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• 
 

 BY THE SAME AUTHOR. 
 
 ALGEBRA FOR SCHOOLS AND 
 COLLEGES. 
 
 ELEMENTS OF GEOMETRY. 
 
 ASTRONOMY. For Students and Gen- 
 eral Readers. By Simon Newcomb and 
 Edward S. Holden. 
 
-^u^VJ»^ 
 
 NEWCOMB-8 MATHEMATICAL SERIES. ^ , 
 
 ::b.q.V- (^ 
 
 ALQEBEA 
 
 FOR 
 
 SCHOOLS AND COLLEGES 
 
 SIMON" NEWCOMB 
 
 Professor of Mathematics, United States Navy 
 SECOND EDITION, ME VISED. 
 
 NEW YORK 
 HENRY HOLT AND COMPANY 
 
 1881 
 
Copyright, 1881, 
 
 BT 
 
 Henry Holt & Co. 
 
 KLKCTROTYPED BY 
 
 SMITH & McDOUGAL, 
 82 Beekman St., N. Y, 
 
PREFACE 
 
 The course of algebra embodied in the present -work is substantially 
 that pursued by students in our best preparatory and scientific schools 
 and colleges, with such extensions as seemed necessary to afford an 
 improved basis for more advanced studies. For the convenience of 
 teachers the work is divided into two parts, the first adapted to well- 
 prepared beginners and comprising about what is commonly required for 
 admission to college ; and the second designed for the more advanced 
 general student. As the work deviates in several points from the models 
 most familiar to our teachers, a statement of the principles on which it is 
 constructed may be deemed appropriate. 
 
 One well-known principle underlying the acquisition of knowledge is 
 that an idea cannot be fully grasped by the youthful mind unless it is 
 presented under a concrete form. Whenever possible an abstract idea 
 must be embodied in some visible representation, and all general theorems 
 must be presented in a variety of special forms in which they may be 
 seen inductively. In accordance with this principle, numerical exam- 
 ples of nearly all algebraic operations and theorems have been presented. 
 For the purpose of illustration, numbers have been preferred to hteral 
 symbols when they would serve the purpose equally well. The relations 
 of positive and negative algebraic quantities have been represented by 
 hues and directions from the beginning in order that the pupil might be 
 able to give, not only a numerical, but a visible, meaning to all algebraic 
 quantities. Should it appear to any one that we thus detract from the 
 generality of algebraic quantities, it is sufficient to reply that the system 
 is the same which mathematicians use to assist their conceptions of 
 advanced algebra, and without which they would never have been able 
 to grasp the complicated relations of imaginary quantities. Algebraic 
 
IV PREFACE. 
 
 operations with pure numbers are made to precede the use of symbols, 
 and the latter are introduced only after the pupil has had a certain 
 amount of familiarity with the distinction between algebraic and numer- 
 ical operations. 
 
 Another, but, unfortunately, a less familiar fact is, that all mathematical 
 conceptions require time to become engrafted upon the mind, and the 
 more time the greater their abstruseness. It is, the author conceives, 
 from a failure to take account of this fact, rather than from any inherent 
 defect in the minds of our youth, that we are to attribute the backward 
 state of mathematical instruction in this country, as compared with the 
 continent of Europe. Let us take for instance the case of the student 
 commencing the calculus. On the system which was almost universal 
 among us a few years ago, and which is still widely prevalent, he is con- 
 fronted at the outset with a number of entirely new conceptions, such 
 as those of variables, functions, increments, infinitesimals and limits. 
 In his first lesson he finds these all combined with a notation so entirely 
 diflferent from that to which he has been accustomed, that before the 
 new ideas and forms of thought can take permanent root in his mind, 
 he is through with the subject, and all that he has learned is apt to vanish 
 from his memory in a few months. 
 
 The author conceives that the true method of meeting this difficulty is 
 to adopt the French and German plan of teaching algebra in a broader 
 way, and of introducing the more advanced conceptions at the earhest 
 practicable period in the course. Accordingly, the attempt is made in the 
 present work to introduce each advanced conception, disguised perhaps 
 under some simple form, in advance of its general enunciation and at as 
 early a period as the student can be expected to understand it. In doing 
 this, logical order is frequently sacrificed to the exigencies of the case, 
 because there are several subjects with which a certain amount of famil- 
 iarity must be acquired before the pupil can even clearly comprehend 
 general statements respecting them. 
 
 A third feature of the work is that of subdividing each subject as 
 minutely as possible, and exercising the pupil on the details preparatory to 
 combining them into a whole. To cite one or two instances : a diflBculty 
 which not only the beginner but the expert mathematician frequently 
 meets is that of stating his conceptions in algebraic language. Exercises 
 in such, statements have therefore been made to precede any solution of 
 
PREFACE. ^fy 
 
 problems. In general each principle which is to be presented or used is 
 stated singly, and the pupil is practiced upon it before proceeding to 
 another. 
 
 Subjects have for the most part been omitted which do not find appli- 
 cation either in the work itself or in subsequent parts of the usual course 
 of mathematics, or which do not conduce to a mathematical training. 
 Many teachers may be disappointed to find the greatest common divisor 
 of polynomials, the square roots of binomial surds, and Stuixn's theorem, 
 omitted. The reason of their omission is, firstly, that they neither have 
 any application in the usual course of mathematical study nor advance 
 the student's conception of algebra, and, secondly, that in studying them, 
 power is expended which can be devoted to more profitable objects. 
 Thoroughness at each step has been aimed at rather than multiplicity 
 of subjects. It is, the author conceives, a great and too common 
 mistake to present a mathematical subject to the mind of the student 
 wk4sout sufficient fulness of explanation and variety of illustration to 
 enable him to comprehend and apply it. If he has not time to master a 
 complete course, it is better to omit entirely what is least necessary than 
 to gain time by going rapidly over a great number of things. Some 
 hints to those who may not have time to master the whole work may 
 therefore be acceptable. 
 
 Part I is essential to every one desiring to make use of algebra. Book 
 VIII, especially the concluding sections on notation, is to be thoroughly 
 mastered, before going farther, as forming the foundation of advanced 
 algebra ; and affording a very easy and valuable discipline in the language 
 of mathematics. Aflerward, a selection may be made according to cir- 
 cumstances. The student who is pursuing the subject for the sole 
 purpose of liberal training, and without intending to advance beyond it, 
 will find the theories of numbers and the combinatory analysis most 
 worthy of study. The theory of probabilities and the method in which 
 it is applied to such practical questions as those connected with insurance 
 will be of especial value in training his judgment to the affairs of life. 
 
 The student who intends to take a full course of mathematics with a 
 view of its application to physics, engineering, or other subjects, may, if 
 necessary, omit the book on the theory of numbers, and portions of the 
 chapter on the summation of series. Functions and the functional notation, 
 the doctrine of limits, and the general theory of equations will claim his 
 
VI PREFACE. 
 
 especial attention, while the theory of imaginary quantities will be studied 
 mainly to secure thoroughness in subsequent parts of his course. 
 
 As it has frequently been a part of the author's duty to ascertain what 
 is really left of a course of mathematical study in the minds of those 
 who have been through college, some hints on the best methods of 
 study in connection with the present work may be excused. If asked 
 to point out the greatest error in our usual system of mathematical 
 instruction from the common school upward, he would reply that it con- 
 sisted in expending too much of the mental power of the student upon 
 problems and exercises above his capacity. With the exception of the 
 fundamental routine-operations, problems and exercises should be confined 
 to insuring a proper understanding of the principles involved : this once 
 ascertained, it is better that the student should go on rather than expend 
 time in doing what it is certain he can do. Problems of some difficulty 
 are found among the exercises of the present work; they are inserted 
 rather to give the teacher a good choice from which to select than to 
 require that any student should do them all. 
 
 It would, the author conceives, be found an improvement on our usual 
 system of teaching algebra and geometry successively if the analytic and 
 the geometric courses of mathematics were pursued simultaneously. The 
 former would include algebra and the calculus, the latter elementary 
 geometry, trigonometry, and analytic geometry. The analytic course 
 would then furnish methods for the geometric one, and the latter would 
 fiimish applications and illustrations for the analytic one. 
 
 The Key to the work, which will be issued as soon as practicable, will 
 contain not only the usual solutions, but the explanations and demonstra- 
 tions of the less familiar theorems, and a number of additional problems. 
 
 The author desires, in conclusion, to express his obligation to the many 
 friends who have given him suggestions respecting the work, and espe- 
 cially to Professor J. Howard G-ore of the Columbian University who 
 has furnished solutions to most of the problems, and given the benefit of 
 his experience on many points of detail 
 
TABLE OF COI^TE^TS. 
 
 PART I. —ELEMENTARY COURSE. 
 
 BOOK I.— THE ALGEBRAIC LANGUAGE. 
 
 Chapter I.— Algebraic Numbers and Operations, 3. General 
 Definitions, 3. Algebraic Numbers, 4. Algebraic Addition, 6. 
 Subtraction, 8. Multiplication, 9. Division, 11. 
 
 Chapter II.— Algebraic Symbols, 12. Symbols of Quantity, 12. 
 Signs of Operation, 13. 
 
 Chapter III,— Formation op Compound Expressions, 17. Funda- 
 mental Principles, 17. Definitions, 19. 
 
 Chapter IV.— Construction op Algebraic Expressions, 22. Exer- 
 cises in Algebraic Language, 25. 
 
 BOOK IL— ALGEBRAIC OPERATIONS. 
 
 General Remarks, 28. Definitions, 28. 
 
 Chapter I. — Algebraic Addition and Subtraction, 30. Algebraic 
 Addition, 30. Algebraic Subtraction, 33. Clearing of Parenthe- 
 ses, 35. Compound Parentheses, 37. 
 
 Chapter II. — Multiplication, 38. General Laws of Multiplication, 38. i- 
 Multiplication of Positive Monomials, 40. Rule of Signs in 
 Multiplication, 41. Products of Polynomials by Monomials, 44. 
 Multiplication of Polynomials by Polynomials, 47. 
 
 A Chapter III. — Division, 52. Division of Monomials by Monomials, 52. 
 Rule of Signs in Division, 53. Division of Polynomials by Mono- 
 
Viii CONTENTS. 
 
 mials, 54. Factors and Multiples, 55. Factors of Binomials, 58. 
 Least Common Multiple, 61. Division of one Polynomial bj 
 another, 62. 
 
 I^JUhapter rv.— Of Algebraic Fractions, 67. Negative Exponents, 71. 
 Dissection of Fractions, 73. Aggregation of Fractions, 74. Factor- 
 ing Fractions, 78. Multiplication and Division of Fractions, 79. 
 Division of one Fraction by another, 83. Reciprocal Belations of 
 Multiplication and Division, 83. 
 
 BOOK III.— OF EQUATIONS. 
 
 Chapter I.— The Reduction of Equations, 85. Axioms, 87. Opera- 
 tions of Addition and Subtraction — transposing Terms, 87. 
 Operation of Multiplication, 89. Reduction to the Normal Form, 
 90. Degree of Equations, 93. 
 
 Chapter II. — Equations of the First Degree with One Un- 
 known Quantity, 94 Problems leading to Simple Equations, 99. 
 Problem of the Couriers, 105. Problems of Circular Motion, 108. 
 
 Chapter III.— Equations op the First Degree with Several 
 Unknown Quantities, 109. Equations with Two Unknown. 
 Quantities, 109, Solution of a pair of Simultaneous Equations 
 containing Two Unknown Quantities, 109. Elimination by Com- 
 parison, 110. Elimination by Substitution, 111. Elimination by 
 Addition or Subtraction, 112. Problem of the Sum and DiflFer- 
 ences, 113. Equations of the First Degree with Three or More 
 Unknown Quantities, 116. Elimination, 116. Equivalent and 
 Inconsistent Equations, 121. 
 
 Chapter IV,— Of Inequalities, 123. 
 
 BOOK IV.— RATIO AND PROPORTION. 
 
 Chapter I.— Nature of a Ratio, 128. Properties of Ratios, 132. 
 
 Chapter II.— Proportion, 133. Theorems of Proportion, 134. The 
 Mean Proportional, 138. Multiple Proportions, 139. 
 
CONTENTS, IX 
 
 BOOK v.— OF POWERS AND ROOTS. 
 
 Chapter I. — Involution, 144. Involution of Products and Quotients, 
 144. Involution of Powers, 145. Case of Negative Exponents, 147. 
 Algebraic Sign of Powers, 148. Involution of Binomials — the 
 Binomial Theorem, 148. Square of a Polynomial, 153. 
 
 Chapter II. — Evolution and Fractionaij Exponents, 155. Powers 
 of Expressions with Fractional Exponents, 157. 
 
 Chapter III— Reduction of Irrational Expressions, 159. Defini- 
 tions, 159. Aggregation of Similar Terms, 160. Factoring Surds, 
 161. Perfect Squares, 166. To Complete the Square, 167. Irra- 
 tional Factors, 169. 
 
 BOOK VI.— EQUATIONS REQUIRING IRRATIONAL OPERATIONS. 
 
 Chapter I. — Equations with Two Terms only, 170. Solution of a 
 Binomial Equation, 170. Special Forms of Binomial Equations, 
 171. Positive and Negative Roots, 172, 
 
 Chapter II. — Quadratic Equations, 174. Solution of a Complete 
 Quadratic Equation, 175. Equations which may be reduced to 
 Quadratics, 179. Factoring a Quadratic Equation, 184. Equations 
 having Imaginary Roots, 188. 
 
 Chapter III. — Reduction of Irrational Equations to the Normal 
 Form, 189. Clearing of Surds, 189. 
 
 Chapter IV.— Simultaneous Quadratic Equations, 193. 
 
 BOOK VII.— PROGRESSIONS. 
 
 Chapter I. — Arithmetical Progression, 200. Problems in Pro- 
 gression, 202. 
 
 Chapter II. — Geometrical Progression, 207. Problems of Geo- 
 metrical Progression, 208. Limit of the Sum of a Progression, 
 211. Compound Interest, 217. 
 
X CONTENTS. 
 
 PART II.-ADVANCED OOUESE. 
 
 BOOK VIII.— RELATIONS BETWEEN ALGEBRAIC QUANTITIES. 
 
 Functions and their Notation, 221. Equations of the First Degree 
 between Two Variables, 224. Notation of Functions, 230. Func- 
 tions of Several Variables, 232. Use of Indices, 233. Miscellaneous 
 Functions of Numbers, 235. 
 
 BOOK IX.— THE THEORY OF NUMBERS. 
 
 Chapter I.— The DiYisrBiLiTY of Numbees, 238. Division into Prime 
 Factors, 239. Common Divisors of two numbers, 240. Relations 
 of numbers to their Digits, 245. Divisibility of Numbers and their 
 Digits, 245. Prime Factors of Numbers, 248. Elementary Theorems, 
 251. Binomial Coefloicients, 251. Divisors of a Number, 254. 
 
 Chapter II.— Op Conths'ued Fractions, 258. Relations of Converging 
 Fractions, 267. Periodic Continued Fractions, 270. 
 
 BOOK X.— THE COMBINATORY ANALYSIS. 
 
 Chapter I.— Permutations, 273. Permutation of Sets, 275. Circular 
 Permutations, 277. Permutations when Several of * the Things 
 are Identical, 279. The two classes of Permutations, 281. Sym- 
 metric Functions, 284. 
 
 Chapter II.— Combinations, 285. Combinations with Repetition, 
 287. Speeial Cases of Combinations, 289. The Binomial Theorem 
 when the Power is a Whole Number, 296. 
 
 Chapter IH.— Theory op Probabilities, 299. Probabilities depend- 
 ing upon Combinations, 300. Compound Events. 305. Cases of 
 Unequal Probability, 310. Application to Life Insurance, 316. 
 Table of Mortality, 318. 
 
 BOOK XL— OF SERIES AND THE DOCTRINE OF LIMITS. 
 Chapter I.— Nature op a Series, 321. Notation of Sums, 324 
 
CONTENTS. Xi 
 
 CiiAPTER II.— Development m Powers of a Variable, 326. 
 Method of Indeterminate Coefficients, 327. Multiplication of Two 
 Infinite Series, 333. 
 
 Chapter III.— Summation op Series. Of Figurate Numbers, 336. 
 Enumeration of Triangular Piles of Shot, 339. Sum of the 
 Similar Powers of an Arithmetical Progression, 341. Other Series, 
 345. Of Differences, 350. Theorems of Differences, 355. 
 
 Chapter IV. — The Doctrine op Limits, 358. Notation of the 
 Method of Limits, 361. Properties of Limits, 364. 
 
 Chapter V. — The Binomial and Exponential Theorems. The 
 Binomial Theorem for all values of the Exponent, 368. The 
 Exponential Theorem, 373. 
 
 Chapter VI. — Logarithms, 378. Properties of Logarithms, 378. Com- 
 * parison of Two Systems of Logarithms, 384. 
 
 BOOK XII.— IMAGINARY QUANTITIES. 
 
 Chapter I.— Operations with the Imaginary Unit, 391. Addi- 
 tion of Imaginary Expressions, 393. Multiplication of Imaginary 
 Quantities, 393. Reduction of Functions of i to the Normal 
 Form, 396. 
 
 Chapter II.— The Geometrical Representation op Imaginary 
 Quantities, 404, 
 
 BOOK XIII.— THE GENERAL THEORY OF EQUATIONS. 
 
 Every Equation has a Root, 416. Number of Roots of General 
 Equation, 418. Relations between Coefficients and Roots, 425. 
 Derived Functions, 427. Significance of the Derived Function, 430. 
 Forms of the Roots of Equation, 431. Decomposition of Rational 
 Fractions, 433. Transformation of Equations, 438. Resolution of 
 Numerical Equations, 443. 
 
FIRST PART. 
 ELEMENTARY COURSE, 
 
THE ALGEBRAIC LANGUAGE. 
 
 CHAPTER I. 
 
 OF ALGEBRAIC NUMBERS AND OPERATIONS. 
 
 General Definitions. 
 
 1. Definition. Mathematics is the science wMcli 
 treats of the relations of magnitudes. 
 
 The magnitudes of mathematics are time, space, force, 
 value, or other things which can be thought of as entirely 
 made up of parts. 
 
 2. Def. A Quantity is a definite portion of any 
 magnitude. 
 
 Example. Any definite number of feet, miles, acres, 
 bushels, years, pounds, or dollars, is a quantity. 
 
 3. Def. Algebra treats of those relations which 
 are true of quantities of every kind of magnitude. 
 
 4. The relations treated of in Algebra are discovered 
 by means of numbers. 
 
 To measure a quantity by number, we take a certain por- 
 tion of the magnitude to be measured as a unit, and express 
 how many of the units the quantity contains. 
 
 Remark. It is obviously essential that the quantity and 
 its unit shall be the same kind of magnitude. 
 
 5. Def. A Concrete Number is one in which the 
 kind of quantity which it measures is expressed or 
 understood ; as 7 mlles^ 3 days, or 10 pounds. 
 
4 THE ALGEBRAIC LANGUAGE. 
 
 6. Def. An Abstract Number is one in whicli no 
 particular kind of unit is expressed ; as 7, 3, or 10. 
 
 Eemark. An abstract number may be considered as a 
 concrete one expressing a certain number of units, without 
 respect to the kind of units. Thus, 7 means 7 units. 
 
 Algebraic Numbers. 
 
 7. In Arithmetic, the numbers begin at 0, and in- 
 crease without limit, as 0, 1, 2, 3, 4, etc. But the 
 quantities we usually measure by numbers, as time 
 and space, do not really begin at any point, but extend 
 without end in opposite directions. 
 
 For example, time has no beginning and no end. An 
 epoch of time 1000 years from Christ may be either 1000 years 
 after Christ, or 1000 years before Christ. 
 
 A heayy body tends to fall to the ground. A body which 
 did not tend to move at all when unsupported would have no 
 weight, or its weight would be 0. If it tended to rise upward, 
 like a balloon, it would have the opposite of weight. 
 
 If we have to measure a distance from any point on a 
 straight line, we may measure out in either direction on the 
 line. If the one direction is east, the other will be west. 
 
 One who measures his wealth is poorer by all that he owes. 
 If he owes more than he possesses, he is worth less than 
 nothing, and there is no limit to the amount he may owe. 
 
 8. In order to measure such quantities on a uni- 
 form system, the numbers of Algebra are considered as 
 increasing from in two opposite directions. Those in 
 one direction are called . Positive ; those in the other 
 direction Negative. 
 
 9. Positive numbers are distinguished by the sign 
 +, plus ; negative ones by the sign — , minus. 
 
 If a positive number measures years after Christ, a negative 
 one will mean years before Christ. 
 
 If a positive number is used to measure toward the right, a 
 negative one will measure toward the left. 
 
ALOEBRAIG NUMBERS. 
 
 If a positive number measures weight, the negative one 
 will imply levity, or tendency to rise from the earth. 
 
 If a positive number measures property, or credit, the nega- 
 tive one will imply debt. 
 
 10. The series of algebraic numbers will therefore 
 be considered as arranged in the following way, the 
 series going out to infinity in both directions. 
 
 -^ NEGATIVE DIRECTION. POSITIVE DIRECTION, ^r 
 
 Before. After. 
 
 Downward, Upward. 
 
 Debt. Credit, 
 
 etc. etc. 
 
 etc. -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5, etc. 
 
 Rem. It matters not which direction we take as the 
 positive one, so long as we take the opposite one as 
 negative. 
 
 If we take time before as positive, time after will be nega- 
 tive ; if we take west as the positive direction, east will be 
 negative; if we take debt as positive, credit will be negative. 
 
 11. Positive and negative numbers may be conceived 
 as measuring distances from a fixed point on a straight 
 line, extending indefinitely in both directions, the dis- 
 tances one way being positive, and the other way 
 negative, as in the following scheme : * 
 
 etc. — 7, -6. -5. -4, —3, -2. -1, 0, +1, +2. +3, +4, +5, +6, +7 . etc. 
 
 I I I I I i I I i I 1 1 i I i 
 
 In this scale, the distance between any two consecu- 
 tive numbers is considered a unit or unit step. 
 
 12. Def. Tlie signs + and — are called the Alge- 
 braic Signs, because they mark the direction in which 
 the numbers following them are to be taken. 
 
 * The student should copy this scale of numbers, and have it before 
 him in studying the present chapter. 
 
6 TEE ALGEBRAIC LANGUAGE. 
 
 The sign + may be omitted before positive numbers, when 
 no ambiguity is thus produced. The numbers 2, 5, 12, taken 
 alone, signify +2, +5, +12. But the negative sign must 
 ahvays be written when a negative number is intended. 
 
 13. Bef. One mimber is said to be Algebraically 
 Greater than another when on the preceding scale it 
 lies to the positive (right hand) side. Thus, 
 
 — 2 is algebraically greater than — 7 ; 
 
 
 
 u 
 
 u 
 
 a 
 
 a 
 
 -2; 
 
 6 
 
 (( 
 
 u 
 
 a 
 
 iC 
 
 -5. 
 
 Alsrebraic Addition. 
 
 ^53 
 
 14. Def. In Algebra, Addition means the combi- 
 nation of quantities according to their algebraic signs, 
 the positive quantities being counted one way or added, 
 and negative ones the opposite way or subtracted. 
 
 15. Def. The Algebraic Sum of several quantities 
 is the surplus of the positive quantities over the nega- 
 tive ones, or of the negative quantities over the positive 
 ones, according as the one or the other is the greater. 
 
 The sum has the same algebraic sign as the prepon- 
 derating quantity. 
 
 Example. 
 
 The sum of 
 
 
 
 + 7 and 
 
 -7 is 0; 
 
 
 + 9 '' 
 
 -7 " +2; 
 
 
 + 5 *' 
 
 -7 " -2. 
 
 The sum of several positive numbers may be represented 
 on the line of numbers, § 11, by the length of the line formed 
 by placing the lengths represented by the several numbers 
 end to end. The total length will be the sum of the partial 
 lengths. 
 
 If any of the numbers are negative, the algebraic sum is 
 represented by laying their lengths oif in the opposite direction. 
 
 Example 1. The algebraic sum of the four numbers 9, 
 — 7, 1, —6, would be represented thus : 
 
ALGEBRAIC ADDITION. 
 
 
 —3 , +9 
 
 ET 
 
 Here, starting from 0, we measure 9 to the right, then 7 
 to the left, then 1 to the right, then 6 to the left. The result 
 would be 3 steps to the left from 0, that is, — 3. Thus, — 3 
 is the algebraic sum of -j-9, —7, -f 1, and — 6. 
 
 Ex. 2. If we imagine a person to walk back and forth 
 along the line of numbers, his distance from the starting- 
 point will always be the algebraic sum of the separate distances 
 he has walked. 
 
 Ex. 3. A man's wealth is the algebraic sum of his posses- 
 sions and credits, the debts which he owes being negative 
 credits. If he has in money $1000, due from A $1200, due to 
 X $500, due to Y $350, his possessions would, in the language 
 of algebra, be summed up as follows : 
 
 Cash, . . - . 
 
 Due from A, .... 
 
 Due from X, .... 
 
 Due from Y, .... 
 
 Sum total, . . . . + $1350 
 
 [In the language of Algebra, the fact that he owes X $500 
 may be expressed by saying that X owes him — $500.] 
 
 16. Def. To distinguish Ibetween ordinary and 
 algebraic addition, the former is called Numerical or 
 Arithmetical addition. 
 
 Hence, the numerical sum of several numbers 
 means their sum as in arithmetic, without regard to 
 their signs. 
 
 17. Rem. In Algebra, whenever the word sum 
 is used without an adjective, the algebraic sum is 
 understood. 
 
 + $1000 
 
 + 
 
 1200 
 
 — 
 
 500 
 
 — 
 
 350 
 
8 THE ALOEBBAIG LANGUAGE. 
 
 Algebraic Subtraction. 
 
 18, Memorandum of arithmetical definitions and operations. 
 
 The Subtrahend is the quantity to be subtracted. 
 
 The Minuend is the quantity from which the subtrahend 
 is taken. 
 
 The Remainder or Difference is what is left. 
 
 If we subtract 4 from 7, the remainder 3 is the number of 
 
 unit steps on the scale of numbers (§ 11) from +4 to +7. 
 
 This is true of any arithmetical difference of numbers. In 
 
 Algebra, the operation is generalized as follows: 
 
 19. Def. The Algebraic Difference of two num- 
 bers is represented by the distance from one to the 
 other on the scale of numbers. 
 
 The number from which we measure is the Subtra- 
 hend. 
 
 That to which we measure is the Minuend. 
 
 If the minuend is algebraically the greater (§ 13), 
 the difference is positive. 
 
 If the minuend is less than the subtrahend, the dif- 
 ference is negative. 
 
 In Arithmetic we cannot subtract a greater number from a 
 less one. But there is no such restriction in Algebra, because 
 algebraic subtraction does not mean taking away, but finding 
 a difference. However the minuend and subtrahend may be 
 situated on the scale, a certain number of spaces toward the 
 right or toward the left will always carry us from the subtra- 
 hend to the minuend, and these spaces make up the difference 
 of the two numbers. 
 
 30. The general rule for algebraic subtraction may be 
 deduced as follows : It is evident that if we pass from the 
 subtrahend to on the scale, and then from to the minuend, 
 the algebraic sum of these two motions will be the entire space 
 between the subtrahend and minuend, and will therefore be 
 the remainder required. But the first motion will be equal to 
 the subtrahend, but positive if that quantity is negative, and 
 vice versa, and the second motion will be equal to the minuend. 
 
ALGEBRAIC MXTLTIP LIGATION. 9 
 
 Hence tlie remainder will be found by changing the algebraic 
 sign of the subtrahend, and then adding it algebraically to the 
 minuend. 
 
 EXAMPLES. 
 
 Subtracting + 5 from + 8, the difference is 8 — 5 = 3. 
 
 + 8 
 
 
 + 5, " 
 
 " 5 - 8 = - 3. 
 
 + 8 
 
 
 -5, " 
 
 ^f_5_8=-13. 
 
 — 8 
 
 
 5, *' 
 
 " 5 H 8 = + 13. 
 
 + 13 
 
 
 0, " 
 
 — 13. 
 
 -13 
 
 
 0, " 
 
 ' " + 13. 
 
 21. By comparing algebraic addition and subtraction, it 
 will be seen that to subtract a positive number is the same 
 thing as to add its negative, and vice versa. Thus, 
 
 To subtract 5 from 8 gives the same result as to add — 5 
 to 8, namely 3. 
 
 To subtract — 5 from 8 gives 8 + 5, namely 13. 
 
 Hence, algebraic subtraction is equivalent to the 
 algebraic addition of a number with the opposite 
 algebraic sign. Algebraists, therefore, do not consider 
 subtraction as an operation distinct from addition. 
 
 Algebraic Multiplication, 
 
 22. Memorandum of arithmetical definitions. 
 
 The Multiplicand is the quantity to be multiplied. 
 The Multiplier is the number by which it is multiplied. 
 The result is called the Product. 
 
 Factors of a number are the multiplicand and multiplier 
 which produce it. 
 
 23. To multiply any algebraic quantity by a posi- 
 tive whole number means, as in Arithmetic, to take it a 
 number of times equal to the multiplier. 
 
 Thus, 4x3= 4 + 4+4=+ 12; 
 
 — 4x3 = — 4 — 4 — 4= — 12. 
 
 The product of a negative multiplicand by a positive 
 multiplier will therefore be negative. 
 
10 THE ALGEBRAIC LANQVAGE. 
 
 24. If the multiplier is negative, the sign of the 
 product will be the opposite of what it would be if the 
 multiplier were positive. 
 
 Thus, 4-4 X —3 = —12; 
 
 -4 X -3 = + 12. 
 
 The product of two negative factors is therefore 
 positive. 
 
 25. The most simple way of mastering the use of algebraic 
 signs in multiplication is to think of the sign — as meaning 
 opposite in direction. Thus, in § 11, — 4 is opposite in 
 direction to + 4, the direction being that from 0. If we mul- 
 tiply this negative factor by a negative multiplier, the direction 
 will be the opposite of negative, that is, it will be positive. A 
 third negative factor will make the product negative again, a 
 fourth one positive, and so on. For example, 
 
 -3 X -4 = +12; 
 
 _4 = — 2 X +12 = — 24; 
 -4 = -3 X -24 = + 72 ; 
 etc. 
 
 -2 X 
 
 -3 X 
 
 —3 X —2.x 
 
 -3 X 
 
 etc. 
 
 
 Hence, 
 
 
 26. Theorem. 
 
 The 
 
 The continued product of an even 
 number of negative factors is positive ; of an odd num- 
 ber, negative. 
 
 Rem. Multiplying a number by —1 simply changes 
 its sign. 
 
 Thus, -1-4 X —1 = —4; 
 
 -4 X -1 = + 4. 
 
 EXERCISES. 
 
 Find the algebraic sums of the following quantities : 
 
 1. 4 — 6 + 12-1—18. 
 
 2. —6-3-8. 
 
 3. _ 6 — 10 — 9 + 34. 
 
 4. Subtract the sum in Ex. 3 from the sum in Ex. 2. 
 
 5. Subtract the sum 5 — 6 + 3—1 — 16, from the sum 
 -2-7-4 + 8. 
 
ALGEBRAIC DIVISION. 11 
 
 6. Subtract the sum 5 — 6 +3—1 — 16, from the sum 
 7 _ 3 _ 8 + 4. 
 
 7. Form the product —7x8. 
 
 8. Form the product —8x7. 
 
 9. Form the product 6x— 5x7x— 4. 
 
 10. Form the product — 6x— llx8x— 2. 
 
 11. Form the product — Ix— Ix— Ix— 1. 
 
 12. Subtract the sum in Ex. 1 from the sum in Ex. 3, and 
 multiply the remainder by the sum in Ex. 2. 
 
 13. Subtract 8 from — 3, — 3 from — 1,-1 from 8, and 
 find the sum of the three remainders. 
 
 14. Subtract 7 from — 9 and the remainder from 2, and 
 multiply the result by the product in Ex. 7. 
 
 Alg^ebraic Division. 
 
 37. Memorandum of arithmetical definitions. 
 
 The Dividend is the quantity to be divided. 
 The Divisor is the number by which it is divided. 
 The Quotient is the result. 
 
 28. Bule of Signs in Division. The requirement 
 of division in Algebra is the same as In Arithmetic ; 
 namely, 
 
 The product of the quotient hy the divisor must he 
 equal to the dividend. 
 
 In Algebra, two quantities are not equal unless they have 
 the same algebraic sign. Therefore the product, 
 
 quotient x divisor 
 
 must have the same algebraic sign as the dividend. From 
 this we can deduce the rule of signs in division. 
 
 Let us divide 6 by 2, giving 6 and 2 both algebraic signs, 
 and find the signs of the quotient 3 : 
 
 + 3x+2=+6; therefore, +6 divided by +2 gives +3. 
 
 + 3x— 2 = — 6; " —6 " "—2 " +3. 
 
 — 3x+2 = — 6; " —6 " " +2 " —3. 
 
 — 3 X — 2 = +6; " +6 " " —2 " —3. 
 
12 THE ALGEBRAIC LANGUAGE. 
 
 Hence, the rule of signs is the same in division as in mul- 
 tiplication, namely : 
 
 Like signs in dividend and divisor give + . Unlike 
 signs give — . 
 
 EXERCI SES. 
 
 Execute the following algebraic divisions, expressing each 
 result as a whole number or vulgar fraction : 
 
 1. Dividend, _ 7 + 10 — 11 + 25 ; divisor, 20 — 3. 
 
 2. Dividend, 12 — 3 + 15 — 10 ; divisor, 3 — 10. 
 
 3. Dividend, 25 — 36 + 6 — 20 ; divisor, —3 + 8. 
 
 4. Dividend, — 7 x — 8 ; divisor, —8 + 4. 
 
 5. Dividend, 56 + 8 x — 3 ; divisor, — 4 — 4. 
 
 6. Dividend, — 24 x — 1 ; divisor, — 3 x — 3. 
 
 7. Dividend, —13 x —10 x — 8; divisor, — 4x5x— 6. 
 
 8. Dividend, — 1 x — 1 ; divisor, — 3 x — 3. 
 
 CHAPTER II. 
 ALGEBRAIC SYMBOLS. 
 
 Symbols of Quantity. 
 
 29. Algebraic quantities may be represented by 
 letters of the alphabet, or other characters. 
 
 The characters of Algebra are caUed Symbols. 
 
 30. Def. The Value of an algebraic symbol is the 
 quantity which it represents or to which it is equal. 
 
 The value of a symbol may be any algebraic quan- 
 tity whatever, positive or negative, wliich we choose to 
 assign to the symbol. 
 
 31. The language of Algebra differs in one respect from 
 ordinary language. In the latter, each special word or sign 
 
SIGIf-a OF OPERATION. 13 
 
 has a definite and invariable meaning, which every one who 
 uses t"he language must learn once for all. But in Algebra a 
 symbol may stand for any quantity which the writer or speaker 
 chooses, and his results must be interpreted according to this 
 meaning. 
 
 33. The same character may be used to represent several 
 quantities by applying accents or attaching numbers to it to 
 distinguish the different quantities. Thus, the four symbols, 
 a, a', a", a'", may represent four different quantities. The 
 symbols «,, a.2, a^, a^, «6, etc., may be used to designate any 
 number of quantities which are distinguished by the small 
 number written after the letter a. 
 
 Si^ns of Operation. 
 
 33. In Algebra, the signs +, — , and x are used, 
 as in Arithmetic, to represent addition, subtraction, and 
 multiplication, these operations being algebraic, not 
 numerical. 
 
 34. Signs of Addition and Svhtr action. The com- 
 bination a + & means the algebraic sum of the quantities 
 a and &, and a — h means their algebraic difference. 
 
 EXAMPLES. 
 
 If rt r= + 4 and J = + 3, then a-{-l = +7, a— I = +1. 
 
 If «= + 5 and 5 = — 7, then a-\-'b——%,a—l—-\-\%. 
 
 If 6?= — 6 and ^>= + 3, then a + 6 == — 3, «— Z> = — 9. 
 
 If «= — 6 and 2»= — 3, then a + Z> = — 9, a— J = —3. 
 
 The signs of addition and subtraction are the same as those 
 used to indicate positive and negative quantities, but the two 
 applications may be made without confusion, because the 
 opposite positive and negative directions correspond to the 
 opposite operations of adding and subtracting. 
 
 35. Sign of Multiplication, The sign of multipli- 
 cation, X , is generally omitted in Algebra, and when 
 different symbols are to be multiplied, the multiplier is 
 
14 THE LANGUAGE OF ALGEBRA. 
 
 written before the multiplicand without any sign be- 
 tween them. 
 
 Thus, 4« means « x 4. 
 
 ax " XX a. 
 Sabmi/ " yxmxixaxS, 
 
 If numbers are used instead of symbols, some sign of mul- 
 tiplication must be inserted between them to avoid confusion. 
 Thus, 34 would be confounded with the number thirty-four. 
 A simple dot is therefore inserted instead of the sign x . 
 
 Thus, 3.4 = 4x3 = 12. 
 
 3-12.2 = 72. 
 1.2.3.4.5 = 120. 
 1.2.3.4.5.6 = 720: 
 The only reason why the point is used instead of x , is 
 that it is more easily written and takes up less space. 
 
 36. Division in Algebra is sometimes represented 
 by the symbol -r-, the dividend being placed to the left 
 and the divisor to the right of this symbol. 
 
 Ex. a -T-h means the quotient of a divided by J. 
 
 But division is more generally represented by writing 
 the dividend as the numerator and the divisor as the 
 denominator of a fraction. 
 
 Ex. The quotient of a divided by l is written v* 
 
 It is shown in Arithmetic that a fraction is equal to the 
 quotient of its numerator divided by its denominator ; hence 
 this expression for a quotient is a vulgar fraction. 
 
 37. Powers and Exponents. A Power of a quan- 
 tity is the product obtained by taking that quantity a 
 certain number of times as a factor. 
 
 Def. The Degree of the power means the number 
 of times the quantity is taken as a factor. 
 
 If a quantity is to be raised to a power, the result 
 may, in accordance with the rule for multiplication, be 
 
8IGJ!^S OF OPERATION. 15 
 
 expressed by writing the quantity the required number 
 of times. 
 
 Examples. The fifth power of a may he written 
 axaxaxaxa or aaaaa ; 
 and the fourth power of 7, 7.7.7-7 = 2401. 
 
 To save repetition, the symbol of which the power is 
 to be expressed is written but once, and the number of 
 times it is taken as a factor is written in small figures 
 after and above it. 
 
 Thus, aaaaa is written a^ ; 
 
 7.7.7.7 " " 7^ 
 
 XXX " " xK 
 
 Def. A figure written to indicate a power is called 
 an Exponent. 
 
 Def. The operation of forming a power is called 
 Involution. 
 
 38. Hoots. A Root is one of the equal factors 
 into which a number can be divided. 
 
 Def. The figure or letter showing the number of 
 equal factors into which a quantity is to be divided is 
 called the Index of the root. 
 
 The square root of a symbol is expressed by writing 
 the sign ^ (called root) before it. 
 
 Ex. I. 'v/io means the square root of 49, that is, 7. 
 
 Ex. 2. Vx means the square root of x. 
 
 Any other root than the square is represented by 
 writing its index before the sign of the root. 
 Ex. I. \^x means the cube root of x. 
 Ex. 2. \/x means the fourth root of x. 
 
 Def. The operation of extracting a root is called 
 Evolution. 
 
 39. The operations of Addition, Subtraction, Multi- 
 plication, Division, Involution, and Evolution, are the 
 six fundamental operations of Algebra. 
 
16 THE ALGEBRAIC LANOUAOE. 
 
 40, Def, An Algebraic Expression is any combi- 
 nation of algebraic symbols made in accordance with 
 the foregoing principles. 
 
 EXERCISES. 
 
 In the following expressions, suppose az= — 7, J = — 5, 
 c = 0, ??i = 3, n = 4, 2) = 9, and compute their numerical 
 Talues. 
 
 I. a -\- b -\- m -{- p. 2. a -\- m + n. 
 
 3. m — n — a — b. 4. n -}- p — m — a. 
 
 5. 3a —m + b — 2n. 6. 2a — 7p + 2b — m, 
 
 7. 3mnp, 8. mncp. 
 
 9. hm7i. 10. hip. 
 
 II. ahnp. 12. 2^abnp. 
 
 13. mn 4- 5w. 14. ^m — J;^. 
 
 15. bp — an. 16. 6/? + an. 
 
 17. w^p + wi^J. 18. 7n^n — ap^, 
 
 19. «2 + J2. 20. a^ + ^3. 
 
 21. a^ — ¥. 22. a^m — b^n, 
 
 23. «3^2 — m^n\ . 24. «2j3 — i^mK 
 
 25. a^ + a^j. 26. ab^ — a^J. 
 
 aJ + mn n €Lc — bp 
 
 27. _ 28. ^. 
 
 ab — mn bn — mp 
 
 2m^n^ — lOwi^ ab — mp 
 
 29. 7 • 30. ^. 
 
 j» — ^cwi m — n 
 
 In the following expressions, suppose a = 8, J = — 3, and 
 
 a; to haye in succession the fifteen values — 7, — 6, — 5, etc., 
 
 to + 7, and compute the fifteen corresponding yalues of each 
 
 expression : 
 
 9 , z , a ■\-bx 
 
 3,. x^ + bx + a. 32. ^--^. 
 
 Arrange the results in a table, thus : 
 
 Expression 31 = 78 ; Exp. 32 = — f|. 
 
 a;= -7 
 a; = —6 
 a; = — 5 
 
 " « = 62 ; etc. 
 
 ** " = 48. 
 
 etc. etc. etc. 
 
COMPOUND EXPRESSIONS. 17 
 
 CHAPTER ill. 
 
 FORMATION OF COMPOUND EXPRESSIONS. 
 
 Fvindamental Principles. 
 
 41. The following are two fundamental principles of 
 the algebraic language : 
 
 First Principle, Every algebraic expression, how- 
 ever complex, represents a quantity, and may be 
 operated upon as if it were a single symbol of that 
 quantity. 
 
 Second Principle. A single symbol may be used 
 to re'present any algebraic expression whatever. 
 
 43. When an expression is to be operated upon as 
 a single quantity, it is enclosed between parentheses, 
 but the parentheses may be omitted, when no ambiguity 
 or error will result from the omission. 
 
 Example. Let us have to subtract h from a, and multiply 
 the remainder by the factor m. The remainder will be ex- 
 pressed by « — h, and if we write the product of this quantity 
 by m, in the way of § 35, the result will be 
 
 ma — h. 
 But this will mean I subtracted from ma, which is not what 
 we want, because it is not «, but a — b which is to be multi- 
 plied by m. To express the required operations, we enclose 
 a — h in brackets or parentheses, and write m outside, thus : 
 m(a — I), 
 
 NUMERICAL EXAMPLES. 
 
 7(8 — 2) = 7-6 = 42; but 7-8 - 2 = 56 - 2 = 54. 
 12(3 +4) = 12-7 = 84. 
 (6 -f3)(2 + 6) = 9-8 = 72. 
 (7 - 4) (1 - 5) (2+ 7) = 3 X -4.9 = - 108. 
 2 
 
18 THE LANGUAGE OF ALGEBRA. 
 
 Example 2. Suppose that the expression a — h-^-c is to 
 be added to m, subtracted from m, multiplied by m, divided 
 by 7n, raised to tlie third power, or have the cube root extracted. 
 The results will be written : 
 
 Added to r/i, m -\- {a — b -\- c). 
 
 Subtracted from m, m— {a — h + c). 
 Multiplied by m, m{a — b-{- c). »■ 
 
 Divided by m, -• 
 
 Cubed, {a — h + cf. 
 
 Cube root extracted, ^/{a — ^_+ c). 
 
 There are two of these six cases in which the parentheses 
 are unnecessary, although they do no harm, namely, addition 
 and division, because in the case of addition, 
 
 m -{■ {a — h -\- c) 
 is the same as m -\- a — l + c, 
 
 [For example, 10 + (8 — 5 + 4) = 10 + 7 = 17, 
 and 10 + 8 — 5 + 4 = 17 also.] 
 
 Again, in the case of the fraction, it will be seen that it has 
 exactly the same meaning with or without the parentheses. 
 
 43. An algebraic expression having parentheses as 
 a part of it may be itself enclosed in parentheses with 
 other expressions, and this may be repeated to any 
 extent. Each order of parentheses must then be made 
 larger or thicker, or different in shape to distinguish it. 
 
 Examples, i. Suppose that we have to subtract a from 
 J, the remainder from c, that remainder from d, and so on. 
 We shall have. 
 
 First remainder, . h — a. 
 
 Second, c — (J — «). 
 
 Third, d-[c-[h-a)\ 
 
 Fourth, e-.\d—[c—{h-aY\\. 
 
 Fifth, /- \_e - \d- [c-ib- a)]\l 
 
DEFINITIONS. 19 
 
 2. Suppose that we have to multiply the difference of tlie 
 quantities a and hhj p and subtract the product from m. The 
 result or remainder will be 
 
 m — p {a — h). 
 
 Suppose now that we have to multiply this result by p-\-q. 
 We must enclose both factors m parentheses, and the result 
 will then be written : 
 
 {p + q) [ni —P (a — J)]. 
 
 EXERCISES. 
 
 In the following expressions, suppose 
 « = — 1, Z> = 3, m = 5. a; = — 3, — 1, + 1, -h 3, 
 and calculate the four values of each expression which result 
 from giving x the above four values in succession. 
 
 x{x — a){x — 2«) {x — 3fl5) 
 
 la{p-x)-l){a-x) Y^ 
 m(b — x) -{-h (m — x) 
 
 /p ^ 
 
 3. [ax + b{x-ay-[-m{x- aYf ^:p^* 
 
 4. [^{mx^ + i) — ^(mx^ — I)] ^/{pll) — a). 
 
 Note. When the square root is not an integer, it will be sufficient 
 to express it without computing it in full. 
 Thus, for X = —3, we shall have 
 
 ^{mx^ + &) - y^imx'' -b)= ^^48 - ^/42. 
 
 This is a sufficient answer without extracting the roots. 
 
 Definitions. 
 
 44. Coefficient. Any numlber which multiplies a 
 quantity is called a Ooefl&cient of that quantity. A 
 coefficient is therefore a multiplier. 
 
 Example. In the expression 4«te, 
 
 4 is the coefficient of abx, 
 4« « « " ix, 
 
 ^ab " " « x. 
 
20 TEE LANGUAGE OF ALGEBRA. 
 
 Def. A Numerical Coefficient is a simple number, 
 as 4, in the above example. 
 
 Def. A Literal Coefficient is one containing one 
 or more letters used as algebraic symbols. 
 
 Eem. Any quantity may be considered as having 
 the coefficient 1, because Ix is the same as x. 
 
 Reciprocal. The Reciprocal of a number is unity 
 divided by that number. In the language of Algebra, 
 
 Reci2irocal of N =: -v^« 
 
 Formula. A Formula is an expression used to 
 show how a quantity is to be expressed or calculated. 
 
 Term. When an expression is made up of several 
 parts connected by the signs + or — , each of these 
 parts is called a Term. 
 
 Example. — In the expression, 
 
 a + I)x -\- S?na^, 
 there are three terms, a, hx^ and ^mx^. 
 
 When several terms are enclosed between parentheses, so 
 as to be operated on as a single symbol, they form a single 
 term. 
 
 Thus, the expression 
 
 {a^lx + ^mx>) (a + h) 
 
 (^ + y) (^ - y) 
 
 forms but a single term, though both numerator and denom- 
 inator are each a product of several terms. Such expressions 
 may be called compound terms. 
 
 Aggregate. A sum of several terms enclosed be- 
 tween parentheses in order to be operated upon as a 
 single quantity is called an Aggregate. 
 
 Algebraic expressions are divided into monomials 
 and polynomials. 
 
 A Monomial consists of a single term. 
 
DEFINITIONS. 21 
 
 A Polynomial consists of more than one term. 
 
 A Binomial is a polynomial of two terms. 
 
 A Trinomial is 2i, polynomial of three terms. 
 
 Note. The last three words are commonly applied only 
 to sums of simple terms, formed of single symbols or products 
 of single symbols. 
 
 Entire. An Entire Quantity is one which is ex- 
 pressed without any denominator or divisor, as 2, 3, 4, 
 etc. ; a, &, rr, etc. ; 2a&, 2mp^ ah {x — y\ etc. 
 
 A Theorem is the statement of any general truth. 
 
 45. Other Algebraic Signs. Besides the signs al- 
 ready defined, others are of occasional use in Algebra. 
 
 > , the Sign of Inequality, shows when placed be- 
 tween two quantities, that the one at the open end of 
 the angle is the greater. 
 
 Ex. I. a > 5 means a is greater than h. 
 
 Ex. 2. m < a; <w means x is greater than m, but less 
 than n. 
 
 : , another Sign of Division, is placed between two 
 quantities to express their ratio. 
 
 Thus, a : l means the ratio of a to l, or the quotient of a 
 divided by l. 
 
 .'. means Hence, or Consequently; as, 
 
 « + 2 = 5 ; /. a = d. 
 
 QC means a quantity infinitely great, or Infinity. 
 
 , the Vinculum, is sometimes placed over an 
 aggregate to include it in one mass, in lieu of paren- 
 theses. 
 
 Ex. a — b c — d is the same as {a — l)(c — d). 
 It is mostly used with the radical sign. We often write 
 ^/a -\-h ■\- c instead of V{cl + 5 + c). 
 
22 THE LANGUAGE OF ALGEBRA. 
 
 CHAPTER IV. 
 
 CONSTRUCTION OF ALGEBRAIC EXPRESSIONS. 
 
 46. All operations upon algebraic quantities, however 
 complex, consist in combinations of the elementary operations 
 already described. The result of. each single operation will be 
 an aggregate, a product, a quotient, or a root, and every such 
 result may, in subsequent operations, be operated upon as a 
 single symbol. There are only three cases in which an expres- 
 sion needs any modification in order to be operated upon, 
 namely : 
 
 Case I. An aggregate must be enclosed in parentheses, if 
 any other operations than addition or division are to be per- 
 formed upon it. (§ 42.) 
 
 Case II. When a product is to be raised to a power, or to 
 have a root extracted, it may be enclosed in parentheses in 
 order to show that the operation extends to all the factors. 
 
 If we take the product ale, and write an exponent, 2 for 
 instance, after it thus, abc^, it would apply only to c, and 
 would mean a x b X cK So with the radical sign ; A^abc 
 might mean only A^axhxc, To indicate that the power 
 or root is that of the product as a whole, we may enclose it 
 in parentheses, thus : 
 
 Square root of dbc = ^{adc). 
 Square of abc = (abcy. 
 
 But a root sign is commonly made to include the whole 
 product by simply extending a vinculum over all the factors 
 of the product, thus : Square root of abc = Vctbc. 
 
 Case III. If negative quantities are to be multiplied, 
 merely writing them after each other would lead to mistakes. 
 Thus, the product ax—bx—c, if written without the x 
 sign, would be a — b — c, and would not mean a product at 
 all. But, by enclosing —b and — c in parentheses, we have 
 
 a{-h){-c), 
 which would correctly express the product required. 
 
c 
 
 CONSTRUCriON^OF ALOEBBAIC EXPRESSIONS. 23 
 
 4'7. The following example will show how operations may 
 be combined to any extent. 
 
 The quantity a is to be subtracted from h, and the differ- 
 ence multiplied by y, forming a product P.* The quotient of 
 p — r divided by q is to be multiplied by 7n, and the product 
 subtracted from P. The difference is to form the numerator 
 iV" of a fraction. To form the denominator, h is to be added 
 to a and subtracted from it, and the product Q of the sum and 
 difference formed. The quantity q is to be added to and sub- 
 tracted from p, and the product R of the sum and difference 
 formed. The quotient of Q divided by R is to form the de- 
 nominator of the fraction of which the numerator is P. 
 
 The quantity l subtracted from a leaves h — a. 
 
 Multiplying it by y, the product P is y {b — a). 
 
 Quotient of jp — r divided by q — • 
 
 Multiplying it by W2, m 
 
 [If instead of multiplying the fraction as a whole by m, 
 we had multiplied its numerator, we should have had to 
 
 enclose the p — r in parentheses, thus: --• But 
 
 when the multiplier is written at the end of the line, between 
 the terms of the fraction, as above, it indicates that the frac- 
 tion, as a whole, is multiplied by m.'\ 
 
 V — T 
 
 Subtracting the last product from P, it is y(J) —a) — m — • 
 
 Adding d to a, « + 5. 
 
 Subtracting b from a, a — K 
 
 The product Q of the sum and difference, {a -\-l)){a — h). 
 The product B of p -\- q hy p — q, (P + q) {p — q)' 
 
 The quotient of Q divided by R, (a + b)(a--bl^ 
 
 ^ ^ ' {p-^Q)(p-q) 
 
 * In mathematical language, when a substantive is followed by a 
 symbol in this manner, the latter is used as a sort of proper name to 
 designate the substantive, so that the latter can be afterward referred to 
 by the letter without ambiguity. 
 
 In the present case, the capital letters are used in accordance with 
 the second general principle, § 41. 
 
24 TUE LANGUAGE OF ALGEBBA. 
 
 The fraction having JVfor its numerator and this quotient 
 for its denominator is 
 
 y{h — a) — m ^ ~ 
 
 {a -\-l)){a — b) 
 (P + g) (i? - q) 
 
 48. By the second general principle, § 41, a single sym- 
 bol may be written in place of any algebraic expression whatever. 
 When several symbols indicating such expressions are com- 
 bined, the original expressions may be substituted for them, 
 and be treated in accordance with the first principle. 
 
 EXAMPLES. 
 
 a — dx 
 
 Suppose F = a -\- ix; Q = 
 
 T = X — y; V = mpq. 
 
 It is required to form the expression 
 
 PQ-TV 
 
 PT— QV' 
 The -answer is 
 
 (a + ix) ^-^-^ — i^-y) mpq 
 (a + Ix) {x — y)- ^~ mpq 
 
 EXERCISES. 
 
 Eorm the expressions: 
 
 I. P—T, 2, T—P. 
 
 3. P-Q' 4. Q-V, 
 
 5. VP. 6. VCP + r). 
 
 7. ^(P _ T). 8. P2T2. 
 
 9. V\ 10. T^VK 
 
 VP—QT PT 
 
 ^ 12. - 
 
 (P + T){P- T) {3P-2TY 
 
 '^* {Q + V)(Q-Vy '^- (402 
 
 pg- T 2(P-f T)^ 
 
 ^5- :;y(pZrff ^^' (2T-Vr' 
 
EXERCISES. 25 
 
 2r 
 
 [(r+w + p]r. ... 9i^f 
 
 
 EXERCISES IN ALGEBRAIC LANGUAGE. 
 
 The following questions are proposed to practice the student in ex- 
 pressing the relations of quantities in algebraic language. Should any 
 of them offer difficulties, he is recommended to substitute numbers for 
 the algebraic letters, examine the process by which he proceeds, and then 
 apply the same process to the letters that he applied to the numbers. No 
 solutions of equations are required. 
 
 1. How many cents are there in m dollars ? 
 
 2. How many dollars in m cents? 
 
 3. A man had a dollars in one pocket, and h cents in the 
 other ; how many cents had he in all ? How many dollars ? 
 
 4. The sum of the quantities a and h is to be multiplied 
 by m. Express the product, and its square. 
 
 5. A man having h dollars paid out m dollars to one per- 
 son and n dollars to another. Express what he had left in 
 two ways ? 
 
 6. How many chickens at h cents a piece can be purchased 
 for m dollars ? 
 
 7. A man walked from home a distance of m miles at 4 
 miles an hour, and returned at the rate of 3 miles an hour. 
 How long did it take him to go and come ? 
 
 8. A man going to market bought tomatoes at h cents per 
 peck and potatoes at k cents a peck, of each an equal number. 
 They cost him m cents. How many pecks of each did he buy ? 
 
 9. How many minutes will it require to go a miles, at the 
 rate of h miles an hour ? 
 
 10. A man bought from his grocer a pounds of tea at x 
 cents a pound, h pounds of sugar at y cents a pound, and c 
 pounds of coffee at z cents a pound. How many cents will 
 the whole amount to ? How many dollars ? How many mills ? 
 
 11. A man bought / pounds of flour at m cents a pound. 
 
26 THE ALOEBBAIG LANGUAGE. 
 
 and handed the grocer an ic-dollar bill to be changed ? How 
 many cents ought he to receive in change ? 
 
 12. From two cities a miles apart two men started out at 
 the same time to meet each other, one going m miles an hcur 
 and the other n miles an hour. How long before they will 
 meet ? How far will the first one have gone ? How far will 
 the second one have gone ? 
 
 13. A man left his n children a bonds worth x dollars 
 each, and b acres of land worth y dollars an acre ; but he 
 owed m dollars to each of q creditors. What was each child's 
 share of the estate ? 
 
 14. Two numbers, x and y, are to be added together, their 
 sum multiplied by s, that product divided by a-\-b, and the 
 quotient subtracted from Jc. Express the result. 
 
 15. The sum of the numbers p and q is to be divided by 
 the sum of the numbers a and h, forming one quotient. The 
 difference of the numbers p and q is to be divided by the dif- 
 ference of the numbers a and b, forming another quotient. 
 The sum of the two quotients is to be multiplied by r -\- s. 
 Express the product. 
 
 1 6. The quotient of x divided by a is to be subtracted 
 from the quotient of y divided by b, and the remainder multi- 
 plied by the sum of x and y divided by the difference between 
 X and y. Express the result. 
 
 17. The number x is to be increased by 6, the sum is to be 
 multiplied by a-\-b, q is to be added to the .product, and the 
 sum is to be divided hj r — s. Express the result. 
 
 18. A family of brothers a in number each had a house 
 worth a thousand dollars each. What was the total value of 
 all the houses in dollars ? What was it in cents ? 
 
 19. A grocer mixed a pounds of tea worth x cents a pound, 
 and b pounds worth y cents a pound. How much a pound 
 was the mixture worth ? 
 
 20. x-\-y houses each had a-\-b rooms, and each room 
 m-\-n pieces of furniture. How many pieces of furniture were 
 there in all ? 
 
 21. In a library were p + q volumes, each volume hsLdp-\-q 
 pages, each page p + q words, and each word on the average 
 8 letters. How many letters were there in all the books of the 
 library ? 
 
 22. A post-boy started out from a station, travelling h 
 miles an hour. Three hours afterward, another one started 
 after him, riding m miles an hour. How far was the first one 
 
EXSBCI8E8. 27 
 
 ahead of the second at the end of x hours after the second 
 started ? 
 
 23. Two men started to make the same journey of m miles, 
 one going r miles an hour, and the other s miles an hour. 
 How much sooner will the man going r miles an hour make 
 his journey than the one going s miles an hour ? IIow much 
 sooner will the one going s miles an hour make his journey 
 than the one going r miles an hour ? 
 
 24. One train runs from Boston to New York in h hours, 
 at the rate of n miles an hour. How long will it take another 
 train running 5 miles an hour faster to perform the journey ? 
 
 25. If a man bought h horses for f dollars, and n yoke of 
 oxen for 7n dollars, how much more did one horse cost than one 
 yoke of oxen ? How much more did one yoke of oxen cost 
 than one horse ? 
 
 26. A train making a journey of 2m miles goes the first 
 half of the way at the rate of r miles an hour, and the second 
 half at the rate of s miles an hour. How long did it take it to 
 go ? What was the average speed for the journey ? 
 
 27. Two men, A and B, started to walk from Hartford to 
 New Haven aud back, the distance between the two cities 
 being a miles. A goes p miles an hour and B q miles an hour. 
 How far will A have got on his return journey when B reaches 
 Hartford? 
 
 28. A man having k dollars bought i books at $6 each. 
 How many books at $4 each can he buy with the balance of 
 his money ? 
 
 29. A man going to his grocer with m dollars, bought s 
 pounds of sugar at a cents a pound, and r pounds of coffee at 
 b cents a pound. How many barrels of flour at q dollars a 
 barrel can he buy with the balance of his money ? 
 
 30. A man divided m dollars equally among a poor Chinese 
 and n dollars equally among h orphans. Two of the Chinese 
 and three of the orphans put their shares together and bought 
 X Bibles for the heathen. How much did each Bible cost ? 
 
 31. A pedestrian having agTced to walk the a miles from 
 Boston to Natick in h hours, travels the first k hours at the 
 rate of m miles an hour. At what rate must he travel the 
 remainder of the time ? 
 
 32. A train having to make a journey of x miles in 7i hours, 
 ran for k hours at the rate of r miles an hour, and then made 
 a stop of 7)1 minutes. How fast must it go during the remain- 
 der of its journey to arrive on time ? 
 
BOOK II. 
 ALGEBRAIC OPERA TIONS, 
 
 General Remarks 
 
 The algebraic expressions formed in accordance with the 
 rules of the preceding book admit of being transformed and 
 simplified in a variety of ways. This transformation is effected 
 by operations which have some resemblance to the arithmetical 
 operations of addition, subtraction, multiplication, and division, 
 and which are therefore called by the same names. 
 
 In performing these algebraic operations, the student is not, 
 as in Arithmetic, seeking for a result which can be written in 
 only one way, but is selecting out of a great variety of forms of 
 expression some one form which is the simplest or the best for 
 certain purposes. Sometimes one form and sometimes another 
 is the best for a particular problem. Hence, it is essential 
 that the algebraist, in studying an expression, should be able to 
 see the different ways in which it may be written. 
 
 Definitions. 
 
 49. Function. An algebraic expression containing 
 any symbol is called a Function of the quantity repre- 
 sented by that symbol. 
 
 Ex. I. The expression Zx^ is a function of x, 
 
 Cl -I— X 
 
 2. The expression is a function of x and also a 
 
 ^ ft qi' 
 
 function of a. 
 
 When an expression contains several symbols, we may 
 select one of them for special consideration, and call the ex- 
 pression a function of that particular one. For instance, 
 although the expressions. 
 
DEFINITIONS. ' 29 
 
 a + bx^ + cx% 
 
 contain other symbols besides x, they are both functions 
 of X. 
 
 50. An Entire Function is one in wMcli the quan- 
 tity is used only in the operations of addition, subtrac- 
 tion and multiplication. 
 
 Example. The expressions 
 
 ax + y, 
 (a^ — if)a^ — (Z)2 ^y)x^ — x + d, 
 
 are entire functions of x. But the expressions 
 
 ax -\- y -, o /- 
 
 -^ and dVx 
 
 ax — y 
 
 are not entire functions of x, because in the one x appears as 
 part of a divisor, and in the other its square root is extracted. 
 An entire function of x can always be expressed as a sum 
 of terms, arranged according to the powers of x which they 
 contain as factors. The form of the expression will then be 
 
 A -\- Bx + Cx^ -{- Da^ -{- Ux^ -i- etc., 
 where A, B, G, etc., may represent any algebraic expressions 
 which do not contain x. 
 
 51. Like Terms are those wliicli are formed of the 
 same algebraic symbols, combined in the same way, 
 and differ only in their numerical coefficients. 
 
 Ex. The terms ax, 2ax, —hax are like terms. 
 
 52. Tlie Degree of any term is the number of its 
 literal factors. 
 
 Examples. The expression alxy is of the fourth degree, 
 because it contains four literal factors. 
 
 The expression a? is of the third degree, because the letter 
 X is taken three times as a factor. 
 
 The expression ah^x^ is of the sixth degree, because it con- 
 tains a once, l twice, and x three times as a factor. 
 
 When an expression consists of several terms, its 
 degree is that of its highest term. 
 
30 ALGEBRAIC OPERATIONS. 
 
 CHAPTER I. 
 
 ALGEBRAIC ADDITION AND SUBTRACTION. 
 
 Algebraic Addition. 
 
 53. By the language of Algebra, the sum of any number 
 of quantities, positive or negative, may be expressed by writing 
 them in a row, with the sign + before all the positive quan- 
 tities, and the sign — before the negative ones. 
 
 Ex. A-\-B—D—X-{- Y, etc., is the algebraic sum of the 
 several quantities A, B, —D, —X, Y, etc, 
 
 54. To simplify an expression of the sum of several 
 quantities. 
 
 1. 'VYhen dissimilar terms are to Ibe added, no sim- 
 plification can be effected. 
 
 Ex. If we require the sum of the five expressions, «, —xy, 
 mp, nq, and —ihs, we can only write, 
 
 a — xy -\- mp -^ nq — hhs, 
 according to the language of Algebra, and cannot reduce the 
 expression to a simpler form. 
 
 2. If mere numbers are among the quantities to be 
 added, their algebraic sum may be formed. 
 
 Ex. The sum of the five quantities —8, ab, 5, m7ip, —15, 
 is found to be — 18 + «5 -|- mnp. 
 
 3. When several terms are similar, add the coeffi- 
 cients and affix the common symbol to the sum. 
 
 When no numerical coefficient is written, the coefficient 
 + 1 or —1 is understood. (§ 44.) 
 
 EXAMPLES. 
 
 a + a = 2a [because 1 + 1 = 2]. 
 2a — a = a [because 2 — 1 = 1]. 
 
ALGEBRAIC ADDITION. 31 
 
 3rt + 4« — 7« = [because 3 + 4 — 7 = 0]. 
 
 a -\-'^x — ^a — bx= —2a—dx [adding the a's and the x's}. 
 
 — 3axy + 4:bm — %axy -^ bm = — baxy -{- bhm. 
 
 Add the expressions, 
 
 1. "ix + bhy\ 2x — dby% — ix — 5I)y% hx — hj^ x — hy\ 
 
 WORK. 
 
 For convenience, the several terms may be w^ . Kj.yi 
 
 written under each other, as in the margin. The n^ o iV.2 
 
 coefficients of x are 7, 2, —4, 5, and 1, of which a r,jj\2 
 
 the algebraic sum is 11. The coefficients of y^ k i^ ^ 
 
 are 5, —3, —5, —1, —1 ; the sum is —5. Hence 7^2 
 
 the result. . . ^- 
 
 Sum, llo; — 5%2 
 
 2. ^ax —y — 2y-\-5, 7ax—y—9-\-am, 2ax—y—'d-\-bp, 
 Here 2a!, am, and p, wokk. 
 
 all being different sym- ^ax^ _ y _^x + b 
 
 bols, the terms contain- tv ^ 'i^ g , ^^^^ 
 
 ing them do not admit _ ax^ — y — 3 + 5» 
 
 of simplification (§ 54, ■ '- 
 
 1). The numbers 5, Sum, — Sy — 2x — 7 + am + 5p 
 
 —9. —3, are added by 
 
 the rule (§ 54, 2). The coefficients of ax^ cancel each other (8— 7— 1 = 0). 
 
 3. Add Q{x + y), b{x + y)-{-a, 2{x + y)- 3a. 
 
 Here the aggregate, x + y, enclosed in 
 
 parentheses, is treated as a simple symbol. . \ 
 
 Note. When the student can add . " (a; -f- ?/} 
 
 the coefficients mentally, it is not neces- ^ "^ o 
 
 sary to write the expressions under each _2 ^ 
 
 other. Nor is it necessary to repeat the SuM 13 ix -\- 11^ 2(1 
 
 symbol after each coefficient. 
 
 EXERCISES. 
 
 1. 3a + 75 — 8c + J, 3a — 2b + c — e, —a — b — c—d. 
 
 2. 7a — (a; + y), 8a — (a; + y), 3 {x + y) — 16a. 
 
 3. W — 2x — b, 2x^ — dx + 8, — 9^2 _^ 5^ + 3. 
 
 4. x^ -\-2x — y, 4:X^ + llx — 2y, — 2x^' -^x — 9y, —3x^ 
 
 — X — y. 
 
 5. 9 (a + by, 10 (a + ^Y, {a + by, 2 {a + bY, -x-y-z. 
 
 6. 2 (m + ^0 + 3 (a + b), (a -{- b) — {ni + n), (a + b) 
 
 — (m + n), 
 
 7. 7a3— 2a2 + 3ax, — d^ — a^ — ax, — U^ + 3a2 — 2ax. 
 
32 ALGEBRAIC OPERATIONS. 
 
 8. {m + nf + ir, 2 (m + nf — y, 3 (m + nf — 2x, 
 (m + nY — y. 
 
 9- {P + ^)' - 6, (i? + qf + «, (j» + (7)^ + h {p + qf^c. 
 
 10. 6«(a; — 2/)> ^ai^ — y), ^(^(^ — y), ci{x — y). 
 
 11. 2(w — 7^^ + ^? 3(m + ^)^ — 5, b{m + n)x — G, 
 7 (m + w) a; — 8. 
 
 ''• ^^a' '^« + '^^»' 6J~^' ^~7' «"?• 
 
 13. ,2 2 — , 3 3— ,4 4 — 
 
 ^ y n y n y 'i^ y n 
 
 14. i+1 + 3-^^^, 5^i-^- + 7--±^. 
 wj + w m + ^^ m + ?2 m + /J 
 
 15. Of two farmers, the first had 2x — Zy acres, and the 
 second had x — y acres more than the first. IIow many acres 
 had they both? 
 
 16. A had 2a; dollars, B had y dollars less than A, and C 
 had 2y dollars more than A and B together. How many had 
 they all ? 
 
 17. A father gave his eldest son x dollars, his second 5 dol- 
 lars less than the first, his third 5 dollars less than his second, 
 and his fourth 5 dollars less than his third. How much did 
 he give them all ? 
 
 55. Addition with Literal Coefficients, When dif- 
 ferent terms contain the same symbol, mnltiplied Iby 
 diflferent literal coefficients, these coefficients may be 
 added and the common symbol be affixed to their 
 aggregate. 
 
 EXAMPLES. 
 
 1. As we reduce the polynomial 
 
 ijx -f- bx — 2a; 
 to the single term (6 + 5 — 2) a; = 3a;, 
 so we may reduce the polynomial 
 
 ax -{-Ix — ex 
 to the single term, (a -\-h — c)x. 
 
 2. The expression 
 
 mx -\- ny — Ix -{- dy -\- a + h 
 may be expressed in the form 
 
 {m — h) X -\- {n ■\- d) y -\- a -\- h* 
 
SUBTRACTION. 33 
 
 EXERCISES. 
 
 Collect the coefficients of x and y in the following ex- 
 pressions : 
 
 1. ax -^-hy -{■ mx + ny. 
 
 2. mux + "iliy -\-pqx — 4:iy. 
 
 3. 3x — 2y-{- 6bx — 4:y -\- '^ax -\- m -\- n. 
 
 4. Sax -f 8bx -\- by -\- 7x — 6y -{- X — 5y. 
 
 5. ax -\- hy -\- cz — mx — ny — 'pz. 
 
 6. %dx + Zey + 4/2 — %fx — Zdy + te. 
 2 3 
 
 8. %ax —by — ^bx — ^ay. 
 12, 1 3 
 
 2 1 
 
 10. 4tmx + 2y — Sax — Gcx + ay — -mx + - dx, 
 
 11. 5abx — S^nny — «Ja; + 4cf/?/ — dx. 
 
 12. 3«?/ + 2bx — -<7:7; + 2«?/ — 3bx. 
 1 3 
 
 14. 3;?7:t — ax — -ay -\- x -}- dx — y. 
 
 15. ^aix — my + 2c\/x — dy -\- Vx. 
 
 16. 6mVy — Gx -{- 4.Vy — 3 A/a; — y + V^. 
 
 17. 4^/^ — 6^ + aVy + ex— 's/y — ^a^/y + a/S. 
 
 Algebraic Subtraction. 
 
 56. Bef. Algebraic Subtraction consists in ex- 
 pressing the difference of two algebraic quantities. 
 
 Rule of Subtraction. It has been shown (§ 21) that 
 to subtract a positive quantity, &, is the same as to 
 add, algebraically, the negative quantity, —h. Also, 
 that to subtract — & is equivalent to adding +&. Hence 
 the rule : 
 
 Change the algebraio sign of all the teinns of the 
 subtrahend, or conceive them to be changed, and then 
 proceed as in addition. 
 
34 ALGEBBAIG OPERATIONS. 
 
 NUMERICAL EXAMPLES. 
 
 Min., 10 + 6 = 16 10+ 6=16 10+ 6=16 10+ 6 = 16 
 
 Subt., _9 _=_9 9— 4= 5 9- 8= 1 9 — 12 = -3 
 
 Eem., 1 + 6= 7 1 + 10=11 1 + 14=15 1 + 18= I9 
 
 ALGEBRAIC EXERCISES. 
 
 I. From dx — 4«?/ + 5^> + c, 
 
 Subtract x — '^ay — 8b-\-d. 
 
 WORK. 
 
 Minuend, Sx — 4:ay + 55 + c 
 
 Subtrahend with signs changed, — a; + Hay + 8b — d 
 
 Difference, ^x + day + 135 + c — 6? 
 
 Next we may simply imagine the signs changed. 
 
 2. From "^x — Uxy — l^cy + 85 + dac 
 Take 2x + Ihxy + 8cy — 55 — ^d 
 
 Diff., bx — llbxy — 4:cy + 135 + Sac + 2d 
 
 3. From 8a + 95 — 12c — 18d — 4a; + dcy 
 Take 19^ — 75— 8g — 25</ + 3a; - 4y 
 
 4. From 2572; + 2OI22 + 92y + 35«a; — 6 
 Take 140^ — 82^^ + 20,y + 92ax + 14 
 
 5. From Sa + 145 subtract Ga + 205. 
 
 6. From ^5 — 5 + c — t? take — « + 5 — c + 6?. 
 
 7. From 8rt — 25 + 3c subtract 4« — 65 — c — 2d, 
 
 8. From 2a;2 _ 8a; — 1 subtract 5x^ — Qx + 3. 
 
 9. From 4a;4 _ 3^ _ ^x^ _ r^^ ^ 9 subtract x^—2:t?—2x^ 
 + 7a; — 9. 
 
 10. From 2x^ — 2fl!a; + 3«2 subtract x^ — ax -\- .a\ 
 
 11. From a^ — da^ + 3«52 -- 5^ subtract — a^ + 3flr2^,. 
 
 12. From 7a;3 _ 2:^ -I- 2a; + 2 subtract 4a;3_2a;2— 2a;— 14 
 
 13. From 6{x — y)-{-7{x~z) + 9 (z—x) take 9 (a; — ?/) 
 + 7 (a; — ;?) + 5 (;? - a;). 
 
 14. From 12 (rt — 5) — 3 (« + 5) + 7a — 25 take 7 (a—b) 
 — 5 (a + 5). 
 
 15. From 7---ll^^~15- take _ 5- + 6 ^ - 7 - + 8^- 
 
 y z x y z x b 
 
SUBTRACTION. 35 
 
 Clearing of Parentheses. 
 
 57. In § 42, 2, it was shown that an aggregate of terms in- 
 cluded between jiarentheses might be added or subtracted by 
 simply writing + or — before the parentheses. 
 
 When an aggregate not multiplied by a factor is to be added 
 or subtracted, the parentheses may be removed by the rules 
 for addition and subtraction, as follows : 
 
 5S. Plus Sign he/ore ParentJieses. If the paren- 
 theses are preceded by the sign +, they may be 
 removed, and all the terms added without change. 
 
 Example i. 27 + (8-5-4 + 7) =27 + 8-5-4 + 7 = 33. 
 
 2. m-{-{a — X— y-\-z)z=im-\-a — X — y-\-z. 
 
 3. 2x + (_ 3a; - by) + {^y~4.a) + (2y-2«) 
 
 =. 2a; — Bic — 5?/ + 3y — 4fl + 2y — 2« 
 
 = — a: — 6a. 
 The sign + which precedes the parentheses should also be 
 considered as removed, but if the first term within the paren- 
 thesis has no sign, the sign + is understood, and must be 
 written after removing the parentheses. 
 
 EXERCISES. 
 
 Clear of parentheses and simplify 
 
 1. X — y + (x -{- y). 
 
 2. x + y + {y — x), 
 
 3. dab — 2rnp + (ab — 3x — 2mp). 
 
 4. 2ax — dby + (7nx — 2ax — pz + ^by). 
 
 59. Minus Sign be/ore ParentJieses. If the paren- 
 theses are preceded by the sign — , they may be 
 removed and the algebraic sign of each of the included 
 terms changed, according to the rule for subtraction in 
 § 56. 
 
 EXAM PLES. 
 
 I. 27 -(8 -5-4+ 7) =27 -8 + 5 + 4-7 = 21; 
 that is, 27 — 6 = 21. 
 
36 ALGEBRAIC OPERATIONS. 
 
 2. m— {— a— p-\-y + x)^m-{-a-\-p — y — X, 
 
 3. Za -\- X — {2a — 5a;) — (dx — a) = 3a -{- x — 2a + 5x 
 — {)x -{- a. 
 
 Simplifying as in § 54, this reduces to 2a — dx. 
 
 EXERCISES. 
 
 Clear the following expressions of parentheses and re4nc6 
 the results to the simplest form by the method of § 54. 
 
 1. ah — {m — dab + 2ax) — 7ab. 
 
 2. X — {a — x) -}• {x — a). 
 
 3. 25 -t- (J - 2c) - (5 + 2c). 
 
 4. 4:X — Sy -f 22; — (— 7ir 4- 5?/ — Sz) — (x — y), 
 
 5. "iax — '2dy — (Sax + 3by) — (Sax — 3%). 
 
 6. (a — x) — (a -h x) -\- 2x. 
 
 7. — (a — 1)) — (h — c) — (c — a). 
 
 8. — (3m + 2n) — (3m — 2n) + 9w. 
 
 60. We may reverse the process of clearing of parentheses 
 by collecting several terms into a single aggregate, and chang- 
 ing their signs when we wish the parentheses to be preceded 
 by the minus sign. The proof of the operation is to clear the 
 parentheses introduced, and thus obtain the original expression. 
 
 EXERCISES. 
 
 Eeduce the following expressions to the form 
 X — (an aggregate). 
 
 1. X — a — 5. Ans. x — (a -}- h), 
 
 2. X — m — n. 
 
 3. a + X — 3x -\- 2y. Ans. x — (— a -\- 3x — 2y), 
 
 4. —3b-\-x + 2c-\- bd. 
 
 5. 2x — 2a + 2b, Ans. x — (— x + 2a — 2b). 
 
 6. 2x -{■ a — b. 
 
 7. 3x — 2m + 2n. 
 
 8. 3x -\- ab — m — 3ab + 2m. ■ 
 
 9. x — 2m — (3a — 2b). Ans. x — (2m + 3a — 2b), 
 
 10. X -{■ 3 — (a -\- b). 
 
 11. X -^ a — (b — c) + (m — n). 
 
 12. X — (am + b) — (p — q) — (am — n), 
 
 13. X— (a + b) — (p — q) — (m — n). 
 
8UBTMAGTI0N, 37 
 
 Compound Parentheses. 
 
 61. When parentheses of addition or subtraction arc en- 
 closed between others, they may be separately removed by the 
 preceding rules. 
 
 We may either begin with the outer ones and go inward, 
 or begin with the inner ones and go outward. 
 
 It is common to begin with the inner ones. 
 
 E X AMPLES. 
 
 Clear of parentheses: 
 
 I. /_[«_{rf-[c_(j-«)]n. 
 
 Beginning with the inner parentheses, the expression takes, 
 in succession, the following forms : 
 
 f-\e- \d-{c-h + a-\\'] 
 = f—\_e—{d — c-\-h — aY\ 
 z=f—[e — d-\-c — h-{-a\ 
 = f— e + d — c-^-l) — a, 
 
 2. x—[—{a-\-l)-\- {m -hn) — {x — y)]. 
 Kemoving the inner parentheses, one by one, we have, 
 X — [— a — b + m -\- n — x -\- y] 
 = x + a-\-b—m — n -\- X — y. 
 
 EXERCISES. 
 
 Eemove the parentheses in the following expressions, and 
 combine terms containing x and y, as in §§ 54 and 55. 
 
 1. m+ l-(p-q) -^ (a-b) + (-c-\- d)]. 
 
 2. m ^[— (a — b) — (p + q) -{- {n — Jc)]. 
 
 3. Ilax — [(2ax + by) — (dax — by) + (— '7ax + 2%)]. 
 
 4. a — [a — \a — [a — (a — a)]\^. 
 
 5. p — [a — b— {s -{- t + a) + (—m — n)]. 
 
 6. 2ax — [3ax — by — {7ax + 2by) — (6ax — 3%)]. 
 
 7. ax + by-\- cz + [2ax—dcz — {2cz-\-5ax) — {7by—dcz)], 
 
 8. X— \2x — y — [Sx — 2y — (ix — 3y)] \. 
 
 9. ax — bz — "I ax + bz — [ax — bz — (ax + bz)] } . 
 10. my — {x + 3y + [2my — 3 {x — y) — 4a^] + 5 }. 
 
38 ALGEBRAIC OPERATIONS. 
 
 II 
 
 12 
 
 14 
 IS 
 
 ax + ^cx — {mx -\- ex — y) + ['>yix — {ex + ^)]. 
 dax — Sbx — ( — day — 3^^; + dby) — 3bz. 
 Idax + 2xy — d— ['7ad f (.r«/ -}- ^)] — -ia:?^; 
 m + 4a; — [— 4^ H- 2a; + (ay — x) -\- p]. 
 2aVy — 3m — [bVx — 6/^ 4- (V^ — ^a/^)]- 
 
 CHAPTER II. 
 
 MULTIPLICATION. 
 
 62. The product of several factors can always be 
 expressed by writing them after each other, and enclos- 
 ing those which are aggregates within parentheses. 
 
 EXAMPLES. 
 
 The product ol a -\- i hj c = c (a -j- b). 
 The product of —^ hj x — y = (x — y) --—-' 
 The product of « + 5 by c + ^ = [e ■\- d)(a -^ b). 
 Such products may be transformed and simphfied by the 
 operation of algebraic multiplication. 
 
 General Laws of Multiplication. 
 
 63. Law of Commutation. Multiplier and multi- 
 plicand may be interchanged without altering the 
 product. 
 
 This law is proved for whole numbers in the following way. 
 Form several rows of quantities, each represented by the 
 letter a, with an equal number in each row, thus, 
 
 a a a a a a 
 
 a a a a a a 
 
 a a a a a a 
 
 a a a a a a 
 
 a a a a a a 
 
MULTIPLICATION, 39 
 
 Let m be the number of rows, and n the number of a's in 
 each row. Then, counting by rows there will be 
 m X w quantities. 
 Counting by columns, there will be 
 n X tn quantities. 
 Therefore, m x n = n x m, 
 
 or nm = mn, 
 
 64. Law of Association, Wlien there are three 
 factors, m, n, and a, 
 
 m {no) = (mn) a. 
 
 Example. 3 x (5 x 8) = 3 x 40 = 120. 
 (3x5)x8 = 15 + 8 r= 120. 
 
 Proof for Wliole Numhers, If a in the above scheme 
 represents a number, the sum of each row will be na. Because 
 there are m rows, the whole sum will be m {no). 
 
 But the whole number of a's is inn. Therefore, 
 
 m {na) = (mn) a. 
 
 65. The Distributive Law. The product of an ag- 
 gregate by a factor is equal to the sum of the products 
 of each of the parts which form the aggregate, by the 
 same factor. That is, 
 
 m {p -\- q_ -\- r) = mp -\- mq + mr, (1) 
 
 Proof for Wiole Numhers. Let us write each of the quan- 
 tities jo, q, r, etc., m times in a horizontal line, thus, 
 
 p + p + p ■{- etc., m times = mp. 
 
 q + q + q -\- etc., m times = mq. 
 
 r -{• r + r -\- etc., 7n times = mr. 
 
 etc. etc. etc. 
 
 i 
 If we add up each vertical column on the left-hand side, 
 
 the sum of each will \)q p ■\- q -\- r -\- etc., the columns being 
 all ahke. 
 
 Therefore the sum of the m columns, or of all the quanti- 
 ties, will be 
 
 m{p -\- q -\- r^ etc.). 
 
40 ALGEBBAIG OPERATIONS. 
 
 The first horizontal line of ^'s being mp, the second mq, 
 etc., the sum of the right-hand column will be 
 
 mp + mq ■{- mr, etc. 
 
 Since these two expressions are the sums of the same quan- 
 tities, they are equal, as asserted in the equation (1). 
 
 Multiplication of Positive Monomials. 
 
 66. Rule of Exponents. Let us form the product 
 
 By § 37, x"^ means xxx, etc., taken m times as factor, 
 cc" means xxx, etc., taken n times as factor. 
 
 The product is xxxxx, etc., taken {m-\-n) times as factor. 
 Therefore, a;"' x ^" = x'^'^\ 
 
 Hence, 
 
 Theorem. The exponent of tlie product of like sym- 
 Ibols is the sum of the exponents of the factors. 
 
 67. As a result of the laws of commutation and 
 association, the factors of a product may be arranged 
 and multiplied in such order as will give the product 
 the simplest form. 
 
 68. Any product of monomials may be formed by 
 combining these principles. 
 
 Example. Multiply bmn^x^y^ by "^Inx^y. 
 
 By the rules of algebraic language, the product may be put 
 into the form 
 
 ^mn^x^ifl hnx^y. 
 
 By interchanging the factors so as to bring identical sym- 
 bols together, 
 
 h-'^hmn^noi^x^y^y. 
 
 Multiplying the numerical factors and adding the exponents, 
 the product becomes 
 
 d6lmn^3i^y^. 
 
MULTIPLICATION. 41 
 
 69. "We thus derive the following 
 
 Rule. Mivltiply the numerical coefficients of the 
 factors, a,ffix all the literal parts of the factors, and give 
 to each the sum of its exponents in the separate factors. 
 
 EXERCISES. 
 
 1. Multiply xy by x^y. Ans. x^y\ 
 
 2. Multiply dax by 2abx^. 3. Multiply 5m^y by dm^x. 
 4. Multiply 21my by 2a^m. 5. Multiply 2am. by 2ma. 
 6. Multiply 6x'^yh by x^yh. 7. Multiply 3xyz by Sxyz. 
 
 8. Multiply 2abm by 2ml)a. 9. Multiply Sab^x^ by da^^x, 
 
 TO. Multiply 2'Qmpqr by 2-6pqrs. 
 
 11. Multiply 12axy by 122:?/;2. 
 
 3 2 3 
 
 12. Multiply - m^ii^ by -??i^^'. 13. Multiply -r/i^ by 4m^. 
 
 7 
 14. Multiply - adcd by 4(^e/^. 
 
 70. When we have to find the product of three or more 
 quantities, we multiply two of them, then that product by the 
 third, that product again by the fourth, and so on. 
 
 Ex. 2al) X 2a^ x Sab'^ X dbmxy = 36a^b^mxy. 
 Exercises. Multiply 
 
 15. mxxmyxmz. 16. axxixxcxxdx. 
 
 1 7. Sa^m X ^Wn x mn. 18. abx 2bc x 7c«. 
 
 19. 3m7i^ X 5np^ X 9pm^'. 
 
 20. abxacxadx arnS xyx 2yz x zx. 
 
 21. amx X anx X amxy X anxy x amxyz. 
 
 22. a^x X a^y X ax^ X ay'^ X a^x^ X a^y"^ X 2 
 
 23. 2am X dan xa^x m^ x ^mx x 2nx. 
 
 Rule of Signs in Multiplication. 
 
 *71. It was shown in § 25 that a product of two factors is 
 positive when the factors have like signs, and negative when 
 they have unlike signs. Hence the rule of signs, 
 
 + X + makes +, 
 
 + X - " -, 
 
 - X + " -, 
 
 — X - " +. 
 
42 
 
 ALGEBBAIG 0PEBATI0N8, 
 
 Examples. The quantity a 
 
 Multiplied by 3 makes + 3a. 
 
 The quantity — a 
 
 (( 
 
 • (( 
 
 2 
 
 i( 
 
 H-2«. 
 
 a 
 
 tt 
 
 1 
 
 a 
 
 + a. 
 
 tt 
 
 t( 
 
 
 
 it 
 
 0. 
 
 ti 
 
 a 
 
 -1 
 
 a 
 
 — a. 
 
 tt 
 
 (( 
 
 — 2 
 
 a 
 
 — 2a. 
 
 — a 
 
 
 
 
 
 tiplie( 
 
 iby 
 
 3 
 
 makes 
 
 -3«. 
 
 i( 
 
 (( 
 
 2 
 
 a 
 
 — 2a. 
 
 tt 
 
 ti 
 
 1 
 
 a 
 
 — a. 
 
 tt 
 
 (I 
 
 
 
 a 
 
 0. 
 
 tt 
 
 it 
 
 — 1 
 
 it 
 
 + a. 
 
 tt 
 
 ti 
 
 -2 
 
 ti 
 
 ■{-2a. 
 
 72. Geometrical Illustration of the Rule of Signs. Suppose 
 
 the quantity a to represent a length of one centimetre from 
 
 the zero point toward the right on the scale of § 11. 
 
 Then we shall have 
 
 
 
 a = this line | j 
 
 The product of the line by the factors from -4-3 to —3 
 will be 
 
 a X 3, 
 
 a X 2, 
 
 « X 1, 
 
 « X 0, 
 
 a X — 1, 
 
 « X — 2, 
 
 a X — 3, i" 
 
 We shall also have 
 
 a = this line 
 
MULTIPLICATION. 43 
 
 The products by the same factors will be 
 
 
 
 — « X 3, 
 
 — « X 3, 
 
 — « X 1, 
 
 — « X 0, 
 
 
 
 
 
 — « X — 2, I I I 
 
 
 
 — a X — 3, I I I i 
 
 These results are embodied in the following two theorems : 
 
 1. Multiplying a magnitude by a negative factor, 
 multiplies it by the factor and turns it in the opposite 
 direction. 
 
 2. Multiplying by —1 turns it in the opposite direc- 
 tion without altering its length. 
 
 Note. When more than two factors enter a product, the 
 sign may be determined by the theorem, § 26. 
 
 EXERCISES. 
 
 am X db X ac X ad. 2. ax x ^hx x ex x dx, 
 
 X X —ax X — ahx x — ahcx. 
 
 dax X —2a^^ x — 5a^mx. 
 
 — '7m^y X — 3c^y x 5 ax. 
 
 —22izn X —5n^x'^ x —n^yz — a?^, 
 
 2m X n X —a x —2b. 
 
 —Sax X —2km x —Ix x —4:hmx. 
 
 —nyx gy x —2y x 3bm. 
 
 xy X 2y^ X y^x x 2ayxK 
 
 6y^ X —Sgy x —2z^ x —ax^z. 
 
 6ax X anx x 3^; x Ir^xy. 
 
 —4tl)z X —xz X —yz X agz. 
 
 2(^n X 2x^z X —z^ X —bgz^ 
 
 —e^x X Sx X eW- x ay, • 
 
44 ALGEBRAIC OPERATIONS. 
 
 i6. — 2e X — 2y X a X Ix. 
 
 17. — 4a:» X ^ay x —%c?y x — ^y. 
 
 18. «^:?; X — ay^ x rf X —x^y. 
 
 19. aa;2 X — «/^ X —1 X ^ax x — «^?/. 
 
 20. wi2:c X — w^^ X —mn^ x wi.t x —m\ 
 
 21. — fl^Sa; X — ay^ x ax x a^x^. ' 
 
 22. px^ X qy^ X xy X —ax. 
 
 23. aic X —cP X ax^ X —1 X Sax. 
 
 24. 2^x X Sex X —-^mx X — 4?/2 x 6m. 
 
 25. —6m:?; X —'^n^x x -^ac x —-zm\ 
 '^ 6 5 
 
 26. —a X ic X —1 X 2 X da^ X A.xy x y, 
 
 27. —1 X ax X a^x X a^x^ x hx x d. 
 
 28. —an X 2am^ x —Smn x ^n^y x —m, 
 
 29. —mx X nx X —mn x —xy x — 1. 
 
 30. —2px X -Sqx X -xmH X ^y^ X — 1. 
 
 Products of Polynomials by Monomials. 
 
 73. The rule for multiplying a polynomial is given by the 
 distributive law (§ 65). 
 
 KuLE. Multiply each temn of fhe polynoTnial hy the 
 monoinial, and talce the algebraic sum of the products. 
 
 Exercises. Multiply 
 
 1. dx^ — 4:xy — 5y^ by — 4:ax. 
 
 Ans. — 12ax^ + IQax^y + 20^2:/. 
 
 2. dx^ — xy -}- y^ by Sx. 
 
 3. x^ -\- xy -\- y^ by Sx. 4. ax -\- ly ■\- cz by axyz. 
 5. Saa^-bay'^—'l by ^cibx. 6. 4mj? — Qnq by — Smq. 
 7. ha^y^ — 7ay — la^y by ^ah. 
 
 74. The products of aggregates by factors are formed 
 in the same way, the parentheses being removed, and 
 each tgrm of the aggregate multiplied by the factor. 
 
MULTIPLICATION. 45 
 
 Example. Clear the following expression of parentheses : 
 
 am {a — h -{- c) — p[a — (h — Ic) — m(a — b)]. 
 
 By the rule of § 73, the first term will be reduced to 
 
 a^m — amd -\- amc. (1) 
 
 The aggregate of the second term within the large paren- 
 theses will be 
 
 a — 7i-\-k — m(a — I)) 
 
 = a — h -{- k — ma + mb, (2) 
 
 because, by the rule of signs in multiplication, 
 
 — mia — h) = — m x a — 7n x —b = — ma + ink 
 
 Multiplying the sum (2) by —p and adding it to (1), we 
 have for the result required : 
 
 a^m — a7nb + amc — pa + ph — ph -\- pma —pmb. 
 
 EXE RCI SES. 
 
 Clear the following expressions of parentheses : 
 
 1. p(a -^m — jj) -\- q{b — c) — r{b -{- c). 
 
 2. (m — an) x — {m + an) y + {an — m) z. 
 
 3. a{x — y)c — b{x — y)d -{-f{x + y) ccl. 
 
 Here note that tlie cocflBcient of aj — y in the first term is ae. 
 
 4. am \x — a{b — c)] — bn [ax + b{c -\- d)]. 
 
 5. p[—a {m-\-n) + b(m—n)] — q [b (m—n) —a {m + n)]. 
 
 6. dx {2q — 7ic) + 2y {5x — 3c)—z (2m + Hn). 
 
 7. am [m (a — b)c — 3h (2k — 4:d) + 4w]. 
 
 8. 2pq[da-'5b — 6c—pq(2m — 3n)]. 
 
 9. bn [-7a-7b{a-c)-{3-a- b)]. 
 10. p(q — r)-\-q{r — p)-\-r{p — q). 
 
 75. The reverse operation, of summing several terms into 
 one or more aggregates, each multiplied by a factor, is of fre- 
 quent application. Thus, in § 65, having given 
 
 mp + mq + mVf 
 we express the sum in the form 
 
 m (p -\- q -^ r). 
 
46 ALGEBRAIC OPERATIONS. 
 
 The rule for the operation is 
 
 If the sum of several terms having a eomfnon factor 
 is to be formed, the coefficients of this factor may he 
 added, and their aggregate he multiplied hy the factor. 
 
 Note. This operation is, in principle, identical with that of § 55. 
 
 EXAMPLES. 
 
 alx — hex — ady -\-3dt/—dI}X-\-4:ady-{- ?nt/ — amy —danx 4- imx. 
 
 Collecting the coefficients of x and y as directed, we have 
 (a^ — ic — db — 3cm + bm) x + (—ad-\-3d + 4^ad-\-m—am) y. 
 
 Applying the same rule to the terms within the parentheses, 
 
 we find 
 
 aI) — hc — 3b = b{a — c — 3). 
 
 — 3cm + bm = m.(b — 3c). 
 
 — ad-{-3d + 4:ad = 3ad + 3d 
 
 = {3a -{-3)d 
 
 = 3 (a -h 1) d 
 
 m — am =^ m (1 — a). 
 
 Substituting these expressions, the reduced expression 
 becomes 
 Yb{a — c — 3)^7n(l) — 3c)] x + \3 (a -\-l)d + m{l — «)] y. 
 
 ' The student should now be able to reverse the process, and 
 reduce this last expression to its original form by the method 
 
 of § 74. 
 
 EXERCISES. 
 
 In the following exercises, the coefficients of y, z, and 
 their products are to be aggregated, so tliat the results shall 
 be expressed as entire functions of x, y, and z,. as in § 55. 
 
 1. ax + bx — 3ax + 3bx -{- Qx — 7:r. 
 
 Arts. (—2a + U — l)x. 
 
 2. my + py — my — 2py — 3gy. 
 
 3. mx — ny + px ~ gy -\- rx — sy. 
 
 Ans. (m + p + r) X — (n + g + s) y. 
 
 4. 3az — «/ — 2az -{- z — az ■\- y. 
 
MULTIPLICATION. 47 
 
 5. dbxy — Icxy -\- hdxy. 
 
 6. 36abxy — 24rr — ax — 7xt/. 
 
 7. ay — by — may — nby -f 3^. 
 
 8. aw?/ — Z^w?/ + any — hiy. 
 
 9. ^r^; — ^qrz — 4ppz + Sqhz. 
 10. c?ia; -I- Jwa: — amy — 2b7iy, 
 
 76. An entire function of two quantities can be regarded 
 as an entire function of either of them (§§ 49, 50), and when 
 expressed as a function of one may be transformed into a func- 
 tion of the other. 
 
 Example. The expression 
 
 (2a + d)a^— (4^2 _ 2a) x^ ^ (d^ — 2a -\-l)x — a^ 
 
 has the form of an entire function of x. It is required to 
 express it as an entire function of a. 
 Clearing of parentheses, it becomes 
 
 2aa^ + 32;3 — U^x^ + 2ax^ + a^x — 2ax + x — a\ 
 
 Now, collecting the coefficients of a^, a^, etc., separately, it 
 becomes 
 
 (_ ^x^ ^x — 1) a^ + {2x^ + 2x'^ — 2x) a + 3:^^ + x, 
 which is the required form. 
 
 EXERCISES. 
 
 Express the following as entire functions of y : 
 
 1. (^y^—4:y)x^-}-{y^—2y^ + l) x^+{2y^-^6i/—'7)x—y^—6, 
 
 2. {y^ — y^) x^ -{- (y^ — y) X -{■ y^ — 1. 
 
 3. lys _ 2^/3) a^+ (yi — 2y^) 2-2 + (2/ — 2y) x + y'^ — 2, 
 
 4. (2/5 + 32/^) x^-\-(f^- By') x^ + (y^ + By) x^ + {y^ + B) x. 
 
 Multiplication of Polynomials by Polynomials. 
 
 77. Let us. consider the product 
 
 {a + J) (^ 4- g + r). 
 
 This is of the same form as equation (1) of § 65, {a + I) 
 taking the place of m. Therefore the product just written is 
 equal to 
 
 {a-\-l)p + (a + J)g+ (a + &)r. 
 
48 ALGEBRAIC OPERATIONS. 
 
 But {a + h)p =1 np -{- dp. 
 
 {a + b)q = aq-{- hq. 
 (a -\- b) r =^ ar + b7\ 
 Therefore the product is 
 
 ap -{- bp -{- aq + bq -{- ar -i- br. 
 
 It would have been still shorter to first clear the paren- 
 theses from (a + b), putting the product into the form 
 
 a{P + q + r) + b (p + q + r). 
 
 Clearing the parentheses again, we should get the same 
 result as before. 
 
 We have therefore the following rule for multiplying aggre- 
 gates : 
 
 78. Rule. Multiply each term of the multiplicand 
 hy each term of the multiplier, and add the coefficients 
 with their proper algebraic signs. 
 
 EXERCISES. 
 
 1. {a + b) {2a — bn^ — 2bn% 
 
 2. {a — b) (dm -\- 2n — babmn). 
 
 3. {m^ — n^) {2mn + pm + qn). 
 
 4. {p^ + <?^ + ^2) (pq + qr -t rp), 
 
 5. (2« — U) {2a f 2b). 
 
 6. {mx — ny) {mx + ny). 
 
 T9. It is frequently necessary to multiply polynomials 
 containing powers of the same letter. In this case the begin- 
 ner may find it easier to arrange multiplicand, multiplier, and 
 product under each other, as in arithmetical multiplication. 
 
 Ex. I. Multiply W — ^x^ -\- bx — 4. by Zx^ — 4.x — 5. 
 
 The first line under work. 
 
 the multiplier contains 7a.'^ — 6x^-\-6x — 4 
 
 the products of the sev- g^2 4^; 5 
 
 eral terms of the multipli- 
 
 candby8«2. The second 21a;5_l8.^'4+15.^■3— 12.?;2 
 
 contains the products by —28x*-\-24:X^—20x^-{-l^X 
 
 -Ax, and the third by ~5. —35a;3^ 30^:2— 25a; + 20 
 
 Like terms are placed ; —— ~ rr 
 
 under each other to facil- 21x'-4.Gx'+ 4.X^- 2x^- 9a: + 20 
 itate the addition. 
 
MULTIPLICATION. 49 
 
 Ex. 2. Multiply m -\- nx + px^ hy a — hx. 
 
 m -\- nx -\- px^ 
 a — bx 
 
 am + anx + apx^ 
 
 — dmx — bnx^ — bpa^ 
 
 am + (an — bm) x + {ap — bn)x^ — bp7? 
 
 In the fallowing exercises arrange tlie terms according to 
 the powers and products of the leading letters, a, b, x, y, or Zm 
 
 Multiply 
 
 1. 3«2 -f 5a + 7 by 2«2 _ 3^ ^ 4. 
 
 2. a^ + ab ■\- 1^ \)y a — b. 
 
 3. a^ -\- a^ -\- ax^ -\- a? hj a — x. 
 
 4. a^ — a^ + a — 1 hy a^ — a + 1. 
 
 5. a^ + «a:^ + «%2 + «% 4- «^ by a: — a. 
 
 6. a + J2; + c^2 + di^ by m — 7^2; + pz\ 
 
 7. 3a2 f 5a + 7 by 2«2 + 3a — 4. 
 
 8. a2 _ a^ + Z>2 by « + Z>. 
 
 9. «3 + aH -{■ ax^ -\- 01^ by a — x. 
 
 10. a^ — a^ + a — Ihj a^ + a — 1. 
 
 11. ic^ + fl2:3 + a^rc^ 4- a^a; + a* by a; + ff. 
 
 12. a + ^2 + c;2;2 + ^2;^ by m + W2f — pzK 
 
 13. (a + bx) (m + wa;). 
 
 14. {a -^ bx -^ cx^) [m -{• nx ■{- px^). 
 
 16. (2/'i + |/2 + 2/ + l)(2/2 + 3^ + l) 
 
 17. (2/' - ^'' + 3?/ - 4) (^3 + 2?/2 + 3^ + 4). 
 
 18. 3a2'«a; — ^a^y + 2«2« by «"* — «'* 
 
 19. a2 ^ 6«^> 4- -5 by a — -J. 
 
 20. (a 4- Z») 4- (a _ b) by (a 4- ^) — (^5 — ^)- 
 
 21. «2 _ ^,2 _j. (a _ ^) by fl2 ^ J2 ^ (a + J). 
 
 22. a -{- b -\- c hj a — b -\- c. 
 
 23. «2 ^ j2 _ (3^2 _|_ z^) by 2a 4- 25 — 2 (a - J). 
 
 24. 2 (a — Z*) 4- a: — 2/ by a 4- 5 — (a; 4- 2^). 
 
 25. ax'^ 4- 5a:" — abx by aa:;^ _j_ ^^^ 
 
 26. «»' _ J» by a'" 4- 5". - 
 
 4 
 
60 ALGEBRAIC OPERATIONS. 
 
 27. — l^x^y + ^xy^ — 12y^ by — 6xy, 
 
 28. Ix^ + dax — \a^ by 2o^ — ax — -a\ 
 
 KoTE. Aggregates entering into either factor should be 
 simplified before multiplying. 
 
 Special Forms of Multiplication. 
 
 80. 1. To find the square of a binomial, as a + 5. "We 
 multiply « + J by a + 5. 
 
 a {a ^ V) =^ c^ -\- db 
 bla-^l) = ab -^-W- 
 
 {a + i){a-\-i) = a^ + 2ab + b^ 
 Hence, {a + bf = a^ + 2ab -\- l^ (1) 
 
 2. We find, in the same way, 
 
 (a — by = a^ — 2ab + bK (2) 
 
 These forms may be expressed in words thus : 
 
 Theorem. The square of a binomial is equal to the 
 sum of the squares of its two terms, plus or minus twice 
 their product. 
 
 3. To find the product of « + 5 by a — &. 
 
 a{a -\-b) = a^ -\- ab 
 ^bla + b) = -ab-^ 
 Adding, {a + b){a-b) =a^- bK (3) 
 
 That is : 
 
 Theorem. The product of the sum and difference of 
 two numbers is equal to the difference of their squares. 
 
 The forms (1), (2), and (3) should be memorized by the student, owing 
 to their constant occurrence. 
 
 When ^ — 1, the form (3) becomes 
 
 {a + 1) (« - 1) = a2 - 1. 
 
 The student should test these formulae by examples like 
 the following : 
 
 (9 + 4)3 = 92 + 2.9.4 + 42 = 81 + 72 + 16 = 169. 
 (9 - 4)2 = 92 - 2.9.4 + 42 = 81 - 72 + 16 = 25. 
 
MULTIPLICATION. 51 
 
 (9 _^ 4) (9 _ 4) _ 92 _ 4. = 65. 
 
 Prove these three equations by computing the left-hand 
 member directly. 
 
 EXERCI SES. 
 
 Write on sight the values of 
 
 I. (m + 'inf. 
 
 2. (m — 2nY. 
 
 3. {^a-ibf. 
 
 4. {4.x -6yy. 
 
 5. (2x + y) {2x - y). 
 
 6. {3x + 1) {3x - 1). 
 
 7. (4a;2 + 1) (4a;2 _ i). 
 
 8. {5a^ - 3) (5:^:3 _^ 3). 
 
 81. Because the product of two negative factors is positive, 
 it follows that the square of a negative quantity is positive. 
 
 Examples. {— af = a:^ = (+ af. 
 
 {p _ af =. a^ — 'Zab -^y^ z= {a- h)\ 
 
 Hence, 
 
 The expression c? — %ab -\-l^ is the square both of 
 a — b and of b — a. 
 
 83. We have — « x « = — a^. 
 
 Hence, 
 
 The product of equal factors with opposite signs is a 
 negative square. 
 
 Example. — [a — b){a — b) — — a^ + 2ab — b^, 
 which is the negative of (2). Because — {a — b) z:^ b — a, 
 this equation may be written in the form, 
 
 {l-a){a — b) = —a^ + 2ab — V^, 
 which is readily obtained by direct multiplication. 
 
 EXERCISES. 
 
 Write on sight the values of 
 
 1. — (a + ^) X — (« + b). 
 
 2. {p^ — y){y — ^)- 3. {^ + y){-^ — y)- 
 
 4. (3« — Zb) (3b — 2a). 5. (3b — 2a) (— 3b + 2«). 
 
 6. (am — bn) (bn — am). 7. (xy — 2) (2 — xy). 
 
52 ALGEBRAIC OPERATIONS, 
 
 CHAPTER III. 
 
 . DIVISION. 
 
 83. The problem of algebraic division is to find such an 
 expression that, when multiplied by the divisor, the product 
 shall be the dividend. 
 
 This expression is called the quotient. 
 
 In Algebra, the quotient of two quantities may always be 
 indicated by a fraction, of which the numerator is the divi- 
 dend and the denominator the divisor. 
 
 Sometimes the numerator cannot be exactly divided by the 
 denominator. The expression must then be treated as a frac- 
 tion, by methods to be explained in the next chapter. 
 
 Sometimes the divisor will exactly divide the dividend. 
 Such cases form the subject of the present chapter. 
 
 Division of Monomials by Monomials. 
 
 84. In order that a dividend may be exactly divisi- 
 ble by a divisor, it is necessary that it shall contain the 
 divisor as a factor. 
 
 Ex. I. 15 is exactly divisible by 3, because 3'5 = 15. 
 2. The product alt^c is exactly divisible by ac, because ac is 
 a factor of it. 
 
 To divide one expression by another which is an exact 
 divisor of it; 
 
 EuLE. Remove from the dividend those factors the 
 product of which is equal to the divisor. The remain^ 
 ing factors will he the quotient. 
 
 85. Rule of Exponents. If both dividend and divisor 
 contain the same symbol, with different exponents, say m and 
 n, then, because the dividend contains this symbol m times as 
 a factor, and the divisor n times, the quotient will contain it 
 m — n times. Hence, 
 
DIVISION, m 
 
 In dividing, exponents of like synibols are to he sub- 
 tracted. 
 
 EXERCISES. 
 
 1. Divide 26.^?/ by 2y. Ans. l^x, 
 
 2. Divide na?bc by 75c. 
 
 3. Divide 7^ by x^. Ans. x, 
 
 4. Divide ISa^ by 6«. Ans. 3a. 
 
 5. Divide Iba^m by 3«. Ans. bam. 
 
 6. Divide 14«3y/^2 i^y 7^577^. 
 
 7. Divide 16a5/7^4 by ^ahn\ 8. Divide 36a;t^%3 by Qxyz. 
 9. Divide 40a2a;3^5 by IWxz!^. 10. Divide 35«Z>5 by lab. 
 
 Rule of Signs in Division. 
 
 86. The rule of signs in division corresponds to that in 
 multiplication, namely : 
 
 If dividend and divisor have the same sign, the quo- 
 tient is positive. 
 
 If they have opposite signs, the quotient is negative. 
 
 Proof. 
 
 •\-mx -V- { + m) = -{-X, because 4-a; x {-\-m) = -\-mx. 
 
 •^mx -^ (— m) = — X, " — X X (— ^0 = ^mx. 
 
 — mx -7- i + ni) =■ —X, " —X X ( + w) = — mx. 
 
 — mx -r- (— m) = -\-x, " -i-x X ( — w) = — 7nx. 
 
 The condition to be fulfilled in all four of these cases is 
 that the product, quotient x divisor, shall have the same alge- 
 braic sign as the dividend. 
 
 EXERCISES. 
 
 Divide 
 
 1. + « by + a. 
 
 2. + « by — a. 
 
 3. —ahy+a. 
 
 4. —a by — a. 
 
 5. — 33ahnx by llax. 
 
 6. — 2^xhjz by 12xyz. 
 
 7. 21a??iV" by — 7amx\ 
 
 Ans. 
 
 + 1. 
 
 Ans. 
 
 -1. 
 
 Ans. 
 
 -1. 
 
 Ans. 
 
 + 1. 
 
 Ans. — 
 
 Sam. 
 
 Ans. 
 
 -2x. 
 
 Ans. —3mx^"-\ 
 
54 ALOEBRAIG OPERATIONS. 
 
 8. — 18a^/;^ by — Qa^p. Ans. Sa^-^p"'-K 
 
 9. — IQa^x'^y"' by Aax^y^, 
 
 10. Ub'pi by — 7^>^^s'. 
 
 11. — lUH^'Jc'^ by — ^lH'^lc\ 
 
 12. 12 (a — Vf (^ by 3 (a — Vf c. Ans, 4 (a — J) c^. 
 
 13. 42 (a: - «/)^ by - 7 {x - y)\ 
 
 14. — 44a* {x — yY by lla^ (x — y)K 
 
 15. — 45^*^ (a — Z')'^ by W {a — ly. 
 
 16. — 48 {tn + n)P by — 8 (m + w)9'. 
 
 17. 64 {a + 5)^^ (a; — y)^ by 4 (a + ^) (^ — y)- 
 
 Division of Polynomials by Monomials. 
 
 87. By the distributive law in multiplication, whatever 
 quantities the symbols m, a, h, c, etc., may represent, we have : 
 
 {a -\- h -{■ c -\- etc.) X m = y/za + wz5 + wc + etc. 
 Therefore, by the condition of division, 
 
 (wa + ml + mc + etc.) -T-m = a + 5 + c + etc. 
 We therefore conclude, 
 
 1. In order that a polynomial may l)e exactly divisi- 
 ble by a monomial, each of its terms must be so 
 divisible. 
 
 2. The quotient will be the algebraic sum of the 
 separate quotients found by dividing the different terms 
 of the polynomial. 
 
 EXERCISES. 
 
 Divide 
 
 I. 2^2 + daH — MH^ by 2«2. Ans. 1 + Zax — ^aHK 
 
 Gm^n — 12m^n^ — ISmn^ by 6mn, 
 
 8aW — 16^4^4 4- 8a^b^ by 4:a^^ 
 
 4:xy^ — Sx^y^ -\- 4x^y hj — Axy. 
 
 12aix — 24:al)x^ by — 12aix. 
 
 21am^X^ — Ua^m^a^ + 28a^m^x^ by — Hamxf^. 
 
 12a^x + 24«:c + 4Sax^ by 24^52:. 
 
 a(b — c) + b (c — a) -\- c {a — b) -\- abc by «Jc. 
 
 27 (« - by — 18 (« — by + 9 (a — by by 9 (flj - b). 
 
 am (^ _ J^)n _ gn (^ _ lyn by ^w (^ _ J)n, . 
 
DIVISION. 65 
 
 11. {a + ly (a - hf + {a + lf (a-hy by (a-^l) {a-l). 
 
 12. 10 {x + y)"'{x — yY — 5 (a; + yy {x — yf 
 
 by 5 (a; + y) {x - y), 
 
 13. (« + Z*) (« - I) by ^2 _ J2. 
 
 Factors and Multiples. 
 
 88. As in Arithmetic some numbers are composite and 
 others prime, so in Algebra some expressions admit of being 
 divided into algebraic factors, while others do not. The latter 
 are by analogy called Prime and the former Composite. 
 
 A single symbol, as a or a;, is necessarily prime. 
 
 A product of several symbols is of course composite, and 
 can be divided into factors at sight. 
 
 A binomial or polynomial is sometimes prime and some- 
 times composite, but no universal rule can be given for dis- 
 tinguishing the two cases. 
 
 89. When the same symbol or expression is a factor of all 
 the terms of a polynomial, the latter is divisible by it. 
 
 EXAMPLES. 
 
 1. ax ■\- abx^ + d?cx^ := a {x -\- bx^ -{- acx?), 
 
 2. aWx -\- d^Wy? = aWx (b + ax). 
 
 3. a^ + a"ic« = a^ (a"' + x^). 
 
 EXERCISES. 
 
 Factor 
 
 I. ax^ + a^x. 2. a^^cy + a^c^y + ah^c^y. 
 
 3. fl2n jTi ^ ^n pn^ 4, ^3w ^n _ ^2» ^271 ^ ^n ^n^ 
 
 5. a^¥^c^ -^ a^^b^^c^ + a^^h^c'^^. 
 
 90. There are certain forms of composite expressions 
 which should be memorized, so as to be easily recognized. 
 The following are the inverse of those derived in § 80. 
 
 1. a2 + 2ab + b\= {a -\- b)\ 
 
 2. «2 _ 2ab + J2 = (« - b)\ 
 
 3. a'-b''= {a^ b) {a - b). 
 
 The form (3) can be applied to any difference of even 
 powers ; thus, 
 
66 ALQEBBAIG OPERATIONS, 
 
 a^-b'^ (a^ + IP) (a2 _ h^) . 
 
 fl« - J« = (a3 + J3) («3 _ ^3) . 
 
 and, in general, a^"" — ^>2» = (a^ + l)"") (a^ — z^'*). 
 
 If the exponent is a multiple of 4, the second factor can be 
 again divided. 
 
 EXAM PLES. 
 ^8 _ ^,8 :^ (^44. ^,4) («4 _ J4) ^ (<24 + J4) (^^+^,2) (^ ^ J) {a-V). 
 
 When 5 is equal to 1 or 2, the forms become 
 
 a^-lz= (a + l){a — 1). 
 052 _ 4 z^ (« + 2) {a — 2). 
 
 «2 + 2a + 1 = (« + 1)2. 
 a2 + 4a + 4 = (« + 2)2. 
 
 ^2 _ 2a 4- 1 = (« - 1)2 = (1 - a)\ 
 flr2 — 4a + 4 = (a - 2)2 = (2 — af. 
 
 By putting 2^ for h, they give 
 
 «2 _ 4^2 = (a + 2Z.) (a — 25). 
 a2 + 4a5 + 4&2 = (a + 2^>)2. 
 
 EXERCISES. 
 
 Divide the following expressions into as many factors as 
 possible : 
 
 I. ^ — 16. ^W5. (a;2 + 4) (a: + 2) (a; — 2). 
 
 a;2 + 62: + 9. ^;^5. (re + 3)1 
 
 a;2 _ 6a; + 9. 
 
 4fl2^2 _ 9J2^2. 6, 16^4^ _ 1, 
 
 9a;2 - l^xy + 4^2. g, ^2^2 ^ 2aa:?^ + y\ 
 
 4a2a;2 + 4aJa;«/ + lpy\ 10. a* + 4a2^>2 + 4*2. 
 
 a4 _ 2.'c2?/2 _|. y, 12. a;4 — 4a:2y2 4. 4^. 
 
 ^4 _ 4^2^2 ^ 4J4. 14. ^4 _ ^2^2. 
 
 flSn _ 2a« + 1. 16. a;2« — 4aa;» + 4a2. 
 
 3^Z + 2a;3//3;2 + fz. 
 
 Ans. z (a;« + 2a;3^3 _^ ^) = ^; (a;3 ^ ^3)2. 
 
DIVISION. 67 
 
 19. «3 _ 4^2j ^ 4rtj2, 20. d^ — ¥^. 
 
 21. ^bx^— ^Oa^y H- 1 6xY- 2 2 . ^x^y^ — "dxhf. 
 
 2T^. ^x/^y^ — 12a;y + 9a:l 24. a? — x^y^, 
 
 25. :c4?/i _ 2a;^^^ + ?/2». 26. rc^m _ 2a;2»i + 1. 
 
 27. 2:2 4- a; + J. 28. a:^"* ^ x"^ + -- 
 
 91. By combining the preceding forms, yet other forms 
 may be found. 
 
 For example, the factors 
 
 (^2 ^ab + b^) («2 -ab + b% (1) 
 
 are respectively the sum and difference of the quantities 
 Or + l^ and ab. 
 
 Hence the product (1) is equal to the difference of the 
 squares of these quantities, or to 
 
 (^2 ^ J2)2 _ ^2J2 =: «4 ^ ^2J2 _p J4. 
 
 Hence the latter quantity can be factored as follows : 
 
 fl4 ^ ^2^2 4- ^,4 3^ (^2 _|_ «§ 4_ ^,2) (^2 _ «J 4_ J2). 
 EXERCISES. 
 
 Factor 
 
 I. ar* + xY + y^. 2. a* ^ 8«2J2 _^ 1(3 j4. 
 
 3. ft* 4- 9«2a;2 -I- 81a:*. 4. «*^ + «2» ^2« _^ J4n^ 
 
 5. a^ar^ + 4:aWx^ + le^^^a:^. 6. «« + 8a^b^ + 16a2J*. 
 
 7. a^ + x^^y^^ + a:'*^^ 
 
 8. m^ — 05^ -j_ 2a^ — W. Ans. (m — a + b) {m + a — b). 
 Here the last three terms are a negative square. Compare § 82. 
 
 9. a^ — 4:b^ -}- 4:bc — c^. 10. a^ _ 4^j2 _|_ 4^^^ — ^^^ 
 
 92. The following expression occurs in investigating the 
 area of a triangle of which the sides are given : 
 
 (a + b + c){a + b — c){a — b + c){a — b — e). (1) 
 
 By § 80, 3, the product of the first pair of factors is 
 
 {a + by —c'^ = a^ + 2ab + TJ^— & ; 
 
 and that of the second pair, 
 
 (a _ j)2 _ c2 = a2 - 2ab + b^ - c". 
 
68 ALQEBBAIG OPERATIONS. 
 
 By the same principle, the product of these products is 
 {cfi + ,&2 - c2)2 _ 4.aW, 
 which we readily find to be 
 
 a^^h^ + d^— 2aW — Wg^ — 2c^a\ (2) 
 
 Hence this expression (2) can be divided into the four 
 factors (1). 
 
 Factors of Binomials. 
 
 93. Let us multiply 
 a;»-i + ax^-^ + «2^^-3 + + oT'-^x + ft'^-i by x — a. 
 
 OPERATION. 
 Qfl-^ + «a;^-2 _|_ ^2^71-3 _|_ ^3^-4 ^ ^ ^n-2^ ^ ^«-l 
 
 X — a 
 
 xn 4- ax-^-^ 4- «2^c^-2 + a%^-3 + + a'l-i a; 
 
 _ (^a;^-i — ^2^:^-2 — a^a:^-^ — — a^^'^x — a"' 
 
 Prod., ce'* — a^ 
 
 The intermediate terms all cancel each other in the product, 
 leaving only the two extreme terms. 
 
 The product of the multiplicand hj x — a is therefore 
 afi — a^. Hence, if we divide x^ — a^ by x — a, the quotient 
 will be the above expression. Hence the binomial x"^ — a"- 
 may be factored as follows : 
 
 xr^ — a^= (x — a) (x''-'^ -\- ax"^-^ -}- a^x^'^ -}- -^a'^-^x + a''-^). 
 
 Therefore we have, 
 
 Theorem. The difference of any power of two num- 
 bers is divisible by the difference of the numbers 
 themselves. 
 
 Illustration". The difference between any power of 7 
 and the same power of 2 is divisible by 7 — 2 = 5. For 
 instance, 
 
 72 _ 22 :::z 45 = 5.9. 
 
 7^ — 23 = 335 =. 5.67. 
 
 T — %^ = 2385 = 5.477. 
 
 etc. etc. etc 
 
DIVISION. 69 
 
 94. Let us multiply 
 
 by ic + a = a; — (— «). 
 
 Rem. This expression is exactly like the preceding, except 
 that — « is substituted for a. It will be noticed that the 
 coefficients of the powers of x in the multiplicand are the 
 powers of — a, because 
 
 (-ay=-a, 
 (- ay = + a^ 
 (-.aY=—a% 
 {-ay=+aS 
 etc. etc. 
 
 The sign of the last term will be positive or negative, 
 according as n — 1 is an even or odd number. 
 
 OPEBATIOK. 
 
 a:»-i_fl5a;n-2 _^ a^x»'-^—a^x»'-^ + .... + {—ay-^x-\- {—a^-^ 
 
 X -\- a = X — { — a) 
 
 xn — ax^-^ 4- a2^«-2 — a^x"^-^ . . . + (— «)^~^ 
 
 + a^?"-^ — a^x^-^ + a^x^'^ .... — (— fl?)^~^ x — {—aY 
 Prod., a:'* — {—aY 
 
 The multiplier a; + a is the same as x — (— a) (§ 59). 
 In multiplying the first terms, we use + a, and in the last 
 ones — {— a), because the latter shows the form better. 
 
 Hence, reasoning as in (1), the expression x^ — (— aY 
 admits of being factored thus : 
 x"' — (— aY = {x -^ a) [a;'*-! — ax^-"^ + a^x^-^ — 
 
 + (— aY'^x + (— «)«-!]. 
 
 If n is an even number, then (— aY = «^j and 
 
 xn — (_ aY = X"' — a^ 
 
 If n is an odd number, then (— a^) = — a% and 
 
 x^ — (—aY = x^-\- a^' 
 Therefore, 
 
 Theorem 1. When n is odd, tlie binomial 3?^+ a" is 
 divisible by a; + a. 
 
60 ALOEBRAIG OPERATIONS. 
 
 Theorem 2. When n is even, tlie binomial x^—a^ is 
 divisible by ic+ a. 
 
 Note. These theorems could have been deduced imme- 
 diately from that of § 93, by changing a into — a, because 
 X — a would then have been changed to x -\- a, and x^ — a^ 
 to x^ + a^ or x"- — a^, according as n was odd or even. 
 
 The forms of the factors in the two cases are : 
 
 When n is odd, 
 r^n j^a"" = {x + a) (a;»-i — a7^-^ + ^2^-3 _ . . . . +«n-i). 
 
 When n is even, 
 a;» _ flw _ (a; _j_ a) (.t^-i — cra;^-2 + a^jf'-^ — . . . . ~a^-i). {a) 
 
 In the latter case, the last factor can still be divided, be- 
 cause x^ — a^ is divisible by a; — a as well as by a; + a. We 
 find, by multiplication, 
 (x — a) (a;«-2 + a2a;7i-4 _|_ ^4^n-6 -f .... 4- ^^-2) 
 
 — a;n-i _ rta;n-2 _|_ ^2a;n-3 _ ^3^n-4 -f . . . . + a^-«a; — a^-i. 
 
 Therefore, from the last equation {a) we have : 
 When n is even, 
 ic^ — a^ = (a;H-ff) {x—a) {x^^-^ ■\- a^x''-^ -\- a^x""-^ — .... + a^-2). 
 
 EXERCISES. 
 
 Factor the following expressions, and when they are purely 
 numerical, prove the results. 
 
 2). 
 
 I. 52 - 22. 
 
 
 
 
 Ans. (5 + 2) (5 
 
 [Proof. 
 
 52-22 
 
 = 
 
 25- 
 
 - 4 = 21 ; 
 
 (5 + 2) (5 - 
 
 3) 
 
 = 7-3 = 21.] 
 
 2. 53 — 23. 
 
 
 
 3. 
 
 54 — 24. 
 
 4. 55 - 25. 
 
 
 
 5- 
 
 5« - 26. 
 
 6. 73 + 23. 
 
 
 
 7. 
 
 73 - 23. 
 
 8. 74 - 24. 
 
 
 
 9- 
 
 X^ - «2. 
 
 10. 01? — a\ 
 
 
 
 II. 
 
 x^ — a*. 
 
 12. x^ — a^ 
 
 
 
 IS- 
 
 x^ + a3. 
 
 14. a^ -f a\ 
 
 
 
 IS- 
 
 a3 _ 8^3. 
 
 16. 8^3-27^3. 
 
 
 
 17. 
 
 16«4 - J4. 
 
 18. x^-{- Sf. 
 
 
 
 19. 
 
 x^ - m/. 
 
 20. 8^3 + 27^3. 
 
 
 
 21. 
 
 x^ — Ua\ 
 
DIVISION, 61 
 
 Least Common Multiple. 
 
 95. Def. A Common Multiple of several quanti- v 
 ties is any expression of which all the quantities are 
 factors. 
 
 Example. The expression amhi^ is a common multiple of 
 the quantities a, m, n, am, amn, mn?, m^n^f etc., and finally of 
 the expression itself, am^n\ But it is not a multiple of a^, nor 
 of X, nor of any other symbol which does not enter into it as a 
 factor. 
 
 Def. The Least Common Multiple of several 
 quantities is the common multiple which is of lowest 
 degree. It is written for shortness L. C. M. 
 
 KuLE FOE FINDING THE L. C. M. Factov the several 
 quantities as far as possible. 
 
 If the quantities have no connnon factor, the least 
 common multiple is their product. 
 
 If several of the quantities have a common factor, 
 the m^ultiple required is the product of all the factors, 
 each of them being raised to the highest power which it, 
 has in any of the given quantities. 
 
 Ex. I. Let the given quantities be 
 
 ^ab, Wc, Qac. 
 
 The factors are 2, 3, «, h, and c. The highest power of h is 
 ^, while a and c only enter to the first power. Hence, 
 
 L. C. M.=6a^c. 
 
 Ex. 2. «2 _ j2^ ^2 ^ 2ab + l^, c? — ^ab + ^, a* — b*. 
 
 Factoring, we find the expressions to be, 
 
 (a + b){a- b\ {a + bf, {a - bf, {a^ +1^' {a + b) {a - b). 
 
 By the rule, the L. C. M. required is 
 
 (a-\-bY(a-bf{a^-\-l^),- 
 
62 ALGEBRAIC OPERATIONS. 
 
 EXERCISES. 
 
 FindtheL. CM. of 
 
 I. xy, xZf yz. 2. a^if l^Cf (?d, ^a, 
 
 3. a, dby abc, abed, 4. a^, a¥, Ic^, 
 
 5. x^-y% x^y, x — y. 
 
 6. a^ — 4:, X^ — 4:X -\- 4:, X^ -\- 4rX -\- 4t. 
 
 7. 16a2^ — 4:7n% 2ax + w, 2aa: — 7n. 
 
 8. i2:2 — 1, a;2 + 1, x^ — 2x-{- 1, a;^ + 2ic + 1. 
 
 9. 4« (b + c), 5 (« — c), 2ab. 
 
 10. 2 (« — 5)2, 2{a + bf, 2{a — h){a-\- b). 
 
 11. 3 (x + 2/), 3 (a; - y), 3 (^^^ + ?/3). 
 
 12. « — 2>, a2 _ ^2^ ^3 _ J3^ ^4 _ ^,4. 
 
 15. X + y, X — y, a -{- b, a — b. 
 14. a^ — a*, ic^ + a^, ic^ — a\ x -\- a, 
 1.5. a:^ _ 64^6^ ^ _ 15^4^ ^2 _ 4^2. 
 
 16. a + b, «2 4- 2ab + d^, a^ — b\ 
 
 Division of one Polynomial by another. 
 
 If the dividend and divisor are both polynomials, and entire 
 functions of the same symbol, and if the degree of the numer- 
 ator is not less than that of the denominator, a division may 
 be performed and a remainder obtained. The method of 
 dividing is. similar to long division in Arithmetic. 
 
 96. Case I. Wlien there is only one algebraic sym- 
 hoi in the dividend and divisor. 
 
 Let us perform the division, 
 
 3;^4 _ 4^3 ^ 2:^2 ^ 3a; — 1 -^ t^ — x-^I. 
 
 We first find the quotient of the highest term of the divi- 
 sor x^, into the highest term of the dividend 3a;*, multiply the 
 whole divisor by the quotient 32;^, and subtract the product 
 from the dividend. We repeat the process on the remainder, 
 and continue doing so until the remainder has no power of x 
 so high as the highest term of the divisor. The work is most 
 conveniently arranged as follows : 
 
- x^- 
 
 x^ + 3x- 
 
 X^— X 
 
 -1 
 
 > ~~ 
 
 2x^ + ^x- 
 2x^ -{-2x- 
 
 -1 
 
 -2 
 
 DIVISION. 
 
 Dividend. Divisor. 
 
 3rc* — 4a;3 + 22;2 + 3a; — 1 \ x^ — x + l 
 
 8a;' X Divisor, U"^ — 3x^ + ^ ^^ 32^ _ a; — 2 Quotient. 
 
 First Remainder, 
 —X X Divisor, 
 
 Second Remainder, 
 
 —2 X Divisor, 
 
 Tliird and last Remainder, 2x -\- \ 
 
 The division can be carried no farther without fractions, 
 because x^ will not go into x. We now apply the same rule as 
 in Arithmetic, by adding to the quotient a fraction of which 
 the numerator is the remainder and the denominator the 
 divisor. The result is, 
 
 x^ — x + 1 ^ x^—X-\-l ^' 
 
 This result may now be proved by multiplying the quotient 
 by the divisor and adding the remainder. 
 
 There is an analogy between the result {a) and the cor- 
 responding one of Arithmetic. An algebraic fraction hke (a), 
 in which the degree of the numerator is greater than that of 
 the denominator may be called an improper fraction. As in 
 Arithmetic an improper fraction may be reduced to an e7itire 
 iiumher plus a proper fraction, so in Algebra an improper frac- 
 tion may be reduced to an entire function of a symbol plus a 
 proper fraction. 
 
 EXERCISES. 
 
 Execute the following divisions, and reduce the quotients 
 to the form (a) when there is any remainder. 
 
 1. Divide o? — 2x — 1 by a; + 1. 
 
 2. Divide .^3 -f- 2a;2 — 2a: — 1 by a; — 1. 
 
 3. Divide a;^ _ 3,^ ^ 2a: — 1 by a:^ — x. 
 
 T, , 2x^ — 27?-\-x^ — x — h 
 
 4. Reduce 
 
 a:^ — a: — 1 
 
 5. Divide 24^3 _ 38a2 _ '^2a + 50 by 2a — 3. 
 
 35 5 
 
 Atis. Quot. = 12ft2 — a — — ; Rem. = — «* 
 
 Z At 
 
64 ALQEBBAIG OPERATIONS. 
 
 6. Divide a:^ — 1 by a; — 1. 
 
 When terms are wanting in the dividend, they may be considered as 
 zero. In this last exercise, the terms in o^, x-, and x are wanting. But 
 the beginner may write the dividend and perform the operation thus : 
 
 ic* + 0x3 + oa;2 + Oa; - 
 
 -1 
 
 1 x-\ 
 
 X*- X? 
 
 
 aj3 + «« + a; + 1 
 
 X^- X? 
 
 
 
 X' + Oa; 
 
 
 
 X'^- X 
 
 
 
 x-\ 
 
 
 The operation is thus assimilated to that in which the expression is 
 complete ; but the actual writing of the zero terms in this way is un- 
 necessary, and should be dispensed with as soon as the student is able 
 to do it. 
 
 7. Divide c^ — '%a-\-\ by « — 1. 
 
 8. Divide a;^ + 1 by .'T + 1. 
 
 9. Divide W + 125 by 2rt -+ 5. 
 
 10. Divide a^ + 1 by a + 1. 
 
 1 1. Divide «* + 2flj2 _|_ 9 by ^2 ^ 2« + 3. 
 
 12. Divide «« — 1 by a^ ^ %a^ + 2a -f- 1. 
 
 13. Divide 7^ — \%:f^ + ?>^x^ — 32 by x^ — 2. 
 
 14. Divide (a^ — 2xi- 1) (x^ — 12a; — 16) by x^ — 16. 
 
 For some purposes, we may equally well perform the operation by 
 beginning with the term containing the lowest power of the quantity, 
 or not containing it at all. Take, for instance, Example 9 : 
 
 125 + 8a3 I 5 + 2a 
 
 l^-'j' + 50fl? 25 - 10a + 4a'^ 
 
 — 50a 
 
 -50a -20^2 
 
 20a^ + 8^3 
 20fl^ + 8a3 
 
 15. Divide 1 + 3a; + 3a.'2 + x^ by 1 -f a;. 
 
 16. Divide 1 — 4a; + ix^ — a^ hy 1 — x. 
 
 17. Divide 15 + 2a — 3a^ + a^ + 2a* — a^hyB + ia — a\ 
 
 18. Divide 1 — 1/^ by 1 + 2y + 2?/2 -}- if. 
 
 19. Divide 64 — 64a;-|-16a;2— 8a;3^4^4_^ by — 4 + 2a; + a;2. 
 
 20. Divide 64 — 16a:2 4- 2;6 by 4 — 4a; + xK 
 
DIVISION. 65 
 
 97. Case II. When there are several algebraic sym- 
 hols in the divisor and dividend. 
 
 Let us suppose the dividend and divisor arranged according 
 to powers of some one of the symbols, which we may suppose 
 to be X, as in § 76. 
 
 Let us call A the coefficient of the highest power of x in 
 the dividend, and ^ the term independent of ic, so that the 
 dividend is of the form 
 
 Ax'^ + (terms with lower powers of x) + H, 
 
 Let us call a the coefficient of the highest power of x in 
 the divisor, and Ji the term of the divisor independent of x, so 
 that the divisor is of the form 
 
 ax'>^ + (terms with lower powers of x) + Ti, 
 
 Then we have the following 
 
 Theorem. In order that the dividend may Ibe exact- 
 ly divisible by the divisor, it is necessary : 
 
 1. That the term containing the highest power of x 
 in the dividend shall be exactly divisible by the cor- 
 responding term of the divisor. 
 
 2. That the term independent of x in the dividend 
 shall be exactly divisible by the corresponding term of 
 the divisor. 
 
 Reason. The reason of this theorem is that if we suppose 
 the quotient also arranged according to the powers of x, then, 
 
 1. The highest term of the dividend, Ax^, will be given by 
 multiplying the highest term of the divisor, a^P^, by the high- 
 est term of the quotient. Hence we must have, 
 
 AoIj^ 
 Highest term of quotient = -— -• 
 
 cix? 
 
 2. The lowest term of the dividend will be given by multi- 
 plying the lowest term of the dividend by the lowest term of 
 the quotient. Hence, we must have, 
 
 TT 
 
 Lowest term of quotient = ^• 
 
 Eem. 1. Since we may arrange the dividend and divisor 
 according to the powers of any one of the symbols, the above 
 5 
 
66 ALGEBRAIC OPERATIONS. 
 
 theorem must be true whatever symbol we take in the place 
 of X. 
 
 Rem. 3. It does not follow that when the conditions of 
 the theorem are fulfilled, the division can always be performed. 
 This question can be decided only by trial. 
 
 We now reach the following rule : 
 
 I. Arrange both dividend and divisor according to 
 the ascending or descending powers of some common 
 symhol. 
 
 II. Form the first term of the quotient hy dividing 
 the first term of the dividend hy the first term of the 
 divisor. 
 
 III. Multiply the whole divisor hy the term thus 
 found, and suhtract the product from the dividend. 
 
 IV. Treat the remainder as a new dividend in the 
 same way, and repeat the process until a remainder is 
 found which is not divisible hy the quotient. 
 
 Ex. I. Divide a^ + ^ctx^ + ^ci^x + a^ by a; + a. 
 
 OPEKATION. 
 
 \ X ■]- a 
 
 trS-f- ax^ 
 
 3^ + 2ax + «2 
 
 2ax^ + Sa^x 
 
 
 2ax^ + 2a^x 
 
 
 d'x + «3 
 
 
 a^x + «3 
 
 
 
 
 
 Divide x^ — ax^ -\- a {i -i- c) x — alc—lx- 
 
 Ex. 2. Divide x^ — ax^ -]- a (b -i- c) x — abc—bx^—cx^+bcx 
 by X — a. 
 
 Arranging according to § 76, we have the dividend as follows: 
 x^ — (a + b + c) x^ + {ab -\-bc-{-ca)x — abc \ x — a 
 7? — ax^ 
 
 — {b + c)x^ -{- {ab-\-bc + ca)x 
 
 — (b + c)x^-\- (ah + ac)x 
 
 bcx — abc 
 box — abc 
 
 x^ — (b + c)x + be 
 
FRACTIONS. 67 
 
 EXERCISES. 
 
 1. Divide the dividend of Ex. 2 above by x — h. 
 
 2. Divide the dividend of Ex. 2 above by q: — c. 
 
 3. Divide a^ ■\-¥— (^ + dabc by a -\- b — c, 
 
 4. Divide a^ -^b^ + Sab — l by a -{- b — 1, 
 
 5. Divide a^Iy^ + ^abx^ — (a^ + ^) ^^ by «5 + (« — ^) ^. 
 
 6. Divide (^2 — Jc)3 + S¥c^ by a^ + j^. ' 
 
 7. Divide (« + 5 + c) (ah -{- be -}- ca) — dbc by a + 5. 
 
 8. Divide \a -{-b — c){b ^ c — a){c + a — h) 
 
 by a2 — J2 _ c2 -1- 2bc. 
 
 9. Divide «» + 5^ + c^ — 3«5c hj a -\- b + c, 
 
 10. Divide ic^ + 4a* by a;^ — 2aa; + 2a\ 
 
 11. Divide «2 ( j _|. a;) _ J2 (^j _ ^j) _l_ (cf _ j) a?^ + abx 
 
 by a; + a + 5. 
 
 12. Divide a^ — ax^ — b'^x + aW' by (a: — a) {x -\- b). 
 
 13. Divide 12a*a;^ — l^a^x^ + 6a®a;^ -— a' by Sa^a;^ — a^. 
 
 CHAPTER IV. 
 
 OF ALGEBRAIC FRACTIONS. 
 
 98. Bef. An Algebraic Fraction is the expression 
 of an indicated quotient when the divisor will not ex- 
 actly divide the dividend. 
 
 Example. The quotient of ^ -7- g' is the fraction -• 
 
 Def. The numerator and denominator of a frac- 
 tion are called its two Terms. 
 
 Transformation of Single Fractions. 
 
 99. Reduction to Lowest Terms. If the two terms 
 of a fraction are multiplied or divided by the same 
 quantity, the value of the fraction will not be altered. 
 
68 ALGEBRAIC OPERATIONS. 
 
 CLQC 
 
 Example. Consider the fraction — • If we divide both 
 
 ay 
 
 terms by a, the fraction will become -• 
 
 ax _ X 
 ay ~ y 
 Corollary. If tlie numerator and denominator con- 
 tain common factors, they may be cancelled. 
 
 Def. When all the factors common to the two 
 terms of a fraction are cancelled, the fraction is said to 
 "be reduced to its Lowest Terms. 
 
 To reduce a fraction to its loivest terms, factor both 
 
 terms, when necessary, and cancel all the common 
 
 factors. 
 
 ^ alxy^ hx 
 
 Ex. T. ^ = 
 
 acny^ en 
 
 The factor ay^ common to both terms is cancelled. 
 aW a^ 
 
 Ex. 2. -YfTj = Ti- 
 
 a^^ b^ 
 
 The factor a^b^ common to both terms is cancelled. 
 
 ^ ^ ^ a^x 
 
 Ex. 3. Reduce -^-* 
 ^ d}x 
 
 Here a^x is a divisor of both terms of the fraction. Di- 
 
 1 cfioc 1 
 
 Tiding bj it, the result is -s- Hence -y- = -g- 
 a a X a 
 
 gi ^<jiabj^ yi _ [a 4. j^y g j^ j, 
 
 a^' — H^ ~ {a + b){a — b)~ a — b 
 
 Ex. 4. 
 
 -r, mu — nu (m — n)u u 
 
 Ex. 5. = 7 ^ = -• 
 
 mx — nx {in — n) x x 
 
 EXERCISES. 
 
 Eeduce the following fractions to their lowest terms 
 
 a^l^p^ am 
 
 I. — - — • 2. • 
 
 a^¥p ' ahnx 
 
 lOpqr^ 12axy 
 
5- 
 7. 
 9 
 II 
 
 15 
 17 
 
 FRACTIONS. 69 
 
 'i'i{a — x){b — c) 20 (a + y) (m — ti) 
 
 36 («2 _ 2«a: + a;2)* * 24 (a^— 2aa; + a;2)(m:^)" 
 
 ayj-by^ 8. ^¥ — % ^ 
 
 aa; — 6a; ' (ly — 6^ 
 
 d^ — W' a^ + 4fl;a; + 4^ :^ 
 
 «(^ + F)' ^^* ay^ly 
 
 c? + «J - 
 
 a;2n _ iy2n 
 
 «^ — ^>^ a2 + ab + <^2 
 
 ^^T ^.- i6. 
 
 x" — y^ x^ -\- y"^ 
 
 axm — «a:)« „ wa? — wa; 
 
 1 8. 
 
 bym — byn {a -\- b) (ttt — n) 
 
 100. Rule of Signs in Fractions. Since a fraction is an 
 indicated quotient, the rule of signs corresponds to that for 
 division. The following theorems follow from the laws of 
 multiplication and division : 
 
 1. If the terms are of the same sign, the fraction is 
 positive ; if of opposite signs, it is negative. 
 
 2. Changing the sign of either term changes tlie 
 sign of the fraction. 
 
 3. Changing the signs of both terms leaves the frac- 
 tion with its original sign. 
 
 4. The sign of the fraction may be clianged by 
 changing the sign written before it. 
 
 5. To these may be added the general principle that 
 an even number of changes of sign restores the fraction 
 to its original sign. 
 
 Ex. I. 
 Ex. 2. 
 Ex. 3- 
 
 a — a —a a 
 
 
 b~ -b~~ b ~ -b 
 
 
 a —a —a a 
 
 
 b~ -b~ b ~ -b 
 
 
 a — b 1) — a a — b 
 
 b-a 
 
 
 
 m — n n ^m n — m 
 
 m — n 
 
70 ALGEBRAIC OPERATIONS. 
 
 EXE RCISE S. 
 
 Express the following fractions in four different ways with 
 
 respect to signs : 
 
 X — y X — y 
 I. ^. 2. f- 
 
 m a 
 
 p—q ^ a—l +c 
 
 m — n ^ a + m — x 
 
 p + q — r a — m -\- x 
 
 "Write the following fractions so that the symbols x and y 
 shall be positive in both terms : 
 
 c-y 
 
 a -\- x — b 
 
 8. 
 
 10. 
 12. 
 
 m — X 
 n — y 
 a — x 
 
 a — x + b 
 x — a + b 
 
 b^x 
 a -{-b — X 
 
 b — X a — b + y 
 
 101. When the numerator is a product, any one or 
 
 more of its factors can be removed from the numerator 
 
 and made a multiplier. 
 
 ^ abmx , mx ^ x ,1 
 
 Ex. = ab = abm = abmx 
 
 P + q P -hq 2^ + q P + q 
 
 EXERCI SES. 
 
 Express the following fractions in as many forms as possi- 
 ble with respect to factors : 
 
 pax ab abc 
 
 I. — ^— • 2. — • ^. • 
 
 mn ' c a -\- b 
 
 x^ — y^ a^ — ¥ r ^— 16«^ 
 
 a — b X ' X -\- 2a 
 
 102. deduction to Given Denominator. A quan- 
 tity may be expressed as a fraction with any required 
 denominator, D, by supposing it to have the denomi- 
 nator 1, and then multiplying both terms by D. 
 
 For, if we call a the quantity, we have a = - = -jr- ' 
 
FRACTIONS. 
 
 71 
 
 Ex. If we wish to express the quantity db as a fraction 
 having xy for its denominator, we write 
 
 aixy 
 xy ' 
 
 If the quantity is fractional, both terms of the 
 fraction must be multiplied by that factor which will 
 produce the required denominator. 
 
 Ex. To express t with the denominator nh^, we multiply 
 both members by n¥ -~b = nH^. Thus, 
 a _ anH^ 
 
 This process is the reverse of reduction to lowest terms. 
 
 EXERCISES. 
 
 Express the quantity 
 
 I. a with the denominator b. 
 
 2. 
 
 3- 
 
 4- 
 
 5- 
 6. 
 
 7- 
 8. 
 
 9- — -T 
 
 ax 
 db 
 m 
 n 
 
 — 1 
 
 m (n —p) 
 a + b~ 
 
 X — y 
 x^ + 1 
 X -\-l 
 
 a-1 
 
 a 
 
 ax. 
 
 i( ( 
 
 * «J^ 
 
 a i 
 
 ' n{x — y). 
 
 i( i 
 
 ' X, 
 
 (( 
 
 a? - b\ 
 
 i( ( 
 
 x^ — y\ 
 
 a i 
 
 a;2 + 2a; + 1. 
 
 a ^ 
 
 a^ - 1. 
 
 Negative Exponents. 
 
 103. By the principle of § 85, we have 
 
 % = «"'*• 
 
 If we have ^ > w, the exponent of the second member of 
 the equation will be negative, and the first member, by can- 
 
72 ALOEBBAIG OPERATIONS. 
 
 celling n factors from each term of the fraction, will become 
 -j—-' Hence -j—- = a'^'K 
 
 By putting for shortness h — n== s, the equation will be 
 
 i = a". 
 Hence, 
 
 A negative exponent indicates the reciprocal of the 
 corresponding quantity luith a positive exponent. 
 
 If in the formula a^~^ = — ^ we suppose h z= n, it will 
 
 become «o = — , or a^ = 1. Hence, because a may be any 
 
 quantity whatever, 
 
 Any quantity with the exponent is equal to unity. 
 
 This result may be made more clear by sue- a^ ■=. aaa 
 
 cessive divisions of a power of a by a. Every ^2 __ ^^ 
 
 time we effect this division, we diminish the ex- j 
 ponent by 1, and we may suppose this diminution 
 
 to continue algebraically to negative values of a =: 1 
 
 the exponent. On the left-hand side of the _j 1 
 
 equations in the margin, the division is effected ^ 
 
 symbolically by diminishing the exponents ; on -, 
 
 the right the result is written out in the usual a~^ = — 
 
 way. ^^ 
 
 etc. etc. 
 
 EXERCISES. 
 
 In the following exercises, write the quotients which are 
 fractional both as fractions reduced to their lowest terms, and 
 as entire quantities with negative exponents, on the principle, 
 
 etc. 
 
 Ans, X. 
 Ans. - or cc~i. 
 
 X 
 
 a 
 
 V\\ 
 
 « 7, 1 
 
 Vide 
 
 | = «^^-. 
 
 I. 
 
 x^ by X. 
 
 
 2. 
 
 X by x\ 
 
 
 3- 
 
 - 2¥ by h\ 
 
 
 4. 
 
 4«53 by — 2a^h 
 
 Ans. 
 
 ?f or - U-W, 
 
FRACTIONS. 73 
 
 5. — ^a% by ^al)\ 6. l^aWxij by 4«5ar. 
 
 7. 14fl^^^>V by — 1d?¥c^. 8. Uapqxy by ISa^c. 
 
 9. — ZQa^p^x^y by — 24a%?/. 
 
 10. 48«2 (a; — y)2 by 36 (a; — y). 
 
 11. 42^^ ^4by20^-±-^y. 
 
 \x — yf *' \2; — ?// 
 
 12. 22 (flj — h) (m — n) by 15 {a J^b){m -\- n). 
 
 13. 25 (a2 _ J2) (^2 _ ^2) by 15 {a — i) (m + w). 
 
 14. (2;^ _ 1) (^2 _ 4^2) by (a;2 -l)(a+ 21?). 
 
 15. a:^ — 1 by a;2 + 1. 
 
 16. a^Ki^y^ by a^^x^yl 
 
 17. m^n^y'^z by mn^y^A 
 
 18. m (m + 1) (m + 2) (m + 3) by w (m— 1) (m— 2) (m— 3). 
 
 19. tt"* by a^ 20. «5^c^ by ^J«c»*. 
 
 Dissection of Fractions. 
 
 104. If the numerator is a polynomial, each of its 
 terms may be divided separately by the denominator, 
 and the several fractions connected by the signs + or — . 
 
 The principle is that on which the diyision of polynomials 
 is founded (§ 87). The general form is 
 
 iL±^-±£±^ = ^ + ^ + ^ + etc. (1) 
 
 m m m m 
 
 The separate fractions may then be reduced to their lowest 
 terms. 
 
 Example. Dissect the fraction 
 
 32^^^^^; — \%mmj + 15^^^ — \WnH 
 
 l^abx 
 
 The general form (1) gives for the separate fractions, 
 
 ^%aWx _ ISamy Uhiz _ 12Wu_ 
 
 IQabx 16abx 16abx ~~ IQabx 
 
 Reducing each fraction to its lowest terms, the sum becomes 
 
 h ^^^^ _i_ ^^^^ dbn^u 
 " 8bx 16ax ~~ 4:ax ' 
 
74 ALGEBRAIC OPERATIONS. 
 
 EXERCISES. 
 
 Separate into sums of fractions, 
 abc + hcd -^ cda + dah 
 
 abed 
 — xyzu + x^yzu^ + xyh^u — x^y'^z^u^ 
 
 x^'ifz^u^ 
 a^—W- aH — % 
 
 ab ax 
 
 {m —n){n-^q)~- (m + n)(p—q) ^ 
 
 (m — n) {2^ — q) 
 (x — a)(y — b)-\- (x — y)(a — b) + (x — b) (y — a) 
 
 x^ — y^ 
 (a + b) {m —n) — {a — b) {rn + n) 
 
 Aggregation of Fractions. 
 
 105. When several fractions have equal denomina- 
 tors, their sum may be expressed as a single fraction 
 "by aggregating their numerators and writing the com- 
 mon denominator under them. 
 
 EX.X. ^-^ + ^ = A^±^. 
 m 771 m 771 
 
 -r, a — b b — c c — a 
 
 Ex. 2. = 
 
 X — y y — X X — y 
 _ a — b c—b c — a _2c — 2b__2{c — b) 
 ~x — y x — yx — y~~ x — y ~~ x — y 
 Rem. This process is tlie reverse of that of dissecting a fraction. 
 
 EXERCISES. 
 
 Aggregate 
 
 a ab abc a b 
 
 abc abc abc ' {a — by {a — if 
 
 ^ ^— ^ I y—'b a + b x — y 
 
 -"' d^x "^ a^x "^ a^x "^ a^x 
 
 a . b c d 
 
 + 
 
 a — b b — a a — b b — a 
 a — b a — c c — b c -\- a 
 m — n m — n n — in n — m 
 
FRACTIONS. 75 
 
 106. "When all the fractions have not the same denomina- 
 tor, they must be reduced to a common denominator by the 
 process of § 102. 
 
 Any common multiple of the denominators may be taken 
 as the common denominator, but the least common multiple is 
 the simplest. 
 
 To EEDUCE TO A CoMMOi^T Den"ominator. Choose a 
 connnoiz rnuUiple of the denominators. 
 
 Multiply both terms of each fraction hy the multi- 
 ■plier necessary to change its denominator to the chosen 
 ■multiple. 
 
 Note 1. The required multipliers will be the quotients of 
 the chosen multiple by the denominator of each separate 
 fraction. 
 
 Note 2. "When the denominators have no common fac- 
 tors, the multiplier for each fraction will be the product of the 
 denominators of all the other fractions. 
 
 Note 3. An entire quantity must be regarded as having 
 the denominator 1. (§ 102.) 
 
 EXAMPLES. 
 
 I. Aggregate the sum 
 
 1 2. _ J_ 1 
 
 a ah abc abed 
 in a single fraction. 
 
 The least common multiple of the denominators is abed. 
 The separate multipliers necessary to reduce to this com- 
 mon denominator are 
 
 abed, bed, cd, d, 1. 
 
 The fractions reduced to the common denominator abed are 
 
 abed — bed + ed — d +1 
 abed^ abed ' abed ' abed^ abed 
 
 abed — bed -\- ed — d -\- 1 
 
 The sum is 
 
 abed 
 
 By dissecting this fraction as in § 104, it may be reduced 
 
 to its original form. 
 
76 ALGEBRAIC OPERATIONS. 
 
 2. Reduce the sum 
 
 1 ah c 
 a he d 
 to a single fraction. 
 
 The multipliers are, by Note 2, hcd, acd, ahd, ahc. 
 
 Using these multipliers, the fractions become 
 hcd — d^cd ai^d — abc^ 
 abed' abed ' abed' abed ' 
 
 from which the required sum is readily formed. 
 
 3. Reduce the sum 
 
 ^a;-l^a;+l^a;2 — 1 
 The least common multiple of the denominators is x^ — 1. 
 The multipliers are, by Note 1, 
 
 a;2 — 1, x-\-l, x — \, 1. 
 The sum of the fractions is found to be 
 
 x'^ — \^x-^\-\-x^ — x-\-x^ _ ^_ 
 x^—1 x^ — 1 
 
 EXERCISES. 
 
 Reduce to a single fraction the sums, 
 
 I. 1+ ' 
 
 3 
 
 x — 1 
 1 1 
 
 1 — X 1 -^ X 
 
 ax x^ 
 
 a + X a + X 
 a X 
 
 x{a — x) a{a — x) 
 
 2. 
 
 1 .4-1- 
 
 4. 
 
 l-x ' 1+x 
 
 6. 
 
 a h 
 
 a — h a -\- h 
 
 8. 
 
 2a: — 5 5 3 
 
 4:X^ — 1^2x — l X 
 
 X -\- y a^ — y^ ^ — y 
 
 111 
 
 10. 7 + T H 
 
 a — b b ~e c — a 
 
 a , a a -}- h a — h 
 
 11. — ; \- ■ 12. — '-^ --^. 
 
 X + y x — y a — h a + b 
 
FRACTIONS. 77 
 
 
 14. 
 
 1 1__ _ 1 
 
 2 (a; — 1) 2 (a; + 1) ^ 
 5 
 
 ^5. ^l-(l-^' 
 
 16. m + y^ ^ — y ^ y rn+y 
 
 m — n X -\- y m^ m {m — y) 
 
 j8. 1 ^ ^ 
 
 a — iC «^ — X' 
 
 '^ — T^ 
 
 a — b h — c c — a (a — h) ( b — c) {c — a) 
 ^^' a^+~b "^ V+~c "^ c -{- a "^ (« + <$>) (b + c) (c~+~«) 
 
 \a -b^ b-al 
 
 a 
 
 '°- b 
 
 m — (x — a) m — (x -\- a) 
 
 21. ^ ^^ -• 
 
 X + y x — y 
 
 c ^ a b 
 
 22. — + — H 
 
 ab be ac 
 
 a b c 
 
 ^3- (^ _ ^) (^ _ c) + {b -a){b — c)'^{c — a) (c - b) 
 
 X -\-l x — 1 
 
 24. + 4a;. 
 
 ^a; — la;H-l^ 
 
 ab a^ a {a^ + V^) 
 
 a ■\- b a — b c^ — b'^ 
 a X 
 
 ^5. a-^.-^r—.-V 
 
 26. 1 — 
 
 27. 1 — 
 
 X -\- a X — a 
 0? — 2xy + y^ 
 
 Q? + y^ 
 
 %ay 
 \_ 1_ 1 
 
 ^^' {a + bf ^ {a — bf '^ a^ — ¥ 
 
 a^ — ^ab-\-lf^ 
 30. 1 + ^- 
 
•XS ALOEBBAIG OPERATIONS, 
 
 Factoring Fractions. 
 
 107. If several terms of the numerator contain a 
 common factor, the coefficients of this factor may be 
 added, and their aggregate multiplied by the factor for 
 a new form of the numerator. 
 
 EXAMPLES. 
 
 ax — hx -\- cx -\- dx _ (a—h + c-]- d)x 
 m ~ m 
 
 = {a-i + c+-d)^. (§101.) 
 
 adx + hex + acj/ — ahy _ {ah •}- hc)x (ac — ah) y 
 abn ~ ahn ahn 
 
 = (a + c)— + (c-J)f. 
 ^ ' an ^ 'In 
 
 EXERCISES. 
 
 Eeduce 
 
 ahy — hey — acy mnu + mpu + pnu 
 
 ahc ' mn 
 
 ahq + hcq -f ahr -\- her 
 
 ahc 
 ax — hy — Zhx - 
 
 2ma 
 4:mx -{- 2y — dax — 6cx -f- ay 
 
 xyz 
 a^ + 2a^ + a t^ «^^ — 4:ahc — (3y — 4g) a 
 
 xy ' ^* p + q 
 
 x^y — [4a; + a; (U — 4c) + 3ax'] 
 
 
 a-\- h 
 
 
 
 ax^- 
 
 4:cx — 3 [mx + m (« 
 
 — x) — ami 
 
 
 2a -3* 
 
 
 
 4:aVx 
 
 — 2cVx + 2hVx — 
 
 2 (mnVx - 
 
 -Wx) 
 
 
 3a — U 
 
 
 
FRACTIONS. 7^ 
 
 Multiplication and Division of Fractions, 
 
 108. Fundamental Theorems in the Multiplication 
 and Dimsion of Fractions : 
 
 Theorem I. A fraction may be imiltiplied by any 
 quantity by either multiplying its numerator or dividing 
 its denominator by that quantity. 
 
 Cor. 1. A fraction may be multiplied by its denominator 
 by simply cancelling it. 
 
 Cor. 2. If the denominator of the fraction is a factor in 
 the multiplier, cancel the denominator to multiply by this 
 factor, and then multiply the numerator by the other factors. 
 
 TYl 
 
 Ex. —7 YT X a^ (x^ — b^) = am (x + l\ 
 
 a(x — b) ^ ' ^ 
 
 because the multiplier cfi {o? — W) = a {x — b) a {x -{- b). 
 
 Theorem II, A fraction may be divided by either 
 dividing its numerator or multiplying its denominator. 
 
 Theorem III. To multiply by a fraction, the multi- 
 plicand must be multiplied by the numerator of the 
 fraction, and this product must be divided by its de- 
 nominator. 
 
 Let us multiply ^ by — 
 
 We multiply by m by multiplying the numerator (Th. I), 
 and we divide by n by multiplying the denominator (Th. II). 
 
 Hence the product is -^« 
 
 That is, the product of the numerators is the Jiumer- 
 ator of the required fraction, and the product of the 
 denominators is its denominator. 
 
 EXERCISES. 
 
 Multiply 
 
 ab + y , db , X 
 
 I. m x — a. 2. — by -• 
 
 X — a ^ X "^ a 
 
 ab , ^<^ T. o a 
 
80 ALOEBBAIG OPERATIONS. 
 
 abm , , ^ ^^ 1. ^ ^ m — a 
 5. -^— by xy\ 6. -^ by ax^ H 
 
 7. by — ^ — 8. « H by w H 
 
 'mm n '' m 
 
 X , y—ah m + n , n — m 
 
 o. «J by ay + '^^ 10. by ■ 
 
 ^ y X m — n "^ 7n -{■ n 
 
 Til- li- 1 hx , a h X 
 
 11. Multiply «+- by 3 + - + -. 
 
 1 2. Keduce ( m A 1 ( m ) • 
 
 V m — nf \ m + n/ 
 
 13. Eeduce \a \\b — j-y 
 
 14. Multiply h by — 
 
 15. Divide — by p. Ans. — • 
 ^ n "^ ^ np 
 
 16. Divide 7 by « + J. 
 
 1 7. Divide by :?; — 1. 
 
 X -j- i 
 
 18. Divide 4^4 by 1 + ^2. 
 
 19. Divide _-^^-p-^— by h^ - a\ 
 
 109. Reciprocal of a Fraction. The reciprocal of 
 a fraction is formed by simply inverting its terms. 
 
 For, let T be the fraction. By definition, its reciprocal 
 will be 
 
 a 
 ~b 
 
 Multiplying both terms by h, the numerator will be I and 
 
 the denominator ^ x h, that is, a. 
 
 Hence the reciprocal required will be -, or, in algebraic 
 language, 
 
FRACTIONS. 81 
 
 a ~ a 
 l 
 110. Def, A Complex Fraction is one of whicli 
 either of the terms is itself fractional. 
 
 a 
 
 Example. 
 
 m -\- - 
 
 y 
 
 OL X 
 
 is a complex fraction, of whicli t is the ^umerato^, and m ■{ — 
 the denominator. 
 
 The tenns of the lesser fractions which enter into the 
 nnmerator and denominator of the main fraction may 
 be called Minor Terms. 
 
 Thus, 5 and y are minor denominators, and a and x are 
 minor numerators. 
 
 To reduce a comjilex fraction to a simple one, multi- 
 ply both teims hy a multiple of the minor denominators. 
 
 am 
 
 Example. Reduce ^ ^ , » 
 b h 
 
 y'^ X 
 
 Multiplying both terms by xy"^, the result will be 
 
 amx 
 
 bxy + hy"^* 
 which is a simple fraction. 
 
 EXERCISES, 
 
 Reduce to simple fractions : 
 
 •-? 
 
 1-^ 
 
 y 
 
 a — x 
 
 a -\- X 
 
 n -\- X 
 a — X 
 6 
 
 X 
 
 b 
 a 
 
 X 
 
 ab 
 mn 
 
 km 
 
ALQEBBAIG OPERATIONS, 
 
 -»- 1 
 
 n + 1 
 
 1 - 
 
 n — 1 
 
 n-\-l 
 
 am 
 
 +.^ 
 
 8. 
 
 9- 7^*^^- 
 
 an - 
 
 _b_ 
 
 n 
 
 1 + 
 
 (a - by 
 
 iab 
 
 2ab 
 
 "' + 1 + ' 
 
 II. • 12. 
 
 - + a 
 a 
 
 a + 2b a 
 
 l-\-x 
 1—x 
 
 ^1 + x 
 
 1 + x 
 
 1—x 
 
 l—x 
 
 2x- 
 
 1-^x 
 3 
 
 ' y 
 
 a -]- b 
 
 1 
 
 — X 
 
 ^1-a 
 
 1 
 
 a 
 
 1-a 
 
 b^^ 
 
 1 + a 
 1 
 a 
 
 a 1 
 b b 
 x-y 
 
 x-^y 
 
 -t- y2 _ ^2 
 
 ^^' a + 2^ a * ^^' X-]- y x^ — y^ 
 
 a^b b 
 a 
 b a -\- b X — y x'^ — y 
 
 Division of one Fraction by Another. 
 
 111. Let us divide t by — The result will be expressed 
 
 by the complex fraction 
 
 a 
 
 m 
 
 n 
 
 Eeducing this fraction by the rule of § 110, it becomes 
 
 a7i 
 
 b^' 
 
 fir 'i7 
 
 which is equal to y x — That is, 
 
 ^ b m 
 
 To divide by a fraction, we have only to multiply by 
 its reciprocal. 
 
Divide 
 
 FRACTIONS. 83 
 
 EXERCI SES. 
 
 db . a X -{-1 , 2x 
 
 a—h "^ b 
 X , X a^ — ¥ , a^-{-a b 
 
 5. ^ by -^--r- 6. T H by 
 
 a J) c , m , n p 
 ' X y z ^ X y z 
 o a b , b a 
 
 a — b a -\- b a — b a -\- b 
 
 Reciprocal Relations of Multiplication and 
 Division. 
 
 113. The fundamental principles of the operations upon 
 fractions are included in the following summary, the under- 
 standing of which will afford the student a test of his grasp of 
 the subject. 
 
 1. The reciprocal of the reciprocal of a numlber is 
 equal to the number itself. In the language of Algebra, 
 
 1 
 
 a 
 
 2. The reciprocal of a monomial may be expressed 
 by changing the algebraic sign of its exponent. 
 
 3. To multiply by a number is equivalent to dividing 
 by its reciprocal, and vice versa. That is, 
 
 N -~ a or — = aN, 
 
 and mce versa, Jv x - = — • 
 
 a a 
 
84 ALQEBBAIG OPEBATIONS. 
 
 4. When tlie numerator or denominator of a fraction 
 is a product of several factors, any of these factors may 
 be transferred from one term of the fraction to the other 
 by changing it to its reciprocal. That is. 
 
 dbc 
 
 pqr 
 
 etc. 
 
 Or, = — f^ — = ^ "'^^ , etc. 
 
 pqr ar^pqr 
 
 5. Multiplication by a factor 
 
 greater than unity increases^ 
 less than unity diminishes. 
 Division by a divisor 
 
 greater than unity diminisJies, 
 less than unity increases, 
 
 6. («) When a factor becomes zero, the product also 
 becomes zero. 
 
 {(i) When a denominator becomes zero, the product 
 becomes infinite. That is, 
 
 ^ = infinity. 
 
 Note. The following way of expressing what is meant by 
 this last statement is less simple, but is logically more correct: 
 
 If a fraction has a fixed numerator, no matter how 
 small, we can make the denominator so much smaller 
 that the fraction shall be greater than any quantity we 
 choose to assign. 
 
 EXERCISE. 
 
 If the numerator of a fraction is 2, how small must the 
 denominator be in order that the fraction may exceed one 
 thousand ? That it may exceed one million ? That it may 
 exceed one thousand millions ? 
 
 jB.oTr 
 
BOOK III. 
 OF EQUA TIONS. 
 
 CHAPTER I. 
 THE REDUCTION OF EQUATIONS. 
 
 Definitions. 
 
 113. Def. An Equation is a statement, in the lan- 
 guage of Algebra, that two expressions are equal. 
 
 114. Def, The two equal expressions are called 
 Members of the equation. 
 
 115. Def. An Identical Equation is one which is 
 true for all values of the algebraic symbols which enter 
 into it, or which has numbers only for its members. 
 
 Examples. The equations 
 
 14 + 9 = 29 — 6, 
 5 + 13 — 3 X 4 — 6 = 0, 
 which contain no algebraic symbols, are identical equations. 
 So also are the equations 
 
 X =. X, 
 a; — a; = 0. 
 {x -{■ CL) {x — a) ^=^ 0? — a^, 
 (l + 2/)(l-2/)-l + / = 0, 
 
 because they are necessarily true, whatever values we assign to 
 X, a, and y. 
 
 Rem. All the equations used in the preceding two books 
 to express the relations of algebraic quantities are identical 
 ones, because they are true for all values of these quantities. 
 
86 EQUATIONS. 
 
 116. Def. An Equation of Condition is one whicli 
 can be true only when the algebraic symbols are equal 
 to certain quantities, or have certain relations among 
 themselves. 
 
 Examples. The equation 
 
 a; + 6 = 22 
 
 can be true only when x is equal to 16, and is therefore an 
 equation of condition. 
 The equation 
 
 X -^ h =^ a 
 
 can be true only when x is equal to the difference of the two 
 quantities a and h. 
 
 Rem. In an equation of condition, some of the quantities 
 may be supposed to be known and others to be unknown. 
 
 117. Def. To Solve an equation means to find 
 some number or algebraic expression which, being sub- 
 stituted for the unknown quantity, will render the 
 equation identically true. 
 
 This value of the unknown quantity is called a Root 
 of the equation. 
 
 EXAM PLES. 
 
 1. The number 3 is a root of the equation 
 
 2a;2 _ 18 = 0, 
 
 because when we put 3 in place of x, the equation is satisfied 
 identically. 
 
 2. The expression is a root of the equation 
 
 2ca; — 4flj + 25 = 0, 
 
 when X is the unknown quantity, because when we substitute 
 this expression in place of x, we have 
 
 2c 
 
 '^a — h 
 
 ) — 4^ -f 25 = 0, 
 
 \ c 
 
 or 4« — 2b — 46? + 25 = 0, 
 
 which is identically true. 
 
AXIOMS. 87 
 
 Eem. It is common in Elementary Algebra to represent 
 unknown quantities by the last letters of the alphabet, and 
 quantities supposed to be known by the first letters. But this 
 is not at all necessary, and the student should accustom him- 
 self to regard any one symbol as an unknown quantity. 
 
 Axioms. 
 
 118. Def. An Axiom is a proposition wliicli is 
 taken for granted, without proof. 
 
 Equations are solved by operations founded upon the fol- 
 lowing axioms, which are self-evident, and so need no proof. 
 
 Ax. I. If equal quantities be added to the two 
 members of an equation, the members will still be equal. 
 
 Ax. II. If equal quantities be subtracted from the 
 two members of an equation, they will still be equal. 
 
 Ax. III. If the two members be multiplied by equal 
 factors, they will still be equal. 
 
 Ax. IV. If the two members be divided by equal 
 divisors (the divisors being different from zero), they 
 will still be equal. 
 
 Ax. Y. Similar roots of the two members are equal. 
 
 These axioms may be summed up in the single one. 
 Similar operations upon equal quantities give equal 
 results. 
 
 119. An algebraic equation is solved by performing 
 such similar operations upon its two members that the 
 unknown quantity shall finally stand alone as one 
 member of an equation. 
 
 Operations of Addition and Subtraction— Trans- 
 posing Terms. 
 
 130. Theorem. Any term may be transposed from 
 one member of an equation to the other member, if its 
 sign be changed. 
 
88 EQUATIONS. 
 
 Proof. Let us put, in accordance with § 41, 2d Prin., 
 t, any term of either member of the equation. 
 «, all the other terms of the same member. 
 h, the opposite member. 
 The equation is then 
 
 a-]-t = 1). 
 
 Now subtract t from both sides (Axiom II), 
 a-{-t — t=^h — t; 
 or by reduction, a =z h — t. 
 
 This equation is the same as the one from which we started, 
 except that t has been transposed to the second member, with 
 its sign changed from -f to — . 
 
 If the equation is 
 
 b — t =z a, 
 
 we may add t to both members, which would give 
 
 h = a + t. 
 
 NUMERICAL EXAMPLE. 
 
 The learner will test each side of the following equations : 
 19 + 3_9_t_4 = 7 + 10. 
 Transposing 4, 19 + 3 — 9 = 7 + 10—4. 
 
 9, 19 + 3 = 7 + 10-4 + 9. 
 
 " 19, 3 = 7 + 10-4 + 9-19. 
 
 « 3, = 7 + 10—4 + 9-19—3. 
 
 121. Eem. All the terms of either member of an 
 equation may be transpose'd to the other member, 
 leaving only on one side. 
 
 Example. If in the equation 
 
 1) ^ a -\- t, 
 we transpose h, we have =z a -\- 1 — h. 
 By transposing a and t, we have 
 
 l^a--t = 0. 
 
 132. Changing Signs of Members. If we change the signs 
 of all the terms in both members of an equation, it will still 
 be true. The result will be the same as multiplying both 
 
REDUCTION. 89 
 
 members by — 1, or transposing all the terms of each member 
 to the other side, and then exchanging the terms. 
 
 Example. The equation 
 
 17 + 8 = 11 + 14 
 may be transformed into = 11 + 14 — 17 — 8, 
 
 or, :rz - 11 - 14 + 17 + 8, 
 
 or, - 17 — 8 = - 11 - 14. 
 
 Operation of Multiplication. 
 
 133. Clearing of FracMons. The operation of multipli- 
 cation is usually performed upon the two sides of an equation, 
 in order to clear the equation of fractions. 
 
 To clear an equation of fractions : 
 
 First. Method. Multiply its members hy the least 
 common multiple of all its denominators. 
 
 Secoi^d Method. Multiply its members by each of 
 the denominators in succession. 
 
 Eem. 1. Sometimes the one and sometimes the other of 
 these methods is the more convenient. 
 
 Rem. 3. The operation of clearing of fractions is similar 
 to that of reducing fractions to a common denominator. 
 
 Example of First Method. Clear from fractions the 
 equation 
 
 4 + 6+8 = ^^- 
 
 Here 24 is the least common multiple of the denominators. 
 Multiplpng each term by it, we have, 
 
 6a; H- 4a; + 3a; = 624, 
 or 13a; = 624. 
 
 Example of Second Method. Clear the equation 
 
 X — a X -\- a X 
 Multiplying by x — or, we find 
 
 ax — a^ ex — ca „ 
 
 a -\ 1 = 0. 
 
 X -\- a X 
 
90 EQUATIONS. 
 
 Multiplying by a; + «, 
 
 . 2 > ^ cx^ — ca^ . 
 ax -\- a^ -\- ax — a^ -\ = 0. 
 
 X 
 
 Eeducing and multiplying by x, 
 
 2ax^ + cx^ — ca^ = 0. 
 
 EXERCISES. 
 
 Clear tbe following equations of fractions : 
 
 X x^ _h 
 ^' a'^ a^~~a 
 ^ a h X 
 ^- 3 + 4 = 5- 
 
 ^ =1. 8. 
 
 II 
 
 13 
 
 XX X 
 
 2 "^3~4 "" 
 
 5. 
 
 ah^ a^ h- 
 
 " aW 
 
 X — a X -\- a 
 X -\- a _x^ -\- 2ax 
 X — a X — a 
 
 10. 
 
 X 
 
 2x 
 
 X — a 
 
 ~ x+b 
 
 x-2 
 
 x + 2 
 
 x-5 
 
 ~ X + 5 
 
 x — a 
 
 X -\- a 
 
 X y _Ci x — a ^ + ^,^_rv 
 
 y x~h ' X -[- a x — a a~ 
 
 X y 
 
 a — h h — a 
 i the second term : 
 
 X -\- a X — b 
 
 a — X x — a 
 
 Reduction to the ^N^ormal Porm. 
 
 Here the second term is tlie same as r 
 
 a — b 
 
 14. 
 
 134. Def. An equation is in its Normal Form 
 
 wlien its terms are reduced and arranged according to 
 the powers of tlie unknown quantity. 
 
 In the normal form one member of the equation is expressed 
 as an entire function of the unknown quantity, and the other 
 is zero. (Compare §§ 50, 76.) 
 
 To reduce an equation to the normal form : 
 I. Transpose all the terms to one memder of the equa- 
 tion, so as to leave as the other mejnher. 
 
BEBUOTION, 91 
 
 II. Clear the equation of fractions. 
 
 III. Clear the equation of -parentheses hy performing 
 all the operations indicated. 
 
 IV. Collect each set of terms containing lihe powers 
 of the unknown quantity into a single one. 
 
 V. Divide hy any common factor which does not con- 
 tain the unknown quantity. 
 
 Rem. This order of operations may be deviated from 
 according to circumstances. After a little practice, the student 
 may take the shortest way of reaching the result, without re- 
 spect to rules. 
 
 EXAMPLES. 
 
 I. Reduce to the normal form 
 
 {x — 2) {x — 3) _ {x 4- 2) (:r + 4) 
 x — h ~ a; + 5 
 
 1. Clearing of fractions, 
 
 {x + 5) (a; — 2) {x-^)=.{x — 5) {x + 2) {x + 4). 
 
 2. Performing the indicated operations, 
 
 Q? — \9ix-\-Zf^. — x?^x^ — %2x — 40. 
 
 3. Transposing all the terms to the second member and 
 reducing, 
 
 = a;2 — 3a; — 70, 
 
 which is the normal form of the equation. 
 
 Rem. Had we transposed the terms of the second member 
 to the first one, the result would have been 
 
 _ x^ + 3^; + 70 = 0. 
 Either form of the equation is correct, but, for the sake of 
 uniformity, it is customary to transpose the terms so that the 
 coejfficient of the highest power of x shall be positive. If it 
 comes out negative, it is only necessary to change the signs of 
 all the terms of the equation. 
 
 Ex. 2. Reduce to the normal form, 
 
 hm'3?' 2ax dmx^ ^ ^ 
 
 ■ ^ -„ = 2mx — 5a. 
 
 X — a X -\- a x^ — a'^ 
 
93 EQUATIONS. 
 
 1. Transposing to the first member, 
 
 hmx^ 2ax Smcc^ « ^ ^ 
 
 , r, —, — 2mx + 5a = 0, 
 
 X — a X -\- a x^ — a^ 
 
 2. To clear of fractions, we notice that the least common 
 multiple of the denominators is ic- — a\ Multiplying each 
 term by this factor, we have, 
 
 6mx^{x+a)—2ax(x^a)—dm3^—27nx{x^—ar)-\-6a{x^—a^) = 0. 
 
 3. Performing the indicated operations, 
 6mc(^+5amx^—2ax^+2a^x—3m.x^—2ma^+2a^mx+5ax^—6a^=0, 
 
 4. Collecting like powers of x, as in § 76, 
 
 {da + 5am) x^ + {2a^ + 2a^m) x — 6a^ = 0. 
 
 5. Every terra of the equation contains the factor a. By 
 Axiom IV, § 118, if both members of the equation be divided 
 by a, the equation will still be true. The second member 
 being zero, will remain zero when divided by a. Dividing 
 both members, we have 
 
 (3 + 5m)x^ -{-2a{l + m)x — 6a^ = 0, 
 which is the normal form. 
 
 EXERCISES. 
 
 Eeduce the following equations to the normal form, x, y, 
 or z being the unknown quantity : 
 
 dy^ -j- 2t/ _ y — H x — a_x-\-a 
 
 7 ~ 2 ' ' X -\- a~ X * 
 
 x — 1 _ 2x + Q 
 
 ^' 2a; + 10 ~ Ax — 2 
 
 a;3 _ Za^x + 2a^ , , "la^ — 6ax^ 
 
 4. r — x^ — 6ax = —- . 
 
 2x -^ a 2x — a 
 
 6. -^ + ^-^- + -^-^0. 
 
 a -[- 1) h + z a + z 
 
 z^ ^ a^z 
 
 7. :— - + 
 
 a — z a^ — x^ z^ — a^ 
 
REDUCTION. 93 
 
 8. 
 
 ■•^hh? 
 
 = 0. 
 
 
 
 
 a a^ 
 
 ^ x^-a^~ 
 
 = 1. 
 
 
 9- 
 
 X — a x^ — a^ 
 
 
 lO. 
 
 
 b' 
 
 
 II. 
 
 a b 
 
 12. 
 
 m 
 
 m 
 
 , 1~ x — a 
 
 
 X 
 
 n 
 nx 
 
 X 
 
 -I 
 
 13- 
 
 a a^ 
 
 X x^ 
 
 ~ x^' 
 
 
 
 14. 
 
 3z 5^2 
 
 -1 -r 
 
 _ 1 
 ~~ z 
 
 
 - 
 
 15- 
 
 G5rc 
 
 bx 
 
 
 
 1- 1 "1 
 
 X + a 
 
 , 1 ■ 
 
 
 
 ' x-a 
 
 
 16. 
 
 a b 
 
 X a — x 
 
 a 
 
 
 
 b ~ 
 - a 
 
 X 
 
 b 
 
 X 
 
 
 Degree of Equations. 
 
 125. Bef. An equation is said to Ibe of the n*^ de- 
 gree when n is the highest power of the unknown 
 quantity which appears in the equation after it is re- 
 duced to the normal form. 
 
 E X A MPLE S. 
 
 The equation Ax -\- B = (i is of the first degree. 
 
 Aq? + B = " « second " 
 
 Ax^ + Bx+ G=0 « « third " 
 
 etc. etc. 
 
 An equation of the second degree is also called a 
 Quadratic Equation. 
 
94 EQUATIONS OF THE FIB8T DEGREE, 
 
 An equation of tlie tliird degree is also called a 
 Cubic Equation. 
 
 Example. The equation 
 
 ax^ + dx^y^ -\- y^ + ah = 
 is a quadratic equation in x, because x^ is of tlie highest power 
 of X which enters into it. 
 
 It is a cubic equation in y. 
 
 It is of the first degree in z. 
 
 CHAPTER II. 
 
 EQUATIONS OF THE FIRST DEGREE WITH ONE 
 UNKNOWN QUANTITY, 
 
 126. Remark. By the preceding definition of the degree 
 of an equation, it will be seen that an equation of the first 
 degree, with x as the quantity supposed to be unknown, is one 
 which can be reduced to the form 
 
 Ax-{- B = 0, (a) 
 
 A and B being any numbers or algebraic expressions which 
 do not contain x. 
 
 Such an equation is frequently called a Simple Equation. 
 
 Solution of Equations of tlie First Degree. 
 
 127. If, in the above equation, we transpose the term B 
 to the second member, we have 
 
 Ax = —B. 
 
 If we divide both members by A (§ 118, Ax. lY), we have, 
 
 B 
 
 Here we have attained our object of so transforming the 
 equation that one member shall consist of x alone, and the 
 other member shall not contain x. 
 
 :.^^Mi' 
 
ONE UNKNOWN QUANTITY. 95 
 
 To prove that — — is the required value of x, we substi- 
 tute it for X in the equation {a). The equation then becomes, 
 
 or, by reducing, -—5 + ^ = 0, 
 
 an equation which is identically true. Therefore, — -^ is 
 the required root of the equation (a). (§117, Def.) 
 
 128. In an equation of the first degree, it will be unneces- 
 sary to reduce the equation entirely to the normal form by 
 transposing all the terms to one member. It will generally be 
 more convenient to place the terms which do not contain x in 
 the opposite member from those which are multiplied by it. 
 
 Example. Let the equation be 
 
 mx -\- a =L nx -\- l. (1) 
 
 We may begin by transposing a to the second member and 
 nx to the first, giving at once, 
 
 mx — nx z= I — a, 
 
 or {m — n)x=z'b — «, 
 
 without reducing to the normal form. The final result is the 
 same, whatever course we adopt, and the division of both 
 members hy m — n gives 
 
 h — a 
 
 X = 
 
 m — n 
 
 139. The rule which may be followed in solving equations 
 of the first degree with one unknown quantity is this : 
 
 I. Clear the equation of fractions. 
 
 II. Transpose the terms which are multiplied hy the 
 unhnown quantity to one memher ; those which do not 
 contain it to the other. 
 
 III. Divide hy the total coefficient of the unknown 
 quantity. 
 
96 EQUATIONS OF THE FIRST DEGREE. 
 
 Note. Rules in Algebra are given only to enable the beginner to go 
 to work in a way which will always be sure, though it may not always 
 be the shortest. In solving equations, he should emancipate himself 
 from the rules as soon as possible, and be prepared to solve each equa- 
 tion presented by such process as appears most concise and elegant. No 
 operation upon the two members in accordance with the axioms (§ 118) 
 can lead to incorrect results (provided that no quantity which becomes 
 zero is used as a multiplier or divisor), and the student is therefore free 
 to operate at his own pleasure on every equation presented. . 
 
 . EXAM PLES. 
 
 1. Given ^— = 1. 
 
 % 
 
 It is required to find the value of each of the quantities «, 
 h, X, and y, in terms of the others. 
 Clearing of fractions, we have 
 
 ax = ly. _ 
 
 To find a, we divide by x, which gives 
 
 a — —' 
 
 X 
 
 To find l, we divide by y, which gives 
 ax _ ^ 
 
 'y ~ 
 
 To find Xf we divide by a, which gives 
 
 a 
 To find y, we divide by I, which gives 
 ax 
 
 Thus, when any three of the four quantities a, l, x, and y, 
 are given, the fourth can be found. 
 
 2. Let us take the equation, 
 
 x — ^ _ 2a; + 6 
 2aj + 10 ~ 4i; — % 
 
 Clearing of fractions, we have 
 
 4^2 — 30a; + 14 = 4a;2 ^ 32^; + 60. 
 
ONE UNKNOWN QUANTITY 97 
 
 Transposing and reducing, 
 
 — 62a; = 46. 
 Dividing both members by — 62, 
 
 _ __46 _ _ 46 __ _ 23 
 ^ ~ — 62 — 62 ~" 31* 
 
 TMs result should now be proved by computing the value of botli 
 
 23 
 
 members of the original equation when — — is substituted for x. 
 
 X X __ ax 1 
 ^' m n~ h m 
 
 Proceeding in the regular way, we clear of fractions by 
 multiplying by mnb. This gives 
 
 ntx -\- tnbx = amnx — rib. 
 Transposing and reducing, 
 
 {nb + ml) — amn) x =^ — nl. 
 Dividing by the coefficient of x, 
 
 _ nb _ rib 
 
 ~ nb -\- mb — amn ~~ amn — mb — nb 
 
 These two values are equivalent forms (§ 100). 
 
 But we can obtain a solution without clearing of fractions. 
 ax 
 
 Transposing -7-, we have 
 
 X X ax _ 1 
 m n b ~ m' 
 which may be expressed in the form 
 
 n 1_«\ ^ -i. 
 \m n bl m 
 
 Dividing by the coefficient of x, 
 
 1^ 
 m 
 
 
 1 4.1 _^ 
 mnb 
 
 This expression can be reduced to the other by § 110. 
 
 7 
 
98 EQUATIONS OF THE FIRST DEGREE, 
 
 EXERCISES. 
 
 Find the values of x, y, or u in the following equations : 
 
 5 — 3a; _ a-g — 9 
 ^' 2 "" 3 * 
 
 ^ a I c 
 
 7- 3-4 + 5 = ''-2^- 
 
 w ?^ 1 1 
 
 5- « + * = « + J- 
 
 II. 
 
 12. 
 
 13. 
 14. 
 
 2. 
 
 — X =^ a. 
 
 4. 
 
 ^ + f = 9. 
 
 a; — 1 
 
 6. 
 
 36 45 
 
 8. 
 
 a — hx =: d -\- as 
 
 10. 
 
 3 — x 
 
 dx + —jr- = x. 
 
 a 
 
 C — X ' 
 
 c 
 ~ a — X 
 
 x — 1 
 
 x-2 
 
 x-2 
 
 x-3 
 
 -y = 
 
 ■.a-h 
 
 1 
 
 1 
 
 X—Q x—1 
 
 1 1 
 
 a; — 4 ic — 6 
 
 5- K'-S)-l(-3+l(-l)=»- 
 
 16. - + 
 
 a 1) — a h -\- a 
 
 X 1 
 
 17. ax -\- h =: - + -J' 
 ' a 
 
 u — a u — l u — c _ u — {a -\- 1) -\- c) 
 
 18. — 7 — "1 H — -, • 
 
 oca abc 
 
 m ix-{-a) n(x + V) 
 ^ X -\-b ' x -\- a 
 
 20. {x—aY -\--(^—^Y + {x—cy =z 3 (x—a) (x—l) {x—c). 
 
 Find the values of each of the four quantities, a, h, c, and 
 d, in terms of the other three, from the equations 
 
 ad ^^ , 1 A 
 
 21. 7 h T 7 = 0. 22. — + 1 = 0. 
 
 J) — c h — d cd 
 
ONE UNKNOWN QUANTITY. 99 
 
 Problems leadings to Simple Equations. 
 
 130. The first difficulty which the beginner meets with in 
 the solution of an algebraic problem is to state it in the form 
 of an equation. This is a process in which the student must 
 depend upon his own powers. The following is the general 
 plan of proceeding : 
 
 1. Study the problem, to ascertain what quantities in it 
 are unknown. There may be several such quantities, but the 
 problems of the present chapter are such that all these quan- 
 tities can be expressed in terms of some one of them. Select 
 that one by which this can be most easily done as the unknown 
 quantity. 
 
 2. Kepresent this unknown quantity by any algebraic sym- 
 bol whatever. 
 
 It is common to select one of the last letters of the alpha- 
 bet for the symbol, but the student should accustom himself 
 to work equally well with any symbol. 
 
 3. Perform on and with these symbols the operations re- 
 quired by the problem. These operations are the same that 
 would be necessary to verify the adopted value of the unknown 
 quantity. 
 
 4. Express the conditions stated or implied in the problem 
 by means of an equation. 
 
 5. The solution of this equation by the methods already 
 explained will give the value of the unknown quantity. It is 
 always best to verify the value found for the unknown quan- 
 tity by operating upon it as described in the equation. 
 
 EXAMPLES. 
 
 I. A sum of 440 dollars is to be divided among three people 
 so that the share of the second shall be 30 dollars more than 
 that of the first, and the share of the third 80 dollars less than 
 those of the first and second together. What is the share of 
 each? 
 
 Solution. 1, Hero there are really tliree unknown quantities, but 
 it is only necessary to represent the share of the first by an unknown 
 symbol. 
 
100 EQUATIONS OF THE FIRST DEGREE. 
 
 2 Therefore let us put 
 
 X = share of the first. 
 
 3. Then, by the terms of the statement, the share of the second will he 
 
 X + 30. 
 To find the share of the third we add these two together, which makes 
 
 2x + 30. 
 Subtracting 80, we have 
 
 2a; — 50 
 as the share of the third. 
 
 We now add the three shares together, thus, 
 Share of first, x 
 
 " " second, x -{- 30 
 " « third, 2 rg-50 
 Shares of all, 4^x — 20 
 
 4. By the conditions of the problem, these three shares must together 
 make up 440 dollars. Expressing this in the form of an equation, we 
 have 
 
 4:X — 20 = 440. 
 
 5. Solving, we find 
 
 X = 115 — share of first. 
 Whence, 115 + 30 = 145 = share of second. 
 
 115 + 115 — 80 =252. = ^^^^^'^ ^^ *^^^^'^- 
 Sum =r 440. Proof. 
 
 Ex. 2. Divide the nnmber 90 into four parts, such that 
 the first increased by 2, the second diminished by 2, the third 
 multiplied by 2, and the fourth divided by 2, shall all be equal 
 to the same quantity. 
 
 Here there are really five unknown quantities, namely, the four parts 
 and the quantity to which they are all to be equal when the operation of 
 adding to, subtracting, etc., is performed upon them. It will be most 
 convenient to take this last as the unknown quantitj. Let us therefore 
 put it equal to u. Then, 
 
 Since the first part increased by 2 must be equal to u, its value will 
 be u — 2. 
 
 Since the second part diminished by 2 must be equal to u, its value 
 will be u + 2. 
 
 u 
 Since the third part multiplied by 2 must be u, its value will be ^ • 
 
 Since the fourth part divided by 2 must make u, its value will be 2u. 
 
ONE UNKNOWN QUANTITY. 101 
 
 Adding these four parts up, their sum is found to be — • 
 
 By the conditions of the problem, this sum must make up the num- 
 ber 90. Therefore we have 
 
 Solving this equation, vre find 
 
 U = 20. 
 Therefore 
 
 1st part =i u — % z=lS, 
 
 2d *^ = 2/ + 2 == 22. 
 3d '^ =II^^4-2=:10. 
 4th " =: 2w = 40. 
 
 The sum of the four equals 90 as required, and the first part increased 
 by 2, the second diminished by 2, etc., all make the number 20, as re- 
 quired. , . , 
 
 PROBLEMS FOR EXERCISE, y .- — 
 
 V -> •: - 
 
 1. What number is that from which we obtain the same 
 result whether we multiply it by 4 or subtract it from 100 ? 
 
 2. "What number is that which gives the same result when 
 we divide it by 8 as when we subtract it from 81 ? 
 
 3. Divide 284 dollars among two people so that the share 
 of the first shall be three times that of the second and $16 
 more. 
 
 4. Find a number such that \ of it shall exceed \ of it 
 by- 12. 
 
 5. A shepherd describes the number of his sheep by saying 
 that if he had 10 sheep more, and sold them for 5 dollars each, 
 he would have 6 times as many dollars as he now has sheep. 
 How many sheep has he ? 
 
 6. An applewoman bought a number of apples, of which 
 60 proved to be rotten. She sold the remainder at the rate of 
 2 for 3 cents, and found that they averaged her one cent each 
 for the whole. How many had she at first ? 
 
 7. If you divide my age 10 years hence by my age 20 years 
 ago, you will get the same quotient as if 3'ou should divide my 
 present age by my age 26 years ago. What is my present age ? 
 
 8. Divide $500 among A, B, and C, so that B shall have 
 $20 less than x\, and C $20 more than A and B together. 
 
103 EQUATIONS OF TEE FIRST DEGREE. 
 
 9. A father left $10000 to be divided among his five chil- 
 dren, directing that each should receive $500 more than the 
 next younger one. What was the share of each ? 
 
 10. A man is 6 years older than his wife. After they have 
 been married 12 years, 8 times her age would make 7 times 
 his age. What was their age when married ? 
 
 11. Of three brothers, the youngest is 8 years younger than 
 the second, and the eldest is as old as the other two together. 
 In 10 years the sum of their ages will be 120. What are their 
 present ages ? 
 
 12. The head of a fish is 9 inches long, the tail is as long 
 as the head and half the body, and the body is as long as the 
 head and tail together. What is the whole length of the fish ? 
 
 13. In dividing a year's profits between three partners, A, 
 B, and C, A got one-fourth and $150 more, B got one-third 
 and $300 more, and got one-fifth and $60 more. What was 
 the sum divided ? 
 
 14. A traveller inquiring the distance to a city, was told 
 that after he had gone one-third the distance and one-third 
 the remaining distance, he would still have 36 miles more to 
 go. What was the distance of the city ? 
 
 15. In making a journey, a traveller went on the first day 
 one-fifth of the distance and 8 miles more ; on the second day 
 he went one-fifth the distance that remained and 15 miles 
 more ; on the third day he went one-third the distance that 
 remained and 12 miles more ; on the fourth he went 35 miles 
 and finished his journey. What was the whole distance 
 travelled ? 
 
 16. When two partners divided their profits, A had twice 
 as much as B. If he paid B $300, he would only have half as 
 much again as B had. What was the share of each ? 
 
 17. At noon a ship of war sees an enemy's merchant vessel 
 15 miles away sailing at the rate of 6 miles an hour. How fast 
 must the ship of war sail in order to get within a mile of the 
 vessel by 6 o'clock ? 
 
 18. A train moves away from a station at the rate of h 
 miles an hour. Half an hour afterward another train follows 
 it, running m miles an hour. How long will it take the latter 
 to overtake it ? 
 
 19. What two numbers are they of which the difference is 
 9, and the difference of their squares 351 ? 
 
 20. A man bought 25 horses for $2500, giving $80 a piece 
 
ONE UNKNOWN QUANTITY 103 
 
 for poor horses and 1130 each for good ones. How many of 
 each kind did he buy ? 
 
 21. A man is 5 years older than his wife. In 15 years the 
 sums of their ages will be three times the present age of the 
 wife. What is the age of each ? 
 
 22. How far can a person who has 8 hours to spare ride in 
 a coach at the rate of 6 miles an hour, so that he can return at 
 the rate of 4 miles an hour and arrive home in time ? 
 
 23. A working alone can do a piece of work in 15 days, 
 and B alone can perform it in 12 days. In what time can they 
 perform it if both work together ? 
 
 Method of Solution. In one day A can do ^^ of the whole work 
 and B can do ^^. Hence, both together can do (iV+ 1^5) of it- 
 
 If both together can do it in x days, then they can do - of it in 1 day. 
 
 X 
 TT 111 
 
 Hence, 5 = 13 + 15 
 
 is the equation to be solved. 
 
 24. A cistern can be filled in 12 minutes by two pipes which 
 run into it. One of them alone will fill it in 20 minutes. In 
 what time would the other one alone fill it ? 
 
 25. A cistern can be emptied by three pipes. The second 
 pipe runs twice as much as the first, and the third as much as 
 the first and second together. All three together can empty 
 tlie cistern in one hour. In wh^t time would each one sepa- 
 rately empty it ? 
 
 26. A marketwoman bought apples at the rate of 5 for two 
 cents, and sold half of them at 2 for a cent and the other half 
 at 3 for a cent. Her profits were 50 cents. How many did 
 she buy ? 
 
 27. A grocer having 50 pounds of tea worth 90 cents a 
 pound, mixed with it so much tea at 60 cents a pound that 
 the combined mixture was worth 70 cents. How much did 
 he add ? 
 
 28. A laborer was hired for 40 days, on the condition that 
 every day he worked he should receive $1.50, but should for- 
 feit 50 cents for every day he was idle. At the end of the 
 time $52 were due him. How many days was he idle ? 
 
 29. A father left an estate to his three children, on the 
 condition that the eldest should be paid 11200 and the second 
 $800 for services they had rendered. The remainder was to be 
 equally divided among all three. Under this arrangement. 
 
104 EQUATIONS OF THE FIRST DEOREK 
 
 the youngest got one-fourth of the estate. What was the 
 amount divided ? 
 
 30. A person having a sum of money to divide among 
 three people gave the first one-third and $20 more, the second 
 one-third of what was left and $20 more, and the third one- 
 third of what was then left and $20 more, which exhausted the 
 amount. How much had they to divide ? 
 
 31. One shepherd spent $720 in sheep, and another got the 
 same number of sheep for $480, paying $2 a piece less. What 
 price did each pay ? 
 
 32. A crew which can pull at the rate of 9 miles an hour, 
 finds that it takes twice as long to go up the river as to go 
 down. At what rate does the river flow ? 
 
 33. A person who possesses $12000 employs a portion of 
 the money in building a house. Of the money which remains, 
 he invests one-third at four per cent, and the other two-thirds 
 at five per cent., and obtains from these two investments an 
 annual income of $392. What was the cost of the house ? 
 
 34. An income tax is levied on the condition that the first 
 $600 of every income shall be untaxed, the next $3000 shall 
 be taxed at two per cent, and all incomes in excess of $3600 
 shall be taxed three per cent, on the excess. A person finds 
 that by a uniform tax of two per cent, on all incomes he would 
 save $200. What was his income ? 
 
 35. At what time between 3 and 4 o'clock is the minute- 
 hand 5 minutes ahead of the hour hand? 
 
 2,6. One vase, holding a gallons, is full of water ; a second, 
 holding b gallons, is full of brandy. Find the capacity of a 
 dipper such that whether it is filled from the first vase and the 
 water removed replaced by bi-andy, or filled from the second 
 vase and the latter then filled with water, the strength of the 
 mixture will be the same. 
 
 37. Divide a number m into four such parts that the first 
 part increased by a, the second diminished by a, the third 
 multiplied by a, and the fourth divided by a shall all be equal. 
 
 38. Divide a dollars among five brothers, so that each shall 
 have n dollars more than the next younger. 
 
 39. A courier starts out from his station riding 8 miles an 
 hour. Four hours afterwards he is follov/ed by another riding 
 10 miles an hour. How long will it require for the second to 
 overtake the first, and what will be the distance travelled ? 
 
 If X be the number of hours required, the second will have travelled 
 X hours and the first (« 4- 4) hours when they meet. At this time they 
 must have travelled equal distances. 
 
ONE UNKNOWN QUANTITY. 105 
 
 Problem of the Couriers. 
 
 Let us generalize the preceding problem thus : 
 
 131. A courier starts out from his station riding c 
 miles an hour ; h hours later, he is folloivecl hy another 
 riding a miles a7v hour. How long will the latter he in 
 oveHaking the first, and what will he the distance from 
 the point of departure. 
 
 Let us put t for the time required. Then the first courier 
 will have travelled {t-{-h) hours, and the second t hours. 
 Since the first travelled c miles an hour, his whole distance at 
 the end oit-\-h hours will be {t-\-li) c. In the same way, the 
 distance travelled by the other will be at. When the latter 
 overtakes the former, the distances will be equal ; hence, 
 
 at = c{t + li). (1) 
 
 Solving this equation with respect to t, we find 
 
 _^ = ~ (^) 
 
 Multiplying by a gives us the whole distance travelled, 
 which is 
 
 Distance = • 
 
 a — c 
 
 This equation solves every problem of this kind by substi- 
 tuting for a, c, and h their values in numbers supposed in the 
 problem. For example, in Problem 39, we supposed a = 10, 
 c = S, h = 4:. Substituting these values in equation (2), we 
 
 find 
 
 t = 16, 
 
 which is the number of hours required. 
 
 To illustrate the generality of an algebraic problem, we 
 shall now inquire what values t shall have when we make dif- 
 ferent suppositions respecting a, c, and h. 
 
 (1.) Let us suppose a = c, or « — c == 0, that is, the rates 
 of travelling equal. Then equation (2) will become 
 
 . _ ch 
 
106 EQUATION'S OF THE FIRST DEGREE. 
 
 an expression for infinity (§ 112, 6), showing that the one courier 
 would never overtake the other. This is plain enough. But, 
 
 (2. ) Let us suppose that the second courier does not ride 
 
 so fast as the first, that is, a less than c, and a — c negative. 
 
 ch 
 
 Then the fraction will not be infinite, but will be nega- 
 
 u — c 
 
 tive, because it has a positive numerator and a negative denom- 
 inator. It is plain that the second courier would never overtake 
 the first in this case either, because the latter would gain on 
 him all the time ; yet the fraction is not infinite. 
 
 What does this mean ? 
 
 It means that the problem solved by Algebra is more gen- 
 eral, that is, involves more particular problems than were 
 implied in the statement. If we count the hours after the 
 second courier set out as positive, then a negative time will 
 mean so many hours before he set out, and this will bring out 
 a time when, according to our idea of the problem, the horses 
 were still in the stable. 
 
 The explanation of the difiiculty is this. Suppose S to be 
 the point from which the couriers started, and AB the road 
 along which they travelled from 
 S toward B. Suppose also that i......i.^^^_^..i.^^_ 
 
 the first courier started out 
 
 fi'om S at 8 o'clock and the second at 12 o'clock. By the rule 
 of positive and negative quantities, distances towards A are 
 negative. Now, because algebraic quantities do not commence 
 at 0, but extend in both the negative and positive directions, 
 the algebraic problem does not suppose the couriers to have 
 really commenced their journey at S, but to have come from 
 the direction of A, so that the first one passes S, without stop- 
 ping, at 8 o'clock, and the second at 12. It is plain that if the 
 first courier is travelling the faster, he must have passed the 
 other before reaching S, that is, the time and distance are 
 both negative, just as the problem gives them. 
 
 The general principle here involved may be expressed thus: 
 In Algelra, roads and journeys, like time, have no legin- 
 ning and no end. 
 
ONE UNKNOWN QUANTITY, 107 
 
 (3.) Let us suppose that the couriers start out at the same 
 time and ride with the same speed. Then Ji and a — c are 
 both zero, and the expression for t assumes the form, 
 
 '=?■ 
 
 This is an expression which may have one vahie as well as 
 another, and is therefore indeterminate. The result is correct, [ 
 because the couriers are always together, so that all values of 
 t are equally correct. 
 
 The equation (1) can be used to solve the problem in other 
 forms. In this equation are four quantities, a, c, h, and t, and 
 when any three of these are given, the fourth can be found. 
 There are therefore four problems, all of which can be solved 
 from this equation. 
 
 First Problem, that already given, in which the time 
 required for one courier to overtake the other is the unknown 
 quantity. 
 
 Seco]S"d Problem. A courier sets out from a station, 
 riding c miles an hour. After h hours another follows 
 him fromj the same station, intending to overtake him/ 
 in t hours. Sow fast must he ride ? 
 
 The problem can be put into the form of an equation in 
 the same way as before, and we shall have the equation (1), 
 only a will now be the unknown quantity. If we use the 
 numbers of Prob. 39 instead of the letters, we shall have, in- 
 stead of equation (1), the following : 
 
 16a = 8 (16 + 4) = 8-20 = 160, 
 
 whence a = 10. 
 
 If we use letters, we find from (1), 
 
 _ c(t + h ) 
 a- - , 
 
 and the problem is solved in either case. 
 
 Third Problem. I7ie second courier can Hde just a 
 miles an hour, and the first courier staj^ts out h hours 
 
108 EQUATIONS OF THE FIB8T DEGREE. 
 
 hefore him. How fast must the latter ride in order that 
 the other m^ay take t hours to overtake him ? 
 
 Here c, the rate of the first courier, is the unknown quan- 
 tity, and by solving equation (1), we find 
 
 at 
 
 c = 
 
 t + h 
 
 Fourth Problem. The swiftest of two couriers caw 
 ride a D%iles an hour, and the slower c miles an hour. 
 How long a start i7%ust the latter have in order that the 
 other may require t hours to overtake him? 
 
 Here, in equation (1), h is the unknown quantity. By 
 solying the equation with respect to h, we find, 
 
 , at — d 
 which solves the problem. 
 
 PROBLEMS OF CIRCULAR MOTION. 
 
 40. Two men start from the same point to run repeatedly 
 round a circle one mile in circumference. If A runs 7 miles 
 an hour and B 5, it is required to know : 
 
 1. At what intervals of time will A pass B ? 
 
 2. At how many different points on the circle will they be 
 together ? 
 
 We reason thus : since A runs 2 miles an hour faster than B, he jxeta 
 away from him at the rate of 2 miles an hour. When he overtakes him, 
 he will have gained up )n him one circumference, that is, 1 mile. This 
 will require 80 minutes, which is therefore the required interval. In 
 this interval A will have gone round '6\ and B 2| times, so that they will 
 be together at the point opposite that where they were together 30 
 minutes previous. Hence, they are together at two opposite points of 
 the circle. 
 
 41. Wliat would be the answer to the preceding ques- 
 tion if A should run 8 miles an hour, and B 5 ? 
 
 42. Two i-ace-horses i*un round and round a course, the 
 one making the circuit in 30, the other in 35 seconds. If 
 they start out together, how long before they will be 
 too^ether ao:ain 1 
 
 X X 
 
 Note. In x seconds one will make ^ circuit and the other _-. 
 
 oO 00 
 
 43. If one planet revolves round the sun in T and the 
 other in T' years, what will be the interval between their 
 conjunctions? 
 
TWO UNKNOWN QUANTITIES. 109 
 
 CHAPTER III. 
 
 EQUATIONS OF THE FIRST DEGREE WITH SEVERAL 
 UNKNOWN QUANTITIES. 
 
 Case I. Equations with Two UnJcnown Quan- 
 titles. 
 
 133. Bef. An equation of the first degree with two 
 unknown quantities is one which admits of being re- 
 duced to the form 
 
 ax + l)y = c, 
 
 in which x and y are' the unknown quantities and a, 5, 
 and c represent any numbers or algebraic equations 
 which do not contain either of the unknown quantities. 
 
 Def, A set of several equations containing the same 
 unknown quantities is called a System of Simulta- 
 neous Equations. 
 
 Solution of a Pair of Simultaneous Equations 
 containing Two Unknown Quantities. 
 
 133. To solve two or more simultaneous equations, 
 it is necessary to combine them in such a way as to 
 form an equation containing only one unknown quan- 
 tity. 
 
 134. Def. The process of combining equations so 
 that one or more of the unknown quantities shall dis- 
 appear is called Elimination. 
 
 The term "elimination" is used because the unknown 
 quantities which disappear are eliminated. 
 
 There are three methods of ehminating an unknown quan- 
 tity from two simultaneous equations. 
 
110 EQUATIONS OF THE FIRST DEGREE. 
 
 Elimination by Comparison. 
 
 135. EuLE. Solve each of the equations with respect 
 to one of the unknown quantities and put the two values 
 of the unknown quantity thus obtained equal to each 
 other. 
 
 This luill give an equation with only one unknown 
 quantity, of which the value can he found from the 
 equation. 
 
 Tlie value of the other unknown quantity is then 
 found hy substitution. 
 
 Example. Let the equations be 
 ax-\-l)y = c, \ 
 
 a'x + h'y 
 From the first equation we obtain 
 
 c — I 
 
 (2) 
 (3) 
 
 a 
 From the second we obtain, 
 
 c' — Vy 
 a 
 
 Putting 'these two values equal, we have 
 c — hy _ d — Vy 
 
 Reducing and solving this equation as in Chapter II, we 
 find, 
 
 _ ac' — o!g 
 y - ah' -aH' 
 
 which is the required value of y. Substituting this value of y 
 in either of the equations (1), (2), or (3), and solving, we shall 
 find 
 
 h'c-hc'. 
 ah — ah 
 If the work is correct, the result will be the same in which- 
 ever of the equations we make the substitution. 
 
TWO UNKNOWN QUANTITIES. Ill 
 
 Numerical Example. Let tlie equations be 
 
 ^x — 2y = 29. ) ^^ 
 
 From the first equation we find 
 
 X = 2^ — y, 
 
 and from the second x = — ^— ^, 
 
 o 
 
 from which we have 28 — ?/ = ^ — -, 
 
 o 
 
 2/ = 11. 
 Substituting this value in the first equation in x, it becomes 
 
 a; = 28 - 11 = 17. 
 
 If we substitute it in the second, it becomes 
 
 29 + 22 _ 51 _ 
 X- ^ - _ - _ 17, 
 
 the same value, thus proving the correctness of the work. 
 
 Elimination by Substitution. 
 
 136. Rule. Find the value of one of the iinlcnoivn 
 quantities in terms of the other from either equation, 
 and substitute it in the other equation. The latter will 
 have hut one unknown quantity. 
 
 Example. Taking the same equations as before, 
 ax -\- hj =. c, 
 a'x -\- h'y = c', 
 
 the first equation gives x = -' 
 
 Substituting this value instead of x in the second equation, 
 it becomes 
 
 a'c — a'ly , ^, , 
 
 + III = c. 
 
 a ^ 
 
 Solving this equation with respect to y, we get the same 
 result as before. 
 
112 EQUATIONS OF THE FIRST DEGREE. 
 
 Numerical Example. To solve in this way the last nu- 
 merical example, we have from the first equation (4), 
 
 X = 2^ — y. 
 
 Substituting this value in the second equation, it becomes 
 
 84: — dy — 2y = 29, 
 
 from which we obtain as before, 
 
 84 — 29 
 y = —^- = 11. 
 
 This method may be applied to any pair of equations in 
 four ways : 
 
 1. Find X from the first equation and substitute its value 
 in the second. 
 
 2. Find x from the second equation and substitute its 
 value in the first. 
 
 3. Find y from the first equation and substitute its value 
 in the second. 
 
 4. Find y from the second equation and substitute its 
 value in the first. 
 
 Elimination by Addition or Subtraction. 
 
 137. EuLE. Midtiply each equation by such a factor 
 that the coefficients of one of the unknown quantities 
 shall become numerically equal in the two equations. 
 
 Then, by adding or subtracting the equations, we 
 shall have an equation with but one unhnown quantity. 
 
 Kem. We may always take for the factor of each equation 
 the coefiicient of the unknown quantity to be eliminated in the 
 other equation. 
 
 Example. Let us take once more the general equation 
 
 ax -\- by ■= c, 
 a'x -\- b'y = c'. 
 Multiplying the first equation by a', it becomes 
 
 aa'x + a'by = a'c. 
 Multiplying the second by a, it becomes 
 aa'x + ab'y = ac'. 
 
TWO UNKNOWN QUANTITIES. 113 
 
 The unknown quantity x has the same coefficient in the 
 last two equations. Subtracting them from each other, we 
 obtain 
 
 (a'h — ab') y = a'c — ad, 
 
 n!r, — ac! 
 
 y 
 
 ab — ab 
 
 Kem. TVe shall always obtain the same result, whichever 
 of the above three metliods we use. But as a general rule the 
 last method is the most simple and elegant. 
 
 Problem of the Sum and Diflference. 
 
 The following simple problem is of such wide application 
 that it should be well understood. 
 
 138. Problem. Tlie sinn aizd difference of two num- 
 hers being given, to find the numbers. 
 Let the numbers be x and y. 
 Let s be their sum and d their difference. 
 Then, by the conditions of the problem, 
 x-\-y = s, 
 x — y = d. 
 Adding the two equations, we have 
 2x = s -\-d. 
 Subtracting the second from the first, 
 
 2y ^ s — d. 
 Dividing these equations by 2, 
 
 X = 
 
 s + d 
 2 ~ 
 
 s 
 ~ 2 
 
 4 
 
 y=^ 
 
 s —■ d 
 
 2 ~ 
 
 s 
 
 ~ 2 
 
 d 
 
 2 
 
 We therefore conclude : 
 
 The greater number is found by adding half the dif- 
 ference to half the sum. 
 
 TJ^e lesser number is found by subtracting half the 
 difference from half the sum. 
 8 
 
A 
 
 1 
 
 C 
 
 1 
 
 
 B 
 
 1 
 
 C 
 
 
 
 1 
 P 
 
 
 
 114 EQUATIONS OF THE FIRST DEGREE. 
 
 This result can be illustrated geometrically. Let AB and 
 BO be two lines placed end to end, so that AC is their sum. 
 To find their difference, we 
 cut off from AB a length 
 AC =z BC ; then C'B is the 
 difference of the two lines. 
 
 If P is half way between C and B, it is the middle point 
 of the whole line, so that 
 
 AP = PC = iAO = J sum of lines. 
 C'P = PB = iC'B = i difference of lines. 
 
 If to the half sum AP we add the half difference PB, we 
 have AB, the greater line. 
 
 If from the half sum AP we take the half difference C'P, 
 we have left AC, the lesser line. 
 
 EXERCISES. 
 
 Solve the following equations : 
 
 I. dx — 2y = 33, 
 
 2x-3y = 18. 
 
 2. dx-by = 13, 
 
 2x -\-7y = 81. 
 
 3. "ix + Qy = a, 
 
 6X + 6y =: h. 
 
 4. 2x -\- 3y r= m, 
 
 2x — dy = n. 
 
 5. ax+by = p, 
 
 ax — by = q. 
 
 6. i-^f==.e. 
 
 6-7 -^' 
 
 7- M=-> 
 
 -4-^ -29 
 8 + 2 -'^^' 
 
 8. |.|^. 
 
 2 3 "^ 
 
 9. ^x + y)^d (X -y) = 102, 
 ':!{x + y)-d(x-y) = 66. 
 
 Note, Solve this equation first as if x+y and x—y were single sym- 
 bols, of which the values are to be found. Then find x and y by § 138 
 preceding. 
 
 10. x + y + (x — y) = U, x + y — {x — y) = 10. 
 
 1 1_ 5 1_1_Jl 
 
 X y ~ 12' X y ~ 12 
 
TWO UNKNOWN QUANTITIES. 115 
 
 Note. Equations in this fonn can be best solved as if - and - were 
 the unknown quantities. See next exercise. ^ 
 
 3_2_11 45^ 
 
 Solution. If we multiply the first equation by 4, and the second by 
 3, we have 
 
 12 _ 8 _ 44 _ 22 
 
 X y ~ 10~ 6' 
 
 X y 
 
 Subtracting the first from the second, we have 
 23 _ 23, 
 
 y ~ b' 
 
 whence, 
 
 y = b. 
 
 Again, to eliminate - , we multiply the first equation by 5 and the 
 second by 2 and add. Thus, 
 
 15 _ 1 _ n 
 
 X 7 ~ 2 ' 
 
 8^10 „ 12 
 
 ^ + 7 = ^ = ^' 
 
 23 _ 23, 
 a; "~ 2 ' 
 
 X = ^. 
 
 whence. 
 
 ^^' x'^ y~ 1^' X y~ 12 
 ^^' x'^ y~ VI' X y" 24* 
 
 15- 
 i6. 
 
 17. 
 
 5_3__1 3_1_J^^ 
 X y~ 6' cc y "" 30* 
 
 _5 L___i _J 1 _ 1 
 
 x + 1 y — l~ %' x-\-l «/ — 1~30' 
 
 2 3 _ ^ _2 3_ _ _1^ 
 
 x + i'^ y — ^~~ 1%' 2; + 2—«^ — 3~ 12* 
 
116 EQUATIONS OF THE FIRST DEGREE. 
 
 i8. - + - z= c, = d. 
 
 X y X y 
 
 X + y ^ 2x i- dy j^ 
 
 19. — —^ = 2, — - — -^ = b, 
 
 ^ X — y x -{- a 
 
 20. — —7 ^ ^ = 2a, -r-7^ = 1. 
 
 Case II. Equations of the First I>egree with 
 Three or More Unknown Quantities. 
 
 139. When the values of several unknown quantities are 
 to be found, it is necessary to have as many equations as un- 
 known quantities. 
 
 If there are more unknown quantities than equations, it 
 will be impossible to determine the values of all of them from 
 the equations. All that can be done is to determine the value 
 of some in terms of the others. 
 
 If the number of equations exceeds that of unknown quan- 
 tities, the excess of equations will be superfluous. If there 
 are n unknown quantities, their values can be found from any 
 n of the equations. If any selection of n equations we choose 
 to make gives the same values of the unknown quantities, the 
 equations, though superfluous, will be consistent. If different 
 values are obtained, it will be impossible to satisfy them all. 
 
 Elimination. 
 
 140. When the number of unknown quantities exceeds 
 two, the most convenient method of elimination is generally 
 that by addition or subtraction. The unknown quantities are 
 to be eliminated one at a time by the following method : 
 
 I. Select an unhnown quantity to he first eliminated. 
 It is best to begin with the quantity which appeal's in 
 the fewest equations or has the simplest coefficients. 
 
 II. Select one of the equations containing this un- 
 hnown quantity as an eliminating equation. 
 
 III. Elwninate the quantity between this equation 
 and each of the others in succession. 
 
THREE OR MORE UNKNOWN QUANTITIES. Ill 
 
 We shall then have a second system of eqnations less by 
 one in number than the original system and containing a num- 
 ber of unknown quantities one less. 
 
 IV. Repeat the process on the new system of equations, 
 and continue the repetition until only one equation with 
 one unknown quantity is left. 
 
 V. Having found the value of this last unknown 
 quantity, the values of the others can he found hy^ suc- 
 cessive substitution in one equation of each system. 
 
 Example. Solve the equations 
 
 (1) 4,x — ^y—z-\-u—l = 0,\ 
 
 (2) a; — 2/ + 2^ + 2w — 10 = 0, r 
 
 (3) 2x + ^— z — 2u - 2 = 0, f ^^^ 
 
 (4) x + 2y ■\- z-\- u — 1% = ^.) 
 
 We sliall select x as tlie first quantity to be eliminated, and take the 
 last equation as the eliminating one. We first multiply this equation by 
 three such factors that the coefficient of x shall become equal to the co- 
 efficient of X in each of the other equations. These factors are 4, 1, and 2, 
 We write the products under each of the other equations, thus : 
 
 Eq. (1), 
 (4)X4, 
 
 Eq. (2), 
 
 (4)xl, 
 
 Eq. (3), 
 (4)x3, 
 
 By subtracting the one of each pair from the other, we obtain the 
 equations, 
 
 lly + 5;2 + 3^^ — 69 = 0, J 
 Zy— z— U— ^ = (),)■ {h) 
 
 2y + dz + 4:U — d6 =0.) 
 
 The unknown quantity x is here eliminated, and we have three equa- 
 tions with only three unknown quantities. Now eliminating i/ by means 
 of the last equation, in the same way, and clearing of fractions, we find 
 the two equations. 
 
 4:X — 3y— z -}- u - 
 
 4:X -\- Sy -\- ^z -{- 4u - 
 
 - 7 = 
 -76 = 
 
 = 0, 
 
 = 0. 
 
 X — y + 2z -i- 2u - 
 X -}- 2y -\- z -\- u - 
 
 -10 = 
 -19 = 
 
 = 0, 
 
 = 0. 
 
 2x -{-2y — z — 2u- 
 2x + 4?/ + 2z + 2u - 
 
 - 2 = 
 
 -38 r 
 
 = 0, 
 = 0. 
 
 23^ + 38?^ — 258 = 0, 
 Uz + I4:2i — 90 = 
 
 ;} (o) 
 
118 EQUATIONS OF THE FIRST DEGREE. 
 
 The problem is now reduced to two equations with two unknown 
 quantities, which we have already shown how to solve. We find by 
 solving them, 
 
 ;2 = — 2, 
 
 U = S. 
 
 We next find the value of y by substituting these values of g and u 
 in either of the equations (&). The first of them thus becomes: 
 
 lly -^ 10 + 24 — 69 = 0, 
 
 from which we find, 
 
 y = b. 
 
 We now substitute the values of y, s, and u in either of equations {a). 
 The second of the latter becomes 
 
 a; — 5 — 4 + 16 — 10 = 0, 
 
 and the fourth becomes, 
 
 a; + 10 — 2 + 8 — 19 = 0, 
 
 either of which gives 
 
 X = d. 
 
 We can now prove the results by substituting the values of x, y, z, 
 and u in all four of equations (a), and seeing whether they are all satisfied. 
 
 EXERCISES. 
 
 1. One of the best exercises for the student will be that of 
 resolving the previous equations {a) by taking the last equa- 
 tion as the eliminating one, and performing the elimination 
 in different orders ; that is, begin by eliminating u, then 
 repeat the whole process beginning with z, etc. The final 
 results will always be the same. 
 
 2. Find the values oi x^, x^, x^, and x^, from the equa- 
 tions, 
 
 x^ + x^ +x^ + x^ = 64, 
 
 X -t ~j~ Xn — ~~ *^3 ~^~ 4 ~~™ 9 
 
 X -4 — "^ X 2 "i" X o "— " X 4 — ^ O, 
 fl^j 2^2 '^S ~r -^4 — - ^« 
 
 This example requires no multiplication, but only addition and sub- 
 traction of the different equations. 
 
 3. 2x+ 6y -\-3z = 13, 
 2x-\-2y— z = 12, 
 6x + by — 2z = 29. 
 
PROBLEMS. 119 
 
 ^z-\-2u- 
 
 ■ ^y 
 
 = 18, 
 
 
 Zx-\-y- 
 
 ■4.U 
 
 = 9, 
 
 
 x-\-lz- 
 
 ■ Qy 
 
 = 33, 
 
 
 hz — "Zx — 2>y -^ 
 
 2u 
 
 = 15. 
 
 
 X + y -^ z = a, 
 
 6. 
 
 1 1 
 X y 
 1 1 
 
 = Wj 
 
 Z -\- U + X =z C, 
 
 u + X + y = d. 
 
 
 ~y~z 
 1 1 
 
 = n, 
 
 
 
 -. + -.■ 
 
 = p. 
 
 PROBLEMS FOR SOLUTION. 
 
 1. A man had a saddle worth $75 and two horses. If the 
 saddle be put on horse A he will be double the value of B, but 
 if it be put on B his value will be equal to that of A. What 
 is the value of each horse ? 
 
 2. What number of two digits is equal to 7 times the sum 
 of its digits, and to 21 times the difference of its digits ? 
 
 Let X be the first digit, or the number of tens, and y the units. Then 
 the number itself will be lOx + y. Seven times the sum of the digits are 
 llx + ly, and 21 times the difference are 2>\x—2\y. Uniting and solving 
 the equations, we find x = Q, y = d ; the number is therefore 63. 
 
 3. A number of two digits is equal to 6 times the sum of 
 its digits, and if 9 be subtracted from the number the digits 
 are reversed. What is the number? 
 
 4. Find a number of two digits such that it shall be equal 
 to 6 times the sum of its digits increased by 1, while if 18 be 
 subtracted from the number the digits will be reversed. 
 
 5. Find a number which is greater by 2 than 5 times the 
 sum of its digits, and if 9 be added to it the digits will be 
 reversed. 
 
 6. What number is that which is equal to 9 times the sum 
 of its digits and is 4 greater than 11 times their difference ? 
 
 7. What fraction is that which becomes equal to f when 
 the numerator is increased by 2, and equal to ^ when the de- 
 nominator is increased by 4. 
 
 8. Two drovers A and B went to market with cattle. A 
 sold 50 and then had left half as many as B, who had sold 
 none. Then B sold 54 and had remaining half as many as A. 
 Ilow many did each have ? 
 
120 EQUATIONS OF THE FIRST DEOBEE. 
 
 9. A boy bought 42 apples for a dollar, giving 3 cents each 
 for the good ones and 2 cents each for the poor ones. How 
 many of each kind did he buy ? 
 
 10. Find a fraction which becomes equal to J- when its 
 denominator is increased by 13, and to f when 4 is subtracted 
 from its numerator. 
 
 11. Find a fraction which will become equal to f by adding 
 2 to its numerator, or by adding to its denominator 3, will be-_ 
 come \. 
 
 12. A huckster bought a certain number of chickens at 
 32 cents each and of turkeys at 75 cents each, paying $14 for 
 the whole. He sold the chickens at 48 cents each, and the 
 turkeys at $1 each, realizing $20 for the whole. How many 
 chickens and how many turkeys had he ? 
 
 13. An apple woman bought a lot of apples at 1 cent each, 
 and a lot of pears at 2 cents each, paying $1.70 for the whole. 
 
 11 of the apples and 7 of the pears were bad, but she sold the 
 good apples at 2 cents each and the good pears at 3 cents each, 
 realizing $2.60. How many of each fruit did she buy? 
 
 14. When Mr. Smith was married he was \ older than his 
 wife ; twelve years afterward he was \ older. What were their 
 ages when married ? 
 
 15. A and B together can do a piece of work in 6 days, but 
 A working alone can do it 9 days sooner than B working 
 alone. In what time could each of them do it singly ? 
 
 16. A husband being asked the age of himself and wife, 
 replied: "If you divide my age 6 years hence by her age 
 6 years ago, the quotient will be 2. But if you divide her age 
 
 12 years hence by mine 21 years ago, the quotient will be 5. 
 
 17. The sum of two ages is 9 times their difference, but 
 seven years ago it was only seven times their difference. What 
 are the ages now ? 
 
 18. Two trains set out at the same moment, the one to go 
 from Boston to Springfield, the other from Springfield to Bos- 
 ton. The distance between the two cities is 98 miles. They 
 meet each other at the end of 1 hr. 24 min., and the train from 
 Boston travels as far in 4 hrs. as the other in 3. What was the 
 speed of each train ? 
 
 19. A grocer bought 50 lbs. of tea and 100 lbs. of coffee for 
 $60. He sold the tea at an advance of \ on his price, and the 
 coffee at an advance of \, realizing $77 from both. At what 
 price per pound did he buy and sell each article ? 
 
 Note. If oc and y are the prices at wliicli lie bought, then \x and fy 
 are the prices at which he sold. 
 
INCONSISTENT EQUATIONS. 121 
 
 20. For p dollars I can purchase either a pounds of tea and 
 l pounds of cott'ee, or m. pounds of tea and n pounds of coll'ee. 
 What is the price per pound of each ? 
 
 21. A goldsmith had two ingots. The first is composed of 
 equal parts of gold and silver, while the second contains 5 parts 
 of gold to 1 of silver. He wants to take from them a watch- 
 case having 4 ounces of gold and 1 ounce of silver. How 
 much must he take from each ingot ? 
 
 22. A banker has two kinds of coin, such that a pieces of 
 the first kind or h pieces of the second will make a dollar. If 
 he wants to select c pieces which shall be worth a dollar, how 
 many of each kind must he take ? 
 
 23. A has a sum of money invested at a certain rate of 
 interest. B has $1000 more invested, at a rate 1 per cent, 
 higher, and thus gains $80 more interest than A. C has in- 
 vested $500 more than B, at a rate still higher by 1 per cent., 
 and thus gains $70 more than B. What is the amount each 
 person has invested and the rate of interest ? 
 
 24. A grocer had three casks of wine, containing in all 
 344 gallons. He sells 50 gallons from the first cask; then 
 pours into the first one-third of what is in the second, and 
 then into the second one-fifth of what is in the third, after 
 which the first contains 10 gallons more than the second, 
 and the second 10 more than the third. How much wine did 
 each cask contain at first ? 
 
 Equivalent and Inconsistent Equations. 
 
 141. It is not always the case that values of two unknown 
 quantities can be found from two equations. If, for example, 
 
 we have the equations 
 
 ic -{- 2?/ = 3, 
 ^x-\-4.y = 6, 
 
 we see that the second can be derived from the first by multi- 
 plying both members by 2. Hence every pair of values of x 
 and y which satisfy the one will satisfy the other also, so that 
 the two are equivalent to a single one. 
 If the equations were 
 
 a; + 2y = 5, 
 
 2x-\-^y = 6, 
 there would be no values of x and y which would satisfy both 
 equations. 
 
122 EQUATIONS OF THE FIRST DEGREE. 
 
 For, if we multiply the first by 2 and subtract the second 
 from the product, we shall have, 
 
 1st eq. X 2, 2a; + % = 10 
 
 2d eq., 2a; + 4y — 6 
 
 Remainder, 0=4, 
 
 an impossible result, which shows that the equations are incon- 
 sistent. This will be evident from the equations themselves, 
 because every pair of values of x and y which gives 
 
 2a; + 4^ = 6, 
 
 must also give a; + 2^ := 3, 
 
 and therefore cannot give cc + 2?/ = 5. 
 
 143. Generalization of the preceding result. If we take 
 any two equations of the first degree between x and y which 
 we may represent in the form 
 
 ax + iy = c, \ ,^. 
 
 a'x + h'y — c',) ^ ^ 
 
 and eliminate x by addition or subtraction, as in § 137, we have 
 for the equation in y, 
 
 {al — ah') y =z a'c — ac'. 
 Now it may happen that we have, 
 
 a'b — ai' =: identically. (2) 
 
 In this case y will disappear as well as x, and the result 
 will be 
 
 a'c — ac' = 0. 
 
 If this equation is identically true, the two equations (1) 
 will be equivalent ; if not true, they will be inconsistent. In 
 neither case can we derive any value of y or x. 
 
 If we divide the above equation, (2), by aa' we shall have 
 
 ^ _ ^. 
 a ~~ a' 
 Hence, 
 
 Theorem. If the quotient of the coeflScients of the 
 unknown quantities is the same in the two equations, 
 they will be either equivalent or inconsistent. 
 
INEQUALITIES. 123 
 
 This theorem can be expressed in the following form : 
 
 If the terms containing the unhnown quantity in the 
 one equation can he multiplied hy such a factor that 
 they shall both heconie equal to the corresponding terms 
 of the other equation, the two equations will he either 
 equivalent or inconsistent. 
 
 Proof. If there be such a factor m that multiplying the 
 first equation (1) by it, we shall have 
 ma = a\ 
 mh :=!)', 
 Eliminating w, we find 
 
 al — ah' = 0, 
 the criterion of inconsistency or equivalence. 
 
 143. When two equations are inconsistent, there are no 
 values of the unknown quantities which will satisfy both equa- 
 tions. 
 
 When they are equivalent, it is the same as if we had a 
 single equation ; that is, we may assign any value we please to 
 one of the unknown quantities, and find a corresponding value 
 of the other. 
 
 CHAPTER IV. 
 
 OF IN EQUALITIES. 
 
 144. Def. An Inequality is a statement, in the 
 language of Algebra, that one quantity is algebraically 
 greater or less than another. 
 
 Def. The quantities declared unequal are called 
 Members of the inequality. 
 
 The statement that A is greater than B, or that A — B is 
 positive, is expressed by 
 
 A> B. 
 
124 INEQUALITIES. 
 
 That A is less than B, or that ^ — ^ is negative is 
 expressed by 
 
 A < B. 
 
 The form A> B> G 
 
 indicates that the quantity B is less than A but greater than C, 
 
 The form A^ B ^ 
 
 indicates that A may be either equal to or greater than B, but 
 cannot be less than B, 
 
 Properties of Inequalities. 
 
 145. Theorem I. An inequality will still snlbsist 
 after the same quantity has been added to or subtracted 
 from each member. 
 
 Proof. If the inequality be Ay B, A — B must be posi- 
 tive. If we add the same quantity ffto A and B, or subtract 
 it from them, we shall have A ± II — (B ±: H), which is 
 equal to A — B, and therefore positive. Hence, if 
 
 A> B, 
 
 then A±H> B ±H. 
 
 Cor. If any term of an inequality be transposed 
 and its sign changed, the inequality will remain true. 
 
 Theorem II. An inequality will still subsist after 
 its members have been multiplied or divided by the 
 same positive number. 
 
 Proof, li A — B \% positive, then (m or tz being positive) 
 m {A — B) or mA — mB will be positive, and so will 
 
 B 
 
 n 
 
 
 A 
 
 -B A 
 
 or — ■ 
 
 Qi n 
 
 Hence, if 
 
 
 A> B, 
 
 then 
 and 
 
 
 mA > mB, 
 n n 
 
INEQCIALITIES. 125 
 
 It may be shown in the same way that if 772 or w is negative, 
 
 A B 
 
 mA — mB or will be negative. Hence, 
 
 n n ^ 
 
 Theorem III. If both memlbers of an inequality Ibe 
 multiplied or divided by the same negative number, 
 the direction of the inequality will be reversed. 
 
 That is, if Ay B, 
 
 then — mA < — mB, 
 
 and < 
 
 n 71 
 
 Theorem IV. If the corresponding members of 
 several inequalities be added, the sum of the greater 
 members will exceed the sum of the lesser members. 
 
 Theorem V. If the members of one inequality be 
 subtracted from the non-corresponding members of 
 another, the inequality will still subsist in the direction 
 of the latter. 
 
 That is, if A > B, 
 
 oc> y, 
 then A —y y B — X. 
 
 The proof of the last three theorems is so simple that it may be sup- 
 plied by the student. 
 
 Theorem VI. If two positive members of an in- 
 equality be raised to any power, the inequality will 
 still subsist in the same direction. 
 
 Proof. Let the inequality be 
 
 A> B. {a) 
 
 Because A is positive, we shall have, by multiplying by A 
 (Th. II), 
 
 A^ > AB. (1) 
 
 Also, because B is positive, we have, by multiplying {a) 
 
 by B, 
 
 AB > &. (2) 
 
126 INEQUALITIES. 
 
 Therefore, from (1) and (2), 
 
 A'^ > B\ (3) 
 
 Multiplying the last inequality by A, 
 
 A^ > AE^. (4) 
 
 Multiplying (2) by B, 
 
 AE^ > B\ (5) 
 
 Whence, A^ > BK 
 
 The process may be continued to any extent. 
 
 Examples of the Use of Inequalities. 
 
 146. Ex. I. If a and h be two positive quantities, such 
 that 
 
 a^ + Z>2 = 1, 
 
 •we must have a -\- l y 1. 
 
 Proof. If a + 1^1, 
 
 we should have, by squaring the members (Th. YI), 
 
 a^ + 2«5 + ^»2 = 1 ; 
 and by transposing the product 2a5 (Th. I, Cor.), 
 
 «2 _^ Z»2 = 1 _ ^ah 
 
 Because a and I are positive, 2flJ is positive, and 
 
 l — 2ab< 1. 
 Therefore we should have 
 
 a^-\-y^ < 1, 
 and could not have a^ -\-J?=il^ as was originally supposed. 
 
 Ex. 2. lia, h, m, and n are positive quantities, such that 
 
 l>n' (''> 
 
 then the value of the fraction will be contained between 
 
 the values of t and — ; that is. 
 
INEQUALITIES. 127 
 
 a a-\-m m . . 
 
 h ^ b +n ^ n ^^ 
 
 To prove fhe first inequality, we must show that 
 a a + m 
 
 is positive. Eeducing this expression by § 106, it becomes 
 
 an — hn . . 
 
 T{V+nj' ^^ 
 
 From the original inequality {a) we have, by multiplying 
 by the positive factor In, 
 
 an > hm. 
 
 That is, an — hm is positive ; therefore the fraction (3) 
 with this positive numerator is also positive, and (2) is positive 
 as asserted. 
 
 The second inequality (1) may be proved in the same way. 
 
 EXERCISES. 
 
 1. Prove that if a and h be any quantities different from 
 zero, and 1 > a; > — 1, we must have 
 
 a^ — 2abx + i^ > o. 
 
 2. Prove that ( j > ab. 
 
 3. If 3a; — 5 > 13, then x > 6. 
 
 4. If 6x>Y + 18> then x > i. 
 
 5. If J-y>|-3, then«>5. 
 
 7) — 7)1 
 
 6. li m — nxy p — qx, then x > 
 
 ^ ^ q — n 
 
 7. If ^-^I-^ < 1 , and m is positive, then x <, ij, 
 
 'my 
 
 8. If ft2 _}_ J2 _|_ ^2 = 1, and a, l, and c are not all equal, 
 then ab -^ he ■\- ca <i 1. 
 
 Suggestion. The squares of a — &, 6 — c, and c — a cannot be 
 
 negative. 
 
BOOK IV. 
 RATIO AND PROPORTION. 
 
 CHAPTER I. 
 NATURE OF A RATIO 
 
 147. Bef. The Ratio of a quantity A to another 
 quantity ^ is a number expressing the value of A when 
 compared with B as the standard or unit of measure. 
 
 Examples. Comparing j^ , | , [ [ | | [ j | 
 
 the lengths A, B, C, D, it ^^ 
 
 will be seen that ' ' 
 
 A is 2^ times D', C | i i i 
 
 B is i of D; x> | | | | j 
 C is I of D. 
 
 "We express this relation by saying, 
 
 The ratio ot A to B is 21 or , 
 
 " " B to B is I; ) (1) 
 
 " " C to B is 7. 
 
 4 
 
 148. The ratio of one quantity to another is expressed by 
 writing the unit of measure after the quantity measured, and 
 inserting a colon between them. 
 
 The statements (1) will then be expressed thus : 
 
 A:B = 2i = l; B : B = \', '. B = \ 
 
 Bef. The two quantities compared to form a ratio 
 are called its Terms. 
 
RATIO. 129 
 
 Def. The quantity measured, or the first term of 
 the ratio, is called the Antecedent. 
 
 The unit of measure, or the second term of the ratio, 
 is called the Consequent. 
 
 Eem. When the antecedent is greater than the consequent, 
 the ratio is greater than unity. 
 
 When the antecedent is less than the consequent, the ratio 
 is less than unity. 
 
 149. To find the ratio of a quantity ^4 to a standard U, 
 we imagine ourselves as measuring off the quantity A with C/'as 
 a carpenter measures a board with his foot-rule. 
 
 There are then three cases to be considered, according to 
 the way the measures come out. 
 
 Case I. We may find that, at the end, A comes out an 
 exact number of times U. The ratio is then a whole number, 
 and we say that U exactly measures A, or that ^ is a 
 multiple of U. 
 
 Case II. We may find that, at the end, the measure does 
 not come out exact, but a piece of A less than U is left over. 
 Or, A may itself be less than U. We must then find what 
 fraction of CT the piece left over is equal to. This is done by 
 dividing U up into such a number of equal parts that one of 
 these parts shall exactly measure A or the piece of A which is 
 left over. The ratio will then be a fraction of which the num- 
 ber of parts into which U is divided will be the denominator, 
 and the number of these parts in A the numerator. 
 
 Example. If we find that . i . i ■ _ ^ 
 
 by dividing U into 7 parts, 4 of -._«.^_— , 
 
 these parts will exactly make A, I'll' ~ 
 
 •then A=^ ol U, and we have for the ratio of A to U, 
 
 A:U=t 
 
 If we find that A contains U 3 times, and that there is 
 then a piece equal to ^ of £/" left over, we have 
 
 A:U=H = ^. 
 
 9 
 
130 RATIO. 
 
 The 3 ^J's are equal to V- of U, so that we may also say 
 
 A = '^-oiU, or A:U= -t 
 
 which is simply the result of reducing the ratio 34^ to an im- 
 proper fraction. 
 
 In general, if we find that by dividing U into 7i parts, A 
 will be exactly m of these parts, then 
 
 n 
 whether m is greater or less than n. 
 
 When the magnitude of A measured by U can be exactly 
 expressed by a vulgar fraction, A and U are said to be com- 
 mensurable. 
 
 Case III. It may happen that there is no number or frac- 
 tion which will exactly express the ratio of the.two magnitudes. 
 The latter are then said to be incommensurable. 
 
 150. Theorem. The ratio of two incommensuralble 
 magnitudes may always be expressed as near the true 
 value as we please by means of a fraction, if we only 
 make the denominator large enough. 
 
 Examples. Let us divide the unit of measure into 20 
 parts, and suppose that the antecedent contains more than 28 
 but less than 29 of these parts. Then, by supposing it to con- 
 tain 28 parts, the limit of error will be one part, or ^^ of the 
 standard unit. 
 
 In general, if we wish to express the ratio within 1 n^h of 
 the unit, we can certainly do it by 'dividing the unit into n or 
 more parts, or by taking as the denominator of the fraction a 
 number not less than n. 
 
 Illustration hy Deci7nal Fractions. The square root of 2 
 cannot be rigorously expressed as a vulgar or decimal fraction. ' 
 But, if we suppose 
 
 ^2 = 1.4 = If, the error will be < ^ ; 
 
 V2 = 1.41 =ifi, " " <Ti7r; 
 
 ^2 = 1.414 = IfU, " " <Ti^nr- 
 
 etc. etc. etc. etc. 
 
NATURE OF A RATIO, 131 
 
 Since the decimals may be continued without end, the 
 square root of 2 can be expressed as a decimal fraction with an 
 error less than any assignable quantity. This general fact is 
 expressed by saying : 
 
 Tlie limit of the error which we mahe hy representing 
 an incoimnensurahle ratio as a fraction is zero. 
 
 151 . Ratio as a Quotient, From Case II and the explana- 
 tions which precede it we see that when we say 
 
 A:U = ^, 
 
 we mean the same thing as if we had said, 
 
 ^ is 4- of U, or A = ^U. 
 
 If A and U are numbers, we may divide both sides of this 
 equation by Uy and obtain, 
 
 We therefore conclude that when A and U are numbers, 
 
 A ' TT — A. 
 
 That is, ^ ' "^ - U 
 
 Theorem. The ratio of two numbers is equal to the 
 quotient obtained by dividing the antecedent term by 
 the consequent. 
 
 In the case of magnitudes, the relation of a ratio to a quo- 
 tient may be shown thus : 
 
 Let us have two magnitudes M and F, such that M is 
 4 times V, Then we may write the relation, 
 
 if=4F. 
 
 Dividing by 4, we have 
 
 Since V is not a number, we cannot, strictly speaking, 
 multiply or divide by it. But we may take the ratio of M to 
 V without regard to number, and thus find, 
 
 if : F = 4. 
 
132 BATIO. 
 
 Kem. The theory of ratios the terms of which are magni- 
 tudes and not numbers, is treated in Geometry. 
 
 In Algebra we consider the ratios of numbers, or of mag- 
 nitudes represented by numbers. 
 
 152. Def. If we interchange the terms of a ratio, 
 the result is called the Inverse ratio. 
 
 That is, Ui A is the inverse o^ A :U. 
 
 w 
 If U: A = -, 
 
 n 
 
 then U z= — A, 
 
 n 
 
 and we have, by dividing by — , 
 
 A = ~U, 
 
 m 
 
 A TT ^ 
 
 or A : U = — 
 
 m 
 
 Because — is the reciprocal of — , we conclude : 
 
 Theorem. The inverse ratio is the reciprocal of the 
 direct ratio. 
 
 Properties of Ratios. 
 
 153. Theorem I. K both terms of a ratio Ibe multi- 
 plied by the same factor or divided by the same divisor, 
 the ratio is not altered. 
 
 Proof. Ratio of ^ to ^ = ^ : ^ = -j. 
 
 If m be the factor, then 
 
 Ratio of mB to mA = mB : mA = — -: = -r, 
 
 mA A' 
 
 the same as the ratio of B to A. 
 
 154. Theorem II. If both terms of a ratio be in- 
 creased by the same quantity, the ratio will be increased 
 
PBOPOBTION'. 133 
 
 if it is less than 1, and diminislied if it is greater than 1 ; 
 that is, it will be brought nearer to unity. 
 
 Example. Let the original ratio be 2 : 5 = f . If we repeatedly add 
 
 1 to both numerator and denominator of the fraction, we shall have the 
 
 series of fractions, 
 
 I, f. I-, I, etc., 
 
 each of which is greater than the preceding, because 
 
 f-| = /ir; whence, f > f . 
 
 f - f = t\ ; whence, 4 > f . 
 
 f - 4 = /^ ; whence, | > f . 
 
 etc. etc. 
 
 General Proof, Let « : 5 be the original ratio, and let 
 both terms be increased by the quantity u, making the new 
 ratio a-\-u : b-^u. The new ratio mimis the old one will be 
 
 {h — a)u 
 
 If h is greater than «, this quantity will be positive, show- 
 ing that the ratio is increased by adding u. If h is less than a, 
 the quantity will be negative, showing that the ratio is dimin- 
 ished by adding w. 
 
 ^-*-¥ 
 
 CHAPTER II. 
 PROPORTION. 
 
 155. Def, Proportion is an equality of two or 
 more ratios. 
 
 Since each ratio has two terms, a proportion must have at 
 least four terms. 
 
 Def, The terms which enter into two equal ratios 
 are called Terms of the proportion. 
 
 li a'.h be one of the ratios, and p : q the other, the pro- 
 portion will be, 
 
 a-.h ^p:q. (1) 
 
134 PROPOBTION. 
 
 A proportion is sometimes written, 
 
 a : & : : p : q, 
 which is read, " As « is to 6 so is p to q." The first form is to be pre- 
 ferred, because no other sign than that of equality is necessary, but the 
 equation may be read, ** As a is to 6 so is jp to q" whenever that expres- 
 sion is the clearer. 
 
 Def, The first and fourth terms of a proportion are 
 called the Extremes, the second and third are called 
 the Means. 
 
 Theorems of Proportion. 
 
 156. Theorem I. In a proportion the product of 
 the extremes is equal to the product of the means. 
 
 Proof. Let us write the ratios in the proportion (1) in the 
 form of fractions. It will give the equation, 
 
 Multiplying both sides of this equation by hq, we shall have 
 aq = bj), (3) 
 
 Cor. If there are two unknown terms in a propor- 
 tion, they may be expressed by a single unknown 
 symbol. 
 
 Example. If it be required that one quantity shall be to 
 another as p to q, we may call the first px and the second qx, 
 because 
 
 px : qx =: p : q (identically). 
 
 157. Theorem II. If the means in a proportion be 
 interchanged, the proportion will still be true. 
 
 Proof. Divide the equation (3) by pq. We shall then 
 have, instead of the proportion (1), 
 
 a h 
 
 or a \ p = h : q. 
 
PROPORTION. 135 
 
 Def. The proportion in wMcli the means are inter- 
 changed is called the Alternate of the original pro- 
 portion. 
 
 The following examples of alternate proportions should be studied, 
 and the truth of the equations proved l»y calculation : 
 
 1:2= 4 : 
 
 ; 8; 
 
 alternate, 
 
 1 : 
 
 ; 4 = 2 : 
 
 ; 8. 
 
 2:3= 6 : 
 
 ;9; 
 
 (( 
 
 2 ; 
 
 : 6 = 3 : 
 
 : 9. 
 
 5 : 2 = 25 : 
 
 10; 
 
 (( 
 
 5 : 
 
 : 25 = 2 : 
 
 ; 10. 
 
 158. Theorem III. If, in a proportion, we increase 
 or diminish each antecedent by its consequent, or each 
 consequent by its own antecedent, the proportion will 
 still be true. 
 
 Example. In the proportion, 
 
 5 : 2 = 25 : 10, 
 
 the antecedents are 5 and 25, the consequents 2 and 10 (§ 148). Increasing 
 each antecedent by its own consequent, the proportion will be 
 5 + 2:2 = 25 + 10:10, or 7:2 = 35:10. 
 
 Diminishing each antecedent by its consequent, the proportion will 
 become, 
 
 5 - 2 : 2 = 25 - 10 : 10, or 3 : 2 = 15 : 10. 
 
 Increasing each consequent by its antecedent, the proportion will be 
 
 5 : 2 + 5 = 25 : 10 + 25, or 5 : 7 = 25 : 35. 
 
 These equations are all to be proved numerically. 
 
 General Proof, Let us put the proportion in the form 
 
 1) q ^ ' 
 
 If we add 1 to each side of this equation and reduce each 
 side, it will give 
 
 h - q ^ 
 that is, a -^l \ 1) ■= p -\- q '. q, (5) 
 
 In the same way, by subtracting 1 from each side, it will be 
 a — h '. h = p — q '. q. (6) 
 
136 PROPORTION. 
 
 If we invert the fractions in equation (4), the latter will 
 become 
 
 a p 
 
 By adding or subtracting 1 from each side of this equation, 
 
 and then again inverting the terms of the reduced fractions, 
 
 we shall find, 
 
 a : h -\-a = p : q -\-p; 
 
 a : i — a = p : q —p. 
 
 The form (5) was formerly designated as formed "by composition," 
 and (6) as formed " by division." But these terms are now useless, be- 
 cause all the above forms are only special cases of a more general one to 
 be now explained. 
 
 159. Theorem IV, If four quantities form the pro- 
 portk)n 
 
 a \ l = c : d^ {a) 
 
 and if m, n^ p^ and q be any multipliers whatever, we 
 shall have 
 
 ma + nh : pa + qh — mc + nd : pc + qd. 
 
 Proof. The proportion (a) gives the equation. 
 
 1)~ d 
 
 Multiplying this equation by - and adding 1 to each 
 member, 
 
 Eeducing each member to a fraction and inverting the 
 terms, 
 
 qh _ qd 
 pa -{■ qb ~ pc -\- qd 
 Dividing both members by qy 
 
 ~i— ='— ^-. (7) 
 
 2M -j- qb pc -\- qd ^ ' 
 
 The original proportion (a) also gives, by inversion, 
 
PROPORTION. 137 
 
 h_d 
 a~ c' 
 
 from which we obtain, by multiplying by ^, adding 1, etc., 
 
 ql + p a _ qcl + pc 
 IM ~ pc 
 
 pa -}- qb pc -]- qd ^ ' 
 
 (8) X m + (7) X w gives the equation, 
 
 ma -\- nh mc + nd 
 
 pa -\- qh ~ pc -\- qd ' 
 or ma -\- nb \ pa -\- ql) =^ mc + nd : pc ■{■ qd, (9) 
 
 which is the result to be demonstrated. 
 
 160. Theorem V, If each term of a proportion be 
 raised to the same power, the proportion will still 
 subsist. 
 
 Proof. If a : b = p : q, 
 
 a p 
 
 b q' 
 then, by multiplying each member by itself repeatedly, we 
 shall have 
 
 a' 
 
 
 
 
 etc. etc. 
 Hence, in general, 
 
 a^ : b"' = p'^ : q^. 
 
 Cor, If a : b = p • q, 
 
 then a^ : a^ ^b"- = p^ : p^ ±q^ ; 
 
 and a^ ±b"' : b^ = p"' ± q^ : q^. 
 
 Theorem VI. When three terms of a proportion 
 are given, the fourth can always be found from the 
 theorem that the product of the means is equal to that 
 of the extremes. 
 
138 PROPORTION. 
 
 We have shown that whenever 
 
 a '. 1) = 2^ : q, 
 then aq = bp. 
 
 Considering the different terms in succession as unknown 
 quantities, we find, 
 
 dp 
 
 I = ^^-, 
 P 
 
 V - -^^ 
 
 ip 
 q = —' 
 
 Cor, 1. If, in the general equation of the first 
 degree 
 
 ax ^hy = c, 
 
 the term c vanishes, the equation determines the ratio 
 of the unknown quantities. 
 
 Proof. If ax -\- by = 0, 
 
 then ax =: — by, 
 
 , X b 
 
 and - = , 
 
 y a 
 
 or . • X \ y =. —b \ a. 
 
 Cor. 2. Conversely, if the ratio of two unknown 
 quantities is given, the relation between them may Ibe 
 expressed by an equation of the first degree. 
 
 The Mean Proportional. 
 
 161. Def. When the middle terms of a proportion 
 are equal, either of them is called tlie Meein Propor- 
 tional between the extremes. 
 
 The fact that b is the mean proportional between a and c 
 is expressed in the form, 
 
 a : b = b : c. 
 
proportion: 139 
 
 Theorem I then gives, W' =z ac. 
 
 Extracting the square root of both members, we have 
 
 Hence, 
 
 Theorem YII. The mean proportional of two qnan* 
 tides is equal to the square root of their product. 
 
 Multiple Proportions. 
 
 163. "We may have any number of ratios equal to each 
 other, as 
 
 a \ h ^ G '. d ■= e \ f, etc. 
 
 6 : 4 = 9 : 6 = 3 : 2 = 21 : 14. («) 
 
 Such proportions are sometimes written in the form 
 
 6 : 9 : 3 : 21 == 4 : 6 : 2 : 14. (5) 
 
 In the form (5) the antecedents are all written on one side 
 of the equation, and the consequents on the other. Any two 
 numbers on one side then have the same ratio as the cor- 
 responding two on the other, and the proportions expressed by 
 this equality of ratios are the alternates of the original propor- 
 tions («). For instance, in the proportion (Z*) we have, 
 
 6:9 =4:6, which is the alternate of 6 : 4 = 9:6. 
 
 6:3=4:2, " " " 6:4=3:2. 
 
 6 : 21 = 4 : 14, " " " 6 : 4 = 21 : 14. 
 
 9 : 21 = 6 : 14, " " " 9 : 6 = 21 : 14. 
 
 1G3. A multiple proportion may also be expressed by a 
 number of equations equal to that of the ratios. Since 
 
 a : h' := c \ d =z e : f, etc., 
 
 let us call r the common value of these ratios, so that 
 
 a c , 
 
 * = '■' 1 = '' <=*'=• 
 
 Then a — rd, 
 
 c = rd, (c) 
 
 e — rf, 
 
140 PROPORTION. 
 
 will express the same relations between the quantities a, h, c, 
 d, e, fy etc., that is expressed by 
 
 a : b =z c : d •= e : f, etc., (a) 
 
 or a : c ', 6 : etc. =^ b \ d : f -. etc. (b) 
 
 It will be seen tliat where r enters in the form (c) there is one more 
 equation than in the first form {a). [In this form each = represents an 
 equation.] This is because the additional quantity r is introduced, by 
 eliminating which we diminish the number of equations by one, as in 
 eliminating an unknown quantity. 
 
 164. Theorem, In a mnltiple proportion, the sum 
 of any number of the antecedents is to the sum of the 
 corresponding consequents as any one antecedent is to 
 its consequent. 
 
 T. xnr 1 2 6 10 12 ^. 
 
 Ex. We have - = — =—== — • Then 
 5 15 2o 30 
 
 2 + 6 + 10 + 12 30 
 
 5 + 15 + 25 + 30 75' 
 which has the same value as the other four functions. 
 
 Oeneral Proof. Let A, B, C, etc., be the antecedents, and 
 a, b, c, etc., the corresponding consequents, so that 
 
 A: a = B '.b = : c, etc. (1) 
 
 Let us call r the common ratio A\ a, B -.b, etc., so that 
 
 A = r«, 
 B = rb, 
 
 C = re. 
 etc. etc. 
 
 Adding these equations, we have 
 
 A -{• B -i- G-^ etc. = r{a + b + c-{- etc.), 
 
 A + B -\- C -\- etc. 
 
 or — —^ 1 — = r; 
 
 a -\-b + c + etc. 
 
 that is, the ratio ^H-^+C+etc. : a + ^ + c + etc. is equal to 
 r, the common value of the ratios A : a, B \b, etc. 
 
 PROBLEMS. 
 
 I. A map of a country is made on a scale of 5 miles to 
 3 inches. 
 
PROPORTION, 141 
 
 (1.) What will be the length of 8, 12, 17, 20, 33 miles on 
 the map ? 
 
 (2.) How many miles will be represented by 6, 8, 16, 20, 
 29 inches on the maj) ? 
 
 Rem. 1. If X, y, z, u, 1) be the required spaces on the map, we shall 
 have 
 
 5 : S = 8 : X — 12 : y, etc. 
 
 If a, b, c, etc., be the required number of miles, we shall have 
 3:5 = 6:a = 8:&==16:c, etc. 
 
 Rem. 3. When there are several ratios compared, as in this problem, 
 it will be more convenient to take the inverse of the common ratio, and 
 multiply the antecedent of each following ratio by it to obtain the conse- 
 quent. In the first of the above proportions the inverse ratio is f , and 
 
 a; = I of 8, y = f of 12, etc. 
 In the second, a = f of 6, & = f of 8, etc. 
 
 2. To divide a given quantity A into three parts which 
 shall be proportional to the given quantities a, b, c, that is, 
 into the parts x, y, and z, such that 
 
 X : a ^y '. l = z '. c, 
 or X : y \ z =z a \ h : c. 
 
 Solution. By Theorem IV, 
 
 x_y_z_x-{-y-\-z_ A 
 a~h~c~~a-\-b-\-c~a-\-b-\-c 
 
 Therefore, 
 
 _ aA _ ^-^ _ ^^ 
 
 a-^b+c' ^ a + b-^c' a -\- b -\- c 
 
 3. Divide 102 into three parts which shall be proportional 
 to the numbers 2, 4, 11. 
 
 4. Divide 1000 into five parts which shall be proportional 
 to the numbers 1, 2, 3, 4, 5. 
 
 5. Find two fractions whose ratio shall be that of a : J, and 
 whose sum shall be 1. 
 
 6. What two numbers are those whose ratio is that of 7 : 3 
 and whose difference is 24. 
 
 7. What two numbers are those whose ratio is m : w, and 
 whose difference is unity ? 
 
 8. Find x and y from the conditions, 
 
 X '. y ^ a \ b, 
 ax — by = a -{■ i. 
 
142 PROPORTION. 
 
 9. Show that if a : b = A : B, 
 
 c: d= C : D, 
 we must also have ac : bd = AC : BD. 
 
 10. Having given x = aij, find the value of ^ ^^ 
 
 x — 2y 
 
 11. Having given 
 
 x + '^y ^ 
 = 5, 
 
 x — 'Zy 
 
 x — y 
 12. If a : b = 2^ '. q, 
 
 find the value of 
 
 ci p 
 
 prove O!^ _j_ 52 . — ^^^ _ ^? _]_ ^2 . ^t — 
 
 and «« + j->:^__-_=^. + j™:-^--. 
 
 13. If 
 
 a -\-b — c — d a — b — c -\- d* 
 show that a : b = c : d. 
 
 14. A year's profits were divided among three partners, A, 
 B, and C,* proportional to the numbers 2, 3, and 7. If 
 should pay B $1256, their shares would be equal. What was 
 the amount divided ? 
 
 15. In a first year's partnership between A and B, A had 
 2 shares and B had 5. In the second year, A had 3 and B had 4. 
 In the second year, A's profits were $3200 greater and B's were 
 $1700 greater than they were the first. What was each year's 
 profits ? 
 
 16. In a poultry yard there are 7 chickens to every 2 ducks, 
 and 3 ducks to every 2 geese. How many geese were there to 
 every 42 chickens ? 
 
 17. A drover started with a herd containing 4 horses to' 
 every 9 cattle. He sold 148 horses and 108 cattle, and then 
 had 1 horse to every 3 cattle. How many horses and cattle 
 had he at first ? 
 
 18. If a bowl of punch contains a parts of water and b 
 parts of wine, what is the ratio of the wine to the whole 
 punch ? W^hat is the ratio of the water ? What are the sums 
 of these ratios ? 
 
PROPORTION. 143 
 
 19. One ingot consists of equal parts of gold and silver, 
 while another has two parts of gold to one of silver. If I 
 combine equal weights from these ingots, what proportion of 
 the compound will be gold and what proportion silver ? 
 
 20. What will be the proportions if, in the preceding prob- 
 lem, I combine one ounce from the first ingot with three from 
 the second? 
 
 21. One cask contains a gallons of water and h gallons of 
 alcohol, Avhile another contains m gallons of water and n of 
 alcohol. If I draw one gallon from each cask and mix them, 
 what will be the quantities of alcohol and water ? 
 
 22. What will be the ratio of the liquors in the last case, if 
 I mix two parts from the first cask with one from the second ? 
 
 23. What will it be if I mix p parts from the first with q 
 parts from the second ? 
 
 24. A goldsmith has two ingots, each consisting of an alloy 
 of gold and silver. If he combines two parts from the first 
 ingot with one from the second, he will have equal parts of 
 gold and silver. If he combines one part from the first with 
 two from the second, he will have 3 parts of gold to 5 of silver. 
 What is the composition of each ingot ? 
 
 Suggestion. Call r tlie ratio of tlie weight of gold in the first ingot 
 to the whole weight of the ingot ; then 1 — r will be the ratio of the sil- 
 ver in the first to the whole weight of the ingot. See the following 
 question. 
 
 Note. Problems 18-24 form a graduated series, introductory to the 
 processes of Problem 24. 
 
 25. Point out the mistake which would be made if the 
 solution of the preceding problem were commenced in the fol- 
 lowing way : 
 
 If the first ingot contains p parts of gold to q parts of silver, and the 
 second contains r parts of gold to s of silver, then 
 
 Two parts from the first ingot will have 2^ of gold and 2q of silver. 
 
 One part from the second ingot will have r of gold and s of silver. 
 
 Therefore, the combination will contain 2p + r parts of gold," and 
 2q-{-s parts of silver. 
 
 Show also that if we subject p, q, r, and s to the condition 
 P -\-q = r + s, 
 the process would be correct. 
 
 26. Show that if the second term of a proportion be a 
 mean proportional between the third and fourth, the third 
 will be a mean proportional between the first and second. 
 
BOOK V. 
 OF POWERS AND ROOTS. 
 
 CHAPTER I 
 
 INVOLUTiON. 
 
 Case I. Involvition of Products and Quotients, 
 
 165. Def. The result of taking a quantity, A^ 
 n times as a factor is called the ri*'* power of JL, and 
 as already known may be written either 
 
 AAA, etc., n times, or A"-. 
 
 Def. The number n is called the Index of the 
 power. 
 
 Def. Involution is the operation of finding the 
 powers of algebraic expressions. 
 
 The operation of invohition may always be expressed by 
 the appUcation of the proper exponent, the expression to be 
 involved being inclosed in parentheses. 
 
 Example. The n*^ power oi a + h is (a + h)^. 
 The n^^ power of abc is (abc^. 
 
 166. Involution of Products. The n*^ power of the 
 product of several factors a, h, c, may be expressed without 
 exponents as follows : 
 
 abcabcdbc, etc., 
 
 each factor being repeated n times. 
 
INVOLUTION, 145 
 
 Here there will be altogether n a% n h% and n (fs, so 
 that, using exponents, the whole power will be a^'b^c"' (§ 66, 67). 
 Hence, {abo)"' = a^b'^c^. 
 
 That is, 
 
 Theorem. The power of a product is equal to the 
 product of the powers of the several factors. 
 
 167. Involution of Quotients. Applying the same methods 
 
 X ft^ 
 
 to fractions, we find that the n^^ power of - is — • For 
 
 y yn 
 
 
 IxY" XXX 
 
 (-) = > etc., 
 
 V y y y 
 
 n times 
 
 j 
 
 
 
 XXX, etc., n 
 
 ~~ yyy, etc., n 
 
 times ,g 
 times ^^ 
 
 109); 
 
 
 
 - yn 
 
 
 
 
 
 EXERCISES. 
 
 
 
 Express 
 
 the cubes of 
 
 
 
 
 I. abc. 
 
 ab 
 
 2. — • 
 
 c 
 
 3- 
 
 dbc-\ 
 
 
 mn 
 
 a-\-b 
 ^' a-b 
 
 6. 
 
 mn (a 
 pq{a 
 
 -b) 
 
 Express the n^^ powers of the same quantities, the quanti- 
 ties between parentheses being treated as single symbols. 
 
 Case II. Involution of Powers. 
 
 168. Problem. It is required to raise the quantity a^ to 
 
 the n*^ power. 
 
 Solution. The n^ power of a"^ is, by definition, 
 a"* X a"* X a"*, etc., n times. 
 
 By § 66, the exponents of a are all to be added, and as the 
 exponent m is repeated n times, the sum 
 
 m -\- m -\- m + etc., n times, 
 is mn. Hence the result is a'^^ or, in the language of Algebra, 
 
 10 
 
146 IJS VOLUTION. 
 
 Hence, 
 
 TJieorem. If any power of a quantity is itself to Ibe 
 raised to a power, the indices of the powers must be 
 multiplied together. 
 
 EXAMPLES. 
 
 Note. It will be seen that this theorem coincides with that of Case I 
 when any of the factors have the exponent unity understood. 
 
 EXERCISES. 
 
 Write the cubes of the following quantities: 
 
 I. Zxy\ 2. y. 
 
 3. 
 
 a"^. 
 
 4. W. 5. 2a2m». 
 
 6. 
 
 b 
 
 Write the n^^ powers of 
 
 
 
 7. a. 8. a^K 
 
 9- 
 
 aWc. 
 
 10. «^«^ II. ^p^f. 
 
 12. 
 
 (a^h) {c + d). 
 
 13- {^:^y){^ — y)' 
 
 
 
 14. ^{a-\-})-c){a-'by. 
 
 
 
 Arts, 
 
 7^ (a, + Z* — 
 
 cY {a — lyp. 
 
 a ^ «2 
 .15. ^- 16. p. 
 
 17- 
 
 x + y 
 x-y 
 
 „ . m^dl?' 
 
 
 ra^nanizn 
 
 xy^ x^y^^ 
 
 al {c — d f 
 
 Reduce : 
 
 20. (2aPn^Y. 21. (—dmnx^)^ 
 
 22. 2a{—di^mn^y, 23. (7jt?^V)*. 
 
 24. («Z»^)«. 25. (2fl2:2;8jn 26. (m^)^ 
 
 Note 1. If the student find any of these exponential expressions 
 diflBcult of expression, he may first express them by writing each quantity 
 a number of times indicated by its exponent. 
 
 Note 2. The student is expected to treat the quantities in paren- 
 theses as single symbols. 
 
INVOL UTION. 147 
 
 Eem. The preceding theorem finds a practical application 
 when it is necessary to raise a small number to a high power. 
 If, for example, we have to raise 2 to the 30th power, we 
 should, without this theorem, have to multiply by 2 no less 
 than 29 times. But we may also proceed thus : 
 
 22 = 4, 
 
 2^ = 22.22 = 4-4 = 16, 
 
 28 = 2^.24 =16.16 = 256, 
 ■ 216 ^ 28.28 = 2562 _ 65536, 
 
 224 _ 216.28 r= 216.256 = 16777216, 
 
 230 = 224.26 = 224.64 — 1073741824. 
 
 Case of Negative Exponents. 
 
 169. The preceding theorem may be applied to negative 
 exponents. By the definition of such exponents, 
 nP 
 
 jF = '"'^-'- (1) 
 
 Eaising the first member to the n^^ power, we have. 
 
 This is the same result we should get by applying the 
 theorem to the second member of (1), and proves the proposi- 
 tion. 
 
 EXERCISES. 
 
 Express the 6th powers of 
 
 T. ab-\ 2. a^l-K 
 
 3. amp~^. 4. a~^'b~^. 
 
 5. (« + If {a — l))-\ 6. {x + yY {x + z)-^, 
 ■arP {^±hrrn 
 
 Eeduce : 
 9. [{a + Z>)-i {a - J)]^ 10. (ah-^-'^y, 
 II. («Z>~^c""2)-5. 12. {m^ii-J)-K 
 
 13. (x^y-i)-\ 14. (a^^ci'^Y. 
 
 After forming the expressions, write them all with positive 
 exponents, in the form of fractions. 
 
148 INVOLUTION. 
 
 Algebraic Signs of Powers. 
 
 170. Since the continued product of any number of posi- 
 tive factors is positive, all the powers of a positive quantity are 
 positive. 
 
 By § 26, the product of an odd number of negative fac- 
 tors is negative, and the product of an even number is positive. 
 Hence, 
 
 Theorem, The even powers of negative quantities 
 are positive, and tlie odd powers are negative. 
 
 EXAMPLES. 
 
 (— of = a^ ; (— ay = —a^-, (— ay = «*, etc. 
 
 
 
 EXERCISES. 
 
 
 
 Fin 
 
 a the value of 
 
 
 
 
 
 I. 
 
 (-2)2. 
 
 2. 1 
 
 ;-3)3. 
 
 3. 
 
 44. 
 
 4- 
 
 (-5)V 
 
 5. 
 
 [-5)3. 
 
 6. 
 
 (-by. 
 
 7- 
 
 (-a-l)\ 
 
 8. 
 
 [—onny. 
 
 9- 
 
 (-pqy. 
 
 lo. 
 
 {-afn. 
 
 II. 
 
 [-by^+K 
 
 12. 
 
 (-a- ly^-^ 
 
 13. 
 
 {-Ifn, 
 
 14. 
 
 l-iyn+i. 
 
 15- 
 
 (- 1)^-^ 
 
 Case III. Involution of Binomials— the Bino- 
 niial Theorem. 
 
 171. It is required to find the n^^ poiuer of a binomial. 
 1. Let a -\- bhe the binomial ; its n^^ power may be written 
 {a + by. 
 
 Let us now transform this expression by dividing ifc by «^, 
 and then multiplying it by a% which will reduce it to its orig- 
 inal value. We have (§ 167), 
 
 (fl? + by ^ /a + by^ (. by 
 
 Multiplying this last expression by a^, by writing this 
 power outside the parentheses, it becomes 
 
 a^ 
 
 (>+!)■ « 
 
INYOLUTION. 149 
 
 which is equal to {a + h)^. Next let us put for shortness x to 
 represent - , when the expression will become 
 
 2. Now let us form the successive powers of (1 + xy. We 
 multiply according to the method of § 79 : 
 
 Multiplier, 
 
 (1 + 
 
 xf 
 
 = l+x 
 
 1+x 
 1+x 
 -i-x +x^ 
 
 (1 + 
 
 xy 
 xY 
 
 = l-^2x-\-x^ 
 1+x 
 
 
 l + 2x + x^ 
 
 X + 2xi + a^ 
 
 (1 + 
 
 = l + 3x + dx^-{-a^ 
 1 + x 
 
 
 
 l-\-dx + 3x^ + 0^ 
 
 x + 3x^ -i- dx^ + x^ 
 
 Multiplier, 
 
 Multiplier, 
 
 (1 + xy = 1 + 4:X + 6x^ i- ^a^ -\- x^ 
 
 It will be seen that whenever we multiply one of these 
 powers by 1 + a;, the coefficients of x, x% etc., which we add 
 to form the next higher power are the same as those of the 
 given power, only those in the lower line go one place toward 
 the right. Thus, to form (1 + xy, we took the coefficients of 
 (1 + xy, and wrote and added them thus : 
 
 Coef. of (1 + xy, 1, 3, 3, 1. 
 
 1, 3, 3, 1. 
 
 Coef. of (1 -f xy, 1, 4, 6, 4, 1. 
 
 It is not necessary to write the numbers under each other to add 
 them in tliis way ; we have only to add each number to the one on the 
 left in the same line to form the corresponding number of the line below. 
 Thus we can form the coefficients of the successive powers of x at sight 
 as follows: The first figure in each line is 1 ; the next is the coeJQ&cient 
 of x ; the third the coefficient of x^, etc. 
 
150 
 
 ) 
 
 involution: 
 
 
 
 
 First power, 
 
 n = 1, 
 
 coefficients, 
 
 1, 1. 
 
 
 
 Second " 
 
 n = 2, 
 
 a 
 
 1, 2, 
 
 1. 
 
 
 Third " 
 
 n = 3, 
 
 a 
 
 1, 3, 
 
 3, 1. 
 
 
 Fourth « 
 
 n = 4, 
 
 (( 
 
 1, 4, 
 
 6, 4, 
 
 1. 
 
 Fifth « 
 
 n =. 5, 
 
 « 
 
 1, 5, 
 
 10, 10, 
 
 5, 1. 
 
 Sixth « 
 
 n = 6, 
 
 it 
 
 1, 6, 
 
 15, 20, 
 
 15, 6, 1. 
 
 etc. 
 
 etc. 
 
 
 
 etc. 
 
 
 It is evident that the first quantity is always 1, and that 
 the next coefficient in each line, or the coefficient of x, is n. 
 The third is not evident, but is really equal to 
 n (n — 1) 
 
 2 
 
 (*) 
 
 as will be readily found by trial; because, beginning with 
 n = 3, 
 
 Q 3-2 . 4-3 .. 5.4 , 
 3 = -^, 6 = --, 10 = -^, etc. 
 
 The fourth number on each Hue is 
 n {71 — 1) (n — 2) 
 2.3 
 Thus, beginning as before with the third line, where n = df 
 
 l_3:iJ: ^-^±1 10 -^:i^ etc (.) 
 
 - ^ - 2.3 ' ^ - 2.3 ^ -^^ - 2.3 ' ^^''' ^""^ 
 
 3. These several quantities are called Binomial Coeffi- 
 cients. In writing them, we may multiply all the denomi- 
 nators by the factor 1 without changing them, so that, there 
 will be as many factors in the denominator as in the numerator. 
 The fourth column of coefficients, or (c), will then be written, 
 
 3-2.1 4-3- 2 5-4-3 
 
 1-2-3' 1-2-3' 1-2-3' ^^^• 
 
 4. We can find all the binomial coefficients of any power 
 when we know the value of n. 
 
 The numerator and denominator of the second coefficient 
 will contain two factors, as in (b) ; of the third, three factors, 
 as in (c) ; and of the s^^, s factors, whatever s may be. 
 
 In any coefficient, tlie first factor in the numerator is n, 
 the second n — 1, etc., each factor being less by unity than the 
 
INVOLUTION. 151 
 
 preceding one, until we come to the s^^ or last, which will be 
 ?j — s H- 1. 
 
 Such a product is written, 
 
 n{7i — l){n — 2) . . . .{n — s ■\- 1). 
 
 The dots stand for any number of omitted factors, because 
 s may be any number. We have written 4 of the s factors, so 
 that 5 — 4 are left to be represented by the dots. 
 
 The denominator of the fraction is the product of the 5 
 
 factors, 
 
 1.2.3 s, 
 
 each factor being greater by 1 than the preceding one, and the 
 dots standing for any number of omitted factors, according to 
 the value of s. Thus, the s^^ coefficient in the n^^ line will be 
 
 ' 1.2.3. ..7^^ " ^"^^ 
 
 If s is greater than ■J?^, the last factors will cancel some of 
 the preceding ones, so that as s increases from ^n to ^^, the 
 values of the preceding coefficients will be repeated in the 
 reverse order. Thus, suppose n = 6. Then, by cancelling 
 common factors, 
 
 6.5.4.3 6.5 
 
 1.2.3.4 1.2 
 
 6.5.4.3.2 _ 6 _ 
 1.2.3.4.5 ~ 1 ~ 
 6.5.4.3.2.1 
 
 = 15. 
 = 6. 
 = 1. 
 
 1.2-3.4.5.6 
 
 If we should add one more factor to the numerator, it 
 would be 0, and the whole coefficient would be 0. 
 
 The conclusion we have reached is embodied in the follow- 
 ing equation, which should be perfectly memorized : 
 
 /i , \« 1 , , ^(^ — 1) o . n(n — l)(n—2) _ 
 
 ±' Z 1 .2' o 
 
 + 1.2.3.4 ^+....+z. 
 
t^ INVOLUTION. 
 
 EXERCISES. 
 
 1. Compute from the formula (d) all the binomial coeffi- 
 cients for w = 6, and from them express the development of 
 (1 + xf. 
 
 2. Do the same thing for w = 8, and for n = 10. 
 
 172, To find the development of (a + J)% we replace x 
 by - , and then multiply each term by a^. 
 
 [See equations (1) and (2).] We thus have 
 
 (a + b)^ = a^ -\- na'^-^ + — V^— ^"""^^ + etc. to J». 
 
 The terms of the development are subject to the following 
 rules : 
 
 I. The exponents of b, or the second term of the bino- 
 mial, are 0, 1, 2, etc., to n. 
 
 Because Ifi is simply 1, a" is the same as a^¥. 
 
 II. ITie sum of the exponents of a and hisnin each 
 term. Hence the exponents of a are 
 
 n, n — 1, n — 2, etc., to 0. 
 
 III. The coefficient of the first term is unity, and of 
 the second ft, the index of the power. Each folloiving 
 coefficicjit may he found from the next preceding one by 
 multiplying by the successive factors, 
 
 n—1 n—2 n—S 
 
 3 
 
 , etc. 
 
 IV. If b or a is negative, the sign of its odd powers 
 ivill be changed, but that of its even powers will remain 
 the same. 
 
 (Compare § 170.) Hence, 
 
 {a — ly — aP' — na^-^b + ^^^^ a^-'^W — etc., 
 the terms being alternately positive and negative. 
 
INVOLUTION. 163 
 
 EXERCISE S — Continued. 
 
 3. Compute all the terms of {a + ly, using the binomial 
 coefficients. 
 
 4. What is the coefficient of h^ in the deyelopment of 
 {a + hY^ 
 
 5. What are the first four terms in the development of 
 
 6. What are the first three terms in the development of 
 ? What are the last two terms ? 
 
 (-r' 
 
 7. What are the first three and the last three terms of 
 {a — xy^ ? 
 
 8. What is the development of (« + -)• 
 
 9. What are the first four terms in the development of the 
 following binomials : 
 
 (1 + ^y ; (1 + ^x^y ; (1 - 3a;2)n ; 
 
 10. What are the sum and difference of the two develop- 
 ments, (1 + xy and (1 — xy? 
 
 Case IV. Square of a Polynomial. 
 
 173. 1. Square of any Polynomial. Let 
 a -\- h -\- c -\- d -\- etc., 
 be any polynomial. We may form its square thus : 
 
 a + h-\-c-\-d-\- etc. 
 a-\-'b-\-c-\-d + etc. 
 a^ -\- ah + ac -\- ad + etc. 
 
 ab + h^ -\- he -\- M + etc. 
 
 ac -\- Ic -\- c^ -\- cd + etc. 
 ad + bd -i- cd + d^ + etc. 
 
 «^ + ^ + <J^ + ^^ + etc. 
 + 2ab + 2ac + 2ad + etc. 
 
 + 2bc -f- 2bd + etc. + 2cd + etc. 
 
164 INVOLUTION. 
 
 We thus reach the following conclusion : 
 
 Theorem. The square of a polynomial is equal to 
 the sum of the squares of all its terms plus twice the 
 product of every two terms. 
 
 2. Square of an Entire Function. Sometimes we wish to 
 arrange the polynomial and its square as an entire function of 
 some quantity, for example, of x. 
 
 Let us form the square oi a -{- Ix ■{• cx^ ■{- d:^ + etc. 
 
 a -\-'bx -\- cx'^ + d^ + etc. 
 a -\- hx -\- cx^ ■\- dx? + etc. 
 
 c^ + aljx + acx'' + ad^ + etc. 
 
 dbx + 52^2 _|. icx^ _|. i^rjA _|_ e^c^ 
 
 acx^ + tcx^ + G^x^ + etc. 
 adx? + hdxl^ + etc. 
 a^ + "^abx + {^ac + ^'^j a;^ + i^lad + 2^>c) a:^ + etc. 
 
 "We see that : 
 
 The coefficient of x^ is ac -\- hh -\- ca. 
 
 « " "0:3 is ^f? 4. jc + c^> + fZrt. 
 
 " " " a:* is ae + hd -\- ce + dh + ea. 
 
 etc. etc. 
 
 The law of the products ae, Id, ce, etc., is that the first 
 factor of each product is composed successively of all the co- 
 efficients in regular order up to that of the power of x to which 
 the coefficient belongs, while the second factor is composed 
 successively of the same coefficients in reverse order. 
 
 EXERCISES. 
 
 Form the squares of 
 
 I. 1 + 2ir + ^x\ 2. 1 + 2a; + 3a;2 ^ 4^. 
 
 3. 1 + 2x + Zx^ + 4^3 + 5a;5. 
 
 4. 1 4- 22; + 3a:2 + 4^:3 -\- bs^ -\~ Q^. 
 
 5. 1 — 2a; + 3a;2 — 4o;3. 6. a — 'b-\-c — d. 
 7. ^a + n-c + d. 3_ a + l_j_l. 
 
 a h 
 
EVOLUTION. 155 
 
 CHAPTER II. 
 
 EVOLUTION AND FRACTIONAL EXPONENTS. 
 
 174. Def, The n*'* Root of a quantity q is sii€h a 
 number as, being raised to the n^^ power, will produce q. V 
 
 When ?^ = 2, the root is called the Square Root. 
 
 When 71 = 3, the root is called the Cube Root. 
 
 Examples. 3 is the 4th root of 81, because 
 
 3.3.3.3 = 34 = 81. 
 As the student already knows, we use the notation, 
 
 n^^ root of 5^ = y^q. 
 
 There is another way of expressing roots which we shall 
 now describe. 
 
 175. Division of Exponents. Let us extract the square 
 root of a\ We must find such a quantity as, being multiplied 
 by itself, will produce a^. It is evident that" the required quan- 
 tity is a^, because, by the rule for multiphcation (§§ QQ, 166), 
 
 «3 X a^ — a\ 
 
 n 
 
 The square root of ciP' will be a^, because 
 
 n n n n 
 
 n 
 
 In the same way, the cube root of a^ is a^, because 
 
 n n n 
 
 a^ X a^ X a^ = a^. 
 The following theorem will now be evident : 
 
 TJieorem. The square root of a power may be ex- 
 pressed by dividing its exponent by 2, the cube root by 
 dividing it by 3, and the n^^ root by dividing it by n. 
 
 176. Fractional Exponents. Considering only the origi- 
 nal definition of exponents, such an expression as a^ would 
 
156 EVOLUTION. 
 
 have no meaning, because we cannot write a IJ times. But 
 by what has just been said, we see that a^ may be interpreted 
 to mean the square root of a^, because 
 
 a^ X «^ = a\ 
 Hence, 
 
 A fractional exponent indicates the extraction of a 
 root. If the denominator is 2, a square root is indi- 
 cated ; if 3, a cube root ; if n, an nP' root. 
 
 A fractional exponent has therefore the same meaning as 
 the radical sign V, and may be used in place of it. 
 
 EXE RCI SES. 
 
 Express the following roots by exponents only : 
 
 I. ^m. 2. ^/{m + n). 3. "^(a + Vf* 
 
 4. v^(« + J)2. 5. ^/yyiK 6. ^a^, 
 
 7. v'^s. 8. ^/{a-^bY' 9. ^/{a + h)^. 
 
 177. Since the even powers of negative quantities 
 are positive, it follows that an even root of a positive 
 quantity may be either positive or negative. 
 
 This is expressed by the double sign ±. 
 
 EXERCISES. 
 
 Express the square roots and also the cube roots and the 
 n^^ roots of the following: 
 
 I. (a + hf. 2. (a + hf. 3. a^l. 
 
 4. (^ + #. 5. (^ + #- 6. {x + yf. 
 
 178. If the quantity of which the root is to be ex- 
 tracted is a product of several factors, we extract the 
 root of each factor, and take the product of these roots. 
 
 Example. The n*^ root of arn^p is a^m^p^, because 
 {a^m^p^f — am% by §§ 168 and 176. 
 
 If the quantity is a fraction, we extract the root of 
 both members. 
 
EVOLUTION. 
 
 157 
 
 Proof, 
 
 (lY_a 
 
 (§§167,168.) 
 
 Because -r taken n times as a factor makes t i therefore, 
 by definition, it is the n^^ root of t* 
 
 EXERCISES. 
 
 
 
 Express the square roots of 
 
 
 
 I. 4:X\ 2. -— 
 
 4:9m 
 
 3- 
 
 64«J2c3 
 8l777i7V 
 
 Express the cube roots of 
 
 
 
 4. 27-64. 5. 27^3. 
 
 6. 
 
 64.27a36« 
 
 7. ab^c^dK 8. J^ . 
 
 
 
 Express the n*^ roots of 
 
 
 
 9. 7. lo. 4.7. 
 
 II. 
 
 4.7.10. 
 
 _- 5«J'' .. .-«xo« 
 
 
 6«2^^ 
 
 12. ;; " 
 
 Qmp"' 
 
 13. 6a^Z>2». 
 
 15. 
 
 Qfn+l yTl ^—2 
 
 14. 
 
 c'^dn 
 
 amnym 
 
 16. 35^ a-=^ {a + J)4^ (a; — yY 4^ (5 — c + ^"^^ 
 Eeduce to exponential expressions : 
 
 17. ^ya{b — cy^. 
 19. ^Vam. 
 
 ml (a + &)'» 
 
 18. V«^'C3. 
 
 n/a 
 
 Powers of Expressions with Fractional Expo- 
 nents. 
 
 179. Theorem. The p^^ power of the n^ root is 
 equal to the n^^ root of the j?^^ power. 
 
158 FRACTIONAL EXPONENTS. 
 
 lu algebraic language, 
 
 or (a^y = {a^)i>, 
 
 Example. (^8)^ = 2^ = 4, 
 
 or, in words, tlie square of the cube root of 8 (that is, the 
 square of 2) is the cube root of the square of 8 (that is, of 64). 
 
 General Proof. Let us put x = the n^^ root of a, so that 
 
 x^ — a. (1) 
 
 The p^^ power of this root x will then be xp. (2) 
 
 Raising both sides of the equation (1) to the p^^ power, we 
 have 
 
 x!^P = aP = pP^ power of a. 
 
 The n^^ root of the first member is found by dividing the 
 exponent by n, which gives 
 
 7f^ root of jt?^^ power = xP, 
 
 the same expression (2) just found for the y^^ power of the 
 n^^ root. 
 
 This theorem leads to the following conclusion : 
 
 1. The expression p 
 
 1 
 may mean indifferently the %fl^ power of a^^ or the Tith 
 root of aP^ these quantities being identical. 
 
 2. The powers of expressions having fractional ex- 
 ponents may be formed by multiplying the exponents 
 by the index of the power. 
 
 EXERCISES. 
 
 Express the squares, the cubes, and the n^^ powers of the 
 following expressions : 
 
 I. ofi. 2. a^. 3. a^. 
 
 4. a". 5. air, 6. db^c^. 
 
IRRATIONAL EXPRESSIONS. 159 
 
 mm ^ _? 
 
 7. a^b^. 8. a^b p. 
 
 m 
 
 9. {a + ^'j" {a — b)-^. 10. tf-'^^^. 
 
 II. a^b\ 12. -^ — ^' ^ 
 
 Keduce to simple products and fractions : 
 
 / m _m\p m 
 
 13. \x^y ^) . 14. {aU^c~i)"' 
 
 15. {aH^y^. ■ 16. U"^) 
 
 17. — 5 1 • 18. — n : 
 
 m 
 
 a^-* «^^* 
 
 CHAPTER III. 
 
 REDUCTION OF IRRATIONAL EXPRESSIONS. 
 
 Definitions. 
 
 180. Def. A Rational Expression is one in which 
 the symbols are only added, subtracted, multiplied, or 
 divided. 
 
 All the operations we have hitherto considered, except the extraction 
 of roots, have led to rational expressions. 
 
 Def. An expression which involves the extraction 
 of a root is called Irrational. 
 
 Example. Irrational expressions are 
 
 Va, ^/(a + b), V^l ; 
 or, in the language of exponents, 
 
 «^, {a + b)^, 27^. 
 
 In order that expressions may be really irrational, 
 
160 IRRATIONAL EXPRESSIONS. 
 
 they must be Irreducible, that is, incapable of being 
 expressed without the radical sign. 
 
 Example. The expressions 
 
 are not properly irrational, because they are equal to a -\-h 
 and 6 respectively, which are rational. 
 
 Def. A Surd is the root which enters into .an 
 irrational expression. 
 
 Example. The expression a + iVx is irrational, and the 
 surd is "^x. 
 
 Def, Irrational terms are Similar when they con- 
 tain the same surds. 
 
 Examples. The terms V30, 7 a/30, {x + y) a/30, are 
 similar, because the quantity under the radical sign is 30 in 
 each. 
 
 The terms (a + I) Vx -\- y, sVx -\- y, m^x + y are 
 similar. 
 
 Aggregation of Similar Terms. 
 
 181. Irrational terms may be aggregated by the rules of 
 §§ 54-56, the surds being treated as if they were single sym- 
 bols. Hence : 
 
 'W%erh similar irrational terms are connected hy the 
 signs -\- or — , the coefficients of the similar surds may 
 he added, and the surd itself affixed to their sum. 
 
 Example. The sum 
 
 • aVix + y) — hV{x -\- y) + W(x + y) 
 
 may be transformed into (a — 5 + 3) \/{x + y), 
 
 EXERCISES. 
 
 Eeduce the following expressions to the smallest number 
 of terms : 
 
 I. 7a/3 ~ 5a/2 -f 6a/3 + 7«&a/2. 
 
IRRATIONAL EXPRESSIONS. 161 
 
 2. W{x + y) + aV{x _ 2^) + 2 (« + J) V{x + y) 
 
 - 3 (a + ^») ^/{x - y). 
 
 4. (« + ^) a/^ + (a — J) Viz^z/. 
 
 5. Vx (a — b) + {b — c) Vx -\- (c — a) Vx» 
 
 6. aVx — a/^ + 2aVx — (a ■}- d) Vx. 
 
 q 1 
 
 7. J V^ — aVx + 6A/aJ — cVic + o V^- 
 
 9. -Vx—Vx-\- (a — I) Vx + ^^~ — ^/x. 
 
 10. a/aj — 5a/« — Vx -\ ^^^— ^/a — ^ V^. 
 
 4 /* 
 
 3 , / -2(« — Z*) / 
 
 11. - Vic — V a^ + -^^-g — - y^' 
 
 12. 4A/a; — - \^x -{■ {a — h) Vx, 
 
 o 
 
 Factoring Surds. 
 
 183. Irrational expressions may sometimes be transformed 
 so as to have different expressions under the radical sign, by 
 the method of § 178, applying the following theorem: 
 
 Theorem. A root of the product of several factors 
 is equal to the product of their roots. 
 
 In the language of Algebra, 
 
 "s/abcd, etc. = ^/a Vb v^c y^d, etc. 
 
 = a^b^c^d\ etc. 
 
 Proof. By raising the members of this equation to the 
 fit^ power, we shall get the same result, namely, 
 a X h X c X d, etc. 
 
 Example. VsO = Vo VS. 
 11 
 
163 IRRATIONAL EXPRESSIONS, 
 
 EXERCISES. 
 
 Prove the following equations by computing both sides : 
 
 
 
 A/4V49 = a/4-49 = a/196. 
 
 
 Proof. 
 
 a/4 a/49 = 2-7 = 
 
 14, and a/196 
 
 = 14. 
 
 
 
 a/4 a/9 = 
 
 a/4 a/25 = 
 
 ' a/9 VlQ = 
 
 a/25 a/36 = 
 
 a/36. 
 a/4.25. 
 a/9.16. 
 a/25.36. 
 
 
 Express 
 
 with a single surd the products : 
 
 
 I. 
 
 V{a + b) V{a - b). 
 LUTioiq-. V{a + b) V{a 
 
 Va V(a + y). 
 V{x + l)V{x-l). 
 V{x^ + 1) V{x + 1) Vi 
 '(a + J)^ {a - b)^\ 
 l{x^ + l)Hx^.iyHx- 
 
 
 
 So 
 
 2. 
 4. 
 
 6. 
 
 7- 
 8. 
 
 9- 
 
 -J) = a/(« + ^')(«-&) 
 
 = V{a^ - b% 
 3. V7 a/«. 
 5. a/« a/5 V{a + J). 
 
 :.r-i). 
 
 183. If we can separate the qnantity under the 
 radical sign into two factors, one of which is a perfect 
 square, we may extract its root and affix the surd root 
 of the remaining factor to it. 
 
 EXAMPLES. 
 
 V^ = Va^ Vb = aVb. 
 Vab Vac = Va^bc = aVbc. 
 a/12 a/6 = a/72 = a/36 a/2 = 6 a/2. 
 
 A/(4a3 + Sa^ — IQa^c) = V^d^ (« + 25 — 4«c) 
 — 2a V {a + 25 — 4«c). 
 (x^— ixhj -I- 4a;y^)^ = (x — 2?/) x^. 
 
IBBATIONAL EXPRESSIONS. 163 
 
 EXERCISES. 
 
 Reduce, when possible : 
 
 
 
 I. \/8. 
 
 2. 
 
 a/32. 
 
 3. Visa 
 
 4. 
 
 a/3 a/27. 
 
 5. ^/ab ^/ca V^. 
 
 6. 
 
 a/2 a/72. 
 
 7. V^ V72. 
 
 8. 
 
 V{x + 1) V(:?: + 1). 
 
 9. a/175. 
 
 10. 
 
 a/150. 
 
 II. a/108. 
 
 12. 
 
 A/a;H« + ^»). 
 
 13. A/(a^ic + ^abx + Z^s^^;). 
 
 
 
 Here the quantity under the radical sign is equal to 
 {a' + 2«& + &2) aj = (flj + &)2 a;. 
 
 In questions of this class, the beginner is apt to divide an expression 
 like ^/a + 6 + c into ^/a + y^/h + 's/c, which is wrong. The square 
 root of the sum of several quantities cannot be reduced in this way. 
 
 14. Vd^y + ^ay + %. 15. \^4:7nh + ^mz -f- 4:Z. 
 
 Eeduce and add the following surds : 
 16. 4a/2— 6V'8 + 10a/32. 17. a/12 + a/27 + a/75. 
 18. A/i« — 2a/«. 19. 125^— 45* — 80i 
 
 20. '^81-v^192. 21. (aW)^ ■{- (a^(^)^. 
 
 Multiplication of Irrational Expressions. 
 
 184. Irrational polynomials may be multiplied by com- 
 bining the foregoing principles with the rule of § 78. 
 The following are the forms : 
 
 To multiply a + l^x by w + nVy* 
 a{m -\- nVy) = am x an^/y. 
 'b^/x (m + nVy) = bmVx + bnVxy. 
 The product is am + aiiV.y + bniVx + bnVxy. 
 
 EXERCISES. 
 
 Perform the following multiplications and reduce the 
 results to the simplest form (compare § 80) : 
 
 I. (2 4-3a/5)(5-3V2). 2. (7 + 2 a/32) (9 — 5 a/2). 
 
164 IRBATIONAL EXPRESSIONS. 
 
 3. {a + Vh) {a — Vb). 4. {Va+ Vb + Vc+ Vd)l 
 {m 4- n^) (m + 2n^). 6. {a^ — a^) {a^ + a^). 
 {a + a-J. 8. («^_a-i/. 
 
 [a + Sa/(^ + 2/)] [« - ^ V(^ + y)]. 
 \m + ^VCa -H ^)] \rn — n^/{a — &)]. 
 \x + ^(^2 - 1)] \x - V(x^ - 1)]. 
 
 Expressions may often be transformed and factored by 
 combining the foregoing processes. 
 
 Example. To factor ax^ + bx^ + cx^ + dx^', we notice 
 ^ x^ =z x^x?y x^ = x^x% etc. 
 
 so that the expression may be written, 
 
 aa^x^ 4- ix^x^ + cxx^ -f dx^ = (ax^ -^ bx^ -{- ex + d) xk 
 
 exercis:es. 
 
 Eeduce the following expressions to products: 
 
 13. 2+V2. 14. 3t + 2.3i 
 
 15. (a -\- J)i 16. Vy 4- ««/3 _ ^^5^ 
 
 17. X — y — Vx — y. 
 Eeduce to the lowest terms : 
 
 2 Va"-\n> ax^ + hx^ 
 
 18. — -. 19. -^-. 20. — ^ j-. 
 
 V'* a -\- dx^ _ j^-u 
 
 « — a; + \/« — a; ^^^2 _ ^ 
 
 21. — ' 22. 7— • 
 
 a — x— \a — X a -\- 
 
 185. Rationalizing Fractions. The quotient of 
 two surds may be expressed as a fraction with a rational 
 numerator or a rational denominator, by multiplying 
 both terms by the proper multiplier. 
 
 Example. Consider the fraction -— . 
 
 V7 
 
IRRATIONAL EXPRESSIONS. 165 
 
 Multiplying both terms by V?, the fraction becomes 
 
 ■ , and has the rational denominator 7. 
 
 5 
 Multiplying by y^5, it becomes — - — , and has the rational 
 numerator 5. '^^^ 
 
 The numerator or denominator may also Ibe made 
 rational when they both consist of two terms, one or 
 both of which are irrational. 
 
 Let us haye a fraction of the form 
 A + D^/B 
 
 in which the letters A, D, P, Q, and R stand for any algebraic 
 or numerical expressions whatever. If we multiply both nu- 
 merator and denominator by P — Q^Ry the denominator 
 will become 
 
 P2 _ Qm, 
 
 The numerator will become 
 
 AP + pdVb - aqVr - dqVbr. 
 
 so that the value of the fraction is 
 
 AP + pdVb - A q Vr - dqVbr 
 pi _ qiR 
 
 EXERCISES. 
 
 Reduce the following fractions to others having rational 
 denominators : 
 
 /(a + 6\ x^ /I + x\^ 
 
 9V5* ^* sVe * ' 2V2 ' 
 
 a + Vh a — ^/ x Vx + Vy^ 
 
 a— ^/h ' a-{- Vx ' ^x — ^/y 
 
166 PERFECT SQUARES. 
 
 10. —^ -^ -^' II. ; 7— 
 
 a + V{x + y) V5 — V3 
 
 12. V^-A^G^' + y). ^^ .1 
 
 V^ + V(^ -{- y) ^ — Vx^ — a^ 
 
 1 V^ + « + V a; — a 
 
 14. -r £• 15. ■ , ; 
 
 a^ + {a + ip Vx + a— Vx — a 
 
 Perfect Squares. 
 
 186. Def. A Perfect Square is an expression of 
 wMch tlie square root can Ibe formed without any surds, 
 except such as are already found in the expression. 
 
 Examples. Am^, 4:a^ + 4:a -{- 1 are perfect squares, be- 
 cause their square roots are 2Tn% 2a + 1, expressions without 
 the radical sign. 
 
 The expression a + 2'\/«Z> + b, of which the root is 
 
 may also be regarded as a perfect square, because the surds 
 Va + Vb are in the product 2'^ab. 
 
 Criterion of a Perfect Square. The question whether a 
 trinomial is a perfect square can always be decided by compar- 
 ing it with the forms of § 80, namely : 
 
 «2 + 2al) + Z>2 — (^ 4. ly^ 
 or «2 _ <jiah ^h^ = la — h)\ 
 
 We see that to be a perfect square, a trinomial must fulfil 
 the following conditions : 
 
 (1.) Two of its three terms must be perfect squares. 
 
 (2.) The remaining term must be equal to twice the 
 product of the square roots of the other two terms. 
 
 When these conditions are fulfilled, the square root 
 of the trinomial will be the sum or difference of the 
 square roots of, the terms, according as the product is 
 positive or negative. 
 
 The root may liave either sign, because the squares of posi- 
 tive and negative quantities have the same sign. 
 
COMPLETING THE SQUARE. 167 
 
 If the terms which are perfect squares are loth negative, 
 the trinomial will be the negative of a perfect square. 
 
 EXAMPLES. 
 
 V«^ -\-2ah -\-h^ = a + 1) OY —(a + h). 
 
 Va^ — %ab -{■'W' ^ a — h ov h — a. 
 
 _ ^2 + 2«^' -^ ^ = —{a-hf^: —{b — ay. 
 
 EXERCISES. 
 
 Find which of the following expressions are perfect squares, 
 and extract their square roots : 
 
 I. 
 
 9 + 12 + 4. 
 
 2. 
 
 a;2 -1- 4a; + 4. 
 
 3- 
 
 4a;4 ^ 2x^ + 7. 
 4 
 
 4. 
 
 a^^ab- b\ 
 
 5- 
 
 ^a^^ + 12a^b^ f Q¥^. 
 
 6. 
 
 a^ + 2ab - l\ 
 
 7. 
 
 x^ — a:(^y + ^ay. 
 
 8. 
 
 aW - 2abcd + c^t^. 
 
 9- 
 
 m + 2mJni + n. 
 
 10. 
 
 Qr2 _ 2fl!a; + y\ 
 
 II. 
 
 a + Aah^ +4b. 
 
 12. 
 
 a — 2 + a-\ 
 
 13. 
 
 252J^ + 9^2 _ 30^2^. 
 
 14. 
 
 ejim^n ^ Ji2 _^ Q^yiin^ 
 
 15. 
 
 49icy + 9z^ — A2xyz. 
 
 16. 
 
 9m^^ — 2m^^pq + ^ 
 
 y 
 
 To Complete the Square. 
 
 18')'. If one term of a Ibinomial is a perfect square, 
 such a term can always be added to the binomial that 
 the trinomial thus formed shall be a perfect square. 
 
 This operation is called Completing the Squajre. 
 
 Proof. Call a the root of tlie term v^hich is a perfect 
 square, which term we suppose i\\Q first, and call m the other 
 term, so that the given binomial shall be 
 
 a^ + m. 
 
 Add to this binomial the term — ^ , and it will become 
 
 "' + '" + -&■ 
 
168 COMPLETING THE SQUARE. 
 
 This is a perfect square, namely, the square of 
 
 m 
 
 thatis, „. + ™ + _ = ^« + _j. 
 
 Hence the following 
 
 EuLE. Add to the binomial the square of the second 
 term divided by four times the first term^. 
 
 Example. What term must be added to the expression 
 ar^ — 4:ax (1) 
 
 to make it a perfect square ? 
 
 The rule gives for the term to be added, 
 
 4rX^ 
 
 Therefore the required perfect square is 
 
 a^ — 4:ax + 4^2 z= (x — 2ay. 
 We may now transpose 4:a% so that the left-hand member 
 of the equation shall be the original binomial (1). Thus, 
 
 x^ — 4:ax = (a; — 2ay — 4^2. 
 The original binomial is now expressed as the difference of 
 two squares. Therefore, the above process is a solution of the 
 problem: Having a binomial of which one term is a perfect 
 square, to express it as a difference of two squares. 
 
 EXERCISES. 
 
 Express the following binomials as differences of two 
 squares ; 
 
 I 
 3 
 5 
 7 
 9 
 
 IT 
 
 13 
 
 x^ + 2xy. 2. a;2 + 4xy. 
 
 x^ + Qax. 4. 4:X^ -f- 4:xy, 
 
 4:X^ + 4:xy. 6. 9x^ + ax. 
 
 16a;2 + ^2mx. 8. x^ + 4:X. 
 
 a^x^ 4- 2a^x. 10. J^x^ + 2. 
 
 m^x^ 4- 1. 12. ^p^x? 4- bx. 
 
. IRRATIONAL FACTORS. 169 
 
 Irrational Factors, 
 
 188. When we introduce surds, many expressions can be 
 factored which have no rational factors. The following 
 theorem may be applied for this purpose : 
 
 Theorem. The difference of any two quantities is 
 equal to the product of sum and difference of their 
 square roots. 
 
 In the language of algebra, if a and h be the quantities, we 
 shall have 
 
 which can be proved by multiplying and by § 80, (3). 
 
 EXE RCISES. 
 
 Factor 
 
 I. m — n, 2. m — 1. 
 
 3. am — hn. 4. 4:a^m — 9. 
 
 5. x^ — m. 6. x^ — {771 -h n), 
 
 7. {x — aY — -{m — n). 8. x^ — (m — n), 
 
 9. (a + iy — (4;^^ — q)' 10. x^+2xy+y^— (w + w)!. 
 
 Find the irrational square roots of the following expressions 
 by the principles of § 186 : 
 
 11. a — 2 -f-_a~^ Ans. a^ — orK 
 
 12. X — 2Vxy + y. 13. 4 + 4^/3 + 3. 
 14. 9 + 5 — eVs. 15. 4« -h J — 4:ahk 
 16. a + d-\-2(a + I})^x-{-x^, 17. 3 + 2Vl5 + 5. 
 
 18. 3 + 5 — 2^/15. 
 
 20. a — 2^/a + 1. 
 
 22. a + 2^3 + ~T' 
 
 a ^ a , a 
 
 '^' 4+3 + 9- 
 
 26. ^5 + 2 + a-5. 
 
 28. a + J _ 4 + 
 
 Vxy 
 
 19. 
 
 21. 
 
 4"^4 2 * 
 « — 2«^ + at. 
 
 a^ - « + -. 
 
 25- 
 27. 
 
 -iL + i + ^l 
 16 ^ 4 ^ 4 
 
 4a:3 _ 8 + 4a;-3. 
 
 a + J 
 
 (S> 
 
BOOK VI. 
 
 EQUATIONS REQUIRING IRRA 
 TIONAL OPERA TIONS, 
 
 CHAPTER I. 
 
 EQUATIONS WITH TWO TERMS ONLY. 
 
 189. In the present chapter we consider equations which 
 contain only a single power or root of the unknown quantity. 
 
 Such an equation, when reduced to the normal form, will 
 be of the form 
 
 Ax^ + ^ = 0. 
 
 By transposing B, dividing by A, and putting 
 
 B 
 
 the equation may be written, 
 
 of^ — a = 0. 
 or x^ = a, (1) 
 
 Here n may be an integer, or it may represent some fraction. 
 
 Such an equation is called a Binomial Equation, because 
 the expression a^ — a is a binomial. 
 
 Solution of a Binomial Equation. 
 
 190. 1. When the exponent of x is a tvhole number. If we 
 extract the n^^ root of both members of the equation (1), these 
 roots will, by Axiom V, still be equal. The n^^ root of x^ being 
 X, and that of a being a\ we have 
 
 X = «^, 
 and the equation is solved. 
 
BINOMIAL EQUATIONS. 171 
 
 2. When the exponent is fractional. Let the equation be 
 
 m. 
 
 (^ = a. 
 
 Raising both members to the n^^ power, we have 
 Extracting the m^^ root. 
 
 
 If the numerator of the exponent is unity, we only have to 
 suppose m = 1, which will give 
 
 X = a\ 
 
 Hence the binomial equation always admits of solution by 
 forming powers, extracting roots, or both. 
 
 Special Forms of Binomial Equations. 
 
 Def. When the exponent n is an integer, the equa- 
 tion is called a Pure Equation of the degree n. 
 
 When 71 = 2, the equation is a Pure Quadratic 
 Equation. 
 
 When n = S, the equation is a Pure Cubic Equa- 
 tion. 
 
 
 EXERCISE 
 
 :s. 
 
 Hm 
 
 i the values of x in the 
 
 following equations : 
 
 I. 
 
 £ = ,. 
 
 x^ 
 
 
 Ans. X = '—> 
 
 2. 
 
 a + b 
 xi -'' 
 
 3. 
 
 a b 
 
 
 x^ — b x^ — a 
 
 4- 
 
 9 a^ 
 X ~ 24' 
 
 5- 
 
 x — 2a 2x—b 
 x—a ~ x—b 
 
 6. 
 
 x^ — na wi? — b 
 x^ — a ~ x^ — b' 
 
 7- 
 
 p 
 a -{- b xa. 
 
 
 ^ -«-4 
 
 8. 
 
 ^ yi 
 yi ~ xi 
 
 9- 
 
 ^/x -\- 0? b — a 
 
172 POSITIVE AND NEGATIVE BOOTS. 
 
 In the last example, clearing the equation of fractions, we shall have 
 ^x^ - a^ = ¥- a\ 
 or (a;2 - d'f = 1^ - a\ 
 
 We square both sides of this equation, which gives another in which 
 ir^ only appears. 
 
 lo. {x — a)i = A II. (a;2 — a^)^ = mx. 
 
 12. {Vx — Vb)i = nxk 
 
 Positive and Negative Roots. 
 
 191. Since the square root of a quantity may be either 
 positive or negative, it follows that when we have an equation 
 such as 
 
 a^ z=z a, 
 
 and extract the square root, we may have either 
 
 X = + a^f 
 or X z= — a^. 
 
 Hence there are two roots to every such equation, the one 
 positive and the other negative. We express this pair of roots 
 by writing 
 
 X = ± a^, 
 
 the expression ± a^ meaning either + a^ or — a^. 
 
 It might seem that since the square root of x^ is either + a; or —x, we 
 should write 
 
 ± a; = ± a*, 
 having the four equations, x — a*. 
 
 But the first and fourth of these equations give identical values of x 
 by simply changing the sign, and so do the second and third. 
 
 PROBLEMS LEADING TO PURE EQUATIONS. 
 
 1. Find three numbers, such that the second shall be 
 double the first, the third one-third the second, and the sum 
 of their squares 196. 
 
 2. The sum of the squares of two numbers is 369, and the 
 difference of their squares 81. What are the numbers? 
 
PROBLEMS. 173 
 
 3. A lot of land contains 1645 square feet, and its length 
 exceeds its breadth by 12 feet. What are the length and 
 breadth ? 
 
 To solve tliis equation as a binomial, take the mean of the length 
 and breadth as the unknown quantity, so that the length shall be as much 
 greater than the mean as the breadth is less. 
 
 4. Find a number such that if 9 be added to and subtracted 
 from it, the product of the sum and difference shall be 175. 
 
 5. Find a number such that if a be added to it and sub- 
 tracted from it the product of the sum and difference shall be 
 2a + 1. 
 
 6. One number is double another, and the difference of 
 their squares is 192. What are the numbers ? 
 
 7. One number is 8 times another, and the sum of their 
 cube roots is 12. What are the numbers ? 
 
 8. Find two numbers of which one is 3 times the other, 
 and the square root of their sum, multiplied by the lesser, is 
 equal to 128. 
 
 9. What two numbers are to each other as 2:3, and the 
 sum of their squares = 325 ? 
 
 Note. If we represent one of the numbers by 2x, the other will be Zx. 
 
 10. What two numbers are to each other as m : n, and 
 the square of their difference equal to their sum ? 
 
 11. What two numbers are to each other as 9 to 7, and the 
 cube root of their difference multiplied by the square root of 
 their sum equal to 16 ? 
 
 1 2. Find X and y from the equations 
 
 a'x^ + *y = c'. 
 
 13. The hypothenuse of a right-angled triangle is 26 feet 
 in length, and the sum of the sides is 34 feet. Find each side. 
 
 Note. It is shown in Geometry that the square of the hypothenuse 
 of a right-angled triangle is equal to the sum of the squares of the other 
 two sides. In the present problem, take for the unknown quantity the 
 amount by which each unknown side differs from half their sum. 
 
 14. Two points start out together from the vertex of a 
 right angle along its respective sides, the one moving m feet 
 per second and the other n feet per second. How long will 
 they require to be c feet apart ? 
 
 15. By the law of falling bodies, the distance fallen is pro- 
 portional to the square of the time, and a body falls 16 feet 
 the first second. IIow long will it require to fall h feet ? 
 
174 qUADMATlG EQUATIONS. 
 
 CHAPTER II. 
 QUADRATIC EQUATIONS. 
 
 193. Bef, A Quadratic Equation is one whicli, 
 when reduced to the normal form, contains the second 
 and no higher power of the unknown quantity. 
 
 A quadratic equation is the same as an equation of the second degree. 
 
 Def. A Pure quadratic equation is one which con- 
 tains the second power only of the unknown quantity. 
 
 The treatment of a pure quadratic equation is given in the preceding 
 chapter. 
 
 Def. A Complete quadratic equation is one which 
 contains "both the first and second powers of the un- 
 known quantity. 
 
 The normal form of a complete quadratic equation is 
 
 (1) 
 
 (3) 
 
 ax^ -\- bx -\- c = 
 
 0. 
 
 Jf we divide this equation by a, we 
 
 obtain 
 
 a . a 
 
 0. 
 
 Putting, for brevity, - = ^, 
 
 
 c 
 
 
 the equation will be written in the form, 
 
 x^ +])^-^ ^ = ^' (3) 
 
 Def. The equation 
 
 x^ + px -\- q ~ 
 is called the General Equation of the Second Degree, 
 or the General Quadratic Equation, because it is the 
 form to which all such equations can be reduced. 
 
qUADRATIG EQUATIONS. 175 
 
 Solution of a Complete Quadratic Equation. 
 
 193. A quadratic equation is solved hy adding such 
 a quantity to its two members that the member contain- 
 ing the unknown quantity shall be a perfect square. 
 (§187.) 
 
 We first transpose q in the general equation, obtaining <f 
 
 We tlien add ^- to both members, making ^ L a^ v^ VVT ' ■ 
 
 The first member of the equation is now a perfect square. 
 Extracting the square roots of both sides, we have 
 
 P 
 
 * + a 
 
 \/f 
 
 From this equation we obtain a value of x which may be 
 put in either of the several forms. 
 
 X = 
 
 -f±\/f- 
 
 -<!' 
 
 X = 
 
 P Vp'- 
 
 2^ 2 
 1 . ,^ 
 
 i^. 
 
 ^ = -^{—P± Wp^ — ^q)' 
 
 If instead of substituting^ and q, we treat the equation in 
 the form (2) precisely as we have treated it in the form (3), we 
 shall obtain the several results, 
 
 a A: a} 4«^ a 
 
 ~ 2a 
 
176 QUADRATIC EQUATIONS. 
 
 194. The equation in the normal form, (1), may also be 
 solved by the following process, which is sometimes more con- 
 venient. Transposing c, and multiplying the equation by a, 
 we obtain the result 
 
 aV + dbx ^= — ac. 
 
 To make the first member a perfect square, we add — to 
 each member, giving 
 
 c?7? + dbx + 7- = -J — ac, 
 4 4 
 
 Extracting the square root of both sides, we have 
 aa; + ^ = ^ V( J^ — 4a5c), 
 from which we obtain the same value of x as before. 
 
 195, Since the square root in the expression for x may be 
 either positive or negative, there will be two roots to every 
 quadratic equation, the one formed from the positive and the 
 other from the negative surds. If we distinguish these roots 
 with x^ and x^, their values will be 
 
 — — ^ + ^/{b^ — ^ac) 
 
 We can always find the roots of a given quadratic equation by sub- 
 stituting the coefficients in the preceding expression for x. But the stu- 
 dent is advised to solve each separate equation by the process just given, 
 which is embodied in the following rule : 
 
 I. Reduce the equation to its normal or its general 
 ■ form, as may be most convenient. 
 
 II. Transpose the terms which do not contain x to the 
 second memher. 
 
 III. If the coefficient of oc^ is unity, add one-fourth 
 the square of the coefficient of x to both members of the 
 equation and extract the square root. 
 
 IV. // the coefficient of x^ is not unity, either divide 
 by it so as to reduce it to unity, or multiply all the terms 
 
QUADRATIC EQVATI0N8. 177 
 
 hy such a factor that it shall hecome a perfect square, 
 and cojnplete the square by the rule of § 187. 
 
 EXAM PLE. 
 
 Solve tlie equation ^ 1\ ^ 
 
 x — 1 _ y ' >- 
 
 Clearing of fractions and transposing, we find the equation to become 
 2ic2 — 11a; + 1 = 0, 
 
 Adding \ the square of the coefficient of x to each side, we have • 
 2 41 1681 _ 1681 1 _ 1673 
 
 "^""a^"^ 16 - 16 ~2- 16* 
 Extracting the square root and reducing, we find the values of x to he 
 
 X, =:i(41 + V1673), 
 and 
 
 2^3 = ^(41 — ^1673). 
 
 Using the other method, in order to avoid fractions, we multiply the 
 equation (5) by 2, making tlie equation, 
 
 4a;2 _ 822; = — 2. 
 
 41^^ 1681 
 Adding -j- = —r— to each side of the equation, we have 
 
 . , __ , 412 1681 1673 
 
 4rc2 — 82:c + -— = — 2 = —^ — 
 
 4 4 4 
 
 Extracting the square root, 
 
 \/l673 
 
 ^X 
 
 41 /1673 
 
 4 2 ' 
 
 whence we find 
 
 - 41 ± a/1673 
 x = ^ , 
 
 the same result as before. 
 
 EXERCISES. 
 
 Eeduce and solve the following equations 
 
 a; + 2 _ ^--2 _ 5 y + ^ y — 4 _ 10 
 
 ^' ic — 2 a; + 2~6* ^' y — ^^ y -\- i"" 'd' 
 
 12 
 
178 QUABBATIC EQUATIONS. 
 
 3- 
 
 a:_l 1 x — % ~ 3 
 
 
 4. 
 
 «/2 — %ay + a> — 1)' = 0. 
 
 
 5- 
 
 1 -1,^1^ 
 
 
 ft + Z'-j-o; a ^ h ^ X 
 
 
 6. 
 
 a' h 5 _o 
 
 
 i?;2 — a^ ^ X -\- a x — a 
 
 
 7. 
 
 ! + * + « 
 
 a; — a _ , 
 
 
 1 ^7" 
 
 
 
 
 8. 
 
 2 + ^ 2/'-4'2-^-- 
 
 
 9- 
 
 y -\- a y — a 1 1 
 
 , 1 , 
 
 y — a y + a~ y — a y^ — 
 
 ^ ' y — ^ 
 
 
 ^ ^ 1 S -0 
 
 
 
 a + a; « — a; 
 
 PROBLEMS. 
 
 
 1. Find two numbers such that their difference shall be 
 6 and their product 567. 
 
 2. The difference of two numbers is 6, and the difference 
 of their cubes is 936. What are the numbers ? 
 
 3. Divide the number 34 into two such parts that the 
 sum of their squares shall be double their product? 
 
 4. The sum of two numbers is 60, and the sum of their 
 squares 1872. What are the numbers ? 
 
 5. Find three numbers such that the second shall be 5 
 greater than the first, the third double the second, and the 
 sum of their squares 1225. 
 
 6. Find four numbers such that each shall be 4 greater 
 than the one next smaller, and the product of the two lesser 
 ones added to the product of the two greater shall be 312. \ 
 
 7. A shoe dealer bought a box of boots for $210. If there 
 had been 5 pair of boots less in the box, they would have cost 
 him II per pair more, if he had still paid $210 for the whole. 
 How many pair of boots were in the box ? 
 
 Rem. If we call x the number of pairs, the price paid for each pair 
 must have been 
 
 X 
 
qUADBATIG EQUATIONS. 179 
 
 8. A huckster bought a certain number of chickens for 
 110, and a number of turkeys for $15.75. There were 4 more 
 chickens than turkeys, but they each cost him 35 cents a piece 
 less. How many of each did he buy? 
 
 9. A farmer sold a certain number of sheep for $240. If 
 he had sold a number of sheep 3 greater for the same sum, he 
 would have received $4 a piece less. How many sheep did he 
 sell? 
 
 10. A party having dined together at a hotel, found the 
 bill to be $9. GO. Two of the number having left before pay- 
 ing, each of tlje remainder had to pay 24 cents more to make 
 up the loss. What was the number of the party ? 
 
 11. A pedler bought $10 worth of apples. 80 of them 
 proved to be rotten, but he sold the remainder at an advance 
 of 2 cents each, and made a profit of $3.20. How many did 
 he buy ? 
 
 12. In a certain number of hours a man traveled 48 miles ; 
 if he had traveled one mile more per hour, it would have taken 
 him 4 hours less to perform his journey ; how many miles did 
 he travel per hour ? 
 
 13. The perimeter of a rectangular field is 160 metres, and 
 its area is 1575 square metres. What are its length and 
 breadth ? 
 
 14. The length of a lot 'of land exceeds its breadth by 
 a feet, and it contains w? square feet. What are its dimen- 
 sions ? 
 
 15. A stage leaves town A for town B, driving 8 miles an 
 hour. Three hours afterward a stage leaves B for A at such a 
 speed as to reach A in 18 hours. They meet when the second 
 has driven as many hours as it drives miles per hour. What 
 is the distance between A and B ? 
 
 Note. The solution is very simple when the proper quantity is taken 
 as unknown. 
 
 Equations which may be Reduced to Quad- 
 ratics. 
 
 196. Whenever an equation contains only two 
 powers of the unknown quantity, and the index of one 
 power is double that of the other, the equation can Ibe 
 solved as a quadratic. 
 
180 QUADRATIG MqUATIONS. 
 
 Special Example, Let us take the equation 
 
 af^-\-W + c = (). (1) 
 
 Transposing c and adding - h^ to each side of the equation, 
 it becomes 
 
 4 4 
 
 The first member of this equation is a perfect square, 
 
 namely, the square oi 7^ ■\- -1). Extracting the square roots 
 
 Z 
 
 of both members, we have 
 
 Hence, ^ = \\_-h± V(V — 4c)]. 
 
 Extracting the cube root, we have 
 
 General Form. We now generalize this solution in the 
 following way. Suppose we can reduce an equation to the 
 form 
 
 «a;2« + bx^ + c = 0, 
 
 in which the exponent n may be any quantity whatever, entire 
 
 or fractional. By dividing by «, transposing, and adding 
 
 1 Z>2 
 
 J -2 to both sides of the equation, we find 
 
 ^«^ +4a2- 4a2 a 
 The first side of this equation is the square of 
 
 Hence, by extracting the square root, and reducing as in 
 the general equation, we find 
 
 2:« = ~[-&±\/(52-.4ac)]. 
 
QUADRATIC EQUATIONS. 181 
 
 Extracting the n^^ root of both sides, we have 
 
 
 If the exponent w is a fraction, the same course may be 
 followed. 
 
 Suppose, for example, 
 
 ax^ -\- hz^ -\- c = 0, 
 Dividing by a and transposing, we have 
 
 4,58 c 
 
 a a 
 
 Adding —^ to both sides, 
 
 4,58,5^ Z»2 c 
 
 X^ A X^ A = • 
 
 The left-hand member of this equation is the square of 
 
 - + ^- 
 
 Extracting the square root of both members, 
 I _5^ _ /^ _ _c\i _ ( b^ — 4gc)^ 
 
 , 3 —h±{h^ — 4caS 
 
 whence, x^ = ==^ ^—, 
 
 2a 
 
 Eaising both sides of this equation to the f power, we have 
 
 EXERCISES. 
 
 1. Find a number which, added to twice its square root, 
 will make 99. 
 
 2. What number will leave a remainder of 99 when twice 
 its square root is subtracted from it. 
 
183 QUADBATIG EQUATIONS. 
 
 3. One-fifth of a certain number exceeds its square root 
 by 30. What is the number ? 
 
 4. What number added to its square root makes 306 ? 
 
 5. If from 3 times a certain number we subtract 10 times 
 its square root and 96 more, and divide the remainder by the 
 number, the quotient will be 2. What is the number ? 
 
 Solve the equations : 
 
 6. \y^- %y^ = 15. 7. Sy^ - 7^2 = 25. 
 o 
 
 8. hy^ — 8^i = 13. 
 
 m 
 
 9. {f + aY — 4 (a;2 + «2)2n = ^2 _ 2 + 
 
 1 
 
 1.97. When the unknown quantity appears in the form 
 
 ^ -\ — 2' ^^® square may be completed by simply adding 2 to 
 
 ^ 1 . 
 
 this expression, because a;^ _^ 2 + -^ is a perfect square, 
 
 1 ^ 
 
 namely, the square of re + - • The value of x may then be 
 
 deduced from it by solving another quadratic equation. 
 
 3 
 
 Example. ^x^ + -, = 22- 
 
 x^ 
 
 We first divide by 3 and add 2 to each side of the equation, 
 obtaining . , o . 1 _ 5? . 2 - ??. 
 
 Extracting the square root of both sides, 
 
 , 1 2a/7 2^21 2 /_. 
 a; + - = — — - = — ^— = ^ V21. 
 « V3 3 3 
 
 By multiplying by x, this equation becomes, a quadratic, 
 and can be solved in the usual way. 
 
 Let us now take this equation in the more general form, 
 
 a; + - = e, {a) 
 
 o 
 
 which reduces to the foregoing by putting e = o V^l. Clear- 
 ing of fractions and transposing, 
 
 ic2 — e:r + 1 = ; 
 
QUADRATIC EQUATIONS. 183 
 
 which being solved in the usual way, gives 
 
 _ e±V{e^ — 4) 
 X- ^ 
 
 The two roots are therefore 
 
 — e + V(e^ — 4) 
 <^i - 2 
 
 _ e - V{e^ - 4 ) 
 ^2 - 2 
 
 If in the first of these equations we rationalize the numer- 
 ator by multiplying it by e — V(e^ — 4) (§ 185), we shall find 
 
 2 1 
 
 it to reduce to , that is, to — • Therefore, 
 
 e — V (e^ — 4) ^2 
 
 x^ = — identically. 
 
 Vice versa, x^ is identically the same as — • • 
 This must be the case whenever we solve an equation of the 
 form («), that is, one in which the value of a; + - is given. 
 
 X 
 
 50 
 Let us suppose first that e = — -, so that the equation is 
 
 1 50 
 
 It is evident that ic = 7 is a root of this equation, because 
 when we put 7 for x, the left-hand member becomes 7 -f ^, 
 
 50 1 
 
 which is equal to — • If we put ^ for x, the left-hand mem- 
 ber will become 
 
 7^1 n^ 
 
 Hence x and - exchange values by putting - instead of 7, 
 
 X I 
 
 so that their sum x -\- - remains unaltered by the change. 
 
 X 
 
184 QUADBATIG EQUATIONS. 
 
 The general result may be expressed thus : 
 
 Because the value of the expression x + - remains un- 
 
 1 ^ 
 
 altered when we change x into , therefore the reciprocal of 
 
 any root of the equation 
 
 X + - = e 
 
 X 
 
 is also a root of the same equation. 
 
 EXERCISES. 
 
 Find all the roots of the following equations without clear- 
 ing the given equations from denominators : 
 
 I. -^ + ^. = T- - -'-' + ^j. = -^'-^- 
 
 3. 16f + ^, = 28. 4. j, + f = 2m\ 
 
 5. Show, without solving, that if r be any root of the 
 
 equation 1 
 
 x^-\-^ = a, 
 
 then — r, - , and will also be roots. 
 
 Factoring a Quadratic Equation. 
 198. 1. Special Case. Let us consider the equation 
 xi — 2x—16 = 0, 
 or a;2 — 2a; 4- 1 — 16 = 0, 
 
 or (x — 1)2 _ 42 = 0. 
 
 Factoring, it becomes (§ 90), 
 
 (a; — 1 -f 4) (:r — 1 — 4) = 0, 
 or {x + 3) (x — 6) = 0. 
 
 Therefore the original equation can be transformed into 
 (x + 3) (a; - 5) = 0, 
 
 a result which can be proved by simply performing the multi- 
 plications. 
 
QUADRATIC EQUATIONS. 185 
 
 This last equation may be satisfied by putting either of its 
 factors equal to zero ; that is, by supposing 
 
 a; + 3 = 0, whence x = —d ; 
 
 or 2; — 5 = 0, whence x = + 5. 
 
 These are the same roots which we should obtain by solving 
 the original equation. 
 
 2. Factoring the General Quadratic Equation. Let us con- 
 sider the general quadratic equation, 
 
 x^ -\- px -\- q = 0. (a) 
 
 Now, instead of thinking of x as a root of this equation, 
 let us suppose x to have any yalue whatever, and let us con- 
 sider the expression 
 
 a^ + px + q, (1) 
 
 which for shortness we shall call -Z. Let us also inquire how 
 it can be transformed without changing its value. 
 
 First we add and subtract -p% so as to make part of it a 
 perfect square. It thus becomes, 
 
 X= x^-^px + -pi — ^p^ + q; 
 
 or, which is the same thing. 
 
 Factoring this expression as in § 188, it becomes 
 
 The student should now prove that this expression is really equal to 
 x^ + px + g, by performing the multiplication. 
 
 Let us next put, for brevity, 
 
186 QUADRATIC EQUATIONS. 
 
 The preceding value of X will then become, 
 
 X={x-a){x- (3), (3) 
 
 an expression identically equal to (1), when we put for a and 
 j3 their values in (2). 
 
 Let us return to the supposition that this expression is to 
 be equal to zero, and that a; is a root of the equation. 
 
 The equation {a) will then be • 
 
 {x -a){x-(3) = 0. (4) 
 
 But no product can be equal to zero unless one of the fac- 
 tors is zero. Hence we must have either 
 
 X — a = 0, whence x =: a; 
 or X — (3 = Oj whence x = jX 
 
 Hence, a and § are the two roots of the equation (a). 
 
 The above is another way of solving the quadratic 
 equation. 
 
 To compare the expressions (1) and (3), let us perform the 
 multiplication in the latter. It will become, 
 
 X= a?—{a-}-(3)x-^af3. 
 
 Since this expression is identically the same as x^'{-px-{-q, 
 the coefiicients of the like powers of x must be the same. 
 That is, 
 
 a(3 = q, ] ^ ^ 
 
 which can be readily proved by adding and multiplying the 
 equations (2). 
 
 This result may be expressed as follows : 
 
 Theorem. When a quadratic equation is reduced 
 to the general form 
 
 x^ -\- px ■\- q^ = ^^ 
 the coefficient of x will be equal to the sum of the roots 
 with the sign changed. 
 
 The term independent of x will be equal to the 
 product of the roots. 
 
 The student may ask wliy can we not determine the roots of the 
 quadratic equation from equations (5), regarding a and /3 as the unknown 
 quantities ? 
 
QUADBATIG EQUATIONS. 187 
 
 We can do so, but let us see what tlie result will be. We eliminate 
 either « or |3 by substitution or by comparison. 
 
 From the second equation (5) we have, 
 
 a 
 
 Substituting this in the first equation, we have 
 
 a-\-^ = —p. 
 a 
 
 Clearing of fractions and transposing, 
 
 a^ -{- j)a -\- q = 0, 
 
 We have now the same equation with which we started, only a takes 
 the place of x. If we had eliminated a, we should have had the same 
 equation in j3, namely, 
 
 /32_|-^^4_g = 0. 
 
 So the equations (5), when we try to solve them, only lead us to the 
 original equation. 
 
 199. To form a Quadratic Equation when the Roots are 
 given. The foregoing principles will enable us to form a quad- 
 ratic equation which shall have any given roots. We have 
 only to substitute the roots for a and jS in equation (4), and 
 perform the multiplications. 
 
 Y' 3 
 
 EXERCISES. ' 'i 
 
 Form equations of which the roots shall be : 
 
 I. 
 
 + 1 and — 1. 
 
 2. 
 
 3 and 2. 
 
 3- 
 
 — 3 and — 2. 
 
 4. 
 
 3 + 2Vl0and 3—21/10. 
 
 5- 
 
 7 + 21/3 and 7-2\/3. 
 
 6. 
 
 + 1 and + 2. 
 
 7. 
 
 — 1 and + 2. 
 
 8. 
 
 - 1 and - 2. 
 
 9- 
 
 + 1 and — 2. 
 
 10. 
 
 2 + 1/5 and 2 — V5. 
 
 II. 
 
 4 ^^^ 5- 
 
 12. 
 
 2 "^^ 2- 
 
 13. 
 
 2 + a/2 and 2— V2. 
 
 14. 
 
 9 + 2V2 and 9 - 2V2, 
 
 15. 
 
 5 + 7V5and5-7V5. 
 a + Va^ — W' and a - 
 
 16. 
 
 -Va 
 
 a + h and « — 5. 
 
 • 
 
 17- 
 
 ;2 - J2. 
 
188 IMAGINARY ROOTS. 
 
 Equations having Imaginary Roots. 
 
 200. When we complete the square in order to solve a 
 
 quadratic equation, the quantity on the right-hand side of the 
 
 equation to which that square is equal must be positive, else 
 
 there can be no real root. For if we square either a positive 
 
 i or negative quantity, the result will be positive. Hence, if 
 
 >^ the square of the first member comes out equal to a negative 
 
 quantity, there is no answer, either positive or negative, which 
 
 will fulfil the conditions. Such a result shows that impossible 
 
 •' conditions have been introduced into the problem. 
 
 r\ j-^ EXAMPLES. 
 
 i 1 I. To divide the number 10 into two such parts that thei. 
 
 product shall be 34. 
 V If we proceed with this equation in the usual way, we shall 
 
 have, on completing the square, 
 
 x^ _ 10a; + 25 = — 9, 
 or {x — 5)2 = — 9. 
 
 The square being negative, there is no answer. On con- 
 sidering the question, we shall see that the greatest possible 
 product which the two parts of 10 can have is when they are 
 X each 5. It is therefore impossible to divide the number 10 
 
 into two parts of which the product shall be more than 25 ; and 
 because the question supposes the product to be 34, it is im- 
 possible in ordinary numbers. 
 
 2. Suppose a person to travel on the surface of the earth to 
 any distance ; how far must he go in order that the straight 
 line through the round earth from the point whence he started 
 to the point at which he arrives shall be 8000 miles ? 
 
 It is evident that the greatest possible length of this line is 
 a diameter of the earth, namely, 7,912 miles. Hence he can 
 never get 8,000 miles away, and the answer is impossible. 
 
 In such cases the square root of the negative quantity is 
 considered to be part of a root of the equation, and because it 
 is not equal to any positive or negative algebraic quantity, it 
 is ^called an imaginary root. The theory of such roots will be 
 explained in a subsequent book. 
 
 <r«- 
 
IRBATIONAL EQUATIONS. 189 
 
 CHAPTER III. 
 
 REDUCTION OF IRRATIONAL EQUATIONS TO THE 
 NORMAL FORM. 
 
 201. An Irrational Equation is one in which the 
 unknown quantity appears under the radical sign. 
 
 An irrational equation may be cleared of fractions 
 in the same way as if it were rational. 
 
 Example. Clear from fractions the equation 
 
 '\/x + a + Vx — a 2« 
 
 V^' -\- a — "^x — a Vx^ — a^ 
 
 Multiplying both members by Vx^ — a^ = Vx + a Vx—a, 
 we have 
 
 {x + a) Vx — a -\- (x — a) Vx + cl _ ^ 
 "s/x -\- a — ^/x — a 
 
 Next, multiplying by ^/x ■\- a — Vx — a, we have 
 (x + a) Vx — a -{• (x — a) Vx + a = 2aVx-\-a — 2aVx — a. 
 
 Transposing and reducing, we have 
 
 {x + 3a) Vx — a-\-(x — 3a) Vx + a = 0, 
 and the equation is cleared of denominators. 
 
 Clearing of Surds. 
 
 303. In order that an irrational equation may be solved, 
 it must also be cleared of surds which contain the unknown 
 quantity. In showing how this is done, we shall suppose the 
 equation to be cleared of denominators, and to be composed of 
 terms some or all of which are multiplied by the square roots 
 of given functions of x. 
 
 Let us take, as a first example, the equation just found. 
 Since a surd may be either positive or negative, the equation 
 in question may mean any one of the following four : 
 
190 IRRATIONAL EQUATIONS. 
 
 {x + 3fl) ^/x — a + {x — Za) Vx -^ a = 0, (1) 
 
 (x + da) ^/x — a — {x — Zd) Vx~+~a = 0, (2) 
 
 — (x -{- 3a) Vx — a-\-(x — da) ^x + a = 0, (3) 
 
 — {x + 3a) Vx — a— {x — 3a) ^x + « = 0. (4) 
 
 But the third equation is merely the negative of the second, 
 and the fourth the negative of the first, so that only two have 
 different roots. Let us put, for brevity, 
 
 a,) 
 a, S 
 
 (5) 
 
 P z=z {x -\- 3a) wx — a-\-{x — 3a) Vx + a, 
 Q = {x + 3a) \/x — a—{x — 3a) ^x + 
 and let us consider the equation, 
 
 PQ = 0. (6) 
 
 Since this equation is satisfied when, and only when, we 
 
 have either P = or ^ = 0, it follows that every value of x 
 
 which satisfies either of tlie equations (1) or (2) will satisfy (6). 
 
 Also, every root of (6) must be a root either of (1) or (2). 
 
 If we substitute in (6) the values of P and Q in (5), we 
 shall then have 
 
 {x + 3aY {x — a) — {x — 3aY {x -^ a) = 0, 
 which reduces to bx^ — W = 0, 
 
 and gives x = ±: — — • 
 
 It will be remarked that the process by which we free the 
 equation from surds is similar to that for rationalizing the 
 terms of a fraction employed in § 185. 
 
 As a second example, let us take the equation, 
 
 Va^ + ll + V'x^^ -5 = 0. (a) 
 
 We write the three additional equations formed by combin- 
 ing the positive and negative values of the surds in every way : 
 
 — Vx + 11 + Vx — 4: — 6 = 0, 
 Vx + li — a/^^^^4 — 5 = 0, 
 
 — Vx + 11 — Vx — 4: — 5 = 0. 
 The product of the first two equations is 
 
IRRATIONAL EQUATIONS. 191 
 
 ( V.?; _ 4 _ 5)2 _ (a; ^ 11) ^ 0, 
 
 or . 10 — iW^^^^ = 0. (1) 
 
 The product of the last two is 
 
 10 + lOVa; — 4 = 0. (2) 
 
 The product of these two products is 
 
 100 — 100 {x — 4:) = 0, 
 which gives x = 5. 
 
 It will be remarked that (2) differs from (1) only in having 
 the sign of the surd different. This must be the case, because 
 the second pair of equations formed from (a) differ from the 
 first pair only in having the sign of the surd V^ — 4 different. 
 Hence it is not necessary to write more than one pair of the 
 equations at each step. The general process is as follows : 
 
 I. CJzange the sign of one of the surds in the given 
 equation, and multiply the equation thus formed by the 
 07'iginal equation. 
 
 II. Reduce this product, in it change the sign of an- 
 other of the surds, and form a new product of the two 
 equations thus formed. 
 
 III. Continue the process until an equation without 
 surds is reached. 
 
 Example. Solve 
 
 \^Sx-\- 9 + V^x-\- 6 + Vx~^ = 0. 
 
 Changing the sign of Vx + 4, 
 
 VSx + 9 + V2x + 6 — Vx + 4: z=zO. 
 The product is 
 
 {VSx + 9 + V^x + 6^ — (a; + 4) = 0, 
 or, after reduction, 
 
 9a; + 11 + 2VSx + 9 V^x -}- 6 = 0. 
 Changing the sign of V^x + 6, we have 
 
 92; 4- 11 — 2VSx + 9 V2x + 6 = 0. 
 
192 IRBATIONAL EQUATIONS. 
 
 The product of the last two equations reduces to 
 
 IW — mx - 95 = 0, 
 
 33 H- 52 
 which being solved gives x = — j^ 
 
 Remark. Equations containing surds may often reduce to the form 
 treated in § 196. In this case, the methods of that section may be fol- 
 lowed. 
 
 EXERCISES. 
 
 Solve the equations : 
 
 11 2Va — 2Vx 
 
 Vx + Va Vx — Va x — a 
 
 Vx"" -ha X 
 
 3- Vx-\-3 — \^x — 4: = l. 
 
 4. VaT+H + Vx-U= 14. 
 
 5. (3 — x)^ — (3 + a;2)i = 0. 
 
 6. Va-{- ^x + \/« — ^~x = 2Vx^~- 
 
 Vx + 2 ^ — 4 Vi 
 ' Vox + 3 2 
 
 X 
 
 9. Va^ — 2x-{- —===^^ — I, 
 Va^ — 2x 
 
 X + Vx X (x — 1) 
 
 TO. 
 
 II. 
 
 x-Vx 4 
 
 VlT~a 1__ 
 
 Vx — a -\- Vcix — 1 Vx — 1 
 
SIMULTANEOUS QUADRATIC EQUATIONS, 193 
 
 CHAPTER IV. 
 
 SIMULTANEOUS QUADRATIC EQUATIONS. 
 
 Between a pair of simultaneous general quadratic equations 
 one of the unknown quantities can always be eliminated. The 
 resulting equation, when reduced, will be of the fourth degree 
 with respect to the other unknown quantity, and cannot be 
 solved like a quadratic equation. 
 
 But there are several cases in which a solution of two equa- 
 tions, one of which is of the second or some higher degree, 
 may be effected, owing to some of the terms being wanting in 
 one or both equations. 
 
 203. Case I. When one of the equations is of 
 the first degree only. 
 
 This case may be solved thus : 
 
 EuLE. Find the value of one of the unlcnown quan- 
 tities in terms of the other from the equaMon of the first 
 degree. This value being substituted i^^ the other equa- 
 tion, we shall have a quadratic equation from which the 
 other unknown quantity may be found. 
 
 Example. Solve 
 
 2x-3y = 5: ) ^^^ 
 
 From the second equation we find 
 ^ _ 3?/ + 5 
 
 2 
 
 /2 
 
 (*) 
 
 Whence, ^= = V±3^![+^, 
 
 Substituting this value in the first equation and reducing, 
 we find 
 
 4:1/ + 16y + 10 = 26. 
 
 Solving this quadratic equation, 
 13 
 
194 SIMULTANEOUS QUADBATIG EQUATIONS. 
 
 2/ = — ^ ± VS = — 2 ± 2a/3. 
 
 This value of y being substituted in the equation {h) gives, 
 
 — 1 ± 3a/8 — 1 ± 6\/2 
 x = ^ = ^^ 
 
 The same problem may be solved in tbe reverse order by eliminating 
 y instead of x. The second equation {a) gives 
 
 22; — 5 
 
 If vee substitute this value of y in the first equation, we shall have a 
 quadratic equation in x, from which the value of the latter quantity can 
 be found. 
 
 Solve 
 
 EXERCISES. 
 
 I. 
 
 a;2 _ <^xy + 4?/2 = 21. 
 2a; + 2/ = 1^- 
 
 2. 
 
 3a;2 _ 2^2 ^ 5^ _ 2^ = 28. 
 X + y ^ 4: ■= i). 
 
 3. 
 
 5xy + '7y^ — X — y = 72, 
 a;.+ 2y = 0. 
 
 4. 
 
 3a:2 _p 2y^ = 813, 
 7a; — 4?/ = 17. 
 
 5- 
 
 x + y = 7, 
 
 
 X y _ '^ 
 
 y a; ~ 12 
 
 204. Case II. When each equation contains 
 only one term of the second degree, and that term 
 has the same 2>roduct or square of the unknown 
 quantities in the two equations. 
 
 Such equations are 
 
 ax^ -\- dx -i- ey -\- f = 0, 
 a'x^ + d'x + e'y -\-f' = 0, 
 
 where the only term of the second degree is that in x^. 
 
 If we eliminate x^ from these equations by multiplying the 
 first by a' and the second by a, and subtracting, we have 
 
 ;:} <•' 
 
= 41. [ (^) 
 
 SIMULTANEOUS QUADRATIC EQUATIONS. 195 
 
 (a'd — atZ') 2; + {a'e — ae') y -{- a'f — af = 0. 
 Solving this equation with respect to x, we find 
 ^ ^ (ae^ - a'e) y -\- af - a 'f 
 
 a'd — ad' ' ^ ' 
 
 By substituting this value of x in either of the equations 
 (a), we shall have a quadratic equation in y. Solving the 
 latter, we shall obtain two values of y. Substituting these in 
 {h), we shall have the two corresponding values of x, and the 
 solution will be complete. Hence the rule. 
 
 Eliminate the term of the second degree hy addition 
 or subtraction, and use the resulting equation of the first 
 degree with either of the original equations, as in Case I. 
 
 Example. Solve 
 
 2xy — 4tx + 5y=2 23, 
 dxy + Hx + y 
 
 Multiplying the first equation by 3 and the second by 2, 
 and subtracting, we have 
 
 — 26a; + 13y = — 13 ; (J) 
 
 whence, ^ ~ 2^ ~^ 2* ^^^ 
 
 Substituting this value in the first equation, we find a 
 quadratic equation, which, being solved, gives 
 
 y = —2± V29. 
 
 Substituting these values in (c), the result is 
 
 The two sets of values of the unknown quantities are 
 therefore 
 
 y^ = —2 + V29, y^ = —2— V29. 
 
 We might have obtained the same result by solving the equation (c) 
 with respect to y, and substituting in {a). The student should practice 
 both methods. 
 
196 8IMULTANE0US QUADRATIC EQUATIONS. 
 
 EXERCISES. 
 
 1. Q>x^ — ^x — 4:y = 25, 
 
 x^ -^2x — 3i/ = 18. 
 
 2. 2f -^ y = 28, 
 y2 ^Sx — 4:y = 18. 
 
 3. xtj -}- 6x -\- Hy = 66, 
 Secy ^2x + 5y = 70. 
 
 305. Case III. When neither equation con- 
 tains a term of the first degree in oc or y. 
 
 Rule. Miminate the constant terms hy -multiplying 
 each equation hy the constant term of the other, and 
 adding or subtracting the two products. The result will 
 he a quadratic equation, from which either unhnown 
 quantity can he determined in terms of the other. TJien 
 suhstitute as in Case I. 
 
 Example. Solve x^ -\- xy — y"^ = h, \ , . 
 
 2x^ — 3xy + 2y' = U. \ ^ ^ 
 
 14 X Isfc eq., Ux"^ -\- Uxy — Uy^ = 70. 
 
 5 X 2d eq. , lt)a;2 — 15xy + 10^^ := 70. 
 
 Subtracting, 4a;2 + 29xy — 24:tf = 0. 
 
 This is a quadratic equation, by which one unknown quan- 
 tity can be expressed in terms of the other without the latter 
 being under the radical sign. 
 
 Transposing, ^x^ + 2^xy = 2iy\ (2) 
 
 841 1225 
 
 Completing square, 4a^ + 2^xy + ^-^2/^ = ~t^3/^' 
 
 29 35 
 
 Extracting root, - 2a; + -^^ = ± -j-?/. 
 
 ■wi. — 29 ± 35 3 o 
 
 Whence, x = ~ — y = -y or — 8?/. 
 
 Substituting the first of these values of x in either of ilio 
 original equations, we shall have 
 
 ^2 = 16; 
 
SIMULTANEOUS QUADRATIC EQUATIONS. 197 
 
 whence, 2/=±4:; a;=±3. 
 
 Substituting the second value of x, we have 
 
 ^bence, V = ±^Y ' = "^ -^l' 
 
 Therefore the four possible values of the unknown quanti- 
 ties are, 
 
 8 8 
 
 a; = + 3, - 3, + 
 
 Vll' Vll 
 
 Vll Vll 
 
 Each of these four pairs of values satisfies the original 
 equation. 
 
 A slight change in the mode of proceeding is to divide the 
 equation (2) by either x^ or ^f, and to find the value of the 
 quotient. Dividing by y'^ and putting 
 
 X 
 
 u =z -, 
 
 y 
 
 the equation will become 
 
 4^2 ^ 29u — 24 = 0. 
 
 This quadratic equation, being solved, gives 
 
 — 29 ± 35 3 
 u = g :^ _ or - 8, 
 
 X 
 
 Putting - for u, and multiplying by y, 
 
 
 X = jy or —Sy, as before. 
 
 Solve 
 
 EXERCISES. 
 
 I. 
 
 x^ — 2xy + 42^2 _ 4 _ 0. 
 
 2. 
 
 2x^ + 3xy — 7f — 2 = 0, 
 a-2 4- 3xy — 4?/2 + 1 = 0. 
 
198 SIMULTANEOUS QUADMATIG EQUATIONS. 
 
 306. Case IY. When the expressions contain^ 
 ing the unhnown quantities in the two equations 
 have common factors, 
 
 EuLE. Divide one of the equoMons which can he fac- 
 tored by the other, and cancel the common factors. 
 Then clear of fractions, if necessary, and we shall have 
 an equation of a lower degree. 
 
 EXAMPLES. 
 
 1. x^-\-y^ =z 91, »; + ?/ = 7. 
 
 We have seen (§ 94, Th. 1) that a^-\-y^h divisible hj x-\-y. 
 So dividing the first equation by the second, we have 
 x^ — xy ^ if — 13. 
 This is an equation of the second degree only, and when 
 combined with the second of the original equations, the solu- 
 tion may be effected by Case I. The result is, 
 ic = 3 or 4, ?/ = 4 or 3. 
 
 2. xy + y^ = 133, x^ ^ y"^ = 95. 
 
 Factoring the first member of each equation, the equations 
 become 
 
 y{x + y) = 133, {x + y) {x - y) = 95. 
 
 Dividing one equation by the other, and clearing of fractions, 
 
 7 
 12y = "Ix, or y = —x. 
 
 The problem is now reduced to Case I, this value of y 
 being combined with either of the original equations. 
 
 207. There are many other devices by which simultaneous 
 equations may be solved or brought under one of the above 
 cases, for which no general rule can be given, and in which 
 the solution must be left to the ingenuity of the student. 
 Sometimes, also, an equation which comes under one of the 
 cases can be solved much more expeditiously than by the rule. 
 
 Let us take, for instance, the equations, 
 
 ^2 J^ y2 — 65^ ^y — 28. 
 
 These equations can be solved by Case III, but the work 
 would be long and cumbrous. We see that by adding and 
 
SIMULTANEOUS QUADRATIC EQUATIONS. 199 
 
 subtracting twice the second equation to and from the first, 
 we can form two perfect squares. Extracting the roots of 
 these squares, we shall have two simple equations, which shall 
 give the solution at once. Each unknown quantity will have 
 four values, namely, ± 7 ± 4. 
 
 PROBLEMS AND EXERCISES. 
 
 The following equations can all be solved by some short and expe- 
 ditious combination of tlie equations, or by factoring, without going 
 through the complex process of Case III. The student is recommended 
 not to work upon the equations at random, but to study each pair until 
 he sees how it can be reduced to a simpler equation by addition, multi- 
 plication, or factoring, and then to go through the operations thus sug- 
 gested. 
 
 1. ?/2 _|_ rf^y _ ]^4^ ^2 j^ xy ^= 35. 
 
 2. 4a;2 — 2xy = 208, 2xy — y^ = 39. 
 
 3. x^ -\- y =: 4x, ?/2 -f ^ = 4y. 
 
 If we subtract one of these equations from the other, the difference 
 will be divisible by x — y. 
 
 4. a;3 ^ yz ^ 3^ _|. 3^y _ 378, x^ j^ y^ — Zx — Zy z=l 324. 
 
 5. x^ + 2/2 z= 74, x-\-y =z 12. 
 
 6. a;2 4- icy = 63, 2^ — 2/2 = 77. 
 
 a/^ — "s/y 
 
 4, x^ 
 
 -f 
 
 544. 
 
 xy = a, y^ + xy 
 
 x^ -\-xy^ — 10, 2/3 _^ x^y — 5. 
 X = aVx -\- y, ?/ = hy/x -f y, 
 x\/x + ?/ = 12. ys/x 4- y 
 
 15. 
 
 - X 
 
 2x^ + 2?/2 = X -\- y, x"^ ■}- y^ =: X — y. 
 
 5x^ — 5y^ = X -\- y, 3x^ — 3y^ = x — y. 
 
 a:2-f ?/2 + 2;2 r= 30, xy -\- yz -{- zx = 17 , x — y — z = 2. 
 
 ,, Ac + y 8 
 
 y x — y x — y 
 
200 PROBLEMS. 
 
 1 6. A principal of $5000 amounts, with simple interest, to 
 $7100 after a certain number of years. Had the rate of inter- 
 est been 1 per cent, higher and the time 1 year longer, it would 
 have amounted to $7800. What was the time and rate? 
 
 17. A courier left a station riding at a uniform rate. Five 
 hours afterward, a second followed him, riding 3 miles an 
 hour faster. Two hours after the second, a third started at 
 the rate of 10 miles an hour. They all reach their destination 
 at the same time. What was its distance and the rate of riding ? 
 
 18. In a right-angled triangle there is given the hypothe- 
 nuse = a, and the area = h"^', find the sides. 
 
 19. Find two numbers such that their product, sum, and 
 difference of squares shall be equal to each other. 
 
 20. Find two numbers whose product is 216; and if the 
 greater be diminished by 4, and the less increased by 3, the 
 product of this sum and difference may be 240. 
 
 21. There are two numbers whose sum is 74, and the sum 
 of their square roots is 12. What are the numbers ? 
 
 22. Find two numbers whose sum is 72, and the sum of 
 their cube roots 6. 
 
 23. The sides of a given rectangle are m and n. Find the 
 sides of another which shall have twice the perimeter and twice 
 the area of the given one. 
 
 24. A certain number of workmen require 3 days to com- 
 plete a work. A number 4 less, working 3 hours less per day, 
 will do it in 6 days. A number 6 greater than the original 
 number, working 6 hours less per day, will complete the work 
 in 4 days. What was the original number of workmen, and 
 how long did they work per day ? 
 
 25. Find two numbers whose sum is 18 and the sum of 
 their fourth powers 14096. 
 
 Note. Since the sum of the two numbers is 18, it is evident that 
 the one must be as much less than 9 as the other is greater. The equa^ 
 tions will assume the simplest form when we take, as the unknown quan- 
 tity, the common amount by which the numbers differ from 9, 
 
 26. Find two numbers, x and y, such that 
 
 a? + y^ : a^ — y^ :: 35 : 19, 
 
 xy = 24. 
 
 27. Find two numbers whose sum is 14 and the sum of 
 their fifth powers 1G1204. 
 
BOOK VII. ' 
 PROGRESSIONS, 
 
 CHAPTER I. 
 
 ARITHMETICAL PROGRESSION. 
 
 208. Bef. When we have a series of numbers each 
 of which is greater or less than the preceding Iby a con- 
 stant quantity, the series is said to form an Arithmet- 
 ical Progression. 
 
 Example. The series 
 
 7, 12, 17, 22, 27, 32, etc. ; 
 7, 5, 3, 1, —1, -3, etc.; 
 a -{- h, a, a — h, a — 25, a — Zh, etc., 
 
 are each in arithmetical progression, because, in the first, each 
 number is greater than the preceding by 5 ; in the second, 
 each is less than the preceding by 2 ; in the third, each is less 
 than the preceding by h. 
 
 Bef. The amount by which each term of an arith- 
 metical progression is greater than the preceding one is 
 called the Common Difference. 
 
 Bef. The Arithmetical Mean of two quantities is 
 half their sum. 
 
 All the terms of an arithmetical progression except 
 the first and last are called so many arithmetical means 
 between the first and last as extremes. 
 
 Example. The four numbers, 5, 8, 11, 14, form the four 
 arithmetical means between 2 and 17. 
 
202 ARITHMETICAL PBOGBESSION. 
 
 EXERCISES. 
 
 1. Form four terms of the arithmetical progression of 
 which the first term is 7 and common difference 3. 
 
 2. Write the first seven terms of the progression of which 
 the first term is 11 and the common difference — 3. 
 
 3. Write five terms of the progression of which the first 
 term is a — ^n and the common difference 'Zn. 
 
 Problems in Progression. 
 
 309. Let us put 
 
 ay the first term of a progression. 
 d, the common difference. 
 11, the number of terms. 
 I, the last term. 
 Co 2, the sum of all the terms. 
 The series is then 
 
 a, a-\-cl, a-\-2d, .... 7. 
 Any three of the above five quantities being given, the 
 other two may be found. 
 
 Problem I. Given the first term, the coimnon differ- 
 ence, and the number of terms, to find the last term. 
 The 1st term is here a, 
 
 2d " " a + d, 
 
 3d " " a + 2d. 
 
 The coefficient of d is, in each case, 1 less than the number 
 of the term. Since this coefficient increases by unity for every 
 term we add, it must remain less by unity than the number of 
 the term. Hence, 
 
 The i^^ term is a + (i — 1) d, 
 whatever be i. Hence, when i = n, \ 
 
 I = a-\- (n — 1) d. (1) 
 
 From this equation we can solve the further problems : 
 
 Problem II. Given the last term I, the common dif- 
 ference d, and the number of terms n, to find the first 
 term. 
 
ARITHMETICAL PROGRESSION, 203 
 
 The solution is found by solving (1) with respect to a, 
 
 which gives 
 
 a = l-{n-l)d. (2) 
 
 Problem III. Given the first and last terms, a and I, 
 and the ninnher of terms ii, to find the common dijfer- 
 ence. 
 
 Solution from (1), d being the unknown quantity, 
 
 d = l^. (3) 
 
 Problem IV. Given the first and last terms and the 
 common difference, to find the nujiiber of terms. 
 
 Solution, also from (1), 
 
 I — ct , ^ I — a 4- d ,,. 
 
 n^^- + l = —^—. (4) 
 
 Problem V. To find the sum of all the terms of an 
 arithmetical progression. 
 
 We have, by the definition of 2, 
 
 2 = « + (<? -f 6^) + (fif + 2^) + {l — d) + l, 
 
 the parentheses being used only to distinguish the terms. 
 
 Now let us write the terms in reverse order. The term 
 before the last is I — d, the second one before it I — ^d, etc. 
 
 We therefore have, 
 
 ^ z=zl -{- {l — d) + {l — M) -\- {a + d) -\- a. 
 
 Adding these two values of 2 together, term by term, we 
 find 
 22 = {afl) + (a + + {a + l) + + (« + ?) + (« + Z), 
 
 the quantity {a -f T) being written as often as there are terms, 
 that is, n times. Hence, 
 
 22 = n(a -\- l\ 
 ^ = n^l. (5) 
 
 _i_ 7 
 
 Remark, The expression -——, that is, half the sum of 
 the extreme terms, is the mean value of all the terms. The 
 
204 ARITHMETICAL PBOGRESSIOJV. 
 
 sum of the n terms is therefore the same as if each of them 
 had this value. 
 
 210. In the equation (5) we are supposed to know the 
 jBrst and last terms and the number of terms. If other quan- 
 tities are taken as the known ones, we have to substitute for 
 some one of the quantities in (5) its expression in one of the 
 equations (1), (2), (3), or (4). Suppose, for example, that we 
 have given only the last term, the common difference, and the 
 number of terms, that is, /, d, and 7i. We must then in (5) 
 substitute for a its value in (2). This will give, 
 
 1 = 7iU — d) = nl -^ d, (6) 
 
 EXERCISES. 
 
 In arithmetical progression there are 
 
 1. Given, common difference, + 3; third term = 10. 
 Find first term. Ans. First term r=: 4. 
 
 2. Given 4th term = h, common difference = — c. 
 Find first 7 terms, their sum and product. 
 
 3. Given 3d term =z a -\- h, 4th term = « + 25. 
 Find first 5 terms. 
 
 4. Given 1st term =z a — h, 9th term = 9« -f 75. 
 Find 2d term and common difference. 
 
 5. Given, sum of 9 terms = 108. 
 
 Find middle term and sum of 1st and 9th terms. 
 
 6. Given 5th term = Ix — by, 7th term =z ^x — di/. 
 Find first 7 terms and common difference. 
 
 7. Given 1st term = 12, 50th term = 551. * 
 Find sum of all 50 terms. 
 
 8. To find the sum of the first 100 numbers, namely, 
 
 1 + 2+3 +99 + 100. 
 
 Here the first term a ia 1, the last term I 100, and the number of 
 terms 100. The solution is by Problem V. 
 
 9. Find the sum of the first n entire numbers, namely, 
 
 1 + 2 + 3 -\-n. 
 
ARITHMETICAL PEOGRESSION. 205 
 
 10. Find tlie sum of the first n odd numbers, namely, 
 
 1 + 3 + 5 + 2?2 — 1. 
 
 Here the number of terms is n. 
 
 1 1. Find the sum of the first n even numbers, namely, 
 
 2 + 4 + 6 + 2w. 
 
 12. In a school of m scholars, the highest received 134 
 merit marks, and each succeeding one 6 less than the one next 
 above him. How many did the lowest scholar receive ? How 
 many did they all receive ? 
 
 13. The first term of a series is m, the last term 2m, and 
 the common difi'erence d. What is the number of terms? 
 
 14. The first term is k, the last term 10^ — 1, and the 
 number of terms 9. What is the common difference ? 
 
 15. The middle term of a progression is s, the number of 
 terms 5, and the common difference — li. What are the first 
 and last terms and the sum of the 5 terms ? 
 
 16. The sum of 5 numbers in arithmetical progression is 
 20 and the sum of their squares 120. What are the numbers ? 
 
 Note. In questions like this it is better to take the middle term for 
 one of the unknown quantities. The other unknown quantity will be 
 the common difference. 
 
 17. Find a number consisting of three digits in arithmeti- 
 cal progression, of which the sum is 15. If the number be 
 diminished by 792, the digits will be reversed. 
 
 18. The continued product of three numbers in arithmet- 
 ical progression is 640, and the third is four times the first. 
 What are the numbers? 
 
 19. A traveller has a journey of 132 miles to perform. He 
 goes 27 miles the first day, 24 the second, and so on, travelling 
 3 miles less each day than the day before. In how many days 
 will he complete the journey ? 
 
 Here we have given the first term 27, the common difference —3, and 
 the sura of the terms 132. To solve this, we take equation (5), and sub- 
 stitute for 1 its value in (1). This makes (5) reduced to 
 
 a -{- a -^ (n — 1) d n(n — l)d 
 
 1 = n ^ '— = na + --^—^ 
 
 2, a, and d are ^iven by the problem, and n is the unknown quan- 
 tity. Substituting the numerical value of the unknown quantities, the 
 equation becomes 
 
206 ABITIIMETICAL PROQBBSSION. 
 
 Tliis reduced to a quadratic equation in n, the solution of which gives 
 two values of n. The student should explain this double answer by 
 continuing the progression to 11 terms, and showing what the negative 
 terms indicate. 
 
 20. Taking the same question as the last, only suppose the 
 distance to be 140 miles instead of 132. Show that the answer 
 will be imagniary, and explain this result. 
 
 21. A debtor owing $160 arranged to pay 25 dollars the 
 first month, 23 the second, and so on, 2 dollars less each 
 month, until his debt should be discharged. How many pay- 
 ments must he make, and what is the explanation of the two 
 answers ? 
 
 2 2. A hogshead holding 135 gallons has 3 gallons poured 
 into it the first day, 6 the second, and so on, 3 gallons more 
 every day. How long before it will be filled ? 
 
 23. The continued product of 5 consecutive terms is 12320 
 and their sum 40. What is the progression ? 
 
 24. Show that the condition that three numbers, p, q, and 
 r, are in arithmetical progression may be expressed in the form 
 
 q-p ^ -1. 
 q — r 
 
 25. In a progression consisting of 10 terms, the sum of the 
 1st, 3d, 5th, 7th, and 9th terms is 90, and the sum of the re- 
 maining terms is 110. What is the progression ? 
 
 26. In a progression of an odd number of terms there is 
 given the sum of the odd terms (the first, third, fifth, etc.)^ 
 and the sum of the even terms (the second, fourth, etc.). 
 Show that we can find the middle term and the number of 
 terms, but not the common difference. 
 
 27. In a progression of an even number of terms is given 
 the sum of the even terms = 105, the sum of the odd terms = 
 119, and the excess of the last term over the first = 26. What 
 is the progression ? 
 
 28. Given a and Z, the first and last terms, it is required to 
 insert i arithmetical means between them. Find the expres- 
 sion for the i terms required. 
 
GEOMETBICAL PROGRESSION. 207 
 
 CHAPTER II. 
 GEOMETRICAL PROGRESSION. 
 
 311. Def. A Geometrical Progression consists of 
 a series of terms of which each is formed by multiply- 
 ing the term preceding by a constant factor. 
 
 An arithmefcical progression is formed by continual addi- 
 tion or subtraction; a geometrical progression by repeated 
 multiplication or division. 
 
 Def. The factor by which each term is multiplied 
 to form the next one is called the Common Ratio. 
 
 The common ratio is analogous to tbc common difference 
 in an arithmetical progression. 
 
 In other respects the same definitions apply to both. 
 
 E X AM PLE S. 
 
 2, 6, 18, 54, etc., 
 is a progression in which the first term is 2 and the common 
 ratio 3. 
 
 2 1 ^ ^ ^ etc 
 
 2' 4' 8' 
 
 is a progression in which the ratio is -• 
 
 + 3, — 6, + 12, — 24, etc., ■ 
 is a progression in which the ratio is — 2. 
 
 Note. A progression like the second one above, formed by dividing 
 each term by the same divisor to obtain the next term, is included in the 
 general definition, because dividing by any number is the same as multi- 
 plying by the reciprocal. Geometrical progressions may therefore be . 
 divided into two classes, increasing and decreasing. In the increasing 
 progression the common ratio is greater than 1 and the terms go on in- 
 creasing ; in a diminishing progression the ratio is less than unity and 
 the terms go on diminishing. 
 
 Rem. In a progression in which the ratio is. negative, the 
 terms will be alternately positive and negative. 
 
t i- f^vi-/yf< 
 
 ^r-r 
 
 208 OEOMETBIGAL PBOGRESSIOm 
 
 Def. A Geometrical Mean between two quantities 
 is the square root of tlieir product. 
 
 EXE RCI SE s. 
 
 Form five terms of each of the following geometrical pro- 
 gressions : 
 
 1. First term, 1 ; common ratio, 5. 
 
 2. First term, 7 ; common ratio, — 3. 
 
 3. First term, 1 ; common ratio, — 1. 
 
 2 3 
 
 4. First term, - ; common ratio, -• 
 
 4 1 
 
 5. First term, - ; common ratio, ^' 
 
 Problems of Geometrical Prog:ression. 
 
 212. In a geometrical progression, as in an arithmetical 
 one, there are five quantities, any three of which determine 
 the progression, and enable the other two to be found. They 
 are 
 
 a, the first term. 
 ^ Ty the common ratio. 
 
 n, the number of terms. 
 I, the last term. 
 2, the sum of the n terms. 
 
 The general expression for the geometrical progression 
 will be 
 
 «, ffr, ar^, ar^, etc., 
 
 because each of these terms is formed by multiplying the pre- 
 ceding one by r. 
 
 The same problems present themselves in the two progres- 
 sions. Those for the geometrical one are as follows : 
 
 Problem. I. Given the first term, the coimnoiv ratio, 
 and the number of terms, to find the last term. 
 
 The progression will be 
 
 a, ar, ar^, etc. 
 We see that the exponent of r is less by 1 than the number 
 of the term, and since it increases by 1 for each term added, it 
 
OEOMETBICAL PR0QBE88I0N. 209 
 
 must remain less by 1, how many terms so ever we take. 
 Hence the 7i^^ term is 
 
 I = ar^-K (1) 
 
 Problem II. Given the last term, the corynnon i^atio, 
 and the number of terjivs, to find the first term. 
 
 The sohition is found by dividing both members of (1) by 
 r^-'^, which gives 
 
 Problem HI. Given the first term, the last term, and 
 the number of terms, to find the common ratio. 
 
 From (1) we find r^~'^ = -• 
 
 Extracting the (n — 1)^'' root of each member, we have 
 
 =©•"■ 
 
 [The solution of Problem IV requires us to find n from 
 equation (1), and belongs to a higher department of Algebra.] 
 
 Problem V. To find the sum of all n terms of a geo- 
 metrical progression. 
 
 We have 1 = a + ar ■}- ar^ + etc. + ar^~K 
 Multiply both sides of this equation by r. We then have 
 rS = ar -{- ar^ + ar^ + etc + ar"-. 
 
 Now subtract the first of these equations from the second. 
 It is evident that, in the second equation, each term of the 
 second member is equal to the term of the second member of 
 the first equation which is one place farther to the right. 
 Hence, when we subtract, all the terms will cancel each other 
 except the first of the first equation and the last of the second. 
 Illustration. The following is a case in which a = 2,r = 3,n = Q: 
 2=2 + 6 + 18 + 54 + 162 + 486. 
 32 = 6 + 18 + 54 + 162 + 486 + 1458. 
 Subtracting, 32-2 = 1458 - 2 = 1456, 
 
 or 22 =: 1456, and 2 = 728. 
 14 
 
210 GEOMETRICAL PROGRESSION. 
 
 Eeturning to the general problem, we have 
 
 {r — Vjl. — ar^ — a —a (r^ — 1) ; 
 
 fn — 11 — fn 
 
 whence, 1. = a = a U\ 
 
 r — \ \ — r ^ ' 
 
 It will be most convenient to use the first form when r > 1, 
 and the second when r < 1. 
 
 By this formula we are enabled to compute the sum of the 
 terms of a geometrical progression without actually forming 
 all the terms and adding them. 
 
 EXERCISES 
 
 1. Given 3d term = 9, common ratio = -• 
 Find first 5 terms. 
 
 32 2 
 
 2. Given 5th term = — -, common ratio = — -• 
 
 Find first 5 terms. 
 
 3. Given 5th term = 7:^y^, 1st term = yK 
 Find common ratio. 
 
 4. Given 1st term = 1, 4th term = c^. 
 Find common ratio and first 3 terms. 
 
 5. Given 2d term = m, common ratio = — m. 
 Find first 4 terms. 
 
 6. A farrier having told a coachman that he would charge 
 him $3 for shoeing his horse, the latter objected to the price. 
 The farrier then offered to take 1 cent for the first nail, 2 for 
 the second, 4 for the third, and so on, doubling the amount 
 for each nail, which offer the coachman accepted. There were 
 32 nails. Find how much the coachman had to pay for the 
 last nail, and how much in all. (Compare § 1G8, Eem.) 
 
 7. Find the sum of 11 terms of the series 
 
 2 + 6 + 18 + ^c.^ -f /6 S 1/ V^^ 
 in which the first term is 2 and the common ratio 3. 
 
 8. If the common ratio of a progression is r, what will be 
 the common ratio of the progression formed by taking 
 
 I. Every alternate term of the given progression ? 
 II. Every n^'^ term ? 
 
GEOMETRICAL PBOGBESSION. 211 
 
 9. The same thing being supposed, what will be the com- 
 mon ratio of the progression of which every alternate term is 
 equal to every third term of the given progression ? 
 
 10. Show that if, in a geometrical progression, each term 
 be added to or subtracted from that next following, the sums 
 or remainders will form a geometrical progression. 
 
 11. Show that if the arithmetical and geometrical means 
 of two quantities be given, the quantities themselves may be 
 found, and give the expressions for them. 
 
 12. The sum of the first and fourth terms of a progression 
 is to the sum of the second and third as 21 : 5. What is the 
 common ratio? 
 
 13. Express the continued product of all the terms of a 
 geometrical progression in terms of a, r, and n ? 
 
 Limit of the Sum of a Progression. 
 
 313. Theorem. If the common ratio in a geometri- 
 cal progression is less than unity (more exactly, if it is 
 contained between the limits —1 and +1), then there 
 will be a certain quantity which the sum of all the 
 terms can never exceed, no matter how many terms we 
 take. 
 
 For example, the sum of the progression 
 
 111^ 
 
 - + ^ + ^ + etc., 
 
 in which the common ratio is -, can never amount to 1, no 
 
 matter how many terms we take. To show this, suppose that 
 one person owed another a dollar, and proceeded to pay him a 
 series of fractions of a dollar in geometrical progression, 
 namely, 
 
 1111 
 
 2' 4' 8' 16' ^*^' 
 
 When he paid him the - he would still owe another --, 
 z z 
 
 when he paid the ^ lie would still owe another -, and so on. 
 
212 OEOMETRIGAL PROGRESSION. 
 
 That is, at every payment he would discharge one-half the re- 
 maining debt. Now there are two propositions to be under- 
 stood in reference to this subject. 
 
 I. T1i6 entire debt can never he discharged hy such 
 ■payments. 
 
 For, since the debt is halved at every payment, if there was 
 any payment which discharged the whole remaining debt, tlie 
 half of a thing would be equal to the whole of it, which is 
 impossible. 
 
 II. Tlxe debt can he reduced heloiv any assi^nahle 
 limit hy continuing to pay half of it. 
 
 For, however small the debt may be made, another pay- 
 ment will make it smaller by one-half; hence there is no 
 smallest amount below which it cannot be reduced. 
 
 These two propositions, which seem to oppose each other, hold the 
 truth between them, as it were. They constantly enter into the higher 
 mathematics, and should be well understood. We therefore present 
 another illustration of the same subject. 
 
 A B 
 
 I I I III 
 
 Suppose AB to be a line of given length. Let us go one- 
 half the distance from A to B at one step, one-fourth at the 
 second, one-eighth at the third, etc. It is evident that, at each 
 step, we go half the distance which remains. Hence the two 
 principles just cited apply to this case. That is, 
 
 1. We can never reach B by a series of such steps, because 
 we shall always have a distance equal to the last step left. 
 
 2. But we can come as near B as we please, because every 
 step carries us over half the remaining distance. 
 
 This result is often expressed by saying that we should reach B by 
 taking an infinite number of steps. This is a convenient form of expres- 
 sion, and we may sometimes use it, but it is not logically exact, because 
 no conceivable number can be really infinite. The assumption that in- 
 finity is an alsrebraic (juantity often leads to ambiguities and difficulties 
 in the application of ms,thematics. 
 
GEOMETRICAL PROGRESSION. 213 
 
 Def. The Limit of the sum 2 of a geometrical 
 progression is a quantity which 2 may approach so 
 that its difference shall be less than any quantity we 
 choose to assign, but which 2 can never reach. 
 
 EXAMPLES. 
 
 1. Unity is the hmit of the sum 
 
 1111^ 
 
 2. The point B in tlie preceding figure is the limit of all 
 the steps that can be taken in the manner described. 
 
 The following principle will enable us to find the limit of 
 tlie sum of a progression : 
 
 214. Principle. If r < 1, the power r^ can be made 
 as small as we please by increasing the value of n, but 
 can never be made equal to 0. 
 
 _ 3 _ ^ _1 
 
 ^ ~ 4 - 4" 
 
 Suppose, for instance, that 
 3 
 
 Then every time we multiply by r we diminish r^ by 
 line ; that is, 
 
 = 4^ "^1^-4/^ = ^-4''' 
 
 - of its former value ; that is, 
 4 
 
 
 3 3 „ 
 
 fi- 
 
 ■h 
 
 
 
 
 
 
 r^ = \r^ = 
 
 r3_ 
 
 ■l". 
 
 
 
 
 
 
 etc. etc. 
 
 
 etc. 
 
 
 
 
 
 Now let 
 
 us again take 
 
 the 
 
 expression 
 
 for the 
 
 sum 
 
 of 
 
 a 
 
 series of n terms, namely, 
 
 
 
 
 . 
 
 
 
 
 2 - 
 
 1 
 = a- 
 
 
 
 
 
 
 
 
 
 1 - 
 
 — r 
 
 which 
 
 we 
 
 may put 
 
 into the form 
 
 a 
 
 
 1 — r 
 
214 OEOMETBIGAL PROQIIESSION. 
 
 If r is less than unity, we can, by the principle just cited, 
 make the quantity r^ as small as we please by increasing n 
 indefinitely. From this it follows that we can also make the 
 
 term r^ as small as we please. 
 
 1 — r ^ 
 
 Proof. Let us put, for brevity, 
 Tc 
 
 1-r' 
 
 so that the term under consideration is 
 
 Tcr^. 
 
 If we cannot make kr^ as small as we please, suppose s to 
 be its smallest possible value. Let us divide s by k, and put 
 
 ^ - k 
 No matter how small s may be, and how large k may be, 
 T, or t, will always be greater than zero. Hence, by the pre- 
 ceding principle, we can find a value of n so great that r^ 
 shall be less than t. Tliat is, 
 
 Multiplying both sides of this inequality by k, 
 
 kr^ < s. 
 
 That is, however small we take s, we can ^take n so large 
 that kr^ shall be less than s, and therefore .9 cannot be the 
 smallest value. 
 
 Since S := kr^, 
 
 1 — r 
 
 ^ and since we can make kr^ as small as we please, it follows 
 Limit of 2 = 
 
 1 — r 
 This is sometimes expressed by saying that When r < 1, 
 
 a -\- ar + ar^ + ar^ + etc., ad mfiiiitum = :j-^— .> 
 
 and this is a convenient form of expression, which will not lead 
 us into error in this case. 
 
OEOMMTRICAL PB0ORE8SI0N. 215 
 
 EXERCISES. 
 
 Having given the progression 
 
 1111^ 
 
 2 + 4 + 8 + 16 + ^*^" 
 
 of which the limit is 1, find how many terms we must take in 
 order that the sum may differ from 1 by less than the follow- 
 ing quantities, namely : 
 
 Firstly, .001 ; secondly, .000 001 ; thirdly, .000 000 00,1. 
 
 To do this, we must find what power of - will be less than .001, 
 what power less than .00^0 OQl, etc. 
 
 What are the limits of the sums of the following series: 
 3 + 02 + 03 + ^^^-^ ^^ infinitum, 
 o + q + 27 + ^^^-j ^^ infinitum. 
 
 ^ — --g + -3 — etc., ad infinitum. 
 
 4 42 43 
 
 Q + — 3 + Q-3 + etc., ad infinitum. 
 
 rfl + (irnp + (TT^P + ^tc, ad infinitum. 
 + -pj z-rs — etc., ad infiyiitum. 
 
 I — I (J _ 1)2 ' (^, _ 1) 
 
 2 1 2 1 , 7 . ^ ., 
 
 1 ! 5 7,-\ -. — etc., ad mlinitum. 
 
 m 7)1^ iw m^ -^ 
 
 What is that progression of which the first term is 12 
 and the limit of the sum 8. 
 
 9. On the line AB a man starts from A and goes to the 
 point c, half way to B ; then he re- 
 tni-ns to d, half way back to A ; then 1 ^1 ? 
 turns again and goes half way to c, 
 
 then back half way to d, and so on, going at each turn half 
 way to the point from which he last set out. To what point 
 on the line will he continually approach ? 
 
216 . COMPOUND INTEREST. 
 
 315. As an interesting application of the preceding theory, 
 we may examine the problem of finding the value of a circu- 
 lating decimal. Such a decimal is always equal to a vulgar 
 fraction, which is obtained as in the following examples : 
 
 I. What is the value of the decimal 
 
 .373737 ? 
 
 We find the figures wliich form the period to be 37. Dividing the 
 decimal into periods of these figures, its value is 
 
 _3j^ 37^ ^ , 
 
 100 "^ 1002 "^ 1003 + ^^c- 
 
 ^^(lJo + i + i5o-3 + H 
 
 The quantity in the parenthesis is a geometrical progression, in which 
 a = T^r-r , r = zrj-zr • Tlie limit of its sum is therefore — • Therefore the 
 
 luo lou yy ^. 
 
 37 0iy4% 
 
 value of the decimal is ^ • ,^jr\\ 
 
 This result can be proved by changing this vulgar fraction ^o a 
 decimal. 
 
 2. In the case of a decimal which has one or more figures 
 before the period commences, we cut these figures off, and 
 find the value of them and of the circulating part separately. 
 
 Thus, 
 
 56363 etc. = ^ + ^ + ^ + etc. f 
 
 5 63 /. 1.1 
 
 i + rS-o + i4-^ + ^*^') 
 
 "" 10 ^ 1000 
 
 _ 5 63 100 _ 5 ^3^ _ 558 _ 31 
 
 ~ To "^ 1000 "99" ~~ To "^ 990 ■" 990 ~ 55" 
 
 EXERCISES. 
 
 To what vulgar fractious are the following circulating deci- 
 mals equal : 
 
 I. .111111 ? 2. .2222 ? 
 
 3. .9999 ? 4. .09999 ? 
 
 5. .454545 ? 6. .2454545 ? 
 
 7. .108108 ? 8. 72454545 ? 
 
p 
 
 ^ COMPOUND INTEBEST.)p f |^ " /jiv 217 
 
 Compound Interest. 
 
 216. When one loans or invests money, collects the inter- 
 est at stated intervals, and again loans or invests this interest, 
 and so on, he gains compound interest. 
 
 Compound interest can always be gained by one who con- 
 stantly invests all his income derived from interest, provided 
 that he always collects the interest when due, and is able to 
 loan or invest it at the same rate as he loaned his principal. 
 
 Problem I. To find the amount of x> dollars for n 
 years, at c per cent, compound interest. 
 
 Solution". At the end of one year the interest will be 
 
 ^~, which added to the principal will make ^ (l + tt^)* 
 
 If we put p = --— = the rate of annual gain, 
 
 he amount at the end of the year will be j!? (1 + p). 
 
 Now suppose this whole amount is put out for another 
 t) year at the same rate. The interest will be ^ (1 + p) p, which 
 added to the new principal p{l -f p) will make p {1 -\- py. 
 
 It is evident that, in general, supposing the whole sum 
 J'.ept at interest, the total amount of the investment will be 
 -toultiplied by 1 + p each year. Hence the amount at the ends 
 of successive years will be 
 
 ;j(l + p), p{l-\-pY, ;?(l^-p)^ etc. 
 
 At the end of n years the amount will be -- - 
 
 . i?(l+p)^ 
 
 Problem II. A person puts out p dollars every year, 
 ■ttirig the whole sum constantly accumulate a.t com- 
 pound interest. What will the amount l>e at the end of 
 It years? 
 
 Solution". The first investment will have been out at 
 interest n years, the second n — 1 years, the third 7i — 2 years, 
 anvl so on to the n^\ which will have been out 1 year. Hence, 
 from tjie last formula, the amounts will be : 
 
218 COMPOUND IN2 
 
 Amount of 1st payment, ^:> (1 + p)^. 
 
 <( 
 
 " 2d 
 
 a 
 
 p{l-^pY-K 
 
 a 
 
 « 3d 
 
 a 
 
 p{i + pY'K 
 
 li 
 
 " 4th 
 
 a 
 
 jt? (1 + pY-\ 
 
 a 
 
 " 5th 
 
 a 
 
 p{i + pY-k 
 
 
 etc. 
 
 
 etc. 
 
 The sum of the amounts is : 
 
 i?{l+rt + pi^-\-pf +i? (1+p)^ 4- . . ,.p(l^pY' 
 
 This is a geometrical progression, of which the first term is 
 p (1 + p), the common ratio 1 + p, and the number of terms 7i. 
 So in the formula (4), § 312, we put p (1 -\-p) for «, 1 + p for 
 r, and thus find, 
 
 2 = ^ (1 ^„) (i±p)!L-_i = p (i±P)!lL-zii±p). 
 
 ^^■•^1 + p — 1 ^ p 
 
 EXERCISES. 
 
 1. A man insures his life for $5000 at the age of 30, pays 
 for his insurance a premium of 80 dollars a year for 32 years, 
 and dies at the age of 62, immediately before the 33d payment 
 would have been due. If the company gains 4 per cent, inter- 
 est on all its money, how much does it gain or lose by the 
 insurance ? 
 
 Note. Computations of this class can be made with great facility by 
 the aid of a table of logarithms. 
 
 2. What is the present value of a dollars due n years hence, 
 interest being reckoned at c per cent. ? 
 
 Note. If p be the present value, Problem I gives the equation, 
 
 3. What is the present value of 3 annnal payments, of a 
 dollars each, to be made in one, two, and three years, interest 
 being reckoned at 5 per cent. ? 
 
 4. What is the present value of n annual payments, of a 
 dollars each, the first being due in one year, if the rate of in- 
 terest is c per cent. ? What would it be if the first payment 
 were due immediately ? 
 
SECOND PART. 
 ADVANCED COURSE 
 
BOOK VIII. 
 
 RELATIONS BETWEEN ALGEBRAIC 
 QUANTITIES. 
 
 Of Algebraic Functions. 
 
 217. Bef. When one quantity depends upon an- 
 other in such a way that a change in the value of the 
 one produces a change in the value of the other, the 
 latter is called a Function of the former. 
 
 This is a more general definition of the word " function " than that 
 given in § 49. 
 
 Examples. The time required to perform a journey is a 
 function of the distance because, other things being equal, it 
 varies with the distance. 
 
 The cost of a package of tea is a function of its weight, be- 
 cause the greater the weight the greater the cost. 
 
 An algebraic expression containing any symbol is a func- 
 tion of that symbol, because by giving different values to the 
 symbol we shall obtain different values for the expression. 
 
 Def. An Algebraic Function is one in which the 
 relations of the quantities is expressed by means of an 
 algebraic equation. 
 
 Example. If in a journey we call t the time, 8 the average 
 speed, and d the distance to be travelled, the relation between 
 these quantities may be expressed by the equation, 
 
 d = st. 
 
 Any one of these quantities is a function of the other two, 
 defined by means of this equation. 
 
 An algebraic function generally contains more than one 
 
222 FUNCTIONS. 
 
 letter, and therefore depends upon several quantities. But we 
 may consider it a function of any one of these quantities, se- 
 lected at pleasure, by supposing all the other quantities to 
 remain constant and only this one to yary. For example, the 
 time required for a train to run between two points is a func- 
 tion not only of their distance apart, but of the speed of the 
 train. The speed being supposed constant, the time will be 
 greater the greater the distance. The distance being constant, 
 the time will be greater the less the speed. 
 
 Def. The quantities iDetween which the relation ex- 
 pressed by a function exists are called Variables. 
 
 This term is used because such quantities may vary in value, as in 
 the preceding examples. 
 
 Def. An Independent Variable is one to which we 
 may assign values at pleasure. 
 
 The function is a dependent variable, the value of which is 
 determined by the value assigned to the independent variable. 
 
 Def. A Constant is a quantity which we suppose 
 not to vary. 
 
 Rem. This division of quantities into constant and varia- 
 ble is merely a supposed, not a real one ; we can, in an algebraic 
 expression, suppose any quantities we please to remain constant 
 and any we please to vary. The former are then, for the time 
 being, constants, and the latter variables. 
 
 Illusteatio:n-. If we put 
 
 d, the distance from New York to Chicago ; 
 5, the average speed of a train between the two cities; 
 t, the time required for the train to perform the jour- 
 ney, 
 
 then, if a manager computes the different values of the time t 
 corresponding to all values of the speed s, he regards d as a 
 constant, s as an independent variable, and ^ as a function of s. 
 If he computes how fast the train must run to perform the 
 journey in different given times, he regards t as the independ- 
 ent variable, and s as a function of t. 
 
FUNCTIONS. 223 
 
 When we have any equation between two variables, we 
 may regard either of them as an independent variable and the 
 other as a function. 
 
 Example. From the equation 
 ax -\- by = c, 
 
 we derive x = --{--, 
 
 a a 
 
 ax c 
 
 in one of which x is expressed as a function of 2/, and in the 
 other ?/ as a function of x. 
 
 218. Names are given t6 particular classes of functions, 
 among which the following are the most common. 
 
 1. Def. A Linear Function of several varialbles is 
 one in which each term contains one of the variables, 
 and one only, as a simple factor. 
 
 Example. The expression 
 
 Ax + By -{- Cz 
 
 is a linear function of x, y, and z, when A, B, and C are quan- 
 tities which do not contain these variables. 
 
 A linear function differs from a function of the first degree 
 (§ 52) in having no term not multiplied by one of the varia- 
 bles. For example, the expression 
 
 Ax -{- By -\r O 
 
 is a function of x and y of the first degree, but not a linear 
 function. 
 
 The fundamental property of a linear function is this: 
 
 // all the variables he multiplied hy a coimnon fac- 
 tor, the function will he multiplied hy the same factor. 
 
 Proof. Let Ax -\- By -f Cz be the linear function, and r 
 the factor. Multiplying each of the variables x, y, and z by 
 this factor, the function will become 
 
 Arx + Bry + Crz, 
 which is equal to r (Ax -\- By -\- Cz). 
 
224 FUNCTIONS. 
 
 Moreover, a linear function is the only one which pos- 
 sesses this property. 
 
 2. Def. A Homogeneous Function of several va- 
 riables is one in which each term is of the same degree 
 in the variables. (Compare § 52.) 
 
 Example. The expression ax^ + bx^y + ci/h + dz^ is homo- 
 geneous and of the third degree in the variables x, y, and z. 
 
 Eem. a Hnear function is a homogeneous function of the 
 first degree. 
 
 Fundamental Property of Homogeneous Functions. 
 // all the variables he multiplied by a coinmon factor, 
 any homogeneous function of the ix*^^ degree in those va- 
 riables ivill be multiplied by the «i*'* power of that factor. 
 
 Proof. If we take a homogeneous function and put rx for 
 X, ry for y, rz for z, etc., then, because each term contains x, 
 y, or z, etc., n times in all as a factor, it will contain r n times 
 after the substitution is made, and so will be multiplied by r**. 
 
 3. Def. A Rational Fraction is the quotient of two 
 entire functions of the same variable. 
 
 A rational fraction is of the form, 
 
 a -\- hx -\- cx^ -\- etc. 
 
 m + nx + px^ + etc. 
 
 Any rational function of a variable may be expressed as a 
 rational fraction. Compare § 180. 
 
 Equations of the First Degree between Two 
 Variables. 
 
 219. Since we may assign to an independent variable any 
 values we please, we may suppose it to increase or decrease by 
 regular steps. The difference between two values is then 
 called an increment. That is, 
 
 Def. An Increment is a quantity added to one 
 value of a variable to obtain another value. 
 
INCREMENTS. 225 
 
 Rem. If we diminish the variable, the increment is 
 negative. 
 
 Theorem. In a function of the first degree, eqnal in- 
 crements of the independent variable cause equal incre- 
 ments of the function. 
 
 Example. Let x be an independent variable, and call u 
 
 the function -a; + 11, so that we have 
 
 3 
 u = 2^ + 11- 
 
 If we give x the successive values —2, —1, 0, 1, 2, etc., 
 and find the corresponding values of the function u, they 
 will be 
 
 Values of x, — 2, — 1, 0, 1, 2, 3, 4, etc. 
 " " u, 8, 9i, 11, 12i, 14, loi, 17, etc. 
 
 We see that, the increments of x being all unity, those of 
 y are all IJ. 
 
 General Proof. Let au -{- ix := c be any equation of the 
 first degree between the variable x and the function m. Solving 
 this equation we shall have 
 
 c — bx c h 
 
 U = = X. 
 
 a a a 
 
 Let us assign to x the successive values, 
 r, r -\- h, r + 2/^, etc., 
 
 the increment being h in each case. The corresponding values 
 of the function u will be 
 
 h, c h 2/;. 
 
 etc., 
 
 of which each is less than the preceding by the same amount, 
 
 - h. Hence the increment of u is always h, which proves 
 
 the theorem. 
 
 330. Geometric Construction of a Relation of the First 
 Degree. The relation between a variable x and a function u 
 of this variable may be shown to the eye in the following way- 
 15 
 
 c h 
 a-a'^ 
 
 c h h^ 
 a a a 
 
 c b 2b, 
 
 r h, 
 
 a a a 
 
OEOMETBIG CONSTRUCTION. 
 
 •^ 
 
 V 
 
 i:::^ 
 
 X 
 
 -y -a -I o I 2 
 
 i ^':\4 
 
 Take a base line AX, mark a zero point upon it, and from 
 this zero point lay off any yalues of x we please. Then at each 
 point of the line corresponding to a value of x erect a vertical 
 line equal to the corresponding value of u. If u is positive, the 
 value is measured upward; if negative, downward. The line 
 drawn through the ends of these values of u will show, by the 
 distance of each of its points from the base line AX, the values 
 of u corresponding to all values of x. 
 
 Let us take, as an example, the equation 
 
 5u + 3a; = 10, 
 
 the solution of which gives 
 
 = 2 — -X. 
 5 
 
 Computing the values of w corresponding to values of x 
 from —3 to +6, we find : 
 
 ^= -3, -2, -1, 0, +1, +2, +3, +4, +55 +6. 
 u= +3|, +31 2|, 2, If, i i, -f, -1, -If 
 
 Laying off these values in the way just described, we have 
 the above figure. Wherever we choose to erect a value of w, 
 it will end in the dotted line. 
 
 We note that by the property of functions of the first de- 
 gree just proved, each, value of u is less (shorter) than the pre- 
 ceding one by the same amount ; in the present case by -• It 
 
 is known from geometry that in this case the dotted line 
 through the ends of u will be a straight line. 
 
 We call this line through the ends of y the equation line. 
 
OF EQUATIONS OF THE FIRST DEOBEE. 227 
 
 231. When we can once draw this straight line, we can 
 find the value of y corresponding to every value of x without 
 using the equation. We have only to take the point in the 
 base line corresponding to any value of a:, and by measuring 
 the distance to the line, we shall have the corresponding value 
 of u. 
 
 Now it is an axiom of geometry that one straight line, and 
 only one, can be drawn between any two points. Therefore, 
 to form any relation of the first degree we please between x 
 and u, we may take any two values of x, assign to them any 
 two values of ?^ we please, plot these two pair of values of u in 
 a diagram, draw the equation line through them, and then 
 measure off, by this line, as many more values of y as we 
 please. 
 
 Example. Let it be required that for x-= -\-l we shall 
 have w = +1, and for a; = +5, ?* = + 3. What will be the 
 yalues of y corresponding to a; = — 3, —2, —1, 0, etc. 
 
 Drawing the base line AX below, we lay off from 1 the ver- 
 tical line +1 in length, and from the point 5 the vertical line 
 + 2. Then drawing the dotted Hne through the ends, we 
 measure off different values of w, as follows: 
 
 X = -3, -2, -1, 0, +1, +2, +3, +4, +5, +6, etc. 
 u= -1, -J, 0, -hi, 1, +li, +2, +2i, +3, +^, etc. 
 
 EXERCISES. 
 
 1. Plot the equation 2?^ + Zx = 6. 
 
 2. Plot a line such that 
 
 for a; = — 6 we shall have ^ u = +4, 
 for a: = -f 6 " " w = _ 4, 
 
 and find the values of u for a: = 1, 2, 3, 4, and 5. 
 
228 OEOMETBIG G0N8TBUGTI0N 
 
 323. The algebraic problem corresponding to the con- 
 struction of § 220 is the following: 
 
 Having given two values of y corresponding to two 
 given values of X, it is required to construct an equation 
 of the first degree such that these two pairs of values 
 shall satisfy it. 
 
 Example of Solution. Let the requirement be tliat of the 
 equation plotted in the preceding example, namely, 
 
 for X z= 1 we must have w = 1, 
 for ^ = 5 " " u = d. 
 
 The problem then is to find such values of a, h, and c, that 
 in the equation 
 
 ax -\- hu ^= c, (1) 
 
 we shall have w = 1 for a: = 1, and w = 3 for x = b. Sub- 
 stituting these two pairs of values, we find that we must have 
 
 «x5 +^x3 = c; 
 
 or a -{-h = c, 
 
 ha + ^h = c. 
 
 Here a, J, and c are the unknown quantities whose values 
 are to be found, and as we have only two equations, we cannot 
 find them all. Let us therefore find a and b in terms of c. 
 
 Multiplying the first equation by 3, and subtracting the 
 product from the second, we have 
 
 2a ^ — 2c or a = — c. 
 
 Multiplying the first equation by 5, and subtracting the 
 second from the product, we have 
 
 2b = 4:C or b = 2c. 
 
 Substituting these values of a and b in. (1), we find the re- 
 quired equation to be 
 
 2cu — ex =z c. 
 
 We may divide all the terms of this equation by c (§ 120, 
 Ax. Ill), giving 
 
 2u — x = 1, 
 
OF EQUATIONS OF THE FIRST DEGREE. 229 
 
 thus showing that there is no need of using c. The solution 
 of this equation gives 
 
 1 4- a; 
 
 from which, for x=: —3, —2, —1, etc., we shall find the same 
 values of u which we found from the diagram. 
 
 EXERCISES. 
 
 Write equations between x and y which shall be satisfied 
 by the following pairs of values of x and y. 
 
 1. For a; = 2, y =zl\ and for a; = 5, y = — 1. 
 
 2. For 2:=— 2, 2/=— 1; and for a^ = +2, y = +1. 
 
 3. For a;=— 5, ?/=+2; and for x ■= -}-5, y = —2. 
 
 4. For X = 0, y = — 7 ; and for a; = 15, y = 0, 
 
 5. For X = 25, y = 2 ; and for a; = 30, ?/ = 3. 
 
 233. Geometric Solution of Tivo Equations with Two Un- 
 known Quantities. The solution of two equations with two 
 unknown quantities consists in finding that one pair of values 
 which will satisfy both equations. If we lay off on the base 
 line the required value of x, the two values of y corresponding 
 to this value of x in the two equations must be the same ; that 
 is, the two equation lines must cross each other at the 
 point thus found. Hence the following geometric solution : 
 
 I. Plot the two equations frorn the same base line and 
 zero point. 
 
 II. Continue the equation lines, if necessary, until 
 they intersect. 
 
 III. The distance of the point of intersection from the 
 base line is the value of y which satisfies both equations. 
 
 IV. The distance of the foot of the y line from the 
 zero point is the required value of x. 
 
 EXERCISES. 
 
 Solve the following equations by geometric construction : 
 
 1. X — 2u = 3, 2a; 4- ^ = 5. 
 
 2. 2w + 7a; = 4, 3w + a; = 1. 
 
230 NOTATION OF FUNCTIONS. 
 
 2.24. Geometric Explanation of Equivalent and Inconsist- 
 ent Equations. If we have two equivalent equations (§ 200), 
 each value of x will give the same value of the other quantity 
 u or y. Hence the two lines representing the equation will 
 coincide and no definite point of intersection can be fixed. • 
 If the two equations 
 
 au -{- hx = c, 
 
 a'u -\- Vx = c', 
 
 are inconsistent we shall have (§ 142), 
 
 ^ - ^'. 
 a ~ a' 
 
 If b be any increment of x, the increments of u in the two 
 
 equations (§219) will be and >• Therefore these 
 
 ^ ^"^ ^ a a 
 
 increments will be equal, and the two equation lines will be 
 parallel. Hence, 
 
 To inconsistent equations correspond parallel lines, 
 which have no point of intersection. 
 
 If the two equations are equivalent (§ 141, 143), their lines 
 will coincide. 
 
 ^N^otation of Functions. 
 
 225. In Algebra we use symbols to express any numbers 
 whatever. In the higher Algebra, this system is extended 
 thus : 
 
 We may use a,mj syinhol, JiavUtg a letter attached to 
 it, to express a function of the quantity represented hy 
 that* letter. 
 
 Example. If we have an algebraic expression containing 
 a quantity x, which we consider as a function of x, but do not 
 wish to write in full, we may call it 
 
 F(x), or (l){x), or [x], or Ax, 
 or, in fine, any expression we please which shall contain the 
 symbol x, and shall not be mistaken for any other expression. 
 
 In the first two of the above expressions, the letter x is enclosed in 
 parentheses, in order that the expression may not be mistaken for x mul- 
 tiplied by F, or <p. The parentheses may be omitted when the reader 
 knows that multiplication is not meant. 
 
NOTATION OF FUNCTIONS, 231 
 
 The fundamental principle of the functional notation is 
 this : • 
 
 When a symbol with a letter attached represents a 
 function, then, if we substitute any other quantity for 
 the letter attached, the combination will represent the 
 function found by substituting that other quantity. 
 
 Example. Let us consider the expression ax^ + & as a 
 function of x, and let us call it (t>{x), so that 
 
 (f) {x) = ax^ + b. 
 
 Then, to form (/> {y), we write y in place of x, obtaining 
 
 To form (a: + y), we write x + y m place of x, obtaining 
 
 <1>{^-Ty) = ct(x-\-yY-\- h. 
 To form </> {a), we write a instead of x, obtaining 
 
 (/) {a) =z cfi + b. 
 To form </> {ay^). we put mf in place of x, obtaining 
 {(ly^) = a {ay^f + b = aY + b. 
 The equation (j) (z) =z will mean 
 az^ + b = 0. 
 
 EXERCISES. 
 
 Suppose <p {x) =z ax^ — a^x, and thence form the values of 
 I. (t>(y). 2. c{>{z). 3. 0(%). 
 
 4. 0(^ + 2/). 5- <P{^' + ct). 6. (p(x-~a). 
 
 7. 0(0; + «?/). 8. 4>{x-ay). 9. 0(:?;2). 
 
 Suppose i^ (;r) = xa^, and thence form the values of 
 
 10. F{y). II. F[^). 12. i^(3?/). 
 
 13. F{x + y). 14. F{x-y). 15. i^(l). 
 
 Suppose / (:c) = a;2, and thence form the values of 
 
 16. /(I). 17. /(.7;2). 18. f(xP). 
 
 19. /(:^). 20. f(x% 21. /(a;^). 
 
232 FUNCTIONS OF SEVERAL VARIABLES. 
 
 22. Prove that if we put {x) = a^, we shall have 
 
 <f>{x + y)=<t> {x) X (^), </) (^^) = [0 (^)]y = [0 {iMx. 
 
 Let ns put (m) = m (m — 1) (m — 2) {m — 3) ; thence 
 form the values of 
 
 23. 0(6). 24. 0(5). 25. 0(4). 
 26. 0(3). 27. 0(2). 28. 0(1). 
 29. 0(0). 30. 0(-l). 31- 0(-2). 
 
 Functions of Several Variables. 
 
 236. An algebraic expression containing several 
 quantities may be represented by any symbol having 
 the letters which represent the quantities attached. 
 
 Examples. We may put 
 
 {x, y) = ax— ly, 
 
 the comma heing inserted between x and y, so that their 
 
 product shall not be understood. We shall then have, 
 
 (m, n) = am — bn. 
 
 (y, x) =ay — bx, 
 
 the letters being simply interchanged. 
 
 (f>{x + y, x — y) =z a{x-\-y)—h{x — y) 
 — {a—V)x-\-{a^V) y. 
 (a, b) = a^ — b\ 
 (5, a) = ab — ba = 0. 
 ^{a -\- b, ab) = a {a -\- b) — ab^. 
 {a, a) ■=. cfi — ba. 
 etc. etc. 
 
 If we put {a, b, c) = 2fl5 + 3Z> — 5c, we shall have 
 (a-, z, y) = 2x-j- dz — by. 
 {z, y, x) = 2z + dy — 5x. 
 (m, m, m) = 2m + dm 4- 5m = lOwz. 
 0(3, 8, 6) = 2.3 + 3.8-5.6 = 0. 
 
 EXERCI SES.. 
 
 Let us put {x, y) = Sx — Ay, 
 
 \3 y (^j y-> ^) =^ cix -\- by — abz. 
 
USE OF INDICES. 233 
 
 Thence form the expressions : 
 / I. (i)(ij,x), . 2. (i>(a,l). 3. 0(3,4). 
 
 4. 0(4,3).. 5- 0(10,1). 6. f{a,i). 
 
 Y- 7. /(*,«). --v.' 8- /(2/^^)- 9. /(7, -3). 
 
 X.- 10. f(q,—p). II. f{z,x,y). 12. f(b,a,2). 
 
 I . 13. /(«:, ^, ^). 14. /(«^ ^^^. c^). 
 
 ^ 15. f(—a, —h, —ah). 
 
 . m(m — l) (7n — 2) 
 
 Let us put Cm, 71) = — 7 tvt ^r^' 
 
 ^ ^ ^ /i (^i — 1) (^^ — 2) 
 
 Find the values of 
 
 16. (3, 3). 17. (4, 3). 18. (5, 3). 
 
 19. (6, 3). 20. (7, 3). 21. (8, 3). 
 
 22. (2, -1). 23. (3, -2). 24. (4, -2). 
 
 Use of Indices. 
 
 336a. Any number of different quantities may be 
 represented by a common symbol, the distinction being 
 made by attaching numbers or accents to the symbol. 
 
 EXAMPLES. 
 
 1. Any 71 different quantities may be represented by the 
 symbols, p^, p^, p^, pn- 
 
 2. A producer desires to have an algebraic symbol for the 
 amount of money which he earns on each day of the year. If 
 he calls q what he earns in a day he may put : 
 
 q-y for the amount earned on January 1, 
 
 etc. " " " " etc., 
 
 ^3 3 '' " " February 1; 
 
 and so on to the end of the year, when 
 
 (Zses w^^l ^® ^^^ amount for December 31. 
 
 Def. The distinguisliing numbers 1, 2, 3, etc., are 
 here called Indices. 
 
 A symbol with an index attached may represent a 
 function of tbe index, as in the functional notation. 
 
ib 
 
 234 - V8E OF INDICES. 
 
 EXERCISES. 
 
 Let us put at = ^ (^ + 1). Then find the value of 
 
 1. «o +«1 + ^3 + • • • • + ^10- 
 
 2. Prove the following equations by computing both mem- 
 bers : 
 
 4 
 
 5 
 6 
 
 If we put /Si =r 1 + 3 + 3 . . . . + ^, we shall have 
 
 /S'3=:l + 2 + 3:=6, etc., etc. 
 Using the preceding notation, find the values of the ex- 
 pressions: ,- s ^ 
 
 227. Sometibies the relations between quantities distin- 
 guished by indices are represented by equations of the first 
 degree. The following are examples : ^ . 
 
 Let us have a series of quantities, 
 
 ^0' ^1' ^3> ^3> -^4> ®^^'J 
 
 connected by the general relation, 
 
 - Ji^i = Ai + Ai_i. :.! (a) 
 
 It is required to express them in terms of ^^ and ^j. 
 
 We put, in succession, i = 1, i = 2, i = 3, etc. Then, 
 when i = 1, we have from (a), 
 
 A, 
 
 ■=A^ +^0- 
 
 
 ^3 
 
 = A^ +-4i 
 
 = 2A, + A, 
 
 ^4 
 
 = ^3 + A^ 
 
 = 3A, -\-2A, 
 
 When i = 2, 
 
 * = 3, 
 
 * = 4, A, = A^+ A^ rrz 5:4, + 3^0- 
 i = 5, Jg =: ^5 + ^4 = 8^1 + 5ylo, 
 
 and so on indefinitely. 
 
MISCELLANEOUS FUNCTIONS. 235 
 
 EXERCISES. 
 
 1. If ^i+1 = Ai — Ai^t, 
 
 what will be the values of A^ - . - . Aio, and in what way may 
 all subsequent values be determined ? 
 
 2. If Ai+i =z 2Ai — Aq, 
 find A 2 to A^ in terms oi Aq and A^. 
 
 3. If Ai^i = iAi 4- Ai_t, find A^ to ^5. 
 
 4. If Ai = Ai_i + A, 
 
 find the sum Aq + A^ -\- A2 -\- . . . . + An, in terms of A q, 
 h and n. (Oomp. § 209, Prob. V.) 
 
 5. If ^ihi = rAi, 
 
 find ^j +^^3 + ^3 + ' ' - • + ^nj ii^ terms of J^ and r. 
 
 6. If Ai+i = ikAi + ^j 1, 
 find A2, A^, . . . . Aq, in terms of ^^ and A^. 
 
 Miscellaneous Functions of Numbers. 
 
 328. We present, as interesting exercises, certain elemen- 
 tary forms of algebraic notation much used in Mathematics, 
 and which will be employed in the present work. 
 
 1. When we have a series of symbols the num'ber 
 of which is either indeterminate or too great to be all 
 written ont, we may write only the first tw^o or three 
 and the last, the omitted ones being represented by a 
 row of dots. 
 
 Examples. a, b, c, . . . . t, 
 
 Xy /if Of • m • » liOf 
 
 1, 2, Uf 
 
 n being in the last case any number greater than 2. 
 The number of omitted symbols is entirely arbitrary. 
 
 EXERCISES. 
 
 How many omitted expressions are represented by the dots 
 in the following series : 
 
 I. 1, 2, 3, n. 2. 1, 2, 3, n — 2. 
 
236 MISCELLANEOUS FUNCTIONS 
 
 3. 1, 2, 3, n + 'Z. 
 
 4. n, n — If 71 — 2, .... n — s. 
 
 5. tif n — 1, n — 2y . . . . n — s — 1. 
 
 6. n, n — 1, n — 2, n — s + 1. 
 
 What will be the last term in the series : 
 
 7. 2, 3, 4, etc., to 71 terms. 
 
 8. 71, 71 — 1, 71 — 2, etc., to s terms. 
 
 9. 2, 4, 6, etc., to h terms. 
 
 2. Product of the First n Numbers. The symbol 
 
 n\ 
 is used to express the product of tlie first n numlbers, 
 1-2.3 n. 
 
 Thus, 1! = 1. 
 
 2! = 1-2 = 2." 
 
 3! = 1.2.3 = 6. 
 
 4! = 1.2.3.4 = 24. 
 
 etc. etc. 
 
 It will he seen that 2 ! = 2 • 1 ! 
 
 31 = 3.2! 
 And, in general, n\ = 71 {71 — 1) ! 
 whatever number 7i may represent. 
 
 EXERCISES. 
 
 Compute the values of 
 
 I. 5! 2. 6! 3. 8! 
 
 _7j_. .., . 8! 
 
 ^* 3! 4! " ^* 3! 5! 
 
 6. Prove the equation 2.4-6.8 .... 2/i =: 2^*72! 
 
 7. Prove that, when 7i is even, 
 
 -t — ^ (^ — 2) (^ — 4) . . . . 4.2 
 2 • ~ 21 
 
 3. Binomial Coefficients. The binomial coefficient 
 
 n{n — l){n — 2) . . . .to s term s 
 
 12.3 s 
 
 is expressed in the abbreviated form, 
 
MISCELLANEOUS FUNCTIONS 237 
 
 © 
 
 the parentheses heing used to show that what is meant 
 
 is not the fraction -• 
 
 s 
 
 EXAMPLES. 
 
 \5/ ~ 1.2. 
 
 ""'■^ 21. 
 
 3.4.5 
 
 (^') 
 
 (!) 
 © = 
 
 fn\ _ n{7i — l) 2.1 _ 
 
 W ~" 1.2.3 n ~ 
 
 (n + 4) (n + 3) {n + 2) 
 
 _ n(n — l){n— 2) 
 1.2-3 
 
 1.2.3 
 
 EXERCISES. 
 
 Compute the vahies of the expressions : 
 
 ■■ (MM^©*©^(MMS- 
 ■■ ©-(^©^©+©- 
 
 Prove the formulae : 
 
 l^\ — ^- /^\ _ y^! 
 
 ^' W ~ 2! 3! ^' \sJ ~ s\ {n-s)\ 
 
BOOK IX. 
 THE THEORY OE NUMBERS. 
 
 CHAPTER I. 
 
 THE DIVISIBILITY OF NUMBERS. 
 
 329. Def, The Theory of Numbers is a Ibranch 
 of mathematics which treats of the properties of integers. 
 
 Def. An Integer is anj whole numher, positive or 
 negative. 
 
 In the theory of numbers the word numler is used to ex- 
 press an integer. 
 
 Bef. A Prime Number is one which has no divi- 
 sor except itself and unity. 
 
 The series of prime numbers are 
 
 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, etc. 
 
 Def. A Composite Number is one which may be 
 expressed as a product of two or more factors, all 
 greater than unity. 
 
 Eem. Every number greater than 1 must be either prime 
 or composite. 
 
 Def. Two numbers are prime to each other when 
 they have no common divisor greater than unity. 
 
 Example. The numbers 24 and 35 are prime to each 
 other, though neither of them is a prime number. 
 
 Rem. a vulgar fraction is reduced to its lowest terms when 
 numerator and denominator are prime to each other. 
 
DIVISIBILITY OF NUMBERS. 239 
 
 Division into Prime Factors. 
 
 230. Every composite number may by definition be di- 
 yided into two or more factors. If any of these factors are 
 composite, they may be again divided into other factors. 
 When none of the factors can be further divided, they will all 
 be prime. Hence, 
 
 Theorem. Every composite number may he divided 
 into p?%me factors. 
 
 Example. 180 =: 9-20, 
 
 9 = 3-3, 
 20 = 4.5 = 2.2.5. 
 Whence, 180 = 2-2.3.3.5 = 22.32.5. 
 
 Cor. 1. Because every number not prime is composite, 
 and because every composite number may be divided into 
 prime factors, we conclude: Every niunher is either prime 
 or divisible by a prime. 
 
 Cor. 2. Every number, prime or composite, may be ex- 
 pressed in the form 
 
 paq^j,y etc., (a) 
 
 where p, q, r, etc., are different prime numbers ; 
 
 ^} (^} y? etc., the exponents, are positive integers. 
 
 Eem. If the number is prime there will be but one factor, 
 namely, the number itself, and the exponent will be unity. 
 
 EXERCISES. 
 
 Divide the following numbers or products into their prime 
 
 factors, if any, and thus express the numbers in the form («) : 
 
 I. 24. 2. 72. 3. 260. 4- 169. 5. 225. 
 
 6. 256. 7. 91. 8. 143. 9. 360. 10. 217. 
 
 II. 3072. 12. 1.2. 3. 4. 5. 6.7-8. 9. 
 
 Eem. In seeking for the prime factors of a number, it is 
 
 never necessary to try divisors greater than its square root, for 
 
 if a number is divisible into two factors, one of these factors 
 
 will necessarily not exceed such root. 
 
240 DIVISIBILITY OF NUMBERS. 
 
 Common Divisors of Two Numbers. 
 
 231. Theorem I. // two niunhers have, a cownnon 
 factor, their suirv will have that same factor. 
 
 Proof. Let o^ be the common factor ; 
 
 on, the product of all the other factors in the 
 
 one number; 
 n, the corresponding product in the other 
 number. 
 Then the two numbers will be 
 
 am. and an. 
 Their sum will be a {m + n). 
 
 Because m and n are whole numbers, m-\-n will also be a 
 whole number. Therefore a will be a factor of a?n -f aoi. 
 
 Theorem II. // two numbers have a common factory 
 their difference will have the same factor. 
 
 Proof. Almost the same as in the last theorem. 
 
 Cor. If a number is divisible by a factor, all multiples will 
 be divisible by that factor. 
 
 Rem. The preceding theorems may be expressed as follows : 
 
 If tivo numbers are divisible by the same divisor, 
 their sum, difference, and multiples are all divisible by 
 that divisor. 
 
 Rem. If one number is not exactly divisible by another, a 
 remainder less than the divisor will be left over. If we put 
 
 D, the dividend; 
 d, the divisor; 
 q, the quotient; 
 r, the remainder; 
 
 we shall have, D = dq -\- r, 
 
 or D — dq ^= r. 
 
 Example. 7 goes into 66 9 times and 3 over. Hence 
 
 this means 
 
 66 = 7-9 + 3, or 66-7-9 = 3. 
 
DIVISIBILITY OF NUMBERS. 241 
 
 232. Problem. To find the greatest common divisor 
 of two numbers. 
 
 Let m and n be the numjbers, and let m be the greater. 
 
 1. Divide m by n. If the remainder is zero, n will be the 
 divisor required, because every number divides itself. If there 
 IS a remainder, let q be the quotient and r the remainder. 
 
 Then m — 7iq = r. 
 
 Let d be the common divisor required. 
 Because m and 7i are both divisible by d, m — nq must 
 also be divisible by d (Theorem II). Therefore, 
 
 r is divisible by d. 
 
 Hence every common divisor of m and 7i is also a common 
 divisor of n and r. Conversely, because 
 
 m = nq -\- r, 
 
 every common divisor of n and r is also a divisor of m. There- 
 fore, the greatest common divisor of m and n is the same as 
 the greatest common divisor of n and r, and we proceed with 
 these last two numbers as we did with m and n. 
 
 2. Let r go into n q' times with the remainder r'. 
 Then n = rq' -\- r', 
 
 or n — rq' =^ r'. 
 
 Then it can be shown as before that <? is a divisor of r', and 
 therefore the greatest common divisor of r and r'. 
 
 3. Dividing r by r\ and continuing the process, one of two 
 results must follow. Either, 
 
 a. We at length reach a remainder 1, in which case the 
 two numbers are prime ; or, 
 
 d. We have a remainder which exactly divides the pre- 
 ceJtng divisor, in which case this remainder is the divisor 
 required. 
 
 To clearly exhibit the process, we express the numbers w, 
 n, and the successive remainders in the following form : 
 16 
 
242 GREATEST COMMON DIVISOR, 
 
 m = n-q -\- r, (r < w) ; 
 
 n = r-q' + ^', (^' < ^); 
 
 r = r'-q" + r", (r" < r') ; 
 
 r' = r"'q"' -\-r"', {r'" < r") ; 
 
 etc. etc. etc., 
 until we reach a remainder equal to 1 or 0, when the series 
 terminates. 
 
 EXERCISES. 
 
 I. Find the G. CD.* of 240 and 155. 
 
 Dividend. 
 
 Div. 
 
 Quo. Rem. 
 
 240 =z 
 
 155 
 
 1 + 85. 
 
 155 = 
 
 85 
 
 1 + 70. 
 
 85 = 
 
 70 
 
 1 + 15. 
 
 70 = 
 
 15 
 
 4 + 10. 
 
 15 = 
 
 10 
 
 1+ 5. 
 
 10 = 
 
 5 
 
 2. 
 
 Therefore 5 is the greatest common divisor. 
 
 Note. Let the student arrange all the following exercises in the 
 above form, first dividing in the usual way, if he finds it necessary. 
 
 Find the greatest common divisor of 
 
 2. 399 and 427. 3. 91 and 131. 
 
 4. 8 and 13. 5. 1000 and 212. 
 
 6. 799 and 1232. 7. 800 and 1729. 
 
 8. 250 and 625. 9. 1000 and 370. 
 
 10. If jT? be a number less than n and prime to n, show that 
 71 — ^ is also prime to n. 
 
 11. li p be any number less than n, the greatest common 
 divisor between n and p is the same as that between n and 
 n —p. 
 
 12. If n is any odd number, — - — and — ^ — are both 
 prime to it. 
 
 Corollaries. 1. When two numbers are divided by their 
 greatest common divisor, their quotients will be prime to each 
 other. 
 
 * The letters Q. C. D. are an abbreviation for Greatest Common Divisor. 
 
GEARING OF WHEELS. 
 
 243 
 
 2. Conversely, if two numbers, n and n', prime to each 
 other, are each multiplied by any number d, then d will be the 
 G. C. D. of dn and dn', 
 
 233. Gearing of Wheels. An interesting problem con- 
 nected with the greatest com- 
 mon divisor is afforded by a 
 common pair of gear wheels. 
 Let there be two wheels, the 
 one having m teeth and the 
 other n teeth, gearing into each 
 other. If we start. the wheels 
 with a certain tooth of the one 
 against a certain tooth of the 
 other, then we have the questions : 
 
 (1.) How many revolutions must each wheel make before 
 the same teeth will again come together ? 
 
 (2.) "With how many teeth of the one will each tooth of the 
 other have geared ? 
 
 Let q be the required number of turns of the first wheel, 
 having 7n teeth. 
 
 Let p be the required number of turns of the second, hav- 
 ing n teeth. 
 
 Then, because the first wheel has m teeth, qni teeth will 
 have geared into the other wheel during the q turns. In the 
 same way, pn teeth of the second wheel will have geared into 
 the first. But these numbers must be equal. Therefore, 
 when the two teeth again meet, 
 
 pn = qm. 
 
 Conversely, for every pair of numbers of revolutions p and 
 q, which fulfil the conditions, 
 
 2)n = qm, 
 
 the same teeth will come together, because each wheel will 
 have made an entire number of revolutions. This equation 
 gives 
 
 p _ m 
 
 q ~ 71 
 
244 OEARINQ OF WHEELS. 
 
 Til 
 
 Hence, if we reduce the fraction — to its lowest terms, we 
 
 shall have the smallest number of revolutions of the respective 
 wheels which will bring the teeth together again. 
 
 To answer the second question : 
 
 After the first wheel has made q revolutions, qm of its teeth 
 have passed a fixed point. Any one tooth of the other wheel 
 gears into every n^^ passing tooth of the first wheel. Therefore 
 
 any such tooth has geared into — teeth of the first wheel, 
 
 that is, into ;j teeth, because, from the last equation, 
 
 qm 
 
 ^— —2). 
 
 If d be the G. 0. D. of m and n, then 
 
 m = dp, 
 n = dq ; 
 m 
 
 n 
 
 Therefore each tooth of the one wheel has geared into only 
 every d^^ tooth of the other. 
 
 In the figure on the preceding page, n = 21 and n =: G. 
 Hence, d = 3, and each tooth of the one will gear into every 
 third tooth of the other. The numbers on the large wheel 
 show the order in which the gearing occurs. 
 
 How long soever the wheels run, the same contacts will 
 be repeated in regular order. Hence, if each tooth of the 
 one luheel must gear with every tooth of the other, the 
 nuTubers m and n must be prime to each other. 
 
 EXERCISES. 
 
 1. If one wheel has 40 teeth and the other 10, show how 
 they will run together. 
 
 Show the same thing for the following cases: 
 
 2. m — 12, n =z 15. 3. m =z 24:, n = 18. 
 4. m = 36, n = 25. 5. m = 24, n = 7. 
 
NUMBERS AND TUEIR DIGITS. 245 
 
 Relations of Numbers to their Digits. 
 
 234. In our ordinary method of expressing numbers, tlie 
 second digit toward the right expresses lO's, the third lOO's, 
 etc. That is, each digit expresses a power of 10 correspond- 
 ing to its position. 
 
 Def. The numlber 10 is the Base of our scale of 
 numeration. 
 
 Note. The base 10 is entirely arbitrary, and is supposed 
 to have originated from the number of the thumbs and fingers, 
 these being used by primitive people in counting. 
 
 Any other number might equally well have been chosen as 
 a base, but in any case we should need a number of separate 
 characters (digits) equal to the base, and no more. 
 
 Had 8 been the base, we should have needed only the 
 digits 0, 1, 2, etc., to 7, and different combinations of the 
 digits would have represented numbers as follows : 
 1 = 1, 
 7 = 7, 
 10 = 1-8 + = eight. 
 17 = 1-8 + 7 = fifteen. 
 20 = 2-8 + = sixteen. 
 56 = 5-8+ 6 = forty-six. 
 234 = 2-82 + 3-8 + 4 = one hundred fifty-six, 
 etc. 
 Let us take the arbitrary number z as the base of the scale. 
 As in our scale of lO's we have 
 
 234 = 2-102 + 3.10 + 4, 
 
 so in the scale of 2;'s the digits 234 would mean 
 
 2^2 + 3;^ + 4. 
 
 In general, the combination of digits abed would mean 
 
 az^ + M + cz ■\- d. 
 
 Divisibility of Numbers and their Digits. 
 
 235. Theorem. If the sum of the digits of any num- 
 her Jje sitbtracted from the number itself, the remainder 
 will he divisible by % — 1. 
 
246 DIVISIBILITY OF NUMBERS. 
 
 Proof. Let the digits be a, l, c, d. The number expressed 
 will be az^ + dz^ -\- cz -{- d 
 
 Sum of digits = a +1 + c + d 
 
 Subtracting, rem. = a {z^—l) + h (^^— 1) + c {z—1). 
 
 The factors z^ — 1, z^ — 1, and z — 1 are all divisible by 
 2; _ 1 (§ 93). Hence the theorem is proved. (§ 231.) 
 
 Theorem. In any scale having z as Us base, the suin 
 of the digits of any nujriber, ivhen divided by z — 1, ivill 
 leave the same remainder as ivill the number itself when 
 so divided. 
 
 If we put : n, the number ; s, the sum of the digits ; 
 r, r\ the remainders from dividing by z—\\ 
 q, q', the quotients ; we shall have, 
 Number, n =z q {z — 1) -\- r 
 
 Sum of digits, .s ^ q' (z — 1) + r'- 
 
 Eemainder, {q — (f){z — l) ■\- r — r\ 
 
 Because ^ — s and (q — (f) {z — 1) are both divisible by 
 z — 1, their difference r — r' must be so divisible. Since r 
 and r' are both less than z-l, this remainder can be divided 
 hj z — 1 only when r = r', which proves the theorem. 
 
 Zero is considered divisible by all numbers, because a re- 
 mainder is always left. 
 
 'If a be any factor oi z — 1, the same reasoning will apply 
 to it, and therefore the theorem will be true of it. 
 
 In our system of notation, where z = 10, the above theo- 
 rems may be put in the following well-known form : 
 . // the sum of the digits of any number be divisible 
 by 3 or 9, the number itself will be so divisible. 
 
 These are the only numbers of which the theorem is true, 
 because 3 is the only divisor of 9. 
 
 Theorem. If from any number we subtract the digits 
 of the even powers of z, and add those of the alternate 
 powers, the result will be divisible by z -\- 1. 
 
 Proof, To az^ + bz^ -\- cz -^ d 
 
 Add a — b -\- c — d 
 Result, a(z^ + l) + b(z^—l) + c{z + l). 
 
NUMBERS AND THEIR DIGITS. 247 
 
 The factors of a, h, and c are all divisible by 2 + 1 (§§ 93, 
 94), whence the result itself is so divisible. 
 
 Applying this result to the case of ^ = 10, we conclude": 
 
 If on siibtractijig the sum of the digits in the place 
 of units, hundreds^ tens of thousands, etc., from the sum 
 of the alternate ones, the remainder is divisible hy 11, 
 the number itself is divisible by 11. 
 
 If m be any factor of z, it will divide all the terms of ihe 
 number 
 
 az^ + hz^ -\- cz -\- d, 
 except the last. Hence, if it divide this last also, it will di- 
 vide the number itself. Applying this result to the case of 
 z = 10, we conclude : 
 
 // the last digit of any number is divisible by a fac- 
 tor of 10, the number itself is divisible by that factor. 
 
 The factors of 10 being 2 and 5, this rule is true of these 
 numbers only. 
 
 It will be remarked that if the base of the system had been 
 an odd number, we could not have distinguished even and odd 
 numbers by their last figure, as we habitually do. 
 
 For example, if the base had been 9, the figures 72 would 
 have represented what we call sixty-five, which is odd, and 73 
 would have represented what we call sixty-six, which is even. 
 
 The use of the base 10 makes it easy to detect when a num- 
 ber is divisible by either of the first three prime numbers, 2, 3, 
 and 5. If the last figure is divisible by 2 or 5, the whole num- 
 ber is so divisible. To ascertain whether 3 is a factor, we find 
 whether the sum of the digits is divisible by 3. 
 
 In taking the sura, it is not necessary to include all the digits, but in 
 adding we may omit all 3's and 9's, and drop 3, 6, or 9 from the sum as 
 often as convenient. Thus, if the number were 
 
 921642712, 
 we should perform the operation mentally, thus : 
 
 Drop 9 ; 2 + 1 = 3, which drop ; 6, drop ; 4 + 2 = 6, which drop ; 
 7 + 1 = 8 + 2 = 10, which leaves a remainder 1. 
 
 ' EXERCISES. 
 
 T. Prove that if an even number leaves a remainder 1 when 
 divided by 3, its half will leave a remainder 2 when so divided. 
 
248 DIVISIBILITY OF NUMBERS. 
 
 2. If from any number we subtract the sum of units' digit 
 plus the product of the tens' digit by i, plus the product of 
 the hundreds' digit by t^, etc., the remainder will be divisible 
 by 10 — i. (i may be any integer, positive or negative.) 
 
 Note. When i — 1, this gives the rule of 9's and when i = —1, the 
 rule of ll's. 
 
 Prime Factors of Numbers. 
 
 236. First Fukdamental Theorem. A product can- 
 not Tje divided by a prime number unless one of the fac- 
 tors is divisible by that prime number. 
 
 Note. This theorem is not true of composite divisors. For exam- 
 ple, neither 8 nor 9 is divisible by 6, but the product 8 • 9 = 72 is so 
 divisible. But if we take as many numbers as we please not divisible by 
 7, we shall always find their product to leave a remainder when we try 
 to divide it by 7. 
 
 To make the demonstration better understood, we shall first take a 
 special case: 
 
 T/ie j^roduct 6Ga is not divisible by 7, unless a is divisible 
 by 7. 
 
 Proof, Suppose . 66« div. by 7 
 
 7 goes into QQ 9 times and 3 over, because 7-9 = 63, 63^ div. by 7 
 
 Therefore, by Theorem II, § 231, 3« div. by7 
 
 2 
 
 3 goes into 7 2 times and 1 over. Multiply by 2, ija div. by 7 
 
 Subtracting, ^a div. by 7 
 
 We have left, a div. by 7 
 
 Hence, if 66a is divisible by 7, then a is divisible by 7. 
 
 Gauss's Demonstration. If it be possible, let am be the 
 smallest multiple of w which is divisible by p, when neither a 
 nor 7n is so divisible. If a is greater than p, then let ;j go 
 into a b times and r over, so that 
 
 a = Ip -\- r, 
 or a — bp ^= r. 
 
 Then, am div. by p. 
 
 Subtract bpm " " 
 
 Remainder, {a — bp) ni " " 
 Or rm " " 
 
PRIME FACTORS OF NUMBERS. 249 
 
 That is, if «m is divisible by 2^, so is 7in, where r is less 
 than p. 
 
 Therefore the smallest multiple of m whicli fulfils the- con- 
 ditions must be less than pm. 
 
 Therefore, let a <C P- Let a go into p c times and s over, 
 so that 
 
 p = ca -\- Sy 
 or p ~ ca = s. 
 
 Then pm div. by p. 
 
 cam " " (by hypothesis). 
 Subtracting, {p — ca) m " " 
 Or, sm " 
 
 Therefore, 5 being less than a, a is not the smallest multiple; 
 whence the hypothesis that a is the smallest is impossible. 
 
 General Demonstration. Suppose 
 
 p, a prime number ; 
 a, number not divisible by;;; 
 am, a product divisible by p. 
 
 We have to prove that m must be divisible by p. 
 Let;? go into a q times. Because a is not divisible by^, 
 a remainder r will be left. That is, 
 
 a ^ pq + r, or a — pq =:■ r. 
 
 Let r go into p q' times and leave am div. by p. 
 
 a remainder r'. Then, 
 
 p = q'd-{- r\ 
 
 and because pm and (^rm are both di- 
 visible by^, rm is so divisible. 
 
 In the same way, if / goes into p 
 q" times, and leave the remainder r", 
 r"fn will be divisible hyp. Since each 
 of the remainders r, r\ r", etc., must 
 be less than the preceding, we shall at 
 length reach a remainder 1, which will give 
 
 m divisible by p. Q. E. D. 
 
 pqm 
 
 a a 
 
 rm 
 
 a a 
 
 q'rm 
 
 ii i( 
 
 pm 
 
 a (c 
 
 r'm 
 
 a (( 
 
 q"r'm 
 
 i( i( 
 
 pm 
 
 ii a 
 
 r"m 
 
 a « 
 
250 DIVISIBILITY OF NUMBERS. 
 
 Extension to Several Factors, If ??? is a product I x ?^, and 
 h is not divisible by />, then we may show in the same way that 
 n must be so divisible. If w = cs, and c is not divisible, then 
 s must be divisible, and so on to any number of factors. 
 
 Hence, 
 
 Theorem. If a product of any numher of factors is 
 divisible by a prime number, then one of the factors 
 must be divisible by the same prime. 
 
 This theorem is the logical equivalent of the one just 
 enunciated as the first fundamental theorem. 
 
 Note. The student will remark why the preceding demonstration 
 applies only when the divisor ^ is a prime number. If it were composite, 
 we might reach a remainder which would exactly divide it, and then the 
 conclusion would not follow. 
 
 337. Secoi^d Fuiq'DAMEN'TAL Theoeem. J. number 
 can be divided into prime factors in only one way. 
 
 For, suppose we could express the number N in the two 
 ways (§ 204, Cor. 2), 
 
 JSf z= jj" (/ ry, 
 
 where p, q, r, etc., a, h, c, etc., are all prime numbers. Then 
 
 If common prime factors appeared on both sides of this 
 equation, we could divide them out, leaving an equation in 
 which the prime factors p, q, r, etc., are all diiferent from 
 «, b, c, etc. 
 
 Then, because a, h, c, etc., are all prime, none of them are 
 divisible by^. Therefore, by the first fundamental theorem, 
 their products cannot be so divisible. But the left-hand mem- 
 ber of the equation is divisible by p, because p is one of its 
 factors. Therefore the equation is impossible. 
 
 Eem. This theorem forms the basis of the theory of the 
 divisibility of numbers. 
 
 Ti)e preceding theorems enable us to place the definition 
 of numbers prime to each other in a new shape. 
 
BIN031IAL COEFFICIENTS. 251 
 
 Two numbers are said to Ibe prime to each other 
 when they have no common prime factors. 
 
 Example. If one number is p"-q^ry, and the other is 
 af^i^C" (p, q, r, etc., and a, d, c, etc., being prime numbers), 
 then, if ^, q, r, etc., are all different from a, b, c, etc., the two 
 numbers will be prime to each other. 
 
 Elementary Theorems. 
 
 238. The following' general theorems follow from the two 
 preceding fundamental theorems, and their demonstration is 
 in part left as an exercise for the student. 
 
 I. J^o power of an irreducible vulgar fraction can be 
 a whole number. 
 
 Note. An irreducible vulgar fraction is one which is re- 
 duced to its lowest terms. 
 
 II. Corollary. JVb root of a ivhole number can be a 
 vulgar fraction. 
 
 III. If a numjber is divisible by several divisors, all 
 prime to each other, it is also divisible by their product. 
 
 Cor. To prove that a number N is divisible by a number 
 B =^p'^q^ ry, it is sufficient to prove that it is divisible sepa- 
 rately by /?*, by q^, by ry, etc. 
 
 Example. If a number is divisible separately by 5, 8, and 
 9, it is divisible by 5- 8- 9 = 360. Hence, to prove that a num- 
 ber is divisible by 360, it is sufficient to show that 5, 8, and 9 
 are all factors of it. 
 
 IV. If the numerator and denominator of a vulgar 
 fraction have no common prime factors, it is reduced to 
 its lowest terms. 
 
 Binomial Coefficients. 
 
 239. Theorem. The product of any n consecutive^ 
 numbers is divisible by the product of the numbers 
 1-2.3 . . . . 71, or n ! 
 
252 
 
 BINOMIAL COEFFICIENTS. 
 
 Kem. The theorem implies that all binomial coefficients 
 are whole numbers, because they are quotients formed by di- 
 viding the product of n consecutive numbers by n\ 
 
 Proof. 1. We have first to find the prime factors of the 
 product 
 
 1.2.3.4.5-6 n=:n\ 
 
 beginning with the factor 2. 
 
 I. The numbers divisible by 2 are the even numbers 2, 4, 
 6, etc., to n or n — 1, the number of which is 
 
 Note. The expression 
 
 here means the greatest whole 
 n — \ 
 
 number in -, which is ^ itself when n is even, and 
 when n is odd. 
 
 The quotients of the division are 
 1, 2, 3, 4, ... . 
 
 m 
 
 Of these quotients, j are divisible by 2, leaving the 
 second set of quotients, 
 
 1, 2, 3, 
 
 The next set of quotients will be 
 1, 2, . . . 
 
 The process is to be continued until we have no even num- 
 bers left. 
 
 Therefore, if we put a for the number of times that the 
 factor 2 enters into n ! we have, 
 
 + etc. 
 
 II. The numbers in the series n ! containing 3 as a factor are 
 3, G, 9, 12, etc.. 
 
 _2_ 
 
 + 
 
 4 
 
 + [-' 
 
BINOMIAL COEFFICIENTS. 253 
 
 of which the number is ^ r The quotients obtained by di- 
 viding til em by 3 are L -^ 
 
 1,.,3,....[|} 
 
 Of these quotients, - are again divisible by 3, and so 
 
 on as before. Hence, if we put (i for the number of times n ! 
 contains 3 as a factor, we have 
 
 + 
 
 nm 
 
 f I + I ^ I + etc. 
 
 In the same way, if k be any prime number, n ! will con- 
 tain h as a factor 
 
 [1] + [I] + \V\ + ^^- *^'"^'- 
 
 Note. This elegant process enables us to find all the prime 
 factors of n\ without actually computing it, and thus to ex- 
 hibit w ! as a product of prime factors. If we suppose n = 12, 
 we shall find, 
 
 12! = 1-2.3 12 = 210.35.52.7.11. 
 
 2. Next let us find the prime factors of the product 
 
 (6? + 1) (« + 2) (« + n\ 
 
 which contains n factors. Dividing successively by 2, 3, 5, 7, 
 etc., it is shown in the same way as before that the prime fac- 
 tor 2^ is contained in the product at least 
 
 [a * [?] 
 
 + etc. times. 
 
 whatever prime factor p may be. Therefore the numerator 
 (a-fl) («-f 2) . . . . {a-\-7i) contains all the prime factors found 
 in n\ to at least the same power with which they enter n\ 
 Hence (§ 238, III), the numerator is divisible by n ! 
 
 Cor. If the factor a-\-n in the numerator is a prime 
 number, that prime cannot be contained in n ! because it is 
 
254 DIVISORS OF A NUMBER. 
 
 greater than n. Hence the binomial factor will be divisible 
 by it. 
 
 Example. is divisible by 7. 
 
 "We may show in the same way that the binomial coefficient 
 is divisible by all the prime numbers in its numerator which 
 exceed n. 
 
 Divisors of a Number. 
 
 340. Def. The expression 
 
 is used to express how many numbers not greater than 
 m are prime to m. 
 
 Example. Let us find the value of </> (9). 
 
 1 is prime to 9, because their G. C. D. is 1. 
 
 2 « « « « « « 
 
 3 is not prime to 9, because their G. C. D. is 3. 
 
 4 is prime to 9. 
 
 5 " " 
 
 6 is not, because 6 and 9 have the G. 0. D. 3. 
 
 7 is. 
 
 8 is. 
 
 9 is not. 
 
 Therefore, the numbers less than 9 and prime to it are 
 
 1, 2, 4, 5, 7, 8, 
 
 which are six in number. Hence, * 
 
 (9) = 6. 
 
 The numbers less than 13 and prime to 12 are 1, 5, 7, 11. 
 Ilence, 
 
 (12) = 4. 
 
 We find in this way, 
 
 0(1) =1, 0(2) = 1, 0(3) =2, 
 
 0(4) =2, 0(5) =4, 0(6) = 2, 
 
 (7) = 6, etc., etc. 
 
DIVISORS OF A NUMBER. 255 
 
 Cor. 1. The number 1 is prime to itself, but no other 
 number is prime to itself. 
 
 Cor. 2. If m be a prime number, then 
 {m) = m — 1, 
 
 because the numbers 1, 2, 3, m — 1 are then all prime 
 
 to m. 
 
 The following remarkable theorem is associated with the 
 functions (w). 
 
 241. Theorem. If iV be any number, and ^Z-,, d^, 
 eZg, etc., all its divisors, unity and n included, then 
 
 id^) + 0(<^2) + ^(^3) + etc. -^ N, 
 
 Example. Let the number be 18. 
 
 The divisors are 1, 2, 3, 6, 9, 18. We find, by counting, 
 
 0(1) = 1 
 0(2) = 1 
 0(3) = 2 
 0(6) = 2 
 0(9) = 6 
 (18) = _6 
 Sum, 18. 
 
 To show how this comes about, write down the numbers 
 1 to 18, and under each write the greatest common divisor of 
 that number and 18. Thus, 
 
 Num., 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18. 
 G.C.D., 12321612921612321 18. 
 
 Necessarily the numbers in the second line are all divisors 
 of 18 as well as of the numbers over them. 
 
 The divisor 1 is under all the numbers prime to 18, so 
 that there are 
 
 (18) = divisors 1. 
 
 If n be any number over the divisor 2, then - and — -, or 
 
 2 2 
 
 9, must be prime to each other. (§ 232, Cor. 1.) That is, the 
 
256 DIVISORS OF A NUMBER. 
 
 numbers n are all those which, when divided by 2, are prime 
 to 9. So there are 
 
 0(9) divisors 2. 
 
 The divisor 3 marks all numbers which, when divided by 3, 
 
 18 
 are prime to — = 6. Hence, there are 
 
 o 
 
 (f) (6) divisors 3. 
 
 In the same way there are (3) divisors 6, (2) divisors 9, 
 and (1) divisor 18. 
 
 The total number of these divisors is both 18 and </> (18) 
 + (9) + etc. Hence, 
 
 0(18) + 0(9) + 0(6) +0(3) + 0(2) + 0(1) = 18. 
 
 General Proof. Let m be the given number ; 
 di, ^3, (?3, etc., its divisors; 
 
 5'i? Q2f 5'3' t^6 quotients -r-f -r-f etc. 
 
 The quotients g^, q^, etc., will be the same numbers as d^, 
 dz, etc., only in reverse order. The smallest of each row will 
 be 1 and the greatest m. We shall then have 
 
 m = d^q^ = d^q^ = d^ q^, etc. 
 
 From the list of numbers 1, 2, 3, ... . m, select all those 
 which have d-^ (unity) as the greatest common divisor with m, 
 then those which have d^ as such common divisor, then those 
 which have d^, etc., till we reach the last divisor, whicli will 
 be m itself, and which will correspond to m. 
 
 The numbers having unity as G. C. D. will be those prime 
 to m, by definition. Their number is (m). 
 
 Those having dc^ as G. C. D. with 7n will, when divided by 
 d^, give quotients prime to y- or to (/g. Moreover, such quo- 
 tients will include all the numbers not greater than q^ and 
 prime to it, because each of these numbers, when multiplied 
 by r?2, will give a number not greater than m, and having d^ 
 as its G. C. D. with m. Hence the number of numbers not 
 
FERMAT'S THEOREM. 257 
 
 greater than ???, and having d^ as its G. 0. D. with m will 
 be 0(^2). 
 
 Continuing the process, we shall reach the divisor m, which 
 will have m itself as G. C. D., and which will count as the 
 number corresponding to (1) = 1 in the list. 
 
 The m numbers 1, 2, 3, .... m are therefore equal in num- 
 ber to 
 
 (;&(m) + 0to,) + 0(g3)+ .... +</>(!); 
 
 or, since the quotients and divisors are the same, only in re- 
 verse order, we shall have 
 
 </>(!)+'/» (.d^) + 9 (6Z3) +....+ (m) = m. 
 
 242. Fermat's Theorem. // p he any prime ninn- 
 her, and a he a numher prime to p, then aP-^ — 1 will he 
 divisihle hy p. 
 
 Examples. «^ — 1 is divisible by 5 ; a* — 1 is divisible by 7. 
 
 Proof. Develop a^ in the following way by the binomial 
 theorem, 
 
 «*=[l + (a-l)f 
 
 ==1 +p(a-l) + i^^(a-iy + ....+ (a-iy. 
 
 Because p is prime, all the binomial coefficients, 
 
 p, (I), etc., to (-^), 
 
 are divisible by p (§ 239, Cor.). Transposing the terms of the 
 last member of the equation which are not divisible hj p, we 
 find 
 
 a^ — (a — iy — 1 = ii multiple oip. 
 
 or dF — a— [(« — 1)^ — (« — !)] = a multiple of p. 
 
 Supposing a; = 2, this equation shows that 2^ — 2 is a 
 multiple of p\ then, supposing x = 3, we show by § 231, 
 Th. II, that 3^ — 3 is such a multiple, and so on, indefinitely. 
 
 Hence, a^ — a = a multiple of p, 
 
 whatever be a. But a^ — a = {aP~'^ — 1) a, and because this 
 product is divisible by p, one of its factors must be so divisible 
 (§ 23G). Hence, if a is prime io p, aP-^ — 1 is divisible by^:>. 
 17 
 
258 CONTINUED FRACTIONS. 
 
 CHAPTER II. 
 
 OF CONTINUED FRACTIONS. 
 
 243. Any proper fraction may be represented in the form 
 -;- , where x^ is greater than unity, but is not necessarily a whole 
 number. If a^ be the greatest whole number in x^, we can put 
 
 ^1 = «^i +— ^ 
 
 where x^ will be greater than unity. In the same way we 
 may put 
 
 x^ = Ci2 -{■ — » 
 
 _ 1 
 
 ^ ^ ' X^' 
 
 etc. etc. 
 
 If for each x we substitute its expression, the fraction — 
 will take the form ^ 
 
 11 1 
 
 ^1 + — «i + - 
 
 '"* «2 H > 6^c., etc. 
 
 ^3 
 
 If the substitutions are continued indefinitely, the form 
 will be I 
 
 r 
 
 1 
 
 1 
 
 '5 
 
 a, 
 Such an expression is called a continued fraction. 
 
 Def. A Continued Fraction is one of wliicli tlie 
 denominator is a whole number plus a fraction ; the 
 denominator of this last fraction a whole number plus 
 a fraction, etc. 
 
CONTINUED FRACTIONS. 259 
 
 A continued fraction may either terminate with one of its 
 denominators or it may extend indefinitely. 
 
 Def. When the number of quotients a is finite, the 
 fraction is said to be Terminating. 
 
 244. Peoblem. To find the value of a continued 
 fraction. 
 
 We first find the value when we stop at the first denomina- 
 tor, then at the second, then at the third, etc. 
 
 Using only two denominators, the fraction will be 
 11 x^ 
 
 F = 
 
 Xi , 1 a.x^ + 1' 
 
 •C/n 
 
 F being put for the true value of the fraction. 
 
 To find the expression with three terms, we put, in the 
 preceding expression, a2 -\ in place of x^. This gives 
 
 X r 
 
 «3 + 
 
 F= 
 
 3 
 1^ 
 
 ^3 ^2*^3 "T~ -^ 
 
 a,a,+'^ + l K«. + l)^3 + «. 
 
 To find the result with the fourth denominator, we substi 
 be fraction become 
 
 tute a;3 = «3 H The fraction becomes: 
 
 ^4 
 
 [(a^a^ + 1) «3 + «i] ^4 + «i«s 
 
 (a) 
 
 To investigate the general law according to which the 
 successive expressions proceed, we put 
 
 P, the coefficient of a; in any numerator; 
 P', the coefficient of x in the denominator; 
 
 Q, the terms not multiplied by x in the numerator ; 
 Q', the terms not multiplied by x in the denominator ; 
 
 and we distinguish the various expressions by giving each P 
 and Q the same index as the x to which it belongs. 
 
260 CONTINUED FRACTIONS. 
 
 Then we may represent each value of F in the form, 
 
 where i may take any value necessary to distinguish the frac- 
 tion. Comparing with the fractions as written, we see that : 
 
 p^ = 1, Q^ = 0, p; = «i, 0^2 = 1 ; {o) 
 
 P3 = 6?2, Ca = 1^ ^3 = «i«52 + l? tfs = «i ; 
 
 P4 =r fiTgfljg + l, §4 = a^, P\ == a^P's+a^, Q\ = a^a^ + l. 
 
 To show that this form will continue, how far soever we 
 carry the computation, we put in the expression {b) the general 
 value of Xi, 
 
 _ 1 
 
 Xi — Cti -] , 
 
 Xi+1 
 
 1 • , . T7 (aiPi + Qi) Xi+i + Pi / n 
 
 To show the general law of succession of the terms, let us 
 compare the general equation (b) with {d). Putting i+1 for 
 i in (b), it becomes, 
 
 ^ _ P j+i xin 4- Qi+i / X 
 
 Comparing this with (d), we find 
 
 Pi+i = a«Pi + ft, 
 
 ftM = Pi; 
 whence, ft = Pi-i. 
 
 Substituting this value of Qi in the equation previous, it 
 becomes 
 
 Pi+i = aiPi + Pi-i. if) 
 
 "Working in the same way with the denominators, we find 
 
 Q'u, = P'l- 
 
 By supposing i to take in succession the values 1, 2, 3, etc.. 
 
CONTINUED FRACTIONS. 261 
 
 these formulae show that the successive values of P may be 
 computed thus : 
 
 P4 -^ ^3-^3 I -^2? 
 
 i'e = «5^5 + P4. 
 etc., to any extent. 
 
 Also, P\ = 1, 
 
 Pg = «i, 
 
 p; == «3p; + p;, 
 p; = a,p\ + p;, 
 
 etc. etc. 
 
 Since each value of Q is equal to the value of P having the 
 next smaller index, it is not necessary to compute the C's sep- 
 arately. 
 
 If the fraction terminates at the n*^ value of «, we shall 
 have 
 
 Xn = an, exactly. 
 
 If it does not terminate, we have to neglect all the denom- 
 inators after a certain point ; and calhng the last denominator 
 we use the n^^, we must suppose 
 
 ^n "^^ ^ra- 
 in either case, the expression (b) will give the value of the 
 fraction with which we stop by putting i = n and Xn = an. 
 
 Therefore, F = "^^^^"^ ^f , 
 
 an I^n -\- V» 
 
 or, substituting for Q^ and Q'^ their values in (^), 
 
 jp ^n Pn -\- Pn-1 
 
 an Pn 4" Pn-1 
 
 But the general expressions (/) and {g) give 
 
202 CONTINUED FRACTIONS. 
 
 an 
 
 Pn 
 
 + Pn-1 = 
 
 = Pn+h 
 
 an 
 
 P'n 
 
 4- P'n-X - 
 
 - Pn+1- 
 
 
 
 F = 
 
 Pn^i 
 
 Therefore, 
 
 Therefore, to find the value of the fraction to the n*^ 
 term, we have only to compute the values of Pn+i and 
 PnH} without talcing any account of Q. 
 
 Example. Take the fraction, 
 1 
 
 3 + - 
 
 4 + 
 
 1 
 
 5 etc. 
 Here, a^ =1, a^ ^= 2, «3 = 3, . . . . ai = i. 
 
 We now have, by continuing the formulas (c) and (/), and 
 using those values of a^, a^, etc.: 
 
 P, = 0, 
 
 -Pg = 1, 
 
 P3 = a^Pz + Pi = fl'g = 2, 
 
 P, =a^P^ + P2 =3.2 + 1 = 7, 
 
 Pg =«,P4 + P3 =4.7 + 2 = 30, 
 
 Pg.= a,P, + P4 = 5.30 + 7 = 157, 
 
 etc. etc. etc. 
 
 p; = 1, 
 
 P; -=a,=l,' 
 
 p; = «2P; + p; = 2.1 + 1 = 3, 
 p; = ftgp; + p; = 3.3 + 1 = 10, 
 p; = a^p\ + p; = 4.10 + 3 = 43, 
 p; = ^gp; + p; = 5.43 + 10 = 225. 
 
 Therefore, supposing in succession, w = 1, w = 2, ^ = 3, 
 etc., we have, for the successive approximate values of the 
 fraction. 
 

 CONTINUED FRACTIONS. 
 
 For 71 = 1, 
 For n = 2, 
 
 ^ = J? = 1. 
 
 For 71 — 5, 
 
 ^ ~ i^; ~ 225 
 
 263 
 
 These successive approximate values of the continued frac- 
 tion are called Converging Fractions, or Convergents. 
 
 245. The forms (/) and {g) may be expressed in words as 
 follows : 
 
 The numerator of each convergent is formed hy rrtul- 
 tiplying the preceding jiumerator hy the corresponding 
 a, and adding the second nujnerator preceding to the 
 product. 
 
 The successive denominators are formed in the same way. 
 
 Example. The ratio of the motions of the sun and moon 
 relative to the moon's node is given by the continued fraction: 
 1 
 
 12 4-.^ 
 
 1 + i 
 
 -b 
 
 4 + ^ 
 
 3 + etc. 
 
 Let us find the successive convergents. We put the de- 
 nominators a^ = 12, a^ = 1, etc., in a line, thus: 
 a = 12, 1, 2, 1, 4, 3. 
 
 F' = 1' 12' 13' 38' 51' 242' 777* 
 
 Under a^ we write the fraction -, whicli is always the one 
 
 with which to start, because Fi = and P'l = 1 (§ 244, c). 
 
 Next to the right is — , because Pg = 1 and Pg = a. After 
 
 this, we multiply each term by the multiplier a above it, and 
 
264 CONTINUED FRACTIONS. 
 
 add the term to the left to obtain the term on the right. 
 Thus, 2.1 + 1 = 3, 2.13 + 12 = 38, etc. 
 
 Ex. 2. To compute the convergents of 
 
 1 
 
 4 + i 
 
 2 + ^ 
 
 4 etc. 
 
 a = 2, 4, 2, 4, 2, 4, etc. 
 
 Numerators, 2 i i A i2 ^1 
 
 Denominators, 1' 2' 9' 20' 89' 198' 
 
 etc. 
 
 EXERCISES. 
 
 Reduce the following continued fractions to vulgar frac- 
 tions : 
 
 3 + ^. 3 + 1 
 
 7 + A- ^ + ' 
 
 1« 3 + i. 
 
 4- 
 
 3 + ^!^ T 3 + ^^ a + 
 
 l+L-3 5 + ^. J + ^. 
 
 3 + 
 
 ■-I 
 
 246. Problem. To express a fractional quantity as 
 a continued fraction. 
 
 Let R be the given fraction, less than unity. Compute x^ 
 from the formula, 
 
 _ J_ 
 
 x^ - -^. 
 
 Let «i be the whole number and E' the fraction of x^. 
 Then compute 
 
 _ J^ 
 
 ^3 - A>'* 
 
CONTINUED FRACTIONS, 
 
 265 
 
 Let fl^g be the whole number and R" the fraction of x^. 
 We continue, this process to any extent, unless some value 
 of X comes out a whole number, when we stop. 
 
 26 
 
 Example. Express ^ as a continued fraction. 
 
 R "" 26 "■ "^ 26 
 
 ^^- R' - 
 
 26 
 21 
 
 1 + 
 
 R" 6 ^5 
 
 X 
 
 5; 
 
 _ 1 _ 5 
 * ~ R' ~~ 1 
 
 So the continued fraction is 
 
 1 
 
 «i = 2 
 
 «2 = 1 
 
 «3 = 4 
 
 cf^ = 5 
 
 i2' = 
 
 i2" 
 
 21 
 26' 
 
 21' 
 
 ij'" = ^ 
 
 i2'' = 0. 
 
 2 + 
 
 1 + 
 
 It will be seen that the process is the same as that of find- 
 ing the greatest common divisor of two numbers. 
 
 EXERCISES. 
 
 Develop the following quotients as continued fractions : 
 
 113 1049 628 
 
 ^' 925* 
 
 ^* 355 
 
 3326 
 
 247. The most simple continued fraction is that arising 
 from the geometric problem of cutting a line in extreme and 
 mean ratio. The corresponding numerical problem is: 
 
 To divide unity into two such fractions that the less 
 shall he to the greater as the greater is to unity. 
 
 Let r be the greater fraction. Then 1 — r will be the 
 lesser one. We must then have 
 
 ri 1, 
 
2Q6 CONTINUED FRACTIONS. 
 
 which gives 
 
 r2 = 1 - r. 
 
 or 
 
 r^ + r = 1, 
 
 or 
 
 r(r- + l) = l, 
 
 
 1 
 
 1 + r 
 Now, let us put for r in the last denominator the expression 
 , and repeat the process indefinitely. We shall have, 
 
 1 
 
 r = 
 
 i + i 
 
 i + i 
 
 1 + ^ 
 
 1 etc., ad infinitum. 
 
 Now we may form the successive convergents which 
 approximate to the true value by the rule. As all the denom- 
 inators a are 1, we have no multiplying, but only add each 
 term to the preceding one to obtain the following one. Thus 
 we find: 
 
 0112 3 5^13-3134 
 
 i' i' 2' 3' 5' 8' 13' 21' 34' 55' ^ ^* 
 
 The true value of r may be found by solving the quadratic, 
 
 ^2 -f- r = 1, 
 
 v T, • -1± Vs 
 which gives r = ^ • 
 
 The positive root, with which alone we are concerned, is 
 r = "^t = 0.61803399. 
 
 The values of the first nine convergents, with their errors, 
 
 are: 
 
 1:1 =1.0, error = +0.382. 
 
 1:2= 0.5, " —0.118. 
 
 .2:3= 0.6G6...., " +0.0486. 
 
 3:5= 0.600, " —0.0180. 
 
 5:8= 0.625, « +0.00697. 
 
 J 
 
CONTINUED FRACTIONS. 267 
 
 8 : 13 = 0.61538...., error = —0.00265. 
 
 13 : 21 = 0.61904...., " + 0.00101. 
 
 21 : 34 = 0.617647...., « —0.000397. 
 
 34 : 55^=1 0.618182.... , " + 0.000148. 
 
 etc. etc. etc. 
 
 Relations of Successive Convergents. 
 
 248. Theorem I. The successive convergents are 
 alternately too large and too small. 
 
 Proof. The first convergent is — The true denom- 
 
 inator being a^ -] , the denominator a^ is too small, and 
 
 therefore the fraction is too large. 
 
 In forming the second fraction, we put — instead of — • 
 
 Ct.2 X^ 
 
 Because a^ < x^, this fraction is too large, which makes the 
 
 * 1 
 denominator a. A ■ too small. 
 
 The third denominator ffg is too small, which will make 
 the preceding one too large, the next preceding too small, and 
 so on alternately. 
 
 Theorem II. If — and —. he any two consecutive 
 n n 
 
 convergents, then 
 
 mn' — m'n = ± 1. 
 
 Proof We show : 
 
 («) That the theorem is true of the first pair of convergents. 
 
 (jS) That if true of any pair, it will be true of the pair next 
 following. 
 
 («) The first pair of convergents are 
 
 «i' ^ . 1 «i^2 + 1' 
 ^■^^ 
 which gives 7nn' — m'n = 1, thus proving («). 
 
268 CONTINUED FBACriONS. 
 
 (i3) Let — , — , — 
 
 ^ ^ n n n 
 
 be three consecutive convergents, in which 
 
 7n7i' — m'n = ± 1. (1) 
 
 By (/) and (g) we shall have 
 
 m" = am' + m, 
 n" = an' + n. 
 Multiplying the second equation by m! and subtracting the 
 product of the first by n', we have 
 
 m'n" — m"7i' = m'n — mn\ 
 
 which is the negative of (1), showing that the result is =F 1. 
 
 The theorem being true of the first and second fractions, 
 must therefore be true of thfe second and third ; therefore of 
 the third and fourth, and so on indefinitely. 
 
 Corollaries. Dividing (1) by nn', Ave have 
 
 m m' 1 
 
 ? = ± — ,' Hence, 
 
 n n nn 
 
 I. Tlve difference hetiveen the two successive converg- 
 ents is equal to unity divided hy the product of the 
 denominators. 
 
 Because the denominator of each fraction is greater than 
 that of the preceding one, we conclude : 
 
 II. TJie difference between two consecutive convergents 
 constantly diminishes. 
 
 Combining these conclusions with Th. T, we conclude : 
 
 III. Each value of a convergent always lies between 
 the values of the two preceding convergents. 
 
 For if R^, i?g, i2g be three such fractions, and if i?g is 
 greater than R^, then R^ will be less than R^. But it must 
 be greater than R^, else we should not have R^ — Rq numer- 
 ically less than R^ — R^. Hence, if we arrange the successive 
 convergents in a line in the order of magnitude, their order 
 will be as follows : 
 
 -"'4? ^*^6' -"^8' • • • • ^95 RriJ R$i 
 
 each convergent coming nearer a true central value. Hence, 
 
CONTINUED FRACTIONS. 269 
 
 IV. TIi6 true value of the continued fraction al- 
 ways lies between the values of two consecutive con- 
 vergents. 
 
 Comparing with (I), we conclude : 
 
 V. The error which we mahe hy stopping at any con- 
 vergent can never he greater than unity divided hy the 
 product of the denominators of that convergent and the 
 one next following. 
 
 EXAMPLE. 
 
 Eef erring to the table of values of ^ (VS — 1) in § 247, 
 we see that : 
 
 Error of ^ = 3 < ^ -^ f 
 
 (for .0486 < 1). 
 
 Error of 3 : 5 < ^^^ ; 
 
 0* o 
 
 (for .018 <1). 
 
 etc. 
 
 etc. 
 
 Hence, in general, continued fractions give a very rapid 
 approximation to the true value of a quantity. Their princi- 
 pal use arises from their giving approximate values of irrational 
 numbers by vulgar fractions with the smallest terms. 
 
 EXAMPLE. 
 
 Develop the fractional part of V^ as a continued fraction, 
 and find the values of eight convergents. 
 
 Because 1 is the greatest whole number in V^, we put 
 
 a/2 = i + ^; (1) 
 
 whence, x = 
 
 X 
 
 1 
 
 V2 — 1 
 
 Eationalizing the denominator, § 185, 
 X = V2 + 1. 
 Substituting for a/2 its value in (1), 
 
 X 
 
270 CONTINUED FRACTIONS. 
 
 Putting this value of a; in (1) and again in the denominator, 
 and repeating the substitution indefinitely, we find 
 
 V2 = 1 + '^- 
 
 2 + i 
 
 ^ + 2 etc. 
 Forming the convergents, we find them to be 
 
 1 2 5 12 29 70 169 408 
 2' 5' 12' 29' 70' 169' 408' 985' 
 
 Adding unity to each of them, we find the approximate 
 values of \/2 : 
 
 3 7 17 41 99 239 577 1393 
 
 _ _ . PTf* 
 
 2' 5' 12' 29' 70' 169' 408' 985 ' 
 
 Rem. The square root of 2 may be employed in finding a 
 right angle, because a right angle (by Geometry) can be formed 
 by three pieces of lengths proportional to 1, 1, ^2. If we 
 make the lengths 12, 12, 17, the error will, by Cor. V, be less 
 
 than 7^^, or less than -— of the whole length. 
 
 EXERCISES. 
 
 Develop the following square roots as continued fractions, 
 and find six or more of the partial fractions approximating to 
 each : 
 
 I. a/3. 2. Vs. 3. Ve. 4. Vio. 
 
 5. Develop a root of the quadratic equation 
 x^ — ax — 1 = 0, 
 commencing the operation by dividing the equation by x* 
 
 Periodic Continued Fractions. 
 
 249. Def. A Periodic continued fraction is one in 
 which the denominators repeat themselves in regular 
 order. 
 
CONTINUED FRACTIONS, 271 
 
 Example. . A continued fraction in which the successive 
 denominators are 
 
 2, 3, 5, 2, 3, 5, 2, 3, 5, etc., ad infinitum, 
 is periodic. 
 
 A periodic continued fraction can be expressed as 
 tlie root of a quadratic equation. 
 
 I. 
 
 EXAMPLES. 
 1 
 
 i-fi 
 
 2 + ^ 
 
 1 
 
 "^ 2 + etc. 
 If we put X for the value of this fraction, we liave 
 
 1 
 
 X = 
 
 We find the value thus : 
 
 1, 2 + x. 
 
 1 2 + x 
 
 V 1' 3+x' 
 
 Because this expression is x itself, we have 
 
 _ 2 + x 
 ''- d+x' 
 
 which reduces to the quadratic equation 
 x^ + 2x = 2. 
 
 2. Let us take the fraction of which the successive denom- 
 inators are 2, 3, 5, 2, 3, 5, etc., namely, 
 
 1 
 
 X = 
 
 . + i 
 
 3 + ^- 
 
 5 + i 
 
 2 + -^ 
 
 3 + etc., 
 
272 CONTINUED FRACTIONS. 
 
 1 
 
 or. 
 
 2 + i 
 
 3 + ^ 
 
 b -{-X 
 
 We compute thus : 
 
 2, 3, X + 5. 
 
 1 3 3a; + 1 6 
 
 1' 2' V 7^T37' 
 
 Hence we have, to determine x, the quadratic equation, 
 
 350. Development of the Root of a Quadratic Equation. 
 A root of a quadratic equation may be developed in a continued 
 fraction by the following process. Let the equation in its 
 normal form be (§ 192), 
 
 mx^ + wz + jf7 = 0, (1) 
 
 m, n, and ^ being whole numbers. We shall then have 
 
 — ^ ± ^n? — 4m^ 
 ^' ~ 2w 
 
 Let a be the greatest whole number in x, which we may 
 find either by trial in (1) or by this value of x. Then assume 
 
 and substitute this value of x in the original equation. Then, 
 regarding x^ as the unknown quantity, we reduce to the nor- 
 mal form, which gives 
 
 {ma? + na -\- p)x^ + (2m« -{- n) x ^ -\- m ^= ^. (2) 
 If a^ is the greatest whole number in x^, we put 
 
 ^1 = «^i +— . 
 
 2 
 
 and by substituting this value of x^ in (2), we form an equa- 
 tion in iCg. Continuing the transformations, we find the 
 greatest whole number in .Co, which will be called ^g, and so on. 
 The root will then be expressed as a whole number a plus 
 the continued fraction of which the denominators are «i, ^g, 
 ^3, etc. 
 
 ♦' 
 
('I 
 
 BOOK X. 
 THE COMBINATORY ANALYSIS, 
 
 CHAPTER I. 
 
 PE R M U TATI N S, 
 
 351. Bef, The diiferent orders in which a nnmber 
 of things can be arranged are called their Permuta- 
 tigns. 
 
 Examples. The permutations of the letters a, h, are 
 
 ab, la. 
 The permutations of the numbers 1, 2, and 3 are 
 123, 132, 213, 231, 312, 321. 
 
 Problem. To -find how many -permutations of any 
 given numher of things are possible. 
 
 Let us put 
 
 Pi, the number of permutations of i things. 
 
 It is evident from the first of the above examples that there 
 are two permutations of two things. Hence, 
 
 Pg = 2. 
 
 To find the permutations of three letters, a, h, c, we form 
 three sets of permutations, the first beginning with a, the sec- 
 ond with h, and the third with c. 
 
 In each set the first letter is to be followed by all possible 
 permutations of the remaining letters, namely : 
 
 In 1st set, after a write he, cb, making abc, ach. 
 
 " 2d « « h « ac, ca, " hac, hca. 
 
 " 3d " « c " ah, ha, " cah, cha, 
 IS 
 
274 PERMUTATIONS. 
 
 Hence, P3 =1 3-2 = 6. 
 
 The permutations of n things can be divided into sets. 
 The first set begins with the first thing, followed by all possi- 
 ble permutations of the remaining n — 1 things, of which the 
 number is Pn-i- The second set begins with the second thing, 
 followed by all possible permutations of the remaining n — 1 
 things, of which the number is also Pn-i, and so with all n 
 sets. Hence, whatever be n, there will be n sets of Pn~i per- 
 mutations in each set. Therefore, 
 
 Pj, = nPn~i. 
 
 This equation enables us to find Pn whenever we know 
 Pn-h and thus to form all possible values of Pn, as follows: 
 
 It is evident that 
 
 P,=l. 
 
 We have found 
 
 Pg = 2.1 = 2! 
 
 it a 
 
 P3 = 3-2.1 = d\ = 6. 
 
 Putting 7^ = 4, we have 
 
 P^ = 4:P^ = 4tl = 24. 
 
 '' n = 6, '' " 
 
 Pg = 5P^ = 5 ! = 120. 
 
 etc. 
 
 etc. etc. 
 
 It is evident that the number of permutations of Ji things 
 is equal to the continued product 
 
 1.2.3.4 n, 
 
 which we have represented by the symbol n ! so that 
 
 Pn = n\ 
 
 EXERCISES.* 
 
 1. Write all the permutations of the following letters : 
 
 led, acd, aid, abed. 
 
 2. What proportion of the possible permutations of the 
 letters a, e, m, t, make well-known English words ? 
 
 3. Write all the numbers of four digits each of which can 
 be formed by permuting the four digits 1, 2, 3, 4. 
 
 4. How many numbers is it possible to form by permuting 
 the six figures 1, 2, 3, 4, 5, 6. / 5^0 
 
 * If the student finds any difficulty in reasoning out these exercises, 
 he is recommended to try similar cases in which few symbols are involved 
 by actually forming the permutations, until he clearly sees the general 
 principles involved. 
 
PERMUTATIONS. 275 
 
 5. At a dinner party a row of 6 plates is set for the host 
 and 5 guests. In how many ways may they be seated, subject 
 to the condition that the host must have Mr. Brown on his 
 right and Mr. Hamilton on his left ? ^ *^' 
 
 6. Of all numbers that can be formed by permuting^ the K 
 seven digits, 1, 2 .... 7: 
 
 {a) How many will be even and how many odd ? 
 (J)) In how many will the seven digits be alternately even ■ ^ 
 and odd? 
 
 (c) In how many will the three even digits all be together ? 21 *> 
 
 (d) In how many will the four odd digits all be together ? '^ 
 
 7. In how many permutations of the 8 letters, a, l, c, d, e, . 
 /, g, h, will the letters d, e, /, stand together in alphabetical 
 order ? 
 
 '8. In how many of the above permutations will the word 
 deaf be found ? 
 
 9. In how many of the permutations of the first 9 letters ^^U^i<^ 
 will the words age and bid be both found ? 
 
 10. A party of 5 gentlemen and 5 ladies agree with a math- ^ 
 ematician to dance a set for every way in which he can divide "^ 
 them into couples. How many sets can he make them dance ? 
 
 11. In how many of the permutations of the letters a, h, c, 
 d, e, will d and no other letter be found between a and e. 
 
 12. In how many of the permutations of the six symbols, 
 
 a, h, c, d, e,f, will the letters abc be found together in one "' 
 group, and the letters def in another ? 
 
 13. How many permutations of the seven symbols, a, h, c, , ^ 
 d, e, f, g, are possible, subject to the condition that some per- ^ 
 mutation of the letters abc must come first ? 
 
 14. The same seven symbols being taken, how many per- 
 mutations can be formed in which the letters abc shall remain '-^^ 
 together ? 
 
 Permutations of Sets. 
 
 253. Def. When permutations are formed of only 
 s things out of a whole number 71, they are called Per- 
 mutations of n things taken ^ at a time. 
 
276 PERMUTATIONS. 
 
 Example. The permutations of the three letters a, h, c, 
 taken two at a time, are 
 
 if/J^*^ ab, ha^ ac, ca, he, ch. 
 
 The permutations of 1, 2, 3, 4, taken two at a time, are 
 12, 13, 14, 21, 23, 24, 31, 32, 34, 41, 42, 43. 
 
 Problem. To find the number of permutations of 
 n things taken^s at a time. 
 
 Suppose, first, that we take two things at a time, as in the 
 above examples. We may choose any one of the n things as 
 the first in order. Which one soever we take, we shall have 
 n — 1 left, any one of which may be taken as the second in 
 order. Hence, the permutations taken 2 at a time will be 
 
 n{n — 1). 
 
 [Compare with the last example, where n = 4.] 
 To form the permutations 3 at a time, we add to each per- 
 mutation by 2's any one of the n — 2 things which are left. 
 Hence, the number of permutations 3 things at a time is 
 
 n{n — l)(n — 2). 
 
 In general, the permutations of w things taken 5 at a time 
 will be equal to the continued product of the s factors, 
 
 w (?^ — 1) (/I — 2) {n — s + 1), 
 
 which is equal to the quotient Tj r-^^^^ — , 
 
 It will be remarked that when 5 = w, we shall have the 
 case already considered of the permutations of all n things. 
 
 EXERCISES. 
 
 1. Write all the numbers of two figures each which can be 
 formed from the four digits, 3, 5, 7, 9. 
 
 2. Write all the numbers of three figures, beginning with 
 1, which can be formed from the five digits, 1, 2, 3, 4, 5. 
 
 3. How many different numbers of four figures each can 
 be formed with the digits 1, 2, 3, 4, 5, 6, no figure being re- 
 peated in any number ? 
 
 ;si 
 
PERMUTATIONS. 
 
 277 
 
 4. Explain how all the numbers in the preceding exercise 
 may be written, showing how many numbers begin with 1, 
 how many with 2, etc. 
 
 5. In how many ways can 3 gentlemen select their partners 
 from 5 ladies ? 
 
 6. How many even numbers of 3 different digits each can 
 be formed from tiie seven digits, 1, 2, .... 7 ? 
 
 7. How many of these numbers will consist of an odd 
 digit between two even ones ? 
 
 Circular Permutations. 
 
 2^53. If we have the three letters a, i, c, arranged in a 
 circle, as in the adjoining figure, then, 
 however we arrange them, we shall find 
 them in alphabetical order by beginning 
 with a and reading them in the suitable 
 direction. Hence, there are only two 
 different circular arrangements of three 
 letters instead of six, namely, the two 
 di^ctions in which they may be in al- 
 phabetical order. 
 
 Next suppose any number of symbols, say a, h, c, d, e,f, g, 
 7i, and let there be an equal number of positions around the 
 circle in which they may be placed. These positions are num- 
 bered 1, 2, 3, 4, 5, 6, 7, 8. 
 
 For every arrangement of the sym- 
 bols we may turn them round in a body 
 without changing the arrangement. 
 Each symbol will then pass through all 
 eight positions in succession. 
 
 By performing, this operation with 
 every arrangement, we shall have all 
 possible permutations of the eight things 
 among the eight positions, the number 
 
 of which is 8!, which are therefore eight times as many as 
 the circular arrangements. 
 
278 PEBMUTATIONS. 
 
 Hence the number of different circular arrangements is 
 — ', which is the same as 7! 
 
 o 
 
 In general, if we represent the number of circular arrange- 
 ments of 71 things by (Jn, we shall have 
 
 Cn = {n-1)\ 
 
 The same result maybe reached by the following reasoning. 
 To form a circular arrangement, we may take any one symbol, 
 a for example, put it into a fixed position, say (1), and leave it 
 there. 
 
 All possible arrangements of the symbols will then be 
 formed by permuting the remaining symbols among the re- 
 maining positions. Hence, 
 
 Cn = Pn-i = {n-l)\ j\ 
 
 as before. 
 
 EXERCISES. 
 
 1. In how many orders can a party of 7 persons take their 
 places at a round table? 
 
 2. In how many orders can a host and 7 guests sit at a 
 round table in order that the host may have the guest of high- 
 est rank upon his right and the next in rank on his left ? • 
 
 3. Five works of four volumes each are to be arranged on 
 a circular shelf. How many arrangements are possible which 
 will keep the volumes of each set together and in proper order, 
 it being indifferent in which direction the numbers of the 
 volumes read. 
 
 4. In how many circular arrangements of the 5 letters a, b, 
 c, d, e, will a stand between h and d ? 
 
 5. If the 10 digits are to be arranged in a circle, in how 
 many ways can it be done, subject to the condition that even i 
 and odd digits must alternate ? (Note that is even.) 
 
 6. The same thing being supposed, h.ow many arrange- 
 ments are possible, subject to the condition that the even digits 
 must be all together ? 
 
 7. In how many circular arrangements of the first six let- 
 ters will the word deaf he found? What will be the difference 
 of the results if you are allowed to spell it in either direction ? 
 
PERMUTATIONS. 279 
 
 Permutations when Several of the Things are 
 Identical. 
 
 254. If the same thing appears several times among the 
 things to be permuted, the number of different permutations 
 will be less than when the things are all different. 
 
 Example. The permutations of aahb are 
 
 aabl, abab, abba, baab, baba^ bbaa, (1) 
 
 which are only six in number. 
 
 Proble^i. To fiivcl the nimiber of permutations when 
 several of the things are identieal. 
 
 Let us first examine how all 24 permutations of 4 things 
 may be formed from the above 6 permutations of aabb. Let 
 us distinguish the tAVO <:«'s and the two Z^'s by accenting one of 
 each. Then, from each permutation as written, four may be 
 formed by permuting the similar letters among themselves. 
 For, example, taking abba, and writing it abb' a', we have, by 
 permuting the similar letters, 
 
 ab'ba', a'b'ba. abb'a', a'bb'a. (2) 
 
 In the same way four permutations, differing only in the 
 arrangement of the accents, may be formed from each of the 
 6 permutations (1), making 24 in all, as there ought to be. 
 (§ 351.) 
 
 Generalizing the preceding operation, we reach the follow- 
 ing solution of our problem. Let the symbols to be permuted 
 be a, b, c, etc. 
 
 Suppose that a is repeated r times, 
 
 a a 7) '^ ^' S ^^ 
 
 <( i( n '^ *^ / <* 
 
 etc. etc. etc. 
 
 and let the whole number of symbols, counting repetitions, be 
 n, so that 
 
 n=:r-{-s-\-t-\- etc. 
 
 [In the preceding example (1), n — 4, r = 2, .9 = 2.] 
 Also put Xn, the required number of different permutations 
 of the n s3'mbols. 
 
280 PERMUTATIONS. 
 
 Suppose tliese Xn different permutations all written out, 
 
 and suppose the symbols which are repeated to be distinguished 
 
 by accents. Then : 
 
 J ^/j^ From each of the Xn permutations may be formed Pr=zr\ 
 
 '■' permutations by permuting the «'s among themselves, as in 
 
 (2). We shall then have r ! Xn permutations. 
 f,tJLX' P'rom each of the latter may be formed s\ permutations by 
 ' //{.P^r^i^ting the V^ among themselves. We shall then have 
 ^ ' ^ rs\r\ X Xn permutations. 
 
 '^^/" From each of these maybe found t\ permutations by in- 
 
 ^'> ' /) terchanging the ds, among themselves. 
 "^ , . I Proceeding in the same way, we shall have 
 
 '(Xjlrllr' Xn X r\ X s\ X t\ X etc. 
 
 ' ^ f^Vpossible permutations of all w things. But this number has 
 
 'J . / / been shown to be n\ Therefore, 
 
 ^^/a ■ Xn X r\ X s\ X t\ X etc. = n\ 
 
 ^^' By division, Xn ^ , !"], , , (3) 
 
 ijjfj^.i, r! s! ^! etc' ^ ^ 
 
 ^^! : which is the required expression. 
 
 We remark that if any symbols are not repeated, the for- 
 mula (3) will still be true by supposing the number correspond- 
 ing to r, s, or t to be 1. 
 
 J'^Jk' EXAMPLES. 
 
 i *^'^ I. The number of possible permutations of aahl? are 
 
 * t( 4' 24 
 
 ■ C-O '^ ^rr^f — o"ii — ^> ^^ already found. 
 
 2. The possible permutations of aaabhcd are 
 7 ! 5040 
 
 3! 2! 6-2 
 
 = 420. 
 
 EXERCISES. 
 
 Write all the permutations of the letters: 
 I. aaah. 2. aahc. 3. aaahc. 
 
 4. How many different numbers of seven digits each can 
 be formed by permuting the figures 1112225 ? 
 
PERMUTATIONS. 281 
 
 5. If every different permutation of letters made a word, 
 how many words of 13 letters each could be formed from the 
 word Massachusetts. 
 
 The Two Classes of Permutations. 
 
 ^55, The n\ possible permutations of n things are divisi- 
 ble into two classes, commonly distinguished as even permu- 
 tations and odd permutations in the following way : 
 
 We suppose the n things first arranged in alphabetical or 
 numerical order, 
 
 » a, 5, c, df , . , , or 1, 2, 3, 4, ... . n^ 
 
 and we call this arrangement an even permutation. 
 
 Then, having any other permutation, we count for each 
 thing how many other things of lower order come after it, and 
 take the sum. 
 
 If this sum is even, the permutation, is an even one ; if odd, 
 an odd one. 
 
 EXAMPLES. 
 
 I. Consider the permutation 265143. 
 
 Here 2 is followed by 1 number of lower order, namely, 1. 
 " 6 *^ " 4 " " " " 5,1,4,3. 
 
 " 1,4,3. 
 
 5 
 
 i( 
 
 a 
 
 3 
 
 (( 
 
 1 
 
 (( 
 
 a 
 
 
 
 i( 
 
 4 
 
 <( 
 
 11 
 
 1 
 
 a 
 
 it (( (( g 
 
 Then l + 4 + 3-|-0 + l = 9. Hence the permutation is odd. 
 2. Consider cdiea. 
 
 Here c is followed by 2 letters before it in order, namely, ha. 
 " d '' " 2 " " " " ha. 
 
 Then 2 + 2 + 1 + 1 = 6. Hence the permutation is even. 
 
 Def. The total numlber of times which a thing less 
 in order follows one greater in order is called the 
 Number of Inversions in a permutation. 
 
282 , PERMUTATIONS. 
 
 Example. In the preceding permutation, 265143, the 
 number of inversions is 9. In cdbed it is 6. 
 
 Eem. It will be seen that the class of a permutation is 
 even or odd, according as the number of inversions is even or 
 odd. 
 
 Theorem I. If, in a peinnutation, two things are 
 interchanged, the class will he changed frovi^ even to odd, 
 or from odd to even. 
 
 Proof. Consider first the case in which a pair of adjoining 
 things are interchanged. Let us call : 
 
 ilc, the two things interchanged. 
 
 A, the collection of things which precede i and h, 
 
 C, the collection of things which follow them. 
 
 The first permutation will then be 
 
 ^t^a* (a) 
 
 After interchanging i and k, it will be 
 
 AUG. (h) 
 
 Because the order of things in A remains undisturbed, each 
 thing in A is followed by the same things as before. In the 
 same way, each thing in C is preceded by the same things as 
 before. 
 
 Hence, the number of times that each thing in A or C is 
 followed by a thing less in order remains unchanged, and, 
 leaving out the pair of things, i, h, the number of inversions 
 is unchanged. 
 
 But, by interchanging i and Ic, the new inversion M is in- 
 troduced. Therefore the number of inversions is increased 
 by 1. 
 
 * This form of algebraic notation differs from those already used in 
 that the symbols A and C do not stand for quantities, but mere collec- 
 tions of letters. It is an application of the general principle that a single 
 symbol may be used to represent any set of symbols, but must represent 
 the same set throughout the same question. A and G are here used to 
 show to the eye that in forming the permutations of (6) from {a), all the 
 letters on each side of ik preserve their relative positions unchanged. 
 
PERMUTATIONS. 28^ 
 
 If the first arrangement is hi, tliis one inversion is removed. 
 Hence, in either case the number of inversions is changed by 
 1, and is therefore changed from odd to even, or vice versa. ' 
 
 Illustration. In the permutation 
 265143, 
 the inversions, as already found, are the following nine : 
 21, 65, 61, 64, 63, 51, 54, 53, 43. 
 Let us now interchange 5 and 1, making the permutation 
 
 261543. 
 The inversions now are 
 
 21, 61, 65, 64, 63, 54, 53, 43, 
 the same as before, except that 51 has been removed. 
 
 Next consider the case in which the things interchanged 
 do not adjoin each other. Suppose that in the permutation 
 1) a d e h c f 
 
 we interchange a and Ji. We may do this by successively in- 
 terchanging a with d, with e, and with 7^, making three inter- 
 changes, producing 
 
 h d e h a c f . 
 
 Then we interchange Ti with e and with d, making two 
 interchanges, and producing 
 
 l li d e a c f , 
 
 which effects the required interchange of a with h. 
 
 The number of the neighboring interchanges is 3 + 2 = 5, 
 an odd number. Because the number of inversions is changed 
 from odd to even this same odd number of times, it will end 
 in the opposite class with which it commenced. 
 
 Theorem II. TJ^6 possible permjitations of n tilings 
 are one-half even and one-half odd. 
 
 Proof. Write the n ! possible permutations of the n things. 
 
 Then interchange some one pair of things {e.g., the first 
 two things) in each permutation. We shall have the same 
 permutations as before, only differently arranged. 
 
284 PEBMUTATI0N8. 
 
 By the change, eyery even permutation will be changed to 
 odd, and every odd one to even. 
 
 Because every odd one thus corresponds to. an even one, 
 and vice versa, their numbers must be equal. 
 
 Illustration. The permutations in the second column fol- 
 lowing are formed from those in the first by Interchanging the 
 first two figures : 
 
 12 3 
 
 even. 
 
 2 13 
 
 odd. 
 
 13 2 
 
 odd. 
 
 3 12 
 
 even. 
 
 2 13 
 
 odd, 
 
 12 3 
 
 even. 
 
 2 3 1 
 
 even. 
 
 3 2 1 
 
 odd. 
 
 3 12 
 
 even, 
 
 13 2 
 
 odd. 
 
 3 2 1 
 
 odd. 
 
 2 3 1 
 
 even. 
 
 EXERCISES. 
 
 Count the number of inversions in each of the following 
 permutations : 
 
 I. Icdagef, 2. Icagdef. 3. 325941. 
 
 4. 5432. 5. 82917364. 6. 82971364. 
 
 256. Symmetric Functions. An important application of 
 the laws of permutation occurs in the problem, how many 
 different values a function may acquire by permuting the 
 letters which enter into it. We readily find that the ex- 
 pression a^bc takes only the three values a^ic, Wac, and c^ah by 
 permutation. Other expressions do not change at all by per- 
 muting their symbols. 
 
 Def. A Symmetric Function is one which is not 
 changed by permnting the symbols which enter into it. 
 
 EXERCISES. 
 
 Show that the following functions are symmetric : 
 
 1. a -^h -[- c. 2. ale. 
 
 3. a (J -f- c) + J (c + «) + c(« + h). 
 
 4. a^ {h — c)-Y y {fi -a) -\-c'{a- h). 
 

 COMB IN A TIONB: " 285 
 
 CHAPTER II. 
 
 COMBINATIONS. 
 
 257. Bef, The number of ways in whicli it is pos- 
 sible to select a set of s things out of a collection of n 
 things is called the Number of Combinations of s 
 things in n. 
 
 Ex. I. From the three symbols a, h, c, may be formed the 
 
 couplets, 
 
 al, ac, he. 
 
 Hence there are three combinations of 2 things in 3. 
 
 Ex. 2. From a stud of four horses may be formed six dif- 
 ferent span. If we call the horses A, B, C, D, the different 
 
 span will be 
 
 AB, AC, AD, BO, BD, CD. 
 
 Eem. 1. A set is regarded as different when any one of its 
 separate things is different. 
 
 Kem. 2. Combinations differ from permutations in that, 
 in forming a combination, no account is taken of the order of 
 arrangement of things in a set. For instance, ab and ha are 
 the same combination. Hence, we may always suppose the 
 letters or numbers of a combination to be written in alpha- 
 betical or numerical order. 
 
 Notation. The number of combinations of s things in n 
 is sometimes designated by the symbol, 
 
 Cf. 
 
 Problem. To find the nuvther of combinations of s 
 things in n. 
 
 If we form every possible set of s things out of n things, 
 and then permute the s things of each set in every possible 
 way, we shall have all the permutations of n things taken s at 
 a time (§ 252). That is, 
 
 C^ X Ps 
 
 
286 COMBINATIONS. 
 
 express the number of permutations of n things taken 5 at a 
 time. But we have found this number to be 
 
 n (n — l){n — 2) {71 — s -}- 1). 
 
 We have also found 
 
 P, = s\ = 1.2.3.4 s. 
 
 Hence, CP xsl ^ n{n — 1) (n — 2) . . , . (n — 5 + 1), 
 
 and 
 
 ^n _ n (n — 1) ( n — 2)....(n — s-\-l) 
 ^' - 1.2.3. 4.... "1 
 
 = (§228,3); 
 
 a-= ^• 
 
 which is the required expression. 
 
 Rem. For every comlDination of s things which we 
 can take from n things, a combination of n— s things 
 will be left. 
 
 Hence, (7^ =r C^,. 
 
 This formulae may he readily derived from the expression 
 for the number of combinations. For, if we take the equation 
 
 01^= ''• 
 
 si (?i-s)V 
 
 this formula remains unaltered when we substitute n — s for 
 s, and therefore also represents the combinations of 71 — s 
 things in h. 
 
 Def. Two combinations which together contain all 
 the things to be combined are called two Complement- 
 ary combinations. 
 
 EXERCISES. 
 
 1. Write all combinations of two symbols in the five sym- 
 bols, a, h, c, d, e. 
 
 2. Write all combinations of three symbols in the same 
 letters, and show why the number is the same as in Ex. i. 
 
^^ 
 
 ' COMBINATIONS. 287 
 
 3. A span of horses being different when either horse is 
 changed, how many different span may be formed from a stud 
 of 3? Of 7? Of 9? 
 
 4. If four points are marked on a piece of paper, how many 
 distinct lines can be formed by joining them, two and two? 
 How many in the case of n points ? 
 
 From each one of the points can be drawn n — 1 lines to 
 other points; fhen why are there not ?^ [n — 1) lines? 
 
 5. If live lines, no two of which are parallel, intersect each 
 other, how many points of intersection will there be? How 
 many in the case of n lines ? 
 
 6. If n straight lines all intersect each other, how many f^^fC-k^ 
 different triangles can be found in the figure ? 
 
 7. In how many different ways may a set of four things be 
 divided into two pairs ? 
 
 8. In how many ways can a party of four form partners at 
 whist ? 
 
 9. In how many ways can the following numbers be thrown 
 with three dice : 
 
 {a) 1,1,1; {b) 1,2,2; (c) 1,2,3. 
 
 10. A school of 15 young ladies have the privilege of send- 
 ing a party of 5 every day to a picture gallery, provided they 
 do not send the same party twice. How many visits can they 
 make ? 
 
 « 
 Combinations with Repetition. 
 
 358. Sometimes combinations are formed with the liberty 
 to repeat the same symbol as often as we please in any set. 
 
 Example. From the three things a, l, c, are formed the 
 six combinations of two things with repetition, 
 
 aa, ab, ac, hb^ be, cc. 
 
 Problem. To find the number of combinations of s 
 things in n, when repetition is allowed. 
 
 Solution. Let the n things be the first 71 numbers, 
 1, 2, 3, 4, 71. 
 
288 COMBINATIONS, 
 
 Form all possible sets of s of these numbers with repetition, 
 the numbers of each set being arranged in numerical order. 
 
 Let Rs be the required number of sets. Then, in each set. 
 
 Let the first number stand unchanged. 
 Increase the 2d number by 1. 
 " 3d " " 2. 
 " " 4th '' '' 3. 
 
 <t tt gth « t( g / 
 
 We shall then have Rs sets of s numbers, each without rep- 
 etition. 
 
 Example. From the numbers 1, 2, 3 are formed with repetition, 
 
 111, 13, 13, 33, 33, 33, 
 Then, increasing the second numbers by 1, we have 
 13, 13, 14, 33, 24, 34. 
 
 The greatest possible number in any set after the increase 
 will be n -\- s — 1, because the greatest number from which 
 the selection is made is n, and the greatest quantity added is 
 s —1. Hence all the new sets- will consist of combinations of 
 s numbers each from the n -{- s — 1 numbers, 
 
 1, 2, 3, 4, .... ?i .... /i + 5 — 1. {a) 
 
 No two of these combinations can be the same, because then 
 two of the original combinations would have to be the same. 
 Hence the new sets are all different combinations of s numbers 
 from the n -\- s—1 numbers {a). Therefore the number of 
 combinations cannot exceed the quantity C^. 
 
 Conversely, if we take all possible combinations of s differ- 
 ent numbers in n -{■ s — 1, arrange each in numerical order, 
 and subtract 1 from the second, 2 from the third, etc., we 
 shall have different combinations from the first n numbers 
 with repetitions. Hence the number of combinations in the 
 second class cannot exceed those of the first class. 
 
 Hence we conclude that the number of combinations of s 
 things in n with repetition is the same as the combinations of 
 s things in n -\- s — 1 without repetition, or 
 
C0MBINATI0N8. 289 
 
 ^n ^ Cf^^-'' ^ (-—7-^) 
 
 __ ^ (yt + 1) ('^ + 2) (n + s — 1) 
 
 ~ 1.^.3.4 s 
 
 ** EXERCISES. 
 
 1. Write all possible combinations of 3 numbers with repe- 
 tition out of the three numbers 1, 2, 3 ; then increase the second 
 of each combination by 1 and the third by 2, and show that 
 we have all the combinations of three different numbers out of 
 1, 2, 3, 4, 5. 
 
 2. How many combinations of 4 things in 4 with repeti- 
 tion? Of /i things in ?^ ? 
 
 In the last question and in the . following, reduce the result to its 
 lowest terms. 
 
 3. How many combinations of n-\-l things m n—1 with 
 repetition ? 
 
 Special Cases of Combinations. 
 
 259. It is plain that 
 
 because each of these combinations consist simply of one of the 
 n things. Hence, also, 
 
 Ot-i = n, 
 
 because in every such combination one letter is omitted. 
 It is also plain that 
 
 because the only combination of n letters is that comprising 
 the n letters themselves. Hence we write, by analogy, 
 
 although a combination of nothing does not fall within the 
 original definition of a combination. 
 
 260. The formulae of combinations sometimes enable us 
 to discover curious relations of numbers. 
 
 1. Let us inquire how we may form the. combinations of 
 
290 COMBINATIONS. 
 
 8 -\- 1 things when we have those of s things. Let the n 
 things from which the combinations are to be formed be the 
 
 letters 
 
 a, bf c, df e,f, g, etc (n in number). 
 
 Let all the combinations of s + 1 of these n letters be writ- 
 ten in alphabetical order. Then : 
 
 1. In the combinations beginning with a, the letter a will 
 be followed by all possible combinations of s letters out of the 
 n — 1 letters b, c, d, etc., of which the number is CT^. 
 
 2. In the combinations beginning with b, the letter b is 
 followed by all combinations of s letters out of the w — 2 let- 
 ters c, d, e, /, etc. Therefore there are C?'^ combinations 
 beginning with b. 
 
 3. In the same way it may be shown that there are C^~^ 
 combinations beginning with c, C^" beginning with d, etc. 
 The series will terminate with a single combination of the last 
 s-f-l letters. 
 
 Since we thus have all combinations of s-{-l letters, we 
 find, by summing up those beginning with the several letters 
 a, b, c, etc., 
 
 CT' + Cr' + Cr^ +....+ (7J = C?„. {a) 
 
 Substituting for the combinations their values, we find 
 
 By the notation (§ 228, 3), all the terms of the first member 
 have the common denominator s\, while the numerators are 
 each composed of the factors of s consecutive numbers. Mul- 
 tiplying both sides by s\ and reversing the order of terms iu 
 the first member, we have 
 
 1.2.3 .'. . . 5 + 2.3.4 5 + 1 + etc. 
 
 etc. etc. 
 
 -\- (n — 8 — V) (tj _ 3) (7i _ 2) 
 
 + {n — s) (w_2)(w — 1) 
 
 — (^^ — ^) { n — 'il)(n — \)n 
 
 ~ s+1 
 
COMBINATIONS. 291 
 
 The student is now recommended to go over the preceding process 
 with special simple numerical values of n and « which he maj select for 
 himself. 
 
 EXAMPLES. 
 
 If w = 5 and 5 = 2, we have 
 
 ^ 1.2 + 2.3 + 3-4 = ^. 
 If w = 7 and s = 3. 
 
 1.2. 3 + 2. 3-4 + 3. 4. 5 + 4. 5-6 = —1^-. 
 If w = 7 and s = 4, 
 
 1.2.3.4 + 2.3.4.5 + 3.4.5.6 = ^li:^:^. 
 
 If w = 9 and 5 = 3, 
 1.2.3 + 2.3.4 + 3.4.5 + 4.5.6 + 5.6.7 + 6.7.8 = ^1^-. 
 
 Prove these equations by computing both members. 
 261. Another curious example is the following: 
 
 Let us have p + q things divided into two sets, the one 
 containing p and the other q things. Then, to form all possi- 
 ble combinations of s things out of the whole p -\- q, we may 
 take : 
 
 Any s things in set p ; 
 
 Or any combination of s — 1 things in set p with any one 
 thing of set q ; 
 
 Or any combination of 5 — 2 things in set p with any com- 
 bination of 2 things in q ; 
 
 Or any combination of s — 3 things in p with any 3 out 
 of q, etc. 
 
 We shall at length come to the combinations of all s things 
 out of q alone. Adding up these separate classes, we shall 
 have: 
 
 C? + Cf_i a + (7f-2 eg + . . . . + (7? C? 1 + Cf. 
 
 This sum makes up all combinations of s things in the 
 whole p + q, and is therefore equal to C'?^*. Putting the 
 numerical expressions for the combinations, we have the 
 theorem : 
 
292 COMBINATIONS. 
 
 If, as an example, we put 5 = 3, ^ = 4, q =h, this theo- 
 rem will give 
 
 9 .^7 _ 4^3. 2 , i:3 5 4 5 .^ 5.4.3 
 1.2.3 ~ 1-2.3 "^ 1.2'l "^ l'l.2 "^ 1.2.3' 
 
 the correctness of which is easily proved by computation. 
 
 EXERCISES. 
 
 1. Write all the combinations of three letters out of the 
 five, a, h, c, d, e, and show that C| of them begin with a, CI 
 with h, and 0% with c, according to the reasoning of § 260. 
 
 2. Prove that C» = (7| + Cf, 
 
 Cl = Ci^ CI 
 
 and in general, CT'^ =: C? + (7?-i. 
 
 In the following two ways : 
 
 (1.) Let all combinations of s letters in the n letters 
 
 ttyh, c, . , . , n 
 
 be formed, their number being (7?. Then suppose one letter 
 added, making the number n -J- 1. The combinations of s 
 letters out of these w + 1 will include the C^ formed from 
 the n letters, plus each combination of the additional (7i + l)*^ 
 letter with the combinations of s — 1 out of the first n letters. 
 
 (2.) Prove the same general result from the formula, 
 
 C. 
 
 • = ©■ 
 
 3. If we form all combinations of 3 things out of 7, how ^ ^ 
 
 many of these combinations will contain a 7, and how many 
 will not? 
 
 4. If we form all the combinations of s letters out of the n 
 letters 
 
 Uj i, c, , , , , n, 
 
 ^ 
 
the formal zero 
 
 combination, 1 = Co 
 
 r'3 
 
 blank. 
 
 cl, 
 
 a, i, c. 
 
 cl 
 
 ab, ac, ic. 
 
 cl, 
 
 ahc, 
 
 COMBINATIONS. 293 
 
 how many of these combinations will contain a, and how many 
 will not ? 
 
 5. In the preceding case, how many of the combinations 
 will contain the three letters «, J, c ? 
 
 263. Theoeem I. The total nuiriber of combinations 
 which can he formed from n things, including 1 zero 
 combination, is 2**. 
 
 In the language of Algebra, 
 
 cj + C5f + cj + .... + C^l + C^ = 2^ 
 
 Proof. Let us begin with 3 things, a, h, c, and let us call 
 
 Then we have 
 
 Number = 1 
 = 3 
 = 3 
 
 = 1. 
 
 Sum = 8 = 23. 
 
 Now introduce a fourth letter d. The combinations out of 
 the four things, a, b, c, d, will consist of the above 8, plus the 
 8 additional ones formed by writing d after each of the above 
 eight. Their number will therefore be 16. 
 
 In the same way, it may be shown that we double the pos- 
 sible number of combinations for every thing we add to the 
 set from which they are taken. We have found, for 
 
 n = 3, Sum of combinations = 8 = 2^ 
 
 n = 4:, " « = 2-8 = 2*: 
 
 n = 5, " " =2.24=25; 
 
 etc. etc. 
 
 which shows the theorem to be general. 
 
 Theorem II. If the signs of the alternate combina- 
 tions of n things be changed, the algebraic sum will be 
 zero. 
 
 In algebraic language, 
 
 CI-CI+CI-CI + etc. ± C^ = 0. {a) 
 
294 COMBINATIONS. 
 
 Proof. If in the formula of § 261, Ex. 2, namely, 
 
 V g — L/ s -r ^ s-1 i 
 
 we put n — l for n, it becomes 
 
 Putting s successively equal to 0, 1, 2, ... . n, we have 
 
 ci = cr' + cr' ; 
 
 pn _ nn-\ , pn-1 pn-\ -, 
 
 Substituting these values in the expression {a), it becomes 
 
 1 _ (i + cT') + {cv + ^n - (^r' + cV) + . . . . 
 ,= 1 _ 1 _ cr' + cv + cr^ - cv -. cr' + etc. 
 
 How far soever we carry this process, all the terms cancel 
 each other except the last. Therefore, if we continue the addi- 
 tions and subtractions until we come to Cn-i ? the sum will be 
 
 Cl-C\-\-C\- etc ± C^-i = ± cVi = ± 1. 
 
 The last term will be ^ C5! = T 1, and will therefore 
 just cancel the sum of the preceding terms. 
 
 Note. Theorem I may be demonstrated by these same formulae, 
 since the sum of all the terms taken positively will be duplicated every 
 time we increase n by 1. 
 
 263. Independent ComUnations. There is a system of 
 combinations formed in the following way : 
 
 It is required to form a combination of s things, by 
 taking one out of each of s different collections. How 
 many combinations can be formed ? 
 
 Let the 1st collection contain a things, 
 
 ii 2d " '' I '' 
 
 " 3d " " c " 
 
 etc. etc. 
 
COMBmATIONS. 295 
 
 Then we may take any one of a things from the jBrst col- 
 lection. 
 
 With each of these we may combine any one of the l things 
 in the second collection. 
 
 With each of these we may combine any one of the c things 
 of the third collection. 
 
 Continuing the reasoning, we see that the total number of 
 combinations is the continued product 
 
 abc .... to 5 factors. 
 
 If the number in each collection is equal, and we call it a, 
 the number of combinations will be a*. 
 
 This form of combinations is that which corresponds most 
 nearly to the events of life, and is applicable to many questions 
 concerning probabilities. For example, if any one of five dif- 
 ferent events might occur to a person every day, tlie number 
 of different ways in which his history during a year might turn 
 out is 5^^, a number so enormous that 255 digits would be re- 
 quired to express it. 
 
 EXERCISES. 
 
 1. A man driving a span of horses can choose one from a 
 stud of 10 horses, and the other from a stud of 12. How 
 many different span can he form ? 
 
 2. It is said that in a general examination of the public 
 schools of a county, the pupils spelt the word scholar in 230 
 different ways. If in spelling they might replace 
 
 ch by c or ^; 
 
 by au, mo, or oo\ 
 
 1 by II', 
 
 a by e, o, u, or ott; 
 r by re ; 
 in how many different ways might the word be spelt ? 
 
 3. If a coin is thrown n times in succession, in how many 
 different ways may the throws turn out ? 
 
 4. If there are three routes between each successive two of 
 the five cities, Boston, New York, Philadelphia, Baltimore, 
 Washington, by how many routes could we travel from. Boston 
 to Washington ? 
 
296 G0MBmATI0N8. 
 
 The Binomial Tlieorem wlieii tlie Power is a 
 Wliole Number. 
 
 364, The binomial theorem (§ 172), when the power is a 
 positive integer, can be demonstrated by the doctrine of com- 
 binations, as follows : 
 
 Let it first he required to form the product of the n 
 hinomial factors, 
 
 (a^ + ^i) (<^2 + ^2) (^3 + ^3) K + ^n)' {a) 
 
 To understand tlie form of the product, let us first study the special 
 case when n = d. Performing the multiplication of the first three fac- 
 tors, the product will consist of eight terms : 
 
 a^x^x^ 4- a^x^x^ + x^x^x^. \ 
 
 + 
 This product is the expression {a) developed when 71 = 3. 
 
 We conclude, by induction, that the entire product {a) 
 when developed in this same way, will be composed of a sum 
 of terms, each term being a product of several literal factors. 
 
 When {a) is thus multiplied out, we shall call the result 
 the developed expression. 
 
 The developed expression has the following properties : 
 
 I. Each term contains n literal factoids, a's and jo's, 
 and no more. 
 
 For, suppose Xi=a^, a^g = Cg, to Xn = an. Then the 
 expression (a) will reduce to 
 
 2'*«l«2^3 ' ' - ' cin, (b) 
 
 and the developed expression must assume the same value ; 
 that is, it must consist of terms each of which reduces to the 
 expression 
 
 a^a^a^ an, (c) 
 
 when we change x into a. Now if it contained any term with 
 either more or less than fi factors, it could not assume this 
 form. 
 
THE BINOMIAL THEOREM. 297 
 
 II. The factors of each term have all the n indices 
 
 1, 2, 3, n. 
 
 For, the index figure of no term is altered by changing x 
 into a, as in I. Hence, if in any term any index figure were 
 missing or repeated, that term would' not reduce to the form 
 (c), whence there can be neither omission nor repetition of 
 any index. 
 
 III. Because each terin has n factors, it must either 
 have 
 
 n factors a; 
 
 71 — 1 factors a and one factor x; 
 n — 2 factors a and two factors x ; 
 In general, a term may have the factor a repeated 
 n — i times, and x repeated i times. 
 
 ly. In a term which contains i factors x, these i factors 
 must be affected with some combination of i indices out of the 
 whole number 1, 2, 3, .... w ; and the n — i «'s must be 
 affected by the complementary combination of n — i indices. 
 We next inquire whether there is a term corresponding to 
 every such combination. Let 
 
 1, 3, 4, 7, ... . 
 be any combination of i indices, and 
 
 2, 5, 6, 8, ... . 
 
 the complementary combination of 7^ — i indices. 
 
 Since the developed expression must be true for all values 
 of a and x, let us put in (a), 
 
 ¥) 
 
 The product {a) will then reduce to the single term, 
 
 x^a^x^x^a^a^x^a^ (e) 
 
 By the same change the developed expression must reduce 
 to this same value, and it cannot do this unless the expression 
 (e) is one of its terms. 
 
 a^ =0, 
 
 ajg = 
 
 «3 =0, 
 
 a^g = 
 
 a^ = 0, 
 
 X^=:0 
 
 a^ =0, 
 
 0^8 = 
 
 etc. 
 
 etc. 
 
298 COMBINATIONS. 
 
 Hence the developed expression must contain a term 
 corresponding to every combination. 
 
 V. Since every combination of i figures out of 1, 2, 3, n 
 
 will, in this way, give rise to a term like (e), containing the 
 symbol a i times, and the symbol x n — i times, there will be 
 C^ such terms. . 
 
 Now suppose «j = <73 = «3 = . . . . «7i = «. 
 
 The expression [a) will then reduce to (a + xY. 
 
 In the developed expression, all the C^ terms containing x 
 i times and a n — i times will now be equal and their sum 
 will reduce to Cf a^~* x^. 
 
 Hence, putting in succession i = 0, i = 1, etc., to i — n, 
 we shall have 
 
 {a+xY = a^+ Cia^-^x-^ ClaP'-^x^ + -\-Cl tax^-'^ + x^. 
 
 Substituting for (7? its value, we shall have 
 
 {a + xY -—a'' + naP'-^x + \^\a''-H^ + .... + {-zTa p^"^ + (-)^% 
 
 which is the Binomial TJieorem, enunciated, but not demon- 
 strated, in Book V, Chapter I. 
 
 Note, If the student has any difficulty in understanding the steps 
 of the preceding demonstration, he should suppose n = Z, and refer the 
 demonstration to the developed expression {a'). 
 
PROBABILITIES. 299 
 
 CHAPTER III. 
 
 THEORY OF PROBABILITIES. 
 
 265. Def. The Theory of Probabilities treats of 
 the chances of the occurrence of events which cannot be 
 foreseen with certainty. 
 
 Notation, Let a bag contain 4 balls, of which 1 is white 
 and 3 black. If a ball be drawn at random from the bag, we 
 should, in ordinary language, say tbat the chances were 1 to 3 
 in favor of the ball being white, or 3 to 1 in favor of its being 
 black. 
 
 In the language of probabilities we say that the probability 
 
 1 3 
 
 of a white ball is t> and that of a black one t* 
 4 4 
 
 In general, if there are m chances in favor of an event, and 
 n chances against it, its probability is • Hence, 
 
 Def. The Probability of an event is the ratio of 
 the chances which favor it to the whole number of 
 chances for and against it. 
 
 Illustrations. If an event is certain, its probability is 1. 
 If the chances for and against an event are even, its prob- 
 ability is ^' 
 
 If an event is impossible, its probability is 0. 
 
 Cor. 1. If the probability that an event will occur 
 is^?, the probability that it will fail is 1—p. 
 
 Cor. 2. A probability is always a positive fraction, 
 greater than and less than 1. 
 
 266. Method of Probabilities. To find the probabihty of 
 an event, we must be able to do two things : 
 
300 PROBABILITIES. 
 
 1. Enumerate all possible ways in luhicli the event 
 may occur or fail, it being supposed that these ways 
 are all equally probable. 
 
 2. Determine how many of these ways will lead to 
 the event. 
 
 If n be the total number of ways, and m the number which 
 
 /yvt 
 
 lead to the event, the probability required is — 
 
 EXERCISES. 
 
 1. A die has 2 white and 4 black sides. What is the prob- 
 ability of throwing a white side ? 
 
 2. A bag contains n balls numbered from 1 to w, the even 
 numbers being white and the odd ones black. What is the 
 probability of drawing a black ball when n is an odd number? 
 What, when n is an even number ? 
 
 3. A bag contains 3?^ + 2 balls, of which numbers 1, 4, 7, 
 etc., are white ; 2, 5, 8, etc., are red; 3, 6, 9, etc., are black. 
 What are the respective probabilities of drawing a white, red, 
 and black ball ? 
 
 Rem. In tlie last example the probabilities are all less than ^ ; there- 
 fore, should one attempt to guess the color of the ball to be drawn, he 
 would be more likely to be wrong than right, no matter what color he 
 guessed. This exemplifies a lesson in practical judgment to be drawn 
 from the theory of probabilities. If there are three or more possible re- 
 sults of any cause, it may happen that the best judgment would be more 
 likely to be wrong than right in attempting to jiredict the result. Thus, 
 if there are three presidential candidates with nearly equal chances, the 
 chances would be against the election of any one that might be named. 
 
 Gamblers of the turf are nearly always found betting odds against 
 every horse that may be entered for a race, though it is certain that one 
 of them will win. 
 
 Hence, if a natural event may arise from a number of causes with 
 nearly equal facility, it is unphilosophical to have any theory whatever 
 of the cause, because the chances may be against the most probable 
 cause being the true one. 
 
 Probabilities deponcling^ upon Combinations. 
 267. Problem i. Two coins are thrown. What are the 
 respective probabilities that the result will be : Both heads ? 
 head and tail ? both tails ? 
 
PROBABILITIES. 301 
 
 At first sight it might appear that the chances in favor of 
 these three results were equal, and that therefore the probabil- 
 ity of each was ^' But this would be a mistake. To find the 
 
 probabilities, we must combine the possible throws of the first 
 
 coin (which call A) with the possible throws of the second 
 (which call B), thiis : 
 
 A, head ; B, head. 
 
 A, head ; B, tail. 
 
 A, tail ; B, head. 
 
 A, tail ; B, tail. 
 
 These combinations are all equally probable, and while 
 there are only one each for both heads and both tails, there are 
 
 two for head and tail. Hence the probabilities are -, -, j* 
 
 The sum of these three probabilities is 1, as it ought always 
 to be when all possible results are considered. 
 
 Prol. 2. Five coins are thrown. What are the respective 
 probabilities: q heads, 5 tails? 
 
 1 head, 4 tails? 
 
 2 heads, 3 tails? 
 etc. etc. 
 
 Let the several coins be marked a, h, c, d, e. Coin a may 
 be either head or tail, making two cases. Each of these two 
 cases of coin a may be combined with either case of b (as in the 
 last example), making 4 cases. 
 
 Each of these 4 cases may be combined with either case of 
 coin c, making 8 cases. 
 
 Continuing the process, the total number of cases for five 
 coins is 2^ = 32. 
 
 Of these 32 cases, only one gives no head and 5 tails. 
 
 There are 5 cases of 1 head, namely: a alone head, h alone 
 head, etc., to e. 
 
 2 heads may be thrown by coins a^h\ a, c, etc. ; J, c ; b, d, 
 etc. ; c, d, etc. ; that is, by any combination of two letters out 
 of the five, «, h, c, d, e. Hence the number of cases is 
 
 Cl = 10. 
 
303 PBOBABILiriES. 
 
 In the same way the number of cases corresponding to 3, 
 4, and 5 heads are, respectively, 
 
 Cl = 10, Cl = 5, Cl = 1. 
 
 Dividing by the whole number of cases, we find the respec- 
 tive probabilities to be 
 
 1 ^ S 1^. 1? A ' ^_ 
 32' 32' 32' 32' 32' 32* 
 
 The following general proposition is now to be proved by 
 the student : 
 
 Theorem. // there are n coins, the probability of 
 throwing s heads and n — s tails is 
 
 91, - ^ ' - 
 
 From this result we may prove the theorem in combina- 
 tions of § 262. If we suppose, in succession, .<? = 0, 5 = 1, 
 s = 2, etc., to s = n, the respective probabilities of head, 
 1 head, 2 heads, etc., will be I /^ 
 
 C} CJ Cl , • , ^ 
 2» ' 2^ ' 2" ' 2^ 
 
 Because the sum of all these probabilities musfl^e unity, 
 
 we find ; 
 
 Proh. 3. Two dice are thrown at backgammon. W,hat are 
 the respective probabilities of throwing 5 and 6 and two 6's ? 
 
 If we call the dice a and b, any number from 1 to 6 on aj 
 may be combined with any number from 1 to 6 on b. There- 
 fore, there are in all 36 possible combinations. " * - 
 
 In order to throw two 6's, a must come 6 and 5' also. 
 Therefore there is only one case for this result, so that its 
 
 probability is ^^ • 
 
 To bring 5 and 6, a may be 5 and b 6, or b b and a 6. So 
 there are two cases leading to this result, and its probability is 
 
 A - JL 
 36 ~ 18* 
 
PB0BABILITIE8. 303 
 
 Note. That 5 and 6 are twice as probable as a double 6 may be 
 clearly seen by supposing that the two dice are thrown in succession. If 
 the first throw is either 5 or 6, there is a chance for the combination 5, 6, 
 but there is no chance for a double 6 unless the first throw is 6. 
 
 Froh. 4. If three dice are thrown, what are the respective 
 
 probabilities that the numbers will be : 
 
 1, 1, 1? 1, 1, 2? 1, 2, 3? 
 
 The solution of this case is left as an exercise for the 
 student. 
 
 Proh. 5. From a bag containing 3 white and 2 black balls, 
 2 balls are drawn. What are the respective probabilities of 
 
 Both balls white? 
 
 1 white and 1 black ? / 
 
 Both black ? 
 
 Since any 2 balls out of 5 may be drawn, the total number 
 of cases is Cj. 
 
 Only one of these combinations consists of two white balls. 
 
 C\ of the cases bring both balls black. 
 
 A white and black are formed by combining any one of the 
 three white with any one of the two black. 
 
 The respective probabilities can now be deduced by the 
 student. 
 
 EXERCISES. 7 
 
 1. It takes two keys to unlock a safe. They are on a 
 bunch with two others. The clerk takes three keys at random 
 from the bunch. What is the probability that he has both the 
 safe keys? 
 
 2. A party of three persons, of whom two are brothers, seat 
 themselves at random on a bench. What are the probabilities 
 («) that the brothers will sit together, {h) that they will have 
 the third man between them ? 
 
 3. If two dice are thrown at backgammon, w^hat are the • 
 probabilities 
 
 («) Of two aces ? 
 
 {h) Of one ace and no more ? 
 
 4. In order that a player at backgammon may strike a cer- 
 
304 PROBABILITIES. 
 
 tain point, the sum of the numbers thrown must be 8. What 
 are his chances of succeeding in one throw of his two dice ? 
 
 5. A party of 13 persons sit at a round table. What is the 
 probability that Mr. Taylor and Mr. Williams will be next to 
 each other? (See § 253.) 
 
 6. An illiterate servant puts two works of 2 volumes each 
 upon a shelf at random. What is the probability that both 
 pair of companion volumes are together? 
 
 7. A gentleman having three pair of boots in a closet, sent 
 a blind valet to bring him a pair. The valet took two boots at 
 random. What are the chances that one was right and the 
 other left ? What is the probability that they were one pair ? 
 
 8. If the volumes of a 3-volume book are placed at random 
 on a shelf, what is the probability that they will be in regular 
 order in either direction ? 
 
 9. A man wants a particular span of horses from a stud 
 of 8. His groom brings him 5 horses taken at random. What 
 is the probability that both horses of the span are amongst 
 them? 
 
 10. From a box containing 5 tickets, numbered 1 to 5, 
 3 are drawn at random. What is the probability that numbers 
 2 and 5 are both amongst them ? 
 
 11. The same thing being supposed, what is the probability 
 that the sum of the two numbers remaining in the box is 6 ? 
 
 12. Of two purses, one contains 5 eagles and another 10 
 dollar-pieces. If one of the purses is selected at random, and 
 a coin taken from it, what is the probability that it is an 
 eagle ? 
 
 13. From a bag containing 3 white and 4 black balls 
 2 balls are drawn. What is the probability that they are of 
 the same color ? 
 
 14. The better of two chess players is twice as likely to win 
 as to be beaten in any one game. What chance has his weaker 
 opponent of winning 2 games in a match of 3 ? 
 
 15. From a bag containing m white and n black balls, two 
 balls are drawn at random. What is the probability that one 
 is white and the other black ? 
 
PROBABILITIES. 305 
 
 1 6. From a bag cont;aining 1 white, 2 red, and 3 black 
 balls, 3 balls are drawn. What is the probability that they are 
 all of different colors ? 
 
 17. li n coins are thrown, what is the chance that there 
 will be one head and no more ? 
 
 18. From a Congressional committee of 6 Republicans and 
 5 Democrats, a sub-committee of 3 is chosen by lot. What is 
 the probability that it .will be composed of two Republicans 
 and one Democrat ? 
 
 Compoiiiid Events. 
 
 268. Theorem I. The prohahility that two independ- 
 ent events will both happen is equal to the produet of 
 their separate probabilities. 
 
 Proof. For the first event let there be 7n cases, of which 
 p are favorable ; and for the second n cases, of which q are 
 favorable. Then, by definition, the respective probabilities 
 
 will be — and - • 
 m n 
 
 When both events are tried, any one of the m cases may be 
 combined with any one of the n cases, making in all rn x n 
 combinations of equal probability. 
 
 The combinations favorable to both events will be those 
 only in which one of the p cases favorable to the first is com- 
 bined with one of the q cases favorable to the second. The 
 number of these combinations is p xq. 
 
 Therefore the probabihty that both events will happen is 
 
 m X n m n' 
 which is the product of the individual probabilities. 
 
 If there are three events of which the probabilities are p, q, 
 'and r, and we wish to find the probability that all three will 
 happen, we may by what precedes regard the concurring of the 
 first two events as a single event, of which the probability is 
 pq. Then the probability that the third event will also con- 
 cur is the product of this probability into r, or 
 
 pqr. V 
 
 30 
 
306 PROBABILITIES. 
 
 Proceeding in the same way with 4, 5, 6, ... . events, we 
 reach the general 
 
 Theorem II. The probability that any number of in- 
 dependent events will all occur is equal to the continued 
 product of their individual probabilities. 
 
 Rem. This theorem is of great practical use as a guide to 
 our expectations. It teaches that if success in an enterprise 
 requires the concurrence of a great number of favorable cir- 
 cumstances., the chances may be greatly against it, although 
 each circumstance is more likely than not to occur. 
 
 This is illustrated by the following 
 
 Example i. A traveller on a journey by rail has 8 connec- 
 tions to make, in order that he may go through on time. 
 There are two chances to one in favor of each connection. 
 What is the probability of his keeping on time ? 
 
 2 
 
 The probability of each connection being - , the probabil- 
 
 o 
 
 ity of successfully making the first two connections will, by the 
 preceding theorems, be I - ) , the first three ( - 1 , and all eight 
 
 ©■ 
 
 28 356 1 , 
 
 3-3 = 6561 = 20' '''^'^^- 
 
 Therefore there are 25 chances to 1 against his going 
 through on time. 
 
 On the other hand, if, instead of any one accident being 
 fatal to success, success can be prevented only by the concur- 
 rence of a series of accidents, the probability of failure may 
 become very small. 
 
 Ex. 2. A ship starts on a voyage. It is an even chance 
 
 that she will encounter a heavy gale. The probability that 
 
 9 
 she will not spring a leak in the gale is r^- If a leak occurs, 
 
 9 
 there is a probability of — that the engine will be able to 
 
 3 
 pump her out. If they fail, the probability is -r that the com- 
 
PROBABILITIES. 307 
 
 partments will keep the ship afloat. If she sinks, it is an even 
 chance that any one passenger will be saved by the boats 
 What is the probability that any individual passenger will be 
 lost at sea ? 
 
 The probability that 
 
 the ship will meet a heavy gale is ^ 
 
 the ship will spring a leak in the gale is ^a 
 
 the engines cannot pump her out is — 
 
 the compartments cannot keep her afloat is 
 
 1 
 
 the boats cannot save the passenger is ^ 
 
 The continued product of these probabilities is ttttjt:, 
 which is the probabiHty that the passenger will be lost. 
 
 369. The preceding theorem as enunciated supposes that 
 the several events are independent , that is, that the probability 
 of the occurrence of any one is not affected by the occurrence 
 or non-occurrence of the others. To investigate what modifi- 
 cation is required when the occurrence of one of the events 
 alters the probability of another of the events, let us distinguish 
 the two events as ihQ first and second. We then reason thus : 
 
 Let the total number of equally possible cases be m, and let 
 p of these cases favor the first event. Its probability will 
 
 then be — • 
 m 
 
 It is certain that the events cannot both happen unless the 
 first one happens. Hence the cases which favor both events 
 can be found only among the p cases which favor the first. 
 Let q of these p cases favor the second event. Then the prob- 
 ability of both events will be — • 
 
 In case the first event happens, one of the p cases which 
 
308 PROBABILITIES, 
 
 favor it must occur, and the probability of the second event 
 
 will then be ^. Then 
 p 
 
 Probability of both events = -^ = — x -• Hence, 
 
 Theorem. TJve prohdbility that two events will hoth 
 occur is equal to the prolfdbility of the first event multi- 
 plied by the probability of the second, in case the first 
 occurs. 
 
 By continuing the reasoning to more events, we reach the 
 general 
 
 Theorem. The probability that a number of events 
 will all occur is equal to the product 
 
 i X Prob. of second in case first occurs. 
 Prob. of first \ x Prob. of third in case first two occur. 
 
 ( X Prob. of fourth in case first three occur, 
 etc. etc. etc. 
 
 Example. Prom a bag containing 2 white and 3 black 
 balls, 2 balls are drawn. What are the probabilities (1) that 
 both balls are white, (2) that both are black? 
 
 This problem has already been solved, but we are now to 
 see how the answers may be reached by the last theorem. It 
 is evident that we may suppose the two balls drawn out one 
 after the other, and the probabilities of their being white or 
 black will be the same as if both were drawn together. 
 
 I. Both balls white. The probability that the first ball 
 2 
 drawn is white is -• If it really proves to be white, there will 
 
 be left 1 white and 3 black balls. In this event, the probability 
 
 that the second also will be white is - 
 
 4 
 
 Hence the probability that both are white is 
 
 2 11 K.')' 
 
PROBABILITIES. 309 
 
 II. Both halls hlach. Applying the same reasoning, we 
 find for the probability of this case, 
 3 1 _ ^ 
 5 ^ 2 ~ lO' 
 
 EXERCISES. 
 
 1. Two men embark in separate commercial enterprises. 
 The odds in favor of one are 3 to 2 ; in favor of the other, 2 
 to 1. What are the probabilities (1) that both will succeed? 
 (2) that both will fail? 
 
 2. The probability that a man will die within ten years is 
 
 ~, and that his wife will die is — r- What are the respective 
 
 probabilities that at the end of ten years, 
 
 («) Both are living ? 
 
 l(i) Both are dead? 
 
 (y) Husband living, but wife dead? 
 
 {6) Husband dead, but wife living? 
 
 2 
 
 3. The probability that a certain door is locked is - • The 
 
 o 
 
 key is on a bunch of 4. A man takes 2 of the four keys, and 
 goes to the door. What are the chances that he will be able or 
 unable to go through it ? 
 
 4. Two bags contain each 4 black and 3 white balls. A 
 person draws a ball at random from the first bag, and if it be /*• 
 white he puts it into the second bag, mixes the balls, and then 
 draws a ball at random. What is the probability of drawing 
 a white ball from each of the bags ? 
 
 5. If a Senate consists of m Democrats and n Eepublicans, 
 what is the probability that a committee of three will include 
 2 Democrats and 1 Republican ? 
 
 6. A bag contains 2 white balls and 5 black ones. Six i 
 people. A, B, 0, 1), E, F, are allowed to go to the bag in alpha- 
 betical order and each take one ball out and keep it. The 
 first one who draws a white ball is to receive a prize. What 
 are their respective chances of winning? 
 
 Note. A's chance is easily calculated, because he has the draw from 
 all 7 balls. 
 
 V. 
 
310 PROBABILITIES 
 
 In order that B may win, A must first fail. Therefore, to find B's 
 probability we find (1) the probability that A fails, (2) the probability that 
 if A fails then B will win. We then take the product of these probabili- 
 Ues. 
 
 In order that C may gain the prize, (1) A must fail, (2) B must fail, 
 (3) C himself must gain. So we find the successive probabilities of these 
 occurrences. 
 
 Continuing to F, we find that he cannot win unless the 5 men before 
 him all miss. He is then certain to gain, because only the two white 
 balls would be left. 
 
 7. Two men have one throw each of a coin. X offers a 
 prize if A throws head, and if he fails, but not otherwise, B 
 may try for the prize. If both fail, X keeps the prize himself. 
 What are the respective chances of the three men having the 
 prize ? 
 
 8. A and B are alternately to throw a coin until one of 
 them throws a head and becomes the winner. If A has the 
 first throw, what are their respective chances of winning ? 
 
 9. A crowd of n men are allowed to throw in the same way 
 for a prize, in alphabetical order, the game ceasing as soon as a 
 head is thrown. What are the respective chances of the con- 
 testants ? 
 
 10. Three men take turns in throwing a die, and he who 
 first throws a' 6 wins. What are their respective chances? 
 
 11. If 4 cards are drawn from a pack of 52, show that the 
 probabiUty that there will be one of each of the four suits is 
 
 39 26 13 
 5l'50*49* 
 
 12. One purse contains 5 dimes and 1 dollar, and another 
 contains 6 dimes. 5 pieces are taken from the first purse and 
 put into the second, and after being mixed 5 are taken from 
 the second and put into the first. Which purse is now most 
 likely to contain the dollar ? 
 
 13. Of two purses, one contains 4 eagles and 2 dollars, the 
 other 4 eagles and 6 dollars. One being taken at random, and 
 a coin drawn from it, what are the respective probabilities 
 that it is an eagle or a dollar ? 
 
PROBABILITIES. 311 
 
 Cases of Unequal Probability. 
 
 -^270. Bef. If two or more possible events are so 
 related that only one of them can happen, they are 
 called Mutually Exclusive Events. 
 
 Theorem. The prohdbility that some one of several 
 exclusive events, we care not which, will occur, is equal 
 to the sum of their separate probabilities. 
 
 Proof. Let there be m possible and equally probable cases 
 in all; let p of these cases be favorable to one event, q to the 
 
 P Q T 
 
 second, r to the third, etc., so that — , — , — , are the re- 
 spective probabilities. 
 
 Since only one of the events is possible, the p cases which 
 favor one must be entirely different from the q cases which 
 favor the second, and these cases p + q must be entirely differ- 
 ent from the r which favor the third, etc. 
 
 Hence there will be ^j + 5' -f r + etc. , cases which favor some 
 one or another of the events. Hence the probability that some 
 one of these events will occur is 
 
 p -\- q -\- r -\- etc. 
 ' m ' 
 
 which is equal to the sum of the probabilities, 
 
 par, 
 - + — + --]- etc. 
 m m m 
 
 Rem. If the concurrence of some two events, say the first 
 and second, had been possible, some one or more of the p cases 
 which favor the first would have been found among the q cases 
 which favor the second. Then the whole number of cases 
 which favored either event would have been less than p-\-q, 
 and the probability that one of the two events would happen 
 less than the sum of their respective probabihties. 
 
 271. Ge]!^eral Problem. To find the probability that 
 an event of luhich the probability on any one trial is p, 
 will happen exactly s times in n trials. 
 
313 PROBABILITIES. 
 
 Tliis problem is at the basis of some of the widest applica- 
 tions of the theory of probability to practical questions, espe- 
 cially those associated with life and fire insurance. The con- 
 ditions which it implies are therefore to be fully comprehended. 
 
 We may conceive a trial to mean giving the event an oppor- 
 tunity to happen. The simplest kind of trial is that of throw- 
 ing a coin or die. At each throw, any side has an opportunity 
 to come up. Then, if we throw 50 pieces, or which amounts 
 to the same thing, throw the same piece 50 times, there will 
 be 50 trials; and we may inquire into the probability that a 
 given side will be thrown exactly 9 times in these trials. 
 
 The same conception occurs in another form if we have 50 
 men, each of whom has an equal chance of dying within 
 5 years. Waiting to see if any one man will die in the course 
 of the 5 years is a trial, so that there are 50 trials in all, and 
 we may inquire into the probability that 9 of the men will die 
 during the trials, just as in the case of 50 throws of a die. 
 
 Let us distinguish the several trials by the letters 
 a, h, c, d, e, . , » . n, 
 which must be n in number. 
 
 1. In order that the event may not happen at all, it must 
 fail on every one of the n trials. The probability of this 
 (§ 268, Th. II) is (1 — pY- This is therefore the probability 
 that it will not happen at all. 
 
 Because the probability of the event happening on any one 
 trial is p, the probability of its failing is 1 — p- We now 
 compare the possible results. 
 
 2. The event may happen once on any one of the n trials, 
 a, l, c, etc. In order that it may happen only once, it must 
 fail on the other n — 1 trials. The probability that it will 
 happen on any one trial, say e, and also fail on the remaining 
 n — 1 trials is, by the same theorem, 
 
 p (1 -p)n-l. 
 
 Because there are n trials on which it may equally happen, 
 the probabiHty that it will happen once and only once is 
 np{l — j9)^-i. 
 
I 
 
 PROBABILITIES. 313 
 
 3. The event may happen twice on any two trials ont of tlio 
 n trials. In order that it may happen twice only, it must fail 
 on the other n — % trials. Taking any one combination, say 
 
 Happen on t, d; 
 
 Fail on a, c, e, . , , , n, 
 
 the probability is p^ (1 —jp)"'~K 
 
 But it may happen twice on any combination of two trials 
 out of the n trials, a, h, c, . . . . n. Because these combina- 
 tions are mutually exclusive (§ 270), the total probability of 
 happening twice is 
 
 4. In general, in order that the event may happen just s 
 times, it must happen on some combination of s trials, and fail 
 on the complementary combination of n — s trials. Tlie 
 probability on any one combination is p' (1 — pY~^ and there 
 are C? such combinations. Hence the general probability of 
 happening s times is 
 
 C'sfil-pY-'. (a) 
 
 If there is on each trial an equal chance for and against 
 the event, then J3 = - and 1—p = ^- The probability of 
 the event happening s times then becomes 
 
 2n' 
 
 This case corresponds to that already treated in § 267, 
 Problem 2, and the result is the same there found. 
 
 EXERCISES. 
 
 I. A die having two sides white and four sides black is 
 thrown 5 times. What are the respective probabilities of a 
 white side being thrown 1, 2, 3, 4, and 5 times ? 
 
 Note. Here p, the probability of a white side on one throw, is | , and 
 
 o 
 2 
 
 1 — j9 = ., • The number n of trials is 5. 
 o 
 
314 PBOBABILITIES. 
 
 2. Of 6 healthy men aged 50, the probability that any one 
 will live to 80 is j- What is the probability that three or 
 
 more of them will live to that age ? 
 
 3. A chess-player whose chances of winning any one game 
 from his opponent are as 2 to 1, undertakes to win 3 games 
 out of 4. What is the probability that he will be able to do it? 
 
 Note. It would be a fallacy to suppose that the probability required 
 is that of winning exactly 3 games, because he will equally win if he 
 wins all four games. 
 
 272. Events of Maximum ProlaMlity. Returning to the 
 general expression {a), let us inquire what number of times 
 the event is most likely to occur on n trials. The required 
 number is that value of s for which the probability 
 
 is the greatest. 
 
 If we call Pg the probability that the event will happen 
 exactly s times, and if s is to be the number for which the 
 probability is greatest, we must have 
 
 . -Ps > Ps-\, 
 Ps > Ps+\' 
 
 Substituting for these quantities the corresponding forms 
 of the expression (a), which is equal to Ps, we have 
 
 C^'sps (1 -pY-s > CUp"-^ (1 -pY-'^\ ) . ,^. 
 
 ^?i?*(l -pY-s > (7?+ii?*+i (1 - pY-'-\ ) 
 The general formula for C? in § 257 gives 
 
 ^n tl — S + 1 ^n 
 0« = G5-1, 
 
 
 (0) 
 
 Hence we have, by dividing both terms of the first in- 
 equality {b) by C?-ij[?«-i (1 —pY~', 
 
 ; P> 1-p- 
 
PROBABILITIES. 315 
 
 • 
 Multiplying by s, this becomes 
 
 np — sj) + p > s — sjJ. 
 
 luterchaaging the members and reducing, we have 
 
 s<p(/^ + l). {d)- 
 
 Now divide the second inequality {h) by C^ p^ (1 —pY-^\ 
 and reducing by the second equation (c), we have 
 ^ ^ n — s 
 
 Multiplying by 5 + 1 and reducing, we find 
 
 s> p(n + l)-l. (e)_ 
 
 Comparing the inequalities {d) and (e), we see that s lies 
 between the two quantities p {n -f 1) and p {n + 1) — 1, ; 
 that is, J >* 
 
 s is the greatest whole number in p {n -\- 1). 
 
 If the number of trials 71 is a large number, and p is a small 
 fraction, p{n -{- 1) and pii will differ only by the fraction p. 
 We shall then have, very nearly, 
 
 5 = pn. 
 That is : 
 
 Theorem I. TJ^e most probable number of times that 
 an event will happen on a great number of trials is the 
 product of the number of trials by the probability on 
 each trial. 
 
 Example. If a life insurance company lias 6000 members, 
 and the probability that each member will live one year is on 
 
 the average — , then the most probable number of deaths 
 
 during the year is 100. 
 
 Eem. It must not be supposed that in this case the num- 
 ber of deaths is likely to be exactly 100, but only that they 
 will fall somewhere near it. 
 
 There is a practical rule for determining what deviation 
 must be guarded against, the demonstration of which requires 
 more advanced mathematical methods than those employed in 
 this chapter. It is : 
 
316 PROBABILITIES. 
 
 Theorem II. Deviations from the most probable num- 
 ber of deaths, equal to the square root of that number, 
 will be of frequent occurrence. 
 
 Deviations much greater than this square root will 
 be of infrequent occurrence, and deviations more than 
 twice as great will be rare. 
 
 Examples. In a company of which the probable annual ^ 
 number of deaths is 10, the actual number will commonly fall 
 between 10 — VlO and 10 + VlO, or between 7 and 13. It 
 will very rarely happen that the number of deaths is as small 
 as 4 or as large as 16. 
 
 If the company is so large that the most probable number 
 of deaths is 100, the actual number will commonly fall between 
 100 — VlOO and 100 + VlOO, or between 90 and 110. 
 
 If the most probablQ number of deaths is 1000, the actual 
 number will commonly range between 968 and 1032. 
 
 We now see the following result of this theorem: 
 
 The greater the number of dearths to be expected, the 
 greater will be the probable deviation, but the less will be 
 the ratio of this deviation to the whole number of deaths. 
 
 Examples. The reductions of the cases just cited are 
 shown as follows : 
 
 )ected number 
 of deaths. 
 
 10 
 
 Probable 
 deviation. 
 
 3 
 
 Ratio of deviation 
 to expected number. 
 
 0.33 
 
 100 
 
 10 
 
 0.10 
 
 1000 
 
 32 
 
 0.03 
 
 Application to Life Insurance. 
 
 273. At each age of human life there is a certain proba- 
 bility that a person will live one year. This probability di- 
 minishes as the person advances in age. 
 
 It is learned from observation, on the principle described in 
 the preceding section, that events in a vast number of trials 
 are likely to happen a number of times equal to the product of 
 their probability on each trial, multiplied by the number of 
 trials. 
 
PROBABILITIES. 317 
 
 Therefore, by dividing the whole number of times the event 
 has happened by the whole number of trials, the quotient is 
 the most probable value of the probability on one trial. 
 
 Example. If we take 50,000 people at the age of 25, and 
 record how many of them are alive at the end of one year, this 
 is making 50,000 trials whether a person of that age will live 
 one year. 
 
 If 49,650 of them are alive at the end of the year, and 350 
 are dead, we would conclude : 
 
 Probability of living one year, ..... 0.993 
 Probability of dying within the year, . . 0.007 
 
 The probability for all ages may be determined by taking a 
 great number of infants, say 100,000, and counting how many 
 die in each year until all are dead. If n are living at the age 
 y, and n' at the age y -\-l, then the probability of dying 
 
 within one year after the age y will be , and that of 
 
 living will be — • 
 
 It is not, however, necessary to wait through a lifetime to 
 reach this conclusion. It is sufficient to find from observation 
 what proportion of the people of each age die during any one 
 year. Suppose, for instance, that the census of a city is taken, 
 and it is found that there are 2500 persons aged 30, and 2000 
 aged 50. At the end of a year another inquiry is made to 
 ascertain how many are dead. It is found that 20 of the 30 
 year old people, and 30 of the 50 year old people have died. 
 This would show : 
 
 At age 30, probability of dying within 1 year = 0.008. 
 " 50, " " " " = 0.015. 
 
 This same probability being obtained for every year of life, 
 the probability of living 1 year at all ages would be known. 
 Then a table of mortality could be formed. 
 
 A table of mortality starts out with any arbitrary number 
 of people, generally 100,000, at a certain age, frequently 10 
 years. It then shows how many of these people will be living 
 at the end of each subsequent year until all are dead. The 
 following is a specimen of such a table. 
 
818 
 
 PB0BABILITIE8. 
 
 
 
 
 Table of Mortality. 
 
 
 
 Ages. 
 
 Living. 
 
 Dying. 
 
 Prob. of 
 
 surviving 
 a year. 
 
 Prob. of 
 dying 
 within 
 
 the year. 
 
 Ages. 
 
 Living. 
 
 Dying. 
 
 Prob. of 
 
 surviving 
 a year. 
 
 Prob of 
 1 dying 
 ! within 
 
 the year. 
 
 10 
 
 I 00000 
 
 442 
 
 .99558 
 
 .00442 
 
 60 
 
 58373 
 
 1677 
 
 .97127 
 
 .02872 
 
 II 
 
 99558 
 
 407 
 
 .99591 
 
 .00408 
 
 61 
 
 56696 
 
 1760 
 1849 
 
 .96895 
 
 .o3io4 
 
 12 
 
 9Qi5. 
 9^766 
 
 385 
 
 .99611 
 
 .00388 
 
 62 
 
 54936 
 
 .96634 
 
 .03365 
 
 i3 
 
 376 
 
 .99619 
 
 .oo38o 
 
 63 
 
 53087 
 
 1936 
 
 .96353 
 
 .03646 
 
 14 
 
 98390 
 
 379 
 
 .99614 
 
 .00385 
 
 64 
 
 5ii5i 
 
 2014 
 
 .96062 
 
 .03937 
 
 i5 
 
 980 u 
 
 396 
 
 .99595 
 
 .00404 
 
 65 
 
 49137 
 
 2080 
 
 .95766 
 
 .04233 
 
 i6 
 
 97615 
 
 426 
 
 .99563 
 
 .00436 
 
 66 
 
 47057 
 
 2i38 
 
 .95456 
 
 .04543 
 
 ;? 
 
 97189 
 
 f^ 
 
 .99517 
 
 .00482 
 
 % 
 
 44919 
 42733 
 4o5o9 
 
 2186 
 
 .95133 
 
 .04866 
 
 96720 
 
 .99396 
 
 .00542 
 
 2224 
 
 .94795 
 
 .o52o4 
 
 19 
 
 96195 
 
 58 1 
 
 .oo6o3 
 
 69 
 
 2268 
 
 .94401 
 
 .05598 
 
 20 
 
 95614 
 
 621 
 
 .99350 
 
 .00649 
 
 70 
 
 38241 
 
 233 1 
 
 .93904 
 
 .06095 
 .06686 
 
 21 
 
 94993 
 
 645 
 
 .99321 
 
 .00679 
 
 71 
 
 35910 
 
 2401 
 
 .933.3 
 
 22 
 
 94348 
 
 653 
 
 .99305 
 
 .00692 
 
 72 
 
 33509 
 
 2469 
 
 .92631 
 
 .07368 
 
 23 
 
 93695 
 
 65i 
 
 .00694 
 
 73 
 
 3 1040 
 
 253i 
 
 .91846 
 
 .08154 
 
 24 
 
 93044 
 
 647 
 
 .99304 
 
 .00695 
 
 74 
 
 285o9 
 
 2567 
 
 .90995 
 
 .09004 
 
 25 
 
 92397 
 
 ^^7 
 
 .99299 
 
 .00700 
 
 75 
 
 2594a 
 
 2542 
 
 .90201 
 
 .89418 
 
 .09798 
 
 26 
 
 91750 
 
 65 1 
 
 .99290 
 
 .00709 
 
 76 
 
 23400 
 
 2476 
 
 .io58i 
 
 ^? 
 
 91099 
 90431 
 
 668 
 
 .99266 
 
 .00733 
 
 ]l 
 
 20924 
 
 2369 
 
 .88678 
 
 .11321 
 
 686 
 
 .99241 
 
 .00758 
 
 18555 
 
 2247 
 
 .87890 
 
 .12109 
 
 29 
 
 89745 
 
 703 
 
 .99216 
 
 .00783 
 
 79 
 
 i63o8 
 
 2110 
 
 .87061 
 
 .12938 
 
 3o 
 3i 
 
 89042 
 88324 
 
 726 
 
 .99193 
 .99178 
 
 .00806 
 .00821 
 
 80 
 
 81 
 
 14198 
 12229 
 
 \tS 
 
 .86i3i 
 .85092 
 
 .13868 
 .14907 
 
 32 
 
 lll'^l 
 
 733 
 
 .99163 
 
 .00836 
 
 82 
 
 10406 
 
 1672 
 
 .83032 
 .82573 
 
 .16067 
 
 33 
 
 86865 
 
 743 
 
 .99144 
 
 .00855 
 
 83 
 
 8734 
 
 1522 
 
 .17426 
 .18857 
 
 34 
 
 86122 
 
 754 
 
 .99124 
 
 .00875 
 
 84 
 
 7212 
 
 i36o 
 
 .81142 
 
 35 
 
 85368 
 
 768 
 
 .99100 
 
 .00899 
 .00932 
 
 85 
 
 5852 
 
 1 186 
 
 .79733 
 
 .20266 
 
 36 
 
 84600 
 
 ??? 
 
 .99067 
 
 86 
 
 4666 
 
 1014 
 
 .78268 
 
 .21731 
 
 37 
 38 
 39 
 
 838ii 
 
 .99032 
 
 .00967 
 
 V. 
 
 3652 
 
 849 
 
 .76752 
 
 .23247 
 
 83ooo 
 
 83o 
 
 .99000 
 .98972 
 
 .01000 
 
 2803 
 
 689 
 
 .75419 
 
 .24580 
 
 82170 
 
 844 
 
 .01027 
 
 89 
 
 2114 
 
 548 
 
 •74077 
 
 .25922 
 
 40 
 
 81326 
 
 854 
 
 .98949 
 
 .oio5o 
 
 90 
 
 i566 
 
 435 
 
 .72222 
 
 .27777 
 .29708 
 
 41 
 
 80472 
 
 860 
 
 .98931 
 
 .01068 
 
 91 
 
 ii3i 
 
 336 
 
 .70291 
 
 42 
 43 
 
 79612 
 
 888 
 
 '?X 
 
 .01091 
 .01127 
 
 92 
 93 
 
 It 
 
 181 
 
 .68930 
 .66970 
 .643o5 
 
 .3 1 069 
 .33029 
 
 44 
 
 77855 
 
 913 
 
 .98827 
 
 .01172 
 
 94 
 
 367 
 
 i3i 
 
 .35694 
 
 45 
 
 76942 
 
 948 
 
 :ffl 
 
 .01 232 
 
 95 
 
 236 
 
 86 
 
 .63559 
 
 .36440 
 
 46 
 
 75994 
 
 989 
 
 .oi3oi 
 
 96 
 
 i5o 
 
 56 
 
 .62666 
 
 .37333 
 
 IJ 
 
 75oo5 
 
 1029 
 
 .98628 
 
 .01371 
 
 97 
 
 fo 
 
 44 
 
 .53191 
 
 .46808 
 
 7397b 
 
 1067 
 
 .98557 
 .98488 
 
 .01442 
 
 98 
 
 33 
 
 .34000 
 
 .66000 
 
 49 
 
 72909 
 
 1102 
 
 .oi5i 1 
 
 99 
 
 n 
 
 II 
 
 % 
 
 % 
 
 5o 
 
 71807 
 
 ii33 
 
 .98422 
 
 .01577 
 
 100 
 
 6 
 
 4 
 
 K 
 
 % 
 
 5i 
 
 70674 
 
 1167 
 
 .98348 
 
 .01651 
 
 lOI 
 
 2 
 
 2 
 
 
 
 52 
 
 6?3o] 
 * 67052 
 
 1204 
 
 .98267 
 
 .01732 
 
 102 
 
 
 
 .... 
 
 
 
 53 
 54 
 
 I25l 
 
 i3o4 
 
 .98168 
 .98055 
 
 .oi83i 
 .01944 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 Note. The above table is 
 
 hat of 
 
 55 
 56 
 
 5? 
 
 65748 
 64390 
 62976 
 6i5o5 
 
 i358 
 1414 
 1471 
 
 .97934 
 .97804 
 .97664 
 .075 10 
 
 .02065 
 .02195 
 .02335 
 
 the 
 pre 
 
 English Institute of Acti 
 pared betvireen 1862 and 1869 
 
 larios, 
 ,from 
 
 58 
 
 i53i 
 
 .02489 
 .02669 
 
 the 
 
 continued experience of t 
 
 wenty 
 
 59 
 
 59974 
 
 1 601 
 
 "on 
 .97330 
 
 lea 
 
 ling life insurance compan 
 
 es. 
 
PROBADILiriES. 319 
 
 Problem. To find the probability that a person of age a 
 will live to age y. 
 
 Solution. We take from the table the number living at 
 age y, and divide it by the number living at age a. The quo- 
 tient is the probability. 
 
 374. The principle on which the value of a contingent 
 payment is determined is the following : 
 
 Theorem. TJze value of a probable payment is equal 
 to the sum to be paid, inultiplied by the probability that 
 it will be paid. 
 
 Proof. Let there be n men, for each of whom there is a 
 probability p that he will receive the sum s. Then by § 272, 
 Th. I, pn of the men will probably receive the payment, so that 
 the total sum which all will receive will probably be pns. Now, 
 before they know who is to get the money, the value of each 
 one's share is equal. Therefore, to find this value, we divide 
 the whole amount to be received, namely, pns, by the number 
 of men, n. This gives ps as the value of each one's chance, 
 which proves the theorem. * 
 
 Note. In this proof it is tacitly supposed that the pns 
 dollars are as valuable divided among the pn men as divided 
 among all n men. But this, though supposed in mathematical 
 theory, is not morally true. Morally, the money will do more 
 good when divided among all the men than when divided 
 among a portion selected by chance. All gambling, whether 
 by lotteries or games of chance, is in its total effects upon the 
 pecuniary interests of all parties a source of positive disadvan- 
 tage. This disadvantage is treated mathematically by more 
 advanced methods in special treatises. 
 
 EXERCI SES. 
 
 I. Find from the table the probabilities that a person 
 
 a. Aged 30 will live to 70. 
 h. " 30 " " 80. 
 
 c. 
 
 (( 
 
 50 " " 60. 
 
 d. " 60 " " 70. 
 
320 PROBABILITIES. 
 
 e. Aged 70 will live to 80. 
 /. " 80 " " 90. 
 
 g. " 90 '' " 95. 
 
 n. " 95 " " 100. 
 
 2. What age is that at which it is an even chance whether 
 a person aged 40 will be living or dead ? 
 
 3. Show that the probability that a person aged 30 will live 
 to 70 is equal to the product of the probability that he will live 
 to 60 multiplied by the probability that a man aged 60 will 
 live to 70. (Apply the theorem of § 269.) 
 
 4. What premium ought a man of 65 to pay for insuring 
 his life for $7000 for 1 year ? 
 
 5. Ten young men of 25 form a club. What is the proba- 
 bility that it will be unbroken by death for ten years ? 
 
 6. The probability that a planing mill will burn down 
 
 within any oile year is -• What ought an insurance company 
 o 
 
 to charge to insure it to the amount of $3000 for 1 year, for 
 2 years, for 3 years, and for 4 years, respectively ? 
 
 7. If the probability that a house will burn down in any 
 one year is ^, what ought to be the premium for insuring it 
 for s years to the amount of a dollars ? 
 
 Note. In cases like the last two, it is assumed that only one loss 
 will be paid for. 
 
 8. What is the probability that if a man aged 25 marry a 
 wife of 20, they will live to celebrate their golden wedding? 
 
 9. A company insures the joint lives of a husband aged 70 
 and a wife aged 50 for $5000 for 5 years, the stipulation being 
 that if either of them die within that time the other shall be 
 paid the money. What ought to be the premium, no allow- 
 ance being made for interest ? 
 
 10. A man aged 50 insures the life of his wife, aged 35, for 
 
 $10,000 for 20 years, with the promise that the money is not 
 
 to be paid unless he himself lives to the age of 70. What 
 
 ought to be the premium ? 
 
 Note. In computations relating to the management of life insurance, 
 it is always necessary to allow compound interest on all payments. But 
 the above exercises are intended only to illustrate the. application of the 
 theory of probabilities to the subject, and therefore no allowance for in- 
 terest is expected to be made in the answers. 
 
BOOK XI. 
 
 OF SERIES AND THE DOCTRINE OF 
 LIMITS. 
 
 CHAPTER I. 
 NATURE OF A SERIES. 
 
 2*75. Bef. A Series is a succession of terms follow- 
 ing each other according to some general law. 
 
 Examples. An arithmetical progression is a series deter- 
 mined by the law that each term shall be greater than the 
 preceding one by the same amount. 
 
 A geometrical progression is a series subject to the law 
 that the ratio of every two consecutive terms is the same. 
 
 These two progressions are the sinxplest form of series. . 
 
 A series may terminate at some term, or it may continue 
 indefinitely. 
 
 Bef. A series which continues indefinitely is called 
 an Infinite Series. 
 
 Bef. The Sum of a series is the algebraic sum of 
 all its terms. Hence the sum of an infinite series wOl 
 consist of the sum of an infinite number of terms. 
 
 276. The law of a series is generally such that the n^^ 
 term may be expressed as a function of n. 
 For example, in the series 
 
 + Q + 4 + 5 + etc-. 
 
 1.1.11 
 1 IS — 
 
 n ■ 
 21 
 
 the vP^ term is 
 
 w + 1 
 
322 SERIES. 
 
 In the series ^-^ + o ."3 + 3 ."4 + ^^^'> 
 the n*^ term is 
 
 n {n 4- 1) 
 
 Def. The expression for the n*^ term of a series as 
 a function of 7i is called the General Term of the' 
 series. » 
 
 EXE RCI SES. 
 
 Express the 71*^ term of each of the following series : 
 
 2. 1.2 + 3.4 + 5.6 + etc. c^ ir^>-h)J 
 
 X x^ 
 
 3- ^ + 1:2 "^ ry. 3 
 
 « _«^ 
 
 ^' 2.2 "^3.22 "^4.23 
 
 Write four terms of each of the series having the following 
 general terms : 
 
 4t7l^ — 1 
 
 5. The n^^ term to he -r-:. r* 
 
 6. The ^'^^ term to he i (* + 1) {i + 2) a^. 
 
 7. The (w + iy< term to be ^^-7-^-^ ;, , V., — 
 
 ^ ^ • (;^ + 5) (71 + 6) 
 
 8. The (?z-iy< term to he —^ = — 
 
 ^ ^ 1-2 . . . . n 
 
 277. The most common nse of a series is to enable ns to 
 
 compute, by approximation, the values of expressions which it 
 
 is diflScult or impossible to compute directly. Suppose, for 
 
 1 -I- ^ 
 
 example, that we have to compute the value of when x 
 
 J. — X 
 
 is a small fraction, say ^-, and to have the result accurate to 
 
 eight decimals. We shall see hereafter that when x is less than 
 1, we have 
 
CONVERGENCE OF SERIES. 323 
 
 1 4- iC 
 
 = 1 + 2a; + 2:^2 _j_ 2a;3 + etc., «^ infinitum. 
 
 Suppose X = ■— z=: .02. We compute this series thus; 
 
 1 
 
 2 X .02 = .04 
 
 Multiplying by .02, .0008 
 
 " " .000016 
 
 ** " .00000032 
 
 1 02 
 
 Sum = ^l= 1.04081632 
 
 which IS much more expeditious than dividing 1,02 by .98. 
 
 It will be seen that every term we add makes the quotient 
 accurate to one or two more decimals, so that there is no limit 
 to the precision which may be attained by the use of the series. 
 
 If, however, x had been greater than unity, the series would 
 give no result, because the terms %x, 2x% 2a^, would have gone 
 on increasing indefinitely, whereas the true value of the frac- 
 
 1 + ^ 
 tion would have been negative. 
 
 This example illustrates the following two cases of series : 
 
 I. There may he a certain limit to which the sum of 
 the series shall approach, as we increase the nimxher of 
 terms, but which it can never reach, how great soever the 
 number of terms added. 
 
 For example, the series we have just tried, 
 
 1 A A A A 
 
 "^ 50 "^ 502 + 503 + 50"^ + etc., 
 
 1 02 
 approaches the limit ^- , but never absolutely reaches it. 
 
 II. As we increase the number of terrns, the sum 
 may increase without limit, or may vibrate bach and 
 forth in consequence of some terms being positive and 
 others negative. 
 
 These two classes of series are distinguished as convergent 
 and divergent. 
 
324 SERIES. 
 
 Def. A Convergent Series is one of whicli the sum 
 approaches a limit'^s the number of terms is increased. 
 
 Refer to § 213 for an example of infinite series in geometrical pro- 
 gressions which have limits. 
 
 Drf. A Divergent Series is one of which the sum 
 does not approach a limit. 
 
 Examples. The series l + 2-f3 + 4 + etc., ad infimtumj 
 is divergent, because there is no hmit to the sum of its terms. 
 
 The series 1 — 1+ 1 — 1-fl— etc., is divergent, because 
 its sum continually fluctuates between +1 and 0. 
 
 Eem. When we consider only a limited, number of terms, 
 the question of convergence or divergence is not important. 
 But when the sum of the whole series to infinity is to be con- 
 sidered, only convergent series can be used. 
 
 Notation of Sums. 
 
 274. The sum of a series of terms represented by 
 common symbols may be expressed by the symbol 2, 
 followed by one of the terms. 
 
 Example. The expression 
 
 means "the sum of several terms, each represented by «." 
 
 When it is necessary to distinguish the different 
 terms, different accents or indices are affixed to them, 
 and represented by some common symbol. 
 
 Example. The expression 
 2«i 
 means the sum of several terms represented by the symbol a 
 with indices attached ; that is, the sum of several of the quan- 
 tities «j, «3, rtg, a^, etc. 
 
 When the particular indices included in the summa- 
 tion are to be expressed, the greatest and least of them 
 are written above and below the symbol 2. 
 
SIGN OF SUMMATION, 325 
 
 Examples. The expression 
 
 i=15 
 i=5 
 
 means : " Sum of all the symbols ai formed by giving i all in- 
 tegral values from 2 = 5 to i = 15." That is, 
 
 t=15 
 
 2^4=05 + a6 + «7 + «8 + «9 + «I0 + fln + «12 + «fl3 + «14 + Cfl6' 
 i=5 
 
 i=5 
 
 2im means + m + 2m + 3m + 4m + 5m. 
 
 i=0 
 
 ^2 (i,y) means (l,y) + (2,/) + (3,/) + (4, 7). 
 '~Hhj) = 0, 2) + (i, 3) + (i, 4) + (i, 5) + {i, 6). 
 
 i=2 
 
 "s?!! = l! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33. 
 
 n=l 
 
 ~i:i = 7 + 8 + 9 + 10 + 11 = 45. 
 ~li^ = 22 + 32 + 42 + 5« = 54. 
 
 i=2 
 
 EXERCISES. 
 
 Write out the following summations, and compute their 
 values when they are purely numerical : 
 
 j^l n=Q w=6 
 
 I. 2f. 2. 2w(/^— 1). 3. 2M(n + l). 
 
 J=l «.=! n=l 
 
 t=8 n=7 w=6 
 
 4. 2mi. 5. 2w^. 6. 2(/^ + l)(y— 1). 
 
 i=4 /i=4 n=0 
 
 i=4 n=5 n=5 «i 1 
 
 7. 2mi. 8. 2Az2. 9. 2 — — r- 
 
 i=2 n=2 „=0 ^ + 1 
 
 Express the following sums by the sign 2 : 
 
 10. ^o+^i+^2+/^3+V II- l« + 23 + 33 + 48. 
 
 12. 1.2 + 2.3 + 3.4 + 4.5. 13. I +1 + 1 + ^ + 1 
 
326 SERIES. 
 
 CHAPTER II. 
 
 DEVELOPMENT IN POWERS OF A VARIABLE. 
 
 279, Among the most common series employed in math- 
 ematics are those of which the terms are multiplied by the 
 successive powers of some one quantity. 
 
 An example of such a series is 
 
 l + 2z + dz^ + 4:Z^ + 5z^-\- etc., 
 
 in which each coef&cient is greater by unity than the power of 
 z which it multiplies. 
 
 A geometrical progression, it will be remarked, is a series 
 of this kind, in which the terms contain the successive powers 
 of the common ratio. 
 
 The general form of such a series is 
 
 «o + ^1^ + ^2^^ + ^3^^ + ®^c., 
 
 in which the successive coefficients a^, a^, a^, etc., are formed 
 according to some law, but do not contain z. 
 
 Such a series as this is said to proceed according to the 
 ascending powers of the variable z. 
 
 Eem. The sum of a series is often equal to some algebraic 
 expression containing the variable. Conversely, we may find a 
 series the sum of all the terms of which shall be equal to a 
 given expression. 
 
 Bef. A series equal to a given expression is called 
 the Development of that expression. 
 
 To Develop an expression means to find a series 
 the sum of all the terms of which are equal to the ex- 
 pression. 
 
 The most extensively used method of development is that 
 of indeterminate coefficients. 
 
INDETEBMINATE COEFFICIENTS. 327 
 
 Method of Indeterminate Coefficients. 
 
 280. The method of indeterminate coefficients is based 
 upon the following principles : 
 
 Let us have two equal expressions, each containing a varia- 
 ble z, and one or both containing also certain indeterminate 
 quantities, that is, quantities introduced hypothetically, and not 
 given by the original problem, the values of which are to be 
 subsequently assigned so as to fulfil a certain condition. 
 
 The condition to be fulfilled by the values of the inde- 
 terminate quantities is that the two expressions containing z 
 and these quantities shall be made identically equal. 
 
 Then, because the equations are to be identically equal, we 
 can assign any values we please to z, and thus form as many 
 equations as we please between the indeterminate quantities. 
 
 If these equations can be all satisfied by one set of values of 
 these quantities, then by assigning these values to them in the 
 original equation, the latter will be an identical one, as required. 
 
 The student should trace the above general method in the following 
 examples of its application. 
 
 281. Theorem I. If a series proceeding according 
 to the ascending powers of a quantity is equal to zero for 
 all values of that quantity, the coefficient of each sepa- 
 rate term must he zero. 
 
 Proof. Let the several coefficients be a^, a^, a^, etc., and 
 z the quantity, so that the series, put equal to zero, is 
 
 a^ + a^z + a^z^ -}- a^t^ -{^ etc. = 0. 
 
 Because the equation is true for all values of z, it must be 
 true when z = 0. Putting z =: 0, it becomes 
 
 «o = 0. 
 Dropping Gq, the equation becomes 
 
 a^z + a„z^ + a^z^ + etc. = 0. 
 Dividing by z, a^ -\- a^z -f- a^'^ + etc. = 0. 
 From this we derive, by a repetition of the same reasoning, 
 a. = 0. 
 
328 SERIES. 
 
 Continuing the process, we find 
 
 ttg = 0, «3 = 0, etc., indefinitely. 
 
 Theorem II. // two series proceeding hy ascending 
 powers of a quantity are equal for all values of that 
 quantity, the coefficients of the equal powers must he 
 equal. 
 
 Proof. Let the two equal series be 
 
 aQ + a^z + a^z^-^-Qic, = I)Q+d^z + d2Z^-{-etc. (a) 
 
 Transposing the second member to the left-hand side and 
 collecting the equal powers of z, the equation becomes 
 
 «o — ^0 + («i —bi)z-{- («2 — bz) z^ + etc. = 0. 
 
 Since this equation is to be satisfied for all values of z, the 
 coefficients of the separate powers of z must all be zero. 
 
 Hence, 
 
 a^ — Jq = 0, «i — h^ = 0, «g — Jg = 0, etc. 
 
 or aQ = T)q, «i = Jj, a^ = Jg, etc. 
 
 Exercise. Let the student demonstrate these last equa- 
 tions independently from {a), by supposing ;2! = 0, then sub- 
 tracting from both sides of {a) the quantities found to be equal ; 
 then dividing by z ; then supposing 2; = 0, etc. 
 
 Rem. The hypothesis that {a) is satisfied for all values of 
 z is equivalent to the supposition that it is an ideiitical equa- 
 tion. In general, when we find different expressions for the 
 same functions of a variable quantity, these expressions ought 
 to be identically equal, because they are expected to be true 
 for all values of the variable. 
 
 Theorem HI. A function of a variable can only he 
 developed in a single way in ascending powers of the 
 variable. 
 
 For if we should have 
 
 Fz=z A^+ A^z + A^z^ + A^z^ + etc., 
 and also Fz = B^ -^ B^z + B^z^ + B^^ + etc., 
 
mDETEBMINATE COEFFICIENTS. 329 
 
 these two series, being each identically equal to Fz, must be 
 identically equal to each other. But, by Th. II, this cannot be 
 the case unless we have 
 
 Aq = Bq, A^ z=z B^, A^ = ^3, etc. 
 
 The coefficients being equal, the two series are really one 
 and the same. 
 
 283. Expansion ly Indeterminate Coefficients. The above 
 principle is applied to the development of functions in powers 
 of the variable. The method of doing this will be best seen 
 by an example. 
 
 1. Develop in powers of x. 
 
 A. ~\~ X 
 
 Let us call the coefficients of the powers of x a^, a^, etc. 
 The series will be known as soon as these coefficients are 
 known. Let us then suppose 
 
 — - — = a^ -\- a^x + a^x^ + a^oi^ + etc. 
 i -j- X 
 
 Here we remark that, so far as we have shown, this equa- 
 tion is purely hypothetical. We have not proved that any 
 such equation is possible, and the question whether it is possi- 
 ble must remain open for the present. We must find whether 
 we can assign such values to the indeterminate coefficients, a^^, 
 a^, «3, etc., that the equation shall be identically true. 
 
 Assuming the equation to be true, we multiply both sides 
 by 1 -\- X. It then becomes 
 
 1 = «o + («o + «i) ^ + («i + «2) ^^ + etc. ; 
 or transposing 1, 
 = «o — 1 + («o + ai)a; + {a^-{-a2)x^+ (^g+^s)^:^ + etc. 
 By Theorem I, the coefficients must be identically zero. 
 Hence, 
 
 cIq — 1 =0, which gives a^ z= 1; 
 
 tti = —a^ = —1; 
 ffg = — «! = 1; 
 
 — «2 = —1; 
 etc. etc. 
 
 «l + «o = ^y 
 
 a 
 
 11 
 
 «l 
 
 a^-\-a^ = 0, 
 
 it 
 
 a 
 
 «2 
 
 ^3 + ttg = 0, 
 
 « 
 
 iC 
 
 H 
 
330 SERIES. 
 
 Substituting these values of the coeflacients in the original 
 equation, it becomes 
 
 -i- = 1— a; + a:2_a;8_|_a4_ etc. 
 
 1 + iC 
 
 This same method can be applied to the development of 
 any rational fraction of which the terms are entire functions 
 of some one quantity. Let us, for instance, suppose 
 
 m + nx + px^ " 
 
 Multiplying by the denominator of the fraction, this equa- 
 tion gives 
 
 a + I?x = ttiAq + {nAQ+mA-i)x + {pAQ+nA^-\-mA^)x^ 
 
 -\- {pA 1 4- '/i^ 2 + niA 3) x^ + etc. 
 
 We now see that when i > 1, the coeflScient of x^ in this 
 equation is mAi + nAi_t -\- pAi^^. 
 
 Equating the coefficients of like powers of x, 
 
 mA n = a, whence A^ ■= —i 
 
 1 ^ ^ m m ^ ' 
 
 mA^-\-nA^-\-pA^=0, " A^ = -^^A^ -'^A^-, 
 
 mA, + nA^+pA^=0, " A^ = - -^^A, -'^A^. 
 
 "We have from the general coefficient above written, when 
 
 *' > 1' A ^ ^ Pa 
 
 Ai =z Ai-t — ^ Ai-2. 
 
 m 7n 
 
 That is, each coefficient after the second is the same 
 linear function of the two coefficients next preceding. 
 
 Such a series is called a Recurring Series. 
 
 EXERCISES. 
 
 Develop by indeterminate coefficients: 
 
UNDETERMINED MULTIPLIERS. 331 
 
 1 — x 
 
 1 + x 
 
 1 + x 
 
 l + 2x-{-3a^ 
 
 1 _ 2^ + Sx^ 
 
 6. 
 
 7. ^r-r^-r-o-.' 8. 
 
 - x 
 1-x 
 
 l — 2x-^x^ 
 1 — x 
 
 1 + 2^ + 3a;*^ l-{-x — a^ 
 
 283. The development of a rational fraction may also be 
 effected by division, after the manner of §§ 96, 97, the opera- 
 tion being carried forward to any extent. 
 
 Example. Develop -^ — 
 
 l + x \l — x 
 
 1— ^ 1 + 2x + 2x^ + 2x^ + etc. 
 
 2x 
 
 2x — 2x^ 
 
 2a;2 + 
 2x^ — 2a^ 
 
 2x^, etc. 
 
 EXE RCISES. 
 
 Develop by division the expressions : 
 
 1 — 2a; 1 + ic 
 
 I. 2. 
 
 1 -\- X 1 —X -[• x^ 
 
 384. Elimination hy Undetermined Multipliers. There is 
 an application of the method of undetermined coefficients to 
 the problem of eliminating unknown quantities, which merits 
 special attention on account of its instructiveness. Let any 
 system of simultaneous equations between three unknown 
 quantities be 
 
 ax -\- ly -\- cz =^ liy ' (1) 
 
 a'x + Vy + cz = h', (2) 
 
 a"x + h"y + c"z = h". (3) 
 
 Can we find two such factors that, if we multiply two of 
 the equations by them, and add the results to the third, two of 
 the three unknown quantities shall be eliminated ? 
 
333 SERIES. 
 
 This question is answered in the following way : 
 
 If there be such factors, let us call them m and n. If we 
 
 multiply the first equation by m, the second by n, and add the 
 
 product to the third equation, we shall have 
 
 (am 4- a'ti + «") ^ ) 
 + {hn + h'n + h") y> = hm + h'n + h", (b) 
 
 4- {cm + c'n + c") z ) 
 
 In order that the quantities y and z may disappear from 
 this equation, we must have , ^^ 
 
 im + b'n + 5" = %-- ' '^- 
 cm + c'n + c" = Ol 
 
 Since we have these two equations between the quantities 
 W2 and n, we can determine their values. 
 
 Solving the equations, we find : , 
 
 DC — C 
 
 W = -ITT TT-, 
 
 be — be 
 
 _ rc - b e" 
 
 ^ - be' - b'c ' 
 
 These are the required values of the multipliers. Substi- 
 tuting them in the equation (b), we find that the coefificients 
 of y and z vanish, and that the equation becomes 
 
 r a(b'e"-b"c') + a'(b''e-be") _^ .H ^ 
 L be' -T^ +(i jx 
 
 _ h(b'c"-b"e') + h'{b"c-be") ^ , „ 
 be — be 
 
 Clearing of denominators and dividing by the coeflBicient of 
 tc, we find 
 
 ^ ^ h (b'e" - b"c') + h' {b"c - be") + li" {be' - b'c ) 
 a {b'e" - b"e') + a' {b"e - be") + a" {be' - b'c)' ' 
 
 EXERCISES. 
 
 I. Find the values of y and z by the above process for 
 finding x. 
 
MULTIPLICATION OF SERIES. 333 
 
 For this purpose we may begin with the equation (&) and find values 
 of m and n such that the coefficients of x and z in (b) shall vanish. These 
 values will be different from those given in (c). By substituting them in 
 (6), X and z will be eliminated, and we shall obtain the value of 2^. 
 
 We then find a third set of values of m and n, such that the coeffi- 
 cients of X and y shall vanish, and thus obtain the value of z. 
 
 2. Solve by the method of indeterminate multipliers the 
 exercise 3 of § 140. 
 
 Multiplication of Two Infinite Series. 
 
 284a, Pkoblem. To express the product of the two 
 series 
 
 Uq -f- a^x + a^x^ -f a^x^ + etc., 
 and Iq -\- l)-^x.-\- I^q? -\- b^x^ + etc. 
 
 The method is similar to that by which the square of an 
 entire function is formed (§ 173, 2). 
 
 We readily find the first two terms of the product to be 
 
 The combinations which produce terms in x^ are 
 
 cIqIc^x^ H- aj)^x^ 4- a^h^x^ 
 
 Those which produce terms in x? are 
 
 a^b^T^ + a-J)^x^ + a^l)^o(? + a^h^o^. 
 
 In general, to find the terms in x'^ we begin by multiplying 
 Oq into the term bnX^ of the lower series, and then multiplying 
 each succeeding of the first series by each preceding term of 
 the second, until we end with anioX^. Hence, if we suppose 
 
 Product = Aq + A^x -{- A^x'^ -{- . . . . + AnX^ + etc., 
 
 we shall have, for all values of n, 
 
 An = a^bn + Ct^hn-l -\- a^ln-% + + «n&o- 
 
 By giving nail integral values, we shall form as many values 
 as we choose of An, and so as many terms as we choose of the 
 series. 
 
334 SERIES. 
 
 EXERCISES. 
 
 1. Form the product of the two series: 
 
 oc? oc^ ofi , 
 ^-2! +4!- 61 + "'"•' 
 a? 0^ x^ . 
 
 2. Form the square of each of these series. 
 
 3. Can you, by adding the squares together, show that their 
 sum is equal to unity, whatever be the value of a;? 
 
 To effect this, multiply each coefficient of a;« in the sum of the squares 
 by w ! , substitute for each term its value CT given in § 257, and apply 
 § 262, Th. II. 
 
 285. Series proceeding according to the Powers of Two 
 Variables. Such a series is of the form 
 
 aQ + ^0^ + (^iV + ^0^^ + ^i^y + «2.y^ + etc., 
 
 in which the products of all powers of x and y are combined. 
 
 By collecting the coeflBcients of each power of x, the series will 
 
 become 
 
 «o + f^iV + «22/^ + a^y^ -f 
 
 + (^ + ^2/ + ^22/' + ^32/' + • . . 0^ 
 
 + (^0 + c^y + c^y^ + c^y^ + )x^ 
 
 + etc., etc., etc., etc. 
 
 Hence, the series is one proceeding according to the powers 
 of one variable, in which the coefficients are themselves series, 
 proceeding according to the ascending powers of another 
 variable. 
 
 Let us have the identically equal series proceeding accord- 
 ' ing to the ascending powers of the same variables, 
 
 A^+A^y + A^y^^,.,, 
 -^{B,-\-B,y^B,y^ + ..,.)x 
 + (Co + C,y+ C,f + ,...)x^ 
 + etc., etc., etc. • 
 
 Since these series are to be equal for all values of x, the 
 coefficients of like powers of x must be equal. Hence, 
 
SERIES. 335 
 
 O'o + «!«/ + «22/^ + etc. = ^0 + ^iV + ^22/^ + ^'tc. 
 ^0 + ^1^ + *33/^ + etc. = ^0 + ^iV + B^y'' + etc. 
 etc. etc. 
 
 Again, since these series are to be equal for all values of y, 
 we must have 
 
 ttQ = Aq, tti = A^, a^ = A2, etc. 
 
 ^0 = Bq, h^ = B^, &2 = ^2, etc. 
 
 etc. etc. etc. 
 
 Hence, in order that two series proceeding according 
 to the ascending powers of two variables may he identi- 
 cally equal, the coefficients of every like product of the 
 powers must he equal. 
 
336 SERIES. 
 
 CHAPTER III. 
 SUMMATION OF SERIES 
 
 Of Figurate Numbers. 
 
 286. The numbers in the following columns are formed 
 according to these rules : 
 
 1. The first column is composed of the natural numbers, 
 1, 2, 3, etc. 
 
 2. In every succeeding column each number is the sum of 
 all the numbers above it in the column next preceding. 
 
 Thus, in the second column, the successive numbers are : 
 
 1, 1 + 2 = 3, 1 + 2 + 3 = 6, 1 + 2 + 3+4 = 10, etc. 
 
 In the third column we have 
 
 1, 1 + 3 = 4, 1 + 3 + 6 = 10, etc. 
 
 1 
 
 1 
 
 2 1 
 
 3 1 
 
 3 4 1 
 
 6 5 1 
 
 4 10 6 (A) 
 10 15 7 
 
 5 20 21 
 15 35 
 
 6 35 
 21 
 
 7 etc. etc. etc. 
 
 It is evident from the mode of formation that each number 
 is the difference of the two numbers « 
 
 next above and below it in the col- ^ , 
 
 umn next following. • • • 
 
 The numbers 1, 3, 6, 10, etc., in • • • • 
 
 the second column are called trian- 
 gular numbers, because they repre- jv'= 1+2+3+4+5. 
 
SERIES. 337 
 
 sent numbers of points which can be regularly arranged over 
 triangular surfaces. 
 
 The numbers 1, 4, 10, etc., in the third columns are called 
 pyramidal numbers, because each one is composed of a sum 
 of triangular numbers, which being arranged in layers over 
 each other, will form a triangular pyramid. 
 
 All the numbers of the scheme are called figurate num- 
 bers. 
 
 The numbers in the i^^ column are called figurate numbers 
 of the i^^ order. 
 
 387. If we suppose a column of I's to the left of the first 
 column, and take each line of numbers from left to right in- 
 clined upward, we shall have the successive lines 1, 1 ; 1, 2, 1 ; 
 1, 3, 3, 1, etc. These numbers are formed by addition in the 
 same way as the binomial coefficients in § 171, 2. We may 
 therefore conclude that all the numbers obtained by the pre- 
 ceding process are binomial coefficients, or combinatory expres- 
 sions. This we shall now prove. 
 
 Theorem. The n^ nuniber in the i^^ column is equal 
 to CT^-^ or to 
 
 n{n-\-l)(n + 2) ....(n + i-l) 
 
 1.2-3 i ' ^ ^ 
 
 Proof. Because the com.binations of 1 in any number are 
 equal to that number, we have, when i = 1, 
 
 71*^ number in 1st column = ?^ = Ci, 
 
 which agrees with the theorem. 
 
 When i = 2, we have, by the law of formation of the 
 numbers, 
 
 n^^ number in 2d column = C\ -{- Ci + Ci -^ . . . . -\- Cu 
 which, by equation (a) (§ 260, 3), is equal to C2^\ 
 
 Therefore the successive numbers in the second column, 
 found by supposing n = l, n = 2, etc., are 
 
 ci ci, ct,,... or'. 
 
 23 
 
338 SERIES. 
 
 Since the n*^ number in the third column is equal to the 
 sum of all above it in the second, we have 
 
 ni^ number in 3d column = Cl+ Cl+C\-\- CV^ = CV\ 
 
 which still corresponds to the theorem, because, t^hen i = 3, 
 n-]-i — l = n-\-2. 
 
 To prove that the theorem is true as far as we choose to 
 carry it, we must show that if it is true for any value of i, it is 
 also true for a value 1 greater. Let us then suppose that, in 
 the r^ column the first n numbers are 
 
 y^r /-yV+l ryr+2 ^r+n-1 
 
 Since the n*^ number in the next column is the sum of 
 these numbers, it will be equal to 
 
 which is the expression given by the theorem when we suppose 
 4 = r + 1. 
 
 Now we have proved the theorem true when i = d; there- 
 fore (supposing r = 3) it is true for i = 4. There/ore (sup- 
 posing r = 4) it is true for i = 5, and so on indefinitely. 
 
 If in the general expression (1) we put i = 2, we shall 
 have the values of the triangular numbers ; by putting i = 3, 
 we shall have the pyramidal numbers, etc. Therefore, 
 
 The n^^ triangular number = — ^— 5— • 
 
 The n^ pyramidal number = ^^(^ + 1) (^ + ^ i. 
 
 By supposing n = l,2, 3, 4, etc., in succession, we find 
 the succession of triangular numbers to be 
 
 1-2 2^ 4^ 
 1-2' 1.2' 1.2' ' 
 
 and the pyramidal numbers, 
 
 1-2.3 2;. 3^ 3.4.5 
 1.2.3' 1.2-3' 1:2:3' ^^'' 
 which we readily see correspond to the values in the scheme (A). 
 
8EBIE8. 330 
 
 Enumeration of Triangular Piles of Shot. 
 
 288. An interesting application of the preceding theory is 
 that of finding the number of cannon-shot in a pile. There 
 are two cases in which a pile will con- 
 tain a tigurate number: 
 
 I. Elongated projectiles, in which 
 each rests on two projectiles below it. 
 
 II. Spherical projectiles, each rest- 
 ing on three below it, and the whole 
 forming a pyramid. 
 
 Case I. Elongated Projectiles, Here 
 the Yertex of a pile of one vertical layer will be formed of one 
 shot, the next layer below of two, the third of three, etc. 
 Hence the sum of n layers from the vertex down will be the 
 n*^ triangular number. 
 
 It is evident that the number of shot in the bottom row is 
 equal to the number of rows. Hence, if w be this number, 
 and N the entire number of shot in the pile, we shall have, 
 
 _ n {n + 1) 
 2 * 
 If the pile is incomplete, in consequence of all the layers 
 above a certain one being absent, we first compute how many 
 there would be if the pile were complete, and subtract the 
 number in that part of the pile which is absent. 
 
 Example. The bottom layer has 25 shot, but there are 
 
 only 11 layers in all. How many shot, are there? 
 
 25- 26 
 K the pile were complete, the number would be — - — 
 
 Z 
 
 There being 14 layers wanting from the top, the total number 
 of shot wanting is — ~ — Hence the number in the pile is 
 
 _ 25.26-14-15 _ (14 + 11) (15 + 11) -14. 15 
 
 2 - 2 
 
 = il(ll±iM:_iI) ^ ,,0. 
 
340 SERIES. 
 
 Note. This particular problem could have been solved more briefly 
 by considering the number of shot in the several layers as an arithmetical 
 progression, but we have preferred to apply a general method. 
 
 EXERCISES. 
 
 1. A pile of cylindrical shot has n in its bottom row, and r 
 rows. How many shot are there ? 
 
 2. From a complete pile having Ji layers, s layers are re- ' 
 moved. How many shot are left ? 
 
 3. A pile has n shot in its bottom row, and m in its top 
 row. How many rows and how many shot are there? 
 
 4. A pile has p rows and k shot in its top row. How many 
 shot are there ? 
 
 ' 5. Explain the law of succession 
 of even and odd numbers in the se- 
 ries of triangular numbers. 
 
 6. How many balls are necessary 
 to fill a hexagon, having n balls in 
 each side ^ 
 
 Note. In the adjoining figure, 
 
 289. Case II. Pyramid of Balls. If a course of balls 
 be laid upon the ground so as to fill an equilateral triangle, 
 having n balls on each side, a second course can be laid upon 
 these having n — 1 balls on each side, and so on until we 
 come to a single ball at the vertex. 
 
 Commencing at the top, the first course wiU consist of 1 
 ball, the next of 3, the third of 6, and so on through the tri- 
 angular numbers. Because each pyramidal number is the 
 sum of all the preceding triangular numbers, the whole num- 
 ber of baUs in the n courses will be the n^^ pyramidal number, 
 or 
 
 ^ _ n(n + l)(n-\- 2) 
 - 1.2.3 
 
 EXERCISES. y 
 
 I. How many balls in a triangular pyramid having 9 balls 
 on each side ? 
 
SERIES. 341 
 
 2. If from a triangular pyramid of n courses h courses be 
 removed from the top, how many balls will be left ? 
 
 3. How many balls in the frustum of a triangular pyramid 
 having n balls on each side of the base and m on each side of 
 the upper course ? 
 
 Sum of the Similar Powers of an Arithmetical 
 Progression. 
 
 390. Put a^, the first term of the progression ; 
 £?, the common difference; 
 n, the number of terms ; 
 m, the index of the power. 
 
 It is required to find an expression for the sum, 
 
 < + («i + ^^ + iP'i + ^d)^ + +[a^-\-{n-l)d\^y 
 
 which sum we call Sm' 
 
 Let us put, for brevity, a^^ a^, a^, a^, , , , , an for the sev- 
 eral terms of the progression. Then 
 
 «2 = «i + d, 
 
 a^ = a^ -\- 2d = a^ + d, 
 
 an = a^ + {n — l)d = an-i + d. 
 
 Raising these equations to the (m-f 1)^'^ power, and adding 
 the equation an+i = cin -^ d, we have 
 
 am+i = «m+i 4. (,^ + 1) afd + ^^^^^^^^ a^-^d^ + etc. 
 am+i = «m+i + (m + 1) afd + ^''''^l^'^ af-^d^ + etc. 
 «m+i = fl,m+i j^{rn + l)(^d+ ('^-^^)'^ af-H^ + etc. 
 
 arn^ = «ri +{m + l) ar^d + ^;|^ C"'^ + 
 
 etc. 
 
 If we add these equations together, and cancel the common 
 
 terms, «^+i + «^+i + + a^^\ which appear in both 
 
 members, we shall have 
 
342 SERIES. 
 
 fl^Jl = «m+l + (m + 1) dS^^ + ^-^ d^^Sm-l 
 
 , (m + l)m(m — l) ,.-, , 
 
 + ^ ^^^^3 d^Sm-2, etc. 
 
 From this we obtain, by solving with respect to Sm, 
 
 '^- = ife)V - f '^'^'»-' - tS:/^ <?^^.-ete., (.) 
 
 which will enable us to find S'm when we know S^, S2, . . » . 
 Sm-ij that is, to find the sum of the n*^ powers when we know 
 the sum of all the lower powers. It will be noted that S^^ 
 means the sum of the arithmetical series itself, as found in 
 Book YII, Chap. I ; and that Sq = n, because there are n 
 terms and the zero power of each is 1. 
 By § 209, Prob. V, 
 
 To find the sum of the squares, we put m = 2, which gives 
 
 291. The simplest application of this expression is given 
 by the problem: 
 
 To find the sum of the squares of the first n natural 
 nurrdbers, namely, 
 
 12 + 22 + 32 + 42 + ^n\ 
 
 Here 6? = 1, a^ = n, etc., 8^ = 1+2 ....+n = ^-^^^H}^ 
 
 so that (3) gives 
 
 o _ {n + 1 )» — 1 n{n-^l) n 
 ^2- 3^ ^ 3' 
 
 Noting that n + 1 is a factor of the second member, we 
 may reduce this equation to 
 
 g^^n{n + inzn + l)^ (4) 
 
 which is the required expression for the sum of the squares of 
 the first n numbers. 
 
8EBIE8. 343 
 
 292. To find the sum of the cubes of any progression, 
 we put 7W = 3 in the equation (2), which then gives 
 
 ^^ = —a^ - 1 rf^. - ^s, - \ <ps,. (6) 
 
 Applying this as before to the case in which «i, a 2, «3, 
 etc., are the natural numbers, 1, 2, 3, etc., we find 
 
 (^ + 1)^-1 
 
 ^3- 4 
 
 -2^2-^1-1^0 
 
 
 
 {n + 1)^-1 
 ~ 4 
 
 4 
 
 n{n + 1) 
 2 
 
 n 
 ~4 
 
 Separating the factor 7^ + 1 and then reducing, this equa- 
 tion becomes 
 
 ^ r ^(y^ + l) "] 
 
 But ^ ^ is the sum of the natural numbers 
 
 1 + ^ + 3 + etc., 
 
 and S^ being the sum of the cubes, we have the remarkable 
 relation, 
 
 13 + 23 + 33 + + 723 _ (1 + 2 + 3 -f _^ ^)2. 
 
 That is, the sum of the cubes of the first n numbers is 
 equal to the square of their sum. 
 
 We may verify this relation to any extent, thus : 
 When 71=3, l^ + gs = i + g = 9 = (1 + 2)2. 
 When n = ^, l^+B^ + S^ = 1 + 8+27 = 36 = (1 + 2 + 3)2. 
 When 71 = 4, 13 + 23 + 33 + 43 = 1 + 8 + 27 + 64 = 100 = (1 + 2 + 3 + 4)2. 
 etc etc. etc. etc. 
 
 393. Enumeration of a Rectangular Pile of Balls. The 
 preceding theory may be applied to the enumeration of a pile 
 of balls of which the base is rectangular and each ball rests on 
 four balls below it. Let us put p, q, the number of balls in 
 two adjacent sides of the base. 
 
344 SERIES. 
 
 Then the second course will have p — 1 and q — 1 balls 
 on its sides ; the third p — ^ and q — 2, and so on to the top, 
 which will consist of a single row oi p — q + 1 balls (suppos- 
 ing p > q). The bottom course will contain pq balls, the next 
 course {p — 1) (§' — 1)^ etc. The total number of balls in the 
 pile will be 
 N = pq + {p-l){q-l) + (p-2)(q-'^) + .... + (p-q + l). (6) 
 
 To find the sum of this series, let us first suppose p=-q, 
 and the base therefore a square. We shall then have 
 
 N' = q^-\-{q-lY+(q-^f-\-.,,,+l, 
 which is the sum of the squares of the first q numbers. 
 Therefore, by § 291, (4), 
 
 ^, ^ g(g + l)(2^+l) ^^^ 
 
 Next let us put r for the number by which p exceeds q in 
 the general expression (6). This expression will then become 
 JSr=q{q + r) + (2'-l) {q-l^r) + (q-2) fe-2 + r) + • . • . 
 
 + (1 + ^) 
 = ?' + (g - 1)2 + (g - 2)2 + . . . . + 22 + 1 
 
 + [? + (^ - 1) + to -2) + . . . . + 1] r 
 ^ q(q-,l)i2q-,l) ^qjq_±_l)^ ^^ ^^^^ ^^^ 
 
 __ g(g + l)(3r + 2g + l) 
 6 
 
 EXERCISES. 
 
 1. Find the sum of the first 20 numbers, 1 + 2 + 3+ 
 
 + 20, then the sum of their squares, and the sum of their 
 cubes, by successive substitutions in the general equation (2). 
 
 2. Express the sum and the sum of the squares of the first 
 r odd numbers, namely, 
 
 1 + 3 + 5 + . . . . + (2r - 1), 
 and 13 + 32 _|_ 52 _|_ + (2r — 1)2. 
 
 3. Express the sum of the first r even numbers and the 
 sum of their squares, namely, 
 
 2 + 4 + 6 + + 2r, 
 
 and 22 + 42 + 62 + + (2r)2. 
 
SERIES. 345 
 
 4. A rectangular pile of balls is started with a base of p 
 balls on one side and q on the other. How many balls will 
 there be in the pile after 3 courses have been laid ? How 
 many aiter s courses ? 
 
 5. Find the value of the expression 
 
 2 (a + Ja; + cx% 
 
 6. Find the value of 
 
 "2 (fl^ + ^« + cx% 
 x=l 
 
 2Q4:, To find the sum of n terms of the series 
 
 1 + 1 + 1+....+ 1 
 
 1-2 ^2.3 ^ 3.4 ' n{n-{- 1) 
 
 Each term of this series may be divided into two parts, 
 
 thus: 
 
 _1__1_1 1 _ 1 1 
 
 1-2 ~" 1 2' 2.3 ~ 2 3' 
 
 1 _ 1 1_^ 
 
 n{n -{- 1) ~ n n -\-l 
 
 Therefore the sum of the series is 
 
 in which the second part of every term except the last is can- 
 celled by the first part of the term next following. Therefore 
 the sum of the n terms is 
 
 1 /w 
 
 1 — 
 
 n-\-l n-\-l 
 If we suppose the number of terms n to increase without 
 
 limit, the fraction — -— will reduce to zero, and we shall have 
 n -{-1 
 
 j^ + 273 + 3:4 + etc., ad infinitum = 1. 
 
 This is the same as the sum of the geometrical progression, 0+7+5 
 
 -* 4 o 
 
346 SERIES. 
 
 + etc., ad infinitum. It will be interesting to compare the first few terms 
 of the two series. They are 
 
 1111 I4.I 
 
 111111 
 
 2 ■^4'^ 8 "^16 '^32'^ 64* 
 
 We see that the first term is the same in both, while the next three 
 are larger in the geometrical progression. After the fourth term, the 
 terms of the progression become the smaller, and continue so. 
 
 395. Generalization of the Preceding Result, Let us take 
 the series of which the n^^ term is 
 
 P 
 
 {i + n- 1) (/ + ^ - 1) 
 The series to n terms will then be 
 
 ij ^ {i + 1) (/ + 1) ^ {i + ^) U -f 2) ^ * 
 
 + 
 
 (i + n — 1) 0" + ?i - 1) 
 If we suppose / > i, and put, for brevity, 
 Jc =zj — i, 
 the terms may be put into the form 
 
 ij k \i jr 
 
 P -Pi A 1_V 
 
 (i + 1) (j + 1) - ^ U- + 1 j + ir 
 
 etc. etc. 
 
 , P ^ P I 1 1 \ 
 
 (i-\-n — l){j^n-\-l) ]c\i-\' n — 1 j + n — l/ 
 
 When we add these quantities, the second part of each term 
 will be cancelled by the first part of the ¥^ term next follow- 
 ing, leaving only the first part of the first k terms and the 
 second part of the last Ic terms. Hence the sum will be 
 
 P 
 
 (1 + -1-+ 4._i____l ^L._ 1_4 
 
 \i 4 + 1 y+1 i+ri i+n—l"" j + n—V 
 
SERIES. 347 
 
 Example. To find the sum of n terms of the series 
 
 2-5 ' 3.6 ' 4.7 ' 5-8 (?^ + l)(;^ + 4) 
 
 Each term may be expressed in the form 
 
 2.5~3\2 5/' 
 
 3.6 3\3 6/' 
 
 _L-1A_1) 
 
 4.7~3\4 V' 
 
 1 ^lA ^) 
 
 ^ (7i + 3) 3 \^ w + 3/' 
 
 1 = I (_i L_ V 
 
 (/J + 1) (w + 4) 3 \w + 1 w + 4/ , 
 
 Therefore, separating the positive and negative terms, we 
 find the sum of the series to be 
 
 1/1111 1 
 
 3(2+3+4 + 5+-- + ^ + 
 
 n n -\-l 
 _1_1_ _1 1 1 1 1_\ 
 
 5 6 "** n n + l n + 2 n + 3 w + 4/' 
 
 or, omitting the terms which cancel each other, 
 
 1(1 + 1 + 1 1 1 i_). 
 
 3\2^3^4 n + 2 n -\- 3 n -{- ^ 
 When n is infinite, the sum becomes 
 
 1/1 1 1\ _ 1 13 _ 13 
 3\2 "^3 "^4/ ~ 3*12 "■ 36* 
 
 EXERCISES. 
 
 What is the sum of n terms of the series : 
 
 '' 3^ + 4^+5^ + ^*^- 
 
 111 1 
 
 3.5 ' 5.7 ' 7.9 ^ ^ (2w + 1) (27i + 3} 
 
348 SEBIE8. 
 
 2.2.2 
 
 2-5 + 3-6 + 4.7 + • • • • "^ (?^ + 1) (/^ + 4)" 
 
 '^' 1-3 ^2.4^3-5 • ' ^(?i + 2) 
 
 5. Sum the series 
 
 + 7 ^w — r~Q\ + ®*^-» ^^ *V- 
 
 a(a + 1) ^ (rt + 1) (« + 2) ' (« + 2) (a + 3) 
 
 396. To sum the series 
 
 /S^ = 1 + 2r + 3r2 + 4r3 + etc. 
 Let us first find the sum of n terms, which we shall call 
 8n' Then 
 
 i9n = 1 + 2r + 37-2 + 4r3 ^ ^^»-i. 
 
 Multiplying by r, we have 
 
 r/Sn = r + 2r2 + 3r3 4- 4r* + -f nr^. 
 
 By subtraction, 
 
 (1 — 7-)/^n=:l+r+r2 + r3 + r^-i — ?2r'* 
 
 ■*^ "" '^ - wr^ (§ 212, Prob. V). 
 
 1 — r 
 
 1 — r^ wr^ 
 
 Therefore, 8n=^^_^y ^_^ 
 
 Now suppose w to increase without limit. If r > 1, the 
 sum of the series will evidently increase without limit. 
 
 If r < 1, both r^ and nr^ will converge toward zero as n 
 increases (as we shall show hereafter), and we shall have 
 
 1 
 
 S 
 
 (1-r) 
 
 EXERCISES. 
 
 Find in the above way the sum of the following series to n 
 terms and to infinity, supposing r < 1 : 
 
 1. a + 3ar + 5ar^ + 'Tar^ + (2^^ _ 1) ar'^-t. 
 
 2. 2a + 4«r -f dar^ + Sar^ + 2nar^~\ 
 
 3- ia-\-l))r + {a-\-2b)r^ ■\- + (a + nb) r**. 
 
SEBIE8. 349 
 
 297. Sum the series 
 
 of which the general term is —7 — --tt-? — r—^ * 
 ° w (?^ H- 1) (^^ + 2) 
 
 Let us find whether we can express this series as the sum 
 of two series. Assume 
 
 1 ^ 4. ^ 
 
 n{n + l)(n + 2) n (n -\- 1) ^ {n -\- 1) (?^ + 2) ' 
 
 where, if possible, the values of the indeterminate coefficients 
 
 A and B are to be so chosen that this equation shall be true 
 
 identically. 
 
 Eeducing the second member to a common denominator, 
 
 we have 
 
 1 (A + B) n + 2 A 
 
 n{n-\-l) {n + 2) ~ n{n -Yl){n + 2)' 
 
 In order that these fractions may be identically equal, we 
 must have 
 
 {A-\-B)n-{-2A = 1, identically, 
 
 which requires that we have (§ 281), 
 
 A-\- B = 0, 2A = 1. 
 
 This gives ^ == 2' ^ ~ ~ 2* 
 
 Therefore, 
 
 1 1111 
 
 n{n-{-l){n + 2) 2n{n-\-l) 2 {n -\- 1) {n -\- 2)* 
 
 so that each term of the series {a) may be divided into two 
 terms. The whole senes will then be 
 
 We see on sight, that by cancelling equal terms, the sum of 
 n terms is ^ 1 1 
 
 4 2(/^ + l)(7^ + 2)' 
 
 and the sum to infinity is j* 
 
350 8ERIE8. 
 
 298. Consider the harmonic series 
 
 of which the n*^ term is — This series is divergent, because 
 
 7b 
 
 we may divide it into an unlimited number of parts, each 
 equal to or greater than ^ , as follows : 
 
 1st term : 
 
 = 1, 
 
 >i= 
 
 3d 
 
 term 
 
 1. 
 
 "" 2' 
 
 3d and 4:th terms 
 
 >\: 
 
 etc. etc. 
 
 In general, if we consider the n consecutive terms. 
 
 n + 1^ 71-^%^ ^ 27i' 
 
 the smallest will be ^ , and therefore their sum will be greater 
 
 than — X ^, that is, greater than -• 
 
 Now if in {a) we suppose n to take the successive values, 
 1, 2, 4, 8, 16, etc., we shall divide the series into an unlimited 
 
 number of parts of the form (a), each greater than -• There- 
 fore, the sum has no limit and so is divergent. 
 
 Of Differences. 
 
 299. When we have a series of quantities proceeding ac- 
 cording to any law, we may take the difference of every two 
 consecutive quantities, and thus form a series of differences. 
 The terms of this series are called First Differences. 
 
 Taking the difference of every two consecutive differences, 
 we shall have another series, the terms of which are called 
 Second Differences. 
 
 The process may be continued so long as there are any dif- 
 ferences to write. 
 
SERIES. 351 
 
 Example. In the second column of the following table 
 are given the seven val ues of the expression 
 
 a4 _ i0a;s ^ 30a;2 _ 40a; + 25 = (^ix, 
 for 2; = 0, 1, 2, 3, 4, 5, 6. 
 
 In the third column a' are given the differences, 
 6 _ 25 = — 19, 1 — 6 = — 5, — 14 — 1 = — 15, etc. 
 
 In column a" are given the differences of these differences, 
 namely, 
 
 _ 5 _ (_ 19) = + 14, _ 15 — (- 5) = — 10, etc. 
 
 X (jiX A' A" A'" A«- A- 
 
 +25 
 
 -19 
 1+6 +14 
 
 — 5 —24 
 
 2 +1 —10 +24 
 
 — 15 
 
 3 _14 -—lo +24 
 
 — 25 +24 
 
 4 _39 +14 +24 
 
 — 11 + 48 
 
 5 _50 +62 
 
 + 51 
 6+1 
 
 The process is continued to the fourth order of differences, 
 which are all equal, whence those of the fifth and following 
 orders are all zero. 
 
 It will be noted that the sign of each difference is taken so 
 that it shall express each quantity minus the quantity next 
 preceding. We have therefore the following definitions : 
 
 300. Def, The First Difference of a function of 
 any variable is the increment of the function caused by 
 an increment of unity in the variable. 
 
 The Second Difference is the difference between 
 two consecutive first differences. 
 
 In general, the ii*'* Difference is the difference be- 
 tween two consecutive (n — ly^ differences.- 
 
352 SERIES, 
 
 To investigate the relation among the differences, let us 
 represent the successive numbers in each column by the indices 
 1, 2, 3, etc., and let us put Aj, Ag, Ag, etc., for the values of 
 (jix. We shall then have the following scheme of differences, 
 in which 
 
 a;=Ai-Ao, a;=A2-Ai, a; = A3-A3; 
 
 a: = a;-a;, a- = a;-a';, a- = a';-a'^; 
 
 etc. etc. etc. 
 
 the n^^ order of differences being represented by the symbol A 
 with Qi accents. 
 
 Ao 
 
 A' 
 
 Ai " a; 
 
 a; a; 
 
 A. A'; a; 
 
 A' ^" 
 
 As • a; 
 
 An 
 
 Let us now consider the following problem : 
 To express Ai in terms of A^, Ao, Ao, etc. 
 We have, by the mode of forming the differences, 
 
 Ai = Ao + a;, a; = a; -\- a;, a; = a; + a';, etc. («> 
 
 Ag = Aj + a'i, a; = a'i + Ai, Ag = Ai + a''' etc. 
 
 If in this last system of equations, we substitute the values 
 of Aj, Ai, etc., from the system («), we have 
 
 Ag = Ao + 2a; + a;, a; = a; + 2a; + a;', etc. {h) 
 
 Again, 
 
 A3 = Ag + a;, a; =: a; + a;, a'; = a; + a^, etc. 
 
DIFFERENCES. 353 
 
 Substituting the values of Ag, Ag, etc., from (h), we have 
 
 A3 = Ao -f 2a; + a; 
 
 . + a; + 2a; + ^0 
 
 or A3 = Ao + 3a; + 3a;' + Ao (c) 
 
 A3 
 
 = 
 
 A, 
 
 + 3a; 
 
 + 3a'^ 
 
 +a: 
 
 A's 
 
 = 
 
 A'o 
 
 + 3a: 
 
 + a;' 
 
 
 
 
 
 + a; 
 
 + 2a'; 
 
 + A'J 
 
 a; = a; + 3a; + sa; + aj 
 
 Forming A4 = A3 + A3, etc., we see that the coefficients 
 of Aq, Aq, etc., which we add, are the same as the coefficients 
 of the successive powers of x in raising 1 + cc to the n^^ power 
 by successive multiplication, as in § 171. That is, to form A^, 
 A'^, etc., the coefficients to be added are 
 
 13 3 1 
 1 3 3 1 
 
 14 6 4 1 
 
 and these are to be added in the same way to form Ag, and so 
 on indefinitely. Hence we conclude that if i be any index, the 
 law will be the same as in the binomial theorem, namely, 
 
 Ai = Ao + ^'a; + (I) a; + (i) a;' + etc. ) 
 
 a; = a; + K + (I) a; + (I) A'j + etc. ) 
 
 To show rigorously that this result is true for all values of 
 iy we have to prove that if true for any one value, it must be 
 true for a value one greater. Now we have, by definition, 
 whatever be ^, 
 
 At+i = Ai -f Ai, Aixi = Ai + Ai', etc. 
 Hence, substituting the above value of Aj and Ai, 
 
 At+i = Ao + {i + 1) a; + [(I) + *j a; 
 
 + K)+©]^o+etc. (.) 
 23 
 
354 
 
 SERIES. 
 
 We readily prove that 
 
 ©+(i)=m 
 
 etc. 
 
 3 
 
 etc. 
 
 Substituting these values in (e), the result is the same given 
 by the equation {d) when we put i + 1 for i. 
 
 The form (c) shows the formula to be true for i = 3. 
 
 Therefore it is true for i = 4. 
 
 Therefore it is true for i = 5, etc., indefinitely. 
 
 EXAMPLES AND EXERCISES. 
 
 I. Having given A,, = 7, ^'q = 5, Aq = — 2, and A'", A'% 
 etc. = 0, it is required to find the values of Aj, Ag, A3, etc., 
 indefinitely, both by direct computation and by the formula (d). 
 
 We start the work thus: 
 
 The numbers in column A" are all 
 equal to — 2, because A"' = 0. 
 
 Each number in column A' after 
 the first is found by adding A" or — 3 
 to the one next above it. 
 
 Each value of A^ is then obtained 
 from the one next above it by adding 
 the .appropriate value of A^ , 
 
 This process of addition can be 
 carried to any extent. Continuing it 
 to i = 10, we shall find A,o = —33. 
 
 Next, the general formula (d) gives, by putting A,, = 7, 
 ^0 = ^^ ^0 = ~^> and all following values = 0, 
 
 and the student is now to show that by putting i = 1, i = 2, 
 etc., in this expression, we obtain the same values of Aj, Ag, 
 A3, .... Aio, that we get by addition in the above scheme. 
 
 It is moreover to be remarked that we can reduce the last 
 equation to an entire function of % thus : 
 Ai = 7 + 6i - i2. 
 
 i 
 
 Ai 
 
 Ai 
 
 Ai 
 
 
 
 7 
 
 + 5 
 
 
 1 
 
 + 12 
 
 + 3 
 
 -2 
 
 2 
 
 + 15 
 
 + 1 
 
 -2 
 
 3 
 
 etc. 
 
 — 1 
 
 -2 
 
 4 
 
 
 etc. 
 
 -2 
 
 etc. 
 
 
 
 etc. 
 
DIFFERENCES. 355 
 
 2. Having given Ao = 5, A'^, = —20, A'^ = —30, 
 A'" = + 9, it is required to find in the same way the values 
 of Aj to As, and to express Ai as an entire function of i by 
 formula {d). 
 
 3. On March 1, 1881, at Greenwich noon, the sun's longi- 
 tude was 341° 5' 10". 9 ; on March 2 it was greater by 1° 0' 9".6, 
 but this daily increase was diminishing by 2" each day. It is 
 required to compute the longitude for the first seven days of 
 the month, and to find an expression for its value on the n^^ 
 day of March. 
 
 4. A family had a reservoir containing, on the morning of 
 May 5, 495 gallons of water, to which the city added regularly 
 50 gallons per day. The family used 35 gallons on May 5, 
 and 5 gallons more each subsequent day than it did on the day 
 preceding. Find a general expression for the quantity of 
 water on the n*^ day of May ; and by equating this expression 
 to zero, find at what time the water will all be gone. Also ex- 
 plain the two answers given by the equation. 
 
 Theorems of Differences. 
 
 301. To investigate the general properties of differences, 
 we use a notation slightly different from that just employed. 
 
 If u be any function of x, which we may call (px, so that 
 we put 
 
 u = (px, 
 
 then Au = (p (x -]- 1) — (px. (a) 
 
 Here the symbol A does not represent a multiplier, but 
 merely the words difference of. 
 
 The second difference of u being the difference of the dif- 
 ference, may be represented by AAw. 
 
 For brevity, we put 
 
 A^?^ for AA?/, 
 
 where the index 2 is not an exponent, but a symbol indicating 
 a second difference. 
 
 Continuing the same notation, the n^^ difference will be 
 represented by A^. 
 
356 8EBIE8. 
 
 EXAMPLE. 
 
 To find the successive differences of the function 
 
 u =1 aoi? -\- hx\ 
 
 By the formula {a), we have 
 
 ^u = a(x + lf + l){x + 1)2 — «a;3 - Ix^ ; 
 
 and, by developing. 
 
 Aw = Zax^ + (3« + 2^) 2; + a + J. 
 
 Taking the difference of this last equation, 
 
 A% = 3a (a; + 1)2 + (3a ■\- U) {x -^ 1) + a + b 
 
 — Sax^ — (da + 2b)x-'a — b 
 = 6ax + 6a + 2b, 
 
 Again taking the difference, we have 
 
 A^u = 6a (a; + 1) — 6ax = 6a. 
 
 This expression not containing x, A%, A^u, etc., all vanish. 
 
 EXERCISES. 
 
 Compute the differences of the functions : 
 
 I. a^ -]- mx^ + nx + p, 2. 2x^ + dx^ + 5. 
 
 3. 6a^ + 10a;2 + 15. 
 
 4. In the case of the last expression, prove the agreement 
 of results by computing the values of A^^, Ahi, etc., for x = 0, 
 x=:l, and x = S, and comparing them with those obtained 
 by the method of § 299. The latter are shown in the follow- 
 ing table : 
 
 u = 6:^ + 10a:2 + 15. 
 
 X u Au A^u A^u 
 
 
 
 15 
 
 1 15 
 
 30 50 
 
 2 65 30 
 95 80 
 
 3 145 30 
 240 110 
 
 4 255 
 495 
 
 5 
 
DIFFERENCES. 357 
 
 5. Do tlie same thing for exercise 2, and for tlie function 
 tabulated in § 299. 
 
 302. It will be seen by the preceding examples and exer- 
 cises, that for each difference of an entire function of x which 
 we form, the degree of the function is diminished by unity. 
 This result is generalized in the following theorem : 
 
 The -nP^ differences of the function of' are constant 
 and equal to n ! 
 
 Proof. If ^^ = x'^, we have, by the definition of the sym- 
 bol A, 
 
 i^u = {x -\- 1)^ — a;», 
 
 or Am = nx'^-^ + (^)a:^~^ + etc. 
 
 That is, in talcing the difference, the highest power of 
 x is multiplied by its exponent and the latter is dimin- 
 ished by unity. 
 
 Continuing the process, we shall find the n^^ difference 
 to be 
 
 n{n — l){n — ^) 1 = n\ 
 
 Cor. If we have an entire function of x of the degree n, 
 ax^ + hx^-^ + cx^-2 _|_ etc., 
 
 the {n — ly^ difference of hx!^-^, the {71 — 2Y difference of 
 cx'^~'^, etc., will all be constant, and therefore the n^^ difference 
 of these terms will all vanish. Therefore, the n^^ difference of 
 the entire function will be the same as the n^^ difference of 
 ax"^ ; that is, we have 
 
 A^ {ax"^ + Ix^-'^ + etc.) = a7i ! 
 
 Hence, the vP^ difference of a function of the n*^ de- 1 
 gree is constant, and equal to n ! multiplied by the coeffi- 
 cient of the highest power of the variable. 
 
358 SERIES. 
 
 CHAPTER IV. 
 
 THE DOCTRINE OF LIMITS. 
 
 303. The doctrine of limits embraces a set of principles 
 applicable to cases in which the usual methods of calculation 
 fail, in consequence of some of the quantities to be used van- 
 ishing or increasing without limit. 
 
 We have already made extensive use of some of the princi- 
 ples of this doctrine, and thus familiarized the student with 
 their application, but our further advance requires that they 
 should be rigorously developed. 
 
 Axiom I. Any quantity, however small, may Ibe 
 multiplied so often as to exceed any other fixed quan- 
 tity, however great. 
 
 Ax. II. Conversely, any quantity, however great, 
 may be divided into so many parts that each part shall 
 be less than any other fixed quantity, however small. 
 
 Def. An Independent Variable is a quantity to 
 which we may assign any value we please, however 
 smaU or great. 
 
 Theorem I. If a fraction have any finite numerator, 
 and an independent variable for its denominator, we 
 may assign to this denominator a value so great that 
 the fraction shall he less than any quantity, however 
 small, which we may assign. 
 
 Proof. Let a be the numerator of the fraction, x its de- 
 nominator, and a any quantity, however small, which we may 
 choose to assign. 
 
 Let w be the number of times we must multiply u to make 
 it greater than a. (Axiom I.) We shall then have 
 
 a < na. 
 
 Consequently, - < «. 
 
LIMITS. 369 
 
 Hence, by taking x greater than n, we shall have 
 
 a ^ 
 - < «. 
 
 X 
 
 Example. Let a = 10. Then if we take for a in succes- 
 
 ^^^^' Wo' who' Woo' '^'-^ ^^ ^^'' ^^^^ *" ^^^' 
 
 X > 1,000, X > 100,000, X > 10,000,000, etc., 
 to make — less than a. 
 
 X 
 
 In the language of limits, the above theorem is expressed 
 thus : 
 
 The limit of - , when x is indefinitely increased, is 
 zero. 
 
 Theorem IT. If a fraction have any finite numerator, 
 and an independent variable for its denoininator, we 
 may assign to this denominator a value so small that 
 the fraction shall exceed any quantity, however gi^eat, 
 which we may assign. 
 
 Proof. Put as before - for the fraction, and let A be any 
 
 X 
 
 number however great, which we choose to assign. 
 
 Let ^ be a number greater than ^. Divide a into n parts, 
 and let a be one of these parts ; then 
 
 a = na. 
 
 Consequently, - =z n. 
 
 Therefore, if we take for x a quantity less than a, we shall 
 have 
 
 -> n> A, 
 
 X 
 
 or - y A. 
 
 X 
 
 Eem. If we have two independent variables, x and y: 
 We may make x any number of times greater than y. 
 
360 LIMITS. 
 
 Then we may make y any number of times greater than 
 this value of x. 
 
 Then we may make x any number of times greater than 
 this value of y. 
 
 And we can thus continue, making each variable outstrip 
 the other to any extent in a race toward infinity, without 
 either ever reaching the goal. 
 
 Theorem III. If Tc he any fixed quantity, however 
 great, and a a quantity which we may make as small 
 as we please, we may mahe the product ha less than any 
 assignable quantity. 
 
 Proof. If there is any smallest value of Jca, let it be s. 
 Because we may make a as small as we please, let us put 
 
 «<r 
 
 Multiplying by h, we find 
 
 lea < 5. 
 
 So that ka may be made less than s, and s cannot be the 
 smallest value. 
 
 Def, The Limit of a varialble quantity is a value 
 wliicli it can never reach, but to which it may approach 
 so nearly that the difference shall be less than any 
 assignable quantity. 
 
 Eem. In order that a variable X may have a limit, it must 
 be a function of some other variable, and there must be certain 
 values of this other variable for which the value of JT cannot 
 be directly computed. 
 
 EXAMPLES. 
 
 I. The value of the expression 
 
 X— ^ — ^^ 
 
 ~~ X — a 
 can be computed directly for any pair of numerical values of x 
 and a, except those values which are equal. If we suppose 
 x = a, the expression becomes 
 
LIMITS. 361 
 
 
 
 a — a ~ 0' 
 which, considered by itself, has no meaning. 
 
 2. The sum of any finite number of terms of a geometrical 
 progression may be computed by adding them. But if the 
 number of terms is infinite, an infinite time would be required 
 for the direct calculation, which is therefore impossible. 
 
 3. The area of a polygon of any number of sides, and hav- 
 ing a given apothegm, may be computed. But if the number 
 of sides becomes infinite, and the polygon is thus changed into 
 a circle, the direct computation is not practicable. 
 
 EXERCISE. 
 
 r^x 8 
 
 If we have the fraction, X = ^ _ , show that we may 
 
 7 1 
 
 make x so great that X shall differ from - by less than -— , 
 
 less than — -?^, lesss than j^q^' "^^ '" ^^ indefinitely. 
 
 Notation of the Method of Limits. 
 
 304. Put X, the quantity of which the value is to be 
 found ; 
 
 X, the independent variable on which X de- 
 pends, so that X is a function of x ; 
 
 a, the particular value of x for which we can- 
 not compute X; 
 
 L, the limit of X, or the value to which it 
 approaches as x approaches to a. 
 
 Then the limit L must be a quantity fulfilling these two 
 conditions : 
 
 1st. Supposing X to approach as near as we please to a, we 
 must always be able to find a value of x so near to a that the 
 difference L — X shall become less than any assignable quan- 
 tity. 
 
 2d. X must not become absolutely equal to L, however 
 near x may be to a. 
 
363 
 
 LIMITS. 
 
 Rem. The quantity a, toward which x approaches, may be 
 either zero, infinity, or some finite quantity. 
 
 Example i. Suppose 
 
 X — a 
 
 By § 93, this» expression is equal to 
 
 :c2 + «a; 4- a^, (a) 
 
 except when x-=a. But suppose 6 to be the difference be- 
 tween X and «, so that 
 
 X =: a -\- d. 
 
 Substituting this value in the expression (a), the equation 
 becomes 
 
 x^ — a^ 
 
 = Sa^ + dad + 
 
 Now we may suppose 6 so small that 3aS + 6^ shall be less 
 than any quantity we choose to assign. Hence we may choose 
 
 /jj3 q3 
 
 a value of x so near to a that the value of shall differ 
 
 X — a 
 
 from da^ by less than any assignable quantity. Hence, if 
 
 then 
 
 X 
 L 
 
 a^ 
 
 «« 
 
 X — a 
 3a% 
 
 ^yS /irS 
 
 or Sa^ is the limit of the expression as x approaches a: 
 
 X — a 
 
 Ex. 2. The limit of 
 
 iC+ 1 
 
 , when X becomes indefinitely 
 
 great, is unity. 
 
 For, subtracting this expression from unity, we find the 
 difference to be 
 
 1 
 x-\-\ _ 
 
 By taking x sufficiently great, we may make this expression 
 less than any assignable quantity. (§ 303, Th. I.) Therefore, 
 
 X 
 
 — — r approaches to unity as x increases, whence unity is its 
 
 X — j- i. 
 
 limit. 
 
LIMITS. 363 
 
 Notation. The statement that L is the limit of X as x 
 approaches a is expressed in the form 
 
 Lim. X{x=a) = L. 
 
 The conclusions of the last two examples may be ex- 
 pressed thus : 
 
 />»3 /jf3 oj 
 
 Lim. (a;=o) = 3fl2. Lim. (a;=oo) = 1. 
 
 X — a X -{-1 
 
 Eem. This form of notation is often used for the follow- 
 ing purpose. Having a function of x which we may call X, 
 the form X{x=a) means, " the value of X when a; = a." 
 
 EXAMPLES. 
 
 {x^ + a\x=a) = a^ -\-a. {x^ — aPjicc=a) = 0. 
 
 If we require the limit of a fraction when both terms be- 
 come zero or infinite, divide both terms by some common 
 factor which becomes zero or infinity. 
 
 Rem. If the beginner has any difficulty in understanding the pre- 
 ceding exposition, it will be sufficient for him to think of the limit as 
 simply the value of the expression when the quantity on which it de- 
 pends becomes zero or infinity. 
 
 For instance, Lim. — — :r (x = cc), 
 
 the value of which we have found to be unity, may be regarded as simply 
 the value of the expression, go 
 
 00 +l' 
 Although this way of thinking is convenient, and generally leads to 
 correct results, it is not mathematically rigorous, because neither zero 
 nor infinity are, properly speaking, mathematical quantities, and people 
 are often led into paradoxes by treating them as such. 
 
 EXERCISES. 
 
 Find the limit of 
 
 I. when X approaches infinity. 
 
 X 
 
 2. 
 
 Divide both terms by x. 
 
 T — — — when X approaches infinity. 
 
 7723/ 
 
 when X approaches infinity. 
 
 — ax 
 
364 LIMITS. 
 
 when X approaches infinity. 
 
 1 — ax 
 
 /p2 ^2 
 
 5. when X approaches a. t^ 
 
 X ■^~ cl 
 
 6. when x approaches infinity. 
 
 ct — X 
 
 Properties of liimits. 
 
 305. Theorem I. If two functions are equal, they 
 must have the same limit. 
 
 Proof. If possible, let L and L' be two different limits for 
 the respective functions. Put 
 
 , = \(L-L'), 
 
 so that L and L' differ by 2z. 
 
 Because L is the limit of the one function, the latter may 
 approach this limit so nearly as to differ from it by less than z. 
 
 In the same way, the other function may differ from L' 
 by less than z. Then, because L and L' differ by 2^, the func- 
 tions would differ, which is contrary to the hypothesis. 
 
 Theorem II. The limit of the sum of several func- 
 tions is equal to the sum, of their separate limits. 
 
 Proof. Let the functions be JT, X', X", etc. 
 Let their limits be X, L', L", etc. 
 
 Let their differences from their limits be a, a', a", etc. 
 
 Then X=: L-a, 
 
 X' = L' - a', 
 
 X" = Z" - a", 
 
 etc. etc. 
 
 Adding, we have 
 
 X+X' + X" + etc. = Z+Z' + X" + etc. -(« + «' + «"+ etc.) 
 
 The theorem asserts that we may take the functions so near 
 their limits that the sums of the differences « + «' + ce" + etc. 
 shall be less than any quantity we can assign. 
 
LIMITS. 365 
 
 Let h be this quantity, which may be ever so small ; 
 n, the number of the quantities a, a', a", etc. ; 
 a, the largest of them. 
 
 Because we can bring the functions as near their limits as 
 we please, we may bring them so near as to make 
 
 k 
 « < -, or 7ia <ih. 
 
 Then a + «' + «"+ etc. < na (because a is the largest); 
 
 whence, cc-l- «' + «"+ etc. < h. 
 
 Therefore the sum X+X' -\-X" + Qtc. will approach to 
 the sum L + L' -\- L" + etc., so as to differ from it by less 
 than k. Because this quantity k may be as small as we please, 
 X + i' + Z" + etc. is the limit of X+ X' + X" + etc. 
 
 Theorem III. The limit of the product of two func- 
 tions is equal to the product of their limits. 
 
 Proof. Adopting the same notation as in Th. II, we shall 
 have 
 
 XX' = LL' — aV — a'L + aa\ 
 
 Because L and L' are finite quantities, we may take a and 
 a' so small that aL' + a'L — aa shall be less than any quan- 
 tity we can assign. Hence XX' may approach as near as we 
 please to LL', whence the latter is its limit. 
 
 Cor. 1. The limit of the product of any number of 
 functions is equal to the product of their limits. 
 
 Cor. 2. The limit of any power of a function is equal 
 to the power of its limit. 
 
 Theorem IV. The limit of the quotient of two func- 
 tions is equal to the quotient of their limits. 
 
 Proof. Using the same notation as before, we have for the 
 quotient of the functions, 
 
 X' _ L' -a 
 X - L — a' 
 
 U 
 
 while the quotient of their limits is -j-* 
 
 It 
 
366 LIMITS. 
 
 The dijfference between the two quotients is 
 
 X; __ L' -a' _ La' - L'a 
 L L — a ~~ L{L — a) 
 
 If.L is different from zero, we may make the quantities a 
 and «' so small that this expression shall be less than any 
 
 quantity we choose to assign. Therefore, y- is the limit of 
 
 —p , that is, of -.FF- 
 
 yVl ^71 
 
 306. Problem. To find the limit of as x 
 
 X — Ob 
 
 approaches a. 
 
 Case 1. Whenn is a positive whole number. 
 We have from § 93, when x is different from «, 
 
 Q-n nth 
 
 = ^«-i + a:x^'^ + a^aP-^ + + «^-i. 
 
 x — a 
 
 Now suppose X to approach the limit a. Then a:^~i will 
 approach the limit a'^'\ x^-^ the limit a^-% etc. Multiplying 
 by a, a^, etc., we see that each term of the second member 
 approaches the limit a^~\ Because there are n such terms, 
 we have 
 
 Q^n f^n 
 
 Lim. — ix=a) = naP'~\ 
 
 x — a 
 
 Case II. When n is a positive fraction. 
 
 Suppose n = -, p and q being whole numbers. Then 
 
 p p 
 af^ — a^ _ xQ — a^ 
 X — a ~ X — a 
 
 Let ns put, for convenience In writing, 
 
 x^ = y, a^=b; 
 
 then ic = /, a = b^; 
 
 and 
 
 
 y^-lf 
 
 a^ — a^ yP — h^ 
 
 y — h 
 
 x — a ~ y^—b^ — 
 
 f-m 
 
 y — b 
 
LIMITS, 367 
 
 As X approaches indefinitely near to a, and consequently y 
 to b, the numerator of this fraction (Case I) approaches to 
 pbP~^ as its limit and the denominator to qM'K Hence, the 
 fraction itself approaches to 
 
 . qm-^ q 
 Substituting for b its value or, we have 
 
 Lim. ^Zl^ (^=a) z=z'^-bv-^=^-a^ = ^ oT^ 
 X — a q q q 
 
 Hence the same formulae holds when w is a positive fraction. 
 Case III. When n is negative. 
 
 Suppose n = —p,p itself (without the minus sign) being 
 supposed positive. Then 
 
 of' — aP- x-^ — a-P _« _« 
 
 = = x~ParP 
 
 x — a X — a 
 
 iaP-of\ 
 \ X — a I 
 
 — x~P a~P 
 
 When X approaches a, then x~p approaches a~P, and 
 
 a^ — a^ 
 
 approaches ^«^^ Substituting these limiting values, 
 
 X — a 
 
 we have 
 
 /jflti f^n 
 
 Lim. {x=a) = — a'^paP'^ = —parP~K 
 
 X — d 
 
 Substituting for —p its value n, we have 
 
 Lim. (x=a) = na^'K 
 
 X — a 
 
 Hence, 
 
 Theorem. The formulm 
 
 yU ^W 
 
 Lim. («=«) = naP'^ 
 
 X — a 
 
 is true for all values of n, whether entire or fractional, 
 positive or negative. 
 
368 BINOMIAL THEOREM 
 
 CHAPTER V. 
 THE BINOMIAL AND EXPONENTIAL THEOREMS. 
 
 The Binomial Tlieorem for all Values of the 
 Exponent. 
 
 307. We have shown in §§ 171, 264, how to develop 
 {l-\-xY when /» is a positive whole number. We have now to 
 find the development when n is negative or fractionaL Assume 
 
 (1 H- xY = -^0 + -^1^ + ^8^ + ^3^ + etc., {a) 
 
 Bq, B^y etc., being indeterminate coeJB&cients. Because this 
 equation is by hypothesis true for all values of x, it will remain 
 true when we put another quantity a in place of x. Hence, 
 
 (1 + aY = Bq-\- B^a + B^a^ + B^a^ + etc. (jb) 
 
 Subtracting {h) from (a), and putting for convenience 
 
 X =l-\-Xy A = l-\- a, 
 
 the difference of the two equations {a) and {i) will be 
 
 X^-A'' = B^{x — a) + B., {x^ - a^) +B^(a^- a^) + etc. 
 
 The values we have assumed for X and A give 
 
 X— A = X — a. 
 
 Dividing the left-hand member by X— A, and the right- 
 hand member by the equal quantity x — a, we have 
 
 -x—A = ^i +^«-^-:r^ +^3-,^^^ + etc. 
 
 Kow suppose X to approach a. The limit of the left-hand 
 member will be nA^-\ Taking the sum of the corresponding 
 limits of the right-hand member, we shall have 
 
 nA^-^ = Bi + 2B^a + SB^aJ^ + 4:B^a^ + etc. 
 
 Replace A by its value, 1 + a, and multiply by 1 + a. 
 We then have 
 
BINOMIAL THEOREM. 369 
 
 n{l + aY = ^1 (1 + a) + ^B^a{l + o;) + ^B^a^{l + a) 
 
 + 4^4«3 (!+«) + etc. 
 
 + (3^3 +4^4)«^ + etc. 
 Multiplying the equation (b) by n, we have 
 
 ?2 (1 + «)^ = tiBq + 72^1^ + nB^a^ + ^-53^3. 
 
 Equating the coeflBcients of the like powers of a in these 
 equations (§ 281), we have, first, 
 
 B, = nB,. 
 By putting a = in equation (b)y we find Bq = 1, whence 
 
 Then we find successively, 
 2B, = {n-1) B,, whence B, = "^ B, = ""^^"^^ ^ 
 
 SB, = in-2)B,, <^ B,=-^B,=-J!^;^, 
 
 Substituting these values of B^, B^, Bc^, etc., in the equa- 
 tion (a) and using the abbreviated notation, we obtain the 
 equation 
 
 (1 + xY = 1 + w^ + (I) a;2 + (I) a;3 + etc., (c) 
 which equation is true for all values of n. 
 
 308. There is an important relation between the form of 
 this development when w is a positive integer, as in §§ 171 and 
 264, and when it is negative or fractional. In the former 
 case, when we form the successive factors n — 1, n — 2, 
 n — 3, etc., the n^ factor will vanish, and therefore all the 
 coefficients after that of x"' will vanish. 
 
 But if n is negative or fractional, none of the factors 
 n — If n — 2, etc., can become zero, and, in consequence, the 
 series will go on to infinity. It therefore becomes necessary, 
 in this case, to investigate the convergence of the development. 
 
 If a: > 1, the successive powers of x will go on increasing 
 
 indefinitely, while the coefficients (~), (^V etc., will not go 
 24 
 
370 BINOMIAL THEOBEM. 
 
 on diminishing indefinitely in the same ratio. For, let us 
 consider two successive terms of the development, the ith and 
 the (^ + 1)*^ namely, 
 
 (?).. and {^y^ 
 
 The quotient of the second by the first is 
 
 i 
 
 I n \ in\ _n — I 
 
 X. 
 
 As i increases indefinitely, this coefficient of x will approach 
 the limit —1 (§ 304), while x is by hypothesis as great as 1. 
 Therefore, by continuing the series, a point will be reached 
 from which the terms will no longer diminish. Therefore, 
 
 The development of (1 -\- xY in -powers of x is not con- 
 vergent unless a; < 1. 
 
 In consequence, if we develop {a + ly when n is negative 
 or fractional, we must do so in ascending powers of the lesser 
 of the two quantities, a or h. 
 
 EXAMPLES. 
 
 I. Develop (1 + x)^, or the square root of 1 + a;. 
 Putting n =1 -, we have 
 
 
 ©- 
 
 3V2 / _ _ 1 -i 
 
 3(3 ~ Via -^) 1.1.3 
 
 1.3.3 3.4.6 
 
 3 
 
 \il i V3/ ~ 2.4.6- 
 
 etc. etc. etc. 
 
BINOMIAL THEOREM. 371 
 
 Whence, 
 
 If a; is a small fraction, the terms in x^, a?, etc., will be 
 1 
 
 series will give a result very near the truth. "We therefore ^ 
 conclude : 
 
 much smaller than - x itself, and the first two terms of the 
 
 \ 
 
 The square root of 1 plus a small fraction is approxi' 
 mately equal to 1 plus half that fraction. 
 
 2. To develop VlO. 
 
 We see at once that VlO is between 3 and 4. We put 10 
 in the form 
 
 32 + 1 = 32(1 + i), 
 
 when VlO = 3 (l + ^' 
 
 Then, by the development just performed, 
 
 / 1\^ 111 5 
 
 (^ + 9/ "= -^ "^ 2:9 ~ 8^92 + l6'."9^ ~ 128:9^ + ®*^- 
 
 We now sum the terms : 
 
 1st term, 1.0000000 
 
 2d « = 1st -J- 18, .... + .0555556 
 
 3d « = 2d -^ - 36, ... — .0015432 
 
 4th " = 3d -^ - 18, . . . + .0000857 
 
 5th " = 4th X — 5 -^ 72, - — .0000060 
 
 6th " = 5th X -7-T-90, . + .0000005 
 
 Sum = (1 + ^) = 1.0540926 
 
 Whence, VlO = 3 x sum = 3.1622778 
 
 which may be in error by a few units in the last place, owing 
 to the omission of the decimals past the seventh. 
 
372 EXPONENTIAL THEOREM. 
 
 3. To develop a/8. 
 
 We see that 3 is tlie nearest whole number of the root. So 
 we put 
 
 V8 = V(3' - 1) = 'y/s^ (1 - g) = 3 (1 - s)* 
 
 from which the development may be effected as before. 
 
 EXERCISES. 
 
 1. Compute the square root of 8 to 6 decimals, and from it 
 find the square root of 2 by § 183. 
 
 2. Develop (1 — x)^. 
 
 3. Develop (1 — x)~^, and express the term in rr*. 
 
 Term m x^ = — znrn -w-- — ^ • 
 
 2.4.6 . . . . 2i 
 
 4. Develop r and express the general term. 
 
 (1 + xY 
 
 5. Develop (1 + -) and express the general term. 
 
 6. Develop (1 — x) , and express the general term. 
 
 7. Develop the m^ root oi 1 + m, 
 
 8. Develop {a — l)~% when a <,h, 
 
 9. Develop (1 — x)-^, when ic > 1. 
 
 Because the development will not be convergent in ascend- 
 ing powers of x when a; > 1, we transform thus: 
 
 1-x 
 
 \ xr 
 
 (1\" 
 1 ) 
 
 10. Develop the m^^ power of 1 H 
 
 11. Compute the cube root of 1610 to six decimals. 
 
EXPONENTIAL THEOREM. 373 
 
 12. Develop {Va + V^)" 
 
 13. Using the functional notation, 
 
 ^W = l + (f)- + (f)=>^ + (|)^ + ete., 
 
 multiply the two series, (^ (m) and (w), and show by the for- 
 mulse of § 261 that the product is equal to (p{m -\- n). 
 
 The Exponential Theorem, 
 
 309. Let it be required, if possible, to develop a^ in 
 powers of x, a being any quantity whatever. Assume 
 
 a^=C^ + G^x + C^x' + C^^ + etc. (1) 
 
 to be true for all values of x. Putting any other quantity y in 
 place of X, we shall have 
 
 ay=C, + C^y + C^y^ + C,y^ + etc. (2) 
 
 By the law of exponents we must always have 
 
 a"" X av = a'^+y. 
 Now the value of a^'^v is found by writing x + y for a; in 
 (1), which gives 
 a^^y = C, + G, (x+y) + C^ {x+yY+ 0, {x+yY+eic. (3) 
 
 On the other hand, by multiplying equations (1) and (2), 
 we find 
 
 a-oM = (7o2 + C,C,y + C,C,y^ + C,C,f + etc. 
 
 + Co C^x + C^^xy + Cj O^xy^ + etc. , . 
 
 + C,C,x^ + C,G,x^y + etc. ^^ 
 4- Gq G^a? + etc. 
 
 By § 285, the coefficients of all the products of like powers 
 of X and y must be equal. By equating them, we shall have 
 more equations than there are quantities to be determined, 
 and, unless these equations are all consistent, the development 
 is impossible. To facilitate the process of comparison, we 
 have in equation (4) arranged all terms which are homogeneous 
 in X and y under each other. 
 
374 BINOMIAL THEOREM. 
 
 By putting a; = in (1), we find 
 
 flO = G,y whence C^ = 1. (§ 103.) 
 
 Comparing the tenns of the first degree in x and y in (3) 
 and (4), we find 
 
 Coefficient of x, G^ = 0^0^; 
 
 These two equations are the same, and agree with 0^ = 1; 
 but neither of them gives a value for C\, which must therefore 
 remain undetermined. 
 
 Comparing the terms of the second degree, we find, by de- 
 veloping {x + yY, 
 
 C, {x^ + 2xy + y') = C,x^ + C^^xy + C,y% 
 which gives ^^z = ^-i) 
 
 Whence C^ = -\ C^K 
 
 Comparing the terms of the third order in the same way, 
 we have 
 
 C,{a^+3xiy+dxy^+y^) = C,a^-{-C,C,x^y+C^C^xy^+C,y% 
 
 which gives 30^ = C^C^ = I C^^; 
 
 whence C^ z= ^-^ (7,3. 
 
 If the successive values of C follow the same law, we shall 
 have 
 
 and in general, On = — G^^. (5) 
 
 Let us now investigate whether these values of G render 
 the equations (3) and (4) identically equal. 
 
 Let us consider the corresponding terms of the n^^ degree, 
 n being any positive integer. In (3) this term will be 
 
 On {x + yY, 
 
EXPONENTIAL THEOREM. 375 
 
 Expanding, it will be 
 Cn U" + nx^-^ y + (I) iC»-2 y^ + (|) a:^"^ ^^ + etc. J (6) 
 
 In (4) the sum of the corresponding terms will be, putting 
 Co =1, 
 Cn^ + C^Cn-i3^-'y-^G^Cn-2x^-''y^+C^Cn-zx^-^y^ + ^ic. (7) 
 
 The first terms in the two expressions are identical. 
 
 The comparison of the second terms gives 
 
 C 
 nCn = C^Cn-h whence On = -^ On-i. 
 
 ft 
 
 This corresponds with (5), because (5) gives 
 
 and if we substitute this value of Cn-i in the preceding ex- 
 pression for Cn, it will become 
 
 p cl___cl 
 
 ^''- n(n -1)1 ~ i^i!* 
 which agrees with (5). 
 
 The third terms of (6) and (7) being equated give 
 
 iP^ 
 
 iy o Cn 
 
 Substituting the values of Cn, C^, and Cn-% assumed in the 
 general form (5), we have 
 
 (n\ y^n _ 1 1 nn 
 
 and we wish to know if this equation is true. 
 
 Multiplying both sides by n ! and dropping the common 
 factor C\, it becomes 
 
 (i)= 
 
 2! {n-%)V 
 which is an identical equation. 
 
 In the same way, the comparison of the following terms in 
 (6) and (7) give 
 
 ln\ _ n\ hi\ __ n\ 
 
 Vdl ~ 3! (n-d)V \l) - 4! (7* -4)1' ^^'' 
 
376 
 
 EXPONENTIAL THEOREM. 
 
 all of which are identical equations. Hence the conditions of 
 the development, namely, that (6) and (7), and therefore (3) 
 and (4), shall be identically equal, are all satisfied by the values 
 of the coefficients G in (5). Substituting those values in (1), 
 the development becomes 
 
 ««^ = 1 + G^x + ^ G,^x^ + j^ G^^x^ + etc. (8) 
 
 This development is called the Exponential Theorem, 
 as the development of {a + iY is called the binomial theorem. 
 
 310, The value of G^ is still to be determined. To do 
 
 this, assign to x the particular value ^r • Then the equation 
 (8) becomes * 
 
 111 
 
 ^ + ^ + 1.2 + 1.2.3 "^1.2.3.4 
 
 4- etc., ad inf, (9) 
 
 The second member of this equation is a pure number, 
 without any algebraic symbol. We can readily compute its 
 approximate value, since dividing the third term by 3 gives 
 the fourth term, dividing this by 4 gives the fifth, etc. Then 
 
 1 + 1 = 
 
 2.000000 
 
 
 -1.2 = 
 
 .500000 
 
 
 r- 1-2.3 = 
 
 .166667 
 
 
 - 1.2.3.4 = 
 
 .041667 
 
 
 - 1.2.3.4.5 = 
 
 .008333 
 
 
 -1.2.3.4.5.6 = 
 
 .001389 
 
 
 - 1.2.3.4.5.6.7 = 
 
 .000198 
 
 
 - 1.2.3.4.5.6.7.8 = 
 
 .000025 
 
 
 - 1.2.3.4.5.6.7.8.9 = 
 
 .000003 
 
 Sum ( 
 
 )f the series to 6 decimals. 
 
 2.718282 
 
 This constant number is extensively used in the higher 
 mathematics and is called the Naperian base.* It is repre- 
 sented for shortness by the symbol e, so that e = 2.718282.... 
 
 The last equation is therefore written in the form 
 
 j_ 
 
 * After Baron Napier, the inventor of logarithms. 
 
EXPONMTtit^mEOREM. 377 
 
 Eaising to the C^^^ power, we have a = e^^ Hence : 
 
 JTie quantity C^ is the exponent of the power to which 
 we must raise the constant e to produce the number a. 
 
 We may assign one value to a, namely, e itself, which will 
 lead to an interesting result. Putting a = e,WG have 0^ =1, 
 and the exponential series gives 
 
 e»' = l + x + ^ + j^ + j:^ + etc. (10) 
 
 If we put x=zl, we have the series for e itself, and if we 
 put X = — 1, we have 
 
 '~' = -e = ^-^-^ 1:2- i:2r3-^ 17273:1-'''' 
 
 We thus have the curious result that this series and (9) are 
 the reciprocals of each other. 
 
 EXERCISES. 
 
 1. Substitute in the first four or five terms of the expres- 
 sions (6) and (7) the values of C^, Cg, Cn-2, etc., given by (5), 
 and show that (6) and (7) are thus rendered identically equal. 
 
 Note. This is merely a sliglit modification of the process we have 
 actually followed in comparing the coefficients of like powers of x and p 
 in (6) and (7). 
 
 2. Compute arithmetically the values of 2.71832, 2.7183-1, 
 and 2.7183~2, and show that they are the same numbers, to 
 three places of decimals, that we obtain by putting ^ = 2, 
 a; = — 1, and x= — 2 in (10), and computing the first eight 
 or ten terms of the series. 
 
 3. Since 6^+® = eef^, the equation (10) gives, by substituting 
 the developments of e^'^^ and e^, 
 
 X + l + . + (l + ^Vil + M_V(l + ^ + etc. 
 
 (X^ X^ X^ \ 
 
 1 + ^ + 2! + 3! + 4! + «*°- )■ 
 
 It is required to prove the identity of these developments, 
 by showing that the coefficients of like powers of a; are equal. 
 
378 LOGARITHMS. 
 
 CHAPTER VI. 
 
 LOGARITHMS. 
 
 311. To form the logarithm of a number, a constant num^^ 
 ber is assumed at pleasure and called the hase. \ 
 
 Def. The Logarithm of a number is the exponent 
 of the power to which tlie base must be raised to pro- 
 duce the number. 
 
 The logarithm of x is written log x. 
 
 Let us put a, the base ; 
 
 X, the number ; 
 
 I, the logarithm of x. 
 Then d = x. 
 
 Eem. For every positive value we assign to x there will b& 
 one and only one value of I, so long as the base a remains un- 
 changed. } 
 
 Def. A System of Iiogaxithms means the loga- 
 rithms of all positive numbers to a given base. * The 
 base is then called the base of the system. 
 
 Properties of liOgarithms, 
 
 313. Consider the equations, 
 
 a' = l; ) Mogl =0; 
 
 a^ = a; V whence by definition, \loga = 1 ; 
 
 a^ = a2 : 
 
 log a2 = 2. 
 
 Hence, 
 
 I. The logaritJnn of 1 is zero, whatever he the hase. 
 II. The logarithm of the hase is 1. 
 
 III. The logarithm of any numher hetween 1 and the 
 hase is a positive fraction. 
 
 IV. The logarithms of powers of the hase are integers, 
 but no other logarithms are. 
 
LOGARITHMS. 379 
 
 Again we liave 
 
 a 
 
 cr"' = — 
 a"' 
 
 1 \ /i_„l 
 
 log- =-l; 
 
 whence by definition, { log — ^ = — 2 ; 
 
 log — = — n. 
 
 Hence, 
 
 V. The logarithm; of a number hetween and 1 is 
 negative. 
 
 Again, as we increase n, the value of aP- increases without 
 limit, and that of — approaches zero as its limit. Hence, 
 
 VI. TJie logarithm of is negative infinity. 
 
 VII. Theoeem. The logarithin of a product is equal 
 to the sum of the logarithms of its factors. 
 
 Proof. Let p and q be two factors, and suppose 
 
 h = log p, Ic = log q. 
 
 Then a^ == p, a* = q. 
 
 Multiplying, a^a^ = a^'^^ = pq. 
 
 Whence, by definition, 
 
 ^ + ^ = log {pq), 
 or log p + \ogq =z log {pq). 
 
 The proof may be extended to any number of factors. 
 
 VIII. Theoeem. The logarithm of a quotient is found 
 by subtracting the logarithm of the divisor from that of 
 the dividend. 
 
 Proof. Dividing instead of multiplying the equations in 
 the last theorem, we have 
 
 a* q 
 
380 LOGARITHMS. 
 
 Hence, by definition, 7i — ^ = log - , 
 or log^— log^ = log^. 
 
 IX. Theorem. The logarithm of any power of a num- 
 her is equal to the logarithm of the nuiriber multiplied 
 by the exponent of the power. 
 
 Proof. Let h = log p, and let n be the exponent. 
 
 Then a^ = p. 
 
 Eaising both sides to the n^^ power. 
 
 Whence nh = logp^, 
 
 or n logp = \ogp^. 
 
 X. Theorem. I7ie logarithm^ of a root of a numher 
 is equal to the logarithm of the number divided by the 
 index of the root. 
 
 Proof. Let s be the number, and let p be its n^^ root, so 
 that 
 
 p = \^s and s = p^. 
 Hence, log s = log p"' = n log p. (IX.) 
 
 Therefore, log^ = ^^ , 
 
 or. loa: y^s = — ^— 
 
 Note. We may also apply Th. IX, since p = A Con- 
 sidering - as a power, the theorem gives 
 
 log^ = - log s. 
 
 EXE RCISE S. 
 
 Express the following logarithms in terms of log p^ log </, 
 log Xy and log y, a being the base of the system : 
 
L0OARITEM8. 381 
 
 1. Log p% Ans, 2 log ^ 4- log q. 
 
 2. Log pqK 
 
 3. LogjoY- 4- l^ogpq^xY- 
 
 5. Log - = log xpS and explain the identity. 
 
 6. Log|| = \ogxyp-^q-K 
 
 Ans. Log :c + log 2/ — log jt? — log g'. 
 
 ■7. Log^. 8-Log^,- ■ 
 
 9. Log a/s? (Note, § 123). 10. Log Vx Vy* 
 II. Log a/-' 12. Log V^. 
 
 X 
 
 13. Log«a;. 14. Log-. 
 
 '5- Log-. le.lMg^. 
 
 17. Log Vi?^i^. Ans. Log(. + ^) + log(«-») , 
 
 18. LogVT^a^. 19. 'Log{a^ — x^), 
 
 Logarithmic Series. 
 
 313. Rem. The logarithm of a number cannot be devel- 
 oped in powers of the number. For, if possible, suppose 
 
 log X = Cq -^ C^x + C^x^ + etc. 
 Supposing a; = 0, we have 
 
 C, = log 0, 
 
 which we have found to be negative infinity (§ 312, VI). 
 Hence the development is impossible. 
 
 But we can develop log (1 + y) in powers of y. For this 
 purpose, we develop (1 -\- yY by both the binomial and expo- 
 nential theorems, and compare the coefficients of the first 
 power of X. First, the binomial theorem gives 
 
 /-. . \.>. -* . . ^(^— 1) o ^{^ 1)(^— 2) ««. i. 
 
 (1 + «/)«' = 1 + a;y + -\:^- 1 + -^^ — iT^Tg ' ^'+etc. 
 
382 L0QAB1THM8. 
 
 If we develop the coefficients of y\ y\ etc., by performing 
 the multiplications, we have 
 
 Coef. of y^ = ^^; Pa^t in a; = — |. 
 
 ^ "" 2-3 ' ^3 
 
 In general, in the coefficient of y^, or 
 
 I y\ or fl| 
 ■ ^yx i ^ -v — X 
 
 the term containing the first power of x will fce \^^ 
 
 ±1»2.3 (n—l)x _ , ^ 
 
 1.2.3 n ~ ^n 
 
 Hence, 
 
 (1+^)35 = i+a;(?/ — I+I-— |- + etc. j + terms in x\ a^, etc. 
 
 On the other hand, the exponential development, § 309, (8), 
 gives, by putting 1 + y for a. 
 
 (1 + y)^ = 1 -{■ C^x -\- terms in x% a^, etc. 
 
 Equating the coefficients of x in these two identical series 
 we have 
 
 ^•==2/-|' + |'-|* + etc. (1) 
 
 The value of 0^ is given by the theorem of § 310, putting 
 1 + y for ff ; that is, C^ is here defined by the equation 
 eC, — 1 4_ y. 
 
 Hence, if we take the number e (§ 310) as the base of a 
 system of logarithms, we shall have 
 
 C, = log (1+2/). 
 Comparing with (1), we reach the conclusion : 
 
 Theorem. Assuming the Jfaperian base e as a base, 
 we have 
 
 log (1 + 2/) = 2/ - I' + I' - f-V etc., ad inf. (2) 
 
LOGARITHMS. 383 
 
 Def. Logarithms to the Ibase e are called Naperian 
 Logarithms, or Natural Logaxithms. 
 
 The appellation " natural " is used, because this is the simplest system 
 of logarithms. 
 
 Rem. The series (2) is not convergent when ?/ > 1, and 
 therefore must be transformed for use. 
 
 Putting —y fox y in (2), we have 
 
 log (1-2/) = -^-| -|-etc. 
 Subtracting this from (2), and noticing that 
 log (1 + 2/) - log (1 - 2/) = log ^ (Th. VIII), 
 
 we have log ^ = 22/ + ^' + ^' + etc. (3) 
 
 Now n being any number of which we wish to investigate 
 the logarithm, let us suppose y = - — — r* This will give 
 
 1+y ^ n+1 
 1—y n ' 
 
 whence log ~^^ = log ^— = log {71 + 1) — log n. 
 Substituting these values in (3), we have 
 
 log {n + l)-\ogn = ^^ + ^^lys + ^^^^Jy 
 
 + etc. (4) 
 
 This series enables us to find log {n + 1) when we know 
 log n. To find log 2, we put n = 1, which, because log 1 
 = 0, gives 
 
 Summing five terms of this series, we find 
 log 2 = 0.693147 
 
384 LOGARITHMS. 
 
 Putting w = 2 in (4), we have 
 
 which gives log 3 = 1.098612. . 
 
 Because 9 = 3^, log 9 = 2 log 3 = 2.197224. 
 Putting w = 9 in (4), we have 
 
 log 10 = log 9 + 2(i + 3i-^3 + ^-9, + etc.), 
 
 whence log 10 = 2.302585. 
 
 In this way the Naperian logarithms of all numbers may be 
 computed. It is only necessary to compute the logarithms of 
 the prime numbers from the series, because those of the com- 
 posite numbers can be formed by adding the logarithms of 
 their prime factors. (§ 312, VII.) 
 
 314. Definitive Form of the Exponential Series. We are 
 now prepared to give the exponential series (§ 309, 8) its defi- 
 nite form. Since the coefficient G^ is defined by the equation 
 
 e^> = a, 
 the quantity C is the Naperian logarithm of a. Hence, the 
 exponential series is 
 
 a^ = 1 + 
 
 ^log a , (^ log o f , { x log ay . 
 
 which is a fundamental development in Algebra. 
 
 By putting « = e, we have log « = 1, and the series be- 
 comes that for e« already found. 
 
 By putting a; = 1, we have an expression for any number 
 in terms of its natural logarithm, namely. 
 
 Comparison of Two Systems of Logarithms. 
 
 315. Put e, the base of one system ; 
 a, the base of another ; 
 n, a number of which we take the logarithm 
 in both systems. 
 
LOGARITHMS. 385 
 
 Putting I and T for the logarithms in the two systems, we 
 
 have 
 
 ^ = n, a!'' = n, 
 
 and therefore ef' = a^\ (1) 
 
 Now put k for the logarithm of a to the base e. Then 
 
 e* = «, 
 
 and raising both members to the l'^^ power, 
 
 e^' = ai\ 
 
 Comparing with (1), I = TcV, 
 
 or V = lx\' ^ (2) 
 
 This equation is entirely independent of w, and is therefore 
 the same for all values of 7i. Hence, 
 
 Theorem. // we multiply the logarithm of any 
 numher to the base a by the logarithm of a to the base e, 
 we shall have the logajdthui of the number to the base e. 
 
 316. Although there may be any number of systems 
 of logarithms, only two are used in practice, namely : 
 
 1. The natural or Naperian system, base = e = 
 2.718282 
 
 2. The common system, base = 10. 
 
 The natural system is used for purely algebraic 
 purposes. 
 
 The common system is used to facilitate numerical 
 calculations. 
 
 Assigning these values to e and a in the preceding section, 
 the constant k is the natural logarithm of 10, which we have 
 found to be 2.302585. 
 
 Therefore, by (2), for any number, 
 
 nat. log = common log x 2.302585, 
 , , nat. loff. 
 
 and ^^°^"^«^^l«g = 2X0258^. 
 
 = nat. log X 0.4342944.... 
 Hence, 
 
 25 
 
386 LOGARITHMS. 
 
 Theoeem. The common logarithm of amj number 
 way he found by multiplying its natural logarithm by 
 0.4342944 .... or by the reciprocal of the JYaperian loga- 
 rithm of 10. 
 
 Def, The numlber 0.4342944 is called the Modulus 
 of the commoii system of logarithms. 
 
 EXERCISES. 
 
 1. Show that if a and J be any two bases, the logarithm of 
 a to the base h and the logarithm of b to the base a are the re- 
 ciprocals of each other. 
 
 2. What does this theorem express in the case of the natu- 
 ral and common systems of logarithms ? 
 
 Cominon Logarithms. 
 317. Because 
 
 102 _ 100, X /log 100 = 2, 
 
 101 z= 10, j j log 10 =z 1, 
 
 100 = IW \ log 1 = 0, 
 
 ^ ^ . 1 ) we have to base 10, < , 1 ^ 
 
 etc. etc. 
 
 The following conclusions respecting common logarithms 
 will be evident from an inspection of the above examples : 
 
 I. Tfie logarithm of any number between 1 and 10 is 
 a fraction between and 1. 
 
 II. The logarithm of a nuiJiber with two digits is 1 
 plus some fraction. 
 
 III. In general, the logarithm of a number of i digits 
 is i — If plus some fraction. 
 
 IV. The logarithm^ of a fraction less than unity is 
 negative. 
 
 V. The logarithms of two numbers, the reciprocal of 
 each other, are equal and of opposite signs. 
 
LOGARITHMS. 387 
 
 YI. If one nu-mher is 10 times another, its logarithm 
 will be greater by unitij. 
 
 Proof. If 10^ = n, 
 
 then 10^+^ = 10 X 10' = lOn. 
 
 Hence, if I = log n, 
 
 then Z + 1 = log lOn. 
 
 318. To give an idea of the progression of logarithms, the 
 following table of logarithms of the first 11 numbers should be 
 studied. 
 
 The logarithms are not exact, because all logarithms, ex- 
 cept those of powers of 10, are irrational numbers, and there- 
 fore when expressed as decimals extend out indefinitely. We 
 give only the first two decimals. 
 
 logl = 0.00, log 7 = 0.85, 
 
 log 2==: 0.30, logs =0.90, 
 
 log 3 = 0.48, log 9 = 0.95, 
 
 log 4 = 0.60, log 10 = 1.00, 
 
 log 5 = 0.70, log 11 = 1.04. 
 log 6 = 0.78, 
 
 It will be noticed that the difference between two consecu- 
 tive logarithms continually diminishes as the numbers increase. 
 For instance, the difference between log 20 and log 10 must 
 by § 312, VIII, be the same as between log 1 and log 2. 
 
 319. Computation of Logarithms. Since the logarithms 
 of all composite numbers may be found by adding the loga- 
 rithms of their factors, it is only necessary to show how the 
 logarithms of prime numbers are computed. "We have already 
 shown (§ 313) how the natural logarithms may be computed, 
 and (§ 316) how the common ones may be derived from them 
 by multiplying by the modulus 0.4342944.... It is not how- 
 ever necessary to multiply the whole logarithm by this factor, 
 but we may proceed thus: 
 
 We have, putting if for the modulus, 
 
 com. log n = M nat. log n, 
 
 com. log (w + 1) = M nat. log {n + 1) ; 
 
388 L0QABITHM8. 
 
 whence, by subtraction, 
 
 com. log (n + l) — com. log n = M[natAog {n + l)—nat.\ogn]; 
 
 and, by substituting for nat. log (w + 1) — nat. log n its 
 value, § 313, 
 
 com. log (^ + 1) = com. log n + ^^^^^ + ^2^-1)3 
 
 By means of this series, the computations of the successive 
 logarithms may be carried to any extent. 
 
 Tables of the logarithms of numbers up 100,000, to seven places of 
 decimals, are in common use for astronomical and mathematical calcula- 
 tions. One table to ten decimals was published about the beginning of 
 the present century. The most extended tables ever undertaken were 
 constructed under the auspices of the French government about 1795, and 
 have been known under the name of Les Orandes Tables du Cadastre. 
 Many of the logarithms were carried to nineteen places of decimals. 
 They were never published, but are preserved in manuscript. 
 
 330. It may interest the student who is fond of computa- 
 tion to show how the common logarithms of small numbers 
 may be computed by a method based immediately on first 
 principles. 
 
 Let 71 be a number, and let - be an approximate value of 
 
 its logarithm. We shall then have, 
 
 p 
 n = IDS', 
 or, raising to the q^^ power, 
 
 71^ = lOP. 
 Hence, could we find a power of the number which is also 
 a power of 10, the ratio of the exponents would at once give 
 the logarithm. This can never be exactly done with whole 
 numbers, but, if the condition be approximately fulfilled, we 
 shall have an approximate value of the logarithm. 
 
 Let us seek log 2 in this way. Forming the successive 
 powers of 2, we find 
 
 210 = 1024 = 103(1.024). (1) 
 
LOGARITHMS. 389 
 
 Hence, 3 : 10 = 0.3 is an approximation to log 2. To 
 find a second approximation, we form the powers of 1.024 
 until we reach a number nearly equal to 2 or 10, or the quo- 
 tient of any power of 2 by a power of 10. Suppose, for instance, 
 that we find 
 
 1.024^ = 2. 
 
 Because 1.024 = 2^^ -j- 10^, this equation will give 
 
 )x 
 = 2, or 210^ = 2.103«', or 2^^^ = lOs^^, 
 
 oioxcc 
 
 = 2, or 2^"^ = 2.10= 
 
 Sx 
 
 which will give log 2 
 
 10a; 
 
 Tf we form the powers of 1.024 by the binomial theorem, 
 or in any other way, we shall find that x is between 29 and 30, 
 from which we conclude that log 2 = 0.301 nearly. 
 
 To obtain a yet more exact value, we form the 30th power 
 of 1.024 to six or seven decimals, and put it in the form 
 
 1.02490 = 2 (1 + «), 
 
 "where a will be a small fraction. 
 
 Then we find what power of 1 + cc will make 2. Let y be 
 this power. Eaising the last equation to the yth power, we 
 have 
 
 1.02430y = 2y(l-{- ay = 2y+K 
 
 Putting for 1.024 its value, 2^^ -r- 10^, this equation becomes 
 
 2300y 
 
 -— = 22/+1, or 2^-^ = lO^oy, 
 whence, log 2 = ^^^L. 
 
 By a little care, the value of y can be obtained so accurately 
 that the value of log 2 shall be correct to 8, 9, or 10 places of 
 decimals. 
 
 The power to which we must raise 1 + a to produce 2 will 
 
 be approximately — — — ^— , when « is very small. 
 
390 L0OABITEM8. 
 
 EXE RCISES. 
 
 1. In the common system (« = 10) we have 
 
 log 2 = 0.30103, log 3 =: 0.47712. 
 
 Hence find the logarithms of 4, 5, 6, 8, 9, 12, 12J, 15, 16, 
 16|, 18, 20, 250, 6250. 
 
 Note that 5 = J^, 12^ = ^^, 16| = ^, and apply Th. VIIL 
 
 2. How many digits are there in the hundredth power of 2? 
 
 3. Given log 49 = 1.690196 ; find log 7. 
 
 4. Given log 1331 = 3.124178 ; find log 11. 
 
 5. Find the logarithm of 105 and 1.05 from the above data ? 
 
 6. Find the logarithm of I.O510. 
 
 7. If $1 is put out at 5 per cent, per annum compound 
 interest for 1000 years, how many digits will be required to 
 express the amount? (Compare § 216.) 
 
 8. Prove the equation 
 
 log a; = - log {x + 1) +~ log {x - 1) 
 
 ■^ ^[2^^^^ "^ 3 (22;2 _ 1)3 + 5 (22:2 _ 1)5 + ^^^.J 
 
 9. If 2^ = log n, of what numbers will y + 1, y + '^,y — 1, 
 and 2^ — 2 be the logarithms.? 
 
 10. Find X from the equation d^ z^Ji. 
 
 Solution. Taking the logarithms of both members, we have 
 a; log c = log ^ ; 
 
 whence, x = ^^. 
 
 log 6 
 
 11. c<^ = n, 12. (fi^ = — . 
 
 m 
 
 13. ^ = -' 14. h-^=p. 
 
 Show that the answers to (13) and (14) are and ought to be identical. 
 15. a<^ z= m. 16. b(^ = Jc, 
 
 17. Find X and y from the equations 
 
 a^by = p, a^ix = q. 
 
BOOK XII. 
 IMA GIN A RY QUANTI TIE S, 
 
 CHAPTER I. 
 
 OPERATIONS WITH THE IMAGINARY UNIT.* 
 
 331. Since the square of either a negative or a positive 
 quantity is always positive, it follows that if we have to extract 
 the square root of a negative quantity, no answer is possible, 
 in ordinary positive or negative numbers (§§ 170, 200). 
 
 In order to deal with such cases, mathematicians have been 
 led to suppose or imagine a kind of numbers of which the 
 squares shall be negative. These numbers are called Imagi- 
 nary Quantities, and their units are called Imaginary- 
 Units, to distinguish them from the ordinary positive and 
 negative quantities, which are called real. 
 
 323. The Imaginary Unit. Let us have to extract the 
 square root of — 9. It cannot be equal to + 3 nor to — 3, 
 because the square of each of these quantities is + 9. We 
 may therefore call the root V— 9, just as we put the sign V 
 before any other quantity of which the root cannot be extracted. 
 But the root may be transformed in this way : 
 
 Since — 9 = -|-9 x —1, 
 
 it follows from § 183 that 
 
 V^^ = a/9 V^^ = 3V~T. 
 
 * It is not to be expected that a beginner will fully understand tliis 
 subject at once. But he should be drilled in the mechanical process of 
 operating with imaginaries, even though he does not at first understand 
 their significance, until the subject becomes clear through familiarity. 
 
392 IMAOmABY QUANTITIES. 
 
 Def. The surd V— 1 is the Imaginary Unit. The 
 imaginary unit is commonly expressed by the symbol i. 
 
 This symbol is used because it is easier to write i than 
 
 The unit i is a supposed quantity such that, when squared, 
 the result is — 1. 
 
 That is, i is defined by the equation 
 
 ^2= -1. 
 
 Theorem. The square root of any negative quantity 
 may he expressed as a nurnber of imaginary units. 
 
 For let — w be the number of which the root is required. 
 
 Then V— n = Vh- n V— 1 = "s/ni. 
 
 Hence, 
 
 To extract the square root of a negative quantity, 
 extract the root as if the quantity were positive, and- 
 affix the symbol i to it. 
 
 323. Complex Quantities. In ordinary algebra, any num- 
 ber may be supp(5sed to mean so many units. 7 or a, for 
 example, is made up of 7 units or a units, and might be writ- 
 ten 7-1 or al. 
 
 When we introduce imaginary quantities, we consider them 
 as made up of a certain number of imaginary units, each repre- 
 sented by the sign i, just as the real unit is represented by the 
 sign 1. A number i of imaginary units is therefore written M, 
 
 A sum of a real units and b imaginary units is written 
 a + bi, 
 and is called a complex quantity. Hence, 
 
 Def. A Complex Quantity consists of the sum of 
 a certain number of real units plus a certain number of 
 imaginary units. 
 
 Def. When any expression containing the symbol 
 of the imaginary unit is reduced to the form of a com- 
 plex quantity, it is said to be expressed in its Normal 
 Form. 
 
IMAQINABT QXTANTITIES. 393 
 
 Addition of Complex Expressions. 
 
 334. The algebraic operations of addition and subtraction 
 are performed on imaginary quantities according to nearly the 
 same rules which govern the case of surds (§ 181), the surd 
 being replaced by i. Thus, 
 
 aV— 1 + h^/— 1 = a^ + hi =z (a + h) i. 
 Hence the following rule for the addition and subtraction 
 of imaginary quantities : 
 
 Add or subtract all the real terms, as in ordinary 
 algebra. Tl%en add the coefficients of the imaginary 
 unit, and affix the symbol i to their sum. 
 
 Example. Add a + U, 6 + 7i, 5 — IOj, and subtract 
 3a — ^bi + z from the sum. 
 
 We may arrange the work as follows : 
 a + bi 
 .6-+ 7z 
 ' 5 — 10^ 
 — z — ^a -\- 2bi (sign changed). 
 Sum, — z — 2a -^ 11 + (db — 3) L 
 
 EXERCISES. "V^ 
 
 1. Add Zx 4- 4:yi + m, 2m + hni, 6m — 6yi. 
 
 2. Add 4.ai, 17 i, da + Qbi, x + yi. 
 
 3. From the sum a-\-bli-m — ni — p + qi subtract the 
 sum -\- yi — z — ui. 
 
 Keduce to the normal form : • 
 
 4. a -\r bi — {m — ni) — (a; + yi). 
 
 5. m {a — bi) — n{x— yi). 
 
 Multiplication of Complex Quantities. 
 
 335. Theorem. All the even powers of the imagi- 
 nary unit are real units, and all its odd powers are 
 imaginarn^ units, positive or negative. 
 
394 IMAGINABT QUANTITIES. 
 
 Proof. The imaginary unit is by definition such a symbol 
 as when squared will make — 1. Hence, 
 i2 = - 1. 
 
 Now multiply both sides of this equation by i a number of 
 times in succession, and substitute for each power of i its yalue 
 given by the preceding equation. AVe then have 
 
 •^^^=-^, ^^ ^ 
 i^ — —i^ = +1 (because ^2 = — 1), 
 
 ^ i(^ = —i^ = -\-i^ = —1, 
 
 etc. etc. etc. 
 
 It is evident that the successive powers of i will always 
 have one of the four values, *, — 1, — ^, or + 1. 
 
 iy i^, i^, etc., will be equal to i; 
 
 i\ i\ i^% etc., " " -1; 
 
 ^3, i^, *ii, etc., " " — i; 
 
 iS ^8, *i2, etc., " " +1. 
 
 We may express this result thus: . • 
 
 If n is any integer, then: 
 
 To multiply or divide imaginary quantities, we proceed as 
 if they were real and substitute for each power of i its value as 
 a real or imaginary, positive or negative unit. 
 
 Ex. I. Multiply fli by a;/. 
 
 By the ordinary method, we should have the product, 
 axi-^. But i^ = — 1. The product is therefore — ax. 
 
 That is, (fi X xi = — ax. 
 
 Ex. 2. Multiply a + hi by m + 7ii. 
 
 ni (a -I- U) = ani — in (because ni x U = —hn) 
 m. {a -f hi) = hmi + am 
 {m + ni) (a + bi) = am ^ hn -\- (an + hm) i, 
 which is the product required. 
 
IMAOmART QUANTITIES. 395 
 
 EXERCISES. 
 
 Multiply 
 
 I. X -\- yi hj a — b. 2. m + ni by ai. 
 
 3. m — ni by M. 4. 1 + i by 1 — i. 
 
 5. X — yi by 6? + hi. 6. x — yi by a; -f yi. 
 
 7. a — ai — U by a + ai + hi. 
 
 Develop . 
 
 8. {a + hi)\ 9. (m + nif. 
 10. (l + i)2. II. (1 — *?. 
 
 326. Imaginary Factors. The introduction of imaginary 
 units enables us to factor expressions which are prime when 
 only real factors are admitted. The following are the princi- 
 pal forms : 
 
 «2 _[_ J2 — (^ ^ Jjl^ ^(jl^ — ])(^^ 
 
 a^ — h^± 2ahi — {a ± hif. 
 
 The first form shows that the sum of two squares can 
 always be expressed as a product of two complex factors. 
 Por example, 17 = 4^ + 12 = (4 + i) (4 — i). 
 
 EXERCISES. .'— 
 
 Factor the expressions : '^' ^^ 
 
 1. x^ + 4. 2. x^ + 2. S. ,, 
 
 3. rj2 _ 2a; 4- 5 = (a; - 1)» + 4. 
 
 4. x^ — 4:X -}- 13. 5. a + h. 
 
 6. a2 + 2an + 5?i2. 7. x^ -\- 2xy + 2«/2. 
 
 327. Fundamental Principle. A complejo quantity 
 A + Bi cannot be equal to zero unless we have both 
 
 A =:0 and B =: 0. 
 
 Proof. If A and B were not zero, the equation A + Bi=zO 
 would give 
 
 A 
 
 that is, the imaginary unit equal to a real fraction, which is 
 impossible. 
 
 Cor. If both members of an equation containing imagi- 
 
396 IMAGINABT QUANTITIES. 
 
 nary units are reduced to the normal form, so that the equation 
 
 shall be in the form 
 
 A + Bi = M-\- M, 
 
 we must haye the two equations, 
 
 A=z M, B = K 
 
 For, by transposition, we obtain 
 
 A-M+{B-N)i = 0, 
 whence the theorem gives A — M=:0, B — JV=0, Hence, 
 
 Every equation between complex quantities involves 
 two equations between real quantities, formed by equating 
 the numbers of real and imaginary units. 
 
 Reduction of Functions of i to the Normal 
 
 Form. 
 
 338. 1. If we have an entire function of i, 
 
 a -\-U -\- cv^ + di^ + ei^ ■\- fi^ -\- etc., 
 we reduce it by putting 
 
 4^ = — 1, i^ _ _ i^ {i — i^ gtc, etc., 
 and the expression will become 
 
 (a — c + e — etc.) + {h — d +/— etc.) i ; 
 which, when we put 
 
 X = a — c -\- e — etc, ' y — h — d -{-f — etc., 
 becomes x + yi, as required. 
 
 2. To reduce a rational fraction of i to the normal form, 
 we reduce both numerator and denominator. The fraction 
 will then take the form 
 
 a + bi 
 m + ni 
 
 Since this is to be reduced to the form x + yi, let us put 
 a + U 
 
 m -\- m ^ ' 
 
 X and y being indeterminate coefficients. 
 Clearing of fractions, 
 
 a ^U = mx— ny -\- {my -}- nx) i. 
 
IM AGIN ART QUANTITIES. . 397 
 
 Comparing the number of real and imaginary units on 
 each side of the equation, we have the two equations 
 mx — ny = a, nx + my = l. 
 
 Solving them, we find 
 
 ^ma + nh _ mh — n a 
 
 ^ „ a A- hi ma + nh mh — na . 
 
 Therefore, — ; — . = — ^—, — ^ H s— — 5- 1, 
 
 ' m + ni m^ -^ n^ w? -\- n^ 
 
 which is the normal form. 
 
 EXERCISES. 
 
 Eeduce to the normal form : 
 
 I. 7 _ Si _ 6^2 + 2i^ + i^ — i\ 2 
 
 2. 1 _j_ ^ _ 1-3 + ^3 _ ^'4 _ i5 4. ^-6, -^ i_i 
 
 6 + 5^ 1 + *' , fni (x — ai) 
 
 6 — 5i *'l— * a;-|-fifi 
 
 1 — t « + &i (a + U) (a—ii) 
 
 10. What is the value of the exponential series which gives 
 the development of e* ? We put x = * in § 310, Eq. 10. 
 
 11. Develop (1 + xi)^ hj the binomial theorem. 
 
 12. What are the developed values of 
 
 (1 + M)^ -h (1 - M)^ 
 and (1 -f^{)n_(i_ji)np 
 
 13. Write eight terms of the geometrical progression of 
 which the first term is a and the common ratio ^. 
 
 14. Find the limit of the sum of the geometrical progres- 
 
 i 
 sion of which the first term is a and the common ratio -• 
 
 339. To reduce the square root of an imaginary expres- 
 sion to the normal form. 
 
 Let the square root be Va +^. 
 
 We put X -^ yi = a/« + bi. 
 
 Squaring, x^ ^ y^ + 2xyi = a + bi. 
 
398 IMAGINARY QUANTITIES. 
 
 Comparing units, x^ — y"^ = a, 
 2xy = d. 
 Solving this pair of quadratic equations, we find 
 _ V(Va ^+ J^ + a) 
 '^- V2 —' 
 
 ^/ (^/a^-\-¥—a) 
 
 11 = ^^ -y 
 
 Therefore, 
 
 V. + J^ = VI ^ / + V I ^ Y' 
 
 EXERCI SES. 
 
 Eeduce the square roots of the following expressions to the 
 normal form : 
 
 I. 3 + 4^. 2. 4 + 3^, 3. 12 + U. 
 
 4. Find the square roots of the imaginary unit i, and 
 of — iy and prove the results by squaring them. 
 
 Note that this comes under the preceding form when « = and 
 & = ± 1. 
 
 5. Find the fourth roots of the same quantities by extract- 
 ing the square roots of these roots. 
 
 330. Quadratic Equations with Imaginary Roots. The 
 combination of the preceding operations will enable us to solve 
 any quadratic equation, whether it does or does not contain 
 imaginary quantities. 
 
 Example i. Find x from the equation 
 
 a;2 -f 42; + 13 = 0. 
 Completing the square and proceeding as usual, we find 
 a;2 + 4a; + 4 = — 9, 
 whence a; + 2 = \/^9 = ± 3i, 
 
 and X = — 2 ± 3^. 
 
 Ex. 2. 3? -\. hxi — c = 0. 
 
 Completing the square, 
 
 x^ -^-Ixi — - = c— -. 
 4 4 
 
IMAOINABT QUANTITIES. 
 Extracting the root, 
 
 X 
 
 whence 
 
 EXERCISES. 
 
 Solve the quadratic equations : 
 
 I. x^-^x-\-l = 0. 2. x^ — x + 1 — 0. 
 
 3. a;2 + 3a; + 10 = 0. 4. ^^ + 10a; + 34 = 0. 
 
 Form quadratic equations (§ 199) of which the roots shall be 
 
 5. a -{-bi and a — li. 6. ai + h and ai — d. 
 
 331. Exponential Functions. When in the exponential 
 function a^s ^e suppose z to represent an imaginary expression 
 X + yi, it becomes 
 
 This expression could have no meaning in any of our pre- 
 vious definitions of an exponent, because we have not shown 
 what an imaginary exponent could mean. But if we suppose 
 the effect of the exponent to be defined by the exponential 
 theorem (§§ 309, 314), we can develop the above expression. 
 First we have, by the fundamental law of exponents, 
 
 Next, if we put c = Nap. log a, we have 
 a = eP\ 
 whence, «^^ = c^*. 
 
 If we put, for brevity, cy = u, we shall now have 
 
 The value of a^ being already perfectly understood, we 
 may leave it out of consideration for the present, and investi- 
 gate the development of e'^K By the exponential theorem 
 {§ 310, 10), 
 
 . , tf^i^ uH^ uH^ uH^ 
 ,« = ! + „» + __ + _-_ + __ + ^^ + etc. 
 
400 IMAGINARY QUANTITIES. 
 
 Substituting for the powers of i their values (§ 325), 
 
 u^ u^ u^ X . / u^ u^ , \ . 
 ^'* = l-2! + 4!-6! + ^*^- + r-3! + 5!-'N^- 
 
 These two series are each functions of u, to which special 
 names have been given, namely : 
 
 niZ qjij^ f^f^ y^ 
 
 Def. The series 1— 2]+4l""6l"^8!~ ^^^'' ^^ ^^^^^^ 
 the cosine of u, and is written cos u. 
 
 njZ itip nil 7/9 
 
 Def. The series ^ — oi + R-f — y-j + o] — etc., is called 
 the sine of u, and is written sin u. 
 
 Using this notation, the above development becomes, 
 
 QUI _ COS u -{- i sin u, {a) 
 
 which is a fundamental equation of Algebra, and should be 
 memorized. 
 
 Eemaeks. These functions, cos u and sin ?^, have an ex- 
 tensive use in both Trigonometry and Algebra. To familiarize 
 himself with them, it will be well for the student to compute 
 their values from the above series for i = 0.25, i = 0.50, 
 i = 1, i=z 2, to three or four places of decimals. This can 
 be done by a process similar to that employed in computing e 
 in § 310. If the work is done correctly, he will find : 
 
 ^or 
 
 1 
 
 1 
 
 cos - = 
 
 4 
 
 0.969, 
 
 sin I = 0.247. 
 4 
 
 a 
 
 1 
 ^ = 2' 
 
 cos| = 
 
 0.878, 
 
 sin i = 0.479. 
 
 (( 
 
 u = l, 
 
 cos 1 = 
 
 0.540, 
 
 sin 1 = 0.841. 
 
 i( 
 
 u = 2, 
 
 COS 2 = 
 
 — 0.416, 
 
 sin 2 = 0.909. 
 
 332. Let us now investigate the properties of the functions 
 cos u and sin ii, which are defined by the equations, 
 
 2' ' 4' 6' 
 
 cos^ = l--+ +etc. 
 
IMAGINARY QUANTITIES. 401 
 
 Since cos u includes only even powers of w, its value will 
 remain unchanged when we change the sign of u from + to 
 — , or vice versa. Hence, 
 
 cos {— w) = cos u, (1) 
 
 Since sin u contains only odd powers of u, its sign will 
 change with that of u. Hence, 
 
 sin (— u) =. — sin u. (2) 
 
 If in the equation {a) we change the sign of u, we have, 
 by (1) and (2), 
 
 Q-ui — cos (— u) + i sin (— u), 
 or e~w* = cos u — i sin ii. 
 
 Now multiply this equation by {a). Since 
 
 1 
 
 X 
 
 o—ui — 
 
 we have 1 = (cos uY — i?' (sin uY, 
 
 or 1 = (cos uf -h (sin uy. 
 
 It is customary to write cos^ u and sin^ u instead of (cos iif 
 and (sin uy, to express the square of the cosine and of the 
 sine of ti. The last equation will then be written 
 
 cos^ u + sin2 u = 1. {c) 
 
 Although we have deduced this equation with entire rigor, 
 it will be interesting to test it by squaring the equations {h). 
 First squaring cos u, we find (§ 284), 
 
 cos^« = 1 _„» + „4(^^ ^ _^ + ^J _ 
 
 etc. 
 
 The coefficient of ?^^ is found to be 
 
 n\^ V. (n — ^)\^ V. {n - ^y.'^ ^ n\ 
 
 when n is double an even number, and to the negative of this 
 expression when n is double an odd number. 
 Again, taking the square of sin w, we find 
 
 = "' + "1-rr¥!-rr3T) + 
 
 sin^ u = u^ + u^l — ^^--. — :r-r-?.-. I + ctc. 
 26 
 
402 IMAGINABY QUANTITIES. 
 
 the coefficient of W^ being 
 
 1 1 1 
 
 rroT-l)! V.{n-Z)\ b\{n-b)\ 
 
 (^-1)!!!' 
 
 or the negative of this expression, according as ^ 7i is even or 
 odd. 
 
 Adding sin^ u and cos^ u, we see that the terms y? cancel 
 each other, and that the sum of the coefficients of u'^ can be 
 arranged in the form 
 
 4! 1! 3!^2! 2! 3! l! ^4! 
 
 Let ns call this sum A. If we multiply all the terms by 
 4 ! , and note that by the general form of the binomial coeffi- 
 cients, 
 
 n\ _ ln\ 
 
 s\ {n-s)\ ~ \sr 
 
 wefind 4!^ = l-g)+(|)-(|) + g), 
 
 which sum is zero, by § 262, Th. II. Therefore the coefficients 
 of u"' cancel each other. 
 
 Taking the sum of the coefficients of u^, we arrange them 
 in the form 
 
 n\ 1! {71-1)1'^ 2\ (n-2)\ 3! (w - 3)! "^ ^ ^'^ 
 which call A. Then multiplying by w ! , we have 
 
 »'^=^-{9+©-e)+--+e)' 
 
 which sum is zero. Therefore all the coefficients of u^ cancel 
 each other in the sum sin^ u H- cos^ u, leaving only the first 
 term 1 in cos^ u, thus proving the equation (c) independently. 
 This example illustrates the consistency which pervades all 
 branches of mathematics when the reasoning is correct. The 
 conclusion (c) was reached by a very long process, resting on 
 many of the fundamental principles of Algebra ; and on reach- 
 
IMAGINARY QUANTITIES, 403 
 
 ing a simple conclusion of this kind in such a way, the mathe- 
 matician always likes to test its correctness by a direct process, 
 when possible. 
 
 Let us now resume the fundamental equation {a). Since 
 u may here be any quantity whatever, let us put nu for u. 
 The equation then becomes, 
 
 Quui — cos nu + i sin nu. 
 
 But by raising the equation (a) to the n^^ power, we have 
 
 Hence we have the remarkable relation, 
 
 (cos u -\- i sin u)"' = cos nu + i sin nu. 
 
 Supposing n = 2f and developing the first member, we 
 have 
 
 cos^ u — sin^ u + 2i sin u cos u = cos 2u + ^ sin 2u. 
 Equating the real and imaginary parts (§ 327, Cor.), we have 
 cos^ u — sin^ ^f = cos 2zc, 
 2 sin u cos u = sin 2w, 
 
 relations which can be verified from the series representing 
 cos u and sin u, in a way similar to that by which we verified 
 sin^ 2i 4- cos^ u = 1. 
 
 EXERCISES. 
 
 1. Find the values of cos^ w, sin^ qi, cos^ u, and sin^ u by 
 the preceding process. 
 
 2. Write the three equations which we obtain by putting 
 u = a, u = d, and u = a -\- i in equation (a). Then equate 
 the product of the first two to the third, and show that 
 
 cos (a -{- h) =^ cos a cos b — sin a sin J, 
 sin (a -{- i) = sin a cos b + cos a sin b. 
 
 3. "Reduce to the normal form, 
 
 {x — i) (x — 2i) {x — Si) (x — U). 
 
 4. Develop {a + bi)^ by the binomial theorem, and reduce 
 the result to the normal form. 
 
404 QEOMETBIG REPRESENTATION, 
 
 CHAPTER II. 
 
 THE GEOMETRIC REPRESENTATION OF IMAGINARY 
 QUANTITIES. 
 
 333. In Algebra and allied branches of the higher mathe- 
 matics, the fundamental operations of Arithmetic are extended 
 and generalized. In Elementary Algebra we have already had 
 several instances of this extension, and as we are now to have 
 a much wider extension of the operations of addition and mul- 
 tiplication, attention should be directed to the principles 
 involved. 
 
 In the beginning of Algebra, we have seen the operation of 
 addition, which in Arithmetic necessarily implies increase, so 
 used as to produce diminution. 
 
 The reason of this is that Arithmetic does not recognize 
 negative quantities as Algebra does, and therefore in employ- 
 ing the latter we have to extend the meaning of addition, so as 
 to apply it to negative quantities. When thus applied, we 
 have seen that it should mean to subtract the quantity which 
 is negative. 
 
 In its primitive sense, as used in the third operation of 
 Arithmetic, the word multi'ply means to add a quantity to itself 
 a certain number of times. In this sense, there would be no 
 meaning to the words "multiply by a fraction." But we ex- 
 tend the meaning of the word multiply to this case by defining 
 it to mean taking a fraction of the quantity to be multiplied. 
 We then find that the rules of multiplication will all apply to 
 this extended operation. 
 
 This extension of multiplication to fractions does not take 
 account of negative multipliers. In the latter case we can 
 extend the meaning of the operation by providing that the 
 algebraic sign of the quantity shall be changed when the mul- 
 tiplier is negative. We thus have a result for multiplication 
 by every positive or negative algebraic number. 
 
 Now that we have to use imaginary quantities as multi- 
 
GEO METRIC REPRESENTATION: 405 
 
 pliers, a still further extension is necessary. Hitherto our 
 operations with imaginary units have been purely symbolic ; 
 that is, we have used our symbols and performed our operations 
 witliout assigning any definite meaning to them. We shall 
 now assign a geometric signification to operations with imagi- 
 nary units, subject to these three necessary conditions : 
 
 1. The operations must be subject to the same rules as 
 those of real quantities. 
 
 2. The result of operating with an imaginary quantity 
 must be totally different from that of operating with a real one, 
 and the imaginary quantity must signify something which a 
 real quantity does not take account of. 
 
 3. If the imaginary quantity changes into a real one, the 
 operation must change into the corr3Si3onding one with real 
 quantities. 
 
 334. Geometric Representation of Imaginary Units. Cer- 
 tain propositions respecting the geometric representation of 
 multiplication have been fully elucidated in Part I, and are 
 now repeated, to introduce the corresponding representations 
 of complex quantities. 
 
 I. All real numbers, positive and negative, may be arranged 
 along a line, the positive numbers increasing in one direction, 
 the negative ones in the opposite direction from a fixed zero 
 point. Any number may then be represented in magnitude 
 by a line extending from to the place it occupies. 
 
 We call this line a Vector. 
 
 II. If a number a be multiplied by a positive multiplier 
 (for simplicity, suppose +1), the direction of its vector will 
 remain unaltered. If it be multiplied by a negative multiplier 
 (suppose — 1), its vector will be turned in the opposite direc- 
 tion (from — « to 4- «, or vice versa). Compare § 72, 
 where the coarse lines are the vectors of the several quantities. 
 
 — a +a 
 
 I I I 
 
 III. If the number be multiplied twice by — 1, that is, by 
 (— 1)^ its vector will be restored to its first position, being 
 twice turned, and if it be multiplied twice by + 1, that is, by 
 (+ Vf, its vector will not be changed at all. Its vector will 
 
406 
 
 IMAQINABT QUANTITIES. 
 
 ■¥Ul 
 
 therefore be found in its first position, whether we multiply it 
 by the square of a positive or of a negative unit; in other 
 words, both squares are positive. 
 
 IV. To multiply the line + a twice by the imaginary unit 
 i, is the same as multiplying it by i^ or — 1. Hence, 
 
 Multiplying hy the imaginanj unit i must give the 
 vector such a motion as, if i^epeated, will change it froirv 
 -\- a to — a. 
 
 Such a motion is given by turn- 
 ing the vector through a righ t angle, 
 into the position + ia. A second 
 motion brings it to the position 
 — a, the opposite of -f a. A third 
 motion brings it to — ia, a position 
 the opposite of + ia. A fourth 
 motion restores it to the original ~*^ 
 
 position + a. 
 
 If we call each of these motions multiplying hy i, we have, 
 from the diagram, a z= a, ia = ia, i^a = — a, i^a = — ia, 
 i^a = a, which corresponds exactly to the law governing the 
 powers of i (§ 325). Hence : 
 
 // a quantity is represented hy a vector extending 
 from a zero point, the multiplication of this quantity hy 
 the imaginary unit m^ay he represented hy turning the 
 vector through 90°. 
 
 y. In order that multiplier 
 and multiplicand may in this op- 
 eration be interchanged without 
 affecting the product, we must 
 suppose that the vertical line 
 which we have called ia is the 
 same as ai, that is, that this line 
 represents a imaginary units. 
 
 We have therefore to count 
 the imaginary units along a 
 
 vertical line on the same system that we count the real 
 units on a horizontal line. 
 
 —2-1 
 
 1 2 
 
 -i 
 -24 
 
 — 3i 
 
OEOMETRIG REPRESENTATION. 
 
 407 
 
 -a+hi 
 
 U 
 
 -hi 
 
 t-a-tbi 
 
 U 
 
 ^a 
 
 hi 
 
 hi 
 
 ^ 
 
 335* Geometric Representation of a Complex Quantity. 
 We have shown (§ 15) that algebraic addition may be represented 
 by putting lines end to end, the 
 zero point of each line added be- 
 ing at the end of the line next 
 preceding. The distance of the 
 end of the last line from the zero 
 point is the algebraic sum. 
 
 On the same system, to repre- 
 sent the algebraic sum of the real 
 and imaginary quantities a + hi, 
 we lay off a units on the real (horizontal) line, and then h 
 units from the end of this line in a vertical direction. The 
 end of the vertical line will then be the position corresponding 
 to a -\- M. 
 
 It is' evident that we should reach the same point if w^e 
 first laid off h units from on the imaginary line, and then a 
 units horizontally. Hence this system gives 
 
 li -\- a =^ a + hi, 
 as it ought to, to represent addition. 
 
 If a or 5 is negative, it is to be laid off in the opposite di- 
 rection from the positive one. We then have the points cor- 
 responding to — a -^hi, — a — hi, and a — hi, shown in the 
 diagram, which should be carefully studied by the pupil. 
 
 The result we have reached is the following: 
 
 Every complejc quantity a + hi is cojisidered as he- 
 longing to a certain point on the plane, namely, that 
 point which is reached hy laying off from the zero point 
 a units in the horizontal direction and h units in the 
 
 vertical direction. 
 
 If 
 
 336. Addition of Com- 
 plex Quantities.^ If we have 
 several, complex terms to 
 add, as a + hi, m — ni, 
 p + qi, we may lay them 
 off separately in their ap- 
 propriate magnitude and di- 
 
408 IMAOINABY QUANTITIES. 
 
 rectioD, as in the figure, the last line terminating in a 
 point K. 
 
 If we first add the quantities a + U, etc., algebraically 
 (§ 324), the result will be 
 
 a -\- m ■\- p -\- {l — n-^q)i. 
 
 We may lay off this sum in one operation. The sum a-^-m 
 -\-p will carry us from to M, and the sum {h — n + q)i 
 from M to E, because MR — l — n-^-q. Therefore we shall 
 reach the same point R whether we lay the quantities off sepa- 
 rately, or take their sum and lay off its real and imaginary 
 parts separately. 
 
 337. Vectors of Complex Quantities. The question now 
 arises by what straight line or vector shall we represent a sum 
 of complex quantities ? The answer is ; 
 
 TJ%e vector of a sum of sev- 
 eral vectors is the straight line 
 from the heginnhzg of the first 
 to the end of the last vector 
 added. 
 
 For example, the sum of the 
 quantities OX = a and XP = hi is the vector OP. 
 
 It might seem to the student that the length of the vector represent- 
 ing the sum should be equal to the combined lengths of all the separate 
 vectors. This difficulty is of the same kind as that encountered by the 
 beginner in finding the sum of a positive and negative quantity less than 
 either of them. The solution of the difficulty is simply that by addition 
 we now mean something different from both arithmetical and algebraic 
 addition. But the operation reduces to arithmetical addition when the 
 quantities are all real and positive, because the vectors are then all placed 
 end to end in the same straight line. Therefore there is no inconsistency 
 between the two operations. 
 
 Two imaginary quantities are not equal, unless both their 
 real and imaginary parts are equal, so that their sum shall ter- 
 minate at the same point P. Their vectors will then coincide 
 with each other. Hence : 
 
 Two vectors are not considered equal unless they agree 
 in direction as well as length. 
 
OEOMETBIG BEP RESENT ATION. 409 
 
 In other words, in order to determine a vector com- 
 pletely, we must know its direction as well as its lengtJi. 
 
 This result embodies the theorem of the preceding chapter (§ 327), 
 that two complex quantities are not equal unless both their real and 
 imaginary parts are equal. It is only in case of this double equality that 
 the two complex quantities will belong to the same point on the plane. 
 
 Because OXP is a right angle, we have by the Pythagorean 
 theorem of Geometry, 
 
 (length of vector)^ = a^ _j_ j2^ 
 or length of vector = '\/a? + b\ 
 
 "We are careful to say length of vector, and not merely vec- 
 tor, because the vector has direction as well as length, and the 
 direction is as important an element as length. 
 
 To avoid repeating the words " length of," we shall put a 
 dash over the letters representing a vector when we consider 
 only its length. Then OX will mean length of the line OX. 
 
 Def. The length of the vector, or the expression 
 ^/a^ + 6^, is called the Modulus of the complex ex- 
 pression a -\- hi. 
 
 The modulus is the absolute value of the expression, con- 
 sidered without respect to its being positive or negative, real 
 or imaginary. Thus the different expressions, 
 
 — 5, H- 5, 3 + 4i, 4 — U, 5iy 
 
 all have the modulus 5 (because V^^ + 4^ = 5). The points 
 which represent them are all 5 units distant from the zero 
 point, and so lie on a circle, and their vectors are all 5 units in 
 length. 
 
 The German mathematicians therefore call the modulus 
 the ahsohde value of the complex quantity, and this is really 
 a better term than the English expression modulus. 
 
 Def. The Angle of the vector is the angle which it 
 makes with the line along which the real units are 
 measured. 
 
 If OA is this line, and OB the vector, the angle is AOB. 
 
410 IMAOINABT QUANTITIES. 
 
 
 EXERCISES. 
 
 
 Lay off the following complex quantities, draw the vectors 
 
 corresponding to them, and find the modulus both by measure- 
 
 ment and calculation 
 
 
 
 I. 4 + 3*. 
 
 2. 4 — 3/. 
 
 3. _ 4 + 3*. 
 
 4. _ 4 - 3*-. 
 
 5. 3 + 4i. 
 
 6. 3 - U, 
 
 7. - 3 4- 4*. 
 
 8. _ 3 - 4^. 
 
 9. 5 + 7z. 
 
 lo. 6 + 6i. 
 
 II. 5 + 5i. 
 
 12. 5 + 4i. 
 
 13. 3 + 3*. 
 
 14. 3 + u 
 
 15. 3 — ^. 
 
 i6. 3 - 2t. 
 
 
 
 17. Draw a horizontal and vertical 
 
 line; mark several 
 
 points on the plane of these lines, and find by measurement 
 the complex expressions for each point. Also, draw the sev- 
 eral vectors and measure their length. Continue this exercise 
 until the relation between the complex expressions and their 
 points is well apprehended. 
 
 Note. The student may adopt any scale he pleases, but a 
 scale of millimeters will be found convenient. 
 
 338, Geometric Multiplication. The question next arises 
 whether the results we obtain for multiplication of complex 
 quantities follow, in all respects, the usual laws of multiplica- 
 tion, especially the commutative and distributive laws. 
 
 I. To multiply a vector by a real factor. 
 
 Let the vector be a -f di and the Bf 
 
 factor m. The product will be -^ ^.-'''1 
 
 ma + mU. 
 
 In the geometric construction, let 
 OA- = a and AB = bi. We shall 
 then have, by the rule of addition. 
 
 Vector OB = « + di. 
 
 When we multiply a by m, let OA' be the product ma, and 
 A'B' the product mhi. Because the lines OA and AB are both 
 multiplied by the same real factor 7n to form OA' and A'B', we 
 shall have 
 
 OA : AB : OB = OA' : A'B' : OB'. 
 
QEOMETEIG REPRESENTATION. 411 
 
 Therefore the triangles OAB and OA'B' are similar and 
 equiangular, so that 
 
 angle A'OB' = angle AOB. 
 
 This shows that the lines OB and OB' coincide, so that 
 BB' is the continuation of OB in the same straight line. More- 
 over, the above proportion gives 
 
 OB' = mOB, 
 or, from (1), vector OB' = m vector OB. 
 
 Therefore, multiplying a vector hy a real factor 
 changes its length without altering its direction. 
 
 II. To 7)^ultiply a vector by the imaginary unit. 
 
 Multiplying a + ii by i, the ^q 
 result is 
 
 — b + ai. 
 
 The construction of the two '"' 
 vectors being made as in the fig- 
 ure, we have 
 
 OB = « + hi, 
 oq = -J) + ai. 
 
 Because the triangles OPQ and OAB are right-angled at P 
 and B^and have the sides containing the right angle equal in 
 length, they are identically equal, and 
 
 angle POQ = angle OBA = 90° — angle BOA. 
 Hence the sum of the angles POQ and BOA is a right 
 angle, and because POA is a straight line, therefore, 
 angle BOQ = 90°. 
 
 Therefore, the result of multiplying the vector OB hij 
 the imaginary unit is to turn it 90° without changing 
 its length. 
 
 We have assumed this to be the case when the vector represents a 
 real quantity, or lies along the line OB ; we now see that the same thing 
 holds true when the vector represents a complex quantity. 
 
 If instead of the multiplier being simply the imaginary 
 unit, it is of the form ni, then, by (I), in addition to turning 
 the vector through 90°, we multiply it by n. 
 
412 
 
 IMAGINARY QUANTITIES. 
 
 III. To multiply a vector by a complex quantity, 
 m + ni. 
 
 This will consist in multiplying separately by m and ni, 
 and adding the two products. Put OB = a + bi, the vector 
 to be multiplied; ON = 
 7n 4- ni, the multiplier. 
 
 To multiply OB by m, 
 we take a length 00, deter- 
 mined by the proportion, 
 
 00 : OB = m : 1, (I) 
 
 whence by (I), 
 
 00 = m.OB 
 z=i m{a -\- li). 
 
 To multiply OB by ni, we take a length CD determined 
 by the condition, 
 
 length CD = w length OB, 
 
 or CD : OB = w : 1 ; 
 
 and to multiply by i, we place it perpendicular to OB. (II) 
 We then have, 
 
 CD = OB X ni. 
 
 In order to add it to 00, the other product, we place it as 
 in the diagram, and thus find a point D which corresponds to 
 the sum 
 
 00 + CD = OBxm + OBxm; 
 
 that is, to the product 
 
 {m + ni) (a + di). 
 Now because 00 = OB x m and CD = OB x n, we have 
 00" : CD" = m:n = OM : WN, 
 
 and because the angles at M and C are right angles, the tri- 
 angles OOD and OMN are similar. Therefore, 
 
 angle COD = angle MON. 
 
 Hence the angle AOD of the product-vector is equal to the 
 sum of the angles of the multiplier and multiplicand. 
 For the length OD of the product-vector we have. 
 
GEO METRIC UEPRESENTATION, 
 
 413 
 
 length OD^ = 00^ + CD" 
 
 = m^OB^ + -nfOW 
 = (m2 + n^) OW, 
 Extracting the square root, 
 
 length OD = Vm^ + n^ • OB 
 
 Therefore the length of the product-vector is equal to the 
 products of the lengths of the vectors of the factors. 
 
 Combining these two results, we reach the conclusion : 
 
 The modulus of the product of two complejo factors is 
 equal to the product of their moduli. 
 
 The angle of the product is equal to the sum of the 
 angles of the factors, 
 
 339. The Roots of Unity. We 
 have the following curious problem : 
 
 Given, a vector A, which call a ; 
 it is required to find a complex factor 
 X, such that when we multiply a n 
 times by x, the last product shall be a 
 itselt That is, we must have 
 
 xP-a 
 
 a. 
 
 The required factor must be one 
 which will turn the vector round without changing its length. 
 Let us begin with the case of n-=^. 
 
 Since three equal motions must restore OA to its original 
 position, the condition will be satisfied by letting x indicate a 
 motion through 120°, so that OA shall take the position OB 
 when angle AOB = 120°. Then, P being the foot of the per- 
 p^ndicular from B upon AO produced, we shaU have angle 
 POB = 60°, and angle PBO = 30°. Therefore, 
 
 P0 = 1., 
 
 PB = ^, 
 
 and 
 
 vector OB =i xa =. — ^a ■\- —j-ai. 
 
414 IMAGINARY QXIANTITIES. 
 
 Because the factor x has not changed the length of the line, 
 the modulus of x is unity, and because it has turned the lino 
 through 120°, its angle is 120°. Therefore its value is 
 
 — OP + vm 
 
 on a scale of numbers in which OB = 1 ; that is, 
 
 1 , a/3. 
 
 ^ = -2+-2-^- 
 
 Reasoning in the same way with respect to the product x^a, 
 which produces the vector 00, we find 
 
 ^- 2 2 '' 
 
 an equation which we readily prove by squaring the preceding 
 value of X and reducing. 
 
 Multiplying these values of x and x^, v/e find 
 
 iC3= 1, 
 
 which ought to be the case, because x^a = a. Hence, 
 
 1 a/^ 
 The complex quantity — « + —^i is a cube root of 
 
 unity. 
 
 But the vector 00, of which the angle is 240°, also repre- 
 sents a cube root of unity, if we suppose 00 = 1, because 
 three motions of 240° each turn a vector through 720°, or two 
 revolutions, and thus restore it to its original position. This 
 also agrees with the algebraic process, because, by squaring the 
 above value of x% we have 
 
 1 3^'v/3._ l^'V/3._ 
 
 4~4 + ~2"'~ "2 + ~2"*-~*'^' 
 
 1-2— 2- V 
 
 and by repeating the process we find 
 
 Since 1 itself is a cube root of unity, because 1^ = 1, we 
 conclude : 
 
 TJiere are three cube roots of unity. 
 
OEOMETBIC REPRESENTATION'. 415 
 
 We readily find, by the process of § 334, IV, that 
 iy — 1, — i, and 1, 
 
 are all fourth roots of unity. 
 
 By a course of reasoning similar to the above for any value 
 of n, we conclude : 
 
 The vP^ roots of unity are n in number, 
 
 EXERCISES. 
 
 1. Form the first eight powers of the expression 
 
 show that the eighth power is 1, and lay off the vector corre- 
 sponding to each power. 
 
 2. Form the first twelve powers of 
 
 2+2'' 
 
 and show thali the twelfth power is + 1. 
 
 3. Find the fifth and sixth roots of unity by dividing the cir- 
 cle into five and six parts, and either computing or measuring 
 the lengths of the lines which determine the expression. 
 
 Note. The student will remark the similarity of the gen- 
 eral problem of the n^^ roots of unity to that of dividing the 
 circle into n equal parts (Geom., Book VI). 
 
BOOK XIII. 
 
 THE GENERAL THEORY OF EQUA- 
 TIONS. 
 
 Every Equation has a Root. 
 
 340. In BooS* III, equations containing one unknown 
 quantity were reduced to the normal form 
 
 Aoi^ + Buf'-^ + (7^:^-2 -f- -\- F =0. 
 
 If we divide all the terms of this equation by the coefficient 
 A, and put, for brevity, 
 
 Pi 
 
 B 
 "A' 
 
 p% 
 
 G 
 -A' 
 
 etc. 
 
 etc. 
 
 Vn 
 
 F 
 -A' 
 
 the equation will become 
 
 x^ + p^aa^--^ + p^x^-^ + + pn-i x-^pn = 0. (a) 
 
 This equation is called the General Equation of the 
 n"* Degree, because it is the form to which every algebraic 
 equation can be reduced by assigning the proper values to n, 
 and to jPi, p^, p^, etc. 
 
 The n quantities pi, p^, , . . . pn are called the Coeffi- 
 cients of the equation. 
 
 We may consider pn as the coefficient of a:^ = 1. 
 
 341. Theoeem I. Every algebraic equation has a root, 
 real or imaginary. 
 
 That is, whatever numbers we may put in place of p^, p^, 
 Pz, . . . . Pn, there is always some expression, real or imaginary, 
 which, being substituted for x in the equation, will satisfy it. 
 
GENERAL THEORY OF EQUATIONS. 417 
 
 Rem. The theorem that every equation has a root is demonstrated in 
 special treatises on the theory of equations, but the demonstration is too 
 long to be inserted here. 
 
 If we suppose the values of the coefficients Pi,p^, etc., to 
 Tary, the roots will yary also. Hence, 
 
 Theoeem II. Tlxe roots of an algebraic equation are 
 functions of its coefficients. 
 
 Example. In Chapter VI we have shown that the roots 
 of a quadratic equation are functions of the coefficients, because 
 if the equation is •^ 
 
 s^+px + q = 0, 
 
 the root is ^ = :ZZ±_V2E^, 
 
 which is a function of p and q. 
 
 342. Equations wliich can le solved. If the degree of the 
 equation is not higher than the fourth, it is always possible to 
 express the root algebraically as a function of the coefficients. 
 
 But if the equation is of the fifth or any higher degree, it 
 is not possible to express the value of the root of the general 
 equation by any algebraic formulae whatever. 
 
 This important theorem was first demonstrated by Abel in 
 1825. Previous to that time, mathematicians frequently at- 
 tempted to solve the general equation of the fifth degree, but 
 of course never succeeded. 
 
 This restriction applies only to the general equation, in 
 which the coefficients p^, p^, p^, etc., are all represented by 
 separate algebraic symbols. Such special values may be 
 assigned to these coefficients that equations of any degree shall 
 be soluble. 
 
 343. The problem of finding a root of an equation of the 
 lii^';her degrees is generally a very complex one. If, however, 
 the equation has the roots — 1, 0, or + 1, they can easily be 
 discovered by the following rules : 
 
 I. If the algebraic sum of the coefficients in the equor- 
 tion vanishes, then +1 is a root. 
 
 27 
 
418 GENERAL THEORY OF EQUATIONS. 
 
 II. If the sum of the coejficients of the even powers of 
 X is equal to that of the coefficients of the odd powers, 
 then — 1 is a root. 
 
 III. If the absolute term pn is wanting, then is a 
 
 root. 
 
 These rules are readily proved by putting x = +1, then x= —1, 
 then a; — in the general equation (a) and noticing what it then reduces 
 to. The demonstration of II will be a good exercise for the student. 
 
 Number of Roots of General Equation. 
 
 344. In the equation (a), the left-hand number is an en- 
 tire function of x, which is equal to zero when the equation is 
 satisfied. Instead of supposing an equation, let us suppose x 
 to be a variable quantity, which may have any value whatever, 
 and let us study the function of x, 
 
 X^ -j-p^X^-^+p^X^-^ 4- +Pn-iX +pn, 
 
 which for brevity we may call Fx. 
 
 Whatever value we assign to x, there will be a correspond- 
 ing value of Fx. 
 
 Example. Consider the expression 
 
 Fx = a^ — 7x^ + 36. 
 
 Let us suppose x to have in succession the values — 4, 
 
 — 3,-2, — 1, 0, 1, 2, etc., and let us compute the corre- 
 sponding values of Fx. We thus find, 
 
 X =z — 4, — 3, — 2, — i, 0, 
 
 Fx = — 140, — 54, 0, + 28, + 36, 
 
 a; = 1, 2, 3, 4, 5, 6, 7, 8. 
 
 Fx = -h 30, + 16, 0, — 12, - 14, 0, + 36, + 100. 
 
 We see that while x varies from — 4 to +8, the value of, 
 J^a; fluctuates, being first negative, then changing to positive, 
 then back to negative again, and finally becoming positive once 
 more. 
 
 We also see that there are three special values of x, namely, 
 
 — 2, +3, and + 6, which satisfy the equation Fx = 0, and 
 which are therefore roots of this equation. 
 
GENERAL THEORY OF EQUATIONS. 
 
 419 
 
 845. Representation of Fx hy a Curve. In Book VIII it 
 was shown how a function of a variable of the first degree might 
 be represented to the eye by a straight line. The relation 
 between a variable and any function of it may be represented 
 to the eye in the same way by a curve, as shown in Geometry, 
 Book VII. We take a base line, mark a zero point upon it, 
 and lay off any number of equidistant values of x. At each 
 point we erect a perpendicular proportional to the corresponding 1 
 value of Fx at that point, and draw a curve through the ends. 
 
 
 The fluctuations of the vertical ordinates 
 of the curve now show to the eye the corre- 
 sponding fluctuations of Fx. 
 
 When Fx is negative, the curve is below 
 
 the base line. When Fx is positive, the curve 
 
 is above the base line. 
 
 The roots of the equation Fx = are shown by the points 
 
 at which the curve crosses the base line. In the present case 
 
 these points are — 2, +3, +6. 
 
 In order to distinguish the roots from the variable quantity 
 x, we may call them a, (3, y, 6, etc., or x^, x^, X3, etc., or «i, 
 «2, «3, etc., the symbol x being reserved for the variable. 
 
 The distinction between x and the roots will then be this: 
 
 X is an independeA variable, which may have any value 
 whatever. 
 
 Fx is a function of x of which the value is fixed by that of x. 
 
 a, (i, y, etc., or x^, x^, x^, etc., are special values of x which, 
 being substituted for x, satisfy the equation 
 
 Fx = 0. 
 
 Theorem. An equation with real coefficients, of ivhich 
 the degree is an odd number, must have at least one real 
 root. 
 
420 GENERAL THEORY OF EQUATIONS, 
 
 Proof. 1. When n is odd, x'^ will have the same sign (-f 
 or — ) as X. 
 
 2. So large a value, positive or negative, may be assigned to 
 X that the term x'^ shall be greater in absolute magnitude than 
 all the other terms of the expression Fx, For, let us put the 
 expression Fx in the form 
 
 If we suppose x to increase indefinitely either in the posi- 
 tive or negative direction, the terms —, ~, etc., will all 
 approach as their limit (§ 303, Th. I). Therefore the expression 
 1 + — 4- ^ + etc. will approach unity as its limit, and will 
 
 X X 
 
 therefore be positive for large values of x, both positive and 
 negative. The whole expression will then have the same sign 
 as the factor xP-, ^nd, n being odd, will have the same sign as x, 
 
 3. Therefore, between the value of x for which Fx is negative 
 and that for which it is positive there must be some value of x 
 for which Fx = 0, that is, some root of the equation Fx = 0. 
 
 For illustration, take the preceding cubic equation. 
 
 CoK. An equation of odd degree has an odd nuviber 
 of real roots. 
 
 For, as Fx .changes from negative to positive infinity, it 
 must cross zero an odd number of times. 
 
 346. Theoeem I. If we divide the expression Fx hy 
 X — a, the remojinder will he Fa, or 
 
 Kemainder = a^ -\- p^a^~'^ +p^a^~^ + . . . . -\- pn* 
 Special Illustration. Let the student divide 
 
 by X — a, according to the method of § 96. He will find the 
 remainder to come out 
 
GENERAL THEORY OF EQUATIONS. 421 
 
 General Proof, When we divide Fxh^ x— a, let iis put 
 Q, the quotient ; 
 By the remainder. 
 
 Then, because the dividend is equal to the product. Divi- 
 sor X Quotient + Remainder, 
 
 {x-a)Q-\- R = Fx. 
 
 Two things are here supposed: 
 
 1. That this equation is an identical one, true for all values 
 of X, This must be true, because we have made no supposition 
 respecting the value of x. 
 
 2. That we have carried the division so far that the remain- 
 der R does not contain x. 
 
 Because it is true for all values of x, it will remain true 
 when we put x = a on both sides. It thus reduces to 
 
 B = F{a), 
 which is the theorem enunciated. 
 
 The value of x being still unrestricted, let us in dividing 
 take for a a root a of the general equation Fx = 0. Then, 
 by supposing x = a, the equation (a) will be satisfied, or 
 
 Fa = 0. 
 
 Therefore if we divide the general expression Fx hj x — a, 
 the remainder Fa will be zero. Hence. 
 
 Theorem II. // we denote by a a root of the equation 
 Fx = 0, the expression Fx will be exactly divisible by 
 X— a. 
 
 Illustration. One root of the equation 
 a^ — x^ — Ux-i- 16 = 
 is 3. If we divide the expression 
 
 2^3 __ ^ _ 11^ ^ 15 
 
 by r<; — 3, we shall find the remainder to be zero. 
 
 347. When we divide Fxhy x — a, the highest power of 
 X in the quotient will be x^-K Therefore the quotient will be 
 an entire function of x of the degree n — 1. 
 
422 QENEBAL THEORY OF EQUATIONS. 
 
 Illustration. The quotient from the last division was 
 a;2 + 2a! - 5, 
 which is of the second degree, while the original expression was of the 
 third degree. 
 
 If we call this quotient F^x, we shall have, by multiplying 
 divisor and quotient, 
 
 Fx = {x — a) F^x. 
 
 Now suppose (i a root of the equation 
 
 F^x = ^', 
 
 then F^x will, by the preceding theorem, be exactly divisible 
 by x — 13. 
 
 The quotient from this division will be an entire function 
 of X of the degree n — 2. This function may again be divided 
 by X — y, representing by y the root of the equation obtained 
 by putting the function equal to zero, and so on. 
 
 The results of these successive divisions may therefore be 
 expressed in the form 
 
 Fx = (x — a) F^x .... (Degree n — 1), ) 
 
 F^x z=z\x — p) F^x (Degree n — %),\ (1) 
 
 F^x = (x — y) F^x .... (Degree n — S), ) 
 etc. etc. etc. 
 
 Since the degree is diminished by unity with every division, 
 we shall at length have a quotient of the first degree in x, of 
 the form 
 
 X — e, 
 e being a constant. 
 
 Then, by substituting in the equations (1) for each func- 
 tion of x its value in the equation next below, we shall have 
 
 Fx = (x — cc){x — fi){x — y) {x — e), 
 
 the number of factors being equal to the degree of the original 
 equation. Hence, 
 
 Theorem I. Every entire function of x of the nth 
 degree may he divided into n factors, each of the first 
 decree in x. 
 
NUMBER OF ROOTS. 423 
 
 Since a product of several factors becomes zero whenever 
 any of the factors is zero, it follows that the equation 
 
 Fx = 
 will be satisfied by putting x equal to any one of the quantities 
 €ii (^iji ' ' • ' ^} because in either case the product . 
 {x — a)(x — (i){x — y) . , , . {x~e) 
 will vanish. Therefore the quantities 
 
 «, /?, y, e, 
 
 are all roots of the original equation Fx = 0. Hence, 
 
 Theorem II. An algebraic equation of the n^ degree 
 has n roots. 
 
 We have seen (§ 195) that a quadratic equation has two 
 roots. In the same way, a cubic equation has three roots, one 
 of the fourth degree four roots, etc. 
 
 Moreover, a product cannot vanish unless one of the factors 
 vanishes. Hence the product 
 
 Fx or (x — a){x — 0){x — y) . . , . {x — e) 
 
 cannot vanish unless x is equal to some one of the quantities, 
 «, i3, y, e. Hence, 
 
 An equation of the n^ degree can have no more than 
 n roots. 
 
 348. We may form an equation of which the roots shall 
 be any given quantities, a, b, c, etc., by forming the product, 
 
 {x — a)(x — h) {x — c), etc. 
 Example. Form an equation of which the roots shall be 
 
 - 1, +1, 1 + 2/, 1 - 2i. 
 Solution, We form the product 
 
 (x-\-l)(x-l)(x-l- 2i) {x-li- 2i), 
 which we find to be 
 
 x* — 23^-\-4x^ + 2x — 5. 
 Therefore the required equation is 
 
 x^ — 2x^ + 4x^ -\- 2x -^ 6 = 0. 
 
424 GENERAL THEOBT OF EQUATIONS. 
 
 EXERCISES. 
 
 Form equations with the roots : 
 
 1. 2 + a/3, 2 — \/3, — 2, + 1. 
 
 2. 3 +. a/5, 3 — a/5, — 3. 
 
 3. 2, -2, 4+ a/7, 4 -a/7. 
 
 4. 1 + a/3, 1 — a/3, 1 + a/5, 1 -Vs. 
 
 349. When we can find one root of an equation, then, by- 
 dividing the equation by x minus that root, we shall have an 
 equation of lower degree, the roots of which will be the remain- 
 ing roots of the given equation. 
 
 Example. One root of the equation 
 
 iC3 _ ^.2 _ iia; ^ 15 ,^ 
 is 3. Find the other two roots. 
 
 Dividing the given equation by a; — 3, the quotient is 
 
 a? -\-2x — b. 
 
 Equating this to zero, we have a quadratic equation of 
 which the roots are 
 
 — 1 + a/6 and — 1 — a/6. 
 
 Hence the three roots of the original equation are 
 
 3, — 1 + Ve, — 1 - a/6. 
 
 EXERCISES. 
 
 1. One root of the equation 
 
 a;3 _ 3:c2 — 14a; + 12 = 
 is— 3. Find the other two roots. 
 
 2. Find the five roots of the equation 
 
 ic5 _ 4^ _l_ i2a^ + 4a;2 — 13a; = 0. 
 (Compare § 343.) 
 
 350. Equal Roots. Sometimes, in solving an equation, 
 several of the roots may be identical. 
 
 For example, the equation 
 
 a:3 — ea;2 + 12a; — 8 = 
 
COEFFICIENTS AND ROOTS. 425 
 
 has no root except 2. If we divide it by x— 2, and solve the 
 resulting quadratic, its roots will also be 2. Hence, when we 
 factor it the result is 
 
 {x — %)(x — 2) (a; - 2) = 0. 
 
 In this case the equation is said to have three equal roots. 
 Hence, in general. 
 
 The n roots of an equation of the n^^ degree are not all 
 necessarily different from each other, hut two or more of 
 them may he equal. 
 
 Relations between CoeflScients and Roots. 
 
 351. Let us suppose the roots of the general equation of 
 the n^^ degree 
 
 0^-\-p^X^-^ -\-Pz7^-'^ + -{-pri^lX +pn = 
 
 to be «, i3, y, . . . . s. 
 
 "We have shown (§ »341) that these roots are functions of 
 the coefficients p^, p.^, . . . . pn- To find these functions is to 
 solve the equation, which is generally a very difficult problem. 
 
 But the coefficients can also be expressed as functions of 
 the roots, and this is a very simple process which we have 
 already performed in some special cases by forming equations 
 having given roots (§ 348). 
 
 If we form an equation with the two roots, « and i3, the 
 result will be 
 
 z= {x — a){x — (3) = x^— {a + fi)x + «/?. 
 
 Comparing this with the general form, 
 
 x^ -\-p^x +p^ = 0, 
 
 we see that p^ = — (a -\- j3), 
 
 a result already reached (§§ 198, 199). 
 
 Next form an equation with the three roots, a, P, y. 
 Multiplying {x — a) (x — (3) by x — y, we find the equa- 
 tion to be 
 
 a^ — {a+(i + y)x^+(a(i-\-[iy-\-ya)x — «j3y = 0. 
 
426 GENERAL THEORY OF EQUATIONS. 
 
 So in this case, p^ = — (« + jS + y), 
 
 ^3 = «^ + ^y -\- r«, 
 
 Pz = — «/57- 
 Adding another root 6, we find the result to be 
 
 Pt = - {a + (i -^ y -^ 6), 
 
 p^ =, ad + ay •\. ad + (iy + (id + yd, (2) 
 
 p^ =■ — «/3y — a[i6 — ayd — (iyd, 
 
 Pi = «/5y(5. 
 
 Generalizing this process, we reach the following conclu- 
 sions : • 
 
 The coefficient Pi of. the second term of the general equa- 
 tion is equal to the sum of the roots taken negatively. 
 
 The coefficient j^g of the third term is equal to the sum of 
 the products of every combination of two roots. 
 
 The coefficient p^ of the fourth term is equal to the sum 
 of the products of every combination of three roots taken 
 negatively. 
 
 The last term is equal to the continued product of the neg- 
 atives of the roots. 
 
 352, Symmetric Fimctions. It will be remarked that the 
 preceding expressions for the coefficients p^, p^, etc., are all 
 symmetric functions of the roots a, (i, y, etc. (§ 256.) 
 
 The following more extended theorem is true : 
 
 Theorem. Everij rational symmetric function of the 
 roots of an equation may he expressed as a rational 
 function of the coefficients. 
 
 Example. From the equations (2) we find 
 
 jt?j2 - 2j»2 = «? + |Q2 + y2 + (52, 
 ^PlV^ -Pl^ - 3i?3 = «3 + ^3 + ^3 4. (J3. 
 
 We thus reach the curious conclusion that although we- 
 may not be able to find any individual root of an equation, yet 
 there is no difficulty in finding the continued product of the 
 roots, their sum, the sum of their squares, of their cubes, etc. 
 
 The general demonstration of this theorem, and the methods by which 
 any rational symmetrical function of the roots may be determined, are 
 found in more advanced treatises. 
 
FOBM OF ROOTS. 431 
 
 where Bli^ is the sum of the remaining terms of the develop- 
 ment in powers of h. 
 
 We then have 
 
 Increment oix = h. 
 
 Corresponding increment of Fx = F(x -{- h) — Fx 
 
 = h (F'x + Bh), 
 
 Ratio of these increments, — ~ = F'x + Bh. 
 
 If we suppose the increment h to approach zero as its 
 limit, the product Bh will also approach zero, and the ratio will 
 approach F'x as its limit. 
 
 But this ratio of the increments may be considered as the 
 ratio of the average rate of increase of the function F to that 
 of the variable x. 
 
 Hence, when we plot the values of Fx by a curve, as in 
 § 345, the derived function shows the slope of the curve at 
 each point. 
 
 When the derived function is positive, the curve is running 
 upward in the positive direction, as from a;= — 3 to x = 0, 
 and from x= +5 to x= -f-oo. 
 
 When the derived function is negative, the curve slopes 
 downward, as from x = to x= +4. 
 
 When the derived function is zero, the curve at the corre- 
 sponding point runs parallel to the base line, as at and +4f. 
 If this point corresponds to a root of the equation, the curve 
 will coincide with the base line at this point, and will there- 
 fore be tangent to it. Hence, from § 356, Th. II, 
 
 A pair of equal roots of an equation are indicated by 
 the curve touching the "base line without intersecting it. 
 
 Forms of the Roots of Equation. 
 
 358. Theorem I. Imaginary roots enter an equation 
 mith real coefficients in pairs. 
 
 That is, \i a + U be a root of such an equation, then 
 a — hi will also be a root. 
 
432 GENERAL THEORY OF EQUATIONS. 
 
 Proof. Let 
 
 af>' +^1^:^-1 +p^x^-^ + -{-pn-ix -\-pn = (1) 
 
 be the equation with real coefficients, and let us suppose that 
 a + hi is a root of this equation. If we substitute a -f hi for 
 Xy we shall have 
 
 Q^ ^a^ + na''-^ hi — !L(!?Zli) ^n-2 ^2 _ ^|^ ^n-s j3^- ^ etc. 
 
 ^jcc»-i = p^a^-^ + p^a^'^hi — etc. 
 If we substitute all the terms thus formed in equation (1), 
 and collect the real and imaginary terms separately, we shall 
 have a result 
 
 A -\- Bi = 
 
 (§ 324), A signifying the sum of all the real terms, 
 a« ^Vl(:^llan-2i2^ p^a^-i^ etc., 
 
 and Bi the sum of all the imaginary ones. 
 
 In order that this equation may be satisfied, we must have 
 identically 
 
 ^ = 0, ^ = (§ 327). 
 
 Next let us substitute a — hi for x. Since the even powers 
 of hi are all real, and the odd powers aU imaginary, this 
 change of sign will leave all the real terms in (1) unchanged, 
 but will change the signs of all tlie imaginary terms. Hence 
 the result of the substitution will be 
 
 A - Bi. 
 But if a + hi is a root, then, as already shown, A = 
 and ^ = ; whence 
 
 A-Bi = 
 
 also, and therefore a — hi is also a root. 
 
 Def. A pair of imaginary roots which differ only 
 in the sign of the coefficients of the imaginary unit are 
 called a pair of Conjugate Imaginary Roots. 
 
 Theorem II. In the eocpression Fx every pair of conju- 
 gate imaginary factors form a real product of the second 
 decree in x. 
 
DECOMPOSITION OF RATIONAL FRACTIONS. 433 
 
 Proof, If in the expression 
 
 Fx =z {x — a) (x — (i)(x — y) . , . . {x — e), 
 
 we suppose a and /? to be a pair of conjugate imaginary roots, 
 which mQ may represent in the form 
 
 a = a + hi, j3 = a — U, 
 then the product of the terms {x — a) (a; — I) or of 
 
 {x — a — hi) {x — a -{- hi), 
 will be {x — af-\- 1^, 
 
 or a?J — %ax + a^ -j- h^, 
 
 a real expression of the second degree in x. 
 
 Cor. Since Fx can always be separated into factors of the 
 first degree, either real or imaginary (§ 347, Th. I), and since 
 all the imaginary factors enter in pairs of which the product 
 is real, we conclude : 
 
 Every entire function of x with real coefficients may 
 he divided into real factors of the first or second degree. 
 
 Decomposition of Rational Fractions. 
 
 359. Def. A Rational Fraction is one which may 
 be reduced to the form 
 
 ax^ + haf^-^ + g:r^-^ + .«♦»+ ^ _ , . 
 
 ^ -\-p^af'-^ + P2^~^ + +Pn ^^^ 
 
 If the exponent m of the numerator is equal to or greater 
 
 than the exponent n of the denominator, we may divide the 
 
 numerator by the denominator, obtaining a quotient, and a 
 
 remainder of which the highest exponent will not exceed 
 
 91 — 1. If we put 
 
 fx, the numerator of the above fraction ; 
 
 Fz, its denominator ; 
 
 Q, the quotient; 
 
 <px, the remainder ; 
 
 fx (bx 
 
 we shall have, Rational fraction =^ ^ = Q + -^' (§ 9G.) 
 
 ^8 
 
434 OENEBAL THEORY OF EQUATIONS, 
 
 Q will be an entire function of x, with which we need not 
 now further concern ourselves. 
 
 The problem now is, if possible, to reduce the fraction 
 
 -^ to the sum of a series of fractions of the form 
 Fx 
 
 X — a x — li x^y x — e' 
 
 Ay B, C, etc., being constants to be determined, and a, p, y, 
 etc., being the roots of the equation Fx = 0. Let us then 
 suppose 
 
 Fx cc — « x — fi x — y x — 
 
 Multiplying both sides by Fx, we have 
 
 AFx , BFx , CFx . , EFx 
 
 ^ x — a x-~(3 x — y 
 
 We require that this equation shall be an identical one, 
 true for all values of x. Let us then suppose x = a. Then 
 because by hypothesis ce is a root of the equation Fx = 0, we 
 have Fa = 0, and the terms in the second member will all 
 vanish except the first. If there is only one root a, we have 
 (§357), 
 
 Lim. (a;=a) = Fa, 
 
 X — a 
 
 Therefore, changing a; to «, we have 
 
 (t>a = AF'a, 
 
 which gives A = -~— 
 
 If a 
 
 In the same way we may find 
 
 F'ft' 
 
 F'y' 
 etc. etc. 
 
 Substituting these values of A, B, etc., in the equation (J), 
 it becomes 
 
 7?- ^^ iA 
 
DECOMPOSITION OF RATIONAL FRACTIONS. 435 
 
 <t>x _ (f>a (t>(i <l>y . 
 
 Fx- {x-a)F'a'^ {x-fi)F'(i^ (x-y)F'y^^^''' 
 
 Note. The critical student should remark that in the 
 preceding analysis we have not proved that the expression of 
 the rational fraction in the form (b) is always possible, but 
 have only proved that if it be possible, then the coefficients A, 
 B, C must have the values (c). To prove that the form is 
 possible, the second member of (i) may be reduced to a com- 
 mon denominator, which common denominator will be Fx^ 
 and the sum of the numerators equated to (t>x. By equating 
 the coefficients of the separate powers of x, we shall have n 
 equations to determine the n unknown quantities A, B, C, 
 etc. Since n quantities can, in general, be made to satisfy n 
 equations, values of ^, B^ (7, etc., will in general be possible. 
 
 It will be instructive to solve the following exercises, both 
 directly and by the common denominator. 
 
 I. Decompose 
 
 EXAMPLES. 
 
 ics _ 7a;2 + 36 
 
 We have already found the roots of the denominator to be 
 — 2, 3, and 6. Using the formulaB (c), we find 
 
 «^a; = 2a^ — 3a; -f 5, 
 
 Fx z=: u? — W ■\- ^Q =z {x + 2) {x — '^) {x — 6), 
 
 F'X z=zW — l^', 
 
 a=z-2, i3 = 3, 7 = e; 
 
 (f>a = 19, 0i3 = 14, 07 = 59; 
 
 F'a = 40, F'fi = - 15, F'y = 24. 
 
 2a^ — 3a; 4- 5 19 14 59 
 
 a;3 - 7arJ + 36 "" 40 (a; + 2) ~ 15 (a; - 3) ■*■ 24 (a; — 6)* 
 
 2a;2 - 7a; + 3 2a;2 - 7a; + 3 
 
 2. Decompose 
 
 ic3 _ 2a?5 _ a; + 2 ~ (a; + l) (a;-l) (a;-2) 
 
 Here the roots of the denominator are -- 1, 1, and 2. Let 
 US effect the decomposition by the following method. Assume 
 
436 GENERAL THEORY OF EQUATIONS. 
 
 {x + l){x- l)(a;-2)~a; + l x — 1 x — 2 
 Keducing the second member to a common denominator, 
 it becomes 
 
 A {x^ - 3a: + 2) + ^ (^^ - ^ - ^) + ^(^^ — 1) 
 ' {X + 1) (^ - 1) (^ - 2) 
 
 Since both members now have the same denominator, their " 
 numerators must also be equal. Equating them, after arrang- 
 ing the last one according to powers of x, we have 
 
 {A + B+ C) x^ -{^A-\-B)x + 2A^2B—C— ^x^ — 7a; + 3. 
 
 Since this must be true for all values of x, we equate the 
 coeflBcients of x in each member, giving 
 
 ^ + i? + (7 = 2, 
 3^ + ^ = 7, 
 2A-2B-0=^, 
 These equations being solved give 
 
 ^ = 2, ^ ;= 1, C = - 1. 
 
 Substituting in {d), 
 
 2a;2 — 7a; + 3 _ ^ , 1 1 
 
 {x-\-l){x — l){x — 2) x-{-l ' x — 1 
 
 EXERCISES. 
 
 Decompose: 
 x + 10 
 
 2a^ — 12af^ — 8rg + 12 
 
 2a 
 
 6. 
 
 x^ + Sx + 4: 
 
 
 ccs ^ a;2 _ 4^ _ 
 
 4 
 
 X 
 
 
 x^ — a^ 
 
 
 aW 
 
 
 ^ :^ — a^ ' (a;2 _ ^2) (a;2 _ j2) 
 
 360. When the equation i^ic = has two or more equal 
 roots, the preceding form fails, because all the terms of the 
 second member of {b') will then vanish when we suppose x 
 equal to one of the multiple roots. In this case we must pro- 
 ceed as follows : 
 
DECOMPOSITION OF RATIONAL FRACTIONS. 437 
 
 If Fx = {x — a)m {x — (iY{x — y)P, 
 
 we suppose 
 (j)x A ^ A^ Aq Am-i 
 
 "^ (rr ,t\m-l "^ (rr _ „\m-2 + ♦••• + 
 
 Fx (x — a)^ (^ — a)»»-i (a: — a)"^'^ ^ * '" ^ x — a 
 4. ?—^ ^ + .... 4- ^^^ 
 
 4. ^ + ^ 4. 4. ^ ^K 
 
 (x — y)P (x— y)P-^ ^ ^ x — y 
 
 etc. etc. etc. 
 
 In the case of m, n, or p =■ 1, this form will be the same 
 as {b), as it should. 
 
 By reducing the second member to a common denominator, 
 and equating the sum of the numerators to 0:r, we shall have, 
 as before, a number of equations the same as the degree of x 
 in Fx. 
 
 EXAMPLE. 
 
 ^ 8a^ — 9x^ — 2.r — 1 
 
 Decompose ^5 _ 2^ _ 2 I .3 4- 4.;^ + ^ - 2 ^ 
 
 of which the roots of the denominator are — 1, — 1, 1, 1, 2. 
 
 Solution, Because of the roots just given, the expression 
 to which the fraction is to be equal is 
 
 ^ , -^1 . ^ 1 -^^ I ^ - 
 (x — lf^x — l^{x-{-iy^x-{-l^x — 2 
 
 Reducing to a common denominator, and equating the co- 
 efficients of the powers of x to the coefficients of the corre- 
 sponding powers in the numerator Sx^ — Ax^ — 2:^: — 1, we 
 
 have 
 
 A^+B^ + C= 0, 
 
 — A^-{-A — 3B^+B= 8, 
 
 — dA^ -\- B^ — 4.B — 2C = — 9, 
 
 A^—dA-^ 7^1 4. 5^ := _ 2, 
 
 2^1 — 2^ + 2^1 H- 25 + C = — 1. 
 
 Solving these equations, we find, 
 
 A = 1, B = 2, 0=3. 
 
 A, =-2, B^=-l, 
 
438 GENERAL THEOBT OF EQUATIONS. 
 
 The given fraction is therefore equal to 
 
 1 2.2 1 . ^ 
 
 (« - 1)2 x-1^ {x + lf x + 1 
 
 a; — 2 
 
 EXERCISES. 
 
 I. Decompose ^.^^^.^^ ^ns. ^-^ + ^^3;^,. 
 
 x-1 ^-^ 
 
 ^* (a; + 1)2* ^* ic3 — ic2-ic + 1* 
 
 a; + 2 
 
 Transformation of Equations. 
 
 361. Def. An equation is said to be Transformed 
 when a second equation is found whose roots bear a 
 known relation to those of the given equation. 
 
 Eem. Sometimes we may be able to find a root of the 
 transformed equation, and thence the corresponding root of 
 the original equation more easily than by a direct solution. 
 
 Problem. To diminish all the roots of an equation 
 by the same quantity h. 
 
 Solution, If the given equation is 
 
 ic» +p^a^-^-irPz^-^ + +i?n = 0, 
 
 and if y is the unknown quantity of the required equation, we 
 must have 
 
 y = x — h. 
 
 Therefore x ■= y -^h. 
 
 Substituting this value of x in the equation, it will become 
 
 rHPi +^/02/^-H[;?g + (w-l);?iA + (|)A2jyn-2+etc. (a) 
 
 When h, w, and the js's are all given quantities, the coeffi- 
 cients of y become known quantities. 
 
OENEBAL THEOBT OF EQUATIONS. 439 
 
 EXE RCISES. 
 
 1. Transform the equation a;^ — 3a; — 4 = into one in 
 which the roots shall be less by 1. 
 
 2. Transform oi^ — dx^ + blx — '7 = into one in which 
 the roots shall be greater by 5. 
 
 363, Removmg Terms from Equations, The quantity 7i 
 may be so chosen that any required term after the first in the 
 transformed equation shall vanish. For, if we wish the second 
 term of the equation {a) to vanish, we have to suppose 
 
 p^ + nh = 0, 
 which gives A = — ^. 
 
 We then substitute this value of h in the equation («), 
 which gives an equation in which the second term is wanting. 
 
 If we wish the third term to vanish, we must determine h 
 by the condition 
 
 which requires the solution of a quadratic equation. Each 
 consecutive term is one degree higher in the unknown quan- 
 tity h, and the last term is of the same degree as the original 
 equation. 
 
 This method is principally applied to make the second 
 term disappear, which requires that we put 
 
 n 
 
 Example. Make the second term disappear from the fol- 
 lowing equation, 
 
 Q? -\- px -\- q — 0, 
 
 Solution, Hence, w = 2 and p^=p, so that 
 
 P 
 
440 GENERAL THEORY OF EQUATIONS. 
 
 Making this substitution, the equation becomes 
 
 which is the required equation. 
 
 Eem. This process affords an additional elegant method of 
 solving the quadratic equation. 
 
 The last equation gives 
 
 y = Y I - 2' = T^'^P^-^' 
 The value of x, being equal to 2/ + h then becomes 
 
 which is the correct solution. 
 
 EXERCISES. 
 
 Kemove the second term from the following equations : 
 
 1. a:3 — 6a:2 + 6a; — 1 = 0. 
 
 2. a:4 _ 4^3 _|_ 3^2 _ 8 z= 0. 
 
 3. irs — 5^-* + 2a;3 ^ ^x^ _ ^^ -. 0. 
 
 4. a;8 — l%x^ + 22^ — a; = 0. 
 
 Rem. The theory of the above process will be readily com- 
 prehended by recalling that the coeflScients of the second term 
 is equal to the sum of the roots taken negatively, or if «, (i, y, 
 etc., be the roots, 
 
 a + (i + y + +e= —Px- 
 
 It is evident that if we subtract the arithmetical mean of 
 
 all the roots, that is, — — , from each of them, their sum will 
 vanish, because 
 
 «+^+/3+^ + r + ^ + etc. = -p^ +n^ = 0, 
 n n n ^^ n 
 
 Hence, when we put «/ — — for a; in the equation, the sum 
 of the roots, and therefore the second term, vanish. 
 
GENERAL THEORY OF EQUATIONS. 441 
 
 363, Problem. To transform an equation so that the 
 roots shall be Ttiultiplied by a given factor m. 
 
 Solution. Since the roots are to be multiplied by m, the^ 
 new iinknown quantity must be equal to mx. So if we callr 
 
 this quantity y, we have 
 
 y = mx, 
 
 which ffives a; = — • 
 
 *^ m 
 
 Substituting this in the general equation, it becomes 
 
 Multiplying all the terms by m^, the equation becomes 
 
 yn _j_ rnp^if'''^ + m^pc^y^''^ + .... + m'^pn = 0. 
 Hence the rule, 
 
 Multiply the coefficient of the second term by m, that 
 of the third by m^, and so on to the last term, which will 
 be Ttiultiplied by m^. 
 
 If the roots are to be divided, we divide the terms in the 
 same order. 
 
 EXE RCI S ES. 
 
 1. Make the roots of a;^ — 2a; + 3 = four times as great. 
 
 2. Divide the same roots by 2. 
 
 364t Problem. To transform an equation so that its 
 roots shall be squared. 
 
 Solution. Let the given equation be 
 
 X^ + PiC^ + P2'^^+ Ps^ + i?4 = 0- 
 
 If y be the unknown quantity of the new equation, we 
 must have 
 
 y = x^, 
 
 which gives x = ± ^^. 
 
 If we substitute x = y^ in the given equation, it may be. 
 reduced to the form 
 
 y^ + p^y + /?4 + {Piy + p^) r = o. 
 
443 GENERAL THEOBT OF EQUATIONS, 
 
 If we substitute a; = — y^, the result will be 
 
 y^ + P^y +P4.- {Piy + p^) y^ = ^' 
 
 Since the value of y must satisfy one or the other of these 
 equations, it must reduce their product to zero ; we therefore 
 multiply them together. Considering them as the sum and 
 difEerence of a pair of expressions, the product will be 
 
 (y^ +i>2y +p^f - (Pty + Vzfy = o, 
 
 or 
 
 y'+(^P2-i>i')y'+W+^p^-^PiP3)y'H^P2P^-P3')y+p^^ 
 
 = 0. 
 
 EXERCI SES. 
 
 1. Transform the quadratic, 
 
 a^ — ^x + 6, 
 
 of which the roots are 3 and 3, into an equation in which the 
 roots shall be the squares of 2 and 3, using the above process. 
 
 2. Transform in the same way 
 
 a^ 4- 12a;2 + Ux + 48 = 0. 
 
 3. Transform 
 
 a« _ 4a;4 _ iq^ ^ ^q^^ + 9a; — 36 = 0. 
 
 Generalization of the Preceding Problems. 
 
 365. Problem. Given, an equation of any degree 
 in an unknown quantity x ; 
 
 Required, to transform this equation into another of 
 which the root shall he a given function of x, 
 
 Solution. Let ?^ be a root of the required equation, and fx 
 the given function. We must then have 
 
 fx = y. 
 
 Solve this equation so as to obtain a; as a function of y. 
 Substitute this value of x in the original equation, and form as 
 many equations as there are values of y. 
 
 The product of these equations will be the required equa- 
 tion in y. 
 
GENERAL TEEOBT OF EQUATIONS. 443 
 
 EXERCISES. 
 
 1. Transform 
 
 ic2 __ 7a; ^ 10 = 
 
 so that the roots of the new equation shall be ^x^, 
 
 2. Transform a^ — 3a;2 + 2a; = 
 so that the roots shall be «a; + 5. 
 
 3. Transform a;^ _ 9^; ^ 18 = 
 
 so that the roots shall be ^x^ — 3. 
 o 
 
 Resolution of Numerical Equations. 
 
 366. Convenient metliod of computing the numerical value 
 of an entire function of x for an assumed value ofx. 
 
 If we have the entire function of x, 
 
 Fx z= ax^ -{• ba^ + cx^ ■}- dx + e, 
 we may put it in the form 
 
 Fx = \[{ax + b)x + c]x + d\x-\-e. J. 
 
 Therefore, if we put 
 
 ax -\-h =z V, b'x + c = c', 
 
 c'x ■\- d = d', d'x -\- e = e', 
 
 we shall have Fx = e'. 
 
 Numerical Example. Compute the values of 
 i^a; = 2a:s — 3a;* — 6a;3 + 8a; — 9 
 for a; = 3 and a; = — 3. 
 We arrange the work thus : 
 
 Coefficients, 
 :*rod. by (a;=3). 
 
 2 
 
 -3 -6 
 
 + 6 +9 
 + 3 +3 
 
 
 + 9 
 + 9 
 
 + 8 
 + 27 
 + 35 
 
 — 9 
 
 + 105 
 + 96 
 
 Hence, 
 
 
 jP3 = 96. 
 
 
 
 
 ^ov X = — 2, 
 
 2 
 
 -3 - 6 
 —4 +14 
 
 -7 +8 
 
 
 —16 
 —16 
 
 + 8 
 
 + 32 
 
 + 40 
 
 — 9 
 —80 
 -89 
 
 Hence, 
 
 
 F{-^) = - 
 
 89. 
 
 
 
444 GENERAL THEORY OF EQUATIONS, 
 
 This, it will be noticed, is a more convenient process tlian tliat of 
 forming the powers of x and multiplying and adding. 
 
 367. Having an entire function of x, and putting x = r + h, 
 it is required to develop the function in powers ofli. 
 
 It will be remarked that this problem is substantially identical with 
 that of § 362, and the solution of this will be the solution of the former. 
 But in the former case h was supposed to be a given quantity, whereas it 
 is now the unknown quantity corresponding to y in the former problem. 
 
 Example of the Peoblem. If we have the expression 
 
 Fx = 2x^ + 3a^ + 4, 
 
 and put X = 2 + hf it will become, by developing the sepa- 
 rate terms, 
 
 F(2-i-h) = 2¥ + 15/^2 + 36A + 32. 
 
 General Rule for the Process. First compute the 
 value of Fr hy the process employed in § 366. 
 
 Then repeat the process, using the successive sum s ob- 
 tained in the first process instead of the corresponding 
 coefficients, and stopping one term hefore the last. TJve 
 result will he the coefficient ofli. 
 
 Repeat the process with the new sums, stopping yet 
 one term sooner. The result will he the coefficient of h^. 
 
 Continue the repetition until we have the first ter-m 
 only to operate upon, which will itself he the coefficient 
 of the highest power of h. 
 
 Ex. I. The example above given is performed as follows: 
 
 Coefficients, 
 Product by r. 
 
 + 2 
 
 + 3 
 4 
 
 
 14 
 
 + 4 
 28 
 
 First sums, 
 Second products. 
 
 
 7 
 4 
 
 14 
 22 
 
 32 
 
 Second sums, 
 Third product. 
 
 
 11 
 
 4 
 
 15 
 
 36 
 
 
 Result, Fi2 + h] 
 
 ) = 2^3 + 15^2 + 36^ + 32. 
 
 
 Ex. 2. In the function, 
 
 Fx = 2x^ -7^ + 6a^ — 2x^ + 6x — S, 
 let us put X = 3 + hj and express the result in powers of h. 
 
OENEBAL THEORY OF EQUATIONS. 445 
 
 Coefficients, 3 
 Products by 3, 
 
 -7 
 6 
 
 + 5 
 -3 
 
 -3 
 + 6 
 
 + 6 
 
 + 13 
 
 -8 
 
 + 54 
 
 First sums, 
 Second products. 
 Second sums, 
 Third products, 
 
 -1 
 
 + 6 
 
 + 5 
 
 6 
 
 + 3 
 + 15 
 
 + 17 
 33 
 
 + 4 
 + 51 
 
 + 55 
 150. 
 
 + 18 
 + 165 
 
 183 
 
 + 46 
 
 Third suras. 
 
 11 
 6 
 
 17 
 
 6 
 
 23 
 
 50 
 
 51 
 
 101 
 
 305 
 
 
 • 
 
 Result, F{Z + h) = 2h' + 33/4* + 10W + 3057i' + 183A + 46. 
 
 EXERCISES. 
 
 1. Compute 2^5 + 237*4 + 101A3_|. 2057^2 + 183/^ + 46, ^j^en 
 h = x — d, 
 
 2. Compute c(^ —^x -\- 7 for x = — 4 + 7*, — 3 + 7«, etc., 
 to + 3 + A. 
 
 Proof of the Preceding Process. If we develop the ex- 
 pression 
 
 a {h + 7')^ + I (h + ry-^ + c(7i + r)«-2 + ^(7z + r)^-3 + etc., 
 
 and collect the coefficients of like powers of 7i, we shall find 
 
 Coef. of h"" = a, 
 
 h^-^ = (I) ar^ ^(n-l)br + c, {A) 
 
 hn-^ = /^) ar^ + ^C~i^ ^^' + (7^ _ 2) cr + d, 
 
 h^-s = {fjarr+ {"^^^r^' + (^^)^^*"' + etc. 
 
 Now examining Ex. 2 preceding, it will be seen that we can 
 make the computation by columns, first computing the whole 
 left-hand column and thus obtaining the coefficient of h'^~\ 
 then computing the next column, thus obtaining the coeffi- 
 cient of h^~% and so on. Commencing in this way, and using 
 the hteral coefficients, a, b, c, etc., and the literal factor r, we 
 shall have the results : 
 
446 GENERAL THEORY OF EQUATIONS. 
 
 h c 
 
 or ar^ + ^r 
 
 ar + b ar^ + br -{■ c 
 
 ar 2ar^ + br 
 
 2ar + b 3ar^ + 2Jr + c 
 
 Bar -h b 6ar^ + 3br + c 
 
 nar + b ( ^l «?^^ + (w — 1) Jr + c. 
 
 If n is the degree of the equation, then, by the preceding 
 process, we shall add the product ar to b n times, the n sepa- 
 rate sums being 
 
 ar-\-b, 2ar+b, Sar-j-b, .... nar-^b. 
 
 To form the second column, we multiply each of these 
 sums except the last by r, and add them to the coefficient c. 
 The terms in ar added being ar^, 2ar^, dar^, etc., the sum 
 will be (1 + 2+3 + +n—l)ar\ The coeiBficient is a figu- 
 
 rate number equal to ^ ^^ ~ "^^ (§§ 286, 287). The sum of 
 
 the coefficients of br \^ n — 1, because there are n — 1 of 
 them used, each equal to unity. Therefore the final result is 
 
 (|)«r2+(^-l)Jr + c, 
 
 which we have found to be the coefficient of 'h^~\ 
 
 In this second column the partial sums or coefficients of 
 ar^ are 
 
 1, 1 + 2 = 3, 1 + 2 + 3 = 6, etc., to 1 + 2 + 3 + .... +(w— 2). 
 
 Therefore the numbers successively added to form the co- 
 efficients of ar^ in the third column are 1, 1 + 3 = 4, 1 + 3 + 6 
 = 10, etc. The coefficients of br^ will be the same as those of 
 ar^ in the column next preceding. 
 
 Continuing the process, we see that the coefficients are 
 formed by successive addition, as in the following table, where 
 each number is the sum of the one above it plus the one on its 
 
GENERAL THEORY OF EQUATIONS. 447 
 
 
 r« 
 
 r 
 
 r2 
 
 r3 
 
 r^ 
 
 r5 
 
 r^ etc. 
 
 Jl^ 
 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 
 
 1 etc. 
 
 h 
 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 etc. 
 
 ^2 
 
 
 3 
 
 6 
 
 10 
 
 15 
 
 etc. 
 
 
 h^ 
 
 
 4 
 
 10 
 
 20 
 
 etc. 
 
 
 
 U 
 
 
 5 
 
 15 
 
 etc. 
 
 
 
 
 U 
 
 
 6 
 
 etc. 
 
 
 
 
 
 h^ 
 
 
 etc. 
 
 
 
 
 
 
 etc. 
 
 etc. 
 
 
 
 
 
 
 
 left. We have carried the table as far as w = 6, and the ex- 
 pressions at the bottom of each column will, when 7i = 6, be 
 formed from the numbers in this table, taken in reverse order, 
 thus : 
 
 Column under l, Qar + b; 
 
 " « c, 15ar^+ 55r 4- c; 
 
 « " dy 20ar^+10br^ + 4:cr + d; 
 
 « « e, 15ar^+10br^+6cr^-^3dr + e; 
 
 « " /, 6ar5+ 6br^+4^r^+3dr^-\-2er+f; 
 
 « " g, ar^-\- dr^-^ cr^+ dr^-\- er^-\-fr-\-g, 
 
 Now the numbers of the above scheme are the figurate 
 numbers treated in § 287, where it is shown that the n^^ num- 
 ber in the i^ column after the column of units is 
 
 n{n 4- 1) (y^ + 2) {n ^-i — \) 
 
 1-2.3 i 
 
 („_^.). 
 
 Comparing with the coefficients in the equations {A), we 
 see that the two are identical, which proves the correctness of 
 the method. 
 
 368. Application of the Preceding Operation to the Ex- 
 traction of the Roots of Numerical Equations, Let the equa- 
 tion whose root is to be found be 
 
 UT^ + hx"^-^ + cQt^-^ + . . . . -f ^ = 0. 
 
 We find, by trial or otherwise, the greatest whole number 
 in the root x. Let r be this number. We substitute r-]-h for 
 
448 GENERAL THEORY OF EQUATIONS, 
 
 X in the above expression, and, by the preceding process, get 
 an equation in A, which we may put in the form 
 
 ah^ + b'h^-^ + c'h^-'^ + d'h^-^ + + ^' = 0. 
 
 Let r' be the first decimal of h. We put r' + h' for 7i in 
 this equation, and, by repeating the process, get an equation 
 to determine h', which will be less than 0.1. If r" be the 
 greatest number of hundredths in A', we put h' = r"-\-h", and 
 thus get an equation for the thousandths, etc. 
 
 369. The first operation is to find the number and approx- 
 imate values of the real roots. There are several ways of doing 
 this, among which SturrrCs Theorem is the most celebrated, 
 but all are so laborious in application that in ordinary cases it 
 will be found easiest to proceed by trial, substituting all entire 
 numbers for x in the equation, until we find two consecutive 
 numbers between which one or more roots must lie, and in 
 difficult cases plotting the results by § 345. 
 
 It is, however, necessary to be able to set some limits be- 
 tween which the roots must be found, and this may be done 
 by the following rules : 
 
 I. An equation in which all the coefficients, including 
 the absolute term, are positive, can have no positive real 
 root. 
 
 For no sum of positive quantities can be zero. 
 
 II. If in computing the value of Fx for any assumed 
 positive value of x, hy the process o/ § 366, i<;e find all the 
 sums positive, there can he no root so great as that 
 assumed. 
 
 For the substitution of any greater number will make all 
 the sums still greater, and so will carry the last sum, or Fx, 
 still further from zero. 
 
 III. If the sums are alternately positive and nega- 
 tive, the value of x we employ is less than any root. 
 
 IV. If two values of x give different signs to Fx, there 
 must he one or some odd numher of roots hetween these 
 values (compare § 345). 
 
OENEBAL THEORY OF EQUATIONS. 449 
 
 V. Two values of x which lead to the same sign of Fx 
 include either no roots or an even nuniber of roots be- 
 tween them. 
 
 Let us take as a first example the equation 
 
 a^-.'^x-^H = 0. 
 
 Let us first assume cc = 4. We compute as follows : 
 
 Coefficients, 1 -7 +7 
 
 Products, __4 _1Q __36 
 
 Sums, +4 +9 +43 
 
 So F {4) = -{- 43, and as all the coefficients are positive, 
 there can be no root as great as 4. 
 
 Putting X = —4:, the sums, including the first coefficient 
 1, are 1, —4, +9, —29. These being alternately positive and 
 negative, there is no root so small as —4. 
 
 Substituting all integers between —4 and +4, we find 
 
 F(-4) = -29, F(0) = + 7, 
 
 F{-3) = + 1, F(l) = + 1, 
 
 F(-2) = +13, F{2)= + 1, 
 
 F{-1) = +13, F{3) = +13. 
 
 If we draw the curve corresponding to these values (§ 345), 
 we shall find one root between —3 and — 4, and very near 
 —3.05, and the curve will dip below the base line between +1 
 and +2, showing that there are two roots between these num- 
 bers ; that is, there are two roots of the form 1+^, h being a 
 positive fraction. Transforming the equation to one in h, 
 by putting 1 + h for x, we find the equation in h to be 
 
 7^3 4. 37^2 _4^j^_^i = 0. (1) 
 
 Substituting h = 0.2, 0.4, 0.6, 0.8, we find that there is 
 one root between 0.3 and 0.4, and one between 0.6 and 0.7. 
 Let us begin with the latter. 
 
 If in the last equation we put 7* = 0. 6 + h'j we find the 
 transformed equation in h' to be 
 
 Fh' = A'3 + 4.8/i'2 + 0.687i' — 0.104 = 0. (2) 
 
 If we substitute different values of /*' in this equation, we 
 29 
 
450 GENERAL THEORY OF EQUATIONS. 
 
 shall find that it must exceed .09, and as it must be less than 
 0.1, we conclude that 9 is the figure sought, and put 
 7i' = .09 + h". 
 Transforming the equation (2), we find the equation in h" 
 
 to be 
 
 A"8 + 5.07A"2 + 1.56837i" - 0.003191 = 0. (3) 
 
 Since h" is necessarily less than 0.01, its first digit, which 
 is all we want, is easily found, because the two first terms of 
 the equation are very small compared with the third. So we 
 simply divide .003191 by 1.5683, and find that .002 is the re- 
 quired digit of h". We now put 
 
 h" = .002 + h'", 
 and transform again. The resulting equation for h'" is 
 
 7t"'8 + 5.076A"'2 + 1.588592A'" — 0.000034112 = 0. (4) 
 
 The digits of rr, h, h', and h" which we have found show 
 the true value of x to be 
 
 X = 1.692 + A'". 
 
 By continuing this process, as many figures as we please 
 may be found. But, after a certain point, the operation may 
 be abbreviated by cutting off the last figures in the coefficients 
 of the powers of h. 
 
 The work, so far as we have performed it, may be arranged 
 in the following form (see next page). 
 
 The numbers under the double lines are the coefficients of 
 the powers of h, h', h", etc. It will be seen that for each digit 
 we add to the root, we add one digit to the coefficient of h^^ 
 two to that of h, and three to the absolute term. We have 
 thus extended the latter to nine places of decimals, which, in 
 most cases, will give nine figures of the root correctly. If this 
 is all we need, we add no more decimals, but cut off one from ■ 
 the coefficient of h, two from that of h% and so on for each 
 decimal we add to the root. 
 
 We shall find the next figure after 1.692 to be zero ; so we 
 cut off the figures without making any change in the coeffi- 
 cients. The next following is 2, so we cut off again for it, and 
 multiply as shown in the following continuation of the process : 
 
GENERAL THEORY OF EQUATION'S. 
 
 451 
 
 
 
 -7 
 
 + 7 1 1.692 
 
 +1 
 
 + 1 
 
 -6 "— 
 
 Ti 
 
 -6 
 
 + 1.000 
 
 +1 
 
 +2 
 
 -1.104 
 
 T3 
 
 —4.00 
 
 - .104000 
 
 +1 
 
 + 2.16 
 
 + .100809 
 
 + 3.0 
 
 -1.84 
 
 - .003191000 
 
 + .6 
 
 + 2.52 
 
 + .003156888 
 -34112 
 
 + 3.6 
 
 + 0.6800 
 
 .6 
 
 + 0.4401 
 
 
 4.2 
 
 + 1.1201 
 
 
 6 
 
 + .4482 
 
 
 +4.80 
 
 + 1.568300 
 
 
 9 
 
 10144 
 
 
 4.89 
 
 + 1.578444 
 
 
 9 
 
 10148 
 
 
 4.98 
 9 
 
 
 
 + 1.588592 
 
 
 + 5.070 
 
 
 
 2 
 
 
 
 5.072 
 
 
 
 2 
 
 
 
 5.074 
 
 
 
 2 
 
 
 
 + 5.076 
 
 
 
 ( 
 
 DONTmUATION OF PROCESS. 
 
 + 15.076 
 
 + 1.58851912 
 
 -34112 1 021471 
 
 
 1 
 
 31774 
 
 
 1.5887 
 
 -2338 
 
 
 1 
 
 1589 
 
 
 1.518j8i8 
 
 — 749 
 
 636 
 
 —113 
 
 111 
 
 -2 
 
 It will be seen that from this point we make no use of the 
 coefficient 1 of h\ and only with the second decimal do we use 
 the coefficient of lA After that, the remaining four figures 
 are obtained by pure division. 
 
 There is one thing, however, which a computer should 
 always attend to in multiplying a number from which he has 
 cut off figures in this way, namely : 
 
 Always carry to the product the number which would 
 have been carried if the figures had not been cut off, and 
 
453 GENERAL TBEOMY OF EQUATIONS. 
 
 increase it hy 1 if the figure following the one carried 
 would have been 5 or greater. 
 
 For instance, we had to multiply by 7 the number 15 1 888. 
 If we entirely omit the figures cut off, the result would be 105. 
 But the correct result is 111|216; we therefore take 111 in- 
 stead of 105. 
 
 Again, in the operation preceding, we had to multiply 
 158|88 by 4. The true product is 635|52. But, instead of 
 using the figures 635, we use 636, because the former is too 
 small by |52, and the latter too great by |48, and therefore the 
 nearer the truth. For the same reason, in multiplying 1.58818 
 by 1, we called the result 1589. 
 
 Joining all the figures computed, we find the root sought 
 to be 1.692021471. 
 
 Let us now find the negative root, which we have found to 
 lie between —3 and —4. Owing to the inconvenience of 
 using negative digits, and thus having to change the sign of 
 every number we multiply, we transform the equation into one 
 having an equal positive root by changing the signs of the 
 alternate terms. The equation then is a:;^ — 7a; — 7 = 0. 
 
 The work, so far as it is necessary to carry it, is now ar- 
 ranged as follows : 
 
 10 -7 -7 1 8.0489173395 
 
 3 ^ 6 
 
 3 3 -1.000000 
 
 3 18 81446 4 
 
 6 20^00 -0.185586000 
 
 9:00 203616 - .19158408 
 
 4 .3633 18791238 
 
 _4 7803 4 ^^^^'^ 
 
 ^•^? • 20.797824 
 
 73088 
 
 ^•1^2 20 .8709113 
 
 Z 833|0 
 
 ^■^^1 20.87914|3 
 
 9.136 
 
 833 
 
 fl 20.8873|7 
 =— 9 
 
 |9-1|44 201.8|8|715 
 
OENEBAL THEORY OF EQUATIONS. 453 
 
 The negative root of the equation is therefore 
 — 3.0489173395. 
 
 EXE RCISES. 
 
 Find the roots of the following equations : 
 
 1. 0^ — ^x^ + 1 = (3 real roots). 
 
 2. cc3 _ 3a; 4_ 1 = (3 real roots). 
 
 3. a;4 _ 4^ _|_ 2 = (2 positive roots). 
 
 4. x^ + X — 1 = 0. 
 
 5. Prove that when we change the algebraic signs of the 
 alternate coefficients of an equation, the sign of the root will 
 be changed. 
 
 370. The preceding method may be applied without 
 change to the solution of numerical quadratic equations, and 
 to the extraction of square and cube roots. In fact, the square 
 root of a number ?^ is a root of the equation a;^ — tj = 0, or 
 . a:2 4- O2: — 7^ = 0, and the cube root is a root of the equation 
 a^ + 0a;2 + O2; — 7i = 0. 
 
 Ex. I. 
 
 To compute y 2. 
 
 
 1 
 
 
 
 -2 I 1.41421356 
 
 
 1 
 
 1 
 
 
 1 
 
 -1.00 
 
 
 1_ 
 
 .96 
 
 
 2^ 
 
 -.0400 
 
 
 0.4 
 
 281 
 
 
 2.4 
 
 -11900 
 
 
 4 
 
 11296 
 
 
 
 -60400 
 
 
 2.80 
 1 
 
 56564 
 
 
 —3836 
 
 
 2.81 
 1 
 
 2828 
 
 
 -1008 
 
 
 2.820 
 
 • 849 
 
 
 4 
 
 -159 
 
 
 2.824 
 
 141 
 
 
 4 
 
 -18 
 
 
 2:8280 
 
 17 
 
 
 2 
 
 1 
 
 
 2.8282 
 
 
 
 2 
 
 
 21.8|218|4 
 
454 
 
 GENERAL THEORY OF EQUATIONS. 
 
 Ex. 2. To compute the cube root of 9842036. 
 
 
 2 
 2 
 2 
 
 
 
 4 
 4 
 8 
 
 4 
 2^ 
 
 60 
 1 
 
 1200 
 61 
 
 1261 
 62 
 
 61 
 1 
 
 62 
 
 1 
 
 630 
 4 
 
 634 
 4 
 
 638 
 4 
 
 132300 
 2536 
 
 134836 
 2552 
 
 137388.00 
 192.69 
 
 137580.69 
 192.78 
 
 137773.47 
 1.93 
 
 642.0 
 .3 
 
 13|7|7|75|4 
 
 642.3 
 3 
 
 
 642.6 
 3 
 
 
 214.30303242 
 
 -581036 
 539344 
 
 41692000 
 
 412j;4207 
 
 -417793 
 
 413326 
 
 4467 
 
 4133 
 
 334 
 
 276 
 
 58 
 
 65 
 
 8 
 
ANSWERS. 
 
 IN the following list, answers to questions which do not require cal- 
 culation or written work, or which it is supposed teachers would 
 prefer to have in a separate Key, are omitted. A Key, to be published 
 for the use of teachers, will contain the complete solutions. 
 
 36. I. -9. 2. -17. 3- +9. 4. -26. 5- +10. 
 
 6. —15. 7. —56. 9. +840 lo. —1056. ii. +1. 
 
 12. —306. 13. 0. 14. —1008. 
 
 24 8 
 28. I. 1. 2.-2. 3.-5. 4.-14. 5. +24. 6. - = 3. 
 
 7. -y 8. -. 
 
 40. I. 0. 2. 0. 3. 11. 4. 17. 5- -37. 6. -90. 
 7. 324. 8. 0. 9. —60. 10. —180. 11. 945. 
 12. 5040. 13. —41. 14. —1. 15. —17. 16. 26. 
 17. 99. 18. 675. 19. 74. 20. —468. 21. —218. 
 22. —529. 23. -9007. 24. —6800. 25. —420. 
 
 47 45 
 26. -840. 27. 1^. 28. — ^. 29. 2. 30. 8. 
 
 31. When ic = 2, Exp. = 6 ; ic = 5, Exp. = 18 ; x = 7, 
 
 23 
 Exp. = 36. 32. When x = —5, Exp. = — — ; a; = 2, 
 
 Exp. = ^; a; = 5, Exp. = - ^• 
 
 43. I. When x = —3, Exp. =0; a; = — 1, Exp. =0; 
 
 a; = 1, Exp. =1 ; a; = 3, Exp. =15. 2. When x = —3, 
 
 144 ^ -p, 16 - -^ 16 
 Exp. = -g^; a; = -1, Exp. = — ; a; = 1, Exp. = — ; 
 
 a; = 3, Exp. = 24. 3. When a; = — 3, Exp. = 46875 ; 
 
 X = —1, Exp.= — I; a; = 1, Exp. = —88434; 
 
 a; = 3, Exp. = — ^ (365)3. 4. ^^gn a; = — 1, Exp. =: 
 
 (Vl4-V2)4; a; = l, Exp. = (\/8 - V2) 4 ; 
 «; = 3, Exp. = (Vis — V42) 4. 
 
456 AN8WER8. 
 
 48. I. a ■\- Ix -^ {x — y). 2, x — y — {a ■]- Ix), 
 
 , , a — hx a—hx , ~ 
 
 3. a + ox 4. mpq. 5. va^hx, 
 
 6. A/(a + hx) ■\- {x — y). 7. ^\a + bx) — (x — y). 
 8. {a + hxy {x — y)\ 9. {mpq)K 10. {x — yY {mpq)\ 
 I , z \ (a — bx)(x — y) 
 
 II. — 
 
 etc., etc., etc. 
 
 54. I. 5a + 45 — 8c — e. 2. — a + (a; + y), 3. 6. 
 
 4. 9a; - 13y. 5. 22 (« + 5)2 - a; - 2^ - ;?. 6. 5 («5). 
 
 7. 0. 8. 7(m + ?02-a;-22^. 9. 4(;? + ^)2 + a + 5 + c-6. 
 
 10. Uaix-^y). II. 15 (m + w)a; + 2(w — 7z)a; — 17. 
 
 a b y n ^ m + n 
 
 15. 5x — 7y. 16. 8a;. 17. 4a; — 30. 
 
 55. I. (« + ?w)a;+(5 + ^)y, 2. {mn + pq) x + {2b — U) y, 
 
 3. (3 + 65 + 7a) a; 4- (- 2 - 4) 2/ + w + tj. 
 
 4. (8a + 85 + 7 + 1) a; + (5 - 5 - 5) 2^. 
 
 5. {a^m)x + {b — n)y + (c—p) z. 
 
 6. (2d-2f) X -h (Se -Sd)y + (4/+ 4e) 0. 
 
 7. (3« + |^)^+ (6a -2) a;. 
 
 8. (2a-35)a: + (-5-4J)y. 
 
 9. ga-i^)a;+(|5 + ?^)y. 
 
 10. (y^ — 3a — 6c + -^j a; + (2 + a)y. 
 
 11. (5a5 — ab — d)x + {4:cd — 3tnn) y. 
 
 12. ^— 5 — -^ja; + 5ay. 13. — 8a; + (3 — ^a)^^. 
 
 14. (3?w + 1 + a — a) a; + (— 1 — I ff) 2^. 
 
 15. 3a5a; + (2c + 1) V^ -|- (— m — a) 3/. 
 
 16. — 6x + {5m + 6) Vy — y — dVx. 
 
 17. ex -\- 6Vx — 6y + {—3a — 1) \/y. 
 
 56. 3. - lla + 165 - 4c + 76? — 7a; + (4 + 3c) y. 
 
AI^SWEBS. 457 
 
 4. 117^ 4- 283^2 + 72y - 57ax — 20. 5. 2a — eh, 
 
 6. ^a —2b -{-2c — 2d. 7. 4fl^ + 45 + 4c + 2d. 
 
 8. _ 3x^ — 2x — 4:. g. 3x^ — 0^ -\. Ux + 18. 
 
 10. x^ — ax-\- 2a\ 11. 2a^ — 6a^ + 3^?^ — b^ 
 
 12. 3^3 ^ 4a; _|_ 16. 13. — 4:{x — y) + 4:{z — x). 
 
 14. 5 (« — 5) + 2 (« + 5) + 7a — 2^>. 
 
 15. 12--17^-8--8l 
 ^ y z X 
 
 5S, I. 2a:. 2. 2?/. 3. 4«5— 4m^— 3a;. 4. mx-jn. 5. 5t' 
 
 59. I. — 3^5 — m — 2ax. 2. 3x — 2a. 3. 25 — 4c. 
 
 4. IQx—'iy + bz. 5. —9ax—2by. 6. 0. 7. 0. 8. 3m. 
 61. I. m—p-]-q + a—b-\-c-{-d. 2. w + «— J+^ + g' — n-{-Jc. 
 
 3. 15«a; — 4:by. 4. 0. 5. ^ + 5 + 5 + ^ + m + ^j. 
 6. lla.T. 7. —2ax—6by—cz. 8. —2a; + 2?^. 9. —45;?. 
 10. 2a; — 6?/ — my + 4«6 — 5. ir. aa; + 2ca;. 
 
 12. 3fl^a; — 35a; + Say + 3a;2 — 35^ — 35^;. 
 
 13. 13ax — Sxy — 2d — Ifad. 14. m-{-3x-\-4:y—ay — p. 
 15. 2fl5\/^ + Vy — 3m + 67^ — 5\/a;. 
 
 69. 2. 6«25a;3. 3. 15m%y. 4. ^2a^mhj. 5. 4«27^2. 
 6. 5a;y;22. 7, 9a;y;z2. g. 4a2j2?7i2. 9. 9«^54a;4. 
 
 10. 14:4:mp^qh'^s. 11. 144«a;y2;. 12. ni^^x^yK 
 
 13. 3mn^k^. 14. 14:abcd^efg. 
 
 70. 15. m^a;?/^. 16. aJc^Za;*. 17. 12aWm^n\ 18. 14^252^2. 
 
 19. 135m3/iy. 20. Qa'^bcdim/z^. 21. d^m^n^x^y^z. 
 
 22. d}^x^y^. 23. 48«%^?i2^2. 
 
 72. I. a^bcdm. 2. — abcdx^. 3. — aWcx!^. 4. 30a^5%a;2. 
 
 5. 105«3m2a;?/3. 6. 10?^'^a;^+'^^/2;2. 7. 4:abmn. 
 8. 168«5m2^a;2. 9. Qhmngy^. 10. ^ax^-f. 
 
 11. —30agx^yh^. 12. Iba^'^naPyz. 13. —^abgxy^. 
 
 14. 4:bc^gnxh^. 15. —3ai^e^c(^y. 16. 4abcxy. 
 17. — 24aV?/3. 18. «%%3. 19. _3fl^4a;y. 
 
 20. — m''w^a:3. 21. c^bx^if^. 22. —apqx^y^. 
 
 2 
 
 23. 3a%cd?7?. 24. ^acm^y'^x^. 25. — -acmVx^. 
 
 o 
 
 26. 3a%cxy\ 27. —a%dx\ 28. —3Wmhi^y. 
 29. m^n^a^y. 30. — -m^pqoi^y^. 
 
468 AJ:i8WBR8. 
 
 73. 2. Ox^ — Sx^y + 3a:?/2. 3. ^x^ + Sx^t/ + 3a:«/2. 
 
 4. «V?/2; + ttZ'a^?/^^ + acxy^. 
 
 5. 27«'"^Zja:* — 45«2^)a:?/2 _ 63«^a;. 6. —12m^pq-\-lSninql 
 7. 40«%3 _ 56«%2 _ 5Qa^l,y. 
 
 74. I. «i? + mp —p^ + iq — cq — ir — or, 
 
 2. ?W2; — anx — tny — any -[- anz — 77iz. 
 
 3. acx — acy — Mx + My + /(?<^a; + fcdy, 
 
 4. «wa; — c^hm + «^cwz — a&w:K — V^nc — hhid. 
 
 5 . — ajim — apn + hym — dpn — hqm + Jg7^ 4- «2'7?j + a^;j. 
 
 6. G^':?; — 3wca; + lOxy — (jcy — 2zm — Izn. 
 
 7. a^m?c — aiv^hc — Qamlik + 12amJid + 4a/7^/^. 
 
 8. ea^*/ — 105^5' — 12cpq — 4^2^^ + 6np^q^. 
 
 9. — 7flr57^ — 7ab^}i + 7^^^^ — 3bn + ad7i + bhi. 10. 0. 
 76. I. (a'2 + 2%3_,_(3^;3_2:r2— 1 + 5%^— 4a:3^ + a;2— 7.T— 6. 
 
 -.9 . .;! 1 Q . / ^ — 0\ - .0 -I 
 
 78 
 
 2. i^y 4- xy^ + (1 — x^) y^ — xy — 1. 
 
 3. xY + xY + (x — 2x^) 2/3 4- (1 _ 2x^) ?/2 — 2xy — 2 
 
 4. x^y^ + a;Y + (S.t^ _|_ ^2) ^3 _^ (334.^3) y2_^ 2x^y-\-3x 
 
 1. 2«.2 — abn^ — 2abn^ + 2«^> — l>^n^ — 2bhi^ 
 
 2. dam + 2an — ba?bmn — Sbm — 2bn + babhnn. 
 
 3. 2m3/^ 4- /?m3 4- qm^n — 2mn2 — pmii^ 4- 5'?^^. 
 
 4. ;:>3^ 4- p^qr 4- jt^^r 4- pq^ 4- g-V 4- jogV 4. pqr^ 4- g^^^ 4. pj^. 
 
 5. 4fl^2 _ 2«J _ 6^2. 6. m2a;2 — ^y. 
 
 79. I. 6^4 4- «3 ^ 11^2 _ ^ _|_ 28. 2. a3 _ j3. 
 
 3. a^ 4- ^3 _|_ ^2^2 _ ^a^ — ^2^ _ ^4^ 
 
 4. «5 — 2^4 4- 3^3 _ 3«2 4. 2^ — 1. 5. 0:5 — ^5. • 
 
 6. flrW 4- &?^^ 4-^/72^2 4- ^/7i;23, 7. 6^*4- 19ff3 4- 17^2^ «_28. 
 8. «3 4. ^3. 9, ^4 _ ^^ 10^ 0^5 _ q3 _|_ «2 _ 2a 4- 1. 
 
 11. a^ + 2fl!o:* 4- 2a^x^ + 2a^x^ 4- 2a% 4- a\ 
 
 12. am 4- (a;i 4- bm) z -]- {bn ■}- cm — ap) z^ 
 
 4- {dm -{- en — bp) z^ 4- (dn — cj)) 2^ — dp^. 
 
 13. «m 4- («?i 4- Jm) a: 4- J?2a;2. 
 
 14. am 4- (a?j + 5m)a; + {ap + bn + cm)x^ + {bp + cn)x^ 4- cpx^. 
 
 15. y5 _ 5^3 + 2y^-\-6y - 4. 16. ^/^ 4- 2?/H 3?/^ 4- 2/H 1. 
 
 17. «/« 4- 2?/^ - 72/2 — 16. 
 
 18. (3«3m_3^2»»+«)^_j_(_3^;n+2^3^«+2) ^^2a'»+2»_2a3». 
 
 19. «^ + ^«25 4- ^ttJ - 2«Z*2 _ 1 j2. 20. 4«5. 
 
 06 9 
 
 21. a44-26f3_|_^2 _ 54 _ 22»3_52. 22. a2 4- 2ac i-c^ — i^. 
 
ANSWERS, 459 
 
 22. ^2 ^ 2cfc + c2 — 52. 23. — 8«2J. 
 
 24. _ a;2 + (3Z. - «) a; -f 2/2 + (5 - 3ff) ?/ + 2^2 -- 251 
 
 25. a2^^"2 ^ «j2;«+2 _ a2^>a;3 _^ aJa;m+3 ^ ^^^m+s _ ^^a:4^ 
 
 26. a3m_J2n. 
 
 96. I. Q., a: - 3 + — ^- 2. a;2 + 3a; + 1. 
 
 3. Q.--^ + ^^- 4. Q.,2.2 + 3 + -,-^-. 
 
 .7. «2 + «_l. 8. Q.,a;-H-^-j-^. 
 
 9. 4^2 _ 10^5 4. 25. 10. d^ — a^ ■\- a^ — a -\- 1. 
 
 II. a2— 2« + 3. 12. a^—^a?-\-2a—l. 13. a:4_ 10:^2^1 6. 
 
 220 
 
 14. Q., :c4 ^ 2a;2 - 15^ + 56 ^— • 
 
 15. 1 + 22; + ^l 16. 1 — 3a; + 2;2. 17. 3 — 2a + a\ 
 18. l — 2y + 2y^-'y\ 19. — 16 + 8:r— 42;2 + 2a;3— ic*. 
 
 20. Q., 16 + 16a; + 8arJ + 4a;3 + 2a;* + j-——.- 
 
 97. I. x^ — (a -{- c) X -\- ac. 2. a;2 — (a^ + J) a; + ab. 
 
 5. al + J.r — ax. 6. ^j* _ 4^2^^ _,_ 7^^. 
 
 7. ff J + «c + c2 ^ j^, 8. c + J — a. 
 
 9. a^ — al) + ¥ — ac — U + c2. 10. a;2 + 'jtax + 2^2. 
 
 II. ab -\- ax — hx, 12. x—h 13. 6«2a;6 _ 4a3^3 + «*. 
 m — n m 4- n 
 
 104. 7. -T- ^1: 
 
 « — £> « + 
 
 l-h + lc 
 
 9. 0. 
 
 2fl'a; 
 
 a* — 0" ax X y^x'' — 1; 
 
 db-\-lc -\'Ca — (^2 + 52 ^ ^2) J 
 
 (rt — J) (J — c) (c — rt) * * a;2 
 
 4g5 2a 1 
 
 ^^- «2 _ yi ^3. ^ _^ J- 14. ^2 (^2 _ !)• 
 
 ^' a — b ' {^ — n) {x -\- y) '* in^ {m — 
 
460 ANSWERS. 
 
 i8. \ -r-' '9- 0. 20. — ^ — 
 Q^ — a^ 
 
 2 («a; — my-\- xy) a^ ■}-¥ + c^ 
 
 21. — ^^ 5 "^ '^' 22. Y • 
 
 x^ — y^ abc 
 
 4:X^ ,x ^ —2ax 
 
 ^3.0. 24.^-^. 25.0. 26.^-^,. 
 
 2xy —{a — yY + o;^ 3 ^^ + ^^ 
 
 27- a:2 + 2/2- 2S. ^^^ • 29. ^^2_^^j2 
 
 30. (^±A). 
 
 m 
 
 107. I. Jl_i_l). ..« + ^ + 
 
 108. I. db •\- y. 12. -y •• 13. db -^x^ — x\- ^ -)" 
 
 T (a \ , a 2a -{- 3m 
 
 14. i(^-ij. i6.^-j, i9.„w:rjis- 
 110. ,. y+1. .. "^. 3. (^' 4. "^- 5. ». 
 
 y — x ax — b ^ {a -[- x)^ dn ^ 
 
 1 + x^ n (am^ + b) 2xy — d 
 
 2x m{an^ — b) ' y{a-\-b — x) 
 
 {a + bf ^2(^2 + 2) + ! 
 
 ^* 2 («2 + 2ab - ^y ^° ^^' a{l+ a^) 
 
 a^ + ¥ 
 
 {x^ + y^) (a; - yY +{x + y ) (x^ + y ^) 
 
 ^' {x + yf(x'^-^y^)-{x^'+y^) 
 
 111. I. =-. 4. — - — 5. ic + 1. 6. -T— 71 — (• 
 
 a — b a '^ bn(bm — an) 
 
 133. I. a:— 27=0. 2. 7a;— 5a; rr 2450. 3. 6a; + 4a:— 3a; =60. 
 
 4. a;2 4- «ic = «5. 5. abx + «% + 7«2Z> = c. 
 
 6. 5 (4« + 3J) = 12a;. 7. a;^ — ^2 _ 2«a;. 
 
 8. a; + 5 = 2a; — 2«. 9. a; + « = ar^ + 2ax. 
 
 10. a;2 + 3a;— 10 = a.-^ — 3a; — 10. 1 1. hx^ — %2 _- axy, 
 
 12. a;3 _ 5^2^ _ Q^ 13. a; — 2/ = «;s — ^2;. 
 
 14. 2x^ — ax — bx = a^ — ab. 
 
ANSWERS. 461 
 
 124. I. 6?^-2-3?/ + 49 = 0. 2. Zax^a^=0. 3. 31a; + 23=0. 
 
 4. 8^ + 9«a;3 — 6«2^:2 ^ ^4 _ q. 
 
 5. 6«y + 3 («3 _ 1) y% _ 7^4^ 4. 3^2 = 0. 
 
 7. 2^3 + «^2 + fl!2;^ =z 0. 8. 7^/3 + 6?/2 + 52/ + 4 = 0. 
 
 9. ^4 _ ^,;^;3 _ 2^V — a^x — 4^4 = 0. 
 
 10. z^ j^(hJ^ c) z^ + (^z -^-W + h(^ + (^ = 0. 
 
 11. ax^ — fl2a; — h^x + b — 0. 12. (1— ?^):c2 + ^ + l = 0. 
 
 13. 2«2a:5 + aa:4 _ ^^x^ _|_ ^3 _ q^ 
 
 14. 10^5 — 13^ + 6:^3 ^ 21^2 — 6;^ — 3 = 0. 
 
 15. [a—l) x^ + (a + h) x^ + («2 — ^3 _f_ ^2j _ ^5) ^ _ 0. 
 
 16. «V 4. (_ ^3 ^ ^2j ^ a5 - Z>2) a; + a% = 0. 
 
 129. I. ^— • 2. — fl^. 3. 12. 4. 4. 5. T — ^ -.• 
 
 2o ^ oc -\- ac — ah 
 
 6. 25. 7. 36^. 8. ^^. 9. 1. 10. -^ 
 
 c + «. 12. 4J. 
 
 « + ^,- ^- -• — 5 
 
 « (^ + c) + 5c 
 
 «3 + c3 + 53 — 
 
 -1 
 
 3a5c 
 
 8a , «2 (S — a) 
 13. h~a. 14.5. 15.35. '^•TF + "*)' 
 
 a (1 _ J2) a (ac + 52 _ 1) + Jc2 _ J _ a 
 
 bn — am 
 ^^' m — 7i" ^°' 3 {a^ +5^ + c^ — ab — ac — be) 
 die — h) , die -\- a) 
 b — d a + d 
 
 _a{b — d) -\- db , _ ab 
 
 d a — b -{- c 
 
 cd -, cd ab ^ ab 
 
 ~ "T' ~ ~ a^ ~ ~ d^ ~~ ~~c' 
 
 130. I. 20. 2. 72. 3. I, $67; II, $217. 4. ^10. 5. 50. 
 
 6. 180. 7. 65. 8. A, 1130; B, $110; C, $260. 
 
 9. $1000, $1500, $2000, $2500, $3000. 
 10. Man, 36; wife, 30. 11. I, 18| ; II, 26|^ ; III, 45. 
 12. 6 ft. 13. $23531^. 14. 81m. 15. 143^1 m. 
 16. A, $600; B, $1200. 17. 8^ m. per h. 
 
 18. —. pr h. 19. 15 and 24. 20. 15, 10. 
 
 2 {in — li)'^ 
 
 21. Man, 40 ; wife, 35. 22. 19-^. 23. 6| days. 
 
462 ANSWERS. 
 
 24. 30 m. 25. I, 6 ; II, 3; III, 2. 26. 3000. 27. 100. 
 
 28. 4. 29. 18000. 30. $142.50. 31. I, $6 ; II, $4. 
 
 32. 3 m. an h. ZZ- $3600. 34. $24800. 
 
 35. 3 h. 21^ m. 
 
 in "^ . ly ^^ 
 
 ' «(1 + 26? 4- «') ' ' 1 + 2« + «2 
 38. I, !fj:ii^; 11, !^; III, ^; lY, !^^; 
 
 V,^^4i5^. 39. 16 h.; 160 m. 
 
 131. 40. 30 m. ; 2 points. 
 
 41. 20 m.; 3 points. 42. 3|- m. 
 T'T 
 
 43* /7T mi' 
 
 138. I. y = 2|, a; = 12|. 2. y = 7, x = 16. 
 
 $. x = a-b, y = — -a, 4. 2/ = — g— > ^=— 4—' 
 
 3 
 
 7. a; = 32, y = 50. ?>. x ^=. a + h, y =z -{a — h). 
 
 2 
 
 9. a; = 9, ^ = 3. 10. a: = 7, ?/ = 5. 
 
 II. y = 6, X = 4. 13. 2/ = 9, 2; = 8. 
 
 14. ?/ = 8, a; = 6. 15. 1/ z= 6, a: = 15. 
 
 16. y = 7, X = 14. 17. 3/ = 12, cc = 6. 
 
 2^ . _ ^^ 
 c — d' c + 6^ 
 
 20. y = a^ — 2ab + b% x = a^ + 2ai + 2*2. 
 140. 2. a^i = 27, x^ = 22, ajg = 8, a;^ = 7. 
 
 3. a? = 2, 2/ = 3, 2 = — 2. 
 
 4. a; = 6, 2/ = — 1, 2; = 3, ?^ = 2. 
 
 a — 2h + c-}- d a -\-'b—2c + d 
 
 5. a; = , y = , 
 
 a -{- b -h c — 2d —2a-\'b + c + d 
 z = 3 ,u = -3 
 
 t. X =z , 1/ = , Z = ■ 
 
 p + m + n ^ p -\- n — m p — n — m 
 
 ^^' y = Z -^^ ^ = T-TTy- 19. 2/ = 2. ^ = 6. 
 
ANSWERS. 463 
 
 I. A, $225; B, $150. 3. 54. 4. 42. 5. 67. 6. 81. 
 7. ~ 8. A, 86; B, 72. 9. 16 good, 26 poor. 
 
 4:0 
 
 lo- ^- II- ^iT- 12. 25, 8. 13. 96,37. 14. 34,18, 
 15 lo 
 
 15. A in 9, and B in 18 d. 16. 28, 23. 17. 35, 28. 
 
 18. I, 40 ; II, 30. 
 
 19. Bought, 72^ and 24^; sold, 90^ and 32?^, 
 
 20. Coffee, ^^—"-^; tea, ^^^:::|^. 
 
 mb — an an — om 
 
 .,.I,i;II,4i. ...^„-^- 
 
 23. A, 13000 @ 4^; B, $4000 @ 5^ ; 0, $4500 @ 6^. 
 
 24. I, 120 ; II, 114 ; III, 110. 
 
 164. 3. 12, 24, 66. 4. 66|, 1334, ^^O, 266|, 333J. 
 
 «;. =- and 7* 6. 42. 18. 7. and • 
 
 a -{- a -\- m — n m—n 
 
 8. a; = -, y = — -— . 10. — ^^. II. 2. 
 
 a—y^ a + b a — % 
 
 14. 17536. 15. I, $7700 ; II, $12600. 16. 8. 
 17. 448 and 1008. 18. — ^, — ^. 
 
 19- "^ P- gol<i^ 5 P- silver. 20. 5 p. gold, 3 p. silver. 
 
 2am-i-an-\-bm , bm + 2bn-\-an . , , 
 
 2 1 . 7 — ■ — ry— r^ > water ; -, — —^-, — ; — r , alcohol. 
 
 22. ^am + ^an + bm. : Zb7i-\-%bm-\-an. 
 
 23- (p + gi) am-\-pan + qbm : {p-\-q)bn-\-pbm-\-qan, 
 24. I, 5 : 3 ; II, 1 : 3. 
 173. T. 1 + 4.T + 10:i;2 + 12a;3 + 9a^. 
 
 2. 1 + 4:^ + 10^' + 20;r3 + 25:f* + 24x5 _^ ig^jG 
 
 3. 1 4. 4:^ + 10a;2 + 20a;3 + 25a:^ + 34a;5 + 36a:6 + 30^-"^ + 405;8 
 + 25a:io. 4. 1 + 4:2: + lOa:" + 20:^3 + 2bx^ + Ux^ + 48a;« 
 + 54x7 _|. 76^8 + 48x9 + 25x10 + 60xii + 36xi2. 
 
 5. 1 _ 4^; + 10x2 — 20x3 _|_ 25x^ — 24x3 + 16x«. 
 
 177. I. (a + b% {a + 2»), (a + b)K 
 
 4. (x + «/)s (x + 2/)i (^ + y)^' 
 178.17. a^ib-cf. 
 
464 ANSWERS. 
 
 184. I. 10+3 (5^/5 - 2\/2 - 3 Vio)^ 2. 37a/2 - 17. 
 4. a + J + c + ^4-2 (V«^ + V«c + "s/ad + V^ 
 
 + V^ + 's/cd). 
 8. a2 _ 4a + 6 - - + i . 9. 0^2 _ ^ (a; + ^). n. 1. 
 
 12. 1. 13. 'v/2(V2 4-l). 17. (^-«/)n(^-2/)^-l]. 
 1 ax-\-h {a — x)i-\-l 
 
 19. 
 
 21. 
 
 ^a + i ax—b {a + x)^ — l 
 
 185. 1.4^6^- --^. Z^-YZZ^- 4.-45- 
 
 5. ^. 6. 5V3- 7. ^^^ 8. i^^?. 
 ^3 ' a^ — b a^ — X 
 
 (Vi+Vy)^ ^^ 0^3 + ^ (a; + .2/)^ - 2 (a; + y) 
 ^* X — y ' a — x — y 
 
 9\/l5 + 41 (v^ — Vx-\- yf 
 
 II. ^ 12. _y - 
 
 X + (x^ — a^)^ / . -, nI i 
 
 13- ^ ^2 -' 14. (a + 1)^- at 
 
 X + V 2^2 
 
 a 
 187. I. x^ + 2a:?/ = (a; - ?/)2 - tf. 
 
 2. a;2 + 4:xy = (a; + 2?/)2 — 4«/2. 
 
 3. x^ + 6aa; = (a; + Saf — 9a\ 
 
 4. 4a;2 + 4:xy = (2x + yY — y\ 
 
 190. I. ^,' 2. i^^Jl^)-. 3. (a + by. 4. 6. 5. V^. 
 
 q c 
 
 nq 
 6. ahi 7. («2 _ b^)mq+np^ p. ( J4 _ 2^2^? + 2«^)i 
 
 lo. 0^ + a. II. -. 12. 
 
 (1 - m2)i (1 - ^2)2 
 
 191. I. 6, 12, 4. 2. 15, 12. 3. 47, 35. 4. 16. 5. a + 1. 
 
 6. 8, 16. 7. 64, 512. 8. 16, 48. 
 
 9. 10, 15. 10. -^-I— — Z_ II. 28, 36. 
 
AK8WBB8, 465 
 
 12. 
 
 lac' — dc fb'c — hc' m o/i 
 
 14. r- ^5- jVh, 
 
 195. I. a; = 10, - ^- 2. 1/ = ± 8. 4. 2/ = «^ ± *• , 
 
 5. a; = — a or — 5. 7. a: = a(l ± a/2). 
 8. ^ = -8-(- 1 ± V129). 9. 2/ = |- 
 
 10. a: = ± «a/|- I. ± 21, ± 27. 2. 4 and 10. 
 
 3. ± 17. 4. 36, 24. 5. 10, 15, 30. 
 
 6. 6, 10, 14, 18, or —18, —14, —10, —6. 
 
 7. 35. 8. 21 turkeys, 25 chickens. 9. 12. 10. 10. 
 
 11. 250. 12. 3. 13. Length, 45; breadth, 35. 
 
 14. -(VimM^ + a) and -{V^m^a^ — a). 
 
 15. 72 or 108. > 
 
 196. 1.81. 2.121. 3.225. 4.289. 5.256. 6. V3±3V6 
 
 203. I. X = j- 2. X — —a. 3. a; = 13. 4. a; = 50. 
 
 5. a; = 1. 6. a; = —-^ — 7. a; = 16. 
 
 8. a; = 5 or |. <), x = a^ — W± Wb^ — a\ 
 
 ^ ^ 1 
 
 10. a; = 4. II. a; = a or -• 
 
 a 
 
 37 10 
 
 203. I. a; = y or 5, 2/ = y or 2. 
 
 2. a; = — 4 or + 13, 2^ = or — 17. 
 
 Z. x=-\ ± I V~863; y = |(l T a/^=^863). 
 
 4. a: = 11 or - 7^, «/ = 15 or - 174|. 
 
 5. a; = — 21 or 4 ; 1/ = 28 or 3. 
 
 204. I. a;=1.37... or —0.156...; y=— 4.46... or —6.096... 
 
 7 , 79 14 
 
 2.^ = ^or-4;a: = -or-~. 
 
 3. a; = 2 or 5 ; 2/ = 6 or 3. 
 
466 ANSWERS. 
 
 305. I. y = ± 1 or ± \/\\ x = 2y ov -4y. 
 1 4 ^3,3 
 
 ^07. I. a; = ± 5; 2/ = ± ^- 2. a; = ± 8; 2/ = ± 3. 
 
 3. a; = ^(5±V5);t/ = |(5=FV5). 
 
 4. y = 7or2; a; = 2 or 7. 5. a; = 5 or 7 ; ^ = 7 or 5 
 6. a:= ±9; 2/ = =F2. 7. a; = ± 25 ; «/ = ± 9. 
 
 5. x= ±--===^;y= ±-7=====' 9- x=2; y=l 
 
 10. X = a(a±b); y = b{a±i). 11. a; = 4 ; «/ = 5 
 
 31 4 1 
 
 i2,x = py = -^' 13. ^ = 15; 2/ = 15' 
 
 14. a; = 5; 2/ = lor 2; ^2= 2 or 1. 15, x = 6; y = d 
 16. Time, 6 or 7; rate, 7 or 6. 17. Dist, 30 or 46|. 
 
 18. X = UVa^ + ^b^ + V«2-4^); 
 
 19. |(l± Vs) and ^(3±a/5). 
 
 20. 24 and 9, or —12 and —18. 21. 49 and 25. 
 
 22. 64 and 8. 
 
 23. m -{- n^ Vni^ + # and m + n ± Vm^ + ^^' 
 
 24. 12 men working 12 h. 25. 8 ; 10. 
 26. X = ±Q; y = ±4:. 27. 11 ; 3. 
 
 210. 7. 14075. 8. 5050. 10. n\ 11. n^ + n. 
 
 12. Lowest, 140 — 6m ; all, 137m — dm% 
 
 16, 0, 2, 4, 6, 8. 17. 951. 18. 4, 10, 16. 19. 11 or 8 
 
 21. 10 or 16 d. 22. 9 days. 23. 2, 5, 8, 11, 14. 
 
 ' 25. 2, 6, 10, 14, 18, 22, 26, 30, 34, 38. 27. 3, 5, .... 2S 
 
 28. a, a + frri' ^ + i + 1 ' 
 
 313. 6. Last nail, $21474836.48 ; all, $42949672.95. 7. 246 
 12. 5 or -• 
 
 „.. 1 ^ 1 4 1 ^ a w2-2m 
 
 314. I. -. 2. 2. 3. 10- 4. 5- 5. y 6. -^. 7. --^1- 
 
ANSWERS, 467 
 
 3 1 
 
 8. 12 — 6 + 3— ^ + etc., ad inf. 9. ^ from A to B. 
 
 ^.^ 1 3 , 1 5 ^ 27 108 
 
 215. I. ^. 2. -. 3. 1. 4. ^' 5. n* ^- lio* 7- 999* 
 
 1100* 
 
 216. I. $216.74. 3. 2.72325a. 
 
 (' + 4)"-' , 
 
 (1 + 1^) -(1+4) 
 
 (^-x^r- 
 
 «- 
 
 o ^^-^ 
 
 226 («). 1. 440. 3. 74. 4. 148. 5. 0. 6. 0. 
 
 16. 1. 17. 4. 18. 10. 19. 20. 20. 35. 21.56. 22. 0. 
 
 23. _ 1. 24. - 1. 3. '^4. 4. 148. 5. 0. 6. 0. 
 
 227. I. A^ = A^-A^]A^ = -A, ; A^ = -A^ ; etc. ; 
 
 -4io = — A^. 
 2. A^ = 16^1 - 15^0- 3- ^6 = 43^1 + 30^0- 
 
 6. A^ =:kA,-\-A,',,,.,A, = (120/{;5 + 96P+9^)^i 
 
 + (120^ + 36A;2 + 1)^0. 
 
 228. 1. n — 4: terms are omitted. 5. s — 2. 
 
 I. 120. 2. 720. 3. 40320. 4. 35. 5. 56. 
 
 I. 247. 2. 70. . 
 
 230. I. 23.3. 2. 23.32. 3. 22.5.13. 4. 132. 5. 32.52. 
 
 6. 28. 
 
 232. 2. 7. 3. 1. 4. 1. 5. 4. 
 
 233. 6. First wheel = 7, second = 5, third = 3 turns. 
 
 „._ 113 17 19 5a; + 1 
 
 345. I. 5^^. 2. — . 3. — . 4 
 
 355 * 58 ^* 72 ^* 3 (52; + 1) + a; 
 bc + 1 
 ^' a{dc + l) +c* 
 
 251. 2. i. 4. 720. 5. 24. 
 
468 ANSWERS. 
 
 6. (a) 2160 even, 2880 odd ; {h) 144; (c) 720 ; {d) 576. 
 
 7. 720. 8. 120. 9. 120. 10. 120. 11. 12. 12. 72. 
 13. 144. 14. 720. 
 
 252. 3. 360. 5. 60. 6. 90 7. 34. 
 
 253. I. 720. 2. 120. 3. 48. 4. 4. 5. ^880. 6. 14400. 
 
 254. 4. 140. 5- 6486480. 
 
 255. I. 5. 2. 5. 3. 8. 4. 6. 5- 16- 6. 17. 
 257. 3. 3, 21, 36. 5. 10. 7. 3. 8. 3. 
 
 9. (a) 1 ; (5) 3 ; (c) 6 ways. 10. 3003. 
 
 261. 3. 15 and 20. 5. (^-^I)' 
 
 263. I. 120. 2. 240. 3. 2^ 4- 81 routes. 
 
 „„^ 1 31 15 A i^i 
 
 ^bT. I. ^- 2. -, -. 3. ^, jg. 4. 3g- 5. g- 0. 3* 
 
 3 „ 1 5 3-1 3 7 
 
 7- 5* ^-r 9-u ^^-lO* "T ^^•.7- ^4.^- 
 2wm f, — — 
 
 ^^' o^M^(w2T^r^* • 10' '^* 2«* 
 
 2 2 63 1 7 9 
 
 269. .. g and j.- 2. «, -; ft g^; y, g^; <5, g^- 
 
 3 
 
 3. 2 to 1 in favor. 4. -y 
 
 5- 
 
 3m?^ (m — 1) 
 
 (m + w) (m + ^ — 1) (?72 + w — 2) 
 
 1 2^ 1 2^* 1 2» 1 
 
 ^' 2 2'* — 1 ' 22 2'* — 1 ' 2^ 2^ — 1 2« — 1 
 
 36 30 25 
 ^°" 91' 91' 91* 
 12. The chances are 41 to 25 in favor of the first purse. 
 
 I. 2^3, ^^3, 2^3, ^^3, ^^3. 2. ^^. 3. ^^. 
 
 274. I. (a) 0.429; (J) 0.159; (c) 0.813; (<Z) 0.655; 
 {e) 0.371; (/) 0.110; (g) 0.151; (Vi) 0.025. 
 
AI^SWBBS. 469 
 
 2. 69. 4. $296.30. 5. 0.4533. 
 
 6. $1000; $1666.67; $2111.11. 7. a[l — {l—py]. 
 8. 0.112^. 9. $1894. 10. $1224. 
 
 278. I. 140. 2. 70. 3. 112, 5. 22Jc. 6. 28 (7 - 1). 
 
 8. 54m2. 9. — • 
 
 283. I. l + :r + a:2 + ^ + etc. 2. 1 -\-2x + 2^x^ + 2^x^-{-etc. 
 
 3. i_22J + 2a;2— 2:«3_,.etc. 4. l + 2ic + 2a;2 4.2a;3 + etc. 
 
 5. 1— a;— .^'2 + 5a:3— 7a;4— etc. 6. l+a; + a;2 + i^ + ^ + etc. 
 
 7. 1--4X + 8x^ — 4a;3 _ 1(3^4 _f_ etc. 
 
 8. 1 _ 2a: + 2x^ — x^ — x^-i- etc. 
 
 283. I. 1— 3a: + 3a;2— 3a^ + etc. 2. l-|.2a; + a;2— a;^— 2a;^— etc. 
 
 388. I. ^^(^^-^ + 1) . ^^ ^(^ + l)-^(^ + l) 
 
 n(n ■{- iy—m{m — l) p{p + 2Z; -- 1) 
 
 3. 2 ' "^^ 2 ' 
 
 6. 3n^ — dn + 1. 
 
 289. I. 165. 2. ^(^ + l)(^ + 2)-^(^ + l)(^ + 2) 
 
 1 • /*• o 
 
 ?z (w + 1) (w -4-_2)_— _(w_— ■ 1) m (m 4- 1) 
 3. __ ^ 
 
 293. I. S^ = 210; iS'g = 2870; ^3 = 42665. 
 
 2. ^,=r-.<^,==^-^. 
 
 3. ^^^,(,^,),^^^ 2^(^ + l)(2^ Jll). 
 
 4. iVa =3i7g-3(;? + 9)+5; 
 
 3^- 
 
 iVp = 5;;^ - ^^\ - (Sp + 3^ - 25 + 1). 
 5. 5«-hi55 + 55c. 6. J L -I- ^^^ /x . ^^ + 
 
 ^^^' '• 3~;rr3* ^- 2(3"~2;r+~3)' 
 
 2/111 1 1 1_\ 
 
 ^•3\2"^3"^4 n + 2 w + 3 w + 4/ 
 
 3/- 1 1 r_\ 1 
 
 "^^ 2\ "^ 2 7^ + l n + 2/' a 
 
470 
 
 AN8WEB8. 
 
 oar . '^ar(l - r'^-i) a; [1 - (2^2 - 1) r^] , ^(1 + r) 
 
 2. 
 
 (1-rY 
 
 1-r 
 
 and 
 
 2a 
 
 300. 2. Ag 
 
 (l-r)-^ 
 
 Jr (1 — r") ar — {a-j-nb)r^ -, (J + «)r— ar* 
 3- - (1 ^ ry + T^^r ^"""^ -JiZL-Ty"' 
 
 305; z^i = ^^■3-^^2_2^+5. 
 
 3. 341° 5' 10".9 + (^ - 1) (1° 0' 9".6) - (n-l){n--2)". 
 (n _ 5) (^ _ 6) 
 
 2 ' 
 
 4. 495 + 15 {n - 5) - 5 
 
 Morning of May 23 or Apr. 24. 
 
 304. I. 1. 2. ^. 3. -• 4. -• 5. ^«. 
 
 308, I. a/8 = 2.828427 ; V2 = 1.414214 
 
 1 1.1 3 1-1-3 3 1.1-3-5 
 
 ^* ■^~2^~2.4^'~2.4.6'' "2.4.6-8 
 
 3. General term = 
 
 1.1.3.5.7. ...2i 
 
 2.4.6.8. ...2i 
 
 1. 
 
 x^ — etc. 
 
 a^. 
 
 4 (_lu l-3.5..,.(2^- l) /^Xl^^ 
 
 4- ^ ^ 2.4.6....2i ^/ 5- U'/a;i 
 6 (- 1) 0^ - 1) (2m - 1) . . . . [(t - 1) m^^ 1] 
 
 7. 1 + 1 + 
 
 — m (1 — w) (1 — 27n) 
 2T~ "^ ~3! "~ 
 
 + (l-m)(l-2m)(l-3m) ^ ^^^^ 
 
 -. a , sa 3.4 «2 
 
 3.4.5....i + 2 g^ 
 
 X 
 
 m (m + 1) (m + 2) 
 1.2.3a;''»+3 
 
 + 
 
 ....), 
 
HINTS ON A COURSE IN ADVANCED ALGEBRA. 
 
 For the benefit of students wlio may contemplate a course of reading 
 in the various branches of Advanced Algebra, the following list of sub- 
 jects and books has been prepared. As a general rule, the most extended 
 and thorough treatises are in the German Language, while the French 
 works are noted for elegance and simplicity in treatment. 
 
 To pursue any of these subjects to advantage, the student should be 
 familiar with the Differential Calculus. 
 
 I. THE GENERAL THEORY OF EQUATIONS.— In English, Tod- 
 
 hunter's is the work most read. 
 Serret, Algebre Superieure, 2 vols., 8vo, is the standard French work, 
 
 covering all the collateral subjects. 
 Jordan, Theorie des Substitutions et des Equations Algebriques, 1 vol., 4to. 
 
 is the largest and most exhaustive treatise, but is too abstruse for 
 
 any but experts. 
 
 II. DETERMINANTS.— Baltzer, Theorie der Determinanten, is the 
 
 standard treatise. There is a French but no English translation. 
 A recent English work is Robert F. Scott, The Theory of Deter- 
 minants and their Applications in Analysis and Geometry. 
 
 III. THE MODERN HIGHER ALGEBRA, resting on the theories of 
 
 Invariants and Covariants. 
 Salmon, Lessons introductory to the Modern Higher Algebra, is the 
 
 standard English work, and is especially adapted for instruction. ' 
 Clebsch, Theorie der bindren Algcbraischen Formen, is more exhaustive 
 
 in its special branch and requires more familiarity with advanced 
 
 systems of notation. 
 
 IV. THE THEORY OF NUMBERS. There is no recent treatise in 
 
 English. Gauss, Disquisitiones Arithmeticm, and Legendre, 
 Theorie des Nombres, are the cid standards, but the latter is rare 
 and costly. Lejeune Dirichlet, Vorlesungen iiber Zahlen theorie, 
 is a good German Work. There is also a chapter on the subject in 
 Serret, Algebre Superieure. 
 
 V. SERIES. — This subject belongs for the most part to the Calculus, but 
 
 CATALAN, Traite elementaire des Series, is a very convenient little 
 French work on those Series'wliieh can be treated by Elementary 
 Algebra. 
 
 VI. QUATERNIONS. — Tait, Elementary Treatise on Quaternions, is 
 
 prepared especially for students, and contains many exercises. The 
 original works of Hamilton, Lectures on Quaternions and Elements 
 of Quaternions, are more extended, and the latter will be found 
 valuable for both reading and reference. 
 
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