nternis^ bj Art. 190. = a""*. Hence (a"*)" = a"*". POWERS AND ROOTS. 299 Thus, to raise any power of a quantity to any other power, its original index must he multiplied by the index of the power to which it is to be raised. 199. To find (ab)^. (a6)"'= ab X ab X ab... to m factors, by definition, = (aaa. . . to m factors) x {bbb. . . to m factors) [Art. 52.] = a"* X 6"*, by definition. Hence (ab)"" = a'^b"'. Similarly (^abc)'^= abc x abc x abc... to m factors, = (aaa... to m factors) x {bbb... to m factors) X {ccc... to m factors) ==a"* X 6"* X C*. Hence (abc)"" = a'^b'^d^, and so on, however many factors there may be in the ex- pression whose mth power is required. Thus, the mth power of a product is the product of the mth powers of its factors. 200. The most general monomial expression is of the form a"'^>^c^.. To find (a^'&^c^...)'". {a'b^(f...Y = {a''Y{l)>'Y{(fY... [Art. 199.] = a^6^'"(f'"... [Art. 198.] Thus, any power of a monomial expression is obtained by taking each of its factors to a power whose index is the product of its original index and the index of the power to which the whole expression is to be raised. 300 POWERS AND ROOTS. 201. The following is an important case. To find ©• 7)=-x-x----tom factors, by definition, bj b b b ' ^ ' _ a X a X a... to m factors p » , 1 fi? i b X b X b... to m factors -^. * \b) ~ b' 202. It should be noticed that all powers of a positive quantity are positive, and that successive powers of a negative quantity are alternately positive and negative. This follows at once from the Law of Signs ; for we have (-ay = {-a)(-a) = + a'', (-a)« = (-a)2(-a) = (+a-0(-a) = -a«; (-ay = {-ayi-a) = {-a')(-a) = + a'', and so on. Thus ( - a) 2" = + a2«^ and ( - a) 2«+i = - a-'^+K From the above it is clear that all even powers, whether of positive or of negative quantities, are positive, and that all odd powers of any quantity have the same sign as the original quantity. 203. We have already proved the following cases of the involution of binomial expressions. POWERS AND ROOTS. 301 and {a-^hY = a^ + ^o?h + 3aW + h\ If we multiply again by a + 6, we shall have (a + hy = a' -\-4.a'b + 6a'b' -{-4.ab^-{- b\ By multiplying the last result by a + 6 we should obtain (a + by, and by continuing the process we could obtain any required power of a + 6 ; but to find in this way any high power, for instance to find (a + b)^, would clearly be very laborious. We shall shortly prove a theorem, called the Binomial Theorem, which will enable us to write down at once any power of a binomial expression. The above formulae are identities, and are true for all values of a and b ; hence we can write down the squaros and the cubes of any binomial expressions. Thus (a* - ¥y = (a^) 2+ 2(a*) ( _ M) + ( - b^y = a^-2a^h*-^b\ and (5 a2 - 3 b^-y = (5 a2)3 + 3(5 a^y(- 3 52) + 3(5 a2)(_ 3 62)2 4. (_ 3^,2)8 = 125 a^ - 225 a*&2 + 135 a'^¥ - 27 b^. Also (a + 6 + cy={a + (6 + c)f = a^ + 3a2(6 + c)+ 3a(6 + c)2 +(6 + c)*. 204. An important case of involution is considered in Art. 70, where the square of a multimonial expression is obtained. 302 POWERS AND ROOTS. EXAMPLES LXI. Write down the value of each of the following : 1. (a^y. 21. (a3 + &3)3. 2. (x5)3. 22. (2a2- 362)8. 3. (-a2)3. 23. (3 a2 - 2 62)8. 4. (-a3)2. 24. (a2 + 62 + c2)2. 5. (-2a5)4. 25. (a3_2 63 + 3c8)2. 6. (-3a*)5. 26. (a2_4 62-3c2)2. 7. (-a62)*. 27. (a;2_3a;-6)2. 8. (a^b^y. 28. (3x2 -a; -5)2. 9. (- a6*y. 29. (2x2 + 5x- 1)2. 10. (-Sa'b^cy. 30. (3x2 -6x- 6)2. 11. {-ab-^&y. 31. (1 + ic + a:2 4- x3)2. 12. {-ba'bH^y. 32. (X3-X2 + X-1)2. 13. i-X' 33. (x3 + X'^ - 2 X - 2)2. Wl 34. (a + 26 + 3c + 4d)2. 14. (-a- 35. (2a-6-fc-2 (Z)2. 15. ( 6%0 36. 37. (x2 + X + \y. (X2 _ X + 2)3. 16. (2 a* + 3 &3)2. 38. (3x2- 5a; + 1)3. 17. 18. 39. i("ir*K-=)' 19. 20. (-a2 + 2a&)2. (a2+&2)3. 40. -C-^)' THE BINOMIAL THEOREM. 205. The numerical coefficients on the right side of the formulae of Art. 203 may be so constructed as to exhibit the law of their formation. Thus, written in the POWERS AND ROOTS. 303 order of their occurrence in the formulae, they and their reconstructed equivalents are 2 2^ 1' 1.2' 3 3-2 3-2.1 1, 2, 1 = 1, 1, 3, 3, 1 = 1, f 1.2' 1 2.3' ^ A a A 1_1 4 4_^ 4-3.2 4.3.2.1 1, 4, b, 4, 1-1, ^, ^ ^, ^ ^ ^, -^2.3-4* If this law of their formation obtains for higher powers of {a-\-b), we shall be able to write out, by a very simple rule, the entire series of terms in the expan- sion of (a + 6)", wherein n is any positive integer. Thus, if the law holds for all positive integral values of n, (a + 6)-^ = a^ -j- 5 a'b + |^ a'b' + f^«'&'' -mi--'*'- and in general (a + by = a^-\- na^-'b -f '^i'^-^l a^-^b' + ... • n(n-l)-.(n-r + l)^„_.^ .. 1-2. 3.. .r Here r is some integer less than n, and y,(n-l)... (n-r+1) n-r7.r 1-2. 3... r which is called the general term, may be made to assume in succession the form and value of every term in the series, by giving to r the successive values 0, 1, 2, 3, 4, ...71-1, n. 304 POWERS AND ROOTS. Thus, if r = 5, this general term becomes n{n-l)(n-2)(n-3)(n-4:) ,, 1.2.3.4.5 We proceed to prove that the above formula is true for all positive integral values of n. The proof here given is by mathematical induction (monomorphic trans- formation) explained in Art. 145. For brevity's sake we write the formula thus : in which the letters c have the respective values _1 _ _n(n — 1) Co — 1, Ci — n, C2 — — - — - — ^, •.. _ n(n-l) ... (n-r + 1) _^ We assume provisionally that the formula is true when the index is n. Can we justify this assumption? Multiplying both sides of it by (a + b), we obtain (a + 6)"+^ = Coa"+^ + (Co + Ci) a-b + (c, + c^) a^ 'b^ + (C2 + C3) a"--6^ + - + (c,_, + c,) a"-'-+^6^-f ... But Co 4- Ci = n + 1, Ci 4- C2 = w + ?i (n — 1) _ n(n + 1) 1 . 2 ~ 1.2' , n(n-l) n (n-l)(n-2) C2 + C3- ^2 +- 1.2.3 _ rt(yi — l)(n + l) 1.2.3 POWERS AND ROOTS. 305 and in general _ n(7i—l)"'(n—r-{-2) n(n — 1) "♦ (n— r+ 1) Cr-i-{-c^- 1.2. 3. ..,,_! + 1.2.3...r _ n{n-l) ••» {n-r-\-2){r + n-r-{-l) ~ 1.2.3...r ^ (y^ + l)n(yi-l) -'■ (n + l-r + 1) ~ l'2'3"-r Substituting these results for Cq, Cq -f- Cj, Ci + Cg, etc., in the formula, we have 1 • ^ . (n + l)n{n - 1) ... (n + 1 - r + 1) ^n+i-r^r 1.2.3...r The series thus arranged is exactly the same in form as that for (a + 6)**, n-\-l having now taken the place of n. Hence, if the formula be true for any positive inte- gral value of n, it is true when n is replaced by the next higher integer. But it is obviously true when w = 1 ; hence it is true when n = 2. And, being true when n = 2, it is true when n = 3, and so on indefinitely. Therefore it is true for every positive integral value of n. [Art. 145.] The proposition thus established is known as the Binomial Theorem for positive integral exponents. It is one of the most important theorems of algebra, and will be discussed more fully in a subsequent chapter to be devoted to this special topic. 306 POWERS AND ROOTS. The special numerical coefficients which have been designated above by Cq, c^, c.2, Cg, • • • c^, • • • c„ are called the binomial coefficients. They are frequently used in other algebraic formulae and should be committed to memory. Ex. 1. If in the binomial formula we write w = 3, we have (a + by = a^ + -a'h + — a62 + 63 1 1 • ^ = a3 + 3a-2& + 3a6-^+ &*. Ex. 2. If a = 2, 6 = ?/2j and n = 3, we have 1 1 • ^ Ex. 3. If a = -, 6 = -, and w = 4, we have 2' y /i + iV^l + i 1 i + iii 1,1. 4.3.2 111 V2%y 2* 1 '23" y"^1.2*22' y2 1.2.3' 2 "ys ^4 ^1 + 1.1 + 3.1+2.1+1 206. The formula for the expansion of {a — hy is obtained by writing — 6 in place of h in the previously written formula (a + ft)" = a'' 4- Cia^-^6 + Cstt"" ^W + Cga^-^ft^ _,_..._,_ ^^n^ the even terms of which become negative by this change, because odd powers of a negative quantity are negative. Thus, (a - ft)** = a- - Cia'-ift + c^aP-^W - c^a^'-^W -f ••• ± 6^ and the last term has the positive or the negative sign according as n is odd or even. POWERS AND ROOTS. 307 Ex. 1. If a = X, b = 2y, and n = 4, the result of substitution in the formula for the expansion of (a — 6)« is {x-2yy = x^-4 x\2 y) + \^x\2 yy -\^^x(2yY^{2yY = a^ - 8 x3y + 24 x^y'^ - 32 xy^ + 16 y*. Ex. 2. If a = 2 a;, 6 = 3 y^, and w = 5, the result of substitu- tion in the formula for the expansion of (a — 6)" is (2x - 3y2)5 =(2a;)5 - 6(2x)4(3 ?/2) + ^ (2x)3(3?/2)2 1 • £i -(32,2)5 = 32 x5 - 240 a;*?/2 + 720 ajSy* _ 1080 x^^ + 810 a;y8 -243yio. 207. General Term. By inspection of the formula, it is at once obvious that the number of any term in the expansion of (« ± 6)" is one greater than the suffix of c (or number of factors in the denominator of c) in that term. Thus c^a''^^ is the fourth term. Hence, in gen- eral, the (r + l)th term is and by means of this formula any isolated term may be written down. Observe that r is at once the exponent of 6, the number of factors in the denominator and the number of factors in the numerator, and that the sum of the expo- nents of a and h is n. 808 POWERS AND ROOTS. Ex. 1. The eleventh term in the expansion of (1 + x'^y^ is 16. 14. 13. 12 -11. 10. 9. 8. 7- 6 1.2. 3. 4. 5. 6. 7. 8. 9. 10 (x2)io = 3003x20. Ex. 2. The sixth term in the expansion of (x — \ y^y^ is ^ _ 11.10.9.8.7^e(^ , , ^ _ 2_31 1.2.3-4.5 ^^^^ 16 ^ The sign of this term is negative because even terms in the ex- pansion of (a — 6)" are negative. EXAMPLES LXII. Write out the following expansions : 1. (x+ay. 3. (Sx-2yy. 6. (2x2-1)6. 2. (l-x2)5. 4. (2a + 3rt2)4. 6. {y - xy. 7. Find the third term of (a - 3 6)^. 8. Find the fifth term of (2x - x^y^. 9. Find the sixth term of (2 a — |)^. 10. Find the seventh term of (1 — xy^. 11. Find the eighteenth term of (1 + x)^. 12. Find the twenty-first term of (1 — x)^. 13. Expand (a + ^by + (a - V^)*- 14. Expand (a + ^hy + (a - ^by. 15. Expand (a + y/by 4- (« - V^)*- 16. Find the middle term of (1 + xy. 17. Find the middle term of (1 + xy^. 18. Find the middle term of (2 x - 3 yy. 19. Show that in the expansion of (1 + x)'»+'», the coeflBcients* of x* and x» are equal. 20. Expand (^ + - ) * 21. Expand I ^x + ^J-i^a;-^)*- POWERS AKD KOOTS. S09 EVOLUTION. 208. We know that there are two square roots of a?, namely ± a ; we also know [Art. 188] that there are three cube roots of a^, of which a is one, and the other two are imaginary. There is therefore an important difference between powers and roots ; for there is only one nth power, but there is more than one nth root, of a given expression. 209. An expression which when raised to the 9ith power, where n is any positive integer, becomes equal to a given expression, is called an ?ith root of the given expression. We have shown in Art. 199 that the mth power of a product is the product of the mth powers of its factors ; hence, conversely, the mth root of a product is the product of the mth roots of its factors.* Thus -y/ahc = y'a ^b -^c, and Va6 = ^a ^b. Again, we have shown in Art. 200 that the nth power of a monomial expression is obtained by multiplying the index of each of its factors by n. It follows conversely that if we divide the index of each factor of a given expression by n, we shall obtain an 7ith root of the expression. For by raising to the nth * It should be noticed that the proof, in Art. 52, that the factors of a product may be taken in any order, only holds good when those factors represent integral or fractional numbers, and does not enable us to assert that VaX Vb = y/b X v^a, when Va or \/& is really a surd. For a proof that surds obey the Fundamental Laws of Algebra see TrecUise on Algebra, Art. 162. 310 POWERS AND ROOTS. power the result obtained by such division of the indices, we must clearly get the original expression. Thus one value of ^a* is a^, one value of VoFb^^ is a^h^c^^ and one value of \/a"^6"« is a^h^. When the square root of an expression which is not a perfect square, or the cube root of an expression which is not a perfect cube, is required, the operation cannot be performed. We can, for example, only write the square root of a as ^a, and the cube root of a? as -^/o?, and similarly in other cases. SQUARE ROOT. 210. We now proceed to consider the square root of multimonial expressions. In Art. 114 we have shown how to write down the square root of any trinomial expression which is a com- plete square. Having arranged the expression according to ascending or descending powers of some letter, the square root of the whole expression is then found by taking the square roots of the extreme terms with the same or with diff - ent signs according as the sign of the middle term is positive or negative. Thus, to find the square root of 4a«-12a^63 + 96«. The square roots of the extreme terms are ± 2a* and ±36^ Hence, the middle term being negative, the required square root is ± (2 a'' — 36"*). %:. POWERS AND KOOTS. 311 Note. — In future only one of the square roots of an expression will be given, namely that one for which the sign of the first term is positive : to find the other root all the signs must be changed. As other examples V(49 aio + 28 a« + 4) = 7 a5 + 2 V(l + 5x1/3 + -2^xV)= 1 + f a;2/3, and y/{a^ + 2 a(6 + c) + (& + c)2}=: a -^ b + c. 211. When an expression which contains only two different powers of a particular letter is arranged accord- ing to ascending or descending powers of the letter, it will only contain three terms. For example, the ex- pression a- + 62 + c^ + 2 6c -f 2 ca + 2 a6, which contains no other power of a but a^ and a, when arranged according to powers of a, is a^ + 2a (6 + c) + (62 4- c2 4- 26c). Thus any expression which only contains two different powers of a particular letter can be written as a trinomial expression ; and since we can write down the square root of any trimonial expression which is a complete square, it follows that the square root of any expression which is a complete square can be written down by inspection, provided that the expression only contains two different powers of some particular letter. For example, to find the square root of a2 + 62 + c2 + 2 6c + 2 ac + 2 a6. Arranging the expression according to the descending powers of a, we have a^ + 2 a(6 + c) + (5^ + 2 6c + c2), that is, a2 + 2a(6 + c) + (6 + c)2, which is (a 4 (6 4- c)]2. 312 POWERS AND ROOTS. Thus V(«^ + &2 + c2 + 2 6c + 2 ca + 2 a6) = (a + 6 + c). Also, to find the square root of «* 4- 4 y* + 9 2;4 + 4 xhj^ - G x'^z'^ - 12 y^z^. Arrange the expression according to descending powers of ic ; we then have x^ + 2x2(2 2/2 _ 3^2)+ 4^4 + 9^4 _ I2y^z^^ that is, x* + 2x2(2?/2-3;s2-)_,.(2?/2-3s2)2, which is {x2 + (2 y'^ - 3 z'^)}'^. % Thus VCic* + 4 ?/4 + 9 2r4 + 4a-2i/2 _ 6 x^z^ - 12 ?/22r2)^(a;2 +2^/2- 3 ^2). Again, to find the square root of a2 -\-2abx-h (62 + 2 ac)x2 + 2 6cx3 + c2a;^ Arrange the expression according to powers of a ; we then have a2 + 2 «(6x + ex'') + 62^2 + 2 bcx^ + c2x4, that is, a2 ^ 2 a(6x + cx2) + (6x + cx2)2, which is {a + (6x + cx2)}2. Hence the required square root is (a, -f 6x + cx2) . And to find the square root of x6 - 2 x5 + 3 x* + 2 x^(y - 1) + x2(l - 2 y) -\- 2 xy -\- y^. The expression only contains y'^ and y ; we therefore arrange it according to powers of y ; we then have ?/2 + 2 y(x3 - x2 4- a;)+ a^ - 2x5 + 3 x* - 2x3 + x2. Now if the expression is a complete square at all, the last of the three terms must be the square of half the coefficient of y ; and it is easy to verify that a^ _ 2x5 4- 3x* - 2x3 + x2 is (x' - x2 -j- x)2. Thus the given expression is 2/2 + 2 y{X^ - X2 + X) + (X3 - X2 + X)2. The required square root is therefore 2/ + x8-x2 + x. POWERS AND ROOTS. 313 From the above it will be seen that however many terms there may be in an expression which is a perfect square, the square root can be written down hy inspection, provided only that the expression contains only two dif> f event powers of some particular letter. EXAMPLES LXIII. Write down the square roots of the following expressions. 1. 9ic2- 30a;y + 252/2. 5. a;8 - 6 a; V + 9 yS. 2. 25 X* - 30 x'V + 9 y*. 6. 9 xi2 - 6 7^y^ + ^. 3. 4x* - 12xV + 92/^- 7. ix6-ix3y3 + 1^6. 4. 4x10 -12x52/3+ 9 j^ 8. 1 x8?/6 _ I a^ys _,. ^. 9. 25 xY' - 40a263xV + 16a*?>6. 10. ^+8xV+ 16 aV- Or 11. 9 ^ - 24 xV + 16 ^• 12. i- 42^+ 49010. x^ x3 13. a2 + 4 52 + 9 c2 4. 12 6c + 6 ca + 4 ah. 14. 4 a2 + 62 _|. 9 e-2 + 6 60 - 12 ca - 4 a6. 15. 4 a* + 6* + c* - 2 6^02 - 4 c2a2 + 4 ^262. 16. 25 a* + 9 6* + 4 c* + 12 62c2 + 20 c2a2 - 30 ^262. 212. In order to show how to find the square root of any algebraical expression, we will take an expression and form its square, and then show how to reverse the process. 314 POWERS AND KOOTS. Consider, for example, x' + 2xy-^'6f (i.), whose square is a;^ + 4ar'2/ + 10a^?/' + 12a/ + 9?y* . . . (ii.), both expressions being arranged according to descending powers of x. We may write the square of x'^ -\- 2 xy -\- ^ y- in either of the following forms : [x" -\- {2xy + Zf)\'=x' + 2x\2xy -\-3f) + {2xy + ZfY (iii-), \{x' + 2xy)^ 2,f\' = {x^ ^-2xyy ^-2{x^ + 2xy)^f + (3/)^ (iv.). Now it is clear that the first term of (ii.) is the square of the first term of (i.). Hence the first term of the root of (ii.) is found by taking the square root of its first term. Again, we see from (iii.) that when we have sub- tracted aj* (the square of the first term of the root), the term in the remainder which contains the highest power of a; is 2 a;^ X 2 xy, which is twice the product of the first and second terms of the root. Hence, after subtracting from (ii.) the square of the first term of the root, the second term is obtained by dividing the first term of the remainder by twice the first term of the root. Again, we see from (iv.) that when we have subtracted {'X^-\-2xyy, that is, the square of the part of the root already found, the term in the remainder which contains POWERS AND ROOTS. 315 the highest power of x is 2x- x 3/,. which is twice the product of the first and third terms of the root. Hence, after subtracting from (ii.) the square of that part of the root already found, the next term of the root is obtained by dividing the first term of the remainder by tivice the first term of the root. If we now subtract the square of x^ -\-2xy -\- 3y^ from the given expression, there will be no remainder; and hence x--{- 2xy -\-3y' is the required root. We will now consider the most general case. Suppose we have to find the square root of {A -f By, where A stands for any number of terms of the root, and B for the rest ; the terms in A and B being arranged according to descending {or ascending) powers of some letter, so that evei^y term in A is of higher (or lower) dimensions than any term in B. Also suppose that the terms in A are known, and that we have to find the terms in B. Subtracting A^ from {A + B)^, we have the remainder (2A-\-B)B. Now from the mode of arrangement it follows that the term of the highest (or lowest) degree in the remainder is twice the product of the first term in A and the first term in B. Hence, to obtain the next term of the required root, that is, to obtain the highest (or lowest) term of B, we subtract from the whole expression the square of that part of the root which is already found, and divide the first term of the remainder by twice the first term of the root. The first term of the root is clearly the square root of the first term of the given expression; and when we 316 POWERS AND ROOTS. have found the first term of the root we can find each of the other terms in succession by the above process. For example, to find the square root of ic* + 4 xhj -\- 10 x^y^ + 12 xy^ + 9y*. 3^+4 a;^^/ -I- 10 xV+ 12 xy^-\-9 y* {x^-\-2xy-\- 3 y'^ (a;2)2 = '^ (x2 + 2 xyy = X* + ix^y + ^x^y^ (x2 + 2 xy + 3 y2)2 - aj4 + 4 aj3y + 10 xV +I2xy^ + 9y*. The given expression must first be arranged according to ascend- ing or descending powers of some letter. We then take the square root of the first term of the given expression : we thus obtain x^, the first term of the required root. Now subtract the square of x^ from the given expression, and divide the first term of the remainder by 2x^: we thus obtain 2 xy, the second term of the root. Now subtract the square of x^ + 2 xy, which is X* + 4 xhj + 4 xhj'^, from the given expression, and divide the first term of the remainder, namely 6 x^?/^, by 2 x^ : we thus obtain 3 y\ the third term of the root. Subtract the square of x^ + 2'xy + 3«/2 from the given expression, and there is no remainder. Hence x^ + 2 xy + 3 ?/2 is the required square root. The squares of x^, x^ + 2 xy, etc. , are placed under the given expression, like terms being placed in the same column : the remainder left after taking away any square is then obvious by inspection. 213. Instead of finding each square independently, some labour can be saved by making use of the previous square. Thus the process of finding the square root of an algebraical expression is generally written as follows, the same example being taken as before : POWERS AND ROOTS. 317 ic* + 4x3?/ + lOxV +12x2/3 + 9^4 (x2 + 2 xy + 3^/2 x* 2x2 + 2x2/) 4x^2/ 4x3y+ 4x2^/2 2x2 + 4x?/ + 3^/2) 0x21/2 6x2?/2+ 12x2/3 + 9?/* The first term of *he root is x2. Having subtracted the square of x^, namely x*, the first term of the remainder is 4 x^y. Now double the part of the root already found and divide the first term of the remainder by the result ; we thus obtain the next term of the root, namely 2 xy Add this term of the root to 2 x2, placing the sum in the ordinary position for the divisor. Now multiply the sum by 2xy and subtract the product from the remainder 4 x^y -{-•••, we then have the remainder 6x22/2+.... Repeat this process as often as may be necessary, [It is of importance to notice that by the above process we have at the end of the second stage subtracted altogether (x2 + 2 X2/)2, since (x2 + 2 xy)- = (x2)2 + (2 x^ + 2 x?/) x 2 xy. Similarly at the end of every stage it will be seen that what is sub- tracted on the whole is the square of that part of the root which has been found up to that stage.] EXAMPLES LXIV. Find the square roots of the following expressions : 1. x4 + 2x3 + 3x2 + 2x + l. 2. 4x* -8x3 + 4x+ 1. 3. 9x*-36x3 + 72x + 36. 4. 1-xy - V- xV + 2 x32/3 + 4 x*y^. 5. 4x4 + 4x3 - ix+ iV 6. x4-2x3 + ix2-|x + xig. 318 POWERS AND ROOTS. 7. x^ -\-2x^y -{- ^x'Y + 2xy^-\-y*. 8. 16 - 96x + 216x2 -216x3 + 81 X*. 9. 1 + 4x+ 10x2 + 12x3 -f9x*. 10. 4x*-4x3 + 3x2-x+ ^. •11. (1 + 2x2)2 - 4x(l -x)(l + 2x). 12. x6-4x5 + 6x*-8x'' + 9x2_4x + 4. 13. 9x5 - 12 x5 + 22 x^ + x2 + 12 x + 4. 14. x/' - 22 X* + 34 xs + 121 x2 _ 374 x + 289. 15. a2_cfx+ |x2 + 8a-4x+ 16. 16. x8 + 2x7+ x8-4x5- 12x* - 8x3+ 4x2+ 16x+ 16. 17. 16x2 -96x + 216 --- + —. X X2 18. x«-6x* + 15x2 -20 + 1^--^- + !. X2 X* X** 19. X* + 2 x\y + z)+ ot'^iy'^ -{- z^ + 4 j/z) + 2 xyz(y + z)-^y^z^, 20. 2x2(2/+0)2+22/2(2;+a;)2+2 02(a;+2/)2+4x*/0(x+!/+0). FRACTIONAL AND NEGATIVE INDICES. 319 CHAPTER XXI. Fractional and Negative Indices. 214. We have hitherto supposed that an index was always a positive integer; and this is necessarily the case so long as we retain the definition of Art. 19; for, with that definition, such an expression as a^, for example, has no meaning whatever. It is however found convenient to extend the meaning of a", so as to include fractional and negative values of n. Now it is essential that algebraical symbols should always obey the same laws whatever their values may be ; we therefore do not begin by assigning any particular meaning to a", when it is not a positive integer, but we first impose the restriction that the meaning of a" must in all cases be such that the fundamental Index Law, namely, a"* X a" = a'""*"", shall always be true; and it will be found that the above restriction is of itself sufficient to define the meaning of a" in all cases, so that there is no further freedom of choice. 1 For example, to find a meaning of a^ consistent with the Index Law. We must have a^ X a- = a- ^ = a^ = a. Thus a- must be such that its square is a, that is a^ must mean ,ya. 320 FRACTIONAL AND NEGATIVE INDICES. Again, to find a meaning of a~i consistent with the Index Law. We must have a^ x a~^ = a^~^ = a^ ; . •. a-i = «V«^ = -• a Thus a"i must mean — a _ We now proceed to consider the most general cases. 1 215. To find the meaning of a% where n is any posi- tive integer. By the index law 111 a" X a" X a" X •••to n factors •+^ + .. to II terms _^n « n _^,n^,jl^^^^ 1 Hence a'* must be such that its nth power is a, that is 1 a" = ya. 216. To find the meaning of a**, where m and n are any positive integers. By the index law m m m a** X a" X a" X •••to ii factors ^ + ^-|.!?+... ton terms ^ M Thus a" ts equal to the nth root of a**, that is m FRACTIONAL AND NEGATIVE INDICES. 321 We have also 2 11 a" X a" X a" X • • • to m factors 1 ^ 1 ^ I _,.... to m terms 'Jt m 1^ Hence ci" may be considered as the mth power of a", and by Art. 215, 2 a" is -ya m Thus we may consider that a" is the nth root of the mth power of a, or that it is the mth power of the nth root of a; which we express by Note. — It should be remarked that it is not strictly true that «/(«»») =:{;y («)}»» unless by the wth root of a quantity is meant only the arithmetical root. For example v^(a*) has two values, namely ± a^^ whereas ( ^a)* has only the value + a^. Examples (i.) 83 = ( ^8)2 = 22 = 4, (ii.) 42 = ^4^ = ^64 = 8, (iii.) 3^ = ^35 = ^43. ^~ 217. To find the meaning of a^. By the index law aP xa"' = a^+"' = a"* ; .-. a« = »"»/«"* = 1. Thus, whatever a may be, a^ = 1. 322 FRACTIONAL AND NEGATIVE INDICES. 218. To find the positive value. By the index law meaning of a-- where m has any a"' X a-^ = a*"-*" = = < and, by the precedin g article, a^ = • . a"" X a~'" = ^1; = 1, so that a-"» = ~ and a^ = a"* 1 Thus we can transfer any factor from the numerator to the denominator, or from the denominator to the mimerator, of ot fraction, provided ive change the sign of its index. For example 2-^1 = 1, 2-^ 4 5 a-^h [ Thf - ff8h-2^-3^,-l —. 1 -''^'^ y a-^h-^xhj and 219. In the preceding articles we have found that in order that the fundamental Index Law, namely a"^ xa"" =z «"*+», may always be obeyed, a*" must have a definite meaning when n has any given positive or nega- tive value. It can be proved* that, with the meanings thus obtained, a"* X a" = a*""*"*, (ory = a'*", and {ahy = a^ft", are true for all values of m and n. It therefore follows that we may deal with quantities containing fractional or negative indices in precisely the same way as if the indices were positive integers. * See Treatise on Algebra, Art. 168. FRACTIONAL AND NEGATIVE INDICES. For example 323 («V)2^«fx2^*x2 a^b. -k 1 1 _2 V a"^ X a'' X a ^ = a a"^ X a ^ = a^ - = a^ = «/a. 1. -|X6 1 a* ^(a35-3c*) = ^(a^) • ^(6-3) . ^(c4)= a^fo scs ^aft-VA EXAMPLES LXV. Find the numerical values of 1. si 3. 1()~2-. ^ _3 5- (A) '■ 2. 4 ^ 4. (,V) '• 6. (f|)-3. Simplify V 7. (27)-^. 9. (T^^)--f. 8. (lOO)-l 10. (t\\)-^. -f 11. «3 X « 3, 14. a X a' 17. (a^5"2)2. 12. a3 X g3.i ^ 15. ah 3 x a h\ 18. (a^ft 1)3. 13. a •' X rt. 20. (a-»6~3)!. 16. (a25)2. 21. {(a-^)2}T. 22. {(a"^)2}-^. , 23. {(a"^)3fi 24. a^ X a"^' X a^^, ::^ (:L-' - 25. J X a~^ x(a2)"'>(aT2)5. 26. xi'+« X x^-9 H- x'-p. -2 13 19. a 3 6^) .J. 27. {x^-'^)p X (xr-p)i X (xt'-iy 1 2 a;'y3 y 7?Z 29. (xyy^^ x^y" x:y-^*y'+':!^+y 5 ? 324 FRACTIONAL AND NEGATIVE INDICES. 32. a¥x(4)^4- 33. X^P+1 X XP-4^ X (x2)9-2r h- X*P-^. J P^ YJ q-r^ -tp r r~j^ "» j p-? g-r r-p — 84. ta;^M la;''- /ta;'^ / -Ha; »■ p « a; *• I ^. J ' X X *• P « . ^^^Express with fractional indices and simplify 36. > X > X >. ^ ^^, ^ ^^, ^ ^4 ^ «l. 37. ^(«-V) X sy(a2/2). ^ ^ ^^c 38. ^(a%5)xV(a«x). ^ "^ ^^^ ^(a'^ftiOcS)^ ^^^^3). (^, 39. ^a3 X «/a7 x a^ ^ a^r. 43, ^(^s^j-a)^ ^(a-^ftS). 7 «• J^_ 44. ^(a66W)x 6""2'x(cV)"i ^ '^ Express the following quantities with radical signs and positive indices. 45. J - a-2. 47. ah-^ - a~h~^. 46. a-'^b'K 48. a-^ft"^ + S-SaVi 220. In the following examples the foregoing princi- ples are applied, but the results of the operations of addition, etc., which have to be performed on the indices are given without any of the intermediate steps. Ex. 1. Multiply a^ + 1 + a~^ by a^ - 1 -1- o~^ a^ + 1 -I- a'^' a^ - 1 + q"^ o^ 4- a^ + 1 1 -I- g"^ -I- g"^ a^ +1 -fa"i FRACTIONAL AND NEGATIVE INDICES. 325 Ex. 2. Divide a + & 4- c - 3 a^b^c^ by a^ + b^ +c^. We perform the work by the synthetic method [Art, 83.]. (63 +C3) a -0 -Sa^b'c^ +(6 + c) - a^(b^ 4- c^) + a^ib^ + c^ + 2 6^ ) - (ft + c) ^3(53 _|. c3)+ 53 4- c3 - 6^c3 0. Ex. 3. Find the square root of (x^y = x3 4 x^ -\- 2 x'^ -\- 4x ■- ix^ + x^" (x^ - 2 x^ + x^ (x^ - 2xfy = x^ - 4x'^ + 4a: (x^-2x^+x^y = x^ - ix^ + 2x^ + 4:X-ix^ -\- xK EXAMPLES LXVI. Multiply : 1. x^-\-y^hjx^ -y^. 3 3 13 2. ic- + y ' by a;2 - y^. 3. xT + 1 by x^ - 1. 4. a;3 + y^ by a;» - i/^. 5. x' + a;3 + 1 by a;3 - 1. 6. x^ -x'y^ -{- y^ by x^ + y^. 7. a^ + a^b^ + 6^ by a-' - bK 4 22 4 4 22 4 -> 8. a3 _|_ ci^i)3 + 5-3- by a3 _ ^353 _|_ 53 5 3 1 _1 3 1 9. x^ - x^ + x^ — X f by x'^ + x+. 735 31 - 10. x^ — x2 + x^ — X by x^ + x'\ r/ll. x» + x2 + 1 by x-« + X 2+1. I 12. X2« + X*y+V2nby X-2n + x-«2/ « + 2/ 2», 13. a^ + 6^ + 0^ - b^c^ - c^a^ - ah^ by a^ + 6^ + c^. ^ .i ^ 326 FRACTIONAL AND NEGATIVE INDICES. 14. i a2" - tJj ah'^ + ^V a^^> - A ^ ' l>y 2 «^" + \ &^- 15. 81 x^ - 27 x^y 3^ + 9 x^y^ - 3 x^y + ?/3 by 3 x^ + 2/^. Divide : ^ 16. a - 6 by a^ - ft^^. 19. x^ + ?/2 by x^ + y^. 17. x3 - 2/3 by a;l _ 2,t. 20. ic-2432/nyx^-32/J. 3» 3n n n _^ 18. x^ -y'^ hyx^-y^. 21. 2/H6V+&* by y-&^y- + ^ 22. arT^/'-f + 2 + x~^?/* by x*?/"* - 1 + x~*y^. 23. a 2 _|_ ^2^1 _ ah^ - ah + ah'^ + 6-* by a^ + 6^. 24. a'f - 2 + a~3 by a 3 - a'^. X 26. (x - x-i) - 2(x* - x"^) + 2(x^ - x~^) by x^ - x"i 26. x^y~'^~ + /^-jc's by x~^y~^ + ?/^x"3. 27. Simplify {ah-^(^)^ x (a^ft^c'i) » x (a-5&c2)*. 28. Simplify (a - 4 6^) (a + 2 aW^ + 4 6^) (a - 2 aW^ + 4 6^) 29. Show that 1 - a^-^h^-a-^-h-^ ^ (g _ a-i)(6 - ^-i)^ ^252 _ a-2ft-2 aft + a-ift-i 2 'J 30. Show that —^ ^^ ^ — + — ^ — = 2 + xl X^ - 1 X3 -f 1 X3 - 1 X3 + 1 31. Multiply 4x2-5x-4-7x-i + 6x-2 by 3x-4 + 2x-i, and divide the product by 3 x — 10 + 10 x-i — 4 x-2. 32. Simplify —^ + -J— + -^— + ^ 1 - X* 1 + x^ 1 + x^ ^ + ^ 33. Multiply (a + 6)^ + (a2 - 62)i ^. (« _ 6)5 by (a + 6) ^ - (a - 6)i FRACTIONAL AND NEGATIVE INDICES. 327 34. Write down the square roots of (i.) x^ + 2 x^ + 1, (ii.) 4x^ - 4x^y^ + y^, (iii.) ax^ — 2a^x^ + a'-^x. 35. Find the square root of 4a:2a-2 _ 12 xa^ + 25 - 24 x-^ a + IQx-'^a^. 36. Find the square root of 25 x^y-^ + i y'^x-- - 20 xy-i - 2 ?/x-i + 9. 37. Find the square root of 8. _3 _1 1 4 4 _6 1 4 17 » x^ — 2a ^x ^ +2 a^x^ + a ^x ^ — 2 a^x^ + a\ 38. Find the square root of 2 5 14 5 x3 — 4x*^ + 4x + 2x*^ — 4x3 + x3. Solve the following equations : 39. x^-2x^ + l=0. 41. x^- 26x4 -27 = 0. 40. X - 8x^ + 12 = 0. 42. x^ + 3x~3" = 4 43. 4x^-3x~^ = 4. 328 SURDS. COMPLEX QUANTITIES. CHAPTER XXII. Surds. Complex Quantities. 221. Definitions. A surd is a root of an arithmetical number which can only be found approximately.. Thus ^2 and ^4 are surds. An algebraical expression such as -^a is also often called a surd, although a may have such a value that ■^ya is not in reality a surd. Surds are said to be of the same order when the same root is required to be taken. Thus ^2 and 6'^ are called surds of tne second order, or quadratic surds ; also ^4 2 and 5 are surds of the third order, or cubic surds ; and -y/5 is a surd of the nth order. Two surds are said to be similar when they can be reduced so as to have the same irrational factors. Thus 2^2 and 5^2 are similar surds. The rules for operations with surds follow at once from the principles already established. Note. — It should be remarked that when a root symbol is placed before an arithmetical number it denotes only the arith- metical root, but when a root symbol is placed before an algebraical expression it denotes any one of the roots. Thus y/a has tioo values, but y/2 is only supposed to denote the arithmetical root, unless it is actually written ± ^2. V (c SURDS. COMPLEX QUANTITIES. 329 222. Any rational quantity can be written in the form of a surd. For example and 2 = ^2 _ ^23 = ^2». Again ^2 = ^l^2^ = ^2\ Also, since ^a x ^b — -\/ab, we have 3 V2 = V3' X V2 = V(3' X 2) = V18, 2^5 = ^2' X ^5 = ^(2« X 5) = sy40, and a-;y6 = ^a** x^h= Va^.. Conversely, we have V18 = V(9 X 2) = V9 X V2 = 3 V2, ^135 + ^40 = ^(3^ X 5) + ^(2« X 5) = 3^5 + 2^5 = 5^5, and ^aP-\--y/ab^ = a^a-{-b-y/a= {a + b)^a. 223. Any two surds can be reduced to surds of the same order. This follows at once from the fact that ^a = "^a"*. Eor example, to reduce -^2 and ^5 to surds of the same order. We have ^2 = ^S^2«J = ^2^ and ^5 = ^l-^5'\ = ^5'; thus the equivalent surds of the same order are -y/S and ^25. Again, to reduce -y/a and ^b to equivalent surds of the same order. 330 SURDS. COMPLEX QUANTITIES. We have -ya = VS V^*"*! ='V^^ and ^b = VS V^"! ="76''. Hence the required surds are "-ya™ and "-^6". Ex. 1. Reduce ^3 and ^5 to surds of the same order. The L. C. M. of 4 and 6 is 12. Hence we have ^3 = ^} ='^25. Thus 1^27 and 1^25 are the required surds. Ex. 2. Which is the greater, ^14 or ^6 ? We must reduce the surds to equivalent surds of the same order. Tlius ^14 = ^142 = ^196, and ^6 = ^63 = ^16. It is now obvious that ^6 is greater than ^14. 224. The product of two surds of the same order can be written down at once from the formula ^ax^b = Vab. When surds are of different orders their product cannot be simplified until they are reduced to equivalent surds of the same order. Ex. 1. Multiply V^ by ^20. We have V^ x \/20 = V(5 x 20) = VlOO = 10. Ex. 2. Multiply ^2 by ^3. ^2x^S= ^23 X ^32 = ^(23 X 32) = ^72. Ex. 3. Divide ^ by y/6. ^2/v6 = ^27^3 = ^(-22/03)= ^.1^. SUEDS. COMl'LEX QUANTITIES. 331 Ex. 4. Multiply 4 v3 + 4.^2 by 2 V3 - 2V2. The process is as under : 4 V3 + 4 V2 2v3- -2V2 8x3 + 8V6 -8V6- -8x2 24 V -16 =8. EXAMPLES LXVII. Simplify - 1. V27+V48. 15. V12 X V24. — 2. V50+V98. 16. 2VI X 3V^' 3. V45-f2Vl25. 17. ^6 X ^36. 4. 2V180-V405. 5. 2 V28 - V63. ' 18. V12 X V27 X y/76. -6. 5V208-3\/325. •^9. ^16 X ^6 X ^. 7. 3^5 + ^625. 20. ^12 X ^75 X ^0. - 8. 3^72 -2^43. 21. */6 X g + V^ (g + V^)(^-V^) »^-^ It is important to notice that a ± ^b is made rational by multiplying by g qp V^ 5 '^^^^ *^^^* V^ ± V^ ^^ made rational by multiplying by ^g =F yyb. When the denominator of a fraction is rationalized its numerical value can be more easily foicnd. 226. The following is an important proposition. If a-\- ^b = « + ^p, where a and a are rational, and ^b and ^p are irratioyial; then will a=: a, and b = ft. For we have a — a-\- ^b — ^p. 0^ SURDS. COMPLEX QUANTITIES. 333 Square both sides ; then, after transformation, we have 2(a-a)^b = (3-b- (a- ay. Hence, unless the coefficient of ^b is zero, we must have an irrational and a rational quantity equal to one another, and this is impossible. The coefficient of -^b in the last equation must there- fore be zero ; hence a = a. And Avhen a = a we have from the given relation, -^^b = ^(3, or b = ^. f Hence, if the sum (or difference) of a rational quantity and a quadratic surd be equal to the sum (or difference) of -s another rational quantity and a quadratic surd, the two rational quantities must be equal to one another, as also the two irrational quantities. ^ Note. — It should be noticed that when a + y/b = a -\- y'/8 we can only conclude that a = a and 6 = /3 provided that y/h and ^/3 are really irrational. We cannot, for example, from the relation 3 + ^4 = 2 + V^, conclude that 3 = 2 and 4 = 9. v 227. The square root of a binomial expression which is the sum of a rational quantity and a quadratic surd can sometimes be found in a simple form. The process is as under. To find ■yj{a + ^b), where -y/b is a surd. Let V(« + V^)=V«+Vi8 (i.) Square both sides ; then Now, since ^b is a surd, we can equate the rational and irrational parts [Art. 226] ; hence a=: a-\- /3 4«/? J , ■ ^-^ 334 SUllDS. COMPLEX QUANTITIES. Hence [Art. 183] a and ^ are roots of the equation a;^ — (la; + - = ; and these roots are ( « + V(«'- 6) ) ^^^ j «-V(a'-6) ) . \ 2 ) I 2 f Thus v(«+v;')=Vr--^^t^^=^| It is clear, that unless ^{o? —6) is rational, the right side of (iii.) is much more complicated than the left. Thus the above process is of no utility unless a^ — 6 is a square number ; and as this condition will not often be satisfied, the process has no great practical utility. From (ii.) we see that we have to find two numbers whose sum is a and whose product is -; and if two rational numbers satisfy these conditions, they can generally be found at once by inspection. Ex. 1. Find ^{Vo + 2y/m). Let V(15 + 2 V56) = V« + V'^• Square both sides ; then 15 + 2 y/m = a + j8 + 2 y/aB. Equating the rational and irrational terms, we have a + i8 = 15, a& = 56. The numbers which satisfy these relations are obviously 7 and 8. Hence y (15 + 2 ^56) = y? + yS. SURDS. COMPLEX QUANTITIES. 335 Ex. 2. Find ^(G-^S6). Let ^((5 - V35) = y/a- y/0. Square both sides ; then 6 - V35 = a-\- 0-2 Va)8. Hence, equating the rational and irrational terms, we have a+)3 = 6, «)3 = 3^. By inspection, or by solving the equations for o and j8, we find that Thus V(6 - V35) = VI - V|. EXAMPLES LXVin. Rationalize the denominators of "^ 1. A. 3. ?^. 5. '^ V7 2 v3 1 + V2 2. A. 4. ^. 6. V^-^ . v/5 \/5 V2 + 1 7. -2^A-. 11. ' V5+V3 1+V2+V3 12. 8 15+14V3 15-2^3' "" 2+V3 + \/S 9. V5 4-3V3, ^18. 1 + 3 , 2 V5 - V3 . ^ - 1 ^ -f 1 10. v^-yi^. ^ 14. ^ + 5 . 2V6+V12 V9-1 ^9 + 1 /;15. Sunplify ?^ + ^16. Simplify + ^ (2-V3)3 (2 + V3)8 17. Simplify (3 + V2)(5-V2) (3-y2)C5 + v2) 336 SURDS. COMPLEX QUANTITIES. 18. Simplify 2V15-3V5 + 2V2 ^ V15 + V2 ^ 21. Show that ^^^2-?i^3 = 2 + V2 + V3 + V6- 22. Simplify 1 ■V- :r + 1+V2 + V3 -1+V2+V'^ 1-V2+V''5 1+V2-V3 23. Simplify ^^^2 . 4^^ V6 V'^ + V 6 V6 + V'2 V2 + V<^ ^ 24. Show that /O _1_ 1 y^ 26. Find the value of y- — : — to three places of decimals. \/2 — 1 27. Find the value of "^ + - — ^„ to three places of decimals. Find the sciuare roots of ^8. 6 + V20. (D82. 101-28 Vl3. . 29. 16 + 6V7. fi)33. 117 + 36^10. 30. 12-6 V3. I>34. 280 + 56 V21. .-^iTTs^SvTSr- ^35. 4| + 2V2. Simplify 36. 3y5-V2+V(7 + 2ylO). 37. 6-4^3+^(16-8^3). SUKDS. COMPLEX QUANTITIES. 337 ^ Find the square roots of ^^ 38. 11+2(1+V5)(1+V7)- Uo. 2a4-2v(«2-a;2). 39. Simplify f41. 3ic-l+2V(2x2+x- -6) 42. 1 V(16 + 6 V7) 43. 1 ^(15 + 2^56) 44. V(7 + 2V10)+V(7-2V10). 45. V(3 + 2v'2)-v2 V2+V(3-2V2) 46. 2 + V3 5 J-V3 V2+V(2+V3) V2- V(2- V3) 47. Show that - — V2 + V^' ) 3-632... y V2+V(7-2V10) 48. Find the value of x^ — 4 ic + 5 when x = 2 -\- y/6. 49. Show that, if x^ = x + 1, then x^ = 2a; + 1 and x^ = 5x + 3. 50. Show that, if x2 = 3 X + 5, then x^ = 14 x + 15 and x* = 57 x + 70. Om-1 coi^iplex quantities. 228. In the chapters on quadratic equations, and on equations of higher degrees, we encountered expressions which involve the square roots of negative quantities. Such square roots present themselves in the form V~^> in which c is a positive number. This is called a pure imaginary as distinguished from the mixed binomial form a-{-^—c, which is called a complex quantity. Such expressions do not enter largely into the applica- tions of elementary algebraic principles, and a detailed study of them would displace other matters of more immediate importance to the student. But some account -t/Ij-^I. m*^ 338 SURDS. COMPLEX QUANTITIES. of their behaviour in the operations of algebra should be given. The following discussion consists, in the main, of explanations, without formal demonstration, of the principal laws of operation with complex quantities. 229. In admitting imaginary forms to the category of algebraic quantity, it becomes necessary to enquire whether the algebraic processes may be applied to them without limitation. And we find in fact at the outset, that if we attempt to apply the rule that the sign -^ is distributive over the factors in a product, and assert that V— ex V— c = ^ {— c) X {— c) = c, this is at once in conflict with the other rule that (V-c)2 = -c, / which asserts that to square an indicated square root \ has merely the effect of removing the radical sign. Driven to a choice of interpretations, we reject the former, adopt the latter, and define outright that and accept the other consequent interpretations to which this definition leads. 230. If now we assume the commutative and associa- tive laws in multiplication [Art. 52], and write axW — c = ^ — c xa, (a X V— c)xh — ax (V^7x 6), we have Vc X V^T X Vc X V^^ = Vc X n/c X V^^ X V^^ SURDS. COMPLEX QUANTITIES. 339 whence (Vc X V— 1)" = — c; and comparing this with we see that it is legitimate to write V— c=a/c X V^^. The form Vc x V— 1, which is the one universally employed, is usually written in the abbreviated form bi, in which i now takes the place of V"~ ^^ ^"^ ^ ^^ V^- The symbol ^—1, or its equivalent i, is called the imag- inary unit. 231. With the interpretation of V~~l we have novr /^adopted, we have f • 7 i^ = — 1, 1^ = — i, i^ = 1, i^ = i, i^ = — l, f z= — i, f = 1, etc., so that i, — 1, — I, 1, are the only values that the inte- gral powers of V~^ ^^^ assume. 232. For the imaginary unit in its combinations with real quantities we have assumed the commutative and associative laws in multiplication, namely a X i = i X a, {axi)xb = ax{i X 6), and it is equally essential that the like laws in addition a-\- i= i-\-a, (a + t) + 6 = a + (i-f6), and the law of distribution i X (a -f 6) = i X a + i X 6 340 SURDS. COMPLEX QUANTITIES. shall also obtain. With these laws and the interpreta- tion of -^— 1 definitely agreed to, we can perform the fundamental algebraic operations upon complex quanti- ties. [See Treatise on Algebra^ pp. 221, 222 ; also Chrystal's Algebra, Vol. I., Chapter XI.] The following examples illustrate the application of the rules of addition, subtraction, multiplication, and division. Ex. 1. The sum of a + hi and c + di is (a -f hi) + (c + di) = (« + c) + (?) + d)i. Ex. 2. The difference of a + hi and c + di is (a + 60 - (c + di) = (a - c) + (& - d)L Ex. 3. The product of a + hi into c + di is (a + hi)(c + di)=ac + ad/ + hci + &(Zi2 = (ac — hd) + (ad + hc)i. Ex. 4. The quotient of a + hi by c + di is a + hi _ (a + hi) (c — d/) c + di (c + di) (c — di) _ ac — gdi + hci — hdt^ c^ — cdi + dci — d'H^ _ (ac + hd) + {he - ad)i _ d^ + d^ Ex. 6. The square of a + hi is >t (a + 60^ = a2 _ 52 ^_ 2 aftt. N Ex. 6. The reciprocal of a + hi is __1 a — hi a + hi~ cfi .+ V^ From these examples it is at once evident that sums, differences, products, and quotients of complex quantities are themselves complex quantities. It may likewise be SURDS. COMPLEX QUANTITIES. 341 shown, that from the other algebraic operations, when applied to complex quantities, only other complex quan- tities (or possibly reals and imaginaries*) can result. [See Stringham's Uniplanar Algebra, Chapter III.] 233. When two complex quantities have their real parts the same and their imaginary parts different only in sign, they are said to be conjugate to one another. For example, —a-\-hi is conjugate to — a — bi. Both the sum and the product of two conjugate complex quantities are real. For, let the expressions be a-\-bi and a — bi, both a and b being real. Their sum is a-\-bi-\-a — bi = 2a, and their product is (a + bi) {a - bi) = a^ - abi + bai - bH- =a--^ b\ Conversely, if the sum and the x)roduct of two complex quantities be both real, the complex quantities must be con- jugate. For, let the expressions be a + bi and c + di, a, b, c, and d being real. Their sum is a-\-bi-\-c + di =a-\-c-\-{b -]-d)i, which cannot be real unless b -\-d = 0; and their prod- uct is (a 4- bi) (c + di) = ac -f- adi + bci + bdi^ = (ac — bd)-^(ad-{-bc)i, which cannot be real unless ad-\-bc = 0. * Reals and imaginaries may be regarded as particular cases of complex quantities, the former being complex quantities with zero imaginary parts, the latter complex quantities with zero real parts. 342 SURDS. COMPLEX QUANTITIES. But if b-\- d== and ad-\-bc = 0, then h(c — a) =0, whence, either c = a, or 5 = 0. Then : (1) If 6 = 0, from b + d = it follows that d is also zero and the original expressions are real. (2) If c = a {b not zero), it follows, because d = — b, that c-\-di = a — bi, which is conjugate to a + bi. q.e.d. -—234. When an expression is written in the form a -f- bij I it is understood that a and b are both real. If a-\-bi = 0, then a = and 6 = 0. For, if a + bi = 0, then a = — bi. But a real quantity cannot be equal to an imaginary one unless both are zero. .-. a = & = 0. This proposition is a lemma to the following. Two comjylex quantities cannot be equal to one another unless their real and their imaginary parts are respectively equal. For, if a-\-bi = c-\- di, then a—c = (d — b)i'j whence, as previously proved, a — c = 0, d — b=0, that is a = c and d = b. q.e.d. EXAMPLES LXIX. Simplify each of the following expressions by reducing it either to a form having no imaginary part, or to a form having no real part ; or if it have both real and imaginary parts, reduce it to the typical form a + 6i. SUHDS. COMPLEX QUANTITIES. 343 JL "4 l±l. t 3 1 * 1-2 14-i 2-5i 1 - ^• 2 1 - 1 .^0. (-3 + 20(3 + 21). 14 1+^* 1 11. (5i + 2)(15 + 6i). l-2i l + 2i 12. (^ + ^*V3)^. 15. 2-3i ^ 2 + 3i 13. (|-aV'3)3. ^"^^^" ^~^* 16. Write the equation whose roots are 1, t, — i, and — 1. 17. Write the equation whose roots are 1, i + g^V^' ^°d 18. If a> = — i + ^ iV3, prove that 1 + w + o'^ = 0. 344 KATIO. PROPOKTION. VAKIATION. CHAPTER XXIII. Ratio. Pkoportion. Variation. 235. Definitions. The relative magnitude of two quan- tities, measured by the number of times which the one contains the other, is called their ratio. Concrete quantities of different kinds can have no ratio to one another : we cannot, for example, compare with respect to magnitude miles and tons, or shillings and weeks. The ratio of a to 6 is expressed by the notation a : 6 ; and a is called the first term, and b the second term of the ratio. Sometimes the first and second terms of a ratio are called respectively the antecedent and the consequent. It is clear that a ratio is greater than, equal to, or less than unity, according as its first term is greater than, equal to, or less than the second. A ratio which is greater than unity is sometimes called a ratio of greater inequality, and a ratio wliich is less than unity is similarly called a ratio of less inequality. 236. Magnitudes must always be expressed by means of numbers; and the number of times which one number is contained in another is found by dividing the one by the other. Hence a:b is equal to • b Thus ratios can be expressed ns fractions. RATIO. PROPORTION. VARIATION. 345 237. A fraction is unaltered in value by multiplying its numerator and denominator by the same number. [Art. 158.] Hence also a ratio is unaltered in value by multiplying each of its terms by the same number. Thus the ratios 2:3, 6:9 and 2 w : 3 m are all equal to one another. Again, the ratios 4:5, 7:9 and 11:15 are equal respectively to 36:45, 35 : 45 and 33 : 45. Hence the ratios 4:5, 7:9 and 11:15 are in descending order of magnitude. 238. A ratio is altered in value when the same quan- tity is added to each of its terms. For example, by adding 1, 10 and 100 to each of the terms of the ratio 4 : 5, we obtain respectively tlie ratios 5 : 6, 14 : 15 and 104 : 105 ; and these new ratios are different from the given ratio and from each other. Since i = l-h l = l-h M = l-TVandig| = l-,.i„ we see that by adding the same quantity to each of the terms of the ratios 4 : 5, a new ratio is obtained which becomes more nearly equal to unity as the quantity added becomes greater. This is a particular case of the following general proposition. 239. Any ratio is made more nearly equal to unity by adding the same positive quantity to each of its terms. By adding x to each term of the ratio a : b, the ratio (a -\- x) : {b + x) is obtained. We have to show that (a + x)/{b -f x) is more nearly equal to 1 than is a/b. 346 RATIO. PROPORTION. VARIATION. Now ^_1 = «LZL^, also ^+^'_1 = «-j5, b -\- X b -{-X and it is clear that the absolute value of (a — &)/(?> + x) is less than that of (a — b)/b, for the numerators are the same and the denominator of the first is greater than that of the second : this proves the proposition. Now when a is greater than b, a/b is greater than 1, and so also is {a -\- x)/{b-\- x)'^ hence, as (a + x)/{b -|- x) is more nearly equal to unity than a/b is, it follows that {a -\- x) / {b -\- x) is less than a/b. Thus a ratio which is greater than unity is diminished by addirig the same positive quantity to each of its terms. If, however, a is less than b, a/b is less than 1, and so also is (a + x)/{b + x); but {a + x)/{b -f x) is more nearly equal to unity than a/b is, and therefore {a -{- x) / (b -\- x) is greater than a/b. Thus a ratio ivhich is less than unity is increased by adding the same positive quantity to each of its terms. Also, when x is very great, the fraction (a—b)/{b-\-x) is very small ; and we can make {a — b)/(b + x), which is the difference between {a + x)/{b + x) and 1, as small as we please by taking x sufficiently great. This is ex- pressed by saying that the limit of {a+x) /{b-\-x) ivhen x is very great, is unity. 240. The following definitions are sometimes required : The ratio of the product of the first terms of any number of ratios to the product of their second terms is called tlie ratio compounded of the given ratios. BATIO. PROPORTION. VARIATION. 347 Thus ac : bd is called the ratio compounded of the ratios a : b and c : d. The ratio a- : b- is called the duplicate ratio of a : b. The ratio a^ : b^ is called the triplicate ratio of a :b. The ratio -y/a : ^b is called the sub-duphoate ratio of a : b. 241. Incommensurable Numbers. The ratio of two quan- tities cannot always be expressed b}^ the ratio of two whole numbers ; for example, the ratio of a diagonal to a side of a square cannot be so expressed, for this ratio is ■^J2, and Ave cannot find any fraction which is exactly equal to -^2. When the ratio of two quantities cannot be exactly expressed by the ratio of two whole numbers, they are said to be incommensurable. Although the ratio of two incommensurable numbers cannot be found exactly, the ratio can be found to any degree of approximation which may be desired ; and the different theorems which are proved with respect to ratios of commensurable numbers can be proved to be true also for the ratios of incommensurable numbers. EXAMPLES LXX. 1. Arrange the ratios 5 : 0, 7 : 8, 41 : 48, and 31 : 36 in descend- ing order of magnitude. 2. For what value of x will the ratio 3 -f a; : 4 + a; be equal to 5:6? ^ 3. For what value of x will the ratio 15 + x : 17 + a; be equal to 1 ? S 4. What must be added to each of the terms of 3:4 to make the ratio equal to 25 : 32 ? ^ 348 RATIO. PROPORTION. VARIATION. 6. Find two numbers whose ratio to one another is 5 : 6, and whose sum is 121. ^ 6. Two numbers are in the ratio 3 to 8, and the sum of their squares is 3577 : tind them. 4 cr -I- t) ?y 7. What is the ratio of x to y, if -t, = 2 ? Sx-y 8. If 4 x2 + ?/2 =z 4 xy, find the ratio of x to y. "yQ. Find x : y, having given x^ -^ 6y'^ = 5 xy. ^ 10. A certain ratio will be equal to 2 : 3 if 2 be added to each of its terms, and it will be equal to 1 : 2 if 1 be subtracted from each of its terms : find the ratio. ^ 11. Find two numbers such that their sum, their difference, and the sum of their squares are as 7 : 1 : 75. '"^ 12. What is the least integer which must be added to the terms of the ratio 9 : 23 in order to make it greater than the ratio 7:11? 13. Write down the ratio compounded of the ratios 2 : 3 and 15 : 16 ; also the ratio compounded of the ratios 5 : 6 and 18 : 25. 14. Write down two quantities which are in the duplicate ratio of 2x:Sy. y 15. Find x in order that x + 1 : x + 4 may be the duplicate ratio of 3:5. 3 16. The ages of two persons are as 3:4 and thirty years ago they were as 1 : 3, what are their present ages ? \ 17. Show that, if from each term of a ratio the inverse of the ' other be taken, the ratio of the differences will be equal to the original ratio. ^ 18. Show that, if a and x be positive and a >x, then a^ - x": a^ + x2 will be greater than a — x : a + Jc. RATIO. PROPORTION. VARIATION. 349 PROPORTION. 242. Definition. Four ({iiantities are said to be propor- tional when the ratio of the tirst to the second is equal to the ratio of the third to the fourth. Thus a, h, c, d are proportional, if This is sometimes expressed by the notation a-.hwG'.d, which is read " a is to 6 as c is to dP The first and fourth, of four quantities in proportion, are sometimes called the extremes, and the second and third of the quantities are called the means. 243. If the four quantities a, 6, c, d are proportional, "we have by definition h~d Multiply each of these equals by hd ; then ad = be. Thus the product of the extremes is equal to the product of the means. Conversely, if ad = be then a, b, c, d will be propor- tional. For, if ad = be, then ad be a e bd bd b d that is a:b = c: d. 350 RATIO. PROPORTION. VARIATION. Tims, if a:h — c:d, then ad = bCf and conversely if ad = be, then a:b = c: d. 244. It follows from the last article that the four rela- tions a:b = c: d, a:c=b : d, b:a = d: c, and b : d = a: c, are all true, provided that ad = be. Hence the above four proportions are all true, when any one of them is true. 245. If a:b = c:d, then will a-{-b:a — b = c-^d:c — d. For a-\-b:a — b = c + d:c — d, if (a + b)ic -d) = (a -b)(c + d), that is, if ac -{-be — ad — bd = ae — bc-\- ad — bd, or, if be = ad. But this condition is satisfied, since a:b = e:d. The above proposition has already been proved in Art. 170, Ex. 1. 246. Definitions. Quantities are said to be in continued proportion when the ratios of the hrst to the second, of the second to the third, of the third to the fourth, etc., are all ecLual. RATIO. PROPORTION. VARIATION. 351 Thus a, 6, c, d, etc., are in continued proportion if a : b = h : c = c : d, etc. , that is, if ^ = ^ = ^ = etc. bed If a:b = b :c, then b is called the mean proportional between a and c ; also c is called the third proportional to a and b. 247. If a, 6j c be in continued proportion, we have a_b .*. b^=ac, OT b=-Vcic. Thus f?ie mean proportional between two given quantities is the square root of their pj'oduct. Also 2x« = ^x^; b b c b , 1 , • a^ a that IS -^ = — b^ G Hence a : c = a^ : b^. Thus, if three quantities are in continued proportion the ratio of the first to the third is the duplicate ratio of the first to the second. 248. It is often very convenient to represent a ratio by a single letter, as in Art. 170. The following are additional examples : Ex. 1. If a:b = c:d, then d^ -i- ab:c^ + cd = b"^ - 2 ab : d^ - 2 cd. liet - = x; then also - = x. b d 352 KATIO. PROPORTION. VARIATION. Hence a = bx, and c = dx. a- + a& ^ b'^x^ + h'^V' ^ h\x'^ + a;) ^ h"^ Also b^-2ab ^ 6^ _ 2 &2a; ^ [>2(i _2x) ^ 6^^ d^-2cd d^-2d^x d%l-2x) d^" Hence ^_±ab ^b^ ~2ab^ c-2 + cd d2 - 2 cd that is a^ + ab:c'^+ cd= b'^ -2ab:d^ -2 cd. Ex. 2. If a : & = c : d = e : /, show that ^3 _,_ c3 + e3 . 53 4. ^3 +y3 ^ ace : 6dr. Let a/& = x ; then c/d = x and e// = x. Hence a = bx, c = dx, and e =fx; «3 + c3 + e» ^ &%3 ^ ^3a;3 _|_ fSx,^ ^ ^g •*• 63 + (^3^_y3 63_^(^3^y3 And ocg^&x-dx./x^^s. 5d/ bdf Hence a^ + c^ + 6^ ^ ac_c, 6^ + d3+/3 ?>d/' that is «» + c3 + e3 . ^,3 4. ^73 4-/3 = ace : 6d/. Ex. 3. Show that, if x:2b + 2c-a = y:2c + 2a-b = z:2a + 2b-Cy then will a:2y + 2z — x = b:2z-\-2x-y = c:2x-\-2y — z, We have X y ^ 26 + 2c-a 2c + 2a-6 2a + 2 6-c Put A for each of these equal fractions ; then x = \(2b -\-2c-d), y = \(2c + 2a-b), z = \(2a + 2b-c). Hence 2y + 2z~x = 9a\, and similarly 2 + 2 a; - y = 9 &A and 2 X + 2 y - 2; = 9 CA. Whence a _ b _ c _ 2y + 2z-x 2z-\-2x-y 2x + 2y RATIO. PROPORTION. VARIATION. 353 249. The definition of proportion given in Euclid is as follows : Four quantities are proportional, when if any equimultiples be taken of the first and the third, and also any equimultiples of the second and the fourth, the multiple of the third is always greater than, equal to, or less than the multiple of the fourth, according as the multiple of the first is greater than, equal to, or less than the multiple of the second. If the four quantities a, b, c, d satisfy the algebraical test of proportionality, we have a _c ^ b~d' therefore, for all values of m and n, ma _ mc nb nd Hence mc>= or = or <7ib. Thus a, b, c, d satisfy also Euclid's test of proportionality. Next, suppose that a, b, c, d satisfy Euclid's definition of proportion. If a and b are commensurable, so that a : b = m : n, where m and n are whole numbers ; then a _ m na = = mb. But, Euclid's definition of the proportion a:b= c: d asserts that nc> = or < md, according as TiCt > = or < w&. z 354 RATIO. PROPORTION. VARIATION. Hence, as na = mb, we must have nc = md ; ^ c _m _a d n h Thus a, b, c, d satisfy the algebraical definition. If a and b are incommensurable we cannot find two whole numbers m and n such thut a\b —m:n. Butj if we take any multiple na of a, this must lie between two consecutive multiples, say mb and {m -h 1)6, of b, so that na > mb and na < {m + 1)6. Hence, by the definition, nc > md and 7ic < (m + 1)(Z ; .-.->— and - <—^^^ — d n d n both a/6 and c/d n lie between n Thus the difference between a/6 nnd c/d is less than 1/n; and as this is the case however great n may be, a/h must be equal to cjd. EXAMPLES LXXI. 1. Show that, if a: h::c: t?, then (i.) ac:bd::c'^:(P. (ii.) ab:cd::a'^:c\ (iii.) a2:c2::rt2_ ;,2.c2_ : 3 c - 5 fZ = 5 rt + 3 6 : 5 c + 3 d, then a : b = c : d. 6. Find a mean proportional to a^b and o6^. 7. Find a mean proportional to (a + by and (a — by. 8. Find a third proportional to a and a- ; also to (a - 6)2 and a-2 - b^. 9. If a:b::c:d, then will aft + cd be a mean proportional between a2 + c2 and 62 4. ^2. ^ 10. Show that, if a : b : : c : d, then (i.) a,: a -\- c : : a + b : a + b + c -\- d. (ii.) a2 ^ ab -^ b- : a"^ - ab + b^ : : c^ -^ cd -\- d^ : c"^ - cd -h dK (iii.) a + 6 : c + d : : Va- + 6^ . y/^^d^. (iv.) VoM^ . Vc2 -f (P : : Va3 4- 6^ : Vc^ + cP. (v.) a^c + ac2 : bH + M^ . . (« + c)^ : (6 + dy. (vi.) \/a"~-fT" : Vc" + d" : : v/a"- - fo*- : a/c" - (/»-. 11. If ? = ?/ = ?, then will £±i/^2Mi^ = i±^, and also a h G a + bb+. cc-\-a (a2 + 6-2 ^_ c2)(a;2 + y-2 ^ ^2) ^^^^ (^^ + 6?/ + C2)2. 12. Show that, if bz — cy _ cx — az _ ay — bx a b c then will x_y_z a b c 356 RATIO. PROPORTION. VARIATION. 2 13. Show that, " hd then will a:h = c:d. 14. Show that, if lx(jiy — mz), my{lz — nx) and nz(inx — ly) be equal and not zero, theii will mn + nl-\- Ini = and yz -{- zx -^^ xy =■ 0. 16. Show that, if ^ _ y b + c — a c + a — b a-fft — c then will (b - c)x + (c - a)y +(a - b)z = 0. 16. Show that, if a:b = c:d = e:f, then * a3 + a^c + ace : b^ + b'^d -\- bdf:: ace + ac^ + c^ : 6fZ/+ bd^ + d^ 17. Ua:x::b:y::c:z, prove that a* 4- a'^6"'^ + &* : X* + XV .f ^4 . . ^4 ^. ^•2^2 ^. c4 : _,yl + y'2^2 _|. ;^. 18. Show that, if a:b = pa — qc: pb — qd, then will c:d = pa-\-qc: pb + qd. 19. Show that, if a : b — c -. d, then a^ , b'^ c/^ , # », J -- 4 - : - + - = aft : cd. bade 20. If 4 a - 6 : 4 a + 6 :: 1 : 2, find the value of 7a + 36:7rt-3&. 21. If xy + 3:a;2 + 1 =y2 + 3:2/2r+ 1, show that x = y or y = 3 e. 22. Show that, if a-\-b— c:c-\-d-\-a = a — c:2d, then b:a — c = a-\-c — d:2d. 23. Show that, if x — z:y — z = x'^-.y'^, then x + 2; : y + 2f = ac^ + 2 x] when the absolute temperature [7'] is constant; the pressure also varies as the absolute temperature when the density is constant ; hence, when both density and temperature change, the pressure will vary as the product of the density and temperature. Thus P^DT. Ex. 1. If J. cc B, and if also A (x. C ; then will P oc C For, since vl x P, we have A = mB, where m is some constant. And, since J. oc (7, we have A — nC\ where n is some constant. Hence B ——G, where - is some constant ; and therefore Pace. '^ ^ Ex. 2. If C (X TI'P, then will W oc ^. P For, since G oc WP., we have G = m • WP, where m is some constant. 1 r' 1 C Hence W = — —, where — is some constant ; therefore iroc — . m P m P Ex. 3. The pressure of a gas varies jointly as its density and it.s absolute temperature ; also when the density is 1 and the tempera- ture 300, the pressure is 15. What is the pressure when the density is 3 and the temperature is 320 ? RATIO. PROPORTION. VARIATION. 361 Since P, a, a — 6, etc. 2. Find the last term of each of the following series : (i.) 3, 6, 9, etc., to 24 terms. (iv.) 14, 46, 78, etc., to 12 terms, (ii.) 5, 9, 13, etc., to 30 terms, (v.) 6, 8|, 11 J, etc., to 14 terms, (iii.) 6, 5, 4, etc., to XO terms, ^(vi.) ^, — .], — ^, etc., to 25 terms, i ARITHMETICAL PROGKESSION. 371 3. The 10th term of an A.P. is 6 and the 6th term is 10. Find the first term. 4. The 12th term of an A.P. is 15 and the 20th term is 25. Find the common difference. 5. The 7th term of an A.P. is 5 and the 12th term is 30. Find the common difference. 6. The 3d term of an A.P. is 40 and the 13th term is 25. Find the first term. 7. What is the 10th term of the A.P. whose first term is 7 and whose third term is 13 ? ^8. What is the 12th terra of the A.P. whose first term is 20 and whose 6 th term is 10 ? 9. The 3d term of an A.P. is 10 and the 14th term is 54. Find the 20th term. 10. The 7th term of an A.P. is 5 and the 5th term is 7. What is the 12th term ? Z^W. Which term of the series 5, 8, 11, etc., is 65 ? 12. Which term of the series f , f , f, etc., is 18 ? 13. Which term of the series 9, 13, 17, etc., is 229 ? 14. Which term of the A.P. 16a -8 6, 15a -7 6, 14a -66, etc., is 8 a ? 15. Write down the arithmetic mean of (i.) 7 and 13; (ii.) 9 and —9, and (iii.) a + 6 and a - 6. 16. Insert 6 arithmetic means between 8 and 29. 17. Insert 8 arithmetic means between 50 and 80. 18. Insert 7 arithmetic means betweeen 269 and 295. 19. Insert 15 arithmetic means between 67 and 43. 20. Insert 25 arithmetic means between 84 and 40 1.. 21. Insert 10 arithmetic means between 5 a — 6 6 and 5 6 — 6 a. t>^ - 372 ARITHMETICAL PROGRESSION. 22. Insert 8 arithmetic means between a — 6b and b — 6a. 23. If a, b, c, d are in A.P., show that a -\- d = b + c. 24. The sum of the 1st and 4th terras of an A. P. is 19, anw the sum of the 3d and 6th terms is 31. What is the first term ? 25. The sum of the 2d and 5th terms of an A. P. is 32, and the sum of the 3d and 8th terms is 48. What is the first term ? 26. The sum of the 3d and 4th terms of an A.P. is 187, and the sum of the 7th and 8th terms is 147. What is the second term ? 27. The sum of the 2d and 20th terms of an A.P. is 2, and the sum of the 9th and 15th terms is 8. What is the sum of the 6th and 7th terms ? 28. Show that, if the same quantity be added to every term of an A. P., tlie sums will be in A.P. . 29. Show that, if every term of an A.P. be multiplied by the same quantity, the products will be in A.P. 30. Show that, if every alternate term of an A.P. be taken away, the remaining terms will be in A.P. 31. Show that, if between every two consecutive terms of an A.P. their arithmetic mean be inserted, the whole will form another arithmetical progression. 32. Show that, if four quantities are in A. P., the product of the 1st and 4th is always less than the product of the 2d and 3d. 259. To find the sum of any number of terms of an arithmetical progression. Let a be the first term and d the common difference. Let n be the number of the terms whose sum is required, and let I be the last term. AKITHMETICAL PROGRESSION. 373 Then, since I is the nth term, we have lz=a,+ {n-l)d. Hence, if S be the required sura, >S = a + (a + (^) + (a + 2 d) + . . . -f (Z - 2 d) + (Z - cZ) + /. Now write the series in the reverse order ; then S = 1 -\- {I - d) + {I -2 d) -\- '-' + {a -{-2 d) + {a + d) ^ a. Hence, by addition of corresponding terms, we have 2/S = ()2+ ... to n terms. 18. The 3d term of an A.l\ is 15, and the 20th term is 23.} ; find the sum of the first 20 terms. 19. The 5th term of an A. P. is 37, and the 13th term is 81 ; find the sum of the first 24 terms. 20. Find the sum of 20 consecutive odd numbers of which the least is 25. 21. Find the sum of 40 consecutive odd numbers of which the greatest is 99. 22. Insert 29 arithmetic means between 5 and 50, and find their sum. 23. Insert 40 arithmetic means between 10 and 100, and find their sum. 24. There are 27 terms of an A. P., of which 3 is the first and 107 is the last. Find the middle term and the sum of all the terms. 25. There are 71 terms of an A. P., of which the first is 7 and the last is 1015. Find the middle term and the sum of all the terms. AKITHMETICAL PROGRESSION. 877 26. There are 19 terms in an A. P., of which the first is 8 and the last — 4. Find the sum of tlie series. 27. rind the sum of 20 terras of the series 3, 5, 7, etc., beginning at the 7th. 28. Find tlie sum of 35 terms of the series 6, 9, 12, etc., begin- ning at tlie 5th. 29. The sum of 10 terms of an A. P., whose first term is 2, is 155. What is the common difference ? 30. The sum of 25 terms of an A. P., whose first term is 6, is 25. What is the common difference ? 31. The sum of 10 terms of. an A.P. is 100, and the 6th term is 11. Wliat is the first term ? —^32. The sum of 28 terms of an A.P. is 133, and the 5th term i.s 0. What is the common difference ? 33. The sum of 10 consecutive terms of the series 3, 8, 13, etc., is 705. Which is the first of them ? 34. The sum of 25 successive terms of the series 5, 8, 11, ••■, is 1025. Which is the first of them ? 35. How many terms of the series |, 1, 5, must be taken in order that the sum may be 'zero ? 36. How many terms of the series 15, 12, 9, •••, must be taken in order that the sum may be 45 ? 1 q C 37. How many terms of the series 1 — , 1 — -, 1 — , •••, a a a must be taken in order that the sum may be —6a? 38. How many terms of the series — 8 — 7 — 6 — ••• will amount to 42 ? 39. The first term of a certain A.P. is 1, the last is 99, and the sum of all the terms is 450. How many terms are there ? 40. The last term of an A.P. of 20 terms is 62, and the sum is 670. What is the common difference ? 378 AEITHMETICAL PROGRESSION. >^, [I. The ninth term of an A. P. is 136, and the sum of the first 19 terms is 2527^ Find the sum of the first 40 terms. 42. Find the sum of 15 terms of an A. P. of which the eighth is 6. 43. Find the sum of 35 terms of an A. P. of which the eighteenth term is 15. 44. Find the sum of 2 w + 1 terms of an A.P. whose (w + l)th term is 100. 45. Show that, if any odd number of quantities are in A. P., the first, the middle, and tlie last are also in A.P. — ^46. Show that, if unity be added to the sum of any number of terms of the series 8, 16, 24, etc., the result will be the square of an odd number. 47. Find the sum of all the odd numbers between 100 and 200. 48. Find the sum of all the even numbers which are between 101 and 999. 49. Find the sum of all the numbers between 100 and 500 which : are divisible by 3. ^^ / -1)^-pcp-i = 1. 31. Show that, if (a^ + b^){b'- + c^-) = {ab + 6c)2, then will a, b, c be in G.P. 267. Sometimes we are not told the law which con- nects successive terms of a series; but when a certain number of the terms are given, the law can in simple cases be at once determined. Suppose, for example, we have the series 3, 9, 15, etc. Here 9 - 3 = 6, and 15 - 9 = 6. Thus, the series is an arithmetical progression^ whose common difference is 6. Again, in the series 3, 9, 27, etc.. Since 9 — 3 = G, and 27 — 9 = 18, the series is not an arith- metical projiression. We then see whether it satisfies the condi- tions for being a geometrical progression, namely f = y, wliich is the case. Thus the series is a geometrical progression^ whose common ratio is 3. GEOMETRICAL PKOGRESSIOK. 39l EXAMPLES LXXVIL Sum the following series : ;;^'l. 3, 2.7, 2.4, etc., to 21 terms. 2. 2, 18, 162, etc., to 7 tLims. 3. 1, .2,' .04, etc., to infinity.- 4. a, fe, — , etc., to n terms. a ^. .3, .03, .003, etc., to infinity. V ^ 6. 3 + 4.3 + 5.6 + ••• to 11 terms. , 2 6, and b -\- c will be in H.P. if 1 I 1^2^ a + 6 b + c 2 b or if b(b + c)+ b{a + ?>) = (a + b){b + c) ; that is, if ?>2 4- 6c + ab ^b''- = ab-\- b'^ -V ac -\r be, or if b'^ = ac ; and this is the case when a, b, c are in G.P. EXAMPLES LXXVIII. 1. Show that if the terms of a harmonic progression be all multiplied by the same quantity, the products will be in harmonical progression. 2. Insert 5 harmonic means between 1 and 7. 3. Insert 4 harmonic means between j and |. 4. Show that, if the arithmetic mean of two quantities is 1, their harmonic mean will be the square of their geometric mean. 5^ V \ HARMONICA L PROGRESSION. 397 5. Show that, if a-, b'^, c^ are in arithmetic progression, & + c, c + a, and a + h will be in harmonical progression. 6. Sliow that, if x, y, z be in H.P., {y^-z- x)\ (2 + X - y)^ {x-\-y - zy will be in A. P. 7. Show that, if x, y, 2r be in H.P., then will ^ ^ ^ , be also in H.P. ^ x + y 8. Show that, if x, y, 5; be in H.P., then will X y z y-\-z-x z-\-x-y x -{- y - z be also in H.P. 9. If a, 6, c, d be in H.P., then will , , c , d v^.TjT, b + c + d c -\- d+ a , and be in H.P. d-h a-\- b a + b -\- c 10. If a, &, c be in H.P., then will 2a — b, b, and 2c — 6 be m G.P. 11. If a, 6, c be in H.P., then will a, a — c, and a — 6 be in H.P. 12. Show that (a2 + &-2) (a2 + a?> + 62)^ «♦ ^ ^2^2 ^ ^,4^ and (a2 + &-2^ (^2 - ab -\- b'^) are in H.P. 13. Show that, if x, ai, a2, ?/ be in A.P., x, gfi, gr2, y ^^ G.P., and X, ^1, h2, y in H.P., then will gig2 _ «i + ^2 . ^l/i2 ~~ ^1 + 7*2 14. If a, ?), c be in H.P., then will b + a b + c _2 6 — a 6 — c 15. If a, ft, c be inG.P., then will ^— , -, and -^ be in A.P. a-\- b b 6 + c 398 HARMONICAL PROGRESSION. 16. If a, b, c be in A.I*., and h, c, d in H.P., then will a : b = c : d. 17. Show that, if a c b — a b — c then & = a + c, or else «, b, c are in harmonical progression. 18. If a, b, c, d be in H.P., then will S{b~a)id-c) = {c-b)(d -a). 19. Show that, if a, 6, c be in H.P., then ? = -J_ + _L_. & b — a b — c 20. Three numbers are in A. P. The product of the extremes is five times the mean, and the sum of the two largest is three times the least. Find the numbers. 21. Show that, if ^ i ^ , 5, A±A are in A.P., then a, \ c I — ab 1 — be b vvill be in H.P. 22. If a, 5, c be in A. P., and a^, 6"2, c^ in H.P., prove that — -, b, c are in G.P., or a z= b = c. 23. Three numbers are in A.P., and if 1 be addcul to the greatest, they will be in G.l*. ; show that the smallest is equal to the sijuare ^of the common difference. 24. Show that, if x, a, 6, ?/ be in A.P., and x^ n, ?', y in H.P., then av = bu = xy. 25. Show that, if r?i, ao, a■A^ «* be in II. P., then 26. Show that, if a, />, c be in A. P., and a, o, 6 be in G.P,, then will 6, a, c be in Il.T. 274. Other Simple Series. Tliere are, besides the pro- gressions, many other series the successive terms of which are formed according to simple laws. The following are examples of such series : ^(^-') SIMPLE SERIES. ^0"'399 12^_22-|-32-f ...+n^ 1.2 + 2.3 + 3.4 + ... + n{n + 1), A 1,1.1. . 1 and + + +...+ 1.22.33.4 n{n-\-l) 275. We proceed to find the sum of n terms of some of these simple series. The sum of n terms of a series is generally denoted by Sn, and the sum of the series when continued without limit by JS^. Ex. 1. Find the sum of n terms of the series 1. 2 + 2. 3 + 3. 4 + 4-5 + ... Here /S^„ =1 • 2 + 2 . 3 + 3 . 4 + ... + (w - l)n + n{7i + 1). Let 2 = 1.2-3 + ,^5j,_ 2.3.4 +^3.4:5 + ... +(m ^l)n(n + 1)+ w(« + l)(n +2). Shifting each term one place to the right, we have S = 1h2.3 + 2-3-4+ --- +(»i-2)(7i -l)u-{-(n - l)n(n-\- 1) ^ ' , +n(n + l)(M + 2). Now subtract the last result from the preceding, taking each term from the one above it ; then we have = 1-2- 3 + 3. 2-3 + 3 -3. 4 + - + 3(n - l)w + 3 w(n + 1) -w(w + l)(n + 2). Hence 3(1 -2 + 2 - 3 + 3 - 4 + - +(w - l)n + «(n + l)}= n(n + l)(w + 2) ; ... ^„ = iw(n + l)(w + 2). Ex, 2. Find the sum of n terms of the series 1-2.3 + 2.3.4 + 3.4-5+ .... Here AS'n=1.2.3+2-3.4 + 3.4.5 + ..- + (n-l)n(n+l) + w(w + l)(w + 2). 400 SIMPLE SEIUES. Let S = 1 . 2 • 3 . 4 + 2 • 3 . 4 . 5 + 3 . 4 . 5 . + ... + (n- 1)m(w + !)(« + 2)+ n{u-\-l)(H-\-2)(n-\-S). Then 2= 1. 2. 3. 4 + 2-3. 4.5+... + (n - 2)(n - l)n(« + 1) + (n - l)«(n + l)(n + 2) + n(«+l) (w + 2)(n-f3). Hence, subtracting each term from the one above it, = 1.2.3.4 + 4.2.3.44-4.3.4.6+... + 4(m- l)ii{n + l)+4«(;i+ l)(n + 2) -n(n + l)(« + 2)(w + 3); .-. 4{1.2.3 + 2.3.4 + 3.4.5+... + /i(n + 1) (« + 2)} = n(M+ l)(n + 2)(M + 3); .-. ^-„ = in(7t + l)(n + 2)(n + 3). Note. — This series and the preceding are examples of an impor- tant type of series in which (i.) each term contains the same number of factors, (ii.) the factors of any term are in A.P., and (iii.) the first factors of the successive terms form the same A. P. as the successive factors of the first term. The S series by means of wrhich the required sum can always be foimd is the series whose terms are formed according to the same law but with an additional factor at the end. Ex. 3. To find the sum of n terms of the series 12.^22 + 32+ ... Here /^„ = P + 22 + 32 + ... + ^A Now w2 = n{n + 1) — n ; .-. ^^ = 1.2 + 2.3+... + n(M + 1) - 1 - 2 n. But, by example 1,' 1.2 + 2.3 + 3.4+ ... + n(n+ l)=jn(n + l)(n'+ 2); also 1 +2 + 3+ ... + n = i7i(n + 1). Hence S„ = | n(n + 1) (n + 2) - i w(n + 1) = iw(n+l)(2n+l). ' ■ SIMPLE SERIES. 4©1 Ex. 4. To find the sum of n terms of the series P + 23 -f 33 + ... + nK Here /^^ = 1^ + 2^ + S^ + ... + n*. Now 4 n3 = {nin + l)f - {(n - l)nf, for aU values of n. Hence we have ^ 4. 13 =(1.2)2, 4 . 23 = (2 . 3)2 _ (1 . 2)2, 4.38=(3.4)«-(2.3)«, 4(» - 1)3 = {(n - 1)»}2 - {(« _ 2)(n - 1)}2, 4 n3 = {«(» + 1)}2 _ {(„ _ i)n}2. Hence, by addition, 4(13 + 23 4- 33 -I- ... + n3} = {nin + 1)}2. ... 13 4_ 23 + 33 + ... + /l3 = J n2(„ _,. 1)2. This result may be expressed differently ; for since 1+2 + 3 + ••• + »* = J»i(7i + l), we have P + 23 + 33 + ... + n3 =(1 + 2 + 3 + ... + n)«. Thus the sum of the cubes of the first n integers is equal to the square of their sum. Ex. 5. Find the sum of n terms of the series _L + ^+J_+.... 1.2 2.3 3.4 Here ^« = _1- + -1- + J_+ ... + 12 2.3 3-4 (n-l)n n(n + 1) Let 2 = 1 + 1 + 1 + 1 + ... + !+ 1 1 234 nn+1*- then 2= 1 + 1 + 1 + .. . + _J_ + 1+ 1 123 n-lnn+l; 2 c 402 SIMPLE SERIES. Hence, by subtraction, 1-2 2-3 3-4 (n-l)w n(n + 1) n + 1 Hence J_ + _i-4....+ ^ " 1 1-2 2-3 n(7i + 1) w + 1 When n is infinite, is zero ; hence the sum to infinity of n+1 the series 1 1 f- -..is 1. 1.2 2-3 3-4 Ex. 6. Find the sum of n terms of the series __L_ + _L_+^_+.... 1-2.3 2-3. 43. 4. 5 Take a series S formed according to the same law but with one factor less in each term, and proceed as before. The result is 2I1.2 (w+l)(w + 2)/ Ex. 7. Find the sum of n terms of the series 1 + 2.r-|- 3a;2 + 4a;3+ .... Let aS = 1 + 2 X + 3 x2 + 4 .r3 + ... + wx" ^ + (w + 1 )x«. Then Sx = x + 2 x2 + 3 x^ + ... + wx'» + (n + l)x»*+i. Hence, by subtraction, /8'(1 - x) = 1 + X + x2 + ac8 + ... + ic« - (w + l)x"H 1 - x«+i But 1 + x + x2+ ... + x« = 1-x Hence, S(l - x)= Ll^^ll^ -(« + l)x«+i; 1 — X (l-x)« ^ '1-x / -f- >v-l MISCELLANEOUS EXAMPLES VI. 403 EXAMPLES LXXIX. Find the sum of the followmg series to n terms, aud when possible to infinity. ^ ^^^^ ^ y^^^ ^ ^ "■ 1. 2'. 4 + 4.6 + 6.8+.... * "'3. 1.4 + 4.7 + 7.10 + .... 2. 3.5+5.7 + 7. 9.4 r2..yL/^4^2 . 5 + 5-8 + 8-11 + 5. 1.3.5 + 3.5.7 + 5.7.9+V:?^t /d^^^^ 6. 2 . 7 . 12 +7 . 12 . 17 + 12 . 17 • 22 + -../f ^r>t 3 ''/ ' \ ^^' ^^^ ^ ^ -Tl.1.1. „ 1.1:1^ 7. 11+ 8 11 1.33.55.7 ■ 2.5 5.8 8.11 9. 1.3.53.5.7 5.7.9 10. 2.5.8 5.8.11 8.11.14 11. ^ + ^ + ^ +. (x+l)(x + 2) (x + 2)(x + 3) (x + 3)(x + 4)^ 12 11,1 (l + x)(l + 2x) ' (l+2x)(l+3a;) (i + 3a:)(l+4a:) 13. '1+ \ + , I o + . '0 . + -.. ^ - - — 1 1 + 2 1+2 + 3 1 + 2 + 3 + 4 /-h'L'hl 14. 12 + 3-^ + 52 + .... 15. 13 + 33 + 53+.... 16. a(a + 6) + (a + 6) (a + 2 6) + (a + 2 6) (a + 3 6) + .... MISCELLANEOUS EXAMPLES VI. A. 1. Simplify (y + 3)(2/2_i)_3(2/+l)(y2_9)+3(y_i)(y2_9)_(y_3)(y2_i^. 2. Prove that (2y-a;)3-(2a.-y)3 ^ (2y-x)«+(2x-y )8^^Q^,_^g , 3(2/-x) a+y \ 2 4-i- ^Y 404 MISCELLANEOUS EXAMPLES VI. 3. Find the H.C.F. of Sx^ - 13rB2 + 23 x - 21 and 6ic8 + x^ -44x4-21. For what value of x will both expressions vanish ? 4. Solve the equations : (i.) 3x + --l = 12x + -+14 = --2x-14. y y y (ii.) x2 - (a + 6 + 2 c)x + (« + & + c)c = 0. 5. A man buys 9 oxen and 20 sheep for £230 ; by selling the oxen at a gain of 25 per cent, and the sheep at a loss of 20 per cent, he gains £ 35 altogether. Find the price he gave for each. 6. If a and )8 be the roots of the equation x'^ + 4 x + 3 = 0, show that the equation whose roots are ° "^ and " "^ is a a 3 x2 - 16 X + 16 = 0. 7. Multiply a^&-i-2 a^6-2+4-8 a-36^ +16 a'^ b by ai + 26i 8. Rationalize the denominator of /^ , 6 — 4y'3 and simplify V25-f 4^34. 9. Sum the series 5 — 3| + 2i| — etc., to 8 terms, and find its limit when continued to infinity. 10. Insert 20 arithmetic means between 100 and 300, and lind their sum. B. 1. Simplify 3{x - 2(y - z)}- [iy + 2{x-y- z}]. 2. Find the factors of mnx^ + rn^xy + n^xy + mny'^, and of x3 - x'^y + x?/2 - y^. 3. If the sum of two numbers be equal to 4, show that their difference is one-quarter of the difference of their squares. 4. Solve (i.)-«_+-^ = l. x + a X + 6 (ii.) V12 X - 3 + Vx + 2 + V7 X - 13 MISCELLANEOUS EXAMPLES VI. 405 5. Show that x^ — 8 x + 22 can never be less than 6. 6. Find the square root of a^ + 2 a^b'^ - 2 a^c + fo* - 2 b'^c + c'^. 7. It a:b::c:d, show that a2 + 62 : c2 + d^ : :(a + &)2 :(c + d)2. 8. Sum the following series : (i.) 5 — 1 — 7 — etc. to 20 terms, (ii.) 2.} 4- 1 + I + ••• to infinity. 9. Find the sum of all the numbers which are less than 1000, and are divisible by 7. 10. A train travelling at the rate of 37.] miles an hour passes a person walking on a road parallel to the railway in 6 seconds ; it also meets another person walking at the same rate as the other, but in the opposite direction, and passes him in 4 seconds. Find the length of the train. C. 1. Find the value of -V(yl±ll ^ ^^izAv^lll, ^{z + 2/) x-z{y- X) when x = 0, ?/ = 1, z = lh 2. Divide 4 x* - 9 x^y^ + 12 xy^ - 4 ^/^ by 2 ^2 + 3 xy - 2 y^-, 3. Show that the square of the sum of two consecutive numbers is equal to four times their product increased by unity. 4. Find the L. C. M. of ic2 - 6 ax + 9 a^, x^-ax-Qa^ and 3ic2-12a2. 5. Simplify _l_ + 3^_+ 1 2 1 -a a+ 1 a+2 6. If 40 minutes would be saved in a journey by increasing the rate of the train by 5 miles an hour, and 1 hour would be lost by diminishing it by the same amount, find the rate of the train and the length of the journey. 7. Simplify 2 + V8 + V2 - V27 - Vl2 + VTS - V(19 + 6 V2). 406 MISCELLANEOUS EXAMPLES VL 8. Prove that any ratio is made nearer to unity by adding the same number to each of its terms. Also show that ^^ "*" ^^^ is „ c mb-\-nd intermediate between - and -• b d 9. Find the sum of the following series : (i.) 21 + 15 + 9 + ••• to 8 terms. (ii.) 5 + 2 + .8 + .-. to infinity. 10. The arithmetic mean of two numbers is 17, and their geo- metric mean is 15. What are the numbers ? D. 1. Simi>lity {x(x-h a)— a(x — a)} {x(x — a)— a(a — x)]. 2. Show that {n + 1)+ - n* = (2 n + 1) (2 n- + 2 « + 1), and that (a + &)* - a* - 6* = 4 cib(a + b)'^ - 2 a^b'^. 3.s« {i-^ff:}{x-iff^}. 4. Solve the equations : (i.) §_§ = 21, ^ + I = -11. X y X y (ii.) Sx^-4xy + 2 y'^ = 33, x'^ - y'^ = 16. 5. A number consists of two digits, of which that in the unit's place is the greater ; the difference between the squares of the digits is equal to the number, and if the digits were inverted the number thus formed would be 7 times the sum of the digits. Fhid the number. 6. Find the equation whose roots are the cubes of the roots of the equation a;^ — 4 ^ + 2 = 0. 7. Find the square root of x*^ - 2x2 + 8 + a;^ - 8x ♦ + 16x-6. 8. If a : 6 : : c : <2, prove that ^-±-^ = l±ldlY. c^-i-d^ \c-d) 9. Sum the following series : (i-) i - 1 + I to 10 terms. (ii.) 9-6 + 4 to infinity. MISCELLANEOUS EXAMPLES VI. 407 10. The series of odd numbers is divided into groups as follows : 1 ; 3, 5 ; 7, 9, 11 ; 13, 15, 17, 19 ; and so on. Show that the sum of the numbers in the nth group is n^. E. 1. Simplify x2 -2 a;y+ 2/2 _(x2+X2/+?/2) -{x(- 2?/+ x)+?/2|. 2. Divide Ga^-\-4b^ - a% + 13 ah^+2a^h-^ by 2 a2 + 4 6^ - 3 ah. 3. Find the highest common divisor and the lowest common multiple of 3 # - 3 d-h + ah^ - ¥ and 4 a- - 5 a& + b'\ 4 I a;-l X — 1 '*' 4. Simplify 1 X— 1 X 5. If a and /3 are roots of x^ + mx -h n = 0, find a'^p + p-a in terms of m and n. Test your result on the equation x2 — 3 a: + 2 = 0. 6. Find the value of x for which 3x2 + 5 x + 3 has its least pos- sible value, and show that the least value is i^. 7. Show that if a, &, c, d be in conthmed proportion, b' (a- by^a^ [b-cj d 8. Find the square root of 4 x^ - 12 xy^ - 7 x^y + 24 x^ y'^ -f 16 y^. 9. Sum the series : (i.) - 3 - 2 - 1 ... to w terms, (ii.) 1 — I + ^ ••• to infinity. 10. A train 72 yards long passed another train 60 yards long, which was going in the opposite direction on a parallel line of rails, in 4 seconds. Had the first-mentioned train been travelling at twice its actual speed, the trains would have passed each other in 3 seconds. Find the number of miles per hour at which the trains were travelling. 408 MrSCELLANEOUS EXAMPLES VI. P. 1. Simplify (a; + 2/ + zy-(- x + y + zy-\- (x - y + z^- (x + y - zy. 2. Divide a^ - 3 a^ft + 3 aft^ _ fts _ c3 by a-b- c. 3. Prove that if x = a -\- d, y = b -{■ d^ z = c + d, then x^ + y"^ + z^ — yz — zx — xy = a^ + b^ + c^ — be — ca — ab. ^ a: ,:*„ 1 , 2a 1 • --^-' a_2x'4x2-, cfi c H-2x 5. Solve the equations : (i.) 2x + 4:y-Sz = 22 4:X-2y-\-6z = 18 -• 5x-\- y-2z = 14 .^x-1 x-2 X — 5 x- 6 X- Li -7 (iii.) ^ + y + ^-y.. x-y x-\-y =ll .. JC2 + y2 , = 90 J 6. Show that, if any integer be put for x in the expression 3fi-4x^+ 14x* - 32x3 + 49x2 -60x + 36, the result will be a square number. 7. Reduce to the best form for numerical cal- V(12-V140) culations, and find its value to 4 places of decimals. 8. Show that the ratio a — x:a + x is greater or less than the ratio a^ — x^-.a^ + x^, according as the ratio x : a is greater or less than unity. 9. If - + _ varies inversely as x + ?/, then x^ + y^ varies as xy. X y 10. If a, &, and c are the sums of n, 2 «, and 3 71 terms respec- tively of a geometric progression, then a^ + t'^ = a(b + c). INEQUALITIES. 409 CHAPTER XXVII. Inequalities. 276. The statement p — g is positive is expressed alge- braically hyp — q > 0, and the statement p — q is negative is similarly expressed by p — q < 0. With this notation the definition of an algebraic inequality becomes p > q, a p — q > 0, [Art. 47.] and it immediately follows that p 0, .-. - 3 > - 5, and since -9-(-5)=-4<0, ... _9<_5. This definition necessarily involves the converse propo- sitions that p-q>0, if p>q, p — q<0, a p0, or in that of its reverse p < ; — an obvious conse- quence of the definition of an inequality. II. If p^q and q > r, then p> r. For, {p — q)-{-(q — r)=p — r>0. Cor. Ifp'>q>r>S'">z, then p>z. III. Any term may he transferred from one side of an inequality to the other, provided its sign he changed. For, if p + r > g, then by definition (p + r) — g > 0, that is, p — {q — r)>(), and therefore p > q — r. Sim- ilarly, if J) > g + s, it may be shown that p — s > q. IV. The signs of all the terms of an inequality may be changed, provided the symhol of inequality he reversed. For, if m-\-p — r>n-{-q — s, then by III., — 71 — g + s>— w —p -f- r, that is, —m—p-{-r q and r = s, then p±r>q±s. For, {p±r) — {q ±s) = {p — q) ±(r — s) = p — q= a positive quantity. VI. If p> q and r = s, then pr > qs and p/r > q/s, if r and s he positive ; pr < qs and p/r < q/s, if r and s he negative. INEQUALITIES. 411 For, since r = s, and 2> — ? is positive by hypothesis, .-. pr-qs, 2)/r-q/s = (p-q)r, {p-q)/r; whence it follows that pr — qs and p/r — q/s are positive when r and s are positive, but negative when r and s are negative. CoR. 1. If q be positive, and p > qr, and r > s, then p>qs. CoR. 2. If the terms of an inequaliti/ be fractional, they can be made integral by the proper multiplications. — 4 5 Thus, from | > | we derive 5 x 5 > 4 x 6, and from > we obtain (-4)x(-6)<5x5. ^ ~^ 279. The terms of two or more inequalities may be associated with one another, but under certain important limitations. VII. // Pi>qi,P2> ^2, P3 > Qs, - Pn > 9«, then Pi-\-P2+P3+"- 4-Pn>9i + Q'2 + g3H h7n- For, since (pj -- qi),{p2 - ^2), •••(!>„ - qn) are all posi- tive and (Pi +P2 + •••iO - (gi + ^2 + ••• -\-qn) = {Pl-(il)-^--+{Pn-gn), so is {pi-\-p2-\ l-i\)-( 9l + 92 + ••• 4- qn- Frompi>gi andp2> Q'2wecannot infer /)i—gi>2>2—5'2 • For example, 8 > 7 and 5 > 2, but 8 — 5 < 7 — 2. VIII. If Pi> qi,P2 > ^2, Pz > Qs^ "■Pn> Qny then P1P2P3 "•Pn>q\ g-i g^ "-gn, provided all the quantities be positive. 412 INEQUALITIES. For, since pi > qi and pgPs '"Pn is positive, ••• PlP2"-Pn>giP2P3"-Pny and since pz > ^2 and QiPs-'-Pn is positive, .-. PlP2"'Pn>qiq2Ps"'Pny and proceeding in this way until all the p's on the right are replaced by the corresponding ^'s, we have finally PiP2"'Pn>Qiq2'-qn' From pi > gi and p^ > gg we cannot infer Pi/qi > p^^lqi- For example, 8 > 6 and 4 > 2, but 8/4 < 6/2. CoR. If p and q be positive and n = a positive integer, and ifp > q, theri p"" > g", IX. If p and q be positive, n= a positive integer, and 2>^/% g^/", denote the positive real n'* roots of p, q ; and if p > q, then y/" > g^^". For, if this be not so, then p^/" < or = g^/" ; but since p^/" and g^/" are both real and positive, from p^/" < or = g^/" it follows, by the corollary of VIII, that (p'/'')"q. CoK. Ifp and q be positive and m,n = positive integers, and if p>q, then p±"'/'» > g±»»/«; provided, if n be not 1, only the positive real n"* roots be taken. 280. The following examples illustrate some of the uses of the method of inequalities : • Ex. 1. The sum of any real positive quantity and its reciprocal is greater than 2. Let the quantity be a, and let a '/^ denote its positive square root. Then (aV^ - 1 /aV2)2 = a -2 -\- l/a>0 ; ,', a+l/a>2. [by TIL] INEQUALITIES. 413 Ex. 2. For what values of xisx^ — 4x + 3> — 1? Completing the square for ic, we have a;2 _ 4a; + 3 = (a; _ 2)2 _ 4 + 3. Hence, the given inequality is satisfied if (a; _ 2)2 _ 1 > _ 1, that is, for all real values of x except the value a; = 2. Ex. 3. For what values of a: is (2 a; - l)/(x + 2) >, =, or < 1 ? Let /=(2a;-l)/(x4-2). We have />or or < 0, [by III. ] according as {(2 x - 1) - (x + 2)}/(x + 2) > or < 0, according as (x — 3) / (x + 2) > or < 0, according as (x - 3) (x + 2) > or < ; [by VL] .•./>or<0, according as x > or < 3, if x be positive, or, according as — x > or < 2, if x be negative ; and / = 1> if ic = 3. Ex. 4. The arithmetic mean of any two positive quantities is greater than their geometric mean. If the two quantities be a and 6, the arithmetic mean is i (a + 6), the geometric mean is ^{ab)^ and we liave to prove that K« + ft) > V(«&)- Since (a - hY > 0, .-. (a-&)2-f 4a6>4a6, [by V.] that is, (a + 6)2 > 4 a& ; .-. a + 6>2v(a6), P>y IX.] and dividing by 2 completes the proof. This example is a particular case of the following : Ex. 5. If X, y, ^, ••• i be n real quantities, (w- l)Sx2>2 2xy. 414 INEQUALITIES. The student will have no difficulty in verifying the indentity, S(x - yy^=(n - l)Xx' - 2 Sa-y, by comparing the like terms of its two members. But 2 (re - yy is essentially positive, and hence, (w- 1)2x2 -2 2a:?/ >0, unless X = y = z=: •■■ = t, for which 2(x - yy = 0. .-. {n - 1)2x2 > 2 2x?/. We have here used the 2 notation, as explained in Art. 152. The student should note its conciseness and its great utility in examples of this kind. 281. The theory of simultaneous inequalities, involv- ing two or more unknown quantities, is most clearly pre- sented with the aid of certain geometrical constructions, which are not available to the student who is not familiar with at least the elements of analytic geometry. But the ordinary process of elimination by addition can be applied to some of the simpler examples of this class. It must be remembered that multiplications, with the corresponding terms of two inequalities, can be per- formed only with limitation, and subtractions and divis- ions not at all. [Art. 279.] Ex. 1. Resolve the simultaneous inequalities 2xfy-G>0, -3x + 2?/ + 6>0, x - ?/ + 1 > 0. We apply the ordinary process of elimination by addition, li'rom the first and second inequalities we obtain 6x + 3y-18>0 -6x-f4y + 12>0 iy- 6 > 0, . •. y > f . From the first and third 2x-)-?/-6>0 X - y -I- 1 > 3x- 6>0, .'. x> fv INEQUALITIES. 416 From the second and third -Sx + 2y + 6>0 2x-2y + 2>0 -x4-8>0, .-. xx>f, 9>y>f. Observe that if x and y are to be limited to finite values, the inequalities must have both positive and negative terms. Ex. 2. Resolve the simultaneous inequalities bx -{- ay — ab > 0, xy < 0. Reverse the first inequality, changing its signs for this purpose, and then add the two inequalities together. We obtain xy — bx — ay -h ab < 0, that is, (x - a)(y — b) < 0. .'. x> a, y b. If a and b are both positive, since x and y are of opposite sense, either a; > a, y < 0, or y>b, x < 0. EXAMPLES LXXX. Determine the range of values of x that satisfy each of the fol- lowing inequalities : 1. (x-5)(x-2)>(a:-l)(x-3). 2. x2-4x-h 3>, =, <0. 3. x2-3x+ 10 >, =, <0. 4. (x-|-5)/(x-3)> or < 1. 5. (2x-3)/(x+3)>, =, or <1. e. (3x-4)/(2x-l)>, =, or 0, _3x + 2y + 6>0, x-y>0. 9. Resolve the simultaneous inequalities x-y + S>0, 2x-4?/ + 8<0, x+5y-5<0. 10. Determine the values of x and ij that make (xy + l)/(x + ?/)>,=, or <1. Discuss the cases x + 2/ > and x + ?/ < 0. 11. If X, y, z be real, prove that Sx^>, =, or <3x?/0, according as Sx>, =, or <0. Use the result of Ex. 5, Art. 280, and the identity of Art. 154. [Chrystal, Algebra, Vol. II., p. 40.] 12. If X, ?/, z be real and not all equal, prove that (Sx)3>27x?/^ and <9 2x3. 13. If rt > ft, show that (a; + a)/v'(a^' + «2)>, =, or <(x + b)/ y/(x^ + b^), according as x>, =, or , c, d be all positive and a/b — b/c = c/d, then (a- fZ)>3(&-c). 15. Prove that, in a geometrical progression having an odd number of positive terms, (sum of all terms)^ {number of all terms) x (middle term). LIMITS. INDETERMINATE FORMS. 417 CHAPTER XXVIII. Limits. Indeterminate Forms. 282. Functional Nomenclature. Any expression which involves x in such a manner as to undergo a change of value whenever x changes is called a function of x. If two or more letters, as x, y, z,--- are involved in it, the expression may be regarded as a function of the sev- eral letters x, y, z,"-. These letters, through whose values or changes the function is itself determined or changed in value, are called the variables, and all other letters (including mere numbers) are then considered, for the time being, as not susceptible of change, and are called constants. If the function does not contain, or can be so reduced as not to contain, fractions whose denominators involve the variables, it is said to be integral ; but if such denom- inators are an indispensable part of the function, it is said to be fractional. The term rational, or irrational, is applied to the function according as its variables do, or do not, appear every- where in it in the rational form. The following examples will sufficiently illustrate this functional nomenclature : Ex. 1. (i.) axy ■\- hx -^ cy -\- d \s 9, rational integral function of x and y of the second degree. 2d 418 LIMITS. INDETERMINATE FORMS. (ii.) ax2+ hxy + cy^ is a rational integral homogeneous function of X and y of the second degree. (iii.) (ax -j- by -\- c) / {I -\- z) is a rational fractional function of X, y, and z. It is, however, a rational integral function of x and y. (iv.) axy + hVx + y is an irrational function of x and y. (v. ) x^ + 2^3 _|_ ^3 _j_ a;^2 is a rational symmetrical function of x, y, and s. [Art. 150.] Ex. 2. Describe the following functions : (i. ) ax^ + by^ -i-S cxyz. (ii.) ax + by -i- Sc(xyzy^^. (iii.) (ax+by-{-c)/\^lT~z, (iv.) ax^ + 2?>a; + c. (V.) a(x'^ + y')/Vxy. In all of the above examples the letters x, y, z are regarded as variables, a, b, c, d as constants ; and, in general, constants are denoted by the letters of the first half of the alphabet, variables by the letters of the second half. 283. It is important to have the means of using the expressions /wnc^iow of x, function of x, y, etc., in mathe- matical formulse, and for this purpose the abbreviations f{x), f{x, y), etc., are employed. Thus, such an expression as ax'^ + bxy + cy- -f d is briefly repre- sented by f{x, y) . And if f{x, y) = ax^ + bxy + cy'^ + d, and we assign x = 1, y = 2, then we write /(I, 2) = a • 1 + 6 • 1 • 2 + c • 22 + d = a + 26 + 4c + d[; similarly /(I, 0)=a-f t?. LIMITS. INDETERMINATE FORMS. 419 284. A function of x is continuous, or varies continuously, if its change, due to an arbitrarily small * change in re, is also arbitrarily small. Thus, f{x) is continuous, at the value x, if the difference fix + h)—f{x) may be made arbitrarily small by making h arbi- trarily small. A function is discontinuous, at any value of x for which it does not fulfil the foregoing condition. 285. Definition of a Limit. If by making x approach the fixed value a, in such a way that x — a becomes arbi- trarily small, we make/(ic) approach as near as we please to the fixed value /, then I is said to be the limit of f{x) as X approaches a. We express this briefly by writing It will be convenient to use the sign = as an abbre- viation for the expression approaches as a limit. The following examples will help to make the mean- ing of this definition clear, Ex. 1. What is the value of (x- — a'^)/(x — a) when x = a? If a be substituted for x in (x- — a^) /{x — a), the result is 0/0, an expression which cannot be interpreted as an algebraic opera- tion. There is therefore no proper answer to the question as above formulated. But expressed in the form, what limit does {x^ — a^) / {x — a) approach when ic = a, the question can be answered in the language of limits as follows : Since (ic- — or) /(x — a) = (x + a)(x — a)/(x — a) = x -\- a, for all values of x not absolutely equal to a, we see at once that * Smaller than a previously assigned quantity whose value we may make as small as we please. 420 LIMITS. INDETERMINATE FORMS. when X differs arbitrarily little from a, {x^ — a'^) / (x — a) differs arbitrarily little from 2 a ; hence, "mit^^^^ga. X = a X- a Ex. 2. Find the sum to infinity of the series By the ordinary process of division we have ?^i^ = a; + a:2 + x3 + - + x^-\ 1 — X that is, a; + x2+ ... +cc«-i = ^^ 1 — X I — X and this formula is true for all finite values of x. Putting X = I, we have i+i + - + ^=i-5^,- If the number of terms of the series be now increased beyond finite range, l/2'*-i = 0, and therefore limit /'l + l+...+J_Ui. n = 00 ^2 4 2«-i/ This series was discussed in Art. 266, and its limit was there determined from geometrical considerations. 286. If, when the variable approaches a limit, the function exceeds every finite value however large, we extend the language of limits to this case and say that the function has the limit oo . This use of the term limit must be understood to be exceptional. For example, limit x^-\-x + 2 ^ limit x + 2 ^ ^ x=lx^ -x:^-x+l X = 1 a;2 - 1 The value may obviously occur as a limit. ■p 1 limit x^-1 _ f. For example, x = l^^^-^' LIMITS. INDETERMINATE FORMS. 421 INDETERMINATE FORMS. 287. Certain combinations of and oo produce what are known as indeterminate forms. Such a form made its appearance in Ex. 1 of Art. 285, where the substi- tution of a for X in (a^ — a^) / (x — a) resulted in the illusory fraction 0/0. The evaluation of indeterminate forms is essentially an application of the method of limits. The simpler forms are 0/0, oo/oo, x oo, oo — oo. The following examples will show how they may arise and exemplify the method of their evaluation. Ex. 1. (1 - x^)/{\ - ic2) becomes 0/0 when x = \. But \-x^ _ \-{-x-\-x^-\-7? + x^ ^ 1 -a; . 1-a;"^ \ + X \ —x' limit 1 — a^ _ 5 " a^ = 1 1 - x2 ~ 2' Ex. 2. (2 x3 + 3 a^s) / (3 x2 + 5 o^) becomes 0/0 when x = 0. But 2x^-\-^yf> _ 2 a; + 3 a;^ x^ . 3x*-^4-5a;*~ 3 + 5a;2 ' a;*' limit 2a;3 + 33;5 _0_Q Ex. 3. (a 4- &")/(c + &") becomes 00/00 when 6 > 1 and n = 00. If numerator and denominator be multiplied by &-«, then limit a + ?)" _ limit a6-» + 1 _ 1 n = 00 c + h'^~ n = cb-n + 1 ~ ' If 6 < 1, say b = l/k, where k > 1, then (a + 6«)/(c + b») = (a + ^-")/(c + A;-«), and ^^™^* q + ^''^ _ a n = CO c + ^-n c * If 6 = 1, no inference can as yet be drawn. 422 LIMITS. INDETERMINATE FORMS. Ex. 4. The function (x — 2)/(V2x — 2) assumes the form 0/0 when X = 2, or the form oo/oo when x = oo. In the first case we have limit X — 2 _ limit ^ x + \/2 _ „ ^ = 2 ^(2x)-2~^ = 2 V2 In the second case limit x — 2 _ limit 1 — 2/x _ 1 _ ^ = ^ ^{2x)-2~^ = ^ ^(2/x)-2/x~'0~ Ex. 5. The function assumes the form oo x when x = l. But 1 _i_ « _ x-\- a — 1 X—l X — t and x'^ + x-2=(x-\-2){x-l). Ex. 6. The function /(x)=a/(l + x)-2&/(l-a:2) assumes the form co — co when x = — 1. But a/(l + x)=a(l- x)/(l-x2), and therefore (i.) Now, in whatever manner x attains the value — 1, the numer- ator of this fraction assumes the uni(iue value 2 a — 2 ft, and the denominator becomes zero. But 1 — x"^ is and remains positive if x^ = 1 through increasing values a little less than 1, or it is and remains negative if x^ = 1 through decreasing values a little greater ' than 1. In the former case /(x) = + 2(a — 6) x co, and if a > 6, limit -. X , ^ LIMITS. INDETERMINATE FORMS. 423 In the latter case /(x; = — 2(« — 6) x co, and if a > 6, limit ., ^ The function has two branches, one for values of x less than — 1, the other for values of x gi-eater than —1, and each branch lias its own limit at the value x = — 1. (ii.) If h = a, the function becomes f(x) = - «(1 + x) _ a I -\-x As an indeterminate form it has been changed to 0/0, and its limit, for x = — 1, is obviously — J «. 288. The fundamental indeterminate form is 0/0, to which all the others may be reduced. This property results from the definition of infinity, as given in Art. 168, in accordance with which, if a and b be any finite quantities, the symbol go may be replaced ad libitum by any such value as a/0 or 6/0. Hfuce we may write ^_a .b _a 0_0.a_0 cr^~b^ i)~0^b~{)'b~ (? a b O-a — O-b CO — CO = •= = — Ex. 1. The function a/(l - x)-2 a/ (I - x-) becomes oo — oo when x = 1. But a _ 2a __ a (l + a;)- 2a 1 - X 1 - x^ ~ 1 - x^ and in this form it becomes 0/0 when x = 1. 424 LIMITS. INDETERMINATE FORMS. Ex. 2. The function (1 — a")a^/'», where a > 1, becomes x oo when w = + 0. But and in this form it becomes 0/0 when « = + 0. 289. That any quantity whatever may appear as the result in the evahiation of any of the indeterminate forms thus far considered is at once evident from the consideration that the one form 0/0 may be taken as the general representative of the entire set of four forms, and that, a being an arbitrary finite quantity, axO = 0, 0x0 = 0, and O/oo =0; for, from these three equations we derive, by the proper multiplications and divisions, a = 0/0, = 0/0, and oo = 0/0, respectively. The multiplications and divisions here performed are, of course, merely symbolic. The important fact to be remembered, in this connec- tion, is that the indeterminate form gives no information concerning the critical value of the function that gives rise to it. EXAMPLES LXXXI. Evaluate the limits of the following functions for the values of X indicated, taking account of positive square roots only. 1. (x2_ic_2)/(x2 + a;-6) whena; = 2. 2. (r.8 - a;2 _ a; + l)/(x« -^x^-bx + Z) when x = 1. 8. («« — a«)/(aj- a) when a; = a. 4. (x3/2 _ a8/2)/(a;V2 - a}n) when. a; = a. 6. (Vic- l)/>/(a;- 1) whena; = l. LIMITS. INDETERMINATE FORMS. 425 6. {2 - ^(x + 3)}/(l - X) when x = l. '• v(^^-;;a^-^^^^=^^• (X - 1)2 ^X-1 2 1 9. when x = 2. 2-x 2- y/2X 10. {2 - V(l + .r)}/{3 - V3^} when a; = 3. 11. . — -\ ^ when ic = 00. 12. — —! — when a; = oo. Ix^ + mx^ + n 13. Solve the equation + ^(x^- _ 9) - (a; - 9) = 0. [See page 248.] 14. Solve the equation +y/(x'^ + 2 ax) -(x + a) = 0. 426 EXPONENTIALS AND LOGARITHMS. CHAPTER XXIX. Exponentials and Logarithms. 290. The inverse operations of raising to powers and extracting of roots, considered in Chapter XX., may be regarded as presenting themselves through the medium of the equation x^ = y in the two inverse problems : (1) Given a known number x and a known integer k, to find y. (^Involution.) (2) Given a known number y a7id a knoivn integer kj to find X. (^Evolution.) In these problems the exponent, as it appears either in a;* or in 2/^/*, is always to be regarded, either as an integer, positive or negative, or as the reciprocal of such an inte- ger ; the case where it is a fraction whose numerator is not unity calling for the application of the processes of involution and evolution in succession. Thus, if the form 6"^/^ presents itself, we first find 6"'^, or 216, and seek the fifth roots of 216; or, we may reverse the order of this work and get first the fifth roots of 6 and then cube each one of them. Involution and evolution, therefore, deal primarily with such expressions as 23, 3^ 8^«, etc., and 2y\ 5V«, \V^, etc. EXPONENTIALS AND LOGARITHMS. 427 291. If now the form of the expression ic* be changed to ¥, for the purpose of indicating that the exponent may become an unknown quantity (the variable), the equation gives rise to the two inverse problems : (1) Given x and b, any known numbers, to find y. (2) Given y and b, any known numbers, to find x. * The operation upon b which produces ¥ is called expo- nentiation, b is called the base, and b'' (sometimes also written expja?) is called the exponential of x with respect to the base b. The inverse operation of producing x, when b and y are given, is called logarithmic operation, and is expressed sym- bolically by the equation ic=nogy, in which ^log y is read logarithm of y with respect to the base b. It is an immediate consequence of this definition that y=})^''^y. EXPONENTIATION. 292. The process of exponentiation seeks the value of expressions of the form 6% in which x and b are any real numbers. f But in thus generalizing the combined * A third case may also present itself ; namely, given x and y,io find h. But this is identical with (1), for the equation b' = y may be replaced by & = y^/'. • t In the most general interpretation, exponentiation also allows b and x, in the expression 6==, to have imaginary and complex values. 428 EXPONENTIALS AND LOGARITHMS. operations of involution and evolution, it so restricts (by arbitrary definition) the interpretation of its results that to an unique pair of values of x and b there corresponds one and only one value of b"", so that while 6'/*, regarded as an example in evolution, has k roots, as the exponen- tial of 1/k to base b it has but one value. The expo- nential will therefore be so defined as to embody this restriction. As already indicated, this restriction is arbitrary, but in the chapter on exponential and logarithmic series, a definite algebraic form will be assigned to b", namely, a series whose terms are integral powers of x having co- efficients that involve only b and certain numerical frac- tions, and in this form the uniqueness of the value of b', for a given value of x, will be apparent. 293. In defining 6"^ for our present purposes we suppose that b is real, positive and greater than 1, and should x be irrational, we replace it by one of the rational frac- tions m/n, or (m + 1)/^)* which, by properly choosing the integers m and n, may be made to differ from x by an arbitrarily small quantity, that is, by a quantity that is smaller than 1/n, where ii is as large as we choose to make it. [See Art. 249.] The values of 6* we have to But the discussion of these more general cases does not fall within the scope of this elementary treatise. For a fuller interpretation of exponentiation and logarithmic operation the student is referred toChrystal's^Zg'e6ra, Vol. II., Chapter XXIX., and to Stringham's Uniplanar Algebra, Chapter III. * It must not be inferred that these restrictions are necessary in the general theory of exponentials. [See foot-note to Art. 292.] EXPONENTIALS AND LOGARITHMS. 429 consider, are therefore of the form 6"*/", in which m and n are integers, and we retain only the real positive n*^ root. For negative values of x, say x = — m/n, 6* becomes 1/6'"/", namely, the reciprocal of an exponential with positive exponent. 294. The Laws of Indices as applied to expressions of the form b"", under the limitations above imposed upon the values of b and x, have been fully explained and justified in Chapters XX., XXI., and when we require their application to exponentials and logarithms we may, therefore, write without hesitation, b'b^b'"' = b'+^+''+-, b'/b^ = b'-'j (b'y = {byy = b^, (abc'-y = a'b'(f": 295. If a series of numbers be in arithmetical progres- sion, their exponentials, with respect to any base, are in geometrical progression. This proposition follows immediately from the equa- tions 6* = b', b'+^ = b'b^, Here d is the common difference of the arithmetical progression X, x-\-d, x-\-2d,"-x + nd, and ¥ is the common ratio of the geometrical progression- b', b'¥, b'b^r-'b'b'^. 430 EXPONENTIALS AND LOGARITHMS. LOGARITHMIC OPERATION. 296. We have already defined the logarithm to be a vahie of x obtained from the equation y=b' when h and y are given, and we denote this value of x by x=nogy. The properties of logarithms follow immediately from the fuudamental laws of indices. We proceed to deduce them and to apply them to a few important examples. 297. Properties of Logarithms. I. Since W =1 for all finite values of h, it follows that nogl = 0; that is : The logarithm of 1 is 0, ivhatever the base may be. II. Since b^ = b for all values of b, it follows that 4og b = l; that is : The loganthm of any number with 7^esj)ect to that num- ber as base is always 1. III. If 4og x=p and 4og y = qi then, by definition, x=bP and y ==¥\ .-. xxy=bP X &9 = 6''+«; .-. 4og(a;?/)=i) + ^ = *loga; + 'log2^. Similarly it oan be proved that Hog(xyZ"-) = ''log.'cH- 4og?/4-*log2;+ •••. That is: The logarithm of a product is the sum of the logarithms of its factors. EXPONENTIALS AND LOGARITHMS. 431 ly. If 4og X =p and *log y^q; then, by definition, x=¥ and y = h'^-^ .-. x/y= 6V&« = 6P-«; .-. ^\o^{x/y) = p-q = nog X — 4og 2^. That is : T7«e logarithm of a quotient is the algebraic difference of the logarithms of the dividend and the divisor. V. If a; = 6^ ; then x"' = 6^ for all values of m. Hence 4og of* = 4og bP'^^p xm = ^og X X m. That is : T%e logarithm of any power of a number is the product of the logarithm of the number by the index of the power. VI. Let "log x=p and *log x = q. Then x = a^, and x = ¥. Hence a^ = 6', and therefore a = b''^p, and ?> = a^/«; and, by definition of a logarithm, 4og a = q/p, and "log 6 =p/q ; .-. 4og a X "log 6 = ^ X ?^ = 1. p q Again, from q=p x *log a, we have ^og x = "log a; x *log a. Hence : T7ie logarithms of a series of numbers to the base b are found by multiplying the logarithms of those numbers to the base a by the constant multiplier Hog a. 432 EXPONENTIALS AND LOGARITHMS. VII. Since 6+"= = + x) , and 6^ = 0, for all positive tinite values of h, [b > 1], 4og( + Go) = -f-oo, and *logO = — co. Hence : In every system of logarithms the logarithm of -f-oo is -fGo, and the logarithm ofO is — x>. 298. The logarithms in any system whose base is real, positive and greater than 1 [see Art. 293] have the fol- lowing properties. VIII. The logarithm of a negative number cannot be a real number. For, if b be positive, and a; be a real number, b' cannot be negative ; that is, x cannot be the logarithm of a nega- tive number. IX. Numbers greater than 1 have positive logarithms^ numbers less than 1 have negative logarithms. For, if b"" = u and b > 1, then, since for n = a positive number 6" > 1, and 6~" < 1, .-. a; >, or < 0, according as w >, or < 1 ; that is, since x = *log u, 'log 16 >, or < 0, according as w >, or < 1. Note. — In a system whose base is positive and less than 1, numbers greater than 1 have negative logarithms, and numbers less than 1 have positive logarithms. For, by VI., Art. 297, V«log u = .--!l2£^^ = _ '^2E^^ ^ 61og(l/a) ^loga and if a and h be both greater than 1, by the proposition just established, '•log a is positive, and Hogw is positive or negative, according as w is greater or less than 1. EXPONENTIALS AND LOGARITHMS. 433 299. If a series of numbers he in geometrical progression their logarithms are in arithmetical progression. This is the proposition of Art. 295, stated in terms of logarithms instead of exponentials. Thns, the loga- rithms of the several terms of the geometrical progres- sion y, yr, yi^,-'-yr% are ^log y, 4og y + *log r, • • • *log y -{-n 4og r. 300. The following examples will illustrate this part of our subject. Ex. 1. Find 2log8, 81og2, loioglOOO, and ^log ^100. 8 = 23, 2 = 8^ 1000 = 10^ and ^100 = loi Hence '4og8 = 3, 81og2 = i, loiog 1000 = 3, and i^log ^100 = |. Ex. 2. Having given ioiog2 = .3010300 and ^logS = .4771213, findioiog40, wiogl2, ioiogl5, and i^log ^880. By the laws of logarithmic operation we have : ioiog40 = loiog (4 X 10)= ioiog2-^ + i^loglO = 2 X .3010300 + 1 = 1.6020600. ioiogl2 = loiog (22 X 3)= 2 X ioiog2 -f- wiog3 = 2 X .3010300 + .4771213 = 1.0791813. ioiogl5 = nogl ^ ^ ^^ \= loioglO + i'>log3 - ioiog2 = 1 + .4771213 - .3010300 = 1.1760913. loiog ^880 = 1 X 'oiog (10 X 25 X 32) = 1(^*^108 10 + 5 X ioiog2 + 2 X i'^log3) = i(l + 1.5051500 -f .9542426) = 1.1531308. 2e 434 EXPONENTIALS AND LOGARITHMS. Ex. 3. Solve the equation «**+" = d. Taking the logarithms of both sides of the equation, with respect to base a, we have bx -\- c = "log d. "log d- c .. X- ^ Ex. 4. Solve the equation /(x) = 2* . a**-3 - 2* • a*-2 - 4 • a^x-i + 4 = 0. The left member of the equation may be factored thus : f(x) = 2* . a^-2 (^3x-i _ 1) _4(a3^-i - 1) = (2x.a*-2_4)(a3x-i_i). • Hence 2' • a^-2 -4 = 0, or a^-^ -1=0; and, therefore, by logarithmic operation, in any system, X log 2 + (x - 2) log a - log 2^ = 0, (x - 2)(log 2 + log a) = 0, x = 2, or (3 x — 1) log a = 0, X = I-. EXAMPLES LXXXII. Find : 1. loiogi. 5. wiogV-01. 9. V2iog2. 2. loiog 10000. 6. loiog^lOO. 10. 41og8v2. 3. wiog.Ol. 7. 2log32. 11. »log3v3. 4. i'>log.001. 8. *log32. 12. nog ^64. 13. Having given ioiog2 = .3010300 and i'>log3 = .4771213, find loiogf, wiogGO, ioiog4500, and ^log ^25. 14. Given i^log 3 = .4771213 and i"log 5 = .6989700, find i'>log3.75, i'^logl.28 and i'>log(33 x 53/2"). 16. Given i'>log 6 = .6989700 and ^log 6 = .7781513, find i"log 324, wiog 1.458, and ^log. 00432. 16. Given ^log 6 = .6989700 and i^log 7 = .8450980, find ioiogl.25, ioiogl.28, and wiog(23 x 7^/5^), EXPONENTIALS AND LOGARITHMS. 435 17. Given ^log 12 = 1.0791813 and i«log 18 = 1.2552720, find mogS and i^ogO. 18. Given ^log 24 = 1.3802113 and ^log 36 = 1.5563026, find i«log 72 and i'>log ^432. 19. If «, 6, c be the bases of any three systems of logarithms, prove that ''k^g b 20. Prove that, in a series of logarithmic systems, whose bases are respectively a, 6, c, etc., "logx ^ ^logx ^ ^loga; _ ^^^^ "logy Hogy <=logy 21. Given 4og2 = 69315 and «log 5 = 1.60944, find I'^logSO, Slog 64, and 21og 125. 22. Given «log 5 = 1.60944 and 4og 3 = 1.09861, find 5log27, 31og25, and ^log 15. 23. If "log 30.6614 = 1.48659 and ^log 3.1299 = .49553, what is the relation between 30.6614 and 3.1299 ? 24. If loiog 3.0501 = .4843141, ^log 9.458055 = .9758019 and i'>log 3.1009 = .4914878, what is the relation between 3.0501, 3.1009, and 9.458055 ? 25. Given i'^log2 = .301030, find x from the equation 2^ = 10. 26. Given ^log 2 = .301030 and I'Hog 5 = .698970, find x from the ecjnation 2-«-i = 10^. 27. Solve the equation 2'^ - 2^+' + 1=0. 28. Solve the equation 52* — 5^+^ + 6 = 0, obtaining two values of X in terms of the logarithms of 2, 3, and 5. 29. Solve the equation a^-^ - a^^-i _ q^i 4. 1 — 0. 30. Solve the equation a*-i-i + ft • a== — c = 0. 31. Solve the equation 2* ■ 32^ - 3* - 9 • 6^ + 9 = 0. 32. Solve the equation 3==^-2*+2 = 10, obtaining two values of x in terms of the logarithm of 3 to base 10. 33. Solve the simultaneous equations 2*+2y = 10, 5«y-i = 25*+y, obtaining x and y in terms of the logarithm of 2 to base 10. 436 NATURAL LOGARITHMS. CHAPTER XXX. Natural Logarithms. 301. The limit which the ratio {^logy' —^\ogy)/(y' — y) approaches when y' and y both approach the common limit 1 is called the modulus of the system of logarithms whose base is b. Let y' = ry ; then this ratio becomes 4og ry — *log y ^log r '^y-y y{r-iy and when y = l it is (*log r)/(r — l)* Hence the modu- lus may be defined as the limit which (^]ogr)/(r — 1) approaches when r approaches 1 ; or, symbolically, if M denote the modulus, „ limit 'logr r = l r — 1 When the modulus is 1 the corresponding system of log- arithms is called the natural or Napierian system,! and the base of such a system is called natural or Napierian base,t * This modulus is otherwise described as tlie rate of change of ^logr at the instant when r = 1. [See Strlngham's Uniplanar Algebra, Art. 23.] t The logarithms of the natural system must not be confounded with the numbers of Napier's original tables, which were not cal- culated with respect to a base, and are therefore different from the so-called Napierian logarithms of modern tables. NATURAL LOGARITHMS. 4.S7 and is denoted by the letter e. We therefore have, by definition, limit "logr* r = 1 r = 1. 302. From this definition we are able to derive a for- mula by means of which an approximate value of the number e may be calculated, and this being done, the modulus corresponding to any base b will be given by the formula limit '^OS'' M= r=l r-1 = ]:='lS^l«^^5 [Art. 297, VL] that is, since limit ("log r)/(r — 1) = 1, 3/=^loge = l/''log6. In. place of *log we shall henceforth use the shorter symbol In, made up of the initial letters of logarithm and of natural or Napierian. 303. We first derive a formula which determines e as a numerical limit. For this purpose let In r = u, that is, r = e", and let ^*/(e" — 1) = h, or eHi = h-{-u. * This fraction is positive whether 1 be greater or less than 1, for the logarithm of any number less than 1 is negative [Art. 298]. Hence r may approach 1 through values either a little smaller or a little greater than 1. 438 NATURAL LOGARITHMS. Then uz=0 when r = l, and 7i = l when u = 0, and we have [Art. 301] limit Inr limit u v -i. i i 1 — = f. ■ = limit h = 1. But from e'7i = h -\-u we obtain and hence, since h= 1 when it = 0, limit .-, , xw„ « = „ = o (! + ")• We have here supposed u to approach zero from the positive side, but the supposition was not a necessary one. If we assign to u a negative value, say —v, the formula for the value of e becomes limit v = ™'^ (I _„)-■/". 304. Now it may be shown that in whatever way u is made to approach zero, whether positively or negatively, the limit of (1 + i^)^/** is a finite number, and that when u passes continuously towards zero, (1 + ?*)^/" diminishes, and (1 — w)^/" increases, continuously towards a common limit ; and that e is therefore a definite number. For a formal proof of these statements, which is not properly within the scope of this treatise, we refer the student to Chrystal's Algebra, Vol. II., Chapter XXV., § 13, and only present here a few numerical verifications. By ordinary arithmetical computation we easily find : (1 + 1)2 ^2.25. (1-i)-' =4. (1 + ^)3 =2.370. (1-i)-' =3.5. (1+i)' =2.44140625. (l-i)"' =3.1604903... (1 + 1)* =2.48832. (1-i)'' =3.0r)17r)78125. a + j^y = 2.5937424601. (1 - yV)"'" = 2.867971990 ... NATURAL LOGARITHMS. 439 Thus, as the exponents in (! + ?«)'/" and (1 — w)"^/" are made nuinerically larger, the two series of numbers produced by these formulae slowly approach one another, those in the left column by increase, those in the right column by decrease. We can infer from this calculation that e is greater than 2.5 and less than 2.9. None of the decimal figures, however, in any of these numbers are as yet correct for the value of e, and further extensions of this method become rapidly tedious.* We therefore do not use the formula (1 + ?f)^/'* for obtaining close approximations to the value of e, but replace it by a rapidly converging series. 305. Such a series may be obtained by expanding (1 + \lzy by the binomial theorem, on the supposition that z is a positive integer and ultimately becomes infi- nitely large. t In this investigation we shall use the convenient symbol n ! to denote the product of the first n natural numbers ; that is, n ! = 1 • 2 • 3 • • • to ?i factors. With this notation it is obvious that w ! = {n — V)\n— (w — 2) ! (n — 1) n, etc. By the binomial theorem we have, for integral values of z [Art. 205], * The value of e, correct to seven places of decimals, is 2.7182818, and the value of (1 + -001 )'<^ gives only two of these decimal figures. In fact, the first eight figures in the computed value of (1 + .001)i'wo are 2.7169239. t The demonstration is taken from Chrystal's Algebra, Vol. II., Chapter XXVIII., § 2. 440 NATURAL LOGARITHMS. \ z) z 21 z^ 3 ! !^ z{z-l)-'.(z-n + l) l I "^ n! z^ z'' that is, »; ^2! 3! ^ (l-l/,)...(l-n-lA ) ^ ^^ where J. _ (l-lA)...(l-nA) (1-1/.)...(1-^H^/,) (^ + 1)! (n + 2)! ^ • Let n have any j^a;ed integral value and let z be an arbitrarily large integer, at least larger than n. The series R^ will then terminate ; and each of its numera- tors will be less than 1, for 1/z <2/z<3/z<-" 1. Whatever rational positive value x may have we can always = 1, NATURAL LOGARITHMS. 443 express it as the sum of a positive integer n and a positive finite proper fraction /, and we may therefore write of _ a/+" _ f n a^ X /+ n /+ n n' in which / and «/ remain positive and finite for all positive values of X. But, because/ is finite, limit n _ limit 1 n = cof^n~ w = «^l+//n and therefore limit a^_ ^ limit a^ a; = Qo x~^ n = CO ^* in which n, always a positive integer, approaches oo by successive integral steps. Also, since by hypothesis a > 1, we may write and hence, by the binomial theorem for a positive integral index, a»=l -\. nh-\- ^^^ ~ ^ h"^ + 7ix (other positive terms), ji nn 1 h2 — = -+^H — (n — 1)4- other positive terms, n n 2 and limit a" limit A^ ix , .. -x- * ^_^— =^^_^ — (n — 1)+ other positive terms = oo. limit a^ , .*. „ — = a-'^x CO = 00. Ex. 4. Find the limit of y "logy when y = + 0, a>l. Let y = a-* ; then y «log y = — xa-* = — x/a'f and X = + 00 when y = -f 0. But limit ="(-f)=-^«- [^-^-3 limit ^^^02'"^^^?'=^' 444 NATURAL LOGARITHMS. Ex. 6. Find the limit of x"" when x = + 0. By Art. 291, x* - e^J"*; . limit limit ^xinx-^o-i [Ex. 4.] Ex. 6. Find the limit of (x - l)i°== when x = 1. We may write (a; _ l)lnx = ^(a; _ l)x-l|(lnx)/(x-l). But i^^^i(x-l)^- = l, [Ex. 4.] and iTul"^)/(^-l)=l- [Art. 301.] .. ^i^it^(x-l)inx=ii = i. 308. Exponential Indeterminate Porms. In order to dis- cover under what conditions the function u" will assume an indeterminate form, as in Exs. 5 and 6, Art. 307, we write f^u\ log/='ylogw, the logarithms being taken in any system, and seek the conditions that make v\o^u indeterminate, and these will also be the conditions that make /or u" indetermi- nate. Since log 1 = 0, log oo = co, and log = — oo, these conditions obviously are -y log u = ( ± oo) X 0, or w" = 1 ±*, 'ulogt^ = X (+oo), or i^'' = oo^, 'y log w = X ( — co), or u" — ^^. These are known as exponential indeterminate forms, and the list is evidently exhaustive for forms that result from combinations of values of u and v. But of course NATURAL LOGARITHMS. 445 either u ov v may by itself assume one of the algebraic forms 0/0, oo/co, x co, oo — oo. Ex. 1. The limit of (1 + x/nY when n = oo (x = a finite quan- tity) is limit - , x\« limit f /. , xy/'\ =" ^ Ex. 2. Find the limit of xV* when x = co. Let x=\/z^ then 2 = when a; = oo, and •limit J/, ^ limit ^/^ X = CO = 0' Ex. 3. Find the limit of (In x)*-i when x = 1. We proceed as in Ex. 6, Art. 307, and obtain (lnx)='-i = {(lnx)i°*}(^-i)/i°*, i^^\\lnx)«-i = li = L EXAMPLES LXXXIII. Evaluate the limits of the following functions for the values or x indicated : 1. In(x3 — 1) — ln(x — 1) when x = 1. 2. In(x2 _ 1) 4- In f 1 + — ^— ) when x = 1. 3. In (2 - V(l + x)] - In {3 - V(3 x)] when x = 3. 4. (x2 - 1) ln(x - 1) when x = 1. 5. (1 + \/x^Y when x = oo. 6. (1 + 1/x)*^ when x = ». 7. aji/'"* when x = 1, 446 NATURAL LOGARITHMS. 8. xV(*^-i) when x = l. 9. {(x3 - l)/(x3 + 1)} ==-1 when x = l. 10. (xy^ when x = 0. 11. (a^ — ly when x = 0. 12. (a^ - a-==)^ when x = 0. 13. Solve the equation x^+^ -2x^-x-h2 = 0. 14. Solve the equation a,x . ajx+l _ a; . q;2^ - x^ + «"= = 0, in which a is positive and greater than 1. (Take account of the result of Ex. 3, Art. 307.) CONVERGENCY AND DIVERGENCY. 447 CHAPTER XXXI. CONVERGENCY AND DIVERGENCY OF SERIES. 309. A succession of quantities, formed in order ac- cording to some definite law, if finite in number, consti- tute a finite series ; but if their number exceed every finite quantity, however great, they are said to form an infinite series. We have already seen [Art. 266] that, when r is numerically less than unity, the sum of the n terms of the geometrical progression a -f ar -j- ai^ -\ + ar"~^ can be made to differ from a/{l — r) by an arbitrarily small quantity, by sufiiciently increasing n. Many other series have this property, — that the sum of the first n terms approaches a finite limit when n is increased ad infinitum. 310. Definitions. When the sum of a finite number of terms of an infinite series approaches a finite limit (or zero), as the number of terms is increased ad infinitum, the series is said to be convergent, and this limit is called the sum, or limit of the series. If the series have ± co as its limit, it is said to be di- vergent; and if it have neither a finite nor an infinite limit, it is said to be neutral or indeterminate.* * Series of this last class are also called divergent series in many text-l?ooks of algebra, 448 CONVERGENCY AND DIVERGENCY. Ex. 1. The series | + ^ + | + ••• ad inf. is convergent, and has the limit 1. [Art. 285, Ex.2.] Ex. 2. The series 1 + 2 + 3 + ••• ad iw/. has + oo for its limit, and is therefore divergent. Ex. 3. The series 1 — 1 + 1 — 1 + .-. ad inf. has no definite limit, but is 1 or according as the number of terms is odd or even; it is therefore indeterminate. 311. It is evident that an infinite series whose terms are either all positive or all negative cannot be indeter- minate, but must have either a finite or an infinite limit. If each term of an infinite series be finite (not zero) and the terms are either all positive, or all negative, the series must be divergent. For, if each term be not less than a the sum of n terms will be not less than wa, and na can be made larger than any finite quantity, by suffi- ciently increasing n. 312. The successive terms of an infinite series will be denoted by ?^i, Uz^ u.^, -"U^, -", the sum of the first n terms by ?7„, and the limit of the series, if it be conver- gent, in which case alone it has a limit, by U. Thus and U = Wj + ?i2 + ^3 + '"(^d inf. We shall frequently omit the words ad inf., and denote the limit of the entire series by Wi + 1*2 + ^^3 + •••• 313. In order that the series Ui-\-U2-\ hw„-f--»- may be convergent, it is, by definition, necessary and sufficient that each of the finite sums Uny C/n+u U^xn etc., shall approach the common limit U, as n increases. H(?nce U^, U^fiy C4+2> etc., must differ from Uj and CONVERGENCY AND DIVERGENCY. 449 therefore from one another, by quantities that converge to zero, as n increases without limit. But Ur,+^—Ur,= u^i, U— Un = W„+i H- W„+2 + Wn+3 + •-. Hence : In order that a series may be convergent, the (n + 1)^^ terrrij and also the sum of any number of terms beginning with the {n-[-l)thy must have zero as a limit when n is iyicr eased ad infinitum. For example, the series 1/2 + 1/3 + 1/4 + ••• + 1/n + ••• can- not be convergent, although limit n=»(l/w) = ; for the sum of n terms, beginning at the (u + l)th, is n + 1 n + 2 w-f3 2n which is greater than (1/2 n) x n, that is, than 1/2. 314. Tests for Oonvergency. The following are the most important criteria for the convergency of infinite series. It is convenient to state them as applying only to series whose terms are all positive (Criterion IV. and the cases in which negative terms are cha-racteristic of the series excepted). But they are also effective when applied to series whose terms are all negative, and (provided the signs of all negative terms be changed) to that class of series, containing both positive and negative terms, which remain convergent when all the negative terms are made positive. For, if a series be convergent when all its terms are positive, it is obviously convergent when either all or some of its terms are negative, 3f 450 CONVERGENCY AND DIVERGENCY. For example, if Mj + u.^ .f w^ -f ... is convergent, so is - (Mi 4- Wj + % H )» and also u^ — u^-)r u^ — u^ + ••-. The term series means always infinite series unless specific statement to the contrary is made. 315. Test by the difference between the terms of the given series and the corresponding terms of a convergent series. I. A series {whose terms are all positive) is corivergent if, after any particular term, each of its terms is less than the corresponding term of a known convergent series. Let there be two series U'=S + Ui-\-U2 + Us-{-'", V=Vi-\-v,-\-v^-] , in which S is the sum of a finite number of the terms of U'j suppose V to be convergent, and that u^ < v^ for all values of r. Then U—S is certainly less than V, that is, and since Fand S are finite, so also is U', and U must have a limit, for all its terms are positive. II. It needs only a slight modification of this reason- ing (which the student can easily make) to prove that an infinite series {having only positive terms) is divergent if, after any particular term, each of its terms is greater thO/n th^ corresponding terms of a divergent series, CONVERGENCY AKD DIVERGENCY. 451 Ex. 1. Apply the test for convergency to the series 1 1.2 1.2.3 1.2.3.4 The terms of this series are, in the order of occurrence, less than the corresponding terms of the geometrical progression 1 + J_ + _J_ + _J__ + ... 1 1.2 1.2-2 1-2.2.2 ' convergent. The given series must therefore also be convergent. Ex. 2. Test for convergency the series 1 + 3 + 3^+33 3^ 35 _ 2!3!4!5! After the fifth term each term of this series is less than the corresponding term of the geometrical progression 3^ , 3^ I 3^ I ■■ 4.4!42.4!43.4! ' whose common ratio is 3/4. The given series is therefore con- vergent. 316. Test by the ratio of the terms of the given series to the corresponding terms of a convergent series. III. If the ratio of the corresponding terms of two series {of positive terms) he always finite, the series will both he convergent or hoth divergent. Let the series be respectively U'= -Wi + W2 + W3 H J F= Vi + Va + '^sH . Then, since all the terms are supposed to be positive, U/ V must be greater than the least and less than the greatest of the fractions Wi/vj, U2/V2, u^/v^,--- [Art. 170, 452 CONVERGENCE AND DIVERGENCY. Theorem II.]. Hence U/V is finite; and, therefore, U is finite or infinite according as V is finite or infinite. Ex. 1. The two series 2,3,4 n+1 2- 34. 56. 7 2n(2n+l) * 3+5+7+ +2n+l+ ' are both convergent or both divergent. For, the ratio of the w*^ terms is (n + l)/2 n, or 1/2 + 1/w, which cannot be greater than 3/2 or less than 1/2 for values of n between 1 and «. Ex. 2. Prove that the two series of Ex. 1 are divergent by com- paring the terms of the second one with the terms of the divergent series 1 + 1/2 + 1/3 + ... + 1/w -f •••• 317. Test by ratio of convergence. IV. A series is convergent if, after any particular term, the absolute value of the ratio of each term to the preceding is always less than some fixed finite quantity which is itself less than unity. Let the terms be supposed to be all positive, and in accordance with the conditions here assigned, beginning with the (?• -f 1)*^ term, let «*r+l/«*r < ^> ^r+2A^r+l < ^J '*r+3/^r+2 < ^J CtC, whcrC k < 1. Then w^+i < ku,, w^+2 < ^«*r+i < ^X? ^r+s < ^^r? and so on ad infinitum. Hence u, + ^r+i + ^r+2 + - < t*.(l 4- A; 4- A:^ + •••) [Art. 279.] < V(l - k), :• k Ur, Wr+2 > W^+i >U^y '" U^+n > Kj that is, W^+l + Wr+2 H f- ^«r+« > 't^^r- In either case the sum of these n terms can be made greater than any finite quantity by sufficiently increasing n. Hence this part of the series, and therefore the entire series, is divergent. Ex. 1. The ratio of convergence of the series l+2 + 3 + ... + » + ... 2 22 23 2»* is {(« + l)/2'*+i]/(7i/2«) = (w + l)/2 w, and this is less than f for all values of n greater than 2. Hence the series is convergent. 454 CONVERGENCY AND DIVERGENCY. Ex. 2. The ratio of convergence of the series 2 22 2^ 2* 2" 1.22.33.44.5 n(w + l) is2(w — l)/(n + 1), which is greater than 1 for all values of u greater than 2. Hence the series is divergent. Ex. 3. The ratio of convergence of the series 12 + 22a; + 32a;2 -I- ... + w2a;n-l _!_... is x(n + 1)2/ n2, that is, (1 + l/nyx. Now if X be positive and less than 1, and any fixed quantity k be chosen between x and 1 , the ratio of convergence will be less than k for all tenns after the first one that renders (l + l/n)% y/X/{y/k- y/X), the square roots being taken positively. Hence the series is con- vergent if abv X < 1 . If a; = 1 the series is 1^ 4- 2^ 4- 3"^ 4- ..., which is obviously diver- gent, and it is therefore also divergent if x> 1. When an infinite series is such that, after a finite number of terms, its ratio of convergence w^+i/^n is always less than unity, but approaches unity as a limit, the test by ratio of convergence fails. Further special investigation is then necessary. Ex. 4. The ratios of convergence of the series 1 + 1+1+1 + ... + U 1 . 2 3 4 n n-\-\ 2-2 32 42 n^ (n + 1)- are n/(n 4- 1) and n'^/(n + 1)2 respectively, and each of them is less than unity for all finite values of m, but approaches unity as a limit when n is increased without limit. But, as will be innnedi- ately proved in the next article, the first series is divergent, the second convergent. CONVERGENCY AND DIVERGENCY. 455 318. The convergency or divergency of many series may be determined by the methods of comparison, explained in Arts. 315, 316, with the aid of the series 1/1* + 1/2* + 1/3* + ••• as a standard. For this purpose it is important to know for what values of k this standard series is convergent. VI. The series 1/P + 1/2* + 1/3* -f- ••• is convergent when k is greater than unity, and is divergent ichen k is equal to or less than unity. (i.) If k>l, since each term is then less than the preceding, 2' 3' 2'' 4* 5* 6* 7* 4*' 8* 9* 10* 11* 12* 13* 14* 15* 8*' and in general 11 12" — H ~ 1 < -^. 2"* (2" + l)* (2"+' — 1)* 2"* Hence, the entire series is less than 1.2482" 1* ^ 2* 4* 8* 2"* ' that is, it is less than ' Ok-l ' 9-'(A:-l) ' 93(t-l) ' "^ _ 9>i(A-l) ' ' which is a geometrical progression whose common ratio, 1/2*-^, is less than unity, since A; > 1. The given series is therefore convergent. 456 CONVERGENCY AND DIVERGENCY. (ii.) If Tc—1, the series may be written in the form i+i+a+i)+(i+i+i+i)+- ^^2"-' + 1 2"-^ 4- 2 ^ ^ 2^y with its terms so distributed into groups that each group (within brackets) is greater than ^. Hence 2" terms of the given series is greater than n + 1 terms of the series l+^ + |-4-^+ •••, that is, greater than 1 + | n, which may be made larger than any finite quantity by suffi- ciently increasing n. The series ^i is therefore divergent. (iii.) If A;<1, then each term of l/P+l/2*+l/3*4--- is greater than the corresponding term of the divergent series 1 -f- i + i -f •••. The original series is therefore also divergent in this case. This completes the proof of our theorem. Ex, 1. The series '^ ^ 12+1 22+ 1^32 -fl^ is divergent. For since ^n/in^ + 1) > 1/n for all values of n greater than 1, .-. S2n/(w2+ l)>Sl/n, and S 1/n is divergent. Ex. 2. The series S (w + 2)/(w3 + 1) is convergent. For, (n + 2)/(n8 + 1)< (n + 2)/w8 < 3 n/n^ = 3/n2. .-. S(n + 2)/(«8 + l)<3 2 1/n2, and S l/n^ has been shown to be convergent. CONVEKGENCY AND DIVERGENCY. 457 319. Def. If a series having both positive and nega- tive terms remain convergent when all its negative terms are made positive, it is said to be absolutely convergent. The criteria of Arts. 315-318 are effective only when applied to absolutely convergent series. And there is a class of series, designated as semi-convergent, whose con- vergency is due to the presence of negative terras, and which become divergent when all the terms are taken positively. Other tests, not yet considered, are then necessary. For example, as will be shown in the next article, the series 1 — i + i — i + ••• is convergent, although l+|+^+|^+--« is divergent. 320. Test for semi-convergency. VII. A series ivhose terrns are alternately positive and negative is convergent if each term is less than the preceding and the n*" term has zero as a limit when n increases ad infinitum. The general series conforming to these conditions is U=Ui — U2-^Us—U4-\ ±w„T w„+, ± ..., with Wi >U2> u^ >ih> •"> Un > ^^.+1 > ••• > 0, •1 limit ^ and u,, = ; and this we may write in either of the forms Ui — u^-j- (^3 — u^) + (% — Uq) + ..., ^1 - (ih — %) - (^4 - Ws) , from which it is at once evident that the limit of the series is greater than Ui — ii2 and less than w,, and is therefore finite. It is also similarly evident that abv (C/— C7„) is intermediate to abv (?f„_,_, — ?^„_^2) and 458 CONVERGENCY AND DIVERGENCY. abvw„^.i (abv = absolute value of), and that therefore, since limit w„ = 0, limit ^^^ n — ao^ "^ The series must consequently be convergent. Ex. 1. The series 1 - 1/2 + 1/3 - 1/4 +••• satisfies the condi- tions of this test, and is therefore convergent. Ex. 2. The series 2/1 - 3/2 + 4/3 - 5/4 + ••• cannot be less than 1/2 nor greater than 2, but it is not convergent, for the n^^ term, which is (n 4- l)/w, has not 0, but 1, as its limit for n = oo. 321. Power Series. The infinite series whose general term is a^x"" (a„ independent of x) is called a power series in X. By Criterion IV. [Art. 317] this series is absolutely convergent if abva;a„^i/a„ remain less than 1 in the limit when n= cc, that is, if x satisfy the condition abv X < abv limit (a„/a„^.i) when n = oo. Hence : VIII. The power series IS absolutely convergent for all values of x that satisfy the condition abv a; < abv ^™^ («n/««4-i)- 322. If a power series be only semi-convergent, since the successive terms of every convergent series must ulti- mately decrease in absolute value [Art. 313], the ratio abv(a;a„^.i/a„) must ultimately still remain less than 1 for finite values of n. But if abv(.Ta„^i/a„) remain less than 1 in the limit when n = Go, the series is absolutely con- vergent [Criterion VIII.]. Hence, semi-convergence can CONVERGENCY AND DIVERGENCY. 459 only arise by the ratio of convergence being, in absolute value, ultimately less than 1 for finite values of n, but having the limit 1 when n = oo.* Therefore : IX. If the power series tto + ciiX -f a,;pi? + ••• + a„ic" 4- ••• he convergent for a given value of x, it is absolutely conver- gent for every numerically smaller value ofx; and if it be only semi-convergent for the given value, it is divergent for every numerically larger value. Upon the following corollary rests the proof of the principle of indeterminate coefficients which will be dis- cussed in the next chapter. CoR. If the series a^ + ajcc -f a^^ 4- • • • he convergent for any finite value ofx, ^'^0 (^^^ + a^or^ + ... + a„a;« +...) = 0. Proof: From the identity tto + a^x + a^ H = a^^ -\- x{ai -\- a.pc -\- a^ -\ ) it is evident that for finite values of x that render tto + Oiic -|- • • • convergent, the series ai-\-a^-\- "- is also convergent ; hence, by Criterion IX., aj + a^ + ••• is con- vergent, that is, finite when x = 0. Therefore, when a^x H- a^x^ -\ = x {finite quantity)^ 0. 323. The Biuomial, Exponential, and Logarithmic Series are the three infinite series of greatest importance in mathe- matical analysis. We may now test their convergency. * Harnack, Differential and Integral Calculus, Cathcart's trans- lation, § 44, page 73. 460 CONVERGENCY AND DIVERGENCY. Ex. 1. In the binomial series, the number of terms is finite when m is a positive integer; but when m is not a positive integer, no one of the factors m, m — 1, TO — 2, ... can be zero, and tlie series will not terminate. It is with tliis second alternative that we are at present concerned. The ratio of convergence is Un+l n + 1 / w+ 1\ and its absolute value, as n is increased, becomes more and more nearly equal to x. If, therefore, abvic< 1, then, either from the beginning or after a finite number of terms, abv(w„+i/M„) will be and remain less than 1, whether the sign of Un+i/Un be positive or negative, and, by Criterion IV. [Art. 317], the series formed by adding together the absolute values of the terms will be conver- gent. Hence, for all values of x numerically less than unity, the binomial series is absolutely convergent. If abv X be greater than unity, abv (t««+i/wn) will ultimately become and remain greater than 1 and, by Criterion V., the series will be divergent. [It may be shown that this series is convergent when a; = 1, provided to < — 1, and when a; = — 1, provided to > 0. See Treatise on Algebra, Art. 338.] Ex. 2. The exponential series is 2! 3! n\ Its ratio of convergence is x/n, and this is numerically less than unity for all terms after the first one that renders x numerically less than n. This series is therefore absolutely convergent for all finite values of x, Ex. 8. The logarithmic series is CONVERGENCY AND DIVERGENCY. 461 and its ratio of convercence is Un+l Un 71+1 \ 71+1) This is numerically less than unity for all values of x that are numerically less than unity, and for such values the series is there- fore convergent. If ic = 1, the series becomes 1 — i + | — i+-", which is con- vergent by Criterion VII. If X = —1, the series becomes — (1 + I + 3 + 1 + ••OjWhich was shown to be divergent in Art. 318. Thus, the logarithmic series is absolutely convergent when x has any value between — 1 and + 1 and semi-convergent when x = 1, but divergent in all other cases. 324. Product of Series. If finite portions of the two series U= 1^1 + W2 + «^3 + ••• + w„ H- ••• be multiplied together by the algebraic rule for the mul- tiplication of two multinomials, a result will be obtained which may be arranged in the form P = UiVi + {UiV.2 + ^2'"!) + (^I'^S + "^^^2 + UsVi) + ... + (WiV^ -t- W2V„_i H + ^„-lV2 + W»'yi) + •••• Whenever these three series are simultaneously conver- gent, the third is defined as the product of the other two. For our present purpose we assume that the first and second series are absolutely convergent, denote their limits by U and V respectively, and attach the usual interpretations to Un, F„, etc. Let the third series be denoted by P. It is required to prove the following very important theorem : 462 CONVERGENCY AND DIVERGENCY. If U and Vbe absolutely convergent, then P will he a convergent series ivhose limit is equal to the limit of the product Ux V. (i.) If all the terms in (7 and Fare positive, then the product U',,^ X V'in will contain all the terms of Po„ plus others, such as u-iV^n, ^3^'2n-ii '^s^sn? 'fhn'^2ny ^tc, aud P^n will contain all the terms uf the product Un X V^ plus others, such as UiV2n, ^2^2,1-11 '^3^2,1-2 y etc.* Hence U2nX V2n>P2n>U,,X F„, and this inequality remains true, however great a finite value n may have. But in the limit, when 71 = 00, U2n = Un= U, and Fs^ = F„ = F Hence, when all the terms are positive, Ux V=P. (ii.) If U and F contain both positive and negative terms, change the signs of all negative terms and denote the resulting series by U' and F'; and let P be the series formed from U' and F' in the same way as P is formed from CTand F Then t/^^ x Fo,, — P^n cannot be, in absolute vahie, greater than U'2n X F'2„ — -P'^nj for the series which the two expressions stand for differ only in the respect that the former contains some negative terms that are posi* tive in the latter. But, by the proof in (i.), * The student can make this quite evident by writing out all tho terms of these products for small values of n. CONVERGENCY AND DIVERGENCY. 463 hence it follows again that Ux V=P* This completes the proof of the theorem. EXAMPLES LXXXIV. Apply the proper test of convergency to each of the following series : I •*' |_ •'^ I '*' 1 _| [- ••' ■ r-2 2-3 3-4 n(«+l) x+1 x + S iC + 5 x-h2n 3. l + l + l+... + _^_+.... 1* 3* 5* (2 w - 1)* 4. 5 1-2 2-33. 4 4-5 3-4 4-5 5-6 6-7 g 1-2 2.3 3.4 4.5 .. • 32.42 42.52 S--'. 6'^ 62.72 7 a^— 1 y — 2 a; — 3 x — 4 ■ rc-fl a; + 2 a; + 3 a: + 4 The following expressions denote, in each case, the w'^ term of an infinite series. By applying the proper tests, determine whether the series thus indicated are convergent or divergent. 8. 1 9^ n(n+ 1) 10. 2n(2«-l) 2'* 1 + n2 * This result is also true when only one of the factor series 17, F, is absolutely convergent, the other being semi-convergent. If neither U nor V be absolutely convergent, nothing is certain about the convergency of P. [See Chrystal, Algebra, Vol. II., Chapter XXVI., § 14.] 464 CONVERGENCY AND DIVERGENCY. 11. 1 + n2 15. arir + b 12. 1 + w 1 + n^ 16. 7l\ 13. n-h 1 17. a;» 1 + n'^ 14. 1 18. X" 1 19. a* a« + x'» 20. 21. (m + l)x'» n2 22. X" aw + 6 ri^ an -\- b LNDETERMINATE COEFFICIENTS. 465 CHAPTER XXXII. Indeterminate Coefficients. 325. In Art. 148 it was proved that if a rational inte- gral function of x vanish when x= a, it is divisible by x — a without remainder. Let the function be f{x) = aic" + 6a;"-i + cx""-^ + ••• + I, and let the division by a; — a be performed. It is clear that the first term of the quotient, namely, the term of highest degree in x, will be ax''~^ ; and since dividend = divisor x quotient, .-. f(x) = {x-a){ax''~'-h"'). Suppose that/(x) also vanishes when x=p, [/8 not =a], then the product (x —a) x {ax^~^ -^ "-), and therefore either (x —a), or (ax"~^ + ••.), must vanish when x = p. But /3 — a is not zero ; hence (aa?""^ + •••) vanishes when x = p, and is therefore divisible by x — /3; and if the division be performed, the first term of the quotient will be aa^~^. .'.f(x) = (x-a)(x-p)(ax--'+^..). In general, if there be r values a, ft y, 8, • • • of a; for which / (a;) vanishes, r repetitions of this process will obviously produce f{x) = (x-a){x-p)(x-y){x-8)'--(ax^-^ +...). Finally, if ax"" + baf-^ -f- ... vanish for w different values 2o 466 INDETERMINATE COEFFICIENTS. of X, it has n factors x — a, x — jS, etc., and a final factor ax'*'", or a. Therefore, under this hypothesis, f{x) = a{x- a) {x - /?) {x - y)..., in which there are n binomial factors. Two or more of these binomial factors may of course be identical. Should there be a factor of the r^^ degree, say {x — ay, it is evident that the number of remaining factors will be n —r, 326. A rational hitegral function of the n*" degree in x cannot vanish for more than n values of x, unless the co- efficients of all the powers of x are zero. For, if /(re), being of the form ax"" + hx""-^ + cxT-"^ +'•', vanish for the n values a, jS, y ..., it must be equivalent to a{x — a) {x — P) (x — y)'". If now we substitute for x any other value, k suppose, different from each of the n values «, ft, y, etc. ; then since no one of the factors k — a, k — ft, etc., is zero, their continued product cannot be zero, and therefore f(x) cannot vanish for the value x = k, except a itself is zero. But if a be zero, f(x) reduces to bx"'^ -f cx""^-!- •••, and is of the (n — l)**" degree ; and hence it can only vanish for n — 1 values of x, except b is zero. And so on. When all tlie coefficients are zero, the function will clearly vanish for any value whatever of x. 327. The values of x for which the expression ax"" + bx""-^ + cx""^ -\ is equal to zero, are the roots of the equation ax"" + fta?"-^ + cxr-'^ + . . . = 0. [Art. 91.] INDETERMINATE COEFFICIENTS. 467 Hence, by Art. 326, an equation of the n"" degree cannot have more than n roots, except the coefficients of all the different powers of the unknown quantity are zero, in which case any value of x satisfies the equation. 328. If the two expressions of the n^^ degree ax"" + bx^~^ + ex''-- -\ , and px"" 4- qx''-'^ + rx""^ H , be equal to one another for more than n values of x, it follows that the equation aaf* -f bx""-^ 4- cx""-^ -\ =px'' + gx""^ + ra;'*""^ -\ ; that is, the equation {a—p)x^-\- (6 — g)ic"-^+ (c- r)a;''-2-f ... =0 has more than n roots. Hence, by Art. 827, the coefficients of all the different powers of x must be equal to zero. Thus CL=Py ^ = Q, c=r, etc. Hence, iftivo rational integral functions of then** degree in X be equal to one another for more than n values of x, the coefficient of any power of x in one function is equal to the coefficient of the same jiower of x in the other. This is the principle of indeterminate coefficients as applied to rational integral functions. 329. When any two rational integral expressions, which have a limited number of terms, are equal to one another for all values of the letters involved, the condi- tion of the last article is clearly satisfied, for the num- ber of values must be greater than the index of the highest power of any contained letter. 468 IKDETERMIHATE COEITFICIEKTS. Hence when any tioo rational integral expressions, which have a limited number of terms, are equal to one another for all values of the letters involved in them, we may equate the coefficients of the different powers of any letter. 330. If the two infinite series ao + a^x -\-a^^-\-'-' + a^x"" + ••., 60 + M + M^ H h M" H , be equal to one another for all finite values of x for lohich they are convergent, then tto = &0J ^1 = ^15 •'•? ^n = ^nJ 6tC. If there be a finite value of x for which the two series are convergent, they are also convergent for x = [Cri- terion IX., Art. 322], and when ic = 0, liy the corollary of Art. 322, aiX + a-iX^ 4- . . . = 0, and biX + b2X^-\ =0; .-. ao = bo. We now have, for the finite values of x that make the original series convergent, ajX + a^pc^ H = 6^0; + biOiP H , that is, tti H- ttgOJ -f- ttoO^ + .•• = 61+ hjx + b^-\ . But these two series are convergent for all the values of X that make aQ-\-aiX-\- -" and bi + b.;fc-\ conver- gent [Art. 322], and hence, by again putting a; = 0, we obtain tti = &i. INDETERMINATE COEFFICIENTS. 469 By successive repetitions of this fjrocess we find a^ = 62? a^ = ^3, and in general, a„ = b^. This is the principle of indeterminate coefficients as ap- plied to infinite series. APPLICATION TO INTEGRAL FUNCTIONS. 331. The principle of indeterminate coefficients was applied, without formal notice of the fact, to Examples 3 and 4 of Art. 153. The following are other examples of this class : Ex. 1. Detennine the values of c and d tliat will make x^ + bx^ -{- ex -^ d a perfect cube for all values of x and b. Assume x^ -\- bx^ + ex + d={x + ^•)^• Then, since (X -\-A-y = x3 + 3 kx^ + Sk^x-h k^, we must have b = Sk, c = S k% d = h\ [Art. 328.] whence k = A 6, c = 1 52, d = J. 63. Ex. 2. Determine under what conditions ax^ + bx^ + ex -\- dis divisible by px^ + qx + r without remainder. If the division be performed, the quotient will consist of two terms ; namely, (a/p)x, and a second term independent of x. Hence, the quotient is of the form (a/p^x + k, where A; is a quan- tity to be determined by the condition that the division shall be exact. We have, therefore, ax^ 4- bx^ + ex + d = l-x + A; J (px^ + qx -\- r) = ax3 + (^ + pA;)x2 -{■ I— + qkY + rk, 470 INDETERMINATE COEFFICIENTS. and since this is to be true for all values of x, the coefficients of like powers of x in the two members of this equation must be equal [Art. 328] ; whence aq ar b=j +i>^, c = — -\-qk, d = rk. Replacing k in the first two of these equations by its value d/r derived from the third, we have bp = aq -\- djj'^/r^ cp = ar + dpq/r. These are the conditions necessary and sufficient in order that the division may be exact. Ex. 3. Transform x^ + pxy + qy^ into the sum of two squares. Assume u =(x + ^qy)h, v ={x — y/qy)k, and determine h and k by the condition that M- + v'2 = a;2 + pxy + qy"^. For this purpose we make the coefficients of xy and y^ in {X + ^qy)Vi^ + (X - ^qy)-k\ that is, in (a:2 + qy'^) (h^ + k^) + 2 ^qxy^h'' - k^) equal to p and q respectively. We thus obtain h^-^k^=l, 2 y/q(h-^ - k^) = p, or V2\V 2Vg/ V2VV 2y/q) Hence, h and k having these values, a;2 -I- pxy + qy'^ = {{x-\- y/qy)h}^ + {(a; - y/qy)k}^. Ex. 4. Find the factors of (6%2 _ a4)(6 - c)-\-(c'2a'^ - h*)(c - a)-¥(a''~b^ - c*)(a - fo). This expression vanislies if c = 6, or if a = c, or if /> = a, and therefore fc — c, c — a, and a — 6 are factors. [Art. 148. J Also, since the expression is symmetrical and is of five dimcn- INDETERMINATE COEPFICIENTS. 471 sions, there must be a fourth symmetrical factor of two dimensions, which must therefore liave the form L (a2 + 62 _|. c2) + ilf {be -\-ca-\- ah). The given expression is therefore identically equal to (& -c)(c-a)(a- h){L{cfi+ h"^ + d^) -\- M {he ^ ca ^- ah)}, in which L and M are to be determined by equating coefficients in accordance with the principle of indeterminate coefficients. The coefficients of a^ in the two expressions are respectively — (6 - c) and — L{b - c), and the coefficients of a^ are ft^ — c^ and &2 _ q2 _ J/ (^52 _ c2-) . hence i = 1 and Jlf = 0, and the re- quired factors are (6 -c), (c - a), (a - 6), {c(P- + 6^ 4. c^). EXAMPLES LXXXV. 1. Determine the value of k that will make a^ — a;* + 2 x^ + ^ divisible by x^ + x + 1 without remainder. 2. Determine the values, of p, q, and r that will make x^ — 5px^ f t>qx'^ — r divisible by {x — c)'^ without remainder. 3. If ax:^ + bx -\- c and a'x- + h'x + c' have a common factor of the first degree, this factor, and also the remaining (second) factor, in each case, will be rational in all the letters. Prove this by de- termining the three distinct factors. [Compare Ex. 3, Art. 173.] 4. Prove that if ax^ + 2hxy+ hif + 2 ^/.-k + 2fy + c be expressible as the square of a rational integi-al function of x and y of the first degree, then af = gh, hg = hf, and ch = fg. 5. Determine the relation between a and b that will make {x + ay)^ and x^ + bxy -]-y^ have a common factor of the first degree in x and y. 6. Determine k such that x^ — y'^ — x— Sy + k may be the product of two rational factors of the first degree in x and y. 7. What value of k will make the three equations 2x -- y -{- 3=0, x + y— 1 = 0, and x-\- ky -\- 1=0 simultaneous in x and y ? 472 INDETERMINATE COEFFICIENTS. 8. Supposing a and h to be given numbers, determine a bino- mial expression (containing an arbitrary factor) which, when sub- stituted for X, will make ax + 6^ a perfect square (for all values of the arbitrary factor). 9. Prove that if ax^ + 6x'^ -f ex + d is divisible by x^ — A:-, then ad = he. 10. Find the factors of 4a6c(a + & + c)+ he (h^ + c2)+ ca(c2 + a'^)+ ah{a^ + b^). 11. Find the factors of 62c2(6 - c) + c2a2(c _ a) + a262 (a - h). 12. Find the factors of a*(6 - c)+ 6t(c -a)-h c*(a -h). 13. Find the cube root of 1 + 3x + 6x2 + 7 x3 4- 6 .X* + 3xs + x^. APPLICATION TO PARTIAL FRACTIONS. 332. In Art. 165, the process of obtaining a single fraction as the result of adding together any number of given fractions was explained. It is sometimes important to be able to perform the converse operation of decom- posing a complex fraction into the sum of a series of simpler partial fractions, and for this purpose the principle of indeterminate coefficients is employed. 2x — 6 Ex. 1. Resolve into (x-l)ix-2) partial fractions. The sum of the two fractions A/(x - I), B/(x - 2) will pro- duce a fraction whose denominator is (x — 1) (x — 2). Let it there- fore be proposed to determine A and B such that 2x-6 ^ A B ^ A(x-2)-\-B(x-l) _ (x-l)(a;-2) x-1 x-2 (x-l)(x-2) INDETERMINATE COEFFICIENTS. 473 For this purpose it is obviously sufficient that Since the left member of this identity contains no power of x higher than the first, both A and B are assumed to be constants, and therefore, by the principle of indeterminate coefficients, A-h B = 2, &nd 2A-^ B = 5, that is, A = 8, and ^ = — 1. . 2x-5 ^ 3 _ ^ (a:-l)(x-2) x-1 x-2 Ex. 2. Resolve '^^—^ into partial fractions. x(x-l)ix-2) Assume ^ = — H 1 , and reduce the x{x-l){x-2) X x-1 x-2' partial fractions to a common denominator. Then, the denomi- nators being omitted, we have x2 + l=^(x-l)(x-2)+ Bx(x-2)-\- Cx(x- 1). We might now expand and equate coefficients, but it is simpler to proceed as follows : Since the identity is true for all values of x, put x equal to 0, 1, and 2 in succession. From these substitutions the following results are at once obtained : x = 0, 1=2^, A = l; x = l, 2=-B, B = -2; x = 2, 5 = 20, 0=1. . x2 + 1 ^ 1 2 5 _ * ■ x(x-l)(x-2) 2x x-1 2(x-2)* Ex. 3. Resolve — — — — into partial fractions. x^-Sx-\-2 The process of decomposition cannot be applied directly to this fraction, because the numerator is not of lower degree than the denominator. But, by division, any such fraction can be replaced 474 INDETERMINATE COEFFICIENTS. by an integral function plus another fraction whose numerator is of lower degree than the denominator. In the present case we find and we apply the process of decomposition to (a:2 + 3a;-l)/(a;3_3a; + 2). Since x^ - 3x -\- 2 =(x - ly^x + 2), this latter fraction, for aught we as yet know, might be the result of adding together three fractions of the respective forms A/(x-iy, B/(x-l), and C/(x + 2), for (x — l)2(x + 2) is the least common multiple of (x— 1)-, (x — 1), and (x + 2). Hence we assume a;2 + 3x-l_ A , B ^ C ib8-3x + 2 (x-1)2 x-\ x + 2 and determine A^ B, and C by the condition x'i ^ Sx - l = A(x -\- 2)-^ B(x - l)(x -{- 2) + C(x-iy. Then, putting x equal to 1, — 2, and in succession, we obtain the following sets of values : aj = l, 3 = 3v4, ^ = 1; a; = _2, -3 = 9(7, = -\; x = 0, -l = 2A-2B-\-C, B = f . . a;8 + a;'^ + 1 ^ 1 , 1 ■ 4 1 ''x»-Sx + 2 {x-iy 3(x-l) 3(x + 2) Should a factor of the form (x + ky occur in the denominator of the original fraction, three partial fractions A/(x -^^ ky^ B/(x + A:)2, C/(x + k) must be provided for, and so on. /V.2 _ 7 « _ 2 Ex. 4. Resolve into partial fractions. x8 - 8 The denominator of this fraction may be resolved into the two factors X — 2, x* + 2 x 4- 4, or into the three factors x — 2, X + 1 + t V^> X + 1 — iy/S, and observing always the rule that the INDETERMINATE COEFFICIENTS. 475 assumed numerator of any partial fraction must be lower in degree, by one, than the denominator, we may assume either x3-8 x-2 X+1 + i^^ X + 1 - V3 ^j, a;2 - 7 X - 2 ^ A Bx -\- C x3-8 ~"x-2 x2 + 2x + 4' Choosing the latter identity, we have x2-7x-2 = ^(x2 + 2x + 4)+ 5x(x-2)+ C(x - 2), and the following particular equations for the determination of A, B, and C : x=:2, -12 = 12^, A=-\] x = 0, - 2 = 4^-2(7, C = -\] x=l, - 8 = 7^-^-0, ^ = 2. . x2 - 7 X - 2 ^ 1 2x- 1 x3 - 8 X - 2 x2 + 2 X + 4' The calculations for D and Z>' in the first identity will show that The student can easily verify these results. EXAMPLES LXXXVI. Resolve into partial fractions : 1 5x + 1 g a;2 + 1 2 x2 + 3x4-3 ^ 3. 4. x2 + 3x + 2 x + 7 3(x + 3)(xH-5)' x2 + 15x + 8 (X + l)-2(x - 5)* -4x + 6 (x-l)(x-2)(x-3) 5. -^^ + ^ 10. x(x- -1)« 2x + 1 x(x2 + 1) ax + 6 x(x2 + &) X-2 -4x + 5 (X- 1)2(X2 + 1) 7x2 -llx+7 (X- ■ 1)H^ + 2) 11. 5x+l X3-1 12. 4 x^-1 13. 2 x(x2 + X + 2) 14. a;'2 — 05 (x-^ + c)(cx2+l) 15. X2+1 X3+1 x2 + ax + 6 x\x^ + 6) X2+1 x^+1 X2-1 X* + X2 + 1 X2+1 476 INDETERMINATE COEFFICIENTS. 16.^^-1 17. 18. 19. X\X +1) (X - 1)4 (X3 + 1) APPLICATION TO EXPANSION OF FUNCTIONS. 333. In the expansion of functions into infinite series in ascending powers of the variable, only such values of the variable are permissible as will make the series con- vergent. For any other values, the limit of the series cannot be placed equal to the given function, and if there be no values of the variable for which the series is con- vergent, the expansion is impossible. For example, the process of division applied to 1-7- (1 —x) produces the series l + a; + a;^4-a^+..-ad inf., but the limit of this series is not equal to 1/(1 — a;) unless abv x etc., by equat- ing the coefficients of like powers of n. Suppose, for example, that u„ is of a degree in n not higher than 2. Then and &i — &2 + ^3 = cto? 2 62 — 3 63 = tti , 3 63 = ag? and from these equations are derived in succession h = i<^2, &2 = T«i + i«2j &i = «o + i«i + ia2j Sn = h-h («o + i^i + ia2)?i + iiO'i + CI2W + i(hn^' The value of bo may then be found by putting n = 1 in this formula. Observe that if n^ be the highest power of n contained in Sn, the difference S^ — S^-i will contain n^~^, but not ?i^ Hence, the degree (in n) of S^ is higher by 1 than the degree ofu^. Ex. 1. Find the sum of the first n terms of the series whose n^^ term is 1 — w -f n-^. Here ao = 1, ai = — 1, as = 1 ; .-. Sn = ho+{l- i + i) w + K- 1 + 1) w2 + in8 = &o + I w + i w3 ; and since /S'l = 1 = 60 + 1, that is, 60 = 0, Ex. 2. Find the sum of the first n terms of the series 3 + 4 -f 6 + 9 + 13 + 18+ .... The differences of successive pairs of terms of this series form the arithmetical progression 1, 2, 3, 4, ... ; that is, 4-3=1, 6-4 = 2, 9-6 = 3, etc. 2b 482 INDETERMINATE COEFFICIENTS. Hence, denoting the successive terms of the given series by mi, Ui, ?<3, W4, ••• w„, we have U\ = 3, «2 = Ml + 1, W8 = Ml + 1 + 2, M4 = Ml + 1 + 2 + 3, Mn = Mi+l + 2 + ... + (w-1) = Mi + i(w-l)(2 + w-1-1) [Art. 259] = Ml + \n (w — 1). This, the 'nS^ term of the given series, is therefore 3 — \n-\-\ v?. Hence, in the formula for 8n, we have ao = 3, ai = — I, ai — |, and since /S'l = 3 = 60 + 3, &o = 0, .-. ;9„ = in(17 + n2). Let the student verify the results of these two examples by use- ing the method here explained, but without substituting in the formula. 336. The foregoing method may be generalized and made applicable to power series in which the coefficient of a;" is a rational integral function of n. For the summation of such a series it will be found sufficient, when the (n + l)**" term is w„+i = (ao + ^iM H h a;kn*)a;» to assume ^«+i = (^0 + &in + ••• + 6,n*)aj''+i + A; and determine 60 j ^u •••^a ^^^ ^ from the identity An example will best show how the method is applied. INDETERMINATE COEFFICIENTS. 483 Ex. 1. Find the sum of the first ti + 1 terms of the series 1 + 2a; + 3x2 + 4x3 + .... The {n + l)*** term of the series is (?i + 1) x", and if Sn+i denote the sum of the first w + 1 terms, then (71 + 1)X« = ^^+1 - Sn. Assume Sn+i = (a + &n)x«+i + k, where k is independent of n, and a, b are as yet undetermined coefficients. Then Sn = {a + b(n - l)}x« + k, and Sn+i - Sn = {a(x-l) + b + b(x- l)w}x», whence w + 1 = a(x — 1) + & + 6(x — l)n. Equating coefficients in this equation, we now have a(x- 1)+ 6 = 1, 6(x-l)=l, 1 ^ 1-b x-2 X - 1 X - 1 (x-1) 2 Hence ^„+i = j -^^ + -^ | x"+i + k, I (X-l)2 X — 1) for all values of n. We now determine k by putting w = in this expression for Sn+i, and by this substitution we obtain that is, k (x-iy Hence finally, ^„+i = { ^^ + -^ W + -J— -, l(x-l)^ x-1 J (^-1)^ which is the required result. EXAMPLES LXXXVIII. Each of the following expressions is the «th term of a series whose first term is got by putting 7i = 1. Find the sum of the first n terms of each of the series thus indicated. 1. 2w2-l. 3. w(n + l)(ri + 2). 5. p-{.q(n-l). 2. w(n + l). 4. w3- w + 1. 6. ^ 4n2-l 484 INDETERMINATE COEFFICIENTS. Each of the following expressions is the (n + 1)**^ term of a series whose first term is got by putting n = 0. Find the sum of the first n -\- 1 terms of each of the series thus indicated. 7. (p + qn)x'*. 10. (p + qn + rn^)x\ 8. (w2 _ n + l)ic«. 11. (2« - 2 w + \)x\ 9. {2(n + 1)2 - l}x« J2 ^2 + 271 + 2 p^ (w + l)(n + 2) Find the sum of the first n terms of each of the following series : 13. l + 2a; + 5x2+10a;3 + 17x4 4-26a:5+.... 14. 1 . 2 . a; + 2 . 3 . a;2 + 3 . 4 . x3 + 4 . 5 . X* + .... 16. 1 + 2x + 4x2 + 7x3 + llx* + 16x5 + -. 16. 1 + 2x + 6x2 + 13x3 + 23x* + 36x5 + •.-. PERMUTATIONS AND COMBINATIONS. 485 CHAPTER XXXIII. Permutations and Combinations. 337. Definition. The different ways in which r things can be taken from n things, regard being liad to the order of selection or arrangement, are called the permutations of the n things r at a time. Thus two permutations will be different unless they contain the same objects arranged in the same order. For example, suppose there are four objects, represented by the four letters a, 5, c, d. If we take the four objects one at a time, we have the four per- mutations «, 6, c, d. If we take the objects two at a time, we have twelve permu- tations a6, ac, ad, ha, he, hd, ca, ch, cd, da, dh, do. Now to find the whole number of the permutations three at a time we can proceed thus. Take any one of the permutations two at a time and place after it either of the two letters which it does not contain ; then all the permutations so obtained will be different, for either we shall have used a different permutation of two letters or else the final letters will be different ; moreover, since all the possible permutations two together have been used, no permutation of three letters can have been omitted. Hence the number of permutations of 4 things 3 together = number of permutations of 4 things 2 together x 2 = 12 x 2 = 24. We can by similar reasoning find the permutations of n things ' r together. 486 l*ERMtJTATIONS AND COMBINATIONS. The number of permutations of n different things taken r at a time is denoted by the symbol „P^. 338. To find the number of permutations of n different things taken r at a time, where r is any integer not greater than n. Let the different things be represented by the letters a, b, c, '••. It is obvious that there are n permutations of the n things when taken one at a time, so that „Pi = n. Now, if beside any one of the different permutations (r — 1) at a time, we place any one of the n— (r — 1) letters which it does not contain, we shall obtain a per- mutation of the n things r at a time. We thus obtain 71 — (?• — 1) = n — r -j- 1 different permutations r at a time from every one of the different permutations (r — 1) at a time. Hence „P^ = „P^_i x(n — r-\-l). Since the above relation is true for all values of r, we have in succession „P,_i=,.Pr_2X(n-rH-2), nPr-2=nPr 3 X (^ - r + 3), nP2 = J\X(n-l). Also „Pi = )). Multiply all the corresponding members of the above equalities, and cancel all the common factors ; then ,P, = n(n-l)(n-2)...(n-r + l), r being the number of factors on the right. PERMtTTATIONS AND COMBINATIONS. 487 If all the n things are to be taken, r is equal to n, and we have „P„ = n(7i-l)(n-2)...3.2.1. Def. The continued product 7i (n — 1 ) (?i — 2) • • • 3 • 2 • 1 is denoted by the symbol [n, or by n ! The symbols [n and n ! are read ' factorial n.^ Thus |^= 4 ! =4 • 3 • 2 • 1. Ex. 1. In how many different ways can 6 boys stand in a row ? The number required = c A = 6-5-4.3-2-l = 720. Ex. 2. How many different numbers can be formed by using three of the figures 1, 2, 3, 4, 5 ? The number required = 5P3 = 5 • 4 • 3 = 60. Ex. 3. Show that 10P4 = 7 A, 10P4 = 10 • 9 . 8 . 7 and 7P7 = 7 • 6 • 5 • 4 • 3 • 2 • 1. 339. To find the number of permutations of n things taken all together, ivhen the things are not all different. Let there be n letters ; and suppose p of them to be a's, q of them to be 6's, r of them to be c's, and so on. Let Pbe the required number of permutations. If in any one of the actual permutations, we suppose that the a's are all changed into p letters different from each other and from all the rest; then, by changing only the arrangement of these |> new letters, we should, instead of a single permutation, have p ! different permutations. Hence, if the a's were all changed into p letters differ- ent from each other and from all the rest, the &'s, c's, etc., being unaltered, there would be P xp\ permutations. Similarly, if in any one of these new permutations we suppose the 6's are all changed into q letters different from each other and from all the rest, we should obtain q ! per- mutations by changing the arrangement of these q new 488 PERMUTATIONS AND COMBINATIONS. letters. Hence the whole number of permutations would now be Pxpl xql By proceeding in this way we see that if all the letters were changed so that no two were alike, the total number of permutations would be Pxpl xql xrl '". But the number of permutations of n different things is n !. Hence Pxpl X ql X rl x ..-=71!; n! .-. P: plqlrl '" Ex. 1. Find the number of permutations of the six letters aaabbc taken all together. The number = — — — = 60. 3!2! 1! Ex. 2. How many different numbers can be formed by the figures 2, 3, 3, 4, 4, 4, 4 ? 7 ' The number required = — ^ = 105. 412! 340. Def. The different ways in which a selection of r things can be made from n things, without regard to the order of selection or arrangement, are called the Oombinations of the n things r at a time. Thus two combinations will be different unless they both contain precisely the same objects. For example, suppose there are four objects, represented by the four letters a, 6, c, d. The different combinations one at a time are a, 6, c, d ; the combinations two at a time are ahy ac, ad, bCj bd, cd ; the combinations three at a time are abc, abd, acd, bed ; l^ERMtJTATlOi^S AKD combikations. 489 and there is only one combination, namely ahcd^ when all the letters are taken. The number of combinations of n things r together is denoted by the symbol JJ^. 341. To find the number of combinations of n different things taken r at a time. Since every combination of r different things would give rise to ?•! different permutations, if the order of the letters were altered in every possible way, it follows that „a X r ! = „P, = n(n _ 1) (n - 2) ... (n - r + !)• Hence C - ^Q^ -^) 0^ -^)'--{n -r -^ 1) By the following method the number of combinations of n different things r at a time can be found indepen- dently of the number of permutations. Let the different things be represented by the n letters a, b, c, d,-". In the combinations of the n letters r together the number in which a particular letter occurs must be equal to the number of ways in which r — 1 of the remaining n — 1 letters may be selected. Hence in the whole number of combinations r together every letter occurs „_iO^_i times, and therefore the total number of letters employed is n_iCV-i X n; but, since there are r letters in each of the ^C^ combinations, the total number of letters employed is „(7^ x r. Hence r x ,.0^ = w X „-iC^-i. Since the above relation is true for all values of n and r, we have in succession 490 PERMUTATIONS AND COMBINATIONS. (r - 1) X n-lC,. 1 = (n - 1) X n-2Cr-2 (r - 2) X n.20r-2 = {n - 2) X r^-sOrS 2 X n-r+2C.2 = {u - T + 2) X „-r+lCi. Also „-r+iC'i = ?i — r + 1. Multiplying corresponding members of the above equalities and cancelling the common factors, we have r!x„(7, = ti(n-l)(n-2)...(n-r+l), that is „C. = n(n-l)(.-2)...(n-r + l). r\ By multiplying the numerator and denominator of the fraction on the right by {n — ?•)!, we have r\{7i — r)\ r\{n—r)\ Ex. 1. How many groups of 3 boys are there in a class of 16 ? The number = uCa = ^^ ' ^^ 'l^ = 455. 1 • 2 ' o Ex. 2. There are 6 candidates for 4 vacancies, and every elector can vote for any number of candidates not greater than the number of vacancies. In how many ways is it possible to vote ? An elector can vote for 4 candidates in eC* ways, for 3 candi- dates in 6^8 ways, for 2 candidates in ^€2 ways, and for 1 candidate in eCi ways. Thus the whole number of ways in which an elector can vote is 6C4 + eCs + cCo + cOi = 15 + 20 -f 15 + 6 = 56. PERMUTATIONS AND COMBINATIONS. 491 342. From the formula obtained in the last article, we have (n —r)lr\ so that the vnmber of combinations of n things r together is equal to the number of combinations of n things n — r together. The above proposition follows, however, at once from the fact that whenever we take r out of n things we must leave n — r, and if every set of r things differs in some particular from every other, the corresponding set of n — r things will also differ in some particular from every other. Hence the number of different sets of r things must be equal to the number of different sets of n — r things. 343. To prove that „(7, + nGri = nH-iC^. The total number of combinations of (n + 1) things r together can be divided into two groups according as they do or do not contain a certain particular thing. The number which do not contain that particular thing is the number of ways in which r of the remaining n things can be taken, which is „Cv; and the number which do contain the particular thing is the number of ways in which (r — 1) of the remaining n things can be taken, which is „(7^_i. Hence „+i(7^ = „(7^ + „C^-i. Thus, if n = 7 and r = 4, we have 7-6.5-4-3-2.1 ^ 6-5.4-3-2-1 6.5-4.3.2-1 4.3.2.1.3-2-1 4.3.2.1.2.1 3.2.1.3.2.1' 492 PERMUTATIONS AND COMBINATIONS. ■ Note. — The above important proposition may also be proved from the general formula. Thus ^ m(w- l)(w-2)--(n-r+ 1) n(n - l)(w - 2)-.-(n - r + 2) 1.2.3..-r 1.2.3...(r-l) ^ nin - l)(n - 2)...(n - r + 2) . . 1.2.3...r ^^ ^ ^ ^ ( w + l)n(n - l)(n - 2)--Cw - r + 2) 1.2.3-.r 344. Greatest Value of JO^, From Art. 341 we have . r - r y-^-^ + l r Hence JO/^jO^-x according as ^ ~^"^ ^ 1; that is, according as r^i(n + l). Thus the number of combinations of n things increases as r increases, so long as r is less than ^(n + l). If r = -J(n-f 1), then „0^ = „Ct._i, but r cannot be equal to i(n + 1) unless n is oM. If r > i(?i + 1), the number of combinations diminishes as r increases. Thus, if n is even, JJ^ is greatest when r = ^w; and if n is odd, „(7^ =„Or_i when r = ^{n + 1), and these values are the greatest. Ex. 1. Find the greatest value of gCV. The greatest value is when r = 4, and the value is 1.2.3.4 PERMUTATIONS AND COMBINATIONS. 493 Ex. 2. Find the greatest value of jjCV. The gi'eatest values are when r = Q and r = 5, and the values are 11.10.9.8.7 1.2.3.4.5 = 462. EXAMPLES LXXXIX. 1. FindigPs. 2. Find ,(,P^. 4. How many different numbers can be formed by using the five figures 1, 2, 3, 4, 5 ? 5. How many different numbers can be formed by using one or more of the figures 1 , 2, 3, 4, 5, 6 ? 6. In how many ways can 10 boys be put in a row so that two particular boys should not be together ? 7. Of how many different things are there 720 permutations when taken all together ? 8. Find the number of permutations of all the letters of each of the words success, Mississippi, and algebraic. 9. In how many ways can the nine letters aaaabbbcc be ar- ranged in a row ? 10. How many different numbers can be formed by using the seven figures 1, 1, 1, 2, 2, 3, 3 ? 11. How many permutations can be made with the letters of the word essences, and how many begin with e and end with s ? 12. Find nPn, having given that nP^ = ^P^ x 2. 13. Find n, having given that nP^ = n-P* x 12. 14. Find ^^C^, 150,2, and ^C^^. 15. n nC, = „Cio, find nC^. 16. If nO,o=nOii, find nOg. 17. How many different numbers of three figures covUd be made by taking three of the digits 0, 1, 2, 3, 4 ? 494 PERMUTATIONS AND COMBINATIONS. 18. If inC, = nC.,x 12, find n. 19. If 2nC.i = nG^ X 24, find n. 20. Numbers are formed by writing the five figures 1, 2, 3, 4, 6 in eveiy possible (.)rder. How many of tliese numbers are greater than 23,000 ? 21. In a certain town there are five letter boxes. In how many ways can a person post three letters ? 22. How many different sums could be made with 2 pennies, 3 shillings, and 4 sovereigns ? 23. How many different sums could be made with 4 pennies, 6 shillings, and 5 sovereigns ? 24. Show that „+20^+i = nC.+i + 2 „a + „a-i. 25. Show that in gC^ the number of combinations in which a particular thing occurs is equal to the number in which it does not occur. 26. Show that in ^^C^ the number of combinations in which a particular thing occurs is one-third of the whole number of the combinations. 27. Show that in 3„Cw the number of the combinations in which a particular thing occurs is one-third of the whole number of the combinations. 28. There are 10 candidates for 6 vacancies in a committee ; in how many ways can a person vote for of the candidates ? 29. At an election there are 5 candidates and 3 members to be elected, and an elector may give 1 vote to each of not more than 3 candidates ; in how many ways can an elector vote ? 30. In how many ways can a picket of 3 men and an officer be chosen out of a company of 80 men and 3 officers ? 31. In how many ways can a cricket eleven be chosen out of 30 players ; and in how many different ways could two elevens be chosen to play a match with one another ? PERMUTATIONS AND COMBINATIONS. 495 32. Out of 9 children, of whom 5 are boys and 4 girls, in how many ways can 4 be chosen, 2 being boys and 2 girls ? 33. In how many ways can 2 ladies and 2 gentlemen be chosen to make a set at lawn tennis from a party of 4 ladies and 6 gentle- men? 34. In how many ways can 8 children form a ring ? 35. There are ii points in a plane, no three of which are in the same straight line ; and the points are joined in paii-s by straight lines which are produced indefinitely. How many straight lines are there, and how many triangles are formed by them ? 496 THE BINOMIAL THEOREM. CHAPTEK XXXIV. The Binomial Theorem. 345. The binomial theorem for a positive integral index was proved by induction in Art. 205. The follow- ing proof makes use of the principles of the combinatorial analysis, explained in Chapter XXXIII. Suppose we have n factors, each of which is a -|- 6. If we take a letter from each of the factors of (a-{-b){a-^b)(a + b)"' and multiply them all together, we shall obtain a term of the continued product ; and if we do this in every possible way we shall obtain all the terms of the continued product. Now we can take the letter a every time, and this can be done in only one way, hence a"* is a term of the product. The letter b can be taken once, and a the remaining (n — 1) times, and the number of ways in which one b can be taken is the number of ways of taking 1 out of n things, so that the number is „Ci; hence we have a second term Again, the letter b can be taken twice, and a the remaining (n — 2) times, and the number of ways in which two b's can. be taken is the number of ways of THE BINOMIAL THEOREM. 497 taking 2 out of n things, so that the number is „02j hence we have a third term And, in general, b can be taken r times (where r is any positive integer not greater than 71) and a the remaining (n — r) times, and the number of wa3^s in which r 6's can be taken is the number of ways of taking r out of n things, so that the number is ^C^ ; hence we have, as the (r_|_l)thterm, The letter b can be taken every time in only one way ; hence we have the term 6**, which agrees with the result obtained by putting r = n in „C^ • a'*~''6'', since „(7„ = 1. Thus {a + b) {a + 6) (a + &)••• to n factors = a^ + nCi • a^-^b + ^C^^-'^b'' + . . . + ,.C, a^~^b' + • • • + 6". Hence, when n is any positive integer, we have (a+6)" = a" 4- nCi ' a^-'b + nCg • a^'-'b' + ••• This proves the binomial theorem for a positive integral index. The series on the right is called the expansion of {a-\-b)\ If in this we substitute the known values [Art. 341] of nC'ij nCs, etc., we obtain the form in which the theorem was written in Art. 205, namely : (a 4- &)" = a" + na^-^b + ^^^^~^K ^-^b^ + ... A. • Z -h ,/• ^ a^-'-b" + -" + b\ rl(n—r)\ 2i 498 THE BINOMIAL THEOREM. 346. General Term. Any term of the expansion of (a-f- b)" will be found by giving a suitable value to r in n{n-l){n-2)-'{n-r + l) ^„_,^, r! This is called the general tenn of the expansion. It should be noticed that it is the (r+ 1)"' from the beginning. 347. Some applications of the binomial theorem : If we put w = 2 in the binomial formula, we have (a + 6)2zz:a2 + 2a6 + &2. If we put n = 3, we have (a + 6)3 = a3 + §^25 + ^«52 _,. 58 1 i. • A = a3 + 3a26 + 3a&2 + 68. If we put n = 4, we have (a + hy = a* + U% + il|a252 + 4^3:|«68 4. 54 1 1 • ^ 1 • 2 • o = a* + 4a36 + Ga^ft^ + 4a6» + 6*. If we put 2 X for a, and — 3 y for 6, and « = 5, we have ' (2x -^yy = (2xy + b{^xy{-^y) + \^{2x)\-ZyY +^^(2x)K-3y)3 + ^;^-^;^ (2a;)(-y)^+(-3y)5 = 32 x5 - 240 x*y + 720 xV _ 1O8O xV + 810xy* - 243 yS. 348. In the expansion of (a + by by the binomial theorem, the (r + 1)*** term from the beginning and the (r + 1)'** term from the end are respectively ^Cror-'^' and nO„_,a'-6"-'-. But A = nGn-t* for all values of r. [Art. 342.] THE BINOMIAL THEOREM. 499 Hence in the expansion of (a -{-by, the coefficients of any two terms equidistant from the beginning and the end are the same, so that the coefficients in order are the same when read backivards as ivhen read forwards. 349. If in the formula of Art. 345 we put a = 1 and b = X, we have ^ 1-2 r\{n—r)\ This is the most convenient form of the binomial theo- rem, and the one which is generally employed. It should be noticed that the above form of the bino- mial theorem includes all cases; if, for example, we want to find (a -f 6)% we have (« + 6)«= I a(l + 3 }"= ."(l + ^)''= a-(l + «| + etc.) = a« + na''-^b + etc. 350. Greatest Term. In the expansion of (l + a;)**by the binomial theorem, the (?* 4-1)*^ term is formed from the 7^^ by multiplying by x{n — r + l)/r. Now x{n — ?•+ l)/r = x\ {n 4- l)/r — IJ, and (n 4-l)/r clearly diminishes as r increases; hence x(7i — r-^l)/r diminishes as r is increased. If x{7i — r-\-l)/rhe less than 1 for any value of r, the {r -{- ly^ term will be less tjian the r*\ In order, therefore, that the r^^ term of the expansion may be the greatest, we must have n — r-{-l 71 — r — 1 + 1 X < 1, and X ^^^—j > 1. {n + l)x {n-[-l)x Hence r>-^^-j-, and r< ^^^ +1. 500 THE BINOMIAL THEOREM. If r = (w + l)x/(n -\- 1), then x{n — r -f l)/r = 1, and there is no one term which is the greatest, but the r*^ and (r* + 1)*^ terms are equal and are greater than any- other terms. The absolute values of the terms in the expansion of (l-\-x)* will not be altered by changing the sign of x; and hence the r^^ term of (1 — a?)" will be greatest when the r*^ term of (1 + a;)" is greatest. Ex. 1. Find the greatest term in the expansion of (I -\- xy^ when x = ^. The r^^ term is the greatest if r>'^^ and r<-i^ + 1. Hence the third term is the greatest. Ex. 2. Find the greatest term in the expansion of (2 -\-Sxy^ when x = ^. (2 + Sxy^= {2(1 + I x)}i2= 212(1 + I xy^. Hence the r*^ term is the greatest if r > ?/ and r < Y + 1- Hence the sixth term is the greatest. The greatest coefficient of a binomial expansion, i. e, the greatest value of „(7^, was found in Art. 344. EXAMPLES XC. 1. Write out the expansions of : (1 \i5 ix ) • 2^x1 3. Find the 17th term in the expansion of f x^ — ] • 4. Find the coefiBcient of x» in the expansion of ( x- j • THE BINOMIAL THEOREM. 601 5. Write down the two middle terms in the expansion of 2n+l ("iy 6. Find the term in the expansion of (a + xy which has the greatest coefficient. 7. Find the terms in the expansion of (a + x)^ which have the greatest coefficients. 8. Find the greatest term in the expansion of (1 + 2 xy^ when 9. Find the greatest term in the expansion of (1 + Sxy^ when x = l 10. Find the greatest term in the expansion of (3 + 4 xy^ when 11. There are two consecutive terms in the expansion of (5x + 7)23 which have the same coefficient : which are they ? 12. Write down the coefficient of re*" in the expansion of {ax - byK 13. Show that the coefficient of x" in the expansion of (1 + x)2» is double tlie coefficient of x** in the expansion of (1 — x)'^"-i. 14. Show that the middle term of (1 + x)-" is 1.3.5...(2n-l) o„^„ n! 16. Employ the binomial theorem to find 99* and 9998. 351. Properties of the Coefficients. We now proceed to consider some properties of the coefficients of a binomial expansion. For this purpose we write it in the form (1 -f xy = Co + CiX + c^x^ H h c.ic'- H h c„a;", where Co = c„ = 1, Ci = c„_i = n, and c, =c„_, = -—^ — -• t-l^n — r)l L Put «= 1 in the above formula ; and we have 2" = Co -f Ci + c, + ••• + c, + ... c„. 602 THE BINOMIAL THEOREM. Thus the sum of the coefficients in the expansion of (1 + a;)" is 2". II. Again, put a; = — 1 ; and we have (1 - 1)" = Co - Ci -f- C2 - C3 + ... ; .-. = (Co + C2 + C4H )-(Ci + C3 + C5 +...). Thus the sum of the coefficients of the odd terms of a bino- mial expansion is equal to the sum of the coefficients of the even terms. III. Since c^ — c,, ^ [Art. 342], we may write the bino- mial theorem in either of the following ways : (1 + xY = Co-i-CiX + c^^-\ h cX 4- ••• c^a;", or (l+a;)"=c„+c„_ia;+c,._2X-+...c„ XH hq-T^-^+Coa;". The coefficient of x^ in the product of the two series on the right is equal to Co^ + Ci^ + C^^+'-'+C^^. Hence* Co^4-Ci^+--* +Cr^ + "-Cn is equal to the coeffi- cient of a;" in (1 + xy X (1 -f a;)", that is in (1 -f- a;)-", and this coefficient is ^ .^^' . nlnl Hence the sum of the squares of the coefficients in the expansion of (1 + a;)" is equal to i— ^. nlnl Ex. 1. Show that C1 + 2C2 + 3c8 + ••• -\-rCr+ ••• -\-nCn = n2«-\ * See Art. 328. THE BINOMIAL THEOREM. bO^ ci + 2 C2 + 3 C3 + ••• + rcr + ••• + ncn 1.2 1-2.3 ^ r!(w-r)! ^„/i+(„_i)+ («-i)('-2) + ...+ ("-D' +... + 1} I 1.2 (r-l)!(w-r)! J = w(l + 1)"-^ = w2'*-i. Ex. 2. Show that 2 3 fi + lw+1 e,_lc.+|c.4c3+...-f(-l)n-^ _^ 1 I n(n-l) l n(n-l)(n-2) ^ . ^., 1 - 2 ^3 1.2 4 1.2.3 ^ ^^ ^ .^+1 = -J_ /(„ + 1) - (^Jti)ii + O^-tiM^LJllI) _...+(_ i)„ I w+ll 1.2 1.2.3 J n+ln + l*- 1-^ -* (1 - l)n+i + 1 n + 1 1 n + 1 w+ 1 ^ 1 n + 1 352. To ^nc? the continued product of n binomial factors of the form x-\- a, x-\-h, x + c, etc. It will be convenient to use the following notation : Si is written f or a + 6 + c + •••? the sum of all the let- ters taken one at a time. S2 is written for ah-\-ac-\-hc-\ , the sum of all the products which can be obtained by taking the letters tit'o at a time. And, in general, S^ is written for the sum of all the products which can be obtained by taking the letters r at a time. 504 THE BINOMIAL THEOREM. Now if we take a letter from each of the binomial factors of (x + a)(x-{- b) {x + c) (a; + <^) -, and multiply them all together, we shall obtain a term of the continued product ; and, if we do this in every pos- sible way, we shall obtain all the terms of the continued product. We can take x every time, and this can be done in only one way ; hence ic" is a term of the continued product. We can take any one of the letters a, b, c, • • •, and x. from all the remaining n — 1 binomial factors ; we thus have the terms ax""^, bx^~^, ca;''"^ and on the whole S, ' x--\ Again, we can take any two of the letters a, b, c, •••, and X from all the remaining n—2 binomial factors ; we thus have the terms abx""'^, acx''-^ etc., and on the whole And, in general, we can take any r of the letters a, b, c, •••, and X from all the remaining n — r binomial fac- tors ; and we thus have S^ • a;""*". Hence (x-\-a){x-{-b)(x-\-c)(x-{-d) -^ ^zx^' + Si' aJ""^ + ^2 • a;""^ H h ^r •«"-'■ 4- •••, the last term being abed--', the product of all the letters a, b, c, d, etc. By changing the signs of a, b, c, etc., the signs of Si, S^, S5, etc. will be changed, but the signs of S2, /S'4, Sq, etc. will be unaltered. Hence we have (x — a) {x — b) {x — c) (x — d) ••' = x''-Si' x^-' + S2 ' x^-^ - Ss • a;"-3 + ... -\-{-iySr'X*'-''-i-'" -\-{—iyabcd"'. THE BINOMIAL THEOREM. 605 353. Tlie Multinomial Theorem. The expansion of (a + b + c+'-r can be found by the method of Art. 345. We know that the continued product (a + 6 + C+ •••)(« + & + C+ •••)(« + ^ + c +•••)••• is the sum of all the partial products which can be obtained by multiplying any term from the first multi- nomial factor, any term from the second, any term from the third, etc. If there are n of the multinomial factors, every term of the required expansion must be of the v}^ degree, and, therefore, all the terms must be of the form a'^b'o^ •••, where each of r, s, t, etc. is zero, or a positive integer, and r -\- S+ t -\- "• =n. Now the term a'"b^& ••• will be obtained by taking a from any r of the n multinomial factors, which can be done in ^G^ different ways ; then taking b from any s of the remaining n — r factors, which can be done in „_^(7, different ways ; then taking c from any t of the remain- ing 71 — r — s factors, which can be done in „_r-sOt differ- ent ways ; and so on. Hence the total number of ways in which the term a'"6*c'--- will be obtained, that is, the coefficient of the term in the required expansion, must be that is, n\ ^^ (n-r)\ ^^ (n-r-s)l ^^ ^ n\ rl(n—r)\ s!(n— r—s)! t\{n—r—s—t)\ r!s!^!... Hence the general term in the expansion of (a-f-6 + c+ "-y is rlsltl"' 606 THE BINOMIAL THEOREM. where each of r, s, t, etc. is zero or a positive integer, and r-\-s-\-t-^'-- = n. Ex. 1. Find the coefficient of abc in (a + & + c)^. Here w = 3, r = s = t = l. Hence the required coefficient is 3! 1!1!1! 6. Ex. 2. Find the coefficients of a*6, a^b^, and a^b'^c in {a+b+cy. "We have the terms .R-a% -^a^b^ and -^L-a^i^c, 4!1! 3! 2! 2!2!1! Hence the required coefficients are respectively 5, 10, and 30. EXAMPLES XCI. In the following examples Co, Ci, C2, are the coefficients of o^, x^, x^ '•• in the expansion of (1 + x)". 1. Prove that Ci - 2 C2 + 3 Cs + (- l)"-inc„ = 0. 2. Prove that Co - 2ci + 3c2 + ( - !)"('* + 2)c„ = 0. 11 1 2«+i — 1 3. Prove that Cq + -Ci + -C2 + 1 -c„ = — • 2 3 n + 1 w + 1 4. Prove that ^ + 2^ + 3^ + - + w-^ = in(« + 1). Co Ci C2 c„_i 2 2« 6. Prove that ^ + | + f+f+-= ,, 13 6 7 n+ 1 6. Prove that Co + cix + 2 C2cc2 4- ... + rCfOf h nc„x« = 1 + nx(l + a;)"-^ 7. Prove that CoCi + ciC2 f C2C3+ h Cn-iCn =, — . s^'y ., • (n+l)!(n-l)! 8. Prove that C0C2+C1C8+C2C4 + ••• + c„_2Cn — ^"—^ (n + 2)l(n-2)! THE BINOMIAL THEOREM. 607 9. Prove that co^ - Ci2 + cg^ + (- l)"c„2 is equal to if n be odd, and to — if n be even. 10. Prove that c„^ + 2 c,^ + 3 c«^ + ... + (n + l)c„^ = (n+2)(^»-l)K {n — l)!n! 11. Prove that ci2 + 2 C22 + 303^ + ..■ + nc„2 = ^^"~^^\,. - (71 — l)!(w — 1)! 12. Prove that e,-le. + lc3-... + (-l)»-'^c,. = j + l + |+...+l. 13. Prove that l__ci_ , _C2 . c_nn_^!!_:^ n\ X x+l a: + 2 ^ ^ x -{■ n x(x-\-\){x-\-2) — {x-\-n) 14. Show that, if there be a middle term in the expansion of (1 + ic)« its coefficient will be even. THE BINOMIAL THEOREM: ANY INDEX. 354. In Arts. 345, 349, we derived the series 1-2 r ! as the expansion of (1 + a?)**, but on the hypothesis that n was a positive integer ; and under this limitation the series necessarily terminates at the {n-\-iy^ term, for the (n + 2)"*^ term and all subsequent terms contain ?i — 71 as a factor. When n is not a positive integer, the series is endless, since no one of the factors n, n — 1, n — 2, etc., can in that case be zero. Hence, to prove the binomial theorem for a fractional or negative index is a proposition in 508 THE BINOMIAL THEOREM. infinite series, and any demonstration that may be pro- posed can only be valid under conditions that render the infinite series convergent. Tliis limitation must he pre- supposed in all the demonstrations that follow. 355. Before proceeding to give a proof of the bino- mial theorem when the index is fractional or negative, we give some examples of its application. In these examples x must be understood to have only such values as make the series convergent. Ex. 1. Expand (l4-x)-i by the binomial theorem. Put n= — 1 in the formula, and we have a + .)-. = 1 +(- i)x + -tiK^x^ + ^^ix^l^x^.. + ... ^(-l)(-2)...(-r)^^... r ! = 1 - CC + X2 - ic3 +...+(- lyar + .... Ex. 2. Expand (1 + x)-\ We have (1 + .)- = 1 + (- 2)x + tL|^.. + C^M^^SKzJO ^ + ... + f-2-|(-3)(-4)...(-r-l)^ ^ ... r\ = 1 - 2 X + 3a;2 - 4x8 -f- - + (- IKr + l)af + -. Ex. 3. Expand (1 + x)-^. (1 + x)-» ^ 1 + (- 3)x + i^^^ »' + <-^>/-%<-^> »^ + - (-3K-4K-6)-.-(-r-2) „, r! = J^n . 2 - 2 . Sx + 3 . 4x2 - 4 . 6a:8 + ... 1.2^ + (-l)'-(r+l)(r + 2)r- +..•}. THE BINOMIAL THEOREM. 509 Ex. 4. Expand (1 - x) ^. (1 - xy^ = 1 + ( _ 1) ( - X) + <^~^^H~'^ (- xy + (-i)(-f)(-f) (_^)3^... + (-i)(-|)-(-^-r+l) ._^W. 1.2-3 ^ ^ r\ ^ J -r 2 2.4 2-4.6 2.4.6.-2r .356. Euler's Proof of the Binomial Theorem. For our present purpose we shall employ the functional symbol f{n) to denote the series ^ , , n(n — T)o 1 n(n — l)(n —2) , , l + nx+ ^^^ ^ x^+ ^ ^.^^^ ^-a^+-", that is, the entire series if n be positive and integral, or its limit [Art. 310] if n be fractional or negative. In this general form, with n unrestricted in value, the series is called the binomial series. Let there be three binomial series formed as follows : /(n) = l+naj+'-^^^f^a;^+- (i.), /(m)= l + mx + -A___^ar' + ... (ii.), /(m+n)=l + (m+^0^+ : » ^a;2 + ...(iii.), and let x be assumed to have only such values as will make them absolutely convergent. We may then multiply (i.) and (ii.) together, and their product will be a converg- ent series [Art. 324]. Now, if the series on the right of (i.) and (ii.) be mul- tiplied together, and the product be arranged according 510 THE BINOMIAL THEOREM. to ascending powers of x, the result must involve m and n in the same manner whatever their values may be. But, when m and n are positive integers, we know that /(y/i) is (1 + ic)*" and that f{n) is(l + x) % and the prod- uct of /(m) and f{n) is therefore in this case (1 + 0?)**+", which again, as m + n is a positive integer, is f{m -f n). Hence when m and n are positive integers, the product /(m) X f{n) is /(m + n) ; and as the form of the product is the same for all values of m and w, it follows that /(m)x/(n) = /(7n + n) (a), for all values of m and n, for which (i.) and (ii.) are abso- lutely convergent. Eor the purposes of this proof we need not enquire what values of x, m and n will render the series under discussion absolutely convergent. But see Art. 323, Ex. 1. By continued application of (a), we have f(m)xf(n) xfip) X ...•=/(m + ri) xf{p) X .•• =/(m +w+i? + •••)• Now let m = n = p = • • • = r/s, where r and s are posi- tive integers ; then taking s factors, we have ff'!^\ Xff-^X •••to s factors =ff'^+'^ + '" tosterms\ JS^ But since r is a positive integer, f{r) = (1 + ^Y- ,-. (l + .r={/g)}• (^^-a;)•=/ THE BINOMIAL THEOREM. 511 This proves our theorem for any positive fractional exponent. Next, assuming that the binomial theorem is true for any positive index, it can be proved to be true also for any negative index. For, from (a), /(-7i)x/W=/(-7i4-7i)=/(0). Hence, as /(O) is clearly 1, we have Thus (1 + a;)~"=/(— n), which proves the theorem for any negative index. Ex. 1. Show that the coefficients of x""^ in the expansions of (1 — cc)-" and (1 + aj)2«-2 are equal, n being any positive integer. The coefficients are (_n)(-n-l)-(-n-n+2) . .,., ^^^ (2n-2)! ^,_, (n-\)\ ^ ^ (w-l)!(n-l)! The former = <^ + l)>-(2 n - 2) ^,_, ^ (2^-2)! (n-1)! (n-l)!(n-l)! Ex. 2. Find V^^^ ^y ^1^^ binomial theorem. Vioi- V{ioo(i + xk)} = 10(1 + Th^)^ =io|i+^x.oi+i^^x.oooi+ ^^~.^^^^~g^-^ x.oooooi+..-| = 10(1 + .005 - .0000125 + .0000000625} = 10.04987562 .... Ex. 3. Find the coefficient of x^ in the expansion of (1 - 2 a; + 4 ic2)-2 according to ascending powers of x. 512 THE BINOMIAL THEOREM. (1 - 2 a; + 4 x2)-2= {1 - 2 a;(l - 2 x)}-^ = l+(-2){-2x(l-2x)} + ^i^^{-2a;(l-2x)}2 = 1 + 4 x(l - 2 a;) 4- 12 x%l -2xf + 32 x\l _ 2 a;)^ + ... = l + 4x + 4x2 -16x3 + .... Ex. 4. Show that, if (1 + x + x2 + x3)-2 be expanded in a series of ascending powers of x, the coefficients of x^ and x'^ will be zero. (1 + a; + x2 + x3)-2 =/l-:^V^=(l - x)2(l - x*)-2 = (l-2x + x2)(l+2x* + 3x8 + 4x12 +...). EXAMPLES XCn. 1. Expand (1 — x)~* to 5 terms. 2. Expand (1 + 2x)~* to 5 terms. 3. Expand (2 — x)-* to 6 terms. 4. Expand (1 — 3x)~"^ to 5 terms. 3 5. Expand (1 — 5x)5^ to 5 terms. Find the general term in the expansion of each of the following : 6. (l-x)-5. 7. (l-x)-«. 8. (1-x) "^. 9. (1 + x) 10. (1+ x)"^ 11. (1 - 2x)"^. 12. (1 + 3x)"^. 13. Find the coefi&cient of x^ in the expansion of according to the ascending powers of x. "^ ^ ^ 14. Expand (1 — 3 x)^ to 6 terms, and write down the general term in its simplest form. 16. Find the coefficients of x^ and x* in the expansion of (1 -2x + 3a;2)-i, THE BINOMIAL THEOREM. ' 513 16. Find the coefficients of x^ and x* in the expansion of 17. Find the first negative term of (1 + 3 x)^. 18. Find the first negative term of (1 + 3 x)'^. 19. Find each of the following to four places of decimals by means of the binomial theorem : (i.) ^110, (ii.) ^130, (iii.) ^630. 20. Find the coefficients of x^ and x^ in the expansion of (1 + X + x2 + x3)3. 21. Show that in the expansion of (1 4- x 4- x^)~^ in a series of ascending powers of x, the coefficients of x^ and x^ are zero. 22. Show that in the expansion of (1 + x +x2 + x'^ + x^ + x*)-^ in a series of ascending powers of x, the coefficients of x* and x^ are zero. 614 EXPONENTIAL AND LOGARITHMIC SERIES. CHAPTER XXXV. Exponential and Logarithmic Series. 357. The Exponential Series is an infinite series whose first term is 1 and whose (n + 1)*^ term is x^'/n I It is absolutely convergent for all finite values of x [Art. 323, Ex. 2]. Let its limit be denoted hj f{x) ; then the two exponential series /(.)=!+. +2^+g+. ..+!;+... (i.) may be multiplied together and their product will be a convergent series [Art. 324]. Compare the terms of this product with the terms of the series /(a;+7/)=l+(a:4-2/) + -^~|Y^+ ••• + ^^^ The coefficient of a;"'y" in f{x) x f{y) is clearly l/(m In I) and in f(x -f- y) the term (x + ?/)'"+'*/ (^ + '0 • is the only one in which x'^y'' can occur, and by the bino- mial theorem for a positive integral index, the coefficient of a5'"y" in (x + y)"'+''/{m -f n)! is l__A ^ + '') \ that IS, (m + n) ! mini min EXPONENTIAL AND LOGARITHMIC SERIES. 515 Hence, whatever x and y may be, the coefficients of afy'* in f(x) x f{y) and in f(x + y) are the same, however large m and n may be, that is, whatever (integral) values, between 1 and + oo, m and n may have. It follows that the series (iii.)? with its terms expanded, is identical with the series formed as the product of the series (i.) and (ii.) ; hence /Wx/(2/)=/(aj + 2/), -^ (iv.) for all values of x and y. By repeated application of (iv.) we obtain /W X/(2/) x/(0 X ... =f(x + y) xf{t) X-' Therefore, if z be any positive integer, /(I) X/(l) X/(1) X to 2 factors =/(l-f 1 + 1 + ... to 2; terms), that is \f(l)l'=f(z). But /(l) = l + l+l + l+...+ l+... J ! o I nl = e. [Art. 305.] Zl nl if 5J be a positive integer. If z he a positive fraction, say z =p/q, where p and q are positive integers, then f(-) X f(^^ X fl^]x to q factors -4 - + - + - + . -tog terms): q q q ^ J' 516 EXPONENTIAL AND LOGARITHMIC SERIES. = e^, •.' p is a positive integer; Hence 6*=/(2;), for all positive values of z T)Oti integral and fractional. If zbe negative, say zz= — z'j then by (iv.), /(-^')x/(O=/(0), and /(O) is obviously equal to 1. •••/(-^')=^^ = -«, ••• 2/ is positive. Hence, for all rational values of z, be they integral or fractional, positive or negative, e' = /(.)=l+. + |;+...+J + .... Should z be irrational, we replace it by a rational fraction, which may be so chosen as to differ from the given irrational value by an arbitrarily small quantity. [See Art. 249.] 358. We may now derive an infinite series, in ascend- ing powers of x, for the exponential function a\ Let A = In a ; then a = e\ and a' = e^'' ; J ! ni where X depends only upon a. This theorem, by means of which a' can be expanded in a series of ascending powers of x, is sometimes called the exponential theorem. EXPONENTIAL AND LOGARITHMIC SERIES. 517 359. The Logarithmic Series. Let a = e\ so that A = In a. Then a== = e^^ = e^i°«. Hence, from Art. 357, we have a' = 6=^'°" = l + a;lna + i (a;lna)'^ + —^ {xlnaY 4- •••• Now put a = 1 + 2/ ; then we have (l+2/)^=l + ajln(l+2/)4-|jSajln(l + 2/)12+.... Now, provided y he numerically less than unity, (1 + ?/)* can be expanded by the binomial theorem; we then have ^ , , x(x — l) 2 I x(x — l)(x — 2) o , •^"^^^"^ 172 ^ "^~lT2T3 y^^'" = l + a;ln(l + 2/) + ijxln(l + 2/)r+..-. Equate the coefficients of x on the two sides of the last equation. [Art. 330.] We thus obtain ln{l + y) = y-t + t-y4 + ": 2 ' 3 4 This is called the logarithmic series. 360. In order to diminish the labour of finding the approximate value of the logarithm of any number, more rapidly converging series are obtained from the funda- mental logarithmic series. Changing the sign of y in the logarithmic series lu(l+2,) = ^-|>J-J + -, (i.) we have ln(l -?/) = - 2/-|--| --l . (ii.) 518 EXPONENTIAL AND LOGARITHMIC SERIES. Hence lni±2/ = ln(l + y)- ln(l-y) [Art. 297, IV.] Put — for , and therefore — ^^^ for y : then we n 1 — ^ m + ?i have from (iii.) In?!^ = 2 i 2?^l^ + l('??L=iiY+ V??ll^^V+ . . . I . (iv. ) n (m + n 3\m + wy 5\mH-n/ j 361. We are now able to calculate logarithms to base e without much labour. Put m = 2, n = 1, in formula (iv.); then ln2 = 2U + l.l+i.i+...l, (33 33^5 35 J' from -which it is easy to obtain the value ln2 = .693147-... Having found In 2, we have from (iv.) 2 163 53 6 65 i' .'. In 3 -In 2 = .405465 .... Hence hi 3 = In 2 + . 405465 . • • = .693147 ... + .406465 ... = 1.098612.... Again, putting m = 6, w = 3, we have 6 « ri . 1 1.1 1 In: 3 2/UI.I4-I. l+...\ 14 3 48 5 46 / = l/l + l + _l_ + _L_+...l 2 1 48 1280 28672 / = .610826...; EXPOISTEI^TIAL AND LOGARITHMIC SERIES. 510 .-. In 5 = 1.098612 ••• + .510826 ••. = 1.609438...; .-. In 10 = ln5 + ln2 = 1.609438... + .693147.. . = 2.302585.... By such processes as these the natural logarithm of any number can be found to any requisite degree of approximation. 362. In Art. 302 it was shown that In 10 is the recip- rocal of the modulus of the system of logarithms whose base is 10. Hence the modulus of common logarithms is . . M= 1/lnlO = 1/2.302585 ... = .434294.... EXAMPLES XCIII. 1. 1 = J- + A + 1 + ± ^. ... ad inf.- 2! 3! 4! 5! 4. ^ = 1+1 + 1 + A+.... 2 3! 5! 7! 6. (l+x)e- = l+|| + |^' + ^ + 6 1 = JL+ J__L J__Li + '2 1.3 3.5 5.7 7-9 * 7. ln2 = -l- + -i-+-l-+ J_+.. 1.23.45.07.8 520 EXPONENTIAL AND LOGARITHMIC SERIES. 8. In2=- + — ^ — + — - — + — - — + .... 21.2.33.4.55.6.7 X 12x4-1 3 (2x+ 1)3 ^5(2 x+ 1)5^ / 10 In ^ =2 1 1 1 1 1 , 1 1 , ) x-l~ l2x-2-l 3(2x2-1)8 5 (2x2 - 1)5 "^'"i 11. lnx = ^^l + l ^^-1 1 ^^-1 • X + 1 2 (X + 1)2 3 (x + 1)3 12 ln^ + ^= ^^^ I ^f ^^^ Vl V ^^^ Vl a - X a^-\- x2"^ 3U2 + xy bW + x2y "^ **• lnalx+1 3Vx + lyl i 14. If M = i(e* + e<-=' + e«**), 10 = i(e« + w2gwa: 4. a>e«^*), where a> = — | -f- iV— 3, prove that u -\- v -\- lo = e""^ u + uiH + w^^? = e<^'*, and that ifi + v^ + lo^ _ 8 ^ujo = 1. 15. Find the expansions in ascending powers of x for m, v, and w of example 14. LOGABITHMIC COMPUTATION. 521 CHAPTER XXXVI. LOGAKITHMIC COMPUTATION. 363. The logarithms used in all theoretical investi- gations are natural logarithms; but for the purpose of making approximate numerical calculations, for reasons that will shortly appear, logarithms to base 10 are always employed. On this account logarithms to base 10 are called common logarithms."* We have shown, in the preceding chapter, how natural logarithms, or logarithms to base e, can be found; and having constructed a table of natural logarithms, the logarithms to base 10 are obtained by multiplying the former by the modulus of the latter, that is, by the con- stant factor ^"^loge [Art. 291, VI.], whose numerical value has been found to be i«loge = -i- = .43429-... [Art. 362.] COMMON LOGARITHMS. 364. In what follows the logarithms must always be supposed to be common logarithms, and the base 10 need not be written. * Also Briggian logarithms in honour of Henry Briggs, of Oxford, who first constructed tables of such logarithms. 522 LOGARITHMIC COMPUTATION. If two numbers have the same figures, and therefore differ only in the position of the decimal point, the one must be the product of the other and some integral power of 10, and hence from Art. 297, III., the logarithms of the numbers will differ by an integer. Thus log 421.5 = log 4.215 + log 100 = 2 + log 4.215. Again, knowing that log 3 = .30103, we have log. 03 = log(3 - 100) = .30103 - 2. On account of the above property, common logarithms are always written with the decimal x>art positive. Thus log .03 is not written in the form -1.69897, but 2.30103, the minus sign referring only to the integral portion of the logarithm, and being written above the figure to which it refers. Definition. When a logarithm is so written that its decimal part is positive, the decimal part of the logarithm is called the mantissa, and the integral part the character- istic. 365. The characteristic of the logarithm of any number can be written down by inspection. First let the number be greater than 1, and let n be the number of figures in its integral part ; then the number is clearly less than 10" but not less than 10" ^ The logarithm of the number is therefore between n and n — 1 ; thus the logarithm is equal to w — 1 + a decimal. Thus the characteristic of the logarithm of any number greater than unity is one less than the number ofjigures in its integral part. For example, 235 is greater tlian 10- but less than lO''. Hence log 236 = 2 + a decimal, so that the characteristic is 2. LOGARITHMIC COMPUTATION. 523 Next, let tlie number be less than 1. Express the number as a decimal, and let n be the number of ciphers before its first significant figure. Then the number is less than 10~" and not less than 10~"~\ Hence, as the decimal jmrt of the logarithm must be positive, the logarithm of the number will be — (n + 1) + a decimal fraction, the characteristic being — (71 4- 1). Thus, if a number less than unity be expressed as a decimal, the characteristic of its logarithm is negative and greater by one than the number of ciphers before the first significant figure. For example, .02 is greater than 10- 2 but less than lO-i ; hence log .02 is — 2 + a decimal, the characteristic being — 2. Also .00042 is greater than 10-'' but less than 10-3 ; hence log .00042 is — 4 + a decimal, the characteristic being — 4. 366. Conversely, if we know the characteristic of the logarithm of any number whose digits form a certain sequence of figures, we know where to place the decimal point. For example, knowing that log 1.1467 = .0594498, we know that the number whose logarithm is 3.0594498 is 1146.7, and that the number whose logarithm is 4.0594498 is .00011467. 367. Tables are published which give the logarithms of all numbers from 1 to 99999 calculated to seven places of decimals : these are called " seven-figure " logarithms. For many purposes it is, however, sufiicient to use five- figure, or even four-figure, logarithms. In all Tables of logarithms the mantissas only are given, for, as we have seen, the characteristics can always be written down by inspection. 524 LOGARITHMIC COMPUTATION. In making use of tables of logarithms we have, I. to find the logarithm of a given number, and II. to find the number which has a given logarithm. I. To find the logarithm of a given number. If the number have no more than five significant fig- ures, its logarithm will be given in the tables. But if the number have more significant figures than are given in the tables, use must be made of the principle that when the difference of two numbers is small compared with either of them, the difference of the numbers is approximately proportional to the difference of their logarithms. An example will show how the above principle, called the Principal of Proportional Differences, is utilized. Ex. 1. To find the logarithm of 1.14673. From the tables we find that the log 1.1467 = .0594498, and that log 1.1468 = .0594877, and the difference of these logarithms is .0000379. Now the difference between 1.14673 and 1.1467 is ^gths of the difference between 1.1468 and 1.1467. Hence to find log 1.14673 we must add to log 1.1467 three-tenths the difference between log 1.1467 and 1.1468, that is we must add .0000379 X yV = .0000113. Hence log 1.14673 = .0594498 + .0000113 = .0594611. II. To find the number which has a given logarithm. Ex. 2. Find the number whose logarithm is 2.0594611. We find from the tables that log 1. 1467 = .0594498 and that log 1.1468 = .0594877, the mantissa of the given logarithm falling between these two. Now the difference between .0594498 and the given logarithm is .0000113, and the difference between the logarithms of 1.1467 LOGARITHMIC COMPUTATION. 525 and 1.1468 is .0000379. Hence, by the principle of proportional differences, the number whose logarithm is .0594611 is 1.1467 + i}f X .0001 = 1.1467 + .00003 = 1.14673. [N.B. The approximation can only be relied upon for 07ie figure,] Thus .0594611 is log 1.14673, and therefore 2.0594611 is log 114.673. Ex. 3. Find ^100, having given log 4.6415 = .6666584 and log 4.6416 = .6666677. log ^100 = pog 100 = I = .6666666. Now log 4.6416 = .6666677, and log 4. 6415 = .6666584. Hence log ^100 - log 4.6415 = .0000082, and log 4.6416 - log 4.6415 = .0000093. Hence ^100 = 4.6415 + f| of .0001 = 4.64159. COMPOUND INTEREST AND ANNUITIES. 368. The approximate calculation of compound inter- est for a long period, and also of the value of an annuity, can be readily found by means of logarithms. All problems of this kind depend upon the three fol- lowing. [The student is supposed to be acquainted with the arithmetical treatment of interest and present worth.] I. To find the amount of a given sum at compound interest, in a given number of years and at a given rate per cent. Let P denote the principal, n the number of years, f r the interest of $ 1 for 1 year, and A the required amount. Then the interest of P for 1 year will be Pr, and there- fore the amount of principal and interest at the end of 526 LOGARITHMIC COMPUTATION. the first year will be P(l-|-r). This last sum is the capital on which interest is to be paid for the second year; and therefore the amount at the end of the second year will be \_P{1 + r)']{l -\- r)= P{1 -\-ry. Similarly, the amount at the end of 3 years will be P(l + r)'\ and at the end of n years will be P(l + r)". Thus A = P{l-{-r)\ Oor. If the interest be paid and capitalized half-yearly, it can be easily seen that the amount at the end of ^i years will be 7^( 1 -f - J - Ex. Find the amount of $100 in 25 years at 5 per cent per annum. Since the interest on .$100 is f 5, the interest on $ 1 is $jV- Hence the amount of $1 at the end of the first year will be $(1 + .^), and at the end of 25 years will be $(1 + ^V)^^- Hence $A, the amount required, will be ^ 100(1 + j^)^- Hence log A = log 100 + 25 log f J. From the tables, we find that log21 = 1.3222193, and log20= 1.3010300. Hence log ^ = 2 + 25(1.3222193 - 1.3010300) = 2.5297325. Now, from the tables, we find log 338.63 = 2.5297254 and log 338. 64 = 2.5297383. Hence log ^ - log 338. 63 = . 000007 1 , and log 338.04 - log 338.63 = .0000129. Hence A = 338.63 + ^^g of .01 = 338.63 + .005 = 338,635, Hence the amount required is $338,635. LOGARITHMIC COMPUTATION. 527 II. To find the present value of a sum of money which is to be jKiicl at the end of a given time. Let A be the sum payable at the end of n years, and let P be its present worth, the interest on $ 1 being f ?• per annum. Tlien the amount of P in n years must be just equal to A. Hence, from L, P= A{1 -\- r)~\ Ex. What is the present worth of $ 1000 payable at the end of 100 years, interest being at the rate of 5 per cent per annum ? F = 1000(1 + 2L)-ioo ^ iOOO(f^)-ioo. Henc« log P = log 1000 - 100 (log 21 - log 20) = 3 - 100(1.3222193 - 1.3010300) = 3 -2. 11893 = .88107. Now, from the tables, log 7.6045 = .8810707. Hence the required present worth is $ 7 .6045. III. To find the jyresent value of an annuity of$A pay- able at the end of each of n successive years. If the interest on $ 1 be ^ ?• ; then from II. the present value of the first payment is ^4(1 + r)~^, second . . . A{l-\-r)-% 71*^ ... . A{l + r)-\ Hence the present value of the whole is ^l(l + r)-^ + (l + r)--+...+(l + r)-"S Ex. Find the present value of an annuity of $ 100 to be paid for 30 years, reckoning interest at 4 per cent. 628 LOGARITHMIC COMPUTATION. The interest on $ 1 is $ j;^^, therefore 1 -|- r = 1.04. Hence P = ^{1 - (L04)-3)} Tiro = 2500 {!-( 1.04) -30}. Now log(1.04)-3') = - 30 log(1.04) = - 30 x .0170333 = - .510999 = 1.489001. From the tables, log 3.0832 = .4890017 ; hence 1.489001 is log .30832. Hence P = 2500 (1 - .30832) = 2500 x .69168 = # 1729.20. EXAMPLES XCIV, 1. Write down the numbers whose logarithms are 5.3010300 and 5.3010300 respectively, knowing that log 2 = .3010300. 2. Write down the numbers whose logarithms are 3.2990931 and i.2990931 respectively, knowing that log 1.9911 = .2990931. 3. Having given log 46854=4.6707467 and log 46855=4.6707559, find log. 0468546. 4. Having given log 58961 =4.7705648 and log 58.962 = 1.7705722, find log .00589614. 6. Find ^29, having given log 29 = 1.4623980, log 19611 = 4.2924776, and log 19612 = 4.2924997. 6. Find ^100, having given log 19307 = .2857148, and log 19308 = .2857373. 7. Having given log 2 = .3010300, log 3 = .4771213, log 17187 = 4.2352001 and log 17188 = 4.2352253, find ^15 to 5 places of decimals. 8. Having given log 3.4277 = .5350028, log 32483 = 4.5116561, and log 32484 = 4.6U6695, find \^034277 to 6 places of decimals. LOGARITHMIC COIMPUTATION. 629 9. Find the amount of $1 in 100 years at 5 per cent com- pound interest, having given log 1.05=. 0211893, log 1.3150 = .1189258, and log 1.3151 = .1189588. 10. Show that a sum of money w^ill be more than doubled in 18 years at 4 per cent compound interest, having given log 1.04 = .0170333 and log 2 = .3010300. 11. Find the amount of §500 in 10 years at 4 per cent com- pound interest, the interest being payable half-yearly. Given log 1.02 = .0086002, log 1.4859 = .17 19896, and log 1.4860=. 1720188. 12. What is the present value of $ 1000 which is to be paid at the end of 1 5 years, reckoning compound interest at 3 per cent ? Given log 1.03 = .0128372, log 64186 = 4.8074403 and log 64187 = 4.8074471. 2l 530 CONTINUED FRACTIONS. CHAPTER XXXVII. Continued Fractions. 369. An expression of the form a H e + etc. is called a continued fraction. This is usually written in the more compact form , b d f In the general case the numerators b, d, /, • • • and the denominators c, e, ^, ••• may be severally either positive or negative ; but the continued fractions of greatest importance, and those to which the present discussion will be confined, have the form 6+ c+ d+ 6+ ' in which b, c, d, e, ••• are positive integers and a is either an integer (positive or negative), or zero. The numbers a, b, c, d, ••• are called the first, second, third, etc., partial quotients respectively. A continued fraction is said to be terminating or non- terminating according as the number of partial quotients is finite or infinite. CONTINUED FRACTIONS. 531 The following examples will show how continued fractions may be produced. Ex. 1. Convert 74/26 into a continued fraction. Reduced to its lowest terms this fraction is 37/13. Tlie process is then as follows : 2Z = 2 + ^ = 2-|—L_=2-f ' 13- '13/11 i,_l_ l+^_ 11/2 ^5 + i Ex. 2. Convert (^—afl+x^+l)/(x'^-{-l) into a continued fraction. By the ordinary process of division we have x-^+l ' X^-\-l x + - X Ex. 3. Convert -y/l into a continued fraction. Since 2 is the largest integer less than ^7, we write V7 = 2+(V7-2) = 2 + ^^^-L__, and we repeat this process upon the successively occurring surd forms as many times as may be necessary. Thus, segregating in each case the largest integer less than the fractional surd form, we obtain in succession 1 V7 + 2 = 4+ a/7 -2 = 4 + V7-2 "^ ^ l/(V7-2) The fractional surds now recur in the order in which they have appeared in this cycle, and the continued fraction is therefore non- terminating. Writing in the successive partial quotients we have /7 =: 2 + J- -i- J- J ^ ad inf. ^ ^14-1+1 + 4+1 + 532 CONTINtJEt) FRACTIONS. By reason of the recurrence of partial quotients in continued fractions of this class, they are called recurriny or periodic continued fractions. 370. Corresponding to every real number, commensu- rable or incommensurable, there exists a continued fraction, either terminating or non-terminating, to which the number is equal. For, let X be the given number. Then, if a be the greatest integer that does not exceed x, we may write ^ = ^1+^,=* (i.) where X^ is a positive number not less than 1, and where tti is a positive or negative integer, or possibly zero, this last case occurring whenever X<1. A repetition of this process with Xj gives X,=:a, + ^, (ii.) where a^, is the greatest integer that does not exceed X^ and X2 ^ 1 ; and n — 1 such repetitions result in the gen- eral form ^n-i = an + ^, (iii-) An * For example 1 14-V6 = 3 + (V5 — 2)=3 + (V6 + 2)/3' where 2 < 1 + V^ < 4, and l-v5 = -2+(3-V5) = -2 + ^3^^^^^^ . where -2 X„ and X„ ^ 1, for all integral values of n. Substituting from (ii.) in (i.) we noAv obtain and replacing X2 in this result by the expression for X^ derived from (iii.) by making n = 3, we have XT ,1 11 «2+ Ct3+ ^3 and in general, by further repetitions of this process, V ,111 1 0^2 + ^3 + «4 + X^ This is the required continued fraction. In this result X„ may be an integer for a finite value of n and the continued fraction therefore terminating, or X^ may never be an integer, however large n may be, in which case the continued fraction will be non-termi- nating. 371. This process of finding partial quotients will be recognized as identical with that of finding the greatest common divisor in arithmetic. Now when this process is applied to two integers that have no common divisor greater than 1, the last divisor is 1, and the last remain- der is 0. Hence : Every commensurable number may be converted into a terminating continued fraction. On the other hand, since every terminating continued fraction may obviously be converted into an ordinary 534 CONTtNtJED FRACTIONS. fraction with commensurable numerator and denomi- nator, and no such fraction can be equal to a surd, it follows that : A surd cannot be equal to a terminating continued fraction. 372. The series of terminating continued fractions «i> <^i H — > «i H y etc., formed, as here indicated, out of one, two, three, etc., of the partial quotients of ,111 ^2 + <*3 + 0^4 + are called convergents ; a^ is the first convergent, a^ + l/ag is the second convergent, and so on. Written as ordinary fractions, the first three convergents are easily found to be ai«2 4- 1 a^a^a.^ + a^ + a^ tt|, , ^ • ag a^^ 4- 1 The first convergent, but none of the others, may be zero. For the determination of convergents of higher order special methods, to be explained presently, are employed. 373. The convergents of a continued fraction of the form «! H (tti, ttg, ttg, ••• all positive) are a2 + ^3 + alternately less and greater than the fraction itself. For, the first convergent is too small because the frac- tional part ••• is omitted; the second convergent . «2H- aiH — is too large because in the fractional part a por- CONTINUED FRACTI0NI3. 635 tionof tlie denominator, namely, , is omitted; and the third convergent is too small because the part omitted belongs again to the numerator. And in general, in con- structing a convergent, the part of the original continued fraction omitted belongs to the numerator or denominator according as the convergent is of odd or even order. Hence the theorem. 374. To prove the law of formation of the successive convergents. Let the continued fraction be ,111 <^2 + Ots + ^4 + The first three convergents are «! ttittg 4- 1 {aia.2 -\-l)as-\- ai 1 a2 a.^^ -j- 1 the third of which may be formed from the first and second by the following rule: To obtain the numerator {or denominator), add the numerator {or denominator) of the first convergent to the product of the numerator {or denominator) of the second by the third partial quotient. Denoting the successive convergents by Pi/qi, p-ilq^i 2^3/^3) •" Pn/^Hj assume that this law applies to the forma- tion of the 71^^ convergent. Then the hypothesis is Pn = «nPn-l + Pn-2 , Qn = ««^«-l + 5'«-2- Now we may pass from the ?i*'' to the (n + 1)'^ con- vergent by changing a„ into a^ -\- l/a„+i ; and hence, if the 71* convergent be ««Pn-l+Pn-2 636 COJSTINUED FRACTIONS. the (n + 1)"* convergent will be and by the hypothesis this is equal to Hence, if the law be valid for the formation of the w*^ convergent, it is also valid for the formation of the (11 + 1)*'^ convergent. But the third convergent is formed in accordance with this law; hence the law is valid for the formation of every convergent. [Art. 145.] 375. Properties of Oonvergents. In proving the proper- ties of convergents we consider the continued fraction to be ,111 and denote the inP^ convergent by p„/g„. I. If Pn/qn &e the n'* convergent to a continued fraction^ then Pnqn-l-Pn-iqn = {-^Y' For, by the law of formation of convergents just proved, Pnqn-l -Pn-iqn = (clnPn-l +i>n-2)gn-l -Pn-l(an5'n-l4- ^n-g) = (- 1) (Pn-iqn-2 -Pn-^n^l) = {-lf{Pn-«-ig« = (-i)". II. Each convergent is in its lowest terms. For, any common divisor of p^ and g„ will divide PnQn-i — Pn-iQnj that is, unltj ; and hence 1 is the only- such divisor. III. The difference between two successive convergents is a fraction whose numerator is unity. For ^n ^ Pn-l ^ PrSln-l ^ Pn-lQn ^ 1 IV. Any convergent is nearer in value to the entire con- tinued fraction than the immediately preceding convergent, and therefore nearer than any preceding convergent. For, the continued fraction itself is obtained from the {n-\-iy^ convergent by putting a„+i -\ in the place of a„+i ; and hence, if F denote the continued fraction (O'n+l + • • • Vn + Pn-l («»+! + - —"\n + qn-l \ ^n+2-r J Pn+1 + ^Pn where X stands for , a positive quantity less than unity. Hence JT Pn+l ^ Pnf 1 + ^Pn Pn+1 Qn+l Qn+l + ^^Qn Qn+l ^ HPnQn+l -Pn+ign) ^ ( - 1)"X 53S CONTtNUED t'EACTlONS. and jT Pn^ Pn+l + ^Pn Pn Qn Qn+1 + Ag Qn ^ (Pn+lQu -PnQn+l) ^ (- l)"""' But X is less than 1 and is positive, and ^^^.i is greater than g„ ; hence Qn+l Qn V. The convergents of odd order progressively increase^ hut are always less than the continued fraction; and the convergents of even order progressively decrease, hut are always greater than the continued fraction. This proposition is a direct consequence of IV. and of Art. 373. Note. — In enumerating the orders of the convergents of the continued fraction ai ^ , it is customary to regard ai as the ai + convergent of first order, even when it is zero. VI. The difference hetween a continued fraction F and its n'* convergent pjq^ is less than l/q\ and greater than For, by IV., F^^ = Qn qn{qn+l + Ag„) ' where X is positive and less than unity. Hence qnqn+i qn qn{qn+i-\-qny and since q^+i > q„, . ^ ^ IP Pf^ q\ qn 2q' n+l CONTINUED FRACTIONS. 539 Another form of this test of error is sometimes useful, Since g„+l = «n+ig.+ gn-l < «n+ign, that is, ^nQn^l < Cln+iq\, . . i* '^ — < ^ • VII. Thus, any convergent which immediately precedes a large partial quotient is a near approximation to the value of the continued fraction. Ex. 1. Assign a limit of the error involved in assuming the fourth convergent as an approximation to the value of J J 1_ J 1_ J_ J__i 1 2+1 + 3+15 + 2+1 + 3+15+*"' The first four convergents are 1 3 Pz^S±l = i l' 2' (/3 2 + 1 3' P± = 3- 4 + 3 ^ 15 34 3.3 + 2 11' and by VI. F-^<—^ — = -!-. 11 16.112 1815 [Art. 374.] Thus, the error in taking |f as the value of the continued fraction is less than 1/1815, or .00055 +. Ex. 2. Find a series of fractions which converge towards the true value of y'2. Converting y/2 into a continued fraction, we have V2 = l + V2-l = l+ ^,\. =1+ ^ V2 + 1 2 + _-L_ V2 + 1 1 + JL^^.... 2+2+2+ 540 CONTINUED FRACTIONS. and the first six convergents of this are 1 3 2-3 + 1 ^7 2-7 + 3 ^17 l' 2' 2-2 + 1 5' 2-5 + 2 12' 2-17 + 7 ^41 2-41 + 17 ^99 2-12 + 5 29' 2 - 29 -t- 12 ~ 70' the last of which differs from ^2 by a quantity that is less than 1/(2 X 702), or 1/9800. 376. A non-terminating continued fraction of the form a, H Jias a true limit, namely, (p^/o^). In V. of the preceding article it was shown that 9.2n ^2n+l for all values of n {PiJ^lin and P2n+i/^2n+i representing the convergents of even and odd orders respectively), and in VI. it was also shown that and ir_-^aiLLi<_J_. Q2n+1 Q 2n+l Adding the corresponding members of the last two ineqaalities together, we have i>2n P2n+l ^ ^ ■ ^ *^ 2 ' 2 * 92 n Q2n+1 9 2n Q. 2n+\ Now, q2n and ggn+i are the sums of products of integers whose number increases without limit with the increase of n, and we may therefore make q^^ and g'2n+i as large as we please; that is, we may make the difference F2n/2 n W^2 n+l) , or, what is the same thing, EXAMPLES XCV. Convert the following numbers into continued fractions : 1. f. 3. Hi- 5. -I- 7. V17. 9. Va2+i. 2. f. 4. .31. 6. -If. 8. 2V3. 10. ^^ - n^ 4- n -f 1 , Calculate the successive convergents of : 11. l + J-^^i. 2+3+2+2 12. i + _LJ__L^l. 2+ 3+ 4+ 5+ 6 ■ 13. i + ^^A_J_JL 3+5+7+9+11 14. l+^^^-^l. 3+1+15+1 + 3 15. K 'Prl(lr denote the r^^ convergent of •••, a+ &+ a+ 5 + show that^2n+2 ^Pin + 6g2n and g2n+2 = apin + (a& + l)g2n- 16. If Prlflr denote the r* convergent to any continued fraction, prove that j>n+l - Pn-l _Pn. 642 CONTINUED FRACTIONS. 377. Eeciirring Continued Fractions. In Art. 369 a con- tinued fraction whose partial quotients recur in a definite order presented itself as the equivalent of a quadratic surd. The following examples show how we may pass from the recurring continued fraction to the quadratic surd. Ex. 1. Convert into a quadratic surd. Let the fraction be denoted by x. Then a + a; whence x^ + a.x — 1 = 0. The roots of this quadratic equation are 2 2 K a be positive, the continued fraction is positive and the posi- tive value of the radical must be chosen. Ex.2. Convert ~ — ~ — into a quadratic surd. 2+1 + 3+2+1 + 3 + Let X denote the value of the continued fraction. Then ^ = _1_J 1_ 2+1+ 3+x' and the three convergents of this are 3+x+l 4+x h h and 9 + 3X + 2 n + 3x 4 -4- X Hence x = ,^ ^ *' i 11 + 3x or 3x2+10x-4 = 0, and the required surd is the positive root of this quadratic equation ; that is, 5 , V37 CONTINUED FRACTIONS. 643 Ex. 3. Convert 2+^-^-^-^-^-^ -^-^- into a 1 + 5+2+1 + 3+2+ 1 + 3 + quadratic surd. Let y denote the entire continued fraction, and x the part which is recuiTing ; then ^ 111 ^-2+1 + 3 + "* and y —2 ■\- 1 + 5 + x The value of x was found to be — f + ^^^ in Ex. 2, and the o value of «/, obtained from the second equation, is ^ 6 + x or, when x is replaced by its numerical value and the denominator is rationalized, 119+ V 37 ^ 44 378. Every recurring continued fraction is equal to one of the roots of a quadratic equation whose coefficients are rational. Let y denote th.e entire continued fraction, and x the recurring part ; then y and x have the respective forms ,11 111 ,. , y=za-\ > (i.) ,11 1 11 r'\ x = ai-] J (ii.) in which all the letters are integers, and all, except possibly a, are positive. lip/q and x>'/q' he the convergents to (i.) correspond- ing to the partial quotients h and k respectively, then p'x-^p ^ q'x-\-q ^ ^ (iv.) 544 CONTINUED FRACTIONS, whence, by solving for x, q'y -p' and if r/s and r'/s' be the convergents to (ii.) corre- sponding to the partial quotients a„..i and a„ respectively, then x=r:^. (V.) Replacing x in (v.) by its value in terms of y in (iv.), we obtain p-qy r'(p —qy) +r{q'y -p') q'y -p' s'{p - qy) + s{q'y -p') a quadratic equation in y. Cleared of fractions, equation (v.) is s'xF + (s — r')x — r = ; and, since (s — r'y — 4(— rs') is positive and — r is neg- ative, its roots are real, one being positive, the other negative. Hence the values of y, as shown by equation (iii.), are both real. Ex. 1. Convert 2 + ^^-1- -L- ^ 1.11111 and — 1 + ;; — :: — r— z — 7i 3+ 1+ 2+ 1+ 2+ into quadratic surds. CONTINUED FRACTIONS. . 545 from which, by writing down the successive convergents, we derive x+ 1 ' 3 + 4x' whence, by solving for oj, x2-2x-2 = 0, and, by solving for y, y2 -2y-2 = 0. Thus X and y are the roots of the quadratic equation «2-2«-2 = 0, and since x must be positive and y negative, . •. x = l+y/Z, 2/ = 1 - V3. Ex. 2. Express the positive root of x^ — 2x — a^ = as a con- tinued fraction, a being a positive integer. If a, — /3 denote the roots of this equation, a its positive, — /3 its negative root, then x'^-2x- a^=ix - a)(x + /3), and this identity makes it at once clear that x"^ — 2,x — a^ will be negative for all positive values of x less than a. Hence the largest integer less than a is the largest integral value of x that will make ic^ — 2 X — rt^ negative. This value is easily seen to be 1 + a, and we therefore write a = l + a + a — 1— a. But a2 — 2 a — a2 = 0, and therefore a2 - 2 a + 1 - a2 = (a - 1 - a)(a - 1 -f- a)= 1, whence a = 1 + a + a — 1 + a Then a-l + a = 2a + ^ , and therefore a = 1 + a+ =1 + a + 2a+ 2a-t- 546 . CONTINUED FRACTIONS. In the larger text-books of algebra it is shown that any- quadratic surd can be developed into a recurring con- tinued fraction. [See Treatise on Algebra, Art. 367.] EXAMPLES XCVI. Convert the following surds into continued fractions : 1. 3+V5. 4. 1-V7. 7. ^v^^-±i. V2 - 1 2. 3-V5. 6. K\/3-l)- 8- 2+VH-a2. 3. 1+V7. 6. 2+-i-. 9. 2-y/\T7\ Convert the following continued fractions into surds : * 1 1 1 1 10. — — '". 2+ 3+ 2+ 3 + jj *J 1 1 1 1 1_^^ 2+2+1+2+2+ 1+*"* 12. 3 + *-l- -i- -i-..-. 5+ 6+ 6 + 13. i + J_*_l_^J_.... ^2+ 1+ 1+ 1 + 14. i+A_*^^_l_JL.... 2+ 2+4+2+ 4 + 16. %-l+ 1111 a+l+ a-l+ a + l+ a-l + Express the positive root of each of the following equations as a continued fraction : 16. x^-2x-\=0. 18. a;2_2a;-9 = 0. 17. a;'* - 2x - 2 = 0. 19. a;^ _ 4^; + 3 _ «« = 0. * The asterisk indicates the beginning of the recurring part. CONTINUED FRACTIONS. 547 20. Prove that f J- ^ ^ JL...Wa + A_ ^ _I-...\ = l. is a root of the quadratic 21. Show that a - 1 1 J. a — a — a - equation x^ — ax+ I = 0. 22. Show that the negative root of the equation x^—2x — a^=0, expressed as a continued fraction, (a = a positive integer) is 1 1 1 1 -a + -2a+ -2a+ — 2a + 23. Express both roots of the equation x^ — ax — a^b^ = in the form of continued fractions, a and b being positive integers. 548 DETERMINANTS. CHAPTER XXXVIII. Determinants. 379. If the matrices h c c a a b V c' c' a' a' b' 1 Art. 173.] be used to denote the cross-product differences 6c' — 6'c, ca' — c'a, ab' — a'b, the values of x and y derived from the simultaneous equations ax -\- by -\- c = 0, a'x -f b'y + c' = 0, may be presented in the form x = a b a' b' y = c a c' a' a b a' b' v-^+^ [Art. 172.] These cross-product differences whose matrices have two rows and two columns, are called determinants of the second order. Ex. 1. The condition that the equations ax + by = 0, a'x + b'y = 0, may be simultaneous in x and y is la 61=0. \a' b'\ DETERMINANTS. 540 For, if the equations be simultaneous, then 7>y^ a a' whence (a6' — a'b)y = 0, and unless ?/ = 0, we must have ab' - a'b = 0. Ex. 2. AVrite down the solution by determinants of the simul- taneous equations x + 2y-l = 0, Sx-y + 2 = 0. Solution : 2 -1 1 2 X = -1 2 "^ 3 -1 -1 1 1 2 y = 2 3 "^ 3 -1 4-1 3-2 ^5 7' 1-6 Ex. 3. Let the three equations aix+ biy + ciz = 0, a2X + b2y + CiZ = 0, asx + bsy + c^z = 0, be regarded as simultaneous in x, y. z; multiply them by &2C3 — 63C2, bsCi — 61C3, 61C2 — 62C1, respectively ; add the resulting equations together ; show that the new coefficients of y and z vanish ; and thus derive, as the condi- tion of simultaneity, ai 62 C2 - ai bi ci 4- as bi ci bs cs bs cs &2 C2 = 0. 380. When its three matrices are replaced by their equivalent cross-product differences, ^2^3 — ^3^2j ^1^3 — ^3^1? ^1*^2 — ^2^1y 550 DETERMINANTS. the expression tti h c. -a^ h Ci + ^3 h Ci h Cs h C3 &2 C2 becomes It is called a determinant of the third order, and its matrix form is 0/2 t>2 C2 ^3 ^3 We shall frequently denote this determinant by the abbreviated notation [ai^aCs]. Let it now be required to solve the simultaneous equa- tions ttiX 4- &i2/ + CiZ + di = 0, a^x + bsyA-CsZ + ds = 0. When the left members of these equations are multi- plied by &2C3 — ^3^2) —hcs + h^i, and ftiC^ — 62^1 respectively, and the resultant equations are added together, as in Ex. 3, Art. 379, the new coefficients of y and z become &i(&2C3 — &3C2) — ^2(^1^3 — &3C1) + h{^i<^2 — hci) = and 01(62^3 — &3C2) — C2(6iC3 — 63C1) + C3(5iC2 — b^Ci) = respectively, and there remains 0, «1 &i c + di h Ci a^ 62 C2 ^2 b. C2 as &3 C3 d2 &3 C3 DETERMINANTS. 551 whence, in the abbreviated notation, Repeating this process for the purpose of eliminating in succession z, x and x, y, we obtain z = laiboCsJ ^ [(hboCsJ In these fractional values of the unknown quantities we observe that the common denominator is the deter- minant [ai&^Cs], and that the numerators are three deter- minants which may be derived from [0162^3] by writing (in the matrix form) did^d^ in the place of aiCia^s) ^i^Aj and C1C2C3 respectively. 381. A determinant is fully described as a quantity when its terms with their proper signs are Avritten down; but, for the purpose of discovering its properties, it is regarded as evolved from a j)^'^^^^^p^^ term of the form aih.frfl^"'ln by keeping the letters in fixed order and per- muting the suffixes 1, 2, 3, •••7i in all possible ways, or by keeping the suffixes in fixed order and permuting the letters in all possible ways, and then giving to the terms thus obtained the positive or negative sign according as the arrangement of the suffixes (or letters) in such term is derived from that of the principal term by an even or an odd number of inversions of order ; whereby we define that an inversion of order occurs as often as a suffix (or letter) precedes another suffix (or letter) which in the principal term it follows. 652 DETERMINANTS. For the determinant of the third order this method of evolving the several terms is indicated in tlie following sclieme. The letters are supposed to stand in the order a b c, and only the sufl&xes are written down. Natural order. 1 inversion. 2 inversions. 3 inversions. Sign 123 + 123 132 _ 123 213 123 213 231 + 123 132 312 + 123 132 312 32 1 _ For example, there are three inversions in 3 2 1 because 3 pre- cedes both 2 and 1, and 2 precedes 1. 382. Def. A determinant of the 'n}^ order is the alge- braic sum of the n I products that may be evolved from a principal term aj)>j^^d^"'ln by keeping the letters in fixed order and permuting the suffixes 1, 2, 3, •••w in all possible ways ; and any terra is preceded by the positive or negative sign, according as it is derived from the principal term by an even or an odd number of inver- sions of the natural order of the suffixes 1, 2, 3, •••n. The matrix of this determinant of the ?i"' order is «i h. G,'"k «2 62 C^'-k ttg 63 Cs' 'h : : ': \ «n ^'n Cn'"ln Tliis determinant is frequently denoted by the abbrevi- ation [aA^s • • • ^„] ; sometimes also by 2±aA>C3**-C The n^ quantities ai, bi, Ci, ag, h^t-'-ln are called the elements of the determinant, the n! products aibiC^-'-ln) DETERMINANTS. 653 a,J)iC^-"ln, etc., are its terms, the diagonal line cbfi^fi^--'lni extending downwards from the upper left-liand corner, is the principal diagonal, and the product aib.c.^-'-l„, whose factors are the elements of the principal diagonal, is the principal term. In the present discussion we shall be chiefly concerned with determinants of the second and third orders. 383. It follows from our definition that every term of a determinant of the n*'^ order contains all of the n letters a, 6, c, --l, and all of the n suffixes 1, 2, 3, •••n, but none of the letters or suffixes twice repeated; hence, every term contains one and only one element from each row, and one and only one element from each column of the matrix. 384. If in a matrix of a determinant the letters a, h, c, ••• I and the numbers 1, 2, 3, ••• n be interchanged, statements concerning rows become statements concern- ing columns, and vice versa; hence, a determinant in the matrix form is not altered by changing its rows into col- umns and its columns into rows. Thus «2 ^2 ctj a^ and Unless the contrary is explicitly stated, determinants will henceforth be thought of as given in the matrix form. tti 5i Ci = tti as ag ^2 ^2 ^2 61 62 ^3 «3 h C3 Cl C2 Cg 654 DETERMINANTS. 385. When given in their matrix forms, determinants of the second and third orders are easily expanded into the algebraic sums of their terms by the rule which is expressed in the following formulae : a L &1 = = ai^a ■ -aA, 0^2 62 tti 61 Ci = «! &2 C2 -a^ bi Ci +a3 61 Ci ttj 62 C2 &3 C3 h C3 62 C2 a., 63 C3 ' Ex. 1. Expand 2 3 1 6 2 3 3 12 By the rule of expansion this determinant is equal to 2 3 1 2 -5 3 1 1 2 + 3 3 1 2 3 = 2(4 - 3)- 6(6 - 1)+ 3(9 - 2) = 2 -26 + 21 =-2. Ex. 2. Expand a h 9 h b f 9 f c = A. By the rule of expansion we have A = aibc -p)-h{hc-fg) + g{hf^ bg) = abc -\-2fgh- ap - bg"^ - ch\ EXAMPLES XCVII. 1. Count the number of inversions in 2143, 3421, 4321, and 63412. 2. If abcAe be regarded as the standard order, count the number of inversions in acebd., cdeba, and dbace. DETERMINANTS. 555 3. What are the signs of the terms hdk^ ahf, and hdc in a h c d e / g h k 4. Write, with their appropriate signs, all the terms of the deter- minant [aib^Czdi]. 6. In the determinant of the fifth order [^ayh^czdi^e^'] determine the signs of the terms azbiC^d^Ci^ a2?>3Cif?465» rt5&3Ci(?4^2» and a^b^czd^ei. Evaluate the following determinants : ^ 6. 1 2 3 - 4 5 6 7 8 9 • 7. 1 -2 3 -2 3 -4 3 -4 5 8. 1 3 5 3 5 7 5 7 9 . 9. 1 3 5 7 9 . 10. 1 -1 -1 11. a b b a a b • 12. a b c b c a c a b • 13. a b c b b b c b a ^ '/ 14. Show that as bs cs = - «i 6i Ci = 6l «! Ci a2 &2 (^2 052 &2 C2 62 ^2 C2 ai bi ci as bs Cs bs as Cs 15. Show that 1 i i 1 i i = 1, where /^ ^^*Z^^/ 5b6 DETERMINANTS. 386. Properties of Determinants. An interchange of any two adjacent roivs {or columns) of a determinant changes its sign, hut affects its value in no other way. For, in the determinant [aA^^s •••/„], the interchange of any two adjacent rows (or columns) merely inter- changes two adjacent suffixes (or letters) in every term, that is, produces one inversion of suffixes (or letters) in every term without disturbing the order of the letters (or suffixes), that is, changes the sign of every term, and affects the terms in no other way. 387. An interchange of any two rows {or columns) of a determinant changes its sign, hut affects its value in no other way. For, suppose k rows to lie between the two rows that are to be interchanged. These h rows, together with the two rows to be interchanged, form a series of A: -f 2 rows. Then we may make the interchange by passing the upper- most row of the series downwards over k rows, and then passing the lowest row upwards over k-\-l rows. The total number of interchanges of adjacent rows is there- fore 2 fc -f- 1, an odd number, and each such interchange changes the sign of the determinant. Hence the total effect is a change of the sign of the determinant. The proof for the effect of the interchange of any two columns is identical with the foregoing. 388. A determinant, in which any two rows {or columns) are identical, is equal to zero. For, if two rows (or columns) be identical, the deter- minant is unaltered, either in sign or magnitude, by the DETERMINANTS. 667 interchange of these two rows (or cokimns). But, by- Art. 387, the interchange of any two rows (or columns) of a determinant changes its sign. Hence the determi- nant in question is not altered in value by changing its sign, and its value must therefore be zero. Ex. 1. Determine the value of 1 a cfi 1 b &2 1 C C2 Two rows of this determinant become identical and its value zero when a=b. Hence it must have a — b as a factor [Art. 148]. For a like reason it contains b — c and c — a as factors. But the determinant is of the third degree in a, 6, c. Hence, denoting its value by A, A = i(6 - c)(ic - a)ia - b) where JC is a number. But the principal term of the determinant is bc^, and this is the only term in which bc^ occurs, and the coefficient of bc^ in L{b — c) (c — a)(a — b) is L ; .-. L = l, 'A = (6-c)(c-a)(a-6). Prove that and Ex. 2. Ex. 3. Prove that 1 p 7 1 ^ 7^ 1 i33 7« a /3 7 a2 ^ y2 aS /33 yZ /37(l-/S)(^-7)(7-l). = ai37(a-^)(/3-7)(7-a). 389. If all the elements of one row, or of one column, of a determinant be multiplied by the same quantity, the determinant itself will be multiplied by that quantity. 558 DETERMINANTS. For, every term of the determinant contains one element, and only one, from each row and from each column [Art. 383]; and hence, if all the elements of one row, or of one column, be multiplied by the same quantity, every term and, therefore, the sum of all the terms, will be multiplied by that quantity. 390. A determinant in which any two rows, or any two columns, differ only by a constant factor, is equal to zero. This theorem is a direct consequence of Arts. 388, 389. For example : = 0. ma na d = mn a a d mb nb e b b e mc nc f c c f 391. Def. When any number of columns and the same number of rows of a determinant are suppressed, the determinant formed by the remaining elements is called a minor determinant. The minor determinant obtained by* suppressing one column and one row is said to be of the first order, or to be a first minor ; similarly the result of suppressing two columns and two rows is called a second minor ; and so on. The result of suppressing the column and row through any element a; of a determinant A is called the minor corresponding to that element and is denoted by A^. Thus the first minors of A respectively. «l bx Cl a-i b2 Ga as ^ cs corresponding to the elements ai, 6i, ci, are Aa, 62 C2 ao d as ^2 bs Cs , A,^ = as Cs , A.^ = as ba determhstants. 569 It is evident that when, in the first row or first column, all the elements except tlie first vanish the determinant reduces to the product of the non- vanishing element and its corresponding minor. ai 61 ci = ai 62 C2 62 C2 63 Cs &3 C3 Thus 392. To expand a determinant of the ?i'* order in terms of n of its first minors. Since every term of the determinant [^aib^s '" ^n] con- tains one and only one of the elements a^ a^, «3, ••• a^ [Art. 383], we may write tti^i 4- 02^2 + (^zA -\ V ttnAj («) ai 61 Cl a.2 h^ c? as bs Ci " -4« contain 62C1). in which none of the coefficients Au A2, any of the elements aj, ag, ••• a„. For example, ai(&2C3 - &3C2) + a2(&3Ci - biCs) + asC&iCa The quantities Ai, A^^ -" A^ are called the co-factors of «!, a^, ••• a„ respectively. We require to determine these co-factors in terms of the minors of a^, ^2, ••• a„. For this purpose let the several rows a^ b^' "h^ (^s^s"' hi • • • a„ 6„ • • • ?„, in the matrix form of the determinant [ai^gCs-'-y, be alternately moved into the place of the 560 DETERMINANTS. first row. Then the determinant, without being changed in magnitude, assumes successively the forms ...(_l)-l[aAC2C^3-^n-l]; for, when any row, say the k^^, is moved into the place of the first row, it passes over k — 1 rows above it, and by such a transference the determinant does or does not change sign according as k is even or odd [Art. 386]. We may therefore write the equation (a) in the several other forms : - [a2&lC3^4 "•L'}= CtlA + «2^2 H h «n Aj (P) la^bic^d^ '"ln]= (hA + «2^2 H h «« A (y) - [a.46iC2d3 "'ln]= Otl^l + ^2^2 H h O-nAt, (S) Now, in (a) put 0.2= %= ••• = a„ = ; then \_aib>fis"-l„'] becomes a^lbiC^-'-ln], that is, ctjA^^ [Art. 391], and there- fore In (j8) put «!= a^= a^— ••• = a„= ; then [a2^iC3---ZJ becomes a2[biCs"'ln], that is, a2'^a2j ^"^^ therefore a2^2 = — «2^a,- From (y), (8), and (X), we obtain in like manner, a4 A = - ^4^^, DETERMINANTS. 561 Hence, if A stand for the original determinant If columns and rows be interchanged, we obtain A = oiA^ - b,\ + cAc, — ( - ^r-\\, or, if the first and second columns be interchanged, A = - bA,^ + b2\ - b,\ +•••+(- iyK\, ; and it is now evident that the expansion may be made in terms of the minors of any row or any column. 393. If the elements of one column (or row) of a deter- minant be multiplied in order by the co-factors of the corre- sponding elements of any other column (or row), the sum of these products will be zero. In the determinant an O2 • • • A/g ' «3 bs"'h"' let the elements k^ k^^ k^,-" be multiplied in order by the co-factors Ai, A^, A^, ••• of the elements ai, ag, %,•••, and the products added together. The result is kyAi -f- k^A.^ -f" ^^3-^3 4" 2n 562 DETERMINANTS. whose matrix form is k. h- -kr- k. b,. ■ k,... h h- ■h- [Art. 392.] But this is zero, since two of its columns are identical. [Art. 388]. The result is the same when these combinations are made with any two columns or rows ; for any column or row may be moved into the place of the first column without changing the magnitude of the determinant. 394. If each element of any column (or rozv) be a bino- mial the determinant can be expressed as the sum of two determinants of the same order. For, if A-i, A2, A^, • • • be the co-factors of the elements of the first column of tti + ai 61 Ci . aa -f a2 h C2 . a3 + «3 h C3 ' = A, then, by Art. 392 A = (tti + ai)Ai -f (0^2 + «2)^2 + (^3 + a3)^3 + ••• = aiAi + diA + «3^3 H 1- «l^l + «2^2 + «3^3 + ai h. Ci ^2 b. C2 as bs C3 + «1 61 Ci . «2 &2 <^2 ♦ «3 ^3 C3 . DETERMINANTS. 563 ^ 395. A determinant is not altered in value by adding to the respective elements of one column (or row) the same multiples of the corresponding elements of any other column (or row) . For, if in the expanded form Oi^i + ^2^2 -f- asA -f ••• = A, of the determinant A = tti 6i-. ■ k,... a^ 62- ■ k,- ttg 63.- ■h- . . . we replace ai, a2, a^,--- by ax + mk^, ag -f mk^, % + mfcg, m being any number, this expanded form becomes («! + mkx)Ai + (tta + mA;2)^2 + («3 + ^1^:3)^3 H = A -f- m(kxAx 4- ^2^2 + ^'3^3 -\ ). But by Art. 393 k^Ax 4- A^sA + Ms 4- ••• = ; ai + mki hi'-'ki Og 4- wi^2 &2 • • • ^2 " a3 4-wfc3 63. ../jo. = A. Ex. 1. Show that 1 1 3 12 2 1 5 -1 0. 664 DETERMINANTS. By adding the second column to the third the determinant becomes = 0. 1 1 4 = 4 1 1 1 12 4 1 2 1 1 5 4 1 5 1 Ex. 2. Show that a 6 1=0. h a 1 a + & 1 By adding the second column to the first the determinant becomes 0. a + & h 1 = (a + 6) 1 h 1 fe + a a 1 1 a 1 a + 6 1 1 1 Ex. 3. Show that 6 + c c a — c = ab(b — c). 26 b a-b b-\- a a a Subtract the second column from the first and add it to the third : the determinant becomes b c a = ab 1 c 1 b b a 1 b 1 b a 2a 1 a 2 Subtract the second row from the first ; then ab 1 c 1 = ab 1 b 1 1 a 2 c-b 1 b 1 1 a 2 ab{b - c). 396. To express the product of two determinants {of the second order) as a third determinant of the same order. As a preliminary step in the demonstration, we first establish the following lemma: DETERMINANTS. 665 If a^ hi m n tta ^2 P <1 ai ^1 ao ySa then, whatever he the values of m, n, p, q, A = Ai • A^. = A, a, hi a^ 6, ai /3, 0-2 P2 = A, For, A = a &2 p q ai (3i a2 i82 aih2^2 — f^'I>\^2 -a.. hi m n ai A a2 ft Ai . A2. Q.E.D. Now in A put m = q= —1 and n =p = 0, so that ai hi -1 fta &2 ^ — 1 ai ft a. ft Then multiply the first two rows by ai, ft respectively, and add the products to the third row; and again, multiply the first two rows by as, ft respectively, and add the products to the fourth row. The result of these operations is tti hi -10 tta h.2 — J «!«! + a^fSi hiUi + 62ft ^ ttitta + ^^2/^2 ^1«2 + ^2)^2 ^ -1 -1 by the lemma. But • aiUi + tta/Si &i«i + ^2/Si -1 -1 = 1; 666 DETERMINANTS. 0,2 62 «2 ^2 I ai«2 + cf'^i b^cu H- ^2iS2 Note. — This method of proof is general, and may be applied to determinants of any order. For determinants of the third order the result is as follows : If Ai = ai 61 ci and A2 = ai /3i 7i a2 62 C2 02 /82 72 as 63 C3 03 /Ss 73 then A1A2 = aitti + a2pi + a37i ^lai + 1 + &37i Ciai + C2P1 -\- C371 aia2 + a2/32 + «372 haz + ^2^2 + &372 Cia2 + C2i32 + C372 aias + + a373 bias + 62/33 + &373 Cias + €2^3 + C373 For the proof in detail of this more general case the student is referred to Smith's Treatise on Algebra, Art. 430. By virtue of the fact that a determinant is not altered in magni- tude by changing the order of its rows or columns, or by inter- changing columns and rows, the product of two determinants may be exhibited in various forms. 397. To solve the simultaneous equations. a^x 4- biy 4- CiZ + dit = ki , a2X -\- biy -{- c^js + d^ = k2, a.^x + 632/ + c^z + d.t = A^a, a^x -}- 64?/ + C42 + d^t = h^. . Multiply these equations in order by ^,, ^2> -^3> ^4> the co-factors of ai, a2, a^, a^ respectively in the determinant [«i^A^4]> ^^d ^^d. the resulting equations together. Then, by Art. 393, the new coefficients of y, z, and t vanish, and there remains {aiAi 4- a2^2 + f^s^s + 014^4)0; = kiAi -f hyA,^ + k^As+kiA^, that is, [aibiCgjd^'jx^lkib.jCsdi]. DETERMINANTS. 567 By using as multipliers the co-factors of b^, 63? ^37 ^4? we obtain in a similar manner and the symmetry of these relations shows that we shall have for the values of z and t laib2Csdi']z = laj).jcsd^'], laJ).2Csd^']t = lafi.,c.Jc,']. This method is obviously general, and readily appli- cable to simultaneous equations with any number of unknown quantities. 398. Eesultants. A resnltant, or an eliminant as it is also called, is a determinant the vanishing of which is the necessary and sufficient condition in order that two or more equations may be satisfied b}^ the same value or values of the unknown quantities involved. Thus the resultant of the equations ax -\-hy = 0, a^x -^b'i/ = is ab' — a'b [Ex. 1, Art. 379], and the resultant of the equations a^x + biy -f c^z = 0, a.;fc 4- 5.22/ + ('2^ = ^} a.^r. -f &32/ + C32; = 0, is «! bi Ci 0^2 ^2 2 ttg 63 Cg [Ex. 3, Art. 379.] 568 DETERMINANTS. Ex. 1. Find the resultant of the equations ax^+bx+ c = 0, a'x2 + b'x -he' = 0. Multiplying each equation by x we obtain the set of four equa- tions. ax^ + bx^ + ex + = 0, Ox^ + ax2 4- 6x + c = 0, a'x^ + b'x^ + c'x + = 0, 0x3 + a'x2 + b'x-\- c' = 0, which may be treated as simultaneous equations in the four quan- tities X^, X^, X, 1. Multiply these equations in turn by the co-factors of the fourth column of the determinant formed by the coefficients a, b, c, 0, etc., and add the resulting equations together. Then, by Art. 393, the new coefficients of x^, x'^, and x all vanish and there remains a b c =0. a b G a' b' c' a' b' c' This determinant is the resultant required and its vanishing is the condition necessary (it may also be proved sufficient) in order that the equations ax2 4- 6x + c = 0, a'x- + b'x + c' = may have a eommon root. [Compare Ex. 3, Art. 173.] Ex. 2. Find the condition necessary and sufficient in order that the equations ax^ + &x2 -f ex -f fZ = 0, px2 -f gx + r may have a common root. DETERMINANTS. 569 Multiplying the first equation by x and the second by x''^ and j we obtain the set of five equations ao^ -i- hx^ + cx2 + (Zx =0, ax3 + 6x2 + ex + (Z = 0, px* + qx^ + rx"^ = 0, px^ + ^x^ + rx =0, j)x'-^ + gx + r = 0. K these are to be satisfied by the same value of x we must have 1 a b c d = 0, a b c d p q r p q r p q r which is the required condition. The method of elimination employed in these examples is called Sylvester's dialytic method. 1. Transform ES a b I d e J :amples xoviii. f -i g h / c i into an equivalent determinant having : (1) de/as its first column, (2) cjCfe- as its first row, (3) adg as its third row. *— 2. Solve by determinants 3x + 2?/ + 2 = 0, 5x+12y-l=0. 3. Solve by determinants 5x + 2/ - 42-3=0, 2x+Sy-{-7 z + 4: = 0y 3x-2y + 50-7 = 0. 3> 7 570 DETERMINANTS. 4. Find a value of k such that the equations 2x-3y + l = 0, Zx - by -\-k^0, jc + y - 7 = 0, may be simultaneous in x and y. 6. Solve by determinants x-^y-Zz + 2< = 2, Qx-Zy -\- z^ 10^ = 0, 7ic-2y4-22 + 2«=-2 Evaluate the following determinants : 6. 3 2 8 7 5 6 9 4 1. h h f g / v^ ^"l \\'^ 4 3 7 2 5 2 1 6 -2 10 3 5 4 3 X a - b — a X c h — c X 10. 11. 12. 13. a — b X X b ~ c X 1 + a 1 1 X a 1 1 1 + & 1 1 1 + c 1 a a2 a^ 1 b b-' 68 1 C C2 c3 1 d d^ # Prove the following identities : 14. ^ b + c a c a — b b — c c + a b — a c — b c — a a -\- b b — c c b — c c — a a — b — a a — b = 8 abc. DETERMINANTS. 571 -6 ' 15. 16. 17. 6+c+2a h c a c + rt + 26 c a b a + b + 2 c = 2(a + 6 + n)3 aiXi + &ia;2 aiyi + &i?/2 1 X a;2 • 1 y y' 1 2 ^2 «ia:i + &12/1 aiX2 + biyz a^Xl + &22/1 «2a;2 + ^2^2 (W + X)2 {V + Xy (M> + a!)2 (M + y)2 (V + y)2 (wj + yY {U + ZY (V + 2)2 {W + 2)2 18. rtl 02 0^3 6l 62 ^8 Ci C2 C3 ^1 B, Ci = ila B2 02 Az Bz 'C3 ai a2 as &i 62 &3 Ci C2 C3 ai 61 ci 2 = 02 62 C2 as &s C3 ^1 ^1 Ci A2 Bi O2 Az Bz O3 and therefore, where A\, Bi, A2, etc., are the co-factors of ai, &i, 02, etc., in the determinant [ai62C3]. 19. It X + y -\- z = nx -[- by + cz = 0, prove that = 0. X y z c a b b c a 20. Find the condition necessary and suflBcient in order that ax2 4- &x + c = and x^ = 1 may have a common root. 21. Determine X and Y in terras of a, b, x, and y, such that a b . X y = X Y b a y X Y X 572 SCALES OF NOTATION. CHAPTER XXXIX. Scales of Notation. 399. In arithmetic, any number whatever is repre- sented by one or more of the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, called figures or digits, by means of the convention that every figure placed to the left of another represents ten times as much as if it were in the place of that other. The cipher 0, which stands for nothing, is necessary because one or more of the denominations units, tens, hundreds, etc., may be wanting. The above mode of representing numbers is called the common scale of notation, and 10 is called the radix or base. 400. Instead of 10, any other number might be used as the base of a system of numeration^ that is, of a sys- tejn by which numbers are named according to some definite plan, and of the corresponding scale of notation, that is, of a system by which numbers are represented by a few signs according to some definite plan. Thus 4235 is said to be written in the scale of seven, if 3 stands for 3 x 7, 2 for 2x7x7, and 4 for 4 x 7 x 7 X 7, so that every figure placed to the left of another repre- sents seven times as much as if it were in the place of that other. SCALES OF l^OTATTON. 573 In general, any number iVis expressed in the scale of ?• when it is written in the form -"d^d.^didQ, where each of the digits d(flid2d^-" is zero, or a positive integer less than T, and where d^ stands for d^ units, d^ stands for dy X r, d^ stands for ^2 X r x r, and so on. Thus iV= c?o + div + d,/^ H . 401. Any positive integer can be expressed in any scale of notation. Let N be the number, and let r be the radix of the required scale. Divide N by r, and let Q^ be the quotient and d^ the remainder- Then N=^ ^0 + r X Qi. Now divide Qy by ?*, and let Qo ^e the quotient and d^ the remainder. Then Q, = d^ + r x Qi] .: N= d^ + rd^ + i^Q^. By proceeding in this way we must sooner or later come to a quotient which is less than r ; let this be after n divisions by r. The process is now complete, and we have JSr= do 4- chr + dy^ + • • • + d,,r% so that the number would in the scale of r be written d^-'-dsd^^dido. Each of the digits d^, di, f?2? ••• i^ ^ positive integer less than r, and any one or more of them, except the last, d„, may be zero. 574 SCALES OF NOTATION. Ex. 1. Express 1062 in the scale of 7. Tlie quotients and remainders of the successive divisions by 7 are as under : 1062 151 remainder 5 21 4 = di 3 = d., Thus, 1062 vi^hen expressed in the scale of 7 is 3045. Ex. 2. Change 2645 from the scale of 8 to the scale of 10. Since 2645 = 2x83 + 6x 8-^+4x8 + 5 = {(2x8 + 6)8 + 4}8 + 5, the required result may be obtained as follows : Multiply 2 by 8, and add 6 ; multiply this result by 8, and add 4 ; then multiply again by 8, and add 5. This process is clearly applicable in all cases. Ex. 3. Express 2156 in the scale of 5. Ans. 32,111. Ex. 4. Express 34239 in the scale of 11. Ans. 237^7, where t is put for 10. Ex. 5. Express j^- as a fraction in the scale of 4. Ans. jYt* Ex. 6. Change 31426 from the scale of 8 to the scale of 4. Ajis. 3030112. We may first express 31426 in the scale of 10, as in Ex. 1, .and reduce the result to the scale of 4, as in Ex. 2. The result may, however, be obtained by one process, as under : 31426 6305 remainder 2 ' 1461 1 314 1 63 14 3 3 Thus, the number required is 3030112, SCALES OF NOTATION. 575 Explanation. We first divide 8 eights plus 1 by four, giving quotient (>, and remainder 1 ; we tlien divide 1 eight plus 4 by four, giving quotient 3, and remainder ; and so on. 402. It would be a good exercise to perform all the ordinary rules of arithmetic with numbers expressed in various scales. Ex. 1. Add 2345, 6127, and 1503. [Scale 8.] Ex. 2. Subtract 3154 from 4021. [Scale 6.] Ex. 3. Multiply 234 by 456. [Scale 7.] Ex. 4. Divide 22326 by 315. [Scale 8.] Answers. 12177, 423, 150663, 56. 403. Kadix Fractions. Kadix fractions in any scale correspond to decimal fractions in the ordinary scale ; thus .abc '•', in the scale of ?*, stands for -+*+-.+ •••• To shoiv that any given fraction may he expressed by a series of radix fractions in any proposed scale. Let F be the given fraction; and suppose that when expressed by radix fractions in the scale of r, we have F=.abC"', that is F=--j--^-^ -^+ ..., where each of a, 6, c, ••• is zero or a positive integer less than r. Multiply by r ; then 576 SCALES OF NOTATION. Hence a must be equal to the integral part, and h c _ _| 1_ ... must be equal to the fractional part oi F x r. r r Let Fy be the fractional part of Fr ; then r r- Multiply by r again, then as before h must be equal to the integral part of F^ X r. Thus a, 6, c, ••• can be found in succession. Ex. 1. Express y%\ by a series of radix fractions in the scale ''^^' t¥8x6 = 3 + J^; T-Vx6 = + i; i x 6 = 2. Hence -302 is the required result. Ex. 2. Change 431.45 from the scale of 10 to the scale of 4. The integral and fractional parts must be done by separate processes. 4 4 431 107 .... 3 .45 4 4 26 .... 3 1.80 4 6 .... 2 1 ... .2 4 3.2 Thus the requ ired result is 12233.130. 4 0.8 404. Theorem. Any number expressed in the scale of r is divisible by r — 1, if the sum of its digits is divisible by r — 1. Let ^be the number, S the sum of the digits, and let the digits be do, di, d^, etc. Then N= d^ + dxr + d^r" + • • • + d„r", and S = fZ„ + dx -\-d. H f- d^. Hence iV-iS=di(r-l)+(72(?'2-l) + ... + (Z„(r"-l). SCALES OF NOTATION. 577 Now each of the terms on the right is divisible by r-1. [Art. 147.] Hence N— S is divisible by r — 1, and therefore when S is divisible by r — 1, so also is ^. As a particular case of the above, any number expressed in the ordinary scale is divisible by 9 when the sum of its digits is divisible by 9. 405. Theorem. Any number expressed in the scale of r is divisible by r -{- 1 when the difference between the sum of the odd digits and the sum of the even digits is divisible by r-f-1. Let N= do + d,r + d^r"" + d^r^ + •.•, and D = dQ — di-\-d2 — ds-{ . Then N-D=d,{r + 1) + d.,{r'' - 1) + d^{r' + 1) + -. Now each of the terms on the right is divisible by r + 1. [Art. 147.] Hence, A^— Z> is always divisible by r + 1, and, there- fore, when D is divisible by r + 1, so also is N. As a particular case, any number expressed in the ordinary scale is divisible by 11 when the difference between the sum of the odd and the sum of the even digits is divisible by 11. EXAMPLES XCIX. 1. Express 2156, 7213, and 192457 in the scale of 6. 2. Change the following numbers from the scale of 7 to the ordinary scale : 2135, 4210, 30012, 123456. 3. Change 3152 and 23678 from the scale of to the scale of 12. 4. Express 23.42 and 123.45 in the scale of 5, 20 678 MISCELLANEOUS EXAMPLES VII. 6. Multiply 2.31 by 1.25, the numbers being expressed in the scale of C. 6. Show that the numbers represented in any scale by 121, 12321, and 1234321 are perfect squares. 7. Find the values of a and h in order that 215al463 may be divisible by 9 and by 11. 8. Find the values of a and h in order that 516a72456 may be divisible by 99. 9. Find the scale in w^hich 314 is represented by 626. 10. Find a number of two digits in the scale of 5 which is doubled by reversing its digits. 11. Find a number of two digits in the scale of 8 which is doubled by reversing its digits. 12. Find a number of two digits in the scale of 7 which is trebled by reversing the digits. MISCELLANEOUS EXAMPLES VII. 1. Solve the equation x^a = h + y/(ax^ + c). 2. Show that (x + y + 0)2«+i - x«+i - y^n+i _ ^zn+i ig divisi- ble by {y + z)(z + ic)(x + y), and find the quotient when n = 1, and when n = 2. 3. Show that (x + yy-x^ -y^ = S xy(x + y), and that (x + yy - x^ - y^ = ^ xy{x + y) (x-^ ■]- xy + y"^). 4. Simplify -—^ -+ --^_— + ^ 5. Simplify 6. Simplify - (a-b)ia-c) (h-c){b-a) ic-a){c-b) 6c , CO I ab ^ ia-b)ia-c) ib-c){b-a) ic-a)ic-b) 1.1.1 aia-bXa-c) 6(6-c)(6-tt) c{c-a){c-b) MISCELLANEOUS EXAMPLES Vil. 579 7. Simplify 1 + _ 1, + 1 a2(^a-b)ia-c) b'\b-c)ib-a) c\c-d){c-b) _ o- ^^t b + c , c+a , « + & "■ '""P"*^ («-6)(a-c) + ib-Oa-a) + (c-«)(c-t) - _ „. y. b + c c-\-a I a-\-h ' ^^^^^ a{a-b){_a-c) bib-c){b-a) c(^c-a){c-b) 10. Show that ^ ^— — + ^ + 11. Show that 12. Show that + (a-6)(a-c)(x + «) (b-G)ib-a)(x-^b) 1 1 (c-d)ic-b){x-{-c) {x-\-a){x-\-b){x + c) a , b {a-b){a-c){x + a) {b - c){b - a)(x^ b) c —X {c-a)ic-b)(^x + c) (x+a)(x+ 6)(ic + c) a-2 . 62 (rt-6)(a-c)(x+a) (6-c)(6-a)(x + i) c2 x2 (c-a)(c-?))(x + c) (x+a)(x + 6)(x + c) 13. Simplify (b'^-c^y + (c'--'r+(c^'-b')\ 14. Prove the following : 62 (i.) a + & = -^ + a — 6 6 - a (ii.) a+6 + c=- -^- - + - ^^- - + (iii.) a + b+c+d = + ia-bXa-c) (6-c)(5-a) (c-a)(c-6) (a-6)(a-c)(a-d) (6-c)(&-(i)(6-a) c4 . d^ (c-d)(c-a){c-b) {d-a)id-b)id-c) 580 MISCELLANEOUS EXAMPLES Vll. &8 16. Show that + ia-b)ia-c)(,a-d) (b - c){b - d)ib - a) c3 . d^ 16. Show that (c-d)(c-a)(c-6) id-a)id-b)id-c) bed , cda = 1. ia-b)(a-c){a-d) {b - c){b - d)ib - a) dab , abc _ -. (c-d)(c-a)(c-6) id-a)(d-b)id-c) 17. Show that- -^ (^x-ay-h- ^^- -{x-b^ (a — b){a — c) (& - c) (6 — a) + «^ (x-^y = x2. (c-a)(c-&)^ ^ 18. Prove that (6 + c)^ + (c + a)^ + (a + by - {c -{■ a) (a -{■ b) - (a + fo) (6 + c) - (6 + c) (c + a) = a2 + &2 + c2 - 6c - ca - a^. 19. Show that (6 + c)3+(c + a)3+(a + 6)3-a(6 + c)(c + a)(a + 6) = 2(a3+ 63 4.e3-3a6c). 20. Show that (6 + c - a)8 + (c + a - &)^ + (a + & - c)8 - 3(& + c - a) (c + a - 6) (a + 6 - c) = 4(a8 + 63 + c8 - 3 abc). 2\. Show that (x^ -\- 2 yzy -{■ (y^ -\- 2 zxy -\- (z^ + 2 xyy - 3(ic2 + 2 yz) (y^ + 2 2!ic) (z'^ -{-2xy) = (x^ + ?/ + 2* - 3 a;2/2r)2. 22. Show that, if x"^ + y^ -{• z^ = (yz -\- zx + xy), then x = y = z. 23.. Prove the following : (i.) if 2(a2 + 62) = (a + 6)2, then a = 6 ; (ii.) if 3(a2 + 62 + c2) = (a + 6 + c)2, then « = 6 = c ; (iii.) if 4(a2 + 62 + c2 + (?^) = (a + 6 + c + dy, then a = 6 = c = (Z ; and (iv.) if n(a2 + 62 + c2 + ...) = (a + 6 + c + --O^ then a = 6 = c = •••, n being the number of the letters. 24. Prove that, if the sum of two positive quantities be given, their product is greatest when they are equal to one another. MISCELLANEOUS EXAMPLES VII. 581 25. Prove that, if the product of two positive quantities "be given, their sum is least when they are equal to one another. 26. Show that the least value. of a; + - is 2, the least value of X 4 9 a; + - is 4, and the least value of x + - is 6, x being real and X X positive. /V.2 3 ic -I- 4 27. Show that ^^—- cannot be greater than 7 nor less a;2 + 3 a; + 4 than \, for real values of x. 28. If the sum of a given number of positive quantities is fixed, their continued product will be greatest when they are all equal. 29. If the continued product of a given number of positive quan- tities is fixed, their sum will be least when they are all equal. 30. Prove that, if VlJZ^ = zxj-^^ ^^^^ ^^^j ^^^^ fraction be y \z z -\- x equal to ^^ ~ ^ , and also equal to x + y -\- z. x + y 31. Prove that, if a, 5, c, d be all real and if (rt + 6)2 + (6 + c)2 + (c + ciy = 4(a& + ?>c + cd), then a = b = c = d. 32. If ^^^ - ^^ ::= ^" - ^'^ , then will a - 6, or c = ^, a — b — c + d a — b — d + c OT' a + b = c + d. 33. Show that, if x, y, z be determined by the equations : (a _ ayx + (a - ^Yy + (a - i^z = (a - 5)2, ib - a)2x + (7; - ^Yy + (6 - 7)-^ = (& - 5)2, (c - ayx + (c - ^Yy + (c - -iy^z = (c - 5)2 ; then will \d - a^x ^{d- ^Yy + ((? - 7)^2 ={d- 5)2, where d has any value whatever. 34. If the equation -^^- + — ^ = — h — ^ have a pair of X -\- a x-\-b x + c X -\- d equal roots, then either one of the quantities a or b is equal to one 582 MISCELLANEOUS EXAMPLES VII. of the quantities c or d, or else - + - = - + i. Prove also that the a h c d roots are then - a, - a, ; - 6, - 6, ; or 0, 0, - -?^ • a-\-h 36. If 2(x2 + x'-^ - xx')(y^ +V- - VU') = a^V + ^'Y^ then will x = x' and y = y'. 36. Show that ■ (y -\-z){z-\- x)(x + y)-\-xyz = xyz(x + y + z)(--\---\-^A \x y z) 37. If «, 6, c, X are all real quantities, and {cfi + 62)x2 _ 2 6(a 4- c)x + ft-^ + c'^ = 0, then a, &, c, are in geometrical progression and x is their common ratio. 1 — r'^" 38. Show that (1 + a;) (1 -f a;'^) (1 + x*)— to w factors = - — — • 1 — X 39. If s = a + 6 -f c, prove that {as + hc){hs + ca){cs + ah) = {b + c)2(c + rt)2(rt + 6)2. 40. lia-^h-\-c-\-d = 0; then will (a3 -1- 63 _^ c3 + d3)2 = 9(6cd + cda + rZaft -i- rt?>c)2 = 9(6c — ad){ca — bd)(ab — cd) = 9(6 + cy{c + «)-(« + ?>)■-. 41. Show that a(a + ^)(« + 2fZ)(a + 3(Z) + d* is a perfect square. 42. Prove (by the method of indeterminate coeflBcieuts) that the necessary and sufficient condition in order that ax2 ^2hxy-\- by'^ -^ 2 gx + 2fy + c may be resolved into two factors, rational and of the first degree iu X and y, is abc + 2fgh- aP - bg"- - cK^ = 0. 43. Find (by the method of indeterminate coeliicients) the cube root of each of the following expressions : (i.) x3 - 24 x'^y + 192 xy^ - 512 if. (ii.) x6 ^ 9x5 + 33 X* - 03x3 + GOx^ - 30x + 8. (iii.) 8 x« - 36 x5 + 102 x* - 17 1 x/^ + 204 x^ - 144 x + 64. MISCELLANEOUS EXAMPLES Vlt. 588 44. Find the limit of t^ — ^Y- 1 when x = 0. X {-'y 45. Find the limit of [x V^' when x = \. 46. Prove that ^J^l^'^ (1 + a; + 2 ! a:^ + 3 ! a^a + ... ^_ ^ , ^^n) ^ ^ for all finite values of x. 47. Expand ijz jg_+A^I + a;« + a ^ i,^^^^ infinite series. (l+x)-^(l-a^)3 48. Expand (1 + a:)'»/2 and (1 + a-)-"'/2 into infinite series. 49. Expand i (e^ + e-^) and J (e^ - e--^) into infinite series. 50. Resolve 1 + ai + (1 + «i)«2 + (1 + «i)(l + «2)«3 + ••• +(1 + «i)(l + «2) ••• (1 + a«_i)a„ into n binomial factors. 51. Find the sum of the first n+ \ terms of the series 1 4- -^ 4- (« i + ^)^ I (ai + a:)(a2 + a;)x ai ai«2 axa^az _^ (ai + a;)(a2 + a;) (as + x )x ^^^ aifl520!3a4 52. Show that the sum of the first n + \ terms of 1 + ^ + ^^(^+ 1) I ??i(m + l)(m + 2) 12! 3 ! ?7i(m+ l)(?7i + 2)(m + 3) 4! is (m + l)(w ^ 2)(m + 3) ... (m + 7^ - 1) 7i ! 53. Show that the sum of the first w + 1 terms of -J _ w , m{m — 1) _ ?»(m — l)(w — 2) 1 2! 3! m(m - l)(m - 2)(m - 3) 4 ! • (1 — fri) (2 — m)(3 — m) ••• (n — m) m! 584 MISCELLANEOUS EXAMPLES VII. 54. Find the n^'' term and the sum of the first n terms of the series 1 + 3 + 7 + 13 + 21 + 31 + 43 + •••. 55. Find the sum of the first u terms of the series 3 + 4x + 6a;2+ 10r«+ 18 x* + •••• 56. Convert a + V2 + a^ into a continued fraction. 57. Convert the positive root of ax^ — abx — b = into a con- tinued fraction, 58. Show that I a a^ = 1 a be 1 h 62 1 b ca 1 c c2 1 c ab 59. Show that -be bc-\- 62 he + c2 ca + a"' — ca ca + rt^ a6 + a^ a6 + 6^ — ab = 2 C2 62 (c + a)2 «2 (6 + c)2 62 a2 (a _|. 5)2 = 2(6c + ca + a6)3. 60. Find the value of x as determined by the equation X — a X — b X - c =0. X— b X — c X — a X — c X — a X — b TEXT-BOOKS FOR USE IN HIGH SCHOOLS SELECTED FROM THE CATALOGUE OF MACMILLAN & CO. SCIENCE — MA THE MA TICS, BOOKS FOR EXAMINATION. Any book in this List which is intended for class use, will be sent for examination, with a view to introduction, post-paid on receipt of the net price named, or will be charged upon approval subject to return. Whenever payment has been made, we shall be glad, if re- minded to do so, to return the price paid for sample copy, or enclose a free copy, upon the receipt of an order for any reason- able number of such books for introduction. A teacher applying for a specimen copy for examination should give the najne of his school and the subject taught by him. Books not intended for class use are sent at regular rates only. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. THE EFFECTS OF NARCOTICS AND STIMULANTS FULLY TREATED. PHYSIOLOGY FOR BEGINNERS. By MICHAEL FOSTER, M.D., F.R.S., Professor of Physiology in the University of Cambridge, AND LEWIS E. SHORE, M.A., M.D. With Pull Illustrations. New Edition. i6mo. Cloth, pp. 252- Price, 75 cents. ILl.V^ /^\ Fig. 44. — Front view of the heart, great vessels, and lungs. R.V^ right ventricle; L.V, left ventricle; R.A, right auricle; L.A, left auricle; Ao, aorta curving backwards to the left; V.S, superior vena cava; V.I, inferior vena cava; C, carotid arteries; R.J.V, L.f.V, right and left jugular veins; S.C, subclavian vessels, artery and vein on each side; R.L, L.L, right and left lungs; P. A, pulmonary artery dividing into two; P.V, pulmonary veins; y, trachea; B, bronchi. All the vessels except those of the lungs are cut. [SPECIMEN OF ILLUSTRATIONS.] THE MACMILLAN COMPANY, NEW YORK: 66 Fifth Avenue. CHICAGO: The Auditorium. SAN FRANCISCO: 327-331 Sansome Street. COMMENTS. " It seems to be an admirable book, and exceedingly well adapted to the students for whom it is intended. I shall have no hesitation in com- mending the book to teachers and to students in the schools where this grade of instruction is given." — JOSEPH W. Warren, M.D., Dryn Mavjr College. " It seems to me the best book I have seen for the scientific study of the subject, if such study be accompanied by laboratory work ; the plates are remarkable for clearness." — Wm, H. Smiley, Principal Denver High School. " It bears the impress of the master of physiological writing, and appears well adapted as a work for beginners in serious physiological study, not only on account of the admirable proportionate distribution of matter, but also by reason of the clearness and precision of its teaching. It will give me pleasure to recommend it." — JOHN A. ROCKWELL, M.D., Bosk)n Uni- versity School of Medicine. " ' Physiology for Beginners ' is a simple, clear, and compact treatise, designed for those who have no previous knowledge of the subject. The work is well done, and the book ought to find a place in many schools." — E ducal ion. " Nothing at once so scientific and so simple as ' Physiology for Begin- ners,' by Professor Michael Foster and Dr. Shore of the University of Cam- bridge, has appeared on the subject. It is unreservedly commended as a text-book for secondary schools. It is happily without any reference to the half-proved and unprovable nonsense about the effect of stimulants and narcotics on the human system, that American writers and publishers have permitted to disfigure text-books of Physiology, in response to a senti- mental movement for * temperance' teaching." — Educational Review. " ' Physiology for Beginners," by Professor M. Foster and Lewis E. Shore, M.D., is designed as an introductory manual for beginners who wish to make a serious study of the subject. It is more elementary and didactic in style than Huxley's ' Lessons,' and the authors hope it may seive as an introduction to that more extended work. The first two chapters con- tain an outline of chemical facts which is considered to be essential to a sound knowledge of physiology. The high standing of Professor Foster and Dr. Shore is sufficient warrant of the authoritative character of this thoroughly intelligible and well-written little volume." — The Beacon. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. LABORATORY MANUAL AND PRINCIPLES OF CHEMISTRY FOR BEGINNERS. By GEORGE M. RICHARDSON, Associate Professor of Chemistry in the Leland Stanford Jr. University. i2mo. Cloth, pp. 233. ^i.io. FROM THE PREFACE. " The number of quantitative experiments is greater than is commonly found in such laboratory manuals. The writer has found, however, that few of them are too difficult for the ordinary beginning student, and that careful workers can get very acceptable results in all cases. ... In Part II. the writer has attempted to give a clear and concise statement of some of the fundamental theories and principles of chemistry; here he has gone more into detail than is customary in books designed for elementary students. But as the elementary student very soon begins to use these theories and principles, it seems that he should very early have a fairly complete state- ment of them with the evidence on which they are based." COMMENTS. " I am glad to see that Professor Richardson in preparing this little manual has adopted a plan that is now in pretty general use. To require careful experimental and quantitative work from beginners in chemistry is good. Its results are invariably good. I believe that the professor has per- formed a service to teachers of elementary chemical science." — Edgar F. Smith, Professor of Chemistry, University of Pennsylvania. " I took up the book intending to glance hastily through it, but was so charmed with the spirit of the book and the excellent manner in which this is carried out that I made a careful examination of it. The instructions for beginners are delightfully clear, and sufficiently suggestive without telling too much. It contains a good deal of work for the time usually allotted to the beginning year in chemistry, but an earnest student can accomplish it. The main principles of the science are stated with great clearness. I am glad to see so many merely interesting facts about the elements and their compounds relegated to their proper place in the larger reference works. It is, however, the simplicity and clearness with which quantitative work is made prominent that chiefly appeals to me." — John M. Penick, Missouri Valley College. " I am well pleased with the author's arrangement of matter for elemen- tary work. The careful use of the book will certainly furnish an admirable basis for more advanced college work. The choice of, and directions for, the performance of the experiments are most excellent, and well adapted to stimulate the powers of observation and the careful interpretation of phe- nomena." — T. H. Norton, University of Cincinnati. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. INORGANIC CHEMISTRY FOR BEGINNERS. BY SIR HENRY ROSCOE, F.R.S., D.C.L., LL.D., G.P ASSISTED BY JOSEPH LUNT, B.Sc. (Vict.), F.C.S. IViih one hundred and eight Illustrations in the Tcxt- Globe 8vo. Cloth, pp. 245. 75 cents. COMMENTS. " A clear, convenient putting for class use of the most reliable investiga- tions in the Mine of inorganic chemistry. Among the admirable features of the work are the clear-cut directions for a multitude of significant experi- ments and the thousand questions, more or less, for the teachers' use and the child's profit." — The Journal of Education. " Only the elementary principles and certain non-metallic elements are touched upon, but the touch is that of a master hand. The chapters on chemical calculations, equivalent numbers, and physical properties of gases are especially worthy of mention, and the book as a whole is what we ex- pect when we read on the title-page the name of Sir Henry Roscoe." — Dal- housie Gazette. " This work, containing over one hundred illustrations of apparatus and experiments, is admirably adapted for beginners in this fascinating science. There is much omitted that is found in ordinary lessons on non-metallic elements, but the amount of detail given in this book more than compen- sates for the omissions and makes it very useful for the teacher." — Educa- tional Revietv. " The work seems admirably adapted to the uses of secondary schools, and might be used advantageously in colleges, by those students who do not care to take an extended course in science. It is well written, the state- ments being clearly and concisely made, and the principles involved being well explained by the experiments which compose the greater portion of the text. The book is well illustrated and forms a real addition to the great number of chemical text-books." — J. F. McGregory, Colgate University, in The School Review. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. THE OWENS COLLEGE JUNIOR COURSE OF PRACTICAL CHEMISTRY BY FRANCIS JONES. WITH PREFACE BY SIR HENRY ROSCOE. With Illustrations. i8mo. 70 cents. " It is with great pleasure that we announce the appearance of this useful little work, in which the author has cut out a new path of his own, by the exclusively practical character of the lessons and by the style he has adopted." — Chemical Trade Journal. ELEMENTARY LESSONS IN HEAT, LIGHT, AND SOUND. BY D. E. JONES. With Illustrations. 70 cents. " Well arranged, clearly written, and contains many excellent problems for testing the ability of the pupil to apply the principles which he is supposed to have learned, . . . and possesses that rarest of all virtues in a school text-book, scientific accuracy." — Educational Review. LESSONS IN HEAT AND LIGHT. Illustrated. i6mo. $1.00. An expansion of the first two parts of the book named above for more advanceo classes. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NE^AT YORK. MACMILLAN'S SCIENCE CLASS-BOOKS. F'cap 8vo. ELEMENTARY PHYSICAL GEOGRAPHY. cy S.G.F.A., Assistant Professor of Geology and Physical Geography at Cornell University. $1.40. LABORATORY MANUAL AND PRINCIPLES OF CHEMISTRY FOR BEGINNERS. By George M. Richardson, Associate Professor of Chemistry in the Leland Stanford Jr. University. $1.10. PHYSIOLOGY FOR BEGINNERS. By Michael Foster, M.D, F.R.S., and L. E. Shore, M.D. Fully UK. irated. 75 cents. LESSONS IN APPLIED MECHANICS. By J. H. Cotterill and J. H. Slade. $1.25. " Undoubtedly the best rudimentary treatise on the subject that has yet appeared." — Mechanical World. "One of the best little books on the subject that has come under our notice for some time." — Nature. LESSONS IN ELEMENTARY PHYSICS. By Professor Balfour Stew- art, F.R.S. With Illustrations and Colored Diagram. $1,10. " It is the beau ideal of a scientific text-book, clear, accurate, and thor- ough, and withal written in a style so simple and interesting as to impart a real charm to the study." — Educational Times. Questions on the Above for Schools. By T. H. Core. 40 cents. EXAMPLES IN PHYSICS. By Professor D. E. Jones, B.Sc. 90 cents. " About sixty pages of new matter have been added. . . . The chief merit of the book is that it supplies a complete and exhaustive set of prob- lems, and in the solution of these pupils may be trained to apply general principles." — Journal of Education. ELEMENTARY LESSONS IN ELECTRICITY AND MAGNETISM. By Professor Svlvanus P. Thompson. $1.40. " An excellent little text-book. . . . The book contains a large amount of information, clearly stated, assisted by useful figurei^^, and furnished with exercises on the twelve chapters into which it is divided. It is a book which may be commended to the beginner as an excellent introduction to the sub- ject." — Westminster Re7>ieiu. LESSONS ON HEAT, LIGHT, AND SOUND. An Elementary Text- book. By D. E. Jones, B.Sc. With Illustrations. 70 cents. " Well arranged, clearly written, and contains many excellent problems for testing the ability of the pupil to apply the principles which he is sup- posed to have learned, . . . and po ssesses that rarest of all virtues in a school text-book, scientific accur-xcy." — Educatiotial Reineiv. ELEMENTARY LESSONS ON ASTRONOMY. By J. N. Lockyer, F.R.S. With Illustrations. $1.25. "The book is full, clear, sound, and worthy of attention, not only as a popular exposition, but as a scientific index." — Athene^um. Questions on the Above for Schools. By J. Forbes- R6bertson. 40 cents. LESSONS IN ELEMENTARY CHEMISTRY. By Sir H. E. Roscoe, F.R.S. $1.25. " Much new matter has bsen added to keep the book up to date. We have always considered it the best work for those who wish to get a clear and connected knowledge of the outlines of Inorganic and Organic Chemistry." — Journal 0/ Education, SCIENCE CLASS-BOOKS. " It still holds its position among the very best text-books of elementary chemistry."- School Board ChronuU. Problems adapted to the Above. By Professor Thoki-e and W. Tate. With Key. 65 cents, INORGANIC CHEMISTRY FOR BEGINNERS. By Sir Henry Roscoe, F.R.S., assisted by Joseph Lunt, F.C.S. 75 cents. OWENS COLLEGE JUNIOR COURSE OF PRACTICAL CHEMISTRY. By F. Jones. With Preface by Sir H. Roscoe, F.R.S. 70 cents. " It is eminently practical. The text is concise, and at the same time accn- rate. The instructions concerning experiments are clear, and calculated to develop observant habits. At the end of the book there are some well se- lected questions, which should ensure a thorough understanding of the facts to which they refer." — Chemist and Druggist. OWENS COLLEGE COURSE OF PRACTICAL ORGANIC CHEMIS- TRY. By Julius B. Cohen, Ph.D. With Preface by Sir H. E. Roscoe and Professor Schoklemmer. 70 cents. " It is with great pleasure that we announce the appearance of this useful little work, in which the author has cut out a new path of his own, by the exclusively practical character of the lessons and by the style he has adopted." — Chemical Trade Journal. CHEMICAL THEORY FOR BEGINNERS. By L. Dokbin, Ph.D., and J. Walker, Ph.D. 70 cents. " This excellent and useful little work conducts the beginner over the early stagi^, 01" his journey and securely grounds him in chemical theory." — Scotsman. " Elementary students are told just sufficient to enable them to work out a chemical problem. This book is intended to come to the help of the stu- dent in his transition stage, and contains an exceptionally clear concise explanation of chemical theory." — Journal of Education. LESSONS IN ELEMENTARY PHYSIOLOGY. By Rt. Hon. T. H. Huxley, F.R.S. $1.10. " It is an admirable illustration of how the greatest master of a science may teach its elements in the most simple manner." — Medical Press. " A very useful little manual, which should be received with acclama- tion." — Spectator. Questions on the Above for Schools. By Alcock. 40 cents. LESSONS IN ELEMENTARY ANATOMY. By St. G. Mivart, F.R.S. $ 1.75. " It may be questioned whether any other work on anatomy contains in like compass so proportionately great a mass of information." — Lancet. " Its utility to the general reader who desires, in a small space, to be acquainted with the leading facts and generalizatio.is of modern comparative anatomy is manifest." — Pall Mall Gazette. ELEMENTARY LESSONS IN PHYSICAL GEOGRAPHY. By Sir Archibald Geikie, F.R.S. Illustrated. $1.10. " The language is always simple and clear, and the descriptions of the various phenomena are no less vivid than interesting: the lessons are never dull, never wearisome, and they can scarcely fail to make the study of Phys- ical Geography popular wherever tliey arc used." — Academy. Questions on the Same for Schools. i8mo. 40 cents. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. ELEMENTARY LESSONS IN PHYSICAL GEOGRAPHY. ARCHIBALD GEIKIE, LL.D., F.R.S. Illustrated with woodcuts and ten plates. $i.io. " I think it is superior to any other text-book I have seen on the subjec' and shall recommend it for introduction into our schools." — C. DwiGHl Marsh, Leicester Academy. " The language is always simple and clear, and the descriptions ot thtj various phenomena are no less vivid than interesting : the lessons are never dull, never wearisome, and they can scarcely fail to make the study ot Physical Geography popular wherever they are used." — Academy. Questions on the Same, for Use in Schools. i8mo. 40 cents. BY THE SAME AUTHOR. A CLASS-BOOK OF GEOLOGY. Third Edition. Illustrated. lamo. $1.10. " We have no hesitation in declaring the book an excellent one, contain- ing exactly such material as renders it especially fitted for instruction. More than that, to the person with no geological turn of mind the whole matter is so well combined, and the explanation so simple, that by reading the volume, nature's actions in the past, as in the present, can be better understood." — New York Times. " Professor Geikie is recognized in England as one of the leading author- ities in this science, and his interesting work is unusually well adapted to create and foster a taste for this subject. While written in a remarkably plain and simple style, it contains an immense amount of solid and useful information, and is profusely illustrated with engravings of the various fossils and geological formations." — Popular Science News. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. MATHEMATICAL TEXT-BOOKS SUITABLE FOR USE IN PREPARATORY SCHOOLS. SELECTED FROM THE LISTS OF THE MACMILLAN COMPANY, Publishers. ARITHMETIC FOR SCHOOLS. By J. B. LOCK, Author of" Trigonometry for Beginners," "Elementary Trigonometry" etc Edited and Arranged for American Schools By CHARLOTTE ANGAS SCOTT, D.SC, Head of Math. Dept., Bryn Mawr College, Pa. lerno. Cloth. 75 cents. " Evidently the work of a thoroughly good teacher. The elementary truth, that arithmetic is common sense, is the principle which pervades the whole book, and no process, however simple, is deemed unworthy of clear ex planatiom Where it seems advantageous, a rule is given after the explanation. . . . Mr. Ix)ck's admirable Trigonometry ' and the present work are, to our mind, models of what mathematical school books should be." — The Literary World. FOR MORE ADVANCED CLASSES. ARITHMETIC. By CHARLES SMITH, M.A., Author of " Elementary Algebra," "A Treatise on Algebra** AND CHARLES L. HARRINGTON, M.A., Head Master of Dr. J. Sach's School for Boys, New York. 1 6mo. Cloth. 90 cents. A thorough and comprehensive High School Arithmetic, containing many good examples and clear, well-arranged explanations. There are chapters on Stocks and Bonds, and on Exchange, which arc of more than ordinary value, and there is also a useful collection of miscellaneous examples. THP: MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. ALGEBRA FOR BEGINNERS, By H. S. HALL, M.A., and S. R. KNIGHT. 1 6mo. Cloth. 60 cents. American Edition. In Press. " The present work has been undertaken in order to supply a dentiand for an easy introduction to our ' Elementary Aljjrebra for Schools,' and also to meet the wishes of those who, while approvinjr of the order and treatment of the subject there laid down, have felt the want of a beginners' text-book in a cheaper form. As regards the earlier chapters, our order has been determined mainly by two considerations : first, a desire to introduce as early as possible the jtractical side of the subject and some of its most interesting applications, such as easy equations and problems ; and, secondly, the strong opinion that all reference to compound exjtressions and their resolution into factors should be postponed until the usual operations of algebra have been exemplified in the case of simple expressions, \^y this course the beginner soon becomes acquainted with the ordinary algebraical processes without encountering too many of their difficulties ; and he is learning at the same time something of the more attractive parts of the subject." BY THE SAME AUTHORS. ELEMENTARY ALGEBRA FOR SCHOOLS. Ed iiion for American Schools. In Press. " This is, in our opinion, the best Elementary Algebra for school use. It is the combined work of two teachers who have had considerable experience of actual school teaching, . . . and so successfully grapples with difficulties which our pres- ent text-books in use, front their authors lacking such experience, ignore or slightly touch upon. . . . We confidently recommend it to mathematical teachers, who, we feel sure, will find it the best book of its kind for teaching purposes.'''' — Nature. " We will not say that this is the best Elementary Algebra for school use that we have come across, but we can say that toe do not remember to have seen a better. ...Itis the outcome of a long experience of school-teaching, and so is a thor- oughly practical book. All others that we have in our eye are the works of men who have had considerable experience with senior and junior students at the uni- versities, but have had little, if any, acquaintance with the poor creatures who are just stumbling over the threshold of algebra. . . . Buy or borrow the book for yourselves and judge, or write a better. ... A higher text-book is on its way. This occupies sufficient gi-ound for the generality of boys." —.Academy. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. ELEMENTARY ALGEBRA USE OF PREPARATORY SCHOOLS. By CHARLES SMITH, M.A., Author of '■'■ A Treatise on Algebra,'''' '■'An Elementary Treatise on Conic Sections,'''' etc. Revised and adapted to American Schools By IRVING STRINGHAM, Ph.D., Professor of Mathematics and Dean of the College Faculties in the University of California. Briefer edition, $1.10. Complete edition, $1.20. The Complete Edition contains in addition to the material given in the briefer work, chapters on special subjects, a knowledge of which is re(niircd for admission to Harvard, Yale, and other colleges of advanced standing. It should be especially- valuable to students preparing for Harvard, Cornell, the Uiuversity of Michigan, of California, Missouri, etc., in which the " Treatise on Algebra,'' by the same author, is used. PRINCIPLES OF ELEMENTARY ALGEBRA. By N. F. DUPUIS, M.A., F.R.S.C. Price, $1.10. "This is one of the most able expositions of algebraic principles that wo have yet met with. The book is intended to embrace all the ordinary algebraic subjects, b»»t its real value hes in the reliable guidance it offers to sttldeiits who, having hml an ordinary text-book drilling to the end of quadratics, wish to know what It was all about. . . . The concluding chapter contains a very i)racticAl consideration of that ever increasingly iini>ortAnt branch of algebra — determinants. Emphatically a book for teachers, we wish this Algebra the wide sale that it merits." — The St'hotilmanter. "Professor Diipuis has followed up the success achieved by his * Klementarj' Synthetic (leometry ' (Macmlllan, 18s0), and now publislies "this 'intermediate aigebm,' ' a stepping-stone to assist the student in passing IVom the former stage (of absolute beginners) to the latter (of accomplished algebraists).' A rtsnme of the i>reface will indicate the work attenijjted, and carried out in an interesting and satisfactory manner. ProminencHi is given to the formal laws of algebra and to factoring, from whi(!h last the theory of the solution «)f (|uadratic and other etjua- tions is deduced. . . . The inspiration of Chrystal's 'Algebra' is conspicuou* throughout and duly acknowledged." — The Aeiidemy. THE MACMILLAN COMPANY, 66 FIFTH AVENUE. NEW YORK. INTRODUCTORY MODERN GEOMETRY OF THE POINT, RAY, AND CIRCLE. By WILLIAM B. SMITH, Ph.D., Professor of Mathematics in the Tulane University of New Orleans, La. Cloth. $1.10. " To the many of my fellow-teachers in America who have questioned me in regard to the Non-Euclidean Geometry, I would now ^vish to say publicly that Dr. Smith's conception of that profound advance in pure science is entirely sound. . . . Dr. Smith has given us a book of which our country can be proud. I think it the duty of every teacher of geometry to examine it carefully." — i^rom Prof. George Br.rcE I'Ialsted, Ph. D. (Johns Hopkins), Professor of Mathematics, University of Texax. " I cannot see any cogent reason for not introducing the methods of Modern Geometry in te.xt-books intended for first years of a college course. How useful and instructive these methods are, is clearly brought to view in Dr. Smith's admi- rable treatise. This treatise is in the right direction, and is one step in advancing a doctrine which is destined to reconstruct in great measure the whole edifice of Geometry. I shall make ]>rovision for it in the advanced class in this school next term." — l^roOT PrincipalJ ouy M. Colaw, A.M., Monterey, Va. MODERN PLANE GEOMETRY. Being the Proofs of the Theorems in the Syllabus of Modern Plane Geometry issued by the Association for the Improvement of Geometrical Teaching. By G. RICHARDSON, M.A., and A. S. RAMSAY, M.A. Cloth. $1.00. " Intended to be an Introduction to the subject of Modern Plane Geometry and to the more advanced books of Cremona and others. It has a twofold object: to serve, in the first place, as a seijuel to Euclid . . . ; and, secondly, as a systematic means of procedure from Euclidean Geometry to the higher descriptive Geometr}' of Conies and of imaginary points." THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. TEXT-BOOK OF EUCLID'S ELEMENTS. Including Alternative Proofs, together with Additional Theorems and Exercises, classified and arranged By H. S. HALL and F. H. STEVENS. Books I.-VI. and XL $1.10. Also sold separately as follows : Book 1 30 cents. Books I. and II. . . 50 cents. Books I.-IV. ... 75 cents. Books III.-VI. . . . 75 cents. Books V. , VI. , and XI. 70 cents. Book XI 30 cents. "The chief peculiarity of Messrs. Hall and Stevens' edition is the extent and variety of the additions. After each important proposition a large number of exer- cises are given, and at the end of each book additional exercises, theorems, notes, etc., etc., well selected, often ingenious and interesting. . . . There are a great number of minute details about the construction of this edition and its mechanical execution Avhich we have no space to mention, but all showing the care, the patience, and the labor which have been bestowed upon it. On the wliole, we think it tho most usable edition of Euclid that has yet appeared." — The Nation. THE ELEMENTS OF SOLID GEOMETRY. By ROBERT BALDWIN HAYWARD, M.A., F.R.S.. Senior Mathematical Master in Harrow School; Late President of the Association for the Improvement of Geometrical Teaching. 1 6mo. Cloth. 75 cents. " A modification and extension of the first twenty -one propositions of the eleventh book of Euclid, developed out of a Syllabus of Solid Geometry siihniittcd i>y the author to a Committee of the Association for tiie improrciiiciit of (Jfonu-tncAl Teaching, and reported upon by that Committeo with a considerable degree ol fcvor." THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. ELEMENTS SYNTHETIC SOLID GEOMETRY, By NATHAN F. DUPUIS, M.A., F.R.S.C, Professor of Pure Mathematics in the University of Queen's College. Kingston, Canada. 16mo. pp.239. $1.60. FROM THE AUTHOR'S PREFACE. " I have been induced to present the work to the public, partly by receiving from a number of Educationist^^ inquiries as to what work on Solid Geometry I would recommend as a sequel to my Plane Geometry, and partly from the high estimate that I have formed of the value of the study of synthetic solid geometry as a means of mental discipline. . . . " In this work the subject is carried somewhat farther than is customary in those works in which the subject of solid geometry is appended to that of plane geometry, but the extensions thus made are fairly within the scope of an elemen- tary work and are highly interesting and important in themselves as forming valu- able aids to the right understanding of the more transcendental methods." Introductory to the Above, ELEMENTARY SYNTHETIC GEOMETRY OF THE Point, Line, and Circle in the Plane. 16mo. $1.10. " To this valuable wor^ we previously directed special attention. The whole in- tention of the work has been to prepare the student to take up successfully the modern works on analytical geometry. It is safe to say that a student will learn more of the science from this book in one year than he can learn from the old- fashioned translations of a certain ancient Greek treatise in two years. Every mathematical master should study this book in order to learn the logical method of presenting the subject to beginners." — Canada Educational Journal. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NE'W YORK. ELEMENTARY TRIGONOMETRY H. S. HALL, B.A., and S. R. KNIGHT, B.A. Authors of "Algebra for Beginners" "Elementary Algebra for Schools" etc. Cloth. $1.10. " I consider the work as a remarkably clean and clear presentation of the principles of Plane Trigonometry. For the beginner, it is a book that will lead him step by step to grasp its subject matter in a most satisfactory manner." — E. Miller, University 0/ Kansas. " The book is an excellent one. The treatment of the fundamental relations of angles and their functions is clear and easy, the arrangement of the topics such as cannot but commend itself to the experienced teacher. It is, more than any other work on the subject that I just now recall, one which should, I think, give pleasure to the student." — John J. Schobinger, The Harvard School. WORKS BY REV. J. B. LOCK. TRIGONOMETRY FOR BEGINNERS. AS FAR AS THE SOLUTION OF TRIANGLES. 1 6mo. 75 cents. "A very concise and complete little treatise on this somewhat difificult subject for boys; not too childishly simple in its explanations; an incentive to thinking, not a substitute for it. The schoolboy is encouraged, not insulted. The illustrations are clear. Abundant examples are given at every stage, with answers at the end of the book, the general correctness of which we have taken pains to prove. The definitions are good, the arrangement of the work clear and easy, the book itself well orinted " — Jourtial of Education. ELEMENTARY TRIGONOMETRY. 6th edition. (In this edition the chapter on Logarithms has been carefully revised.) 16mo. $1.10. " The work contains a very large collection of good (and not too hard) examples. Mr. Lock is to be congratulated, when so many Trigonometries are in the field, on having produced so good a book; for he has not merely availed himself of the labors of his predecessors, out by the treatment of a well-worn subject has invested the study of it with interest." — Nature. THE MACMILLAN COMPANY, 66 FIFTH AVENUE, NEW YORK. I -0 / ^ ¥ { = c ^^i i^ = -/ ^ y :^ -/' 'r^ i / :r - I / ' 4 ^t. ^ ^ ^ -. V U(= «7^ -v^ /r I LJ ^-/V-' ^ 1 ^^ /2> hh ^^' H