UC-NRLF ■:? »K WW*" b m a^n an PRICE EIGHTEEN- i JWBW, 1 ft ■ ™'.v ! FIBST MNEMONICAL LESSONS IN GEOMETEY, ALGEBEA, AND \ TEIGONOMETBY. * BY THE fe^fe. / REV. THOMAS PEMTOTON KIEKMAN, M.A, RECTOR OF CROFT WITH SOUTHWORTH. JOHN WEALE. S3if \ IN MEMORIAM FLOR1AN CAJOR1 i *-s 7 : fat *tu+ &#-^tiV*- FIRST MNEMONICAL LESSONS. ©ambritrge : Prints at the tUn&crsttg ^3rcss. May be had at Cambridge, of Edinburgh, Glasgow, Aberdeen, Dublin, Belfast, Limerick, Galway, Manchester, Liverpool, Halifax, Newcastle-on-Tyne, Birmingham, Messrs Macmillan & Co. Messrs A. & C. Black. R. Griffin & Co. J. Adam. Mr James M c Glashan. H. Greer. Ledger. T. Croker. Messrs Thompson. G. Philip & Son. Messrs Milner & Sowerbv. F. T. Smith. J. II. Beilby. FIKST MNEMONICAL LESSONS IN GEOMETRY, ALGEBRA, AND TRIGONOMETRY. BY THE REV. THOS. PENYNGTON KIRKMAN, M.A. BECTOB OF CEOFT WITH SOUTHWOKTH. LONDON: JOHN WEALE, 59 HIGH HOLBORN. M.DCCC.LII. SIR JOHN BLUNDEN, BARONET, THIS LITTLE MEMORIAL OF DAYS PLEASANTLY AND THANKFULLY REMEMBERED, IS INSCRIBED, WITH FEELINGS OF HIGH AND MOST DESERVED ESTEEM, BY HIS FAITHFUL SERVANT, AND SINCERE FRIEND, THE AUTHOR. \- PREFACE. It is reckoned a small ambition which is content to ■write a book of rudiments ; and a wise man will hardly do this, unless he knows beforehand, from his fame or scholastic influence, that the profit of the work will compensate its lack of honour. With this reflection, I have ' kept my peace' above ' ten years ;' and, finally, mis- doubting the profit which my obscurity might command, I have saved my paternal feelings, by presenting this little book to the publisher as an addition to his cheap series of elementary mathematics. To this I was moved by my admiration of the remarkable zeal and spirit dis- played by him for the diffusion of scientific knowledge. While it is hoped, that these pages will be easily under- stood by readers who are familiar with arithmetic of whole numbers and fractions, and with the extraction of the square root, it is evident, from the arrangement and treatment of the topics, and particularly from the pau- city of examples, that this little work is intended not as a substitute, but as a companion, for other rudi- mentary treatises. If the experience of others in tuition agrees with my own, I may perhaps look to reap a little praise — not mathematical, on ground like this, but simply didactical — the praise of teaching well, of which I confess myself covetous. It appears to me, that dis- taste for mathematical study often springs, not so much from any abstruseness in the subject at any point, to the student who has mastered the approaches, as from VI PREFACE. the difficulty generally felt in retaining the previous results and reasoning. This difficulty is closely connected with the unpronounceableness of formulae : the memory of the tongue and of the ear are not easily turned to ac- count: nearly everything depends on the thinking faculty, or on the practice of the eye alone. Hence many, who see hardly anything formidable in the study of a lan- guage, look upon mathematical acquirements as beyond their power, when in truth they are very far from being so. My object is to enable the learner to talk to himself, in rapid, rigorous, and suggestive syllables, about the matters which he must digest and remember. I have sought to bring the memory of the vocal organs and of the ear to the assistance of the reasoning faculty, and have never scrupled to sacrifice either good gram- mar or good English, in order to secure the requisites for a useful Mnemonic, which are smoothness, conden- sation, and jingle. I would beg to have judgment pro- nounced upon my method, not from its usefulness or beauty in the eyes of a mathematician, but from its success, good or ill, in the instruction of young per- sons, of ordinary apprehension, who have all their mathe- matics yet to learn. My only apology for the form and colloquial style of these lessons is the fact, that they were at first begun in their present shape, save a few trifling variations, for a juvenile class, which included certain nieces of mine. The readiness, with which school-girls of fair capacity, who had been well taught arithmetic, appre- hended and retained the subject by these aids, strength- ened an impression, which I had cherished for many years, that something of the kind might be generally useful. If my method finds favour with students, it will be easy for me to extend the assistance, here offered PREFACE. VU at the entrance, to the subsequent and higher stages of their mathematical career ; for I have good store of such aids, adapted to most of the leading topics in ma- thematics, both pure and applied. See a Paper " On Mnemonical Aids in the Study of Analysis," in the Ninth Volume (N. S.) of the Memoirs of the Manchester Philo- sophic and Literary Society. The art and mystery of Mnemonics has been brought into disrepute by such writers as Feinagle and Coglan, men worthy of chairs in the university of Laputa. Their cumbrous inventions are about as fit to be compared, for elegance and speed, with the eVea m-epoevra of Richard Grey, as a Dutchman's ox-wagon in Kaffirland with a nobleman's chariot in Middlesex. Concerning Grey's Memoria Technica there are two opinions; one, of those who in their student-days had the good fortune to have the book placed in their hands ; and another, of those who have learned (and forgotten?) their chronology, &c. without it. I am sometimes amused by the readiness of the latter division of persons to pro- nounce judgment on the philosophic old Doctor, with the air of men who have well considered the matter. More than one good scholar and good teacher do I know, who dispose of him coolly thus : * the difference between studying and not studying Grey's book is this, that, in the latter case, you have certain things to learn and remember, and, in the former, you have the same things to learn, and a mass of frightful jargon besides.' • Once upon a time, there was a handy man, who took a fancy to joinering. He went up the town, and bought a complete assortment of carpenter's tools, everything from a wood-man's axe to a sprigbit. As he was scratch- ing his ear in meditation about the best way to convey them home, a simple bystander suggested, ' Why don't Till PREFACE. you look out for a wheelbarrow ?' ' Because I am not an ass/ was the curt reply : then, softening a little, he added, ' Do you see, my good friend, the difference is exactly here : as it is, I have my tools to carry home ; if I took your advice, I should be saddled with both the tools and the wheelbarrow.' CONTENTS. LESSON I. Page 1. Symmetry of the Parallelogram ; Co-ordinates of a Point ; Law of the Right Line, and Forms of its Equation ; Rule of Signs in Multiplication and Division; Symbol of an Infinite Number. LESSON II. P. 16. Sections by Parallels of Diverging Lines; the Height of a Tower ; the Square of the Hypothenuse ; Incommensurable Numbers ; Product of Two Lines. LESSON III. P. 25. Intersection of Two Lines ; Elimination between Two Linear Equations ; Explicit Equation of the Line through Two Given Points; Vinculum Untied. LESSON IV. P. 32. Area of Parallelogram and Triangle ; Expression of the Dis- tance between Two Points; (a=t&) 2 and (a + 6)(a-ft); Rectan- gular Equation of the Circle; Quadratic Equations. LESSON V. P. 42. Examples of Quadratic Equations ; Signification of the Co- efficients ; Imaginary Quantities. LESSON VI. P. 48. Radius and Centre of Circle found from its Equation ; Perpen- dicular from Centre on Chord and Tangent ; Isosceles Triangle ; a5 X CONTENTS. Angles at Centre and Circumference on Equal Arcs; Inscribed Quadrilateral ; Segments of Secant or Chord through a Given Point. LESSON VII. P. 55. Expression of Angles by Numbers ; the Number it ; Definitions and First Notions of Circular Functions. LESSON VIII. P. 64. 6 2 = a 2 + c 2 - 2ac Cos B ; Expansion of Sin (A ± B) and Cos (A * B). LESSON IX. P. 70. Trigonometrical Formulae ; Area of Triangle in Terms of the Sides. LESSON X. P. 78. Similar Arcs of Circles ; Solution of Plane Triangles ; the Sides and Perpendiculars on them from the Angles in Terms of each other. LESSON XL P. 86. The Bisectors of the Sides ; Bisectors of the Angles ; the Per- pendiculars at the Mid-sides ; Simple Theorems on Proportion ; Product of Diagonals of Inscribed Quadrilateral ; to Draw a Circle through Three Given Points ; Radii of Circumscribed, Inscribed, and Escribed Circles ; Area in Terms of these Radii. LESSON XII. P. 96. Similar Polygons; Ratio of Circumference to Diameter; Squaring the Circle. LESSON XIII. P. 103. First Notions of Exponents. LESSON XIV. P. 110. Logarithms and the Tables. CONTENTS. XI LESSON XV. P. 117. Problems in Geometry and Trigonometry. LESSON XVI. P. 124. Area of a Polygon in Terms of the Co-ordinates of its Vertices; Perpendicular from a Given Point on the Line y-ex- b = 0. LESSON XVII. P. 135. Meaning of e in y-ex-b = 0, for any Axes; Distance be- tween Two Points for Oblique Co-ordinates ; Equation of the Line Perpendicular to a Given One ; Equations of the Bisectors of Sides and Angles of a Triangle, and those of the Perpendiculars ; the Intersections of these Lines ; the General Property of Seg- ments of the Sides of a Triangle made by Three Lines from the Angles to a Fourth Point ; the Expression of a Sought Line. LESSON XVIII. P. 144. Arithmetical, Harmonical, and Geometrical Progressions. LESSON XIX. P. 149. Permutations, Combinations, and Variations. LESSON XX. P. 155. The Binomial Theorem. LESSON XXI. P. 162. Explicit Circle through Three Points, and Curve of Second Degree through Five Points. P. 169- Juvenile Conversations on the opening Lessons. MNEMONICAL LINES. [1], [3], &c, mark the Mnemonics; (1), (5), &c., the Sections or Articles in which they occur. Q] (1) Line cutting paralls. makes alter, hits, equal, and ext. equal int. op. And cutter cuts paralls., if equal alter, ints., or ext. equal int. op. Prop. B. [2] ... A pargram. has equal op. angs., or op. sides. 'Tis pargram., if equal op. angs., or op. sides. [_3~] (5) Like sign's give plus, Unlike minus, For sign of pro. or quo. C^G (6) ex is y gives ori. li. [5~] (7) The line (Dif. yb is ex) is pari, to (y is ex). pron. yb, ivybe. \_5'~\ (8) One 's Vi (xl) and vi. (ya) pron. vixle. From 'Or. cuts 1 and a. [6] (13) If paralls. cut legs, vi. segs. is vi. segs., And vi. (parall. cutters) is vi. (corre. segs.) : Measure the segs. from meet, of legs. \J~] (14) Qua. poth is both qua. sides. L 8 1 ( 15 ) P^ r - on P ^ is mean segs ; and side is mea. (poth. nigh seg.) \jf\ (17) ViD(y's) is ViD(x's), pron. vidwise. with 10 o'er 12 — dexes: read ten o'er twelve. ter x'Di(y's) is nil, pron. Dhvise. at 612 round, for gil. dl2 read owe untwo. XIV MNEMONICAL LINES. ART. [10] (18) C's Ax and By; Dot second li : Di(CA)'s by Di.(BA)'s, (dot outs.), is meeting Y. [11 ] (19) After minus untyin' Change every sign. [12] (21) Poth.{Di(.z/) Dl(yd)}, pron. dixie, (ya) a monosyl. joins(^) to (la), xy a dissyl. in co-ords. recta. [13] ... D UQ{T>i(xP) Bi.(yd)} pron. dixie ; yo, one syl. Is r r*, (in Recta) rr and la both dissyll. Gives circ. cen (Id) : [14] (22) QuaSorD(a&) is DUQ(rf6) mol two(ai) : ab one syl. Sum(bd) .Di.(bd) is Sq'.o le sq. a. pron. squibble squa. [15] (24) Ifsq.g' le two 7/ a be 2V, pron. s?ui. y's a mol RooM asj A 77 , pron. ask. [16] (26) If nil be (sq.y' le py' and q') pron. sg»i. j>'s sum, and q is prod, of roo. [17] (29) DUQ(t/'o:) is Ty' yx, pron. cot*. Has tan. nil's y': a. Rad per'p on chor.'s biCh6r : b. Toiich-r'ad is per' on tan : [18] (30) If iVs b, An'g is Bang ; and per'c is bic' is biCang. [19] (31) Rim-angle on arc is arc-dcmi, And right is the angle in semi. [20] . . . (Seg . S.eg) in P - ring, Is mol SU'D (poc wi'ng), For P out or in ring. Pron. pout. MNEMONICAL LINES. XV ART. [21] (33) x and y are Cos and Si of Scale-a'rc from OX ; cen. o'r. Ax Rlgh. [22] . . . So'rCo.Po'th is o'p or ad; H. Si by Cos ta'n, is vi(op. a'd). K. [23] ... Co'rSin a>' 's SinorC(ri'ght min w') G. w , pron. 6. Co'rS is le'm the Co'rS of — ple'm L. CorS(7r et w) is le' CorSw. L'. pron. peta.'. [24] . . . Tan Cos Sin are rec. Cota Sec Cose'c. [25] ... DE7Q(SlCo)'sOne: (M) Co right is none Co none is one. [26] (34) Draw pe'rc: SUD (ba) is SUD (segs of c'). And sq'.b is DUQ ac mol (seg. op) two c, A. sq'.b. pron. squib. As 'tuse or 'cute is Bang op. b ; Or, sq'.b is DUQ ac le CoBang two ac. B. [27] (35) Sin (a mol t) is Sa'.Cor mol Ca'.SiT ; pron. a,T,asa, t. Cos (a mol t) is Cd.Cor lem Sa'.Sir. mol t, one syl. c perc is ba Si Cang ; then put (a mol t) for Cang ; (new — old a) make right ang. [28] (36) HaS (ab) mol HaD (ab) is a or b ; [29] . . . SiMmol SiD's two SorC.CorS, C6D mol CoM's two CorS.CorS. [30] ... Sa mol Sir's two So'rCHaM. Co'rS HaD, Ca' mol Cot's mol two CorSHaM.CorSHaD. SH and CH pron. as in Cheshire. [31] ... CorS is SU'D or two CHa.SH. XVI MNEMONICAL LINES. ART. [32] (36) Roo'P slebi.slec.sla, Is Sine Bang half ca, Is CHaSHca, is Area. Pron.CHand SH asinCheshire Write ' sqb' in quaCH and quaSHaBang, By « CorS is SUD'...and ). [55~] ... e, bas[Di(xl) w D(ay)] joins xy to la, .ry adissyl. Di(xl) pron. dixie. c, LiL Di/3's you know by ' tan(a mol to').' [56] (5G) gil's le cota./3 is perlin's ta/3. [57] (57) a Cos A'ng's b Co.Bang, a Sin Ang's b SiBang, b's a, (in Rom ba) are perc bic and biCang. [v. 18J [5S] (58) P...Ang cuts a' in dotA', &c, AB'.BC'.CA"s AC'.CB'.BA'; dot alterna' Then for lines put oppo. sines. pron. abbicca, acklbba. [59] (59) Di(xe)'s pron. phi. Nil's vle<£u is soughl through(vu) ; <£u pron. fu. (y le of = times. L -^ ) ' \ a le ze by D(une) is Sum Geo. Se. H. One by D(une) is fr8 on e, is Slim, if e's frac. of an infi. Se. [62,2 (6 C 2) If p Clems, have m a's, e b's, i c's, The perms, in p's are p fags by (m fags, e fags. i fags). MNEMONICAL LINES. ART. [6.'5] (64) Comb, non-repea.'s of n in d's, Are d w-backs by d fags. [64] . . . The repe. combs, of n in d's, Are d ;i-ups by d fags. [65] (65) Non-repe. vars. of n in d's Are d n-backs. ^66^ (65) The repe. vars. of n in p's Are n to p tb . [67] (67) Siit6n(un r)? write fr8 on r; Then r to i you multiply By (i n-backs by i fags) : If n has den.e, Put r vi(re); top dits wed e. [68] (70) Round ev. or o. with pin or no For signs you go. SYMBOLS AND ABBREVIATIONS. + is the sign of addition ; plus. — of subtraction ; minus. = of equality ; equals : sometimes, which equals. 3 + 2-1 = 4 is read, 3 plus 2 minus 1 equals 4. x is the sign of multiplication; times: a point is often used instead of x . -j- or : is that of division, as 12:(3 + 1) = 12 + 4 = 3; a:(x + y) is a divided by (x + y). > means is greater than, sometimes greater than. < means is less than, sometimes less than. J_ stands for perpendicular, or perpendicular on. Between algebraic quantities written close together x is always understood; thus (5 -2) (6 + 1) is 3 times 7; (a + 6) (x + y) is (a + b) times {x + y)\ xy is x times y. Quantities collected by a tie or vinculum are not to be treated as if they were untied. The whole tied quantity is affected at once by x preceding or following; as (5 — 2) x (6 + 1) above. But (5-2)x6 + l = 18 + l =19, 5-2 (6 + 1) = 5 -14 = -9. CORRIGENDA. PAGE LINE 7 5 exchange x and y. 7 19 for X"OY" read X"0'Y". 10 3 for q or n read p or m in this locus. 12 12 for a?, read x. 1 5 last line, for + read -. 20 8 from bottom, after BC, insert [1]. 22 16 for A.B read AB. 35 11 for jn,<7, read jd,?,. ... 12 for g 3 7 3 read ;v/ 3 . 32 7 from bottom, for Cf= cf read c/= i.e. a = twice b times c, divided by 3 times d; and from this, multiplying by 3d, I deduce 3ad = 2bc, thrice a times d is twice b times c. All these equations must be true, if the first pair are true. Suppose now that I have the information that the product of a and d is 10, without knowing whether these XXIV TO THE YOUNG READER. are whole numbers or fractions ; and that I am told in addi- tion, that b and c are whole numbers, and b greater than c. The last equation shews me that 30 — c 2bc, whence it follows that I5 = bc; from this I gather that c is either 3 or 1 ; because 15 cannot be the product of any whole numbers, except the pairs 5 and 3, and 15 and 1. All this is premised merely to shew the young reader, to whom algebraic characters are new, that we can often reason with symbols of unknown quantities, and, by easy arithmetical operations upon them, arrive at conclusions that may greatly increase our knowledge. The reader is supposed to know the meaning, and to see the truth of, such assertions as these : 5- -2 = = 5 + (- 2) = = 5-(+ 2) = 3, 5-(- -2) = = 5 + 2 = -7; a- -b = * + (- ■ by. ■ &) = -a- (+ - a + b. *)> See the conversations at the end of this little volume, which are intended to be read by the beginner, as notes on the opening lessons. FIRST MNEMONICAL LESSONS IN GEOMETRY, ALGEBEA, AND TKIGONOMETKY. LESSON I. Uncle Penyngton, Jane, Richard. Uncle Penyngton. 1. You are now, my dear children, familiar with the tour rules of arithmetic in whole numbers and fractions, and with the extraction of the square root ; and Richard can prove, from the first book of Euclid, the pi-opositions following : Prop. A. Two intersecting straight lines, {AD and CF) make either four equal angles, (all right angles), or a pair of acute angles, (each less than a right angle) and a pair of obtuse ones, (each greater than a right angle). The two acute angles are vertically opposite and equal, and so are also the two obtuse ones ; and any unequal pair, an acute with an obtuse angle, make two right angles. Prop. B. A line cutting a pair of pa- /* rallels, (lines which can never be produced Q ty ~ to meet), makes the exterior angle (BAC) g -f~ — equal to the opposite interior {ADFf), and two alternate interior angles, {FAB, ADE) equal to each other : and, conversely, if a line cut two lines so as to make either the two alternate interior angles equal, or the exterior angle equal to the interior and opposite, these two lines are parallel to each other, and can never meet. opposite, on the other parallel; interior, within the parallels ; alter- nate, on different sides of the cutting line. Prop. C. A parallelogram, (which is defined to be a four-sided figure whose opposite sides are parallel lines), has A 2 FIRST MNEMONICAL LESSONS. [§ 1. its opposite sides equal, and its opposite angles equal: and conversely, if a quadrilateral, (i.e. a / / four-sided figure) has either two pairs of equal / opposite sides, or two pairs of equal opposite angles, the figure is a parallelograrn. Prop. D. Any exterior angle (CBD) of a triangle is equal to the sum of the two interior remote, (CAB and ACB), and the three interior angles are together equal to two right angles, i.e. to an exterior with its interior. Observe that when an angle is denoted by three letters, the one stand- ing at the angular point is placed in the middle. ABC, BCA, CAB are the three angles B, C, and A of the triangle. As Jane has never learned Euclid, she may for the pre- sent take it for granted that these propositions are true. They are so nearly self-evident, that it is one of the most difficult things to establish the truth of them all by rigorous demonstration ; and neither Euclid, nor any other geometer, has done this by arguments that you could at present com- prehend. An assumption of some kind is always found necessary, which requires demonstration as much as the propositions to be proved thereby. Jane : — It appears to me that the propositions are much easier to believe and comprehend than to remember. Uncle Pen. : — You know, from good old Dr Richard Grey, the value of contraction and cadence as aids to me- mory ; and I recommend you strongly, before you begin to try to remember these properties, to learn perfectly by heart the following mnemonical aids, referring to Props. B and C, and to teach them to your ear and to your tongue, each of which has a memory of its own, by saying them again and again with a sing-song repetition, marking well the accented syllables. Ql] Line cutting pdralls. makes alter, ints. equal, and ext. equal int. op. Prop. B, And cutter cuts paralls., if equal alter, ints., or ext. equal int. op. [£] A pargram. has equal op. angs., or op. sides. Tis pargram., if equal op. angs., or op. sides. Prop. C. paralls for parallels; alter, ints. for alternate interiors; ext. for exter- nal ; op. for opposite. Observe that in 1. or Prop. 15, there are four ext. angles, and 2 pairs of alter, ints. viz. a pair of equal acute, and a pair of equal obtuse angles. I shall now proceed, according to a promise I recently § 2.] FIRST MNEMONICAL LESSONS. 3 gave you, to show you the easiest way, (for there is no little difference as to difficulty in the approaches) to the de- lightful fields of geometry, through which alone you can be introduced to the sublime regions of Mechanical Science, terrestrial and celestial. And the easiest way in this in- stance is the shortest way. First of all, we must express positions of points by num- bers, and thus reduce questions about points, lines, and areas, to questions of arithmetic. Suppose a group of points before you, how would you contrive this ? Richard: — It would be easy to number them one, two, three, &c. ; would that do ? By this plan we should have a number for every point, and should know what we were talking about. Uncle Pen.: — You might thus name your points arbi- trarily ; but we must have a system, by which all numbers shall determine certain points. I am about to show the happy contrivance of the celebrated Des Cartes, which fixes the position of a point by a pair of numbers. 2. Let two leading lines be chosen in the plane of the paper, supposed unlimited in every direction, which meet in a point 0, called the origin. XOX' is the axis of x, and YOY' the axis of if. ZZl A" is read X dashed. V is Y dashed. If we take a certain length for our unit, and call it an inch, and measure from along OX, say 3 inches to p, and from p in a direction parallel to OY, say 2 inches to q, we have the point (x = 3, y — 2), or, more briefly, the point (3, 2), viz. the point q. Op is the x, and pq is the y, of this point. The point q\ whose x and y are 2 inches and 3 inches, is the point (x = 2, y = 3), or (2, 3). The x and y of a point are called its co-ordinates, and the leading lines OX and OY are called axes of co-ordinates, or co-ordinate axes. If we draw qr parallel to the axis of x, we may call Or, which by \Jl~\ is equal to pq, the y of the point q, and qr [2] the x of it. Thus having given us the co-ordinates A2 r ? •i FIRST MNEMONICAL LESSONS. [§ 2. x = 3 and y = 2 in inches, we can find by measurement the position of the point (3, 2), or q: and if this point had been given us in position, we could have found the co-ordinates, by simply drawing from q a parallel to OY, and then mea- suring the inches in Op and pq. I shall assume once for all that we have the power of joining any two given points by a line, of drawing a parallel to any line, and of measuring lengths to any degree of accuracy, say to the millionth part of an inch, if required. Jane : — This is very ingenious of Monsieur Des Cartes ; but I do not yet see how confusion is always to be avoided about the point (3, 2). If the axes were chosen at right angles as thus, (or indeed at any angle), I cannot see what right q has to be considered the point (3, 2) rather than k or m, or n ; all which are alike determined by three inches from on the axis of x, and by two inches on that of y. Uncle Pen. : — Your objection is well stated. The points q, k, m, n, are called (x = 3, y = 2) or (3, 2), (x = -3, y = 2) or (-3, 2), (x = -3,y = -2) or (-3,-2), (x = 3,y = -2) or (3, -2). The directions OX and OY are positive or above nothing: their opposites OX' and OY' are negative or below nothing. The length — 3 (read minus 3) is 3 negative, or below nothing : and the difference between 3 and — 3 is, that the first denotes a length measured, from whatever point, or on whichever axis, in the positive, while the latter is a length backwards, in the negative direction. If a point moves by an inch at a step from b to c, its distances, mea- sured from or zero, will be after the successive steps, from restate, 3-1 = 2, 3-2 = 1, 3-3 = 0, 3-4 = -l, 3-5=-2, 3-6 = - 3; i.e. it will have a distance from after the fifth and sixth steps, —2 and —3, or 2 and 3 negative, below nothing. It is of no consequence, when we choose our axes, which direction along either we make the positive one ; but we shall in general consider OX to the right, and OY upwards, to be the + (plus) or positive, and their opposites to be the — (minus) or negative directions. Observe that the x of a point is often called its abscissa and the y of it its ordinate. 3. You know, when our axes and our unit length are chosen, what point is given by the two statements, §3.] FIRST MNEMONICAL LESSONS. X - ~ s, read x equals 3, y-- = 2, y equals 2. From this ation pair of assertions, or equations, follows the X 3 ~2' read x by y divided by . equals 3 by 2 ; Consider now the points 3 2 3 * 2= 4 3 * 3= 8 3 ■r 5 = '6 ^ 6 = *9 1 1 1 ^ 3 = 4 l _y 5 ='4 y 6 = '6 &c. Read x x y u tr at 1, y at 1 ; a* a y 2 , x at 2, # at 2, &c. ; these subindiccs serve merely to show to the eye that x and y, Xi and #„ &c, are co-ordinates of different points ; x a y a , x b y b , (x at a, y at a, x at b, y at b, Sec.) would answer the purpose as well. From these six pairs of equations follow in order by division of equals by equals, 3 3 3 Xi _ 3 x 2 _ 4 _ 3 Xs_8_3 4 _ ? . I _ ? ? _ ? V;~2'* 7 2 "l _ 2 ; .%~T~2 ; T = 4"2-4 X l-2- _3 jr 4 - i ~2 0.6 0.4 0.6 Of the x and ?y of any of these six points, all within a short distance from 0, can be affirmed, as of our first x and y, t \ x 3 W - = - , ,7/2 that the quotient or proportion of every pair is the same. And we can find any number of such points, as near to each other as we please. As from such an equation as 9 3 6 follows, if we multiply these equals by 6> 9x2 = 6x3; so from the above equation follows, if we multiply its equal members by yx2, (j/ times 2, or twice y), 2x = 3y, (twice x is 3 times y. 6 FIIIST MNEMONICAL LESSONS. [§ 3. whence dividing these equals by 3, (a') y = ^- , y = two-thirds of X. which is the same law, or condition between x and y, with (a). Richard: — I do not see what information is conveyed by (a) or (a') ; for the x and y in them have no more to do with our point (3, 2) than with any other : x and y may mean anything. Uncle Pen. : — Not exactly. Although (a) and (a') do not state the values of x and y, they both affirm the same property; the former tells us, that if these numbers vary, they must always maintain one .proportion : the latter asserts, that whatever number x may be, y must be two- thirds of it. If we we obtain (a") divide the equals 3 above by 2 read a instead of by 3, 3 3 • is - of y, or - times. the same 1 aw still. Observe here, that when we wish to represent the product of a number and a symbol of a number, or that of two symbols, we write the two quan- tities together, 3.r, ca\ or else with a point between them, as 3.,r, e..r.. You may read these either three a; ex, or 3 times x, c times m ; the latter is the best and safest way. And sometimes the product even of two numbers is represented by writing them together with a point between them ; but care should be taken to place the point in this case at the bottom between the figures, to distinguish it from the decimal point. Thus 3.5 = 3 times 5 = 15 ; but 3-5 = 3 point 5 = 3.3. Observe that the best way to read a decimal frac- tion is to call the"point point ; as 0-1, 6*01, 7200f>, are, point 1, (J point nought 1, 72 point nought nought 6. If then we choose any length upon OX for x in (a), the corresponding y is always | of that length : x ami y so found give a point in the series. Put then for x in order the numbers 0*1, 0*2, 03, &c. and you obtain in order for y ° 1 ° 1 — = — of an inch, — , - , &c. : and thus a series of points 3 15 15' 5 r are found pretty near to each other, viz. VTo' T5J' \5' 15J' Uo' 5)' &c - How will these look when found? Richard: — They will form a dotted trace of some kind. 1 wonder, what would be its shape and direction. § 4.] FIRST MNEMONICAL LESSONS. 7 9.x J ane : — The equation y = — , though it gives no infor- mation about any particular point, may perhaps tell us something about the figure formed by them all. Is it so ? Uncle Pen. : — We shall see that presently. First of all this equation («') shows that if y = 0, x = also, so that the point 0, which is (x = 0, y = 0) is one point in the series. The figure is therefore either a bent line, or a straight line, passing through the origin. I will prove that it is not a bent line. 4. For if it is a bent or curved line, let any right line 00' through the origin, which meets it again, meet it again first at some point 0' : the co-ordinates of 0' can be drawn ; let them be (x = m, y = n). Place yourself now directly opposite to me, and taking O'X" and O'Y" drawn pa- rallel but in opposite directions to OX and OY, for your positive axes of x and y, find by equation («') referred to your axes, a series of points in the locus (a'). The angle X"OY"=XOY, and my origin will be your point (x = ?n, y = ji), by [2] ; and as the conditions determining your curve and mine are ex- actly similar, O'O will meet your curve again first in 0, and your figure Avill appear to you as mine to me, and will be concave or convex to O'X" as mine is to OX; so that the two curves will lie on opposite sides of 00, and will have no point in common between and 0'. Let the point s, whose ordinate meets OX and OX" in I and v, be any point of my series between and 0'. At the points 0' and s, by (a'), I have 0'b = %.0b, 2 s t = _ . Qt^ or subtracting equals from equals, 0'b-st = -. (Ob - Ot), or because O'b = tv, by [2], iv _ st = - . tb, or since tb = O'v by [2], 2 n , sv = -. Ov; Note. Ob, O'b, si, &c. stand here for the number of linear units, or inches, in the designated lines ; and we shall frequently have occasion to substitute lilies for numbers, which express their lengths. 8 FIRST MNEMONICAL LESSONS. [§ 5. from which it appears, comparing this with («') as employed by you, that s, any point in my series, is a point of your series; or the two curves have all their points in common be- tween and 0' ; which is absurd. This contradiction is involved in the supposition that the series of points is a curve in any part of its course ; for to such a curve a line can be drawn from meeting it again first in some point 0'. Hence it is not a curve, but a straight line 00'. You know that (x = - 3, y = -2) is a point within the angle X'OY'. Now as three negative inches are thrice one negative inch, — 3 = 3 x — 1, and - 2 = 2 x - 1, whence by division of equal pairs, -_3_3_x-l -3 _3 ^2 ~ 2 x-1 ' ° r ^2 ~ 2 ' which, compared with (a), proves that (- 3, - 2) is a point of the series formed from (a) or (a'). And as m negative inches, whatever number, whole or fractional, m may be, is m times one negative inch, — m m x — 1 m —n Hx-1 n so that it — = - , = - likewise. n 2 -n 2 This shows that for every point of the series, (w, n), in the angle XOY, there is a point exactly corresponding to it in position, in the opposite angle X'OY', namely, (— m, —71). Hence it appears that the series will have the same figure, as to the axes, in both angles, or the points form a straight line extending in opposite directions from 0. — 33 5. From — = -, it appears that the quotient of two negative numbers is positive. It follows, that the product of two negative numbers is positive ; for — 3x— 2 — — 3-. — — , equal the quotient of two negatives. Here — - cannot be positive, for if — = p, a positive quantity, it follows that l=-2xy, a negative quantity, which is absurd. We have then C] FIRST MNEMONICAL LESSONS. 1 1—1 .in in — in m in — m The product of a positive and a negative quantity is nega- tive ; for 3 x — 2 inches = — 6 inches, evidently. Hence the quotient of a positive by a negative, or of a negative by a posi- 3 1 — 3 1 tive is negative. For — = 3 x — - ; and — = — 3 x - ; both a -2 -2' 2 2' which are products of a negative and a positive. These results are easily retained by the following rule. The product or ywotient of two quantities having like signs (whether both positive or both negative) is positive. The product or quotient of two quantities having unlike signs is negative. Or you may say it thus : [[3] Like sign's give plus, Unlike minus, For sign of pro. or quo. 6. We have made no restriction as to the angle XOY, and whatever this may be, the equation y = ^x represents a line drawn through within that angle, if y and x are the co-ordinates of a point referred to the axes forming it. Let now two other axes OX, OY, be drawn, as above, to the left of our former ones, making the angle XOY = X'OY; and let x and y be the co-ordinates of a point referred to these new axes, while x and y retain their sig- nification as to the old ones. The locus y 2 , . , . 2x | = -, which is y = -g", a, is a right line ; let it be OpmO'. The locus •- = --, which is y = ~- hich is y = -— , A, A. 10 FIRST MNEMONICAL LESSONS. [§ 6. is next to be considered. The quotient of y and x is here 2 negative, being = - - : therefore they cannot have like signs, by fjf]. If x is negative, as at q or n, y must be positive, 2 and in length equal to - of x, by A. Let ° r r°r?\ then will (WW 0« = On J ( wj« = mn, 2 2 for by a, pq = ^ . Oq or = - . Oq ; which by A is = pq, and 2 2 mn-^On or =- On; = mn. Now as the angle X'OY — the angle XOY, the points p and w, which are found by the same measurements along the axes with the points p and m, form with the same figure that p and m form with O, the figure on the right being exactly what the figure on the left is when seen through (he paper. But O, p, m, have been proved to be in a line ; therefore 0, p, m, are in a line, and in the same way it can be shown that every point in the locus A is in this line Opm, which extends in opposite directions from 0, as Opm does from O. Qx 2x We have thus demonstrated that y — — , and y = — — are true of points (xy) that lie in given right lines, and of no other points. These lines pass through the origin, and these equations are called the equations to those lines. If for | any other number be substituted in either of the equations, the ordinate y corresponding to a given value of x, as x = I, will be lengthened or shortened, and a dif- ferent line through the origin will be represented for every different multiplier of x. If for § be substituted, in all the preceding argument, the symbol e, representing the frac- tion |, then will it be proved equally, that // = ex, and y = — ex, are equations of the same lines through the origin : and if e represents any constant number, which docs not vary with x and y, then it is equally proved, that y = ex is the equation to the corresponding line through the origin, and this whether e have a positive or a negative value. We have thus established the following proposition : The locus of the points (xy) whose co-ordinates satisfy the § 7.] FIRST MNEMONICAL LESSONS. 11 equation y = ex, e being any constant number, is a straight \\ne through the origin. You may say this briefly thus: \^4T] ex is y gives ori. li. ex long. 7. We find a point in the locus 1/ = ex by taking in OX a length x x lor x, and raising at the extremity of x x an ordinate of e times that length, ending at a point q. If we carried this ordinate one inch further in the positive direction beyond q, we should arrive at a point p, in the locus y — ex + 1. By adding an inch to the ordinate of any other point q l q 2 q 3 &c, found in the locus y = ex, we should obtain as many points p p x p 2 p 3 &c. in the locus y — ex + 1. The points p x p 2 p 3 , &c, would form a line everywhere an inch distant along the ordinate from the line q q x q 2 q 3 , and would therefore lie in a parallel to the line y — ex; for these lines can evidently never meet. If we add a negative instead of a positive inch to every ordi- nate, i. e. retreat an inch along the ordinate from q q x Sec. in the direction 0Y X , we shall ob- tain a series of points similarly placed on the negative side of / the line q q x q 2 ... , being all * points in the locus y — ex — 1. If qp = q x p x = q 2 p 2 = &c. had been b inches instead of 1, b being any number of either sign, whole or fractional, p p, , &c. would have been points in the locus y = ex + b. We shall often speak of this locus, as the line y = ex + b. We have thus established The locus of the points whose co-ordinates satisfy the equation y = ex + b, where e and b are constants, is a line parallel to the line through the origin whose equation is y = ex. y = ex+b gives, by subtracting b from both sides, y — b = ex, and this is evidently the same locus with y = ex + b. Put Dif. for difference : call y-b Dif. (yb), (pron. wyb), the difference between y and b, or y minus b : let pari, stand for parallel. Then £5] The line (Dif. yb is ex) is pari, to (y is ex). 8. Every line through the origin has an equation of the form y = ex. For let {x-x i ,y=y x ) be a point in a given 12 FIRST MNEMONICAL LESSONS. [§ 9. line, then is y =— a line through the origin, by \jf\, for — is a constant number ; and this locus contains the points (0, 0), and (x l} y } ), two points in the given line, as appears, if for x and y in the equation be put their values at those points ; "wherefore this is the equation to none other than the given line. Here x t and y x are two known numbers. Every line has an equation of the form y = ex + b. For every line is parallel to some line through the origin. Let a given line be parallel to the line y = c x x, and let it meet the axis of y at a distance b x from the origin; e x and b x being known numbers, then is the equation y = e x x + b x that of a line parallel to y — e x x x , \J>~\, and of one which passes through the point {x = 0, y — b x ). Through this point there can only be one line, viz. the given one, drawn parallel to y = e x x ; therefore y = e x x + b x is the equation to the given line. 9. Every line has an equation of the form -^+-=1. Read x by l + y by a equals 1. Let a given line be, e and b being given positive numbers. (c) y = ex + b ; then, dividing equals by b, ye e '■j- = - . x + 1 , or subtracting from both - .r. b b o e for the fraction -j- is unchanged in value by the dividing both the numerator and denominator by the same number e. Thus (c) is reduced in (c') to the form +--=l, in which / and a, being general symbols, may of course have any particular values a = b, and / = --. The equation under this form (c) exhibits visibly the position and course of the line (c): for it shows that if x = 0, y = b; and that §9.] FIRST MNEMONICAL LESSONS. 13 if y = 0, x •= ; i. e. the line meets OF at 6 positive inches from the origin, in the point (0, b), and OX at - negative inches from the origin in the point ( — , 0). Thus we obtain instantly from (c) two points of the line, and can therefore draw it, when our axes are given : and these are always supposed to be given. We can thus draw, for example, the line 3 2 2 y — — x + — ; which is, dividing the equals by - , 4 7 ' 1 4x? 7 7 2=JL, i 2 4 2 7 3*7 7 21 this is now of the form y:a + xd = 1, a being the fraction 2:7, and I being —8:21. The former is the length cut off from on the axis of y ; the latter is that intercepted from the origin on the axis of x. 2 y It will often be convenient to write the fractions -, -, 7 o> &c. in the form 2:7, y.a ; and you may read these either two to 7, y to a, or two by seven, y by a ; the ratio of 2 to 7 is the same number as the quotient of 2 by 7. To remember this, you may add to the last mnemonic [5], the words Q5'] One's Vi (xl) and vi. (ya) pron. vixle : and is +. From 'Or. cuts 1 and a. Here vi. denotes quotient of: division ; One 's means One is, or 0>ie =. The line 1 = quote of {y by /) and quote of (x by a) cuts from Origin (on OX and OY) the intercepts / and a. The constants b, e, I, a, in [5] are not in general whole numbers ; nor would every line be reducible to the forms y = ex + b, and xd+y:a = l, if these four constants were 14 FIRST MNEMONICAL LESSONS. [§ 10. not capable of representing any numbers of either sign. If a line is to be represented in general without fractions, three constants are required in its equation. 10. Every line has an equation of the form Ax + By = C, in which the constants are not fractions. 3 2 Thus the line y = - X + - , by multiplying both sides by 4 x 7, becomes 2Sj/ = 2Lr + 8, or, taking 21* from both equals, -21x4-28^ = 8, which is of the form Ax + By = C, without fractions. Show now that the line 13x + 5y = 8 is parallel to x:y =- 13:5, and that it cuts off from the origin a = 8:5 and / = 8:13. To draw any given line y = ex, it is only necessary to find one point of it, and to join that to the origin. The point (1, e) is such a point, evidently. 11. The axis of y is x = ; the axis of x is y = 0. No point whose x is not nothing is on the former axis ; nor is any whose y is not nothing on the latter. Do you see this, Richard? Richard: — Plainly: and it is evident that every point whose x is nothing is in OY, and that every point whose ordinate is nothing is in OX. Jane: — But I do not see these equations, y = and .r = are of the form y = ex : if e have the value zero, indeed, y = is the result; but what must e be to bring out x= 0? Uncle Pen. : — The equation y = ex is true still when both sides are divided by e : i. e. - = x, or y x - - x, follows from if = ex. Now the fraction - diminishes as c increases : if e J c increases beyond all conceivable limit, - lessens beyond all limit: if e becomes infinite, - dwindles to zero, and the e equation y . - = x is then = x. The equation y = is the case ofy-cx, which arises from the supposition, e = 0: x = § 11.] FIRST MNEMONICAL LESSONS. 15 is that which springs from the supposition, e = - , which is the symbol of an infinite number. Let r be an indefinitely small number; the quotient - is indefinitely great: e.g. OWOOOI is ten millions ' If r = (-0000001) 2 , and e = I , e is 100 billions: if in r the number of zeros between the point and unity be unlimited, r = 0, and e = - is then e = - , which is greater than any finite number. You know that the number - is called the reciprocal of 5. Infinite is the reciprocal of zero. Jane : — I know also that a number times its reciprocal is equal to unity ; 5 x - = 1. Is then x - = 1 ? Uncle Pen. : — The quantity x - or — is called an inde- terminate quantity. Zero by zero gives any quotient you please ; for 5 x 0, and 1x0, are equally 0. This is sufficient for you to know at present : we shall not have any occasion for some time to handle the quantity - . Consider the two lines y — ex + b, y = — ex + b, e and b being the same pair of fixed numbers in both. Both equa- tions are true at (0, b) i.e. the two lines meet in (0, b). When x — n, the same value in both, we obtain y x — en + b, y 2 = — en + b, the subindices serving for distinction between the two ordinates, which are both measured on the same parallel to OY. By addition, y l + y 2 = 2b, whatever be the values of e and n. Draw a parallel to OX through (0, b) : we have proved that if this and our two lines be cut in A, B, C, by a parallel to OF through any point P of OX, PB + PC=2PA. Draw the figure and examine this. 16 FIRST MNEMONICAL LESSONS. [§12. LESSON II. Richard: — When shall we be able to understand the secret that you promised to show us, how to measure the height of a tower without ascending it ? Uncle Pen. : — You shall know it presently, if you will pay close attention to the following argument. 12. Let OP be a line through the origin and any given point P. If we put the co-ordinates PQ, OQ, &c. their lengths in inches, for .(A) PQ we know that y = ,.-„ x, ,..,., y PQ or, dividing by x, J = q- q , is the equation to OP ; for by [4], this is a line through the origin O, and it is true for (y = PQ, x = OQ), so that the line whose equation is (A) contains two points of OP. If p and p' are points in OP, B Pq ~^ and -rt'- fQ r- for (A) must be true at both these points. You may read B either pq by Oq equals PQ by OQ, or pq to Oq as PQ to OQ, and I recommend you to read every equation aloud. If we choose now to consider OP and OY, instead of OX and OY, to be our positive axes of x and y, the point Q referred to these, is (*' = OP, y' = ~ Ob or - QP) [2]. And the equation to OQ is OP (A'). lor (A') is « line which contains tiro points of OQ, and can § 12.] FIRST MNEMONICAL LESSONS. 17 therefore be none other than OQ. The dashes merely show- that y' and x' are co-ordinates referred to the new axes. As (A') must be true at q and q', points of OQ, -qp -PQ , q'p' -PQ From the pair of equations a - Oq'OQ' P 1 _PQ b - Op'OP' comes, multiplying the equal sides of B by the fraction Oq:PQ, and the sides of b by Op:PQ, the pair pq _Oq_ pq__Op_ y B - PQ~OQ' PQ'OP* And by taking the quotient of the left members of B and b, and that of their equal members on the right, we get Bb 9E-9L Bb ' Oq~OQ- lllB ' PQ~ Oq-PQ' Ulli °'Oq Oq-Op Oq . pq From the pair of equations, r j^_PQ U Oq' ~ OQ ' M_PQ c ' Op' ~ OP ' we obtain in the same manner, with q'p' for qp, for B' and Bb. r > tt -°3L p'q^ = 9£ c' u - PQ OQ' PQ OP' r , 01 OP UC - Oq'~OQ' I have omitted the negative signs from b ; because from -3—6 3 6 such a truth as — = — , comes - = - , [3], multiplying both sides by - 1. And I have left out the signs in the left 18 FIRST MNEMONICAL LESSONS. [§ 12. member of C : for — - - = -, [Si : for ^ - in c I put its — 22' — ' — Ou l ii equal J*J, , and then omit negative signs as in b. Do you understand ail this ? Jane : — I think I see that every step of the argument is proved ; but I know not where I am, or what is before me, and cannot see much of what is behind me. It is like plunging into a dark cavern guided by a slender thread: I have just hold of it, and that is all. Uncle Pen. : — It will never break, for the twine is in- destructible ; and there will be light enough presently. If you are convinced that equals divided by equals give equal quotients, you are certain that Bb and Cc are true ; and if equals multiplied by the same quantity remain still equals, B', b', C, c', are true likewise. When you see all the mean- ing and application of these results, you will know that they contain the whole science of geometry. If you mul- tiply both sides of B by OQ, you get PQ = S3. . OQ, read {pq by Oq) times OQ, which contains the secret of the tower. Suppose the axes chosen rectangular, for YOX, as well as POX, ma}' be any angle we please ; and if the co-ordinates are parallel to the axes, all our equations remain unaltered in their truth. Let PQ be the perpendicular tower ; let p be any point which your eye at O sees in a line with the summit P; e.g. Op may be a telescope directed to P, /; being & — the centre of the object-glass. If you know the length of OQ, the IY ' horizontal distance of your eye at from the tower ; and know too the length of the perpendicular Pq, which is parallel to PQ and to the axis of y, and also the length of Oq, you obtain PQ from the last written equation, by mul- tiplying together the lengths pa and OQ, and dividing the product by the length Oq. Adding to this quotient the height of your eye from the ground, you have the number of inches or feet between P and the ground, (supposed there on a level with your foot), according as your other lengths are expressed in inches or in feet. Equation lid, which is § 13.J FIRST MNEMONICAL LESSONS. 19 OP=°£OQ, D. Oq gives the distance OP, if you first know the lengths Op, Oq, and OQ. You will shortly learn how to determine either PQ or OP, when only OQ and the magnitude of the angle POQ are known. Richard : — How very charming ! All this comes out of our first notion of the point, (x = 3, y = 2), and a little divi- sion and multiplication, thus, x=3, giving, |=g, and then, y = — ; this is in fact the height of the tower, if x is OQ, and p happens to be the point (3, 2). I do not despair now of measuring mountains in the Moon, a feat that you were helping cousin Henry last week to perform. Jane : — I see that you are right, if the axes are rectan- gular. How simple, after all! The two first equations written by Richard are true of no x and y but those of the point (3, 2) ; the next expresses a law by which x and y may vary through all the points in the line Op. How delighted ij must feel, in the third of these equations, to be free from his confinement in the first, and to be able to assume any value, positive or negative, that pleases him, compelling x every moment to assume the corresponding value ! 13. Uncle Pen. : — To see now the meaning of B', b', and Bb, look at the figure thus : OP and OQ *^ are any pair of diverging (legs or) lines, P V v which are cut by the parallels PQ and /\ pq, drawn in any direction (OY). Op, y- — V OP are the segments of one leg, and / \ Oq, OQ those of the other, made by the 4 'a parallels. From B' and b' we learn that the ratio or proportion of Op to OP, (i.e. their quotient) is the same number as the ratio of Oq to OQ : and that either of these ratios is the same with that of pq to PQ. From Bb we see that the ratio of the two segments made by pq is equal to that of the two made by PQ. In this case the two cutting parallels are on the same side of 0. If pq moves 20 FIRST MNEMONICAL LESSONS. [§ 14. parallel to itself to any position on the farther side of from PQ, as to p'q', we see from C, c', and Cc, that the same things are still true, putting p' for p and q' for q. Thus is proved that If a pair of parallel lines (PQ, pq) cut a pair of lines meeting in a point (O), the ratio of the two segments (OP, OQ) made by one cutting line is equal to that of the corresponding segments (Op, Oq) made by the other : and the ratio of the intercepted parallels (Pq, pq) is equal to that of the correspond- ing segments cut off by them in cither line, (OQ, Oq) or (OP, Op) ; the segments being measured all from O. This is easily remembered in a condensed shape thus : \fT\ If paralls. cut legs, vi. segs. is vi. segs., And vi. (parall. cutters) is vi. (corre. segs.) : Measure the segs. from meet, of legs. Here vi. means quote of, as in [5']. Segs. stands for segments. Corre. is Corresponding ; meet, for meeting. You are not to lay down PQ : pq as Op : OP. I advise you to learn the mnemonic first, and to meditate afterwards with this at your tongue' s end ; for ready words are instruments of ready thought. 14. You can now easily understand the proof of that most renowned theorem of Pythagoras, (Euclid, i. 47) on the discovery of which he is said to have sacrificed a hecatomb. Let ABC be any triangle right-angled at C : let CD be a perpendicular from C on AB, the hypothenuse, or side opposite the right angle, meeting that side in D. Cut off on AB, AC = AC, and on AC, AD' = AD. Then joining CD', we see plainly that the triangle AD'C is the triangle ADC turned face downwards, so that AD'C is a right angle like ADC and ACB, where- fore D'C is parallel to BC, both being at right angles to the same line AC. Because the parallels BC and D'C cut thelegs^Cand^7i[G], AC AB AD'~ AC" or, since AD = AD' and AC = AC, A£_AB AD~ AC ; whence, multiplying these equals by AC. AD, (read AC times AD), § 14.] FIRST MNEMONICAL LESSONS. 2i (a) AC.AC=AB.AD. We have here proved, that if in any right-angled triangle D is the point in which the perpendicular from the right angle C meets the hypothenuse, the squared length of the base AC is the product of the lengths of the hypothenuse AB and segment AD adjacent to the base. This will there- fore be true of our triangle when made to stand on the base BC, or (b) BC.BC = BA. BD. Hence, by addition of the quantities on the left of (a) and (b) and of their equals on the right, AC. AC + BC.BC = AB.AD + AB. DB or = AB . (AD + DB) = AB . AB. i.e. {AC) 2 + (BCy = (ABy, (c) or the square of the length of the hypothenuse AB is the sum of the squares of the lengths of the sides AC and BC. Hence if the sides about the right angle are 3 and 4, the hypothenuse must be 5 ; for 5 2 = 3 2 + 4 2 : 6 and 8 for the sides would give 10 for the hypothenuse. Every brick- layer knows that 6, 8, and 10, will make a right-angled triangle. Suppose the two sides were each = 1 ; then l 2 + 1 2 = 2, shows that (AB) 2 = 2, or AB - J%. Thus you see there are lines which cannot be measured ; for no fractional num- ber can be found that expresses the number of inches in the diagonal of a square whose side is one inch. By extract- ing the square root of 2 you find for AB that diagonal 1-41421356237 inches, which is correct to the ten thou- sand millionth part of an inch, but not quite correct; in truth, the decimal has no end. The side of a square and its diagonal are called incommensurable quantities : no scale can measure both ; no number can express what part one is of the other. Richard : — But suppose that two divisions of my scale happened to be exactly the length of the diagonal of a square : surely 2 would then express the length of it ? Uncle Pen. : — It would : but you would not be able to measure, or even exactly to calculate the side. If AC and BC are supposed equal, and AB = 2, as you propose ; our equation would give (AC) 2 + (AC) 2 = 2 2 , or 2 (AC) 2 = 4, or (AC) 2 = 2, or AC = j2, a number incommensurable with unity. But although we cannot measure or write out correctly 22 FIRST MNEMONICAL LESSONS. [§ 14. such a quantity as J%, it is nevertheless a number; and we can write down a symbol for it, and reason correctly about it. Thus J% is such a symbol, and if we say, let AC = J 2, or let PQ = J% ; AC or PQ is such a symbol, and we can reason with it, as accurately as with an integer number. By our suppositions thus far, our symbols (AC), (AB), &c. stand for numbers of inches in certain lines : and the conclusion (AC) 2 + (BC) 2 -(AB) 2 is an assertion about numbers only. Let us repeat our argument about the square of the hypothenuse, putting all through for AC AC. I, for BC BC.I &c, and let / denote a linear visible inch; then AC .1 denotes' AC times such a visible line, and becomes a line, although AC is but an abstract number. The conclusion (c) above, will be modified thus, AC.I.AC.I+BC.I.BC.I = AB.I.AB.I being = A.B.I(AD.I + DB.I), or (AC.I) 2 + (BC.iy = (AB.iy (c') Since / is no number, but a line, P is no number, but a unit line multiplied by itself; and in like manner (AC . I) 2 is a line multiplied by itself. This has no obvious mean- ing; it must be defined; and it is in our power to give it any definition consistent with the operations of our arith- metic. We shall define a line multiplied by another line, to be a rectangle, or right-angled parallelogram, ivhosc ad- jacent sides are the two lines : then I 2 is a square inch, and (AC. I) 2 is the square upon AC inches. The last equa- tion is now an assertion about square spaces, and affirms : that the square upon the line AB is an area equal to the two squares on AC and BC. There is nothing to hinder us from interpreting equation (c) at once to mean the same thing without introducing the linear unit /, if we bear in mind the definition just given of the product of two lines. In future, when we are considering certain points A, B, C, -D..., we shall take the symbol AB to represent either the line drawn from A to B, or the number of inches in that line. Arithmetically viewed, the line and the num- ber are the same. If A represented a number, instead of a point, and B were also a number, AB would mean A times B, a product of two numbers; but we know always whether a symbol A or B stands for a point or a num- ber, and there can here be no confusion. When ABCD are points, AB . AD may be either the product of the two § 15.] FIRST MNEMONICAL LESSONS. 23 numbers AB and AD, or the right-angled space AB inches by AD inches, according as we consider AB and AD to be numbers, or visible lines ; and the same may be remarked of A B. CD. The square upon the hypothenuse of a right-angled tri- angle is equal to the sum of the two squares upon the sides. This is the famous Theorem of Pythagoras. By the square upon a line is meant the square whose side is that line : remember that there is no hypothenuse, where there is no right angle. If we put qua. for quadrate or squared ; and poth. for hypothenuse, this proposition may be fastened to the ear and to the tongue in the condensed form following : \J~] qua. poth. is both qua. sides. scared hypothenuse is = both the scared sides. 15. By (a) above (AC) 2 = AD.AB (a) and ( AC) 2 = (AD) 2 + (DC) 2 by [7], since the triangle ADC is right-angled at D. Wherefore (DC) 2 + (ADy = AD.AB, whence, subtracting (AD) 2 from both sides, leaving equal remainders, (DC) 2 =AD.AB-(ADy = AD.(AB-AD), or (DC) 2 = AD.DB. (d) If you multiply together two numbers and then find the square root of the product, this root is called the mean pro- portional between the numbers : thus 6 is the mean pro- portional between 4 and & because Jk x 9 = 6, or 6 3 = 4 x 9. This root happens to be commensurable (with unity) ; but the mean proportional between 5 and 9 is not, and can only be expressed by the symbol J 5 x 9, or ,J^5. These are called mean proportionals, because of the proportions 4 : 6 :: 6 : 9, and 5 : ^45 :: ^45 : 9; both which are true by the Rule of Three. As (d), (a), and (b), give DC = jA~DTDB, AC = jAD.AB, and BC = JBD . AB, we see that the perpendicular CD is the mean proportional between the two segments AD and BD, which it makes of the hypothenuse; and either side, AC, or BC, is the mean 24 FIRST MNEMONICAL LESSONS. [§ 15. proportional between the hypothenuse, and that one of these two segments, which is adjacent to itself. For what is true of the proportions of the lengths of lines in numbers is true of the proportions of the lines. This proves the following: In any right-angled triangle, the perpendicular let fall from the right angle on the hypothenuse is the mean pro- portional between the segments into which it divides the hypothenuse ; and either side of the triangle about the right angle is the mean proportional between the hypothenuse and that so-made segment thereof, which is adjacent to that side. Let us put mean or mea. for {mean proportional be- tween), and per. for perpendicular, and poth. as in \J~]. We may say pe>. on poth. is mean segs ; \Jf\ and side is mea. (poth. nigh seg.) side means either side ; nigh seg. is the segment adjacent. Ex. 1. OPQ is a triangular field whose sides are OP = 1000, OQ = 840, QP = 380 feet. Oq being 300 feet, what is the length of the line qp crossing the field parallel to PQ, and what is the distance Op ? The solution is obtained from \_6~\. Equations B and Bb, page 11, give pq = 135| feet, and Op = 3571. Ex. 2. The perpendicular on the hypothenuse is p, and one segment of the same is s : what are the sides of the triangle ? . 2 .< St) that the hypothenuse is s + p 2 :s. By. [7] the sides are J? + p 2 and Jp 2 + (p 2 :sf. If p = 3 and s = l, the hypo- thenuse is 10, and the two sides are sjyd and J 10. § 16.] FIRST MNEMONICAL LESSONS. 25 LESSON III. 16. I shall now propose and solve an entertaining pro- blem. OX and OY are lines in the same plane in a dense and extensive forest. Two points pq mark one line, and two points PQ mark another, *^-V- which are to be central lines of two level roads to be made in the same plane in the forest. A third road is to be formed in this plane also, parallel to the line OY, from the inter- section of the roads pq, PQ. The engineer requires to know where the third road will cross the line OX, and how far the crossing point is from the intersection of the other two roads pq and PQ. We can answer this by finding the point in which the line pq meets PQ ; the co-ordinates of this point, if our axes are OX and OY, are the lengths required. This point can be found, when we know the equations of the lines pq and PQ. Our first step must be Tojind the equation of a line from two given points in it. The equation sought is of the form (10) Ax + By = C, 1, and if (x x y^){x 2 y 2 ) be the given points, this must be true at both, whatever numbers A, B, and C may be ; or Ax l + By l = C 2. Ax 2 + By 2 = C 3. If we subtract the left member of 2 from that of 1 , and do the same with their right members, i. e. take equals from equals, the remainders are equal, or Ax + By- Ax l - By, = 0, or A .(*- a?,) + B .(y -^) = ^ where the point is to be read, times, or subtracting B . (y — y x ) from these equals, A.(x-x l ) = -B.{y-y l ) 4. B 26 FIRST MNEMONICAL LESSONS. [§ 16. Repeating both these steps with equations 2 and ">, instead of 1 and 2, we obtain A . (.r, - x 2 ) = - B . (y, -y 2 ) 5. Dividing the equal members of 4 by the equals in 5, A.jx-x,) ^ B.jy-y,) A.{x,-x 2 ) B.{y-y. 2 y ° y LuJ ' for in 4 and 5, the quantities on the right have like signs ; therefore their quotient is not negative : this is evidently ;/-//. 6. x x -x 2 yi-JTi Multiplying these equals by (y x -y 2 ), — 'I y „ = J/ - 3xi then by fo - x. 2 ), ■*1 ~ *2 (#i - J/2) • (* - ati) = Oi -x a ).(y-yd 7. Or 6 gives 7 at one step, if you multiply by (.ri - ,r 2 ) . (y l - y 2 ). The assertion (7 -3) x (4 -2) = (9-5) . (8-0), amounts to 7x4-7x2-3x4 + 3x2=9x8-9x0-5x8+5x6; [8]. and 7 gives y x x — y 1 a l -y 8 x + y 2 x x = x t y -x l y l - x 2 y + x, y x , whence xy x — xy 2 + X t y 2 = x x y — x 2 y + x 2 y { , or adding a- 2 y to both sides, xy x — xy 3 + x 2 y + ^1 ?/ 2 = ar,y + x 2 y l} and subtracting a>,«/ + .r.,y l from both, jry, - xy 2 + x 2 y + x : y 2 - x\y - x,y l = 0, which is either x. (y, - y 2 ) + a, (y a -y) + x 2 (y- y x ) = 8, or (#1 - y») -x+y- (* a - a?0 - *«yi + -va- = °> or O/i - y0 •* + O2 - *i).y = a \h - *iy» 9, which has the form, Ax + By = C 1. The equations 6, 7, 8, 9> are all forms of the same equa- tion to the line required. We can .see with our eyes, in 6', that the line passes through the points (.r,#,) and (x.,y.\ If we multiply the equal sides both by - 1, which merely changes the signs, for (x - a-j) x - I = - x + x\ = x% - x, we have •' ■' = - — (6.) yi - yi *i - *j § 17, 18.] TIRST MNEMONICAL LESSONS. 27 Here x m y , still mean the variables xy, the zeros being appended for the sake of symmetry and memory. If you put now a?j and y x for x and y, (6) becomes = 0; if you put x 2 and ?/ 2 for them, it becomes 1 = 1. The equation is thus true, it is satisfied, at both these points; therefore the line represented by it passes through them. It was true also as it stood before ; giving = 0, and - 1 = - 1. Equation 8 is best remembered thus, putting x y for the variables, *o • (* -#D + *i • (y« -.Vo) + x* • G/o -yd = o (8). 17. It is of importance to remember the form of 6; for having this, you have all the following forms by easy transformation, which will soon become familiar. The forms 6 and 8 are easily retained from their sym- metry. We are informed by (6), that the quotient of two differences of y's = the quotient of two differences of .x's, the subindices being on both sides, 10 above and 12 below. In 8 we see that x times a difference of y's, thrice written, = 0; the subindices being in order 012, 120,201. These three terms are made by writing 012, and then carrying the first figure to the last place as often as possible, 012, 120, 201, 012, 120, &c. You cannot thus obtain more than three arrangements, by going round the circle, so to speak. To remember then these forms say, the quotient of differences of y's is the quotient of differences of a?'s, with 10 over 12 for the subindices. This is (6). And for (8) say : Thrice written ,v difference of j/'s = ; thus, x-{y-y) + x-{y-y) + x-{y-y) = Q; then go round the circle 012 for the subindices, 012, 120, 201. Let vi (divide) stand for quotient as in [5]': let vi.D, mean quote of .Differences. You may thus abbreviate (6), putting -dex for subhufer, &: is for =, [9] viD(^/'s) is viD(a-'s), read viD.u-ise. with 10 o'er 12 — dexesr read ten o'er twelve, and thus 8; ter = thrice ; Di = .Difference ; is for = ; nil = ; ter x'Di(;/s) is nil, at 612 round, for gil. S u is given line. pronounce owe un two, vowel for zero, un for unity. 18. We have now to find the intersection of two given lines. They must be of the forms, (10), Ax + By = C, (A) A'x + B'y = C; (B) and these will represent any pair of lines, when the known and proper values are put for the six constants. As equals multiplied by any number are still equals, A'.(Ax+By)=A'C, A . (A'x + B'y) = AC. B2 28 FIRST MNEMONICAL LESSONS. [§ 18. Read these, A' times (the sum of Ax and By) equals A'C, &c. Either of these must be true of every (x, y) in the line represented : they are then both true of (X, Y) the point of intersection. Putting then X and Y for x and y, and per- forming the multiplications indicated in the left members, A'AX+A'BY=A'C, AAX + AB'Y= AC ; whence by subtraction, AB'Y-A'BY=AC- A'C, which must be true, although we know not yet either X or Y; that is, {^AB' — A'B) Y — AC — A'C, whence by division of equals, Y C'A-CA' 1 ~B'A-BA'- (L; This gives us the value of Fthe ordinate of the intersection, since all the numbers on the right are known : we have only to subtract C times A' from C times A, and divide the re- mainder by the difference {B' A minus BA') : the quotient is the value of Y. It remains to find X; and this will of course be given by either (A) or (B), if we put for y in either the number Y. But the symmetry of our equations will help us more expeditiously to the number X. It is clear that in (A) you may put x for y, if you exchange B for A ; and in (B) likewise, if you exchange B' for A': this makes it highly probable that X may be put for Y in (C) if B and D' be exchanged for A and A'. The probability be- comes a certainty if you write Bx+Ay = C, (A) B'x + A'y = C, (B) and then repeat the above process with B for A, Sec, which gives CB-CB' A'B-AB'' It is important that you should remember the above value for Y ; and you may teach your ear the following rhyme : [](f] C's Ax and By; Cs for C =. Dot second li: D\{CA)'s by Di . (BA)'s, (dot outs.), is meeting Y. § 18.] FIRST MNEMONICAL LESSONS. 29 You dot or accent your constants ABC, for your second line. Differ- ence of (CAY'S by ZJifference of (BAYs, dotting the outside letters as in (C), is the Y of the meeting point. )• _ 5 — O) + 2y -6 = 0; for-2x -y = 2y , [3J: _ 5 hence — — x + 2y = 6, by adding 6 to the equal sides ; and - 5x + 8g = 24, by multiplying both by 4 : (pq). or Ax + By =C. (A). Thus - 5x + 81/ = 24, or — ^-~ + '- = 1, is the line (pq). j > 24 3 Next let Q be (x^) and P be (x$ 2 ) in (8) ; and suppose *, = 3-25 = Os, x 2 = 3-9 = OP, ^ = -1-2- Qs, y 2 = 0. Substituting these values in (8) we obtain x . (- 1-2 - 0) + 3'25 (0 -y ) + S'9 . (y + 1*8) = 0, or - 1-2*+ (3-9 - 3-25) y + 4-68 = 0, or -V2x + '65y =-4-68, or mul. by -100, 120r - 65y - 468, (PQ) A'x + B'y = C. (B). 30 FIRST MNEMONICAL LESSONS. [§ 19. This, which is the same with X - + -2 — - i * s the line PQ 3'9 4()8 our y no] is Z A r»i', = 4 ? x - 5 - 24x120 L J B'A-BA' -65 x-5-8x 120 _ 46 8 + 24 x 24 _ 1044 28 -65 + 8x24 ~ 127 ~ 127* This is the distance of the point of intersection of the roads pq and PQ from the line OP, along a parallel to OY, being 8 miles and gL of a mile. We find X from equation (pq) thus, by using this value of Y, or, subtracting from both 8 x , 12/ then dividing by - 5, - 5 X + 8 x = 24, 127 or, -5X = 24-8x^, 127 Y _ 3048 -8352 127 5304 " 127 ■ Y 5304 324 5xl27 _8e35 ' this is the distance in miles from O along OP to the centre of the road which is to cross OP. We could have found X also from equation (PQ). 19. Jane: — I have observed in the deduction of equa- tions (PQ) and (pq), and of (8) and (9) just now, that any quantity can be transposed from one side of an equa- tion to the other, if care is taken to change the sign of the quantity transposed, and that this transposition is always either an addition or a subtraction of that quantity, per- formed in both members of the equation. Uncle Pen. : — Your observation will be quite correct, if you say any term instead of any quantity : a quantity so transposed must carry with it to the other side its multiplier or its divisor, if it has either. For the difference between a term and a quantity, look at equation (7) ; it has one term on each side, each term being a product of two factors, every factor being composed of two quantities tied by a vinculum. When the vincula are untied, which requires the indicated multiplications to be performed, the equation § 19.] FIRST MNEMONICAL LESSONS. 31 has four terms on each side, as you see in the step fol- lowing (7). In (8) you see three terms on the left; if you untie and multiply, you will have six terms in the left member of the equation. Let me now see, Richard, whether you can find the value of X, corresponding to meeting Y of f 10], by put- ting for y in equation (A) the expression for Y. First transpose the term By ; then put Y for y, and next bring the quantities on the right to a common denominator. Richard: — I have done this, and I cannot succeed. . L . (CB'A -CBA') -(BC'A -BCA') 1 get Ax = s j~ — V™ , which means & BA-AB' ' _ CB'A-CBA'-BC'A-BCA' then lUviding both **~ B'A-AB' ~' by,/ _ CB'A - C BA' - BC'A - BCA' X ~ AB'A - AAB' This is not the value X that you found ; where is the blunder ? Uncle Pen. : — You are correct in the first of these three equations ; the second is not what the first means. You are right in saying {CB'A - CBA') = CB'A - CBA' ; and wrong in saying - (BC'A - BCA') = - BC'A - BCA' ; for this lower term on the left, when untied, must have a sign contrary to that of + (BC'A - BCA') or to that of BC'A - BCA'; i. e. it must be - BC'A + BCA'. Make the correction, and your numerator will then be divisible by A, and a- after the division will shew its true value. Con- sider this : 5 + (2- 5)-(7- 3 + 2 + 1-6)= 5 + 2 -5 -7 + 3-2 -1 + 6 = 1. The first + on the left is the sign, not of 2, but of the tied or the vinculated quantity (+2 — 5); the second - is the sign not of J, but of (+ 7 - 3 + 2 + 1 - 6). When these terms are untied, the signs of the ties or vincula disappear: and the rule is; when you untie a term preceded by the sign -, you must change the sign of every quantity in that term. As you are a careless boy, Richard, I shall require you to repeat the following bad rhyme : 32 FIRST MNEMONICAL LESSONS. [§ 20. [[11] After minus untyin' Change every sign. Jane: — That was a pretty device, by which you got rid of a from the equations (A) and (B), (18); by a pair of multiplications and a subtraction — so short and simple! — and then a division equally simple compelled the y to show himself. How surprised he must have been, after see- sawing with his friend x through the whole scale of num- bers, to find himself alone, and nailed to a certain value ! Uncle Pen. : — You will not fail to observe that this suc- ceeded only with the supposition that x and y had the same pair of values in both equations: this being supposed, we eliminated, i. e. expelled, x. You will have occasion, if you pursue this study, to admire still more the devices of elimi- nation. It may happen that B'A = BA' in (C): this gives Y = infinite, and X also : so that the point of intersection is at an infinite distance. Show from this equation and from (A) and (B), by division, that these lines are parallel to the same line, when B'A = BA'. LESSON IV. 20. Let ABba be any rectangle, whose base is Aa, and altitude ba: and let ACca, AC'c'a, be any parallelograms having the same base Aa, and lying within the same parallels, Aaf, Blc. Def. The altitude of a parallelogram or of a triangle is the perpendicular let fall from an angle on the base or base produced. As Cf= cf = AB, by Prop. C. [2], AB is the common altitude of the three parallelograms. Because Bb = Aa = Cc= C'c', by [2l Prop. C, BC = bc, and Be' = bC ; and since ab = AB, and the angle ABC=abc, the triangle ale will exactly cover the triangle ABC', and A lie' will cover abC. Denoting triangles by three and quadrilaterals by four letters ; § 20.] FIRST MNEMONICAL LESSONS. S3 ABC = abc, ABc' = abC, ABC - bdC = ale - hdC, C'BAa = C'BAa, ABdb = Cdac, then by addition, ABdb + Ada = Cdac + Ada, ABba = C'Aac'. ABba^CAac. A. Parallelograms (CAac and C'Aac'), which have the same base and altitude, are equal, being each equal to (ABba) the product or rectangle of that base and altitude : vid. definition in (14). Let ABC be any triangle, whose base is AB. If AD parallel to BC be drawn to meet CD parallel to AB, the sides AD and DC are in order equal to CB and BA, and contain the angle ADC = the angle CBA by [2] Prop. C ; and the triangles ABC and ADC will exactly cover each other, and are equal. Hence the triangle ABC is half the parallelogram ADCB, and its area is half the product of AB and CH, CH, J_ on AB, being the altitude either of ABC, or of ADCB. We have thus proof that, B. The area of a parallelogram is the rectangle (or pro- duct) of its base and altitude ; and the area of a triangle is half the product of its base and altitude. Prob. To divide a triangular plot of ground ABC, whose area is H square feet, into halves, by a line through p a point of the side AB. Let pB = h feet, pB being not less than pA ; then if we suppose the thing done, and that pq is the line required to be drawn, we know that h x | (altitude of q from base AB) is the area of pqB ; call this altitude y ; we have the condition ±.hy = \.H ; giving y = H:h, by division of equals by ^ h. This fraction H:h is the number of feet, y, in the J_ distance of q from the base. If we can draw a J_ through A, p, or B, of the length y feet, and through its extremity a parallel to AB, cutting AC in m and BC in q, either of the triangles pqB or pmB will contain half the area of ABC. Richard: — I think the shortest way to do this would be to make a right angle at A and another at B, by the device B5 34 FIRST MNEMONICAL LESSONS. [§21. that Pythagoras, as I suppose, first taught to the brick- layers ; then measuring y feet on my two perpendiculars, I should readily draw the line mq. Uncle Pen. : — You will observe that an area is always given in squa?-e units ; and if I speak of the area H, I always mean H square inches, unless a different measure is expressly named. The value of ;/ would be perhaps more lumi- nously written 12y . /= H. (12) 2 / 2 :(h . 12/), where / is our linear unit, as in (14). 21. Let the point P be (x = /, y=a), and p, be (jr a = Oq l} y x = ?«, 3 = Oq 3 , y 3 = - q 3 q 3 ), the co-ordinates being rectangular, parallel to the right axes OX and OY. Pm, = Q qi = 0q t -0Q= x,-l; p»»»i = .Mi - »»i<7i =ihqi-PQ=yi- a ; whence by [Jl the triangle P/v»i gives ^~iy + (y 1 -ay=(Pp [ y ; Read, the squared difference (r at 1 minus I) + ttie squared dift*. (y, -a) equals (Ppi) 2 . The subindices under x and y prevent confusion among Pi, Pit Pi- Pm, = Qq 2 = Oq 2 -OQ=x 2 -l; p 2 m 2 = m 2 q 2 + q s p 2 = PQ + q 2 p 2 = a- y 2 ; for y 2 = — qji,, and + qjh ~ —})v Also, (.r 2 - If + (a - y a ) a = (Pp,)', by [7], taking the A Pp 2 m 2 , P7n 3 =Qq 3 = OQ-Oq 3 = l-x 3 , p 3 m 3 = m 3 q 3 + q,p 3 = PQ + q 3 p 3 = a -y 3 ; for y 3 = - q 3 p 3 . (i - *»y + (a - y 3 y = (Pp 3 y, b y m, a p P3 m., § 21.] FIRST MNEMONICAL LESSONS. 35 Now in the value of {Pp 2 ) 2 , {ti-y 2 ) 2 = (y 2 -a) 2 ; [(« -2/2)' (a -2/2) = a 2 -ay 2 -y 2 a +y 2 2 = (y 2 - a) . (y 8 - a)~\ which is merely asserting as in \_3~\ (5) that -nx-?i = n x», Hence (P Pl ) 2 = (*, - l) 2 + ( y x - a) 2 = (Pw*,) 2 + («HPi)"» (Pp.) 9 - (*» - 2 + (y- - «) 2 = ( p? » 2 ) 2 + («^)"» (Pp 3 ) 2 = (* 3 - 0" + (ya - ") 2 = ( Pnj a) 2 + Kp 3 ) 2 ; and generally, if r be the distance required between the point (/, a) and aray giwe/i 'point (x, y), (D) r 8 = (ar-Z) 9 + (y-a) 8 , is the square of the distance, if for x and y be put their proper values with their proper signs. If 0, y) is the point (3, - 4), r 2 = (3 - 1) 2 + (- 4 - a) 2 . If (*, «) is the point (- 3, 4), r 2 = (- 3 - J) 2 + (4 - a) 2 j if O, y) is (- 3, -4), r 2 = (- 3 -0 2 + (- 4-a) 2 , or =(3 + /) 2 + (4 + a) 2 which is the same thing, because m 2 =( r m) 2 by £3]. If then r in (D) be constant, like I and a, and (x, y) be a variable point, the equation affirms that the distance between (I, a) and (x, y) is always r ; this is true of all the points in the circle whose centre is (I, a) and whose radius is r, and of no other points : therefore (D) is the equation to a circle, whose centre is (x =1, y = a) and whose radius is r inches. This equation may be written in any of the forms, v. (22), (* - l Y + Of - «) 2 = r °~> ] x 2 - 2x1 + l 2 +y 2 - 2ya + a 2 = r 2 , I D x 2 + y 2 - 2lx - 2ay = r i -a 2 - I 2 . J You should satisfy yourself by varying the position of P in the angles about 0, that (D) gives exactly the square dis- tance between P and (xy). If, for instance, P were the point (- -3, -4), Cr + -3) 2 + (y--4) 2 = r 2 is the equation of the circle whose centre is (- -3, -4), and radius r, / in (D) having in this the value --3, and P being in the angle X'OY. Required the distance r between the points (2, 3) and (•3 ; _ -4) 7 referred to right axes. By (D) Ave have r 2 = (2 - -3) 2 + (3 + -4) 2 - (1 -7) 2 + (3-4) 2 , 36 FIRST MNEMONIC AL LESSONS. [§21. •whence, extracting the square roots of these equal quantities, r = ± ^2-89 + 11-56 = ± V 14,-45 = * 3-80131556. The sign ± is introduced generally in the solution of such an equation as r 2 = N, thus, r = ± N /]V, to show that r may have either sign you choose to take ; for since m x m = tri\ and — m x — m< = in 2 , m 2 has either ?« or — m for its square root, and 14'45 has either + 3-S0131556 or - 3-80131556 for its square root; and 1 has either + 1 or - 1 for its square root : either root squared will give the same number. Every positive number has two square roots, which differ only in sign. Richard: — But how can r = ± 3-80131556 be a true an- swer to your question ? Can there be two distances between (2, 3) and (-3, -*4), one of them less than nothing? Jane: — The sign ± has just been explained to signify that r may have either sign: you can choose which you please. Uncle Pen. : — There are many questions which can be solved only by the extraction of the square root, in which the negative root is inapplicable to the problem: yet here, if in measuring the distance between two points you take the sign into account at all, the length between the point P and the point P' has as much right to one sign as to the other; for if the direction is positive from P to P' , it must of course be negative from P' to P. From D follows r = ±J{ X -t)» + {y-a)\ (DO If a and b are the sides of any right-angled triangle, and c the hy pot henuse, c 2 = a 2 + b 2 , and c = <±Ja a +b*: call this radical, or surd, poth. ab: and dismissing all thoughts of triangles and lines, let poth. ab. be our abbreviation of the arithmetical square root of the sum of the two squares )% //"' squared algebraic difference of the y's ; the square root of this sum is the distance required. Equation (D') says, §22.] [12] FIRST MNEMONICAL LESSONS. 37 Poth.{Di(x/) Di(j/«)}, pron. dixie. (ya) a monosyl. joins(j;'^) to (la), xy a dissyl. in co-ords. recta. i.e. the square root of the sum of the two squares, {Diff. (x-l)} 2 and {Diff. (y-a)\ 2 joins the point (x, y) to the point (I, a), given in rect- angular co-ordinates. Observe that the algebraic difference {x - 1) may be an arithmetical sum, if x and / have different signs : thus if x = 2, 1 = - 3, (x — I) = 2 + 3 ; if x = - 1, I = 5, (*• - J) = - ( 1 + 5), a negative sum. Let DUQ., (pron. duck) stand for duo guadrata, two squares, then (D) (x - l) 2 + (y — a) 2 = r" = rr, is a given circle. [13] DUQ{Di(xl)Di.(yd)} pron. dixie ; Di = difference; Is r r' , (in Recta) rr a dissyll. ; is for = ; Gives circ. cen {la) : la a dissyll. i.e. Duo ouadrata JDiff. (x—l)) 2 and {Diff. (y-a)} 2 is rr ; gives, in ree/angular co-ordinates, a circle whose centre is (/, a) ; it is needless to add that r is the radius. 22. The following are of great importance to be re- membered. (a + b) 2 = (a + b).(a + b) = a 2 +ab + ba + b 2 = a 2 + b 2 + 2ab (a) {a-b) 2 = (a-b) .(a-b) = a 2 -ab-ba + b 2 = a 2 + b 2 -2ab (b) (b + a){b-a)= b 2 -ba + ab-a 2 = b 2 -a 2 . (c) From (a) we learn that the square of the sum of any two numbers is the two squares of the numbers -t- twice their pro- duct. Thus(4+5) 2 =l6+25 + 40=81. And we may write the same truth thus, /being the linear unit, (4/+5/) 2 = l67 2 + 25/ 2 + 2 x 20/ 2 ; which is evident to the eye, if AB DC be 9\ Ae=4>. From (b) we learn, that the square of the difference of two quan- tities is the sum of the two squares, minus twice their pro- duct. Thus, (9 - 4) 2 = 9 2 + 4 2 - 2 x (9 x 4), as is visible in the figure ; for 5 2 = the whole 9 2 , + 4 2 additional in the corner, diminished by (4 2 + 4 x 5) twice taken. In (c) we are informed, that the sum of two numbers, multiplied by their difference, comes to the difference between their squares. Thus (9 + 5) (9- 5) = 9 2 - 5 a . If BE =5, the rectangle AE x EF = (9 + 5) x (9 - 5) = 9 2 - 5 2 , evidently. 5X4 5- 1, or < - 1, ;/ > 1, and a- = ± J\ - y*, is then the square root of a negative manlier, an imaginary quantity. Let ;/ = 1'lj x 2 then = 1 - 1-21 = - -79- What number squared will come to — ' ? No possible num- ber ; for m x m, whatever sign m may have, must be + m". r c •• 'A v. y V! § 23.] FIRST MNEMONICAL LESSONS. 39 This imaginary value of x, when y 2 >l, shows that there is no point of the locus at which y > 1, or y <- 1. Equation (D) gives by extracting roots of equals, after transposition, x -l = ± Jr 2 - (y - a) 2 , or transposing - /, x = l± Jr 2 — {y — of ', and similarly y = a ± J r 2 -(x-lf. Let the circle, (x + -5) 2 + (y- 3) 2 = 0-4-9, be given for consideration. We see at once that (- -5, 3) is the centre, and that 07 is the radius. We deduce as + '5 = ± V'49 - (tf - 3) 2 , or * = -'5± > /"*9-(y-3) a ; and (y-3) = ± N /-49-(x + -5) 2 , or # = 3 ± 7" 4 9 ~ O + ' 5 f- i . 1 , If y = 0, x = - - ±;s /-49-(-3) 2 = - - ±J- 8-51, a nega- tive real number, added to an imaginary quantity ; such an x has no existence, and this shows us that thei-e is no point of O the circle at which y = 0, or that it no- vkere meets the axis of x. y = 3 gives x = - -5 ± 7, 1^ either *2 or — 1*2, x = gives y = 3± ^-24, either 3'4899, or 2-5101, x = - - gives y '— 3 ± *7 either 3*7 or 2'3. If we arafo'e in (D), we obtain by [14] (Qua D (ab)), x 2 - 2lx + l 2 +y 2 - lay + a 2 = r 2 . If we put for x any given value *,, this is, by trans- posing, y 2 — 2ay = r 2 — a:, 2 + 2lx t — I 2 — a 2 , or, putting N for the known manlier in the right member, y 2 — 2ay = N, from which we have to find the y correspond- ing to x = x r 40 FIRST MNEMONICAL LESSONS. [§ 24. It is very convenient to make such abbreviations, for we can at any moment find the value of N, by adding r 2 to 2lx l} and then subtracting (x^ + I 2 + a 2 ) from their sum: this is all known; for /, a, r, and#,, are given. Innume- rable questions of great beauty and interest reduce them- selves to a result of this form, y 2 -2ay = N, in which a and N are known, and y is the number whose value is to be found. This is called a quadratic equation, containing^ 2 , y quadrate, the square of the unknown quan- tity. To solve this, that is, to find from it the value or values of y, add a 2 to the equals, thus making sums still equal ; y 2 -2ay + a 2 = N+a 2 . The left side is ' DUQ.ay less 2ay,' and is therefore by [14], ' Qua D.(ay),' or (y - a) 2 ; i.e. (y -a) 2 = N + a 2 ; whence extracting the square roots of these equals y — a = ± sJN+ a 2 , or, adding a to equals, y = a±jN + a 2 ; whereby y is given in known quantities, and has plainly two values, which can be found by extracting the square root of the number (N + a 2 ). This is the result before obtained; for putting for JV its value, y * Jr 2 - (x 2 - 2/x, + I 2 ) -a 2 + a 2 or by [14], y = a ± Jr a - (*! - /) a . 24. If then y be any quantity of which we are in search, and we know that the square of y, less twice the product of y and any known number a, is equal to any known number N, we obtain two values of y, called the roots of the equation y 2 -2ya-N=0, by simply adding in turn to a the square roots, positive and negative, of (rt 2 + N). This may be thus remembered, putting sq.y and asq. (pron. squl and ask) for squared // and a square. [15] If sq.y' le two yd be JV, ya a monoeyL y's a mol RooM (isq N', vid. niol. lc, in [14]. § 24.] FIRST MNEMONICAL LESSONS. 41 M, like S, will often stand conveniently for SuM. RooM is the square Hoot of the suiV/ of the two indicated quantities (a 2 , N). If JV is nsga- tive, ltoo3I is in fact the root of a difference. If you should forget the proof of [15], reason thus for a moment upon it. If the assertion y = a ± J a* + N be true, then must y — a ='=*= J a 2 + iV, by subtraction from equals : and the squares of the equals last written must be equal, or (y~a) 2 =a 2 + N, which is by [14] 'QuaD ah' (or QuaD yd), y 2 — 2ay + a 2 = a 2 + N, whence y 2 — 2ya = N, as it ought to be. Thus the equation 3t 2 — 51 — 11 =0, is by transposition and division t 2 - ^t = H ■ of the form t 2 - 2at = N, 5 11 t is here the unknown quantity y ; a = j. , N= — , whence /25 5 6~V 36 25 11 25 + 12x11 157 157 Now 36 + Y = 36 = 16 = *"■ so that /l57 5 for you know by arithmetic, that the square root of a frac- tion is the quote of the roots of its numerator and denomi- nator. Before the solution of a quadratic can be obtained by the formula [15], it is necessary that the square of the unknown should have the coefficient unity ; and we accord- ingly had to divide the above equation by 3. From at 2 — et = c comes € -t = C - , whence by [15], (— for our«j e — ± 2a V 4a : c e = 2a ^ Je 2 + &ac 2a e 2 or 7~i 4a' c , + - = a e 2 4o 2 " t 4a. 4a. c e 2 + 4>ac a (2a; 2 42 FIRST MNEMONICAL LESSONS. [§ 25. LESSON V. 25. Reqvired a number such that the sum of it and its reciprocal shall be 3. We can talk about the number even while it is unknown, if we give it a name : call it the num- ber y. Then it must be true that y + - = S s or multiplying equals by y, y 2 + 1 = 3y, or transposing 1 and 3y, y- - 3y = - 1 of the form y 2 - Stay = N; here our a = ~ and N=-l, so that by [15]], y^-.JI^J^JElJ-^ ^ 2 V 4 2 V 4 2 2 i±^£ or *rJ± i.e. =2-618034 or 0-381966. 2 You may take either of these values for the number sought ; the other is its reciprocal ; and you see that2-6l8034+-381966 = 3. At first sight you would not have suspected that these were reciprocal numbers ; but we can easily test the matter. 3+15 2 If — —- = — it must follow that 2 3-J5' 3 + J 5 ■■= t= , and multiplying equals by 3 - J 5, [14 c], 3 — sj 5 32 ~ (\^) 2 = 4 or 9-5 = 4; which is true. I prefer the more general question, To find a number such that m times the number added to n times its reciprocal shall give a sum equal to r. And you shall choose your own values for m, n, and r. All that I have to do is to say mil + tl — = r, then as before multiplying by y, my' 2 + 11 = ry y whence by transposition, my- — ry — — n, and dividing equals by m, 4r?lin 2m V 4/« 2 m 2m § 25.] FIRST MNEMONICAL LESSONS. 43 Here a = r:2m, N=-n:m, and by [15] yr 2 _ n _ r±Jr r 4m 2 m 2ii If you fix on the values r = 3, m = 1 = n, you have the pre- ceding problem and solution. If you choose m = 1 = n, r - 2, 2 ± /4 — 4, you obtain ?/ = - — — = 1 ± 0. Here is but one value of y, which however solves the problem ; for unity added to its reciprocal gives 2. If you choose m — 2, n = 2, r = 4, you obtain y= 1 ±0 again; and it is true that 2*1 +2:1 =4. If r = 8, m = 8, and n = 2, you obtain y = - ; and it is true that 7« . - + ra . 2 = r. In all these instances the radical vanishes, because r 2 = 4/nrc, and there is but one value for y, which is r:2m. If you take the case of m = r — n = 1, you obtain 1 ± JT^l _ \±J~^S two imaginary values, for J — 3 has no real existence. This shows that there is no number such that ?/ + -=!, or such y that the sum of it, and its reciprocal, is unity. If r = l, l± x /l 2 + 4 _ l+ v /5 1-^5 2 ~ 2 ° r 2 : = 1-618034 or -0-618034. Either of these added to - 1 times its reciprocal, gives unity for the sum. m = 1, n = - 1, # = ^ = — ==- or - Given LM a Zifte A inches in length; it is required to divide it in a point P, so sf- that the rectangle under the line and one part shall be equal to the square of the other part. ' « L * M Let the greater segment LP of the line be y inches in length; the other will be A -y inches. And if / be the linear unit, the condition is (LP) 2 = (LM) x (PM), or 44 first mnemonical lessons. [§ 25. (yI) 2 = AI.(A-y).I, i. e. y 2 P=A 2 P-Ayl\ whence by division of equals, y 2 = A 2 — Ay, or y 2 + Ay = A 2 , which is y s -2ya = N, if a=-±A and jV = ^1 2 ; we have then '[15] 2 V 4 ^4 2 If A = 1, 3/ = \ (-1 ± 2-236068) = -618034 or -1-618034. These are the roots of y 2 + y = 1 ; those of y 2 —y = 1, found above, differ from them only in sign. If LM = 1, and LP = -618034, the square on LP (LPlp') is equal to the rect- angle LM x MP (LMpm); (supposing LP' = LP, and Mp = MP.) There is a second value of y, and therefore a second point, (Q) in which the line LM can be cut so as to solve the problem ; but the distance of Q from L is of the opposite sign to that of LP, and must be measured in the opposite direction. If LQ = - 1 -618034; the square of LQ = the rectangle LM x MQ. The segments made are in either case measured from the point of section to the ex- tremities of the given line PL, PM, and QL, QM. This is Prop. 11 of the second book of Euclid's elements. You will find no difficulty in proving any of the 10 preceding propo- sitions. Thus the 10th merely affirms that (2a + b) 2 + b 2 = 2(a+ b) 2 + 2a 2 , the truth of which is evident if you attend to [14] about QuaS(«6). Jane: — Richard solved mentally last week this question: To find a number such that the square of it added to its half shall = 1 8. Suppose ] 7 put in the place of 18; could you solve the question? Uncle Pen. : — It is too easy only. I prefer finding a number y, such that, "'/ + "•§ = c, i.e. such that m times its square 4 n times its half shall be § 25.] FIRST MNEMONICAL LESSONS. 45 equal to c; m, n, and c, shall have any values you may think of. Dividing these equals by m, it 2 + — . 'i- = — ; which is J m 2 in y 2 - 2ya = iV, if* d = -- ". — = - — , andA r =-. Hence by [15], 2 2m 4?m m n In 2 = ~ 4m ± V 16V c 4m "V 16m' m In your question, m = 1, n = 1, c = 17 ; so that by sub- stitution of these values, 1 /I ~~ 1 A +272 - 1 ± ^273 ^ = -4 ± VT6 +17 = -i ± V~l6- = 4 - + 3-8806779, or - 4-3806779, either of which numbers, squared and increased by half itself, will give 17. Richard's answer was 4; let him find the other value of y, corresponding to c = 18, and he will have completed his solution. If you demand a number such, that thrice its square, diminished by four-Jtfths of its half shall be two; we have 4 m ■= 3, n = — - , c = 2 ; 5 and we obtain from the same expression, 4 5 //1V 2 1 ^ /151 1±V 151 ^ = r2 ± vteJ + 3 = T5 ± VT^ = -T5^ = 0-S858S03, or - "752547- To verify this, 3 x (-8859) 2 = 2-35445643, and | of '——- = -354,36 ; the difference of these = 20009643, somewhat too high, because we called the first value of y -8859, nearly two hundred thousandths too great. Also, 3 x (- -7525) 2 = 1-69876875, and ~ of ~ ' 7525 = - -301, the difference of these is 1-69876875 + '301 = 1*99976875, somewhat too low, because we took our second value of y too great, i. e. too near zero, by nearly half a ten thousandth. 46 FIRST MNEMONICAL LESSONS. [§ 26, 27. 26. Thus every quadratic has two roots: if the quan- tity under the radical Ja'+ N is negative, i.e. if A is negative, and > a 2 , both roots are imaginary ; if the radical is real, both roots are real ; and if the radical vanishes, or a 2 + N=0, which requires N=-a 2 , both the roots are reduced to the same value y = a, which is the only case in which the roots can be equal to each other. If you add together the two roots y 1 = a + J a 2 + N, and y 2 = a — J a 2 + N, you obtain ,?/i+;/ 2 = 2a, which with a changed sign is the coefficient or multiplier of y in the equation y~ -lay = N; if you multiply together the two roots ffi = a + J a 2 + N, and y i = a — J a 2 + A", you obtain y 1 y 2 = a i -(a 2 + N) [14], (sq b le sq a, if b in [14] be put for our a here, and a in [14] for our radical J a 2 + N), i.e. y { y 2 = - 2V, which is the absolute term, (term free from the unknown y) with its proper sign, in the equation y 2 - 2ay-N=0. In any quadratic equation (y 2 — py+q = 0), of which all the terms stand in one member, the coefficient of' (y 2 ) the square of the unknown quantity being unity, the coefficient (— p) of the fust power of the unknown (y) is, with a changed sign, the sum of the roots of the equation, and the absolute term (q) is the product of those roots. [l6] If nil be (sq.;/ le py and q') vid. sq.y in [15] and le. p's sum, and q is prod, of roo. nil for zero ; product of roots. 27- The equation whose roots are y t and y 2 , is and this is the same with . ? /"-(.'/i+;A0/y+;y 1 // 2 = o, neither of which is satisfied by any values of?/ except ;/ = ?/, or y = y. 2 , while both equations are visibly true for either of these values of y. Hence, before we know the roots of y 2 +61/ + 7 = 0, we are certain that their sum is - 6, and their product + 7. § 27.] FIRST MNEMONICAL LESSONS. 47 And we are certain of this before we know whether the equation has any real root at all. Thus the roots of if — 6y + 10 = have a sum = + 6 and a product = 10 ; but from y - 2 x 3y = - 10 comes [1 5] y = 3 ±Jg-lO=3±,J^l, both imaginary values : yet 3 +J-1 added to 3 - J^l gives + 6; and (3+V-l)(3-V-l) = fy[14] S 2 -( N /^l) 2 = 9-(-l) = 10. Jane: — It is a mystery to me that quantities which are impossible and can have no existence, should yet have real and intelligible properties, should have a real sum, and a real product, exactly like two genuine numbers. It looks so like a contradiction. Uncle Pen. : — It would be a contradiction, if these two imaginaries were said to have a sum and a product exacthj like real numbers. Their sum may be real and their pro- duct may be real, but they must be incongruous, such as cannot possibly exist together as sum and product of the same two quantities. If you ask for two numbers whose sum is 6, and whose product is 10, you ask for an im- possibility. You put your question thus: y 2 — 6y + 10 = 0; what then is y ? And the wonderful oracle of Algebra answers, y — 3 ± J— 1 : an answer perfectly correct ; for if you square either of these values, then subtract 6 times that value, then add 10, the result is zero. Jane: — So then whatever mystery or appearance of con- tradiction there may be here, it springs not from the answer of the oracle, but from the ignorance of the interrogator. His duty is not to cavil at the response, but to go away ashamed of himself and wondering. Richard: — I should like to try the verification just hinted at of the truth of your Pythia's arithmetic. If y 2 = (3 + J^iy- = 9+2.3j~i+(J^iy^9 + 6j^i-l, by [14] s + £ +1, -ir The circle (E) has its centre at the point /_ 5 \ /l4>5 ( -J?-, -I) and its radius is N =2*0069324. If we change + 7 for - 7 in (E) it is no longer the same curve, and the result is (~D'+<^>-(^ the radius here is imaginary; therefore this circle has no existence, and there is no point (.r, y) that will satisfy 3x 2 +3y 2 +5x + 6y + 7 = : § 29.] FIRST MNEMONICAL LESSONS. 49 i. e. there are no two numbers possible, x and y, such that thrice the sum of their squares, added to 5 times one num- ber + 6 times the other, shall be equal to — 7. Show now that 3x 2 + 3f + -5x + -6y -7 = is a circle whose centre is V 12' ioy — ), and radius = 1-5306157, and that this locus becomes impossible if the absolute term be + 7, or if it be any positive number > : (> means greater than). The equation to a circle, referred to rectangular co- ordinates, can always he reduced to the form, (x-by+(y-ay = r*, by the method above employed in reducing equation E. Hence the co-ordinates of the centre, and the radius of the circle, can always be found from its equation. 29. Look again at the circle whose centre is the origin of right axes, and radius =r. Its equation (23) is x 2 +y 2 =r 2 , whence follows x = ^ J r 2 — y 2 . This affirms that the two values of x determined by any given value of y are equal and of opposite signs. If y be the posi- tive length Om, measured on OY, the corresponding points of the circle are found by draw- ing pmq parallel to OX. Then the ordinates qQ and pP are each = Om, [2] ; and the values of x are OQ and OP, the first being + Jr 2 - Om 2 and the second — Jr 2 — Om 2 . Hence mq and mp by [%~\ are equal, save the sign : in other words, the chord pq is bisected by the perpendicular Om let fall on it from the centre. This may be any chord ; for Om is any length less than r, and our axis OX may be any line through 0. Hence we can affirm : The perpendicular let fall from the centre of a circle upon any chord of it bisects that chord. (a) C 50 FIRST MNEMONICAL LESSONS. [§ 29. Ax 2 + Ay 2 + Ey = 0, or, putting - T for E: A, (G) x 2 + y 2 -Ty = 0, is simply ( x -oy + (y-±iy = \T*, by adding (i T) 2 to the equal sides. This is therefore (the axes still rectangular) a circle having its centre at (0, + ^ T), and radius = i T. By transposition and extraction of equal roots, which shows that T and y must have like signs, otherwise both -y 2 and + Ty would be negative, and x impossible. When +Ty = +y 2 , or y = + T, x = 0; when + Ty > j/ 2 , .r has two values, which are equal but of contrary signs : as y approaches to the value zero, the values of x continually approach zero from opposite sides; when y = 0, x = ±0, or the two values of x coincide at zero. These considerations supplied by the equation, reveal the position of the circle (G): the axis of x, on which lie all the points having y = (11), meets the circle (G) only in one point, : and the circle will lie en- tirely above or entirely below OX, according as T is positive or nega- tive. The line OX is said to be a tangent at or toucher of the circle, as is every line which meets it in one point only. The point is here the point of contact. The or- ^ dinate \ 7", = — \E:A, of the centre is perpendicular to OX, and passes through O ; and since, by drawing our axes accordingly, OX may be any tangent, we can affirm : The 'perpendicular on the tangent of any circle let fall from its centre passes through the point of contact; or, the line drawn front the centre to any point of the circle is per- pendicular to the tangent at that point; or, the tangent of any circle is perpendicular to the radius of contact. (b) To remember this and (G) from which it is proved, say, [17] DUQ()/.r) is Ty' Has tan. nil's y' : pron. (wix) vid.DUQ [13] (a) Rad per'p on chor.'s biCh6r : P ron ' ,anil s ' is or '' for m - (b) Touch-r'ad is per' on tan : § 30.] FIRST MNEMONICAL LESSONS. 51 i.e. the circle if + a? = Ty, (G) has the tangent o=y, or OX, (11) ; (a) the rarfius perpendicular on any chord is the bisector of the chord; (b) the touch-radius, or rad. of contact, is perpendicular on the tangent. Observe that every curve whose equation has no term free of variables, no absolute term, passes through the origin, (0, 0) being evidently in the locus. The equation H. Af + Ax 2 + By + C = 0, or by division of both sides by A, y 2 +x 2 + (B : A)y = -C:A, is by addition of (^ B:A) 2 to both sides (y+\B-.A) 2 + (x±0) 2 = \B 2 :A 2 -C:A. This is a circle whose centre is in the axis of y at (zero, -\B:A), and whose radius is ±(B 2 :A 2 -4,C:A) h . The circle ax 2 + ay 2 + bx + c = has its centre in the axis of x. 30. The triangle Opq in the last figure but one, having the equal legs Op, Oq, is called an isosceles or equal-legged triangle. If it were folded along Om, the perpendicular on the base pq, the portion Oqm would exactly cover Opm; for the angles at m are equal, being right angles, and by [17, a] mq = pm. Therefore the base angles Oqp and Opq are equal, as are also the two angles mOp and mOq into which the vertical angle O is divided by the perpen- dicular on the base. Hence ; In an isosceles triangle the angles at the base opposite to the equal sides are equal; and the perpendicular on the base from the vertex bisects both the vertical angle and the base. Let a, b, c, be the sides of any tri- angle; then the angles A, B, C, in order opposite to them, are to be called for ear-memory's sake Ang, Bang, Cang. The perpendicular on the base, Cd, may be called perc, per. on c : the bisector of the base, Cf, may be styled bic, oisector of c ; and Ce, the bisector of the vertical angle C, may be called biCang, bisector of Cang. Then say, £18] If as b, An'g is Bang ; and per'c is bic' is blCang. If in any triangle, a = b, then A = B ; and perc, bic, and biCang are all one and the same line. C2 52 FIRST MNEMONICAL LESSONS. [§ 31. Jane: — It appears to me that the most difficult part of this and all the preceding lessons is the arithmetic : the rea- soning is all simple enough to follow, if one can only suc- ceed in remembering it all. Richard: — Do you call the arguments about imaginary quantities so simple ? As to the remembering, I have little fear about that: these mnemonics are nothing compared with Greek conjugations and dialects. Try As in prcesenli and the verbs in fit. Jane: — You have such a taste for hard words, and such a memory for any thing like rhymes, Dickon. By all means astonish the rustics some fine morning by empha- sizing these mnemonics along the fields, as you did the other day by spouting Homer ! Uncle Pen. : — If Richard had as much talent for Science as for languages, he would not require these mnemonical aids; no mathematical genius needs such assistance. Take some pains to master them, my dear Jane, and you will save your time, of which you have very little to spare for mathematics, a study in which I should be sorry to see you deeply engaged. Some of the properties which we have been just proving, seem almost evident enough of themselves without learned demonstrations. I shall next introduce you to something which you would have lived long enough without ima- gining. 31. Let adb be any arc of a circle whose centre is e : and let the chord ab be the common base of any two triangles acb, ac'b, having their vertices in the re- maining portion of the circumference ac'cb. Draw the radii ea, eb, and the diameters dec, d'ec'. Then are aeb, c'eb, aec', ceb, cea, all isosceles triangles. By Prop. D, and [18] the fol- lowing are true of the four external angles deb, dea, d'eb, d'ea. deb — ecb + ebc = 2ecb, d'eb = ec'b + ebc' = Qec'b, dea = eca + eac = 2eca ; d'ea = ec'a + eac' = 2ec'a ; .-. deb + dea^2. {ecb + eca), .-. d'ea - d'eb = 2 (ec'a - ec'b), or aeb = 2. acb ; acb = 2ac'b. That is, since c or c' may be any point in the arc ac'cb : § 31.] FIRST MNEMONICAL LESSONS. 53 (a) In any circle the angle (aeb) at the centre, standing upon any arc (adb) is double any angle (acb) at the circum- ference staiiding on the same arc (adb). When the arc adb is a semicircle, the angle aeb is two right angles, being = aed + deb, (Prop. D.) The demon- stration about deb and dea remains still true, and acb = \ aeb = a right angle. (b) The angle in a semicircle is a right angle. (c) The opposite angles of a quadrilateral inscribed in a circle are supplements. Vid. Def. 7, (32). For their sum is that of half the arcs on which they stand ; i. e. half the circumference ; i. e. two right angles. We shall see that the angle aeb and the arc adb are expressed by the same number : numerically the angle at the centre upon an arc is that arc. You may say then, Qcf] Rim-angle on arc is arc-demi, (a) And right is the angle in semi. (b) i.e. The angle at the rim {= circumference) standing on any arc, is the demi-arc ; semi, for semicircle. This and the following are among the most beautiful of the properties of the circle. Let P be any point without or within a circle whose centre is C, and let any line PAB through P meet the circle in A and B. Then drawing the radius CA = r, and CD perpendicular on AB, we have by £7] in the triangle PCD, 54 FIRST MNEMONICAL LESSONS. [§ 31. (PC) 2 =fPD) 2 +(CD) 2 , r 2 = (AD) 2 + (CD)* {=(CA)*}, whence by subtraction (PC) 2 - r«= (PD) 2 - (AD)', which by [14, c] = (PD+AD).(PD-AD)=(PD + DB).(PD-AD) =(PD+AD).(PD-DB), because AD = BD, by [17, a]; that is, (PC) 2 -r 2 = PA.PB, (a) PA being the sum, and PB the difference, if P is within, and vice versa, if P is without the circle. As this value of the rectangle or product PA . PB de- pends only on PC, which is the distance of P, and not in any wise on the length of (CD) that of the line PAB, from the centre, it will remain the same value for every line drawn through P. From P without the circle, PA . PB = PA 2 .PB 2 = PA 3 .PB 3 =(PC) 2 -r 2 . From P, within it, PA.PB=PA'.PB'=PA.PB = +r 2 -(PCf=-{(PCy-r 2 }; for PA and PB, both measured from P within the circle, have unlike signs, and their product (PC) 2 - r 2 given in (a) must have the negative sign. That this number (PC) 2 — r* is negative is very evident, since r > PC. Every case may be represented thus, PB.PA=±{(PC) 2 -r 2 }; (a) taking the upper sign when P is without, and the lower when P is within the circle. If P is neither within nor without, (a) is still true, being then = 0. The product PB . PA changes sign in passing through zero. Let the distance (PC) of any point from the centre be called poc, (point... C) ; let wing mean radius for rhyme's sake, (it flies when revolving); let Seg.Steg in P ring stand for (segment x segment in any line through P cutting the riflg) ; each segment being measured from P to the ring; and let SUD stand for SUm x Difference, or Differ- ence of squares, of two indicated quantities; thus SUD (ax) is a 2 - **= (a+x).(a-x\ [14, c] SUD (by) is (b 2 -y*) : SUD (poc wing) = (poc) 2 — r 2 . Then you may say to your- self, [203 (Seg : S.eg) in P - ring, pr0 n. s2gsn + 9, 6tr + 8, 2m7t + 6 ; n being any whole number, zero included : 6 and 27r + 6 present the same angular opening to the eye- between OA and OP, but are not the same angle, because they cannot be described in the same time by equable motion, nor are they denoted by the same numbers. The number ir, like J2 and all incommensurable numbers, can never be exactly found : you will learn shortly how to determine ir, and at present may take for granted that ir = 3' 1415926536'. I recommend you always to call this tafa- lout*, after the fashion of Dr Grey, the prince of mnemonici- ans. Every number less than tafalout gives an angle less than • More at length nr is = tafaloudsutuknoint = 8-141692663589793. Grey's ingenious device consists mainly in turning numbers into words. The key is : bdtflspknz 1 2 8 4 6 8 7 8 9 a e i o u au oi ei ou y The number 1861 is akla, or beila, or akub. With this device he com- bines most skilfully cadence and contraction. § 32.] FIRST MNEMONICAL LESSONS. 57 two right angles; twice tafalout (2tt) is the number of inches (to less than one millionth) in the circumference of our circle with radius unity which we may call the scale- circle, and a right angle is | -n = 1 '570796 inches measured on this scale-circle. Whenever an angle or arc 0, or a, is spoken of, or a, unless the contrary is expressly stated, is so many inches, that is, so many radii of the circumference of our scale-circle, and of no other, measured in the direc- tion ABBE, or AEBD, according to the sign of or a. Here or a is a number, not a vague mark of position like A, B, C, (Ang, Bang, Cang.) Def. 1. The sine of any number 0, written sin 0, is the number of units (i. e. on our scale of inches) in the ordi- nate y of the extremity ( P) on the moveable radius of the arc of the scale-circle, x 2 + y 2 = 1, about the origin of co- ordinates : and the cosine of the same 0, cos 0, is the length of the abscissa x (in inches) of the same point (P). Neither of the numbers sin 0, cos 0, can exceed unity the radius of the scale-circle, whatever number positive or negative may be. Def. 2. The tangent of any number 0, tan 0, is the fraction or ratio, sin 0:cos 0. Def. 3. The cotangent of is the reciprocal of the tan- gent, i. e. cot = l:tan = cos 0:sin 0. Def. 4. The secant of is the reciprocal of the cosine, and the cosecant is the reciprocal of the sine, i. e. sec0 = l:cos0, cosec = l:sin 6. Def. 5. The versed sine of 0, versin 0, or ver 0, is = 1 - cos 0. Thus if be the arc AP, and Pp the ordinate of P, Pp = y = sin 0, and OP = x = cos 0. If AP X =-AP or 0, = -0, we see that sin(-0)=-sin0, for P,p=-Pp; and cos (-0) = + cos 0. (E) As P is the same point of the circle, and has the same x and y, for the arc as for 2tt + 0, a whole revolution ADBEA and inches more ; or for (— 27r+0), a whole nega- tive revolution AEBD A and then inches positive; the arcs and (=•= 2tt + 0) have the same sine and cosine ; and so have C5 58 FIRST MNEMONICAL LESSONS. [§ 32. and (=•= 4>n + 0), and and (± 2«7r + 0), rc being any whole number. The sine (or cosine) of any number 6 is the same number with the sine (or cosine) of the number ( c 2mr + 0), n being am integer, positive or negative. (F) There are thus an infinity of angles which have the same sine and cosine. Let AT be the tangent of the scale-circle at A, the point from which angles are measured. By \_\1, h~], OAT is a right angle, and OA, AT, are the x and y of T in the line OP, whose equation is (8) y.x = Pp:Op, TA Pp sin _. . , . , m t a .*. 7T7 = J~ = . ; or OA being = 1, TA = tan 0, OA Op cos ° by definition of the tangent above given. Because of the parallels Pp and TA, we have by [jo'] TO PO . „„ .„ „ — -^ =-pt-, or since PO= AO = l, AO Op T0 = -^- = 7, = sec 0, by definition 4. Op cos Let OD be perpendicular to OA, and let DR be the tangent of the curve at D ; then is ODR f_17]] a right angle. If Pq be parallel to OA, Pq and DR, being both at right angles to OD, are parallels ; whence by QT], DO = 0~q' OX ' ShlCe 0<7 = PP by ^ 120 = tt = - — s = cosec 0, by definition. Pp sin Again, - p =-7y by C^]; whence mult, by Pq:D0, 1)0 = oli' or since Pq = 0;; by ^1 DR = ^= -°-! = cot 0, by definition 3. i'yj sin Also, Ap = l - cos = versin 0, by definition 5. § 32.] FIRST MNEMONICAL LESSONS. 59 Let S be any point of the line OP, and Ss the ordinate of S. Because Ss and Pp are parallel, we have by [62, (A) JY& = ?)p = Op = cosd = cos SOs, and by [6] also Ss SO .... . Pp Pp—pO* ^ multiplying by ^, < B ) J5 - 1£ -i* -■&«-■■»«*■ Since these are true for any position of P between A and Z), they will be true if the triangle SOs be made to stand on the base Ss in the place of Os, and the angle OSs be put in the place of the angle SOs or d ; that is, we should prove inevitably that (C) -^ s = cos(OSs), and~=sin(0^) (D). Def. 6. Two arcs or angles whose sum is a right angle are said to be complements of each other. Def. 7> Two arcs or angles whose sum is t?vo right angles are said to be stipplements of each other. By Prop. D, the three angles of any triangle are equal to two right angles ; whence it follows that (SOs) and (OSs) are complements. We have then proved in (A, B, C, D), since SOs may be any right-angled triangle, that the sine of any acute angle is the cosine of its complement, and the cosine of any acute angle is the sine of its complement, i. e. cos <«) = sin f - — co ] , and sin w = cos f - — w ) G. From (A) and (C) we see that in any right-a?igled triangle, the cosine of an acute angle is the quotient of the adjacent side by the hypothenuse ; and that the si?ie of an acute angle is the quotient of the opposite side by the hy- pothenuse, is proved in (B) and (D). These two properties may be expressed thus. In any right-angled triangle (sine of acute angle) times hypothenuse=opposite J_ ) (cosine of acute angle) times hypothenuse=adjacent _L J ^ ' 60 FIRST MNEMONICAL LESSONS. [§ 33. the two sides about the right angle being perpendicular to each other. By the equation to OP, Ss: Os - Pp: Op = sin 0:cos = tan 8 ; or, in any right-angled triangle, the tangent of an acute angle is the quotient of the opposite perpendicular by the adjacent pcrpcndicidar. K. Produce Pq to meet the circle in Q. In the isosceles a POQ, Oq is biCang by [18], and l POq = iQOq, where- fore the complements of these are equal, or lAOP=s.BOQ; the latter is the supplement of AOQ (by Prop A.) .-. AOP and AOQ are supplements of each other, and AP or 0, and APQ or 7t — 0, are supplemental arcs. By [2] Pp = Qq', the ordinate # at Q, i.e. sin0 = sin(7r — 0) ; which proves that the sine of any arc (AP) is also the sine of its supplemental arc (APQ). 33. As Qq' is the sine, so Oq' is the cosine of APQ; and Oq' = ~Op by the equation to the circle x 2 +y*=l giving x = ±Jl -{Pp)'\ at P and Q; therefore the cosine of any arc (AP) is equal, bid with a contrary sign, to the cosine of its supplemental arc (APQ). These properties are, briefly, sin 8 = sin(7r - 6), cos 6 = - cos (tt - 6). (L.) Hence by (E), — sin 6 = sin (tr + d), — cos0 = cos(7r + 6). L'. The variations in sign and mag- nitude of sin 6 and cos 6 are exactly those of y and x in the circle x 2 + y 2 = l, and should be carefully studied. The sine and cosine of an arc vary with, but not like, the arc 6, and are said to be, as well as tan 0, circular functions of 0. As the moveable radius OP revolves, cos 6 and sin are continually varying whilst the arc is continually in- creasing ; but neither of them can ever exceed unity, the radius of the circle : the tangent and secant vary too con- tinually from to - , from negative to positive infinity. § 33.] FIRST MNEMONICAL LESSONS. 61 7T 1 At D, for = -, tan = - , because its denominator cos = ; and sec = - : as is evident from the figure ; for T, the intersection of the radius OD with AT, is infinitely distant. When P is between D and B, or between J. and E, the sine and cosine have different signs, being the y and x of the point P ; and their quote is negative, i.e. tan is negative ; when P is between .4 and D, or between .B and E, sin and cos have like signs, and their quote, or tan 0, is posi- tive. The tangent of the arc AH, when H is very near to D, is an exceedingly great positive number, being sin0:cos0, the denominator of which, cos 0, is exceedingly small, (11): and exceedingly great positively at H', when it is — sin 0,: — cos 0„ cos 0, being an indefinitely small fraction. At K and K\ supposed very near to D and E on the opposite sides of DE from H and H' , the tangent is an exceedingly great negative number. k. At A and I?, sin and tan 6 are zero : they both change sign with in passing through zero at A, which is the only point at which is zero : and they change sign again in passing through the value 0, when = ir, at B. Sin and sin (— 0) have contrary signs, being the same in both ; thus, sin 1*1 =— sin (— 1*1) ; and generally sin(± 0,) ■= - sin(=F 0,) ; tan (± 0,) = - tan(=j= 0,), taking upper signs together, or lower together. 1. At D and E, on the axis of y, cos = 0, and conse- quently cot is nothing. Cos does not pass through zero with 0, but changes sign when passes through ±tt ; i.e. the cosine changes sign in passing through zero at = ±tt, and the tangent consequently changes sign at the same point of the circle in passing through infinite, as it does again when = -§'"", passing again through -£-• m. The greatest value of sin is unity, when = ^ -rr, its least is - 1, when = f it. The greatest value of cos is + 1, when = 0, or =2mr: the least value is —1, when = (2n + l)7r, n being any whole number. Neither sin nor cos changes sign in passing through +1, or through -1. Sec 0, cosec 0, cot 0, have of course the signs of their reciprocals cos 0, sin 0, and tan 0. 62 FIRST MNEMONICAL LESSONS. [§ 33. At every point of the circle, for all finite values, positive or negative, of 0, since x* +y 2 = I, we have always sin 2 + cos 2 0= 1, M. You will remember that the number -n- is by definition the number of inches or unit radii in the semi-circumference of our scale circle, and that nothing is hereby affirmed or implied concerning the exact length of any other semi- circle. I fancy, Richard, you find some of this rather difficult. Rich.: — Indeed I do, and have left to Jane the honour of following you all through this lesson. How do you feel, dear sister mine, after this feast of trigonometry ? Jane : — A little fatigued ; but more by the variety than the difficulty of the subject. Rich.: — I am waiting for the mnemonics ; when I have mastei-ed them, and all their meaning, I shall know what to think about, and I dare say it will be with me as it has been before; I shall first get a clear notion of what is to be proved, and then obtain the proof by dint of a little thinking and conversation with you. Uncle Pen.: — I advise you both first to make sure of the mnemonics I am about to give you, and of their exact sig- nification : you will find little trouble in the demonstrations. To be able accurately to lay down a proposition is a consi- derable step towards the proof of it. £21] x and y are Cos and Si of Scale-a'rc from OX; cen. o'r Ax Rlgh. Def. 1.(32.) Cosine and ..Vine of arc of scale circle, measured from the line OX; the circle having its centre at origin, and referred to right Axes ; viz. x 2 + y 2 =l. [222 So'rCo.Po'th is o'p or ad ; (H). vid. poth. [7]. ST by Cos ta'n, is vi(op. a'd). Def. 2. (K.) 's for = SorCo is Sin or Cos. Sin x poth = op J- ( II ), and Cos x poth = ad ±; opposite, adjacent, X, Sin 6 by Cos 6 = tan 6 = op. i- by ad. -L, (K). vi for quote. [232 Co'rSin to' 's StnorC(ri'ght min a/) (G.) pron. a>, 0. Co'rS is le'm the Co'rS of-ple'm ( L -) CorS(7r et to) is le' CorSw. (L'.) pron. pStw. CorSin, or CorS, for Cosine or Sine. SinorC. for Sine or Cosine. § 33.] FIRST MNEMONICAL LESSONS. 63 fright min u>) is (right angle minus w) ; min means - under a vinculum ; lem is = less or more ; Cos tf = — Cos (it — 6), Sin — sin (ir - 6) ; ir— is supp/ment of ; (x et to) a dissyll. ; et is + under a vinculum. Cos (it + a>) = - cos a) ; sin (it + a>) = - sin w ; le is -. Q24f] Tan Co's Sin are rec. (pron. reck.) Cota Se'c Cose'c. (Def.3, 4.) Tan 0, Cos 0, Sin 0, are in order the reciprocals of Cotan 0, Sec 0, Cosec 0. Two or's in a line correspond, as to antecedents and conse- quents, in [23J, [22]. * D£7Q(SiCo)'sOne: (M) DUQ.vid. [13.] £25] Co right is none Co none is one. i.e. sin 2 e + cos 2 = 1; Cos (right angle) = 0; Cos(O) = 1. Jane: — The lesson is now brought into a manageable compass, by these five little mnemonics: but what queer numbers these circular functions, as you call them, are ! When I have a number 6 given, I can understand what is meant by its half or its square, or its cube root, and I can find any of these by an arithmetical operation; there is a rule for it. But what sort of a sum in arithmetic would you do to find the sine of a number ? What kind of a rule for finding a sine can be laid down, such that the answer shall never be greater than unity, whatever number 6 you work with, and that thousands of different numbers Q shall give the same proper fraction for the answer ? Uncle Pen.: — At present it is enough for you to know what has been proved, and to be able to find, in the tables of Hutton or others, the value of sin 6 for every number 6 you may choose. You shall have complete and mnemo- nical replies to your questions, and to many others equally curious, when we come to that part of our subject. Ex. If the hypothenuse is 100, the side a60, and 6 = 80, Sin^l-f, CosA=4r, Tan^-^-, Vers. ^1=4-, Sec A = -f-, Cosec A = 4, Cot A = 4, Sin £ = -!-, Cos £ = -§-, &c The external obtuse angles are it — A, -k — B, Sin(7r — A) = -§-, Cos (tt -;!)=--!-, Tan (tt -;!) = --!- 64 FIRST MNEMONICAL LESSONS. [§ 34. LESSON VIII. 34. Let ABC be any triangle; let /F the sides opposite the angles A, B, C, /f be a, b, c; atb — a let p = CD, be the perpendicular from Cone; ... s - BD, be the segment of c between D and B, and(c±s) = AD, D and A ; the lower sign being taken when B is an acute, and the upper when B is an obtuse angle; for when B is obtuse. AD > AB, otherwise the a BCD would contain an obtuse and a right angle, or more than two right angles, which Ls impossible by Prop. D. By [7], b 2 = p*+(c±s)\ a 2 = p 2 + s 2 ; whence by subtraction, b 2 - a 2 = (c ± *) 2 - * 2 - by [14, a, b] c 2 ± 2cs +s 2 - s 2 , or b 2 = a 2 + c 2 ± 2cs = a 2 + c* ± s • 2c. (A) When B is acute, we take the lower sign, or b 2 = a 2 + c 2 - 2cs. (a) Now in the right-angled triangle DBC, Cos (DBC) xCB = DB by [22], (Cos x poth = ad. _L ) i.e., since ABC is in this case DBC, axCos(ABC) = s = aCosB; wherefore the above becomes, putting a Cos B for s, b 2 =a 2 + c 2 -2acCosB. (B) We may consider AB and C to be rmmbers, and not merely symbols of position. When B is obtuse we have b* = a 2 + c s + 2cs. (a') As before, Cos {DBC) x CB = D£; but (Def. 7, 33) DBC is the supplement of >4£C in this case; so that [23, L] Cos DBC = - Cos ABC. § 34.] FIRST MNEMONICAL LESSONS. 65 Therefore - Cos (ABC) xCB = DB, or - Cos B x a = s, by which (a') becomes b 2 = a 2 + c 2 - 2ac Cos B, (B) exactly as before when we considered B acute, so that (B) is true for any angle B of any triangle ; b being always the side opposite B. We may then go round the circle, bcabc, vid. (17), and write c z = b 2 + a 2 -2ba Cos C, ) a 2 =c 2 + b 2 ~2cb Cos A, I In any triangle, the square of any side is obtained by subtracting from the sum oj the squares of the other two sides, twice the product of those two sides multiplied by the Cosine of the angle of the triangle contained by them. Obs. This subtraction is in fact an addition, if the cosine be ne- gative. When the contained angle is a right angle, its Cosine vanishes, and this becomes [J7J the theorem of Pytha- goras ; but you are not to consider this to be our proof of that theorem, although the result (B) includes \J~\ as a particular case ; for we have employed \J~\ continually as a premiss in the reasoning which establishes (B). , The Proposition (A), expressed in words at length, is none other than the 12th and 13th propositions of the second book of Euclid. By (B) we find any side of a triangle if the other two sides and the angle between them be given ; for the tables (of whose use more hereafter) give at a glance the Cosine of the given contained angle, and the sought side is the square root of the known number on the right side of (B). We see from (B) also that if two triangles have two sides a = a x , c = c l} and the angle B between a and c, equal to that between a, and c { ; they have their third sides equal, b = b i ; a truth which has been established in (20) and (30) by the simpler argument that such a pair of triangles will cover each other. Hence The diagonal of a parallelogram bisects it. D. These results are most important, and must be mastered. I will give you a mnemonic both for (A) (B) and for the proof of them. 66 FIRST MNEMONICAL LESSONS. [§35. Draw perc : SUD (ba) is SUD (segs of c'). And sq'.b is DUQ ac mol (seg. op) two c, [26] As 'tuse or 'cute is Bang op. b ; (A) Or, sq'.b is DUQ ac le CoBang'two ac. (B) Draw perc or CD, or, let CD be supposed to be J- to c, [18] ; then we prove b 2 - a 2 = (AD) 2 - (BD) 2 , vid. SUD [20J; segs for segments ; sq.b pron. squib; DUQ[13J ; ± = mol ; seg. op . is the segment of c opposite, not adjacent to b, whose square we are expressing. Pron. two long, to avoid confounding it with to; + or - in mol, with B obtuse or acute ; B = Bang opposite 6, [18J ; le is - ; CoBang two ac means, of course, Cos B times 2ac, x being understood between quantities joined together, as also in ± s2c, (A), in the second line. From (B) we deduce, by transposition and division, (C) Cos B Cos A a- + c 2 - b 2 2ac c* + b 2 - a 2 CosC = &«. 2ba 2cb by which we obtain all the angles of a triangle from knowing its three sides. Thus the triangle whose sides are a = 6, 8, c = 9> gives 36 + 81-64 CosC = 144. 64 + 36-81 96 or Cos B ■■ 53 108 cc-g. «w-g. The tables give the angles /cm /k» /wo r/»7// angles, which have these cosines, and these are evidently the angles of the triangle. 35. Let ABC be any triangle, and let CD be supposed perpendicular from C upon the side c (or AB), and A¥ § 35.] FIRST MNEMONICAL LESSONS. 67 perpendicular from A in the side a (or BC). Then c . (CD) = a (AF) must be true, c and a being the lengths of the sides opposite C and A, and (CD) and (AF) represent- ing the lengths of the perpendiculars on them ; for either of these products is twice the area of the a ABC, by Prop. B, (20). Let the angle DCB = a, DC A = t ; then as AF = b Sin C, [22] c . (CD) = ah . Sin (a ± t), i. e. = ah Sin C, taking the upper or lower sign as CD falls within or with- out ABC. Or (DB * DA ) (CD) = ah Sin (a ± t) ; or DB.CD±DA.CD = ab. Sin (a ± t) ; but [22] DB = a Sin a, CD = h Cos t = a Cos a, DJ = h Sin t ; .-. a . Sin a . b . Cos t ± a . Cos a . h . Sin t = ah . Sin (a ± t), or, div. by at, (vide 77*e Mathematician, Vol. n. p. 134) Sin a . Cos t . ± Cos a . Sin t . = Sin (a ± t). (a) The wpper of these signs is to be taken when a and t are an?/ angles between /;erc falling within am/ triangle and the sides a, 6 ; the lower when a and t are angles between perc falling without,. and the sides a, b. Let CB' be drawn _]_ to CB, so that 5CB' may be a right angle. Then, if z a'= B'CD, it is evident that z B'C A = (a'± t), + or -, according as the LCD falls within or without the a AB'C: and, by the same reasoning with the above, we have Sin a' Cos t ± Cos a' Sin t = Sin (a'± t). Now a'= a, .*. Sin a'= Cos a, and Cos a'= Sin a, by[23,G] and Sin (a'± t) = Sin (£ - a ± tJ = Sin i ^ - (a ^ t)| [11] = Cos(a=pT), by [23, G] /. Cos a Cos r ± Sin a Sin t — Cos (a =f t), which is the same with Cos a . Cos t =j= Sin a . Sin r = Cos (a ± t), (b) the upper signs taken together, or the lower together. G8 FIRST MNEMONICAL LESSONS. [§ 35. We have thus found the expansion of a sine or cosine of a sura or difference of two numbers a and t, in terms of the sines and cosines of those numbers, when neither of them is greater than — ; for our demonstration does not apply to the case of the angles a or t = a right angle, because perc (CD) cannot be at right angles to tivo sides of the triangle. It has yet to be proved that (a) and (b) are true for arcs a and t equal to and greater than a quadrant — . The cases to be considered are, 2' 9. both IT 2' 10. a>7r 7T T = * 5. T>- 6. both 7. a > 7r 12. a>~<2i 13. T >^<2, 2 14. both > 5 The cases of a, or t, or both, greater than 2tt or four quadrants, may be neglected ; for if a = (2«7r 4- 0) and t = (2«V + 0'), a, t, and (a ± t) have the sines and cosines of 0, 0', and (6 ± &') by (F, 32). The equations (a) and (6) are perfectly general, for all values, short of ± infinite, of both a and t. You will find it a most profitable and by no means difficult exercise, to verify them for all the above cases, and for a and t of either sign : all that you have to do is, if you suppose e.g. a > ir, to put for a, on both sides of the equation, (v + a,), a, being supposed less than a quadrant ; and then to shew that each of the equations (o), (b) remains true after the substitution for either upper or lower signs taken together, by some of the following con- siderations. The arcs f - ± j and (=f 0) are complements, and (*■ ± 0) § 35.] FIRST MNEMONICAL LESSONS. 69 and (=f 0) are supplements, as are also ( | ± 6 J and {- =f 6 J, whatever 6 may be, if in these pairs of arcs the upper signs or the lower be taken together. Hence Cos (£**)= Sin (*■ #)> &c - C 23 ' G l > Sin (tt ± 6) = Sin (=f 0) [23, L] = t Sin 0, (32, k) ; whence also Sin («-!•"") = - Sin (1 7r - a )- Remember that if ± terminates at (a?,, #,), ±(ir + 0) terminates at (- ar„ -yj on the same diameter of the circle. As an example I take case 10; let a = -n- + <*! , T=i7r + T,, ai and t, each < 1 7T. Then (a) is Sin (tt + a,) . Cos (1 7T + t,) ± Sin (i 7T + t,) . Cos (*■ + a,) = Sin {tt + a, ± (i tt + Tj)}. (a') Now in the first member of this, Sin (*■ + a,) = Sin (- a,) [23, L] = - Sin a, : Cos (1 ,r + t,) - Sin (- t,) [23, G] = - Sin t, , Sin (i tt + t,) = Cos (- t,) = Cot t, , (33, m) Cos (tt + a,) = - Cos a, , [23, L] Therefore (a') becomes - Sin a, x - Sin t, ± (Cos t,x- Cos a,) = Sin {tt + a, ± (i tt + t,)}. (a') Taking the upper sign in the second member, Sin (tt + a, + 1 tt + tj) = [23, L'] - Sin (a, + t, + A tt) = - Cos (- a, - Tj) = - Cos (aj + t,). Taking the lower sign, Sin (tt + a, - 1 tt -t,) = Cos (t, - a,) a Cos (a, - t,) (33, m). Hence, (a') is reduced to the two equations, attending to [33 'like signs,' Sin a, . Sin r, — Cos a, Cos t, =* — Cos (a, + t,), Sin c*i . Sin r t + Cos c^ Cos t, = Cos (c^ - t,), both which have been proved true for all values of a, t, < ^ TT. 70 FIRST MNEMONICAL LESSONS. [§ 36. The above looks perplexing, but there is here nothing to be committed to memory beyond the propositions (a) and (b), and the reasoning by which they are established. A long multiplication sum is perplexing ; but you are quali- fied to cope with it, if you know and can handle your mul- tiplication table. You cannot see the correctness of a pro- duct of large factors, as you can that of 2 x 3 = 6 ; but you are satisfied if the work step by step is right. Be then equally satisfied with the results of our symbolical arith- metic, although you cannot see your way at a glance from the premises to the conclusion. It is enough, if you are able to find it. I give you now a mnemonic for (a) and (b), and another for their proof, in case you should find any difficulty in remembering the steps of the latter. Pronounce a and r like a and t. [27] («) Sin (a mol t) is Sa'.CoT mol Ca'.SlT ; mol t, one syl. (b) Cos (a mol t) is Cd.CoT lem Sd.Si-r. mol, v. [14]; lem, v. [23]. c perc is ba Si Cang ; then put (a mol t) for Cang ; (new — old a) make right ang. The three last lines contain the steps of the demonstration. Our first step was c.CD = 6aSinC ; CD is J- on c, vid. [18 J : the next is to put for C the sum or difference of angles made by CD, (a ± t) : then we applied [22], and obtained (a) : we then drew a neiv a, CB' making BCB' a right a?igle between old a and new a, and obtained (b). LESSON IX. 36. Richard: — How I long for an amusing question! The last lesson was so dry. I saw the surveyor this morn- ing measuring Holt's farm ; and I wondered how he could find the number of square acres and make a plan of it, by measuring a few lines here and there. Jane : — If there had been any triangular fields, I fancy you would have a perpendicular drawn on a large scale by the theorem of Pythagoras, to get the product of the base and altitude. § 36.] FIRST MNEMONICAL LESSONS. 71 Uncle Pen.: — Most likely not: the surveyor finds it more convenient to form a plan by the aid of a few leading lines, and then draws his perpendiculars on the paper. Neither is it necessary to draw a perpendicular at all, to find the area of a triangle. We will solve presently the two questions following. Two fields contain together llf acres, and one of them contains 3f acres more than the other. What is the area of each field ? A triangular field has its sides 600, 560, and 720 feet in length ; what is the area of the field ? You will be able to solve these yourselves, if you will pay attention to what I have further briefly to say about [27]. Sin (a + t) = Sin a. Cos t + Cos a. Sin t (1), Sin (a - t) = Sin a . Cos t - Cos a . Sin t (2), Cos (ci + t) = Cos a . Cos r - Sin a . Sin t (3), Cos (a - t) = Cos a . Cos r + Sin a . Sin t (4). These four equations are true for any values of a and t : provided that a and t mean the same two numbers on both sides of the same equation. Let a as well as t be the same number in the equations 1 and 2 ; then by addition (1 + 2), and by subtraction 0-2), Sin (a + t) + Sin (a - t) = 2 Sin a . Cos r (5), Sin (a + t) — Sin (a — t) = 2 Cos a . Sin r (6). In like manner from 3 and 4, (4 + 3) and (4 - 3) give Cos (a - t) + Cos (a + t) = 2 Cos a . Cos t (7), Cos (o - t) - Cos (a + t) = 2 Sin a . Sin t (8). The following elegant little transformation will bring you to the solution of our first question. Because h — b = 0, 2a = a + a + b — b = (a + b) + (a - b), 2b = b + b+a-a = (a + b)-(a-b), v. [11], a + b a — b , , a + b a — b , . -'• a= -ir + -ir> and6 = - 2 j- 0); put (a + &):= SuM, (a - 6) = DifF. (9) is, l S ± ^ D = a, or b, as you take + or -. 72 FIRST MNEMONICAL LESSONS. [§ 36- [28] HaS (ab) mol HaD (ab) is a or b ; i.e., hatf the Sum of (a and 6) + Aolf the /)iff. of (a and t) = «, A« S - ha D = t>, whatever two numbers a and b may be ; a, the first named, being usually not necessarily supposed the greater. If then A, B, be the areas of the two fields, y* = ^xllf + £xSf, andS = iof llf-iof 3f, giving v4 and B in acres. Yon may reduce these to a more workmanlike shape yourselves. Put now a+T = S, a-T = D; a = | (S + D), T = l(S-D)by [28]: the equations 5, 6, 7, 8, become Sin S + SinD = 2 Sin ±(S + D). Cos ±(S-D) (10), Sin£-SinD = 2Cosi(S + D).Sinl(.S-D) (11), Cos S + Cos D = 2Cos ^(S+D). Cos ^(S-D) (12), CosD-CosS = 2Sinl(S + Z>).Sini(S-D) (is). Now S and D may be any numbers of either sign, for a and t, whose values are determined [28] by given values of S and D, are general symbols ; we can therefore write a for *S and t for D ; for either S and D, or a and r, stand for any pair of numbers you please : Sin a * Sin r = 2 ^"{i (« + t)| x ^ S |l (a - r)}. (10, 1 1). Cosa ± CoST=±2g i n S ||(a + T)jxg i ° n S {l(a- T )|,(12,13). Suppose a=r in (1) and (3), we have Sin 2a = 2 Sin a Cos a, } Cos 2a - (Cos a) 2 - (Sin a) 2 , / "°' or since a is owj/ number, so that it have the same meaning on both sides, COS a = Cos 2 (la) -Sin 8 ('«),) Sina = 2Cos(4a).Sin(£«), J K } Observe that in (3), Cos a . Cos r does not become Cos 2 a 2 , when a = t, but simply Cos a x Cos a = Cos 2 fl, the squared cosine of a : Cos'a 1 would be the squared cosine of a* a different arc. § 3G.] FIRST MNEMONICAL LESSONS. 73 Our second problem can now be solved in a trice ; but first let me give you mnemonics for 5, 6, 7 5 8, 10... 16. The first four are condensed thus, Sin(a + T ) ± Sin(a- T ) = 4 i o n s( a )sin(^ (5,6) Cos(« - t) ± Cos (a + r) = 2 'stwoS6rC.CorS, (5,6) JwjJ? [15J ' C6D mol CoM's two CorS .CorS. (7, 8) ' s for =. Taking antecedents of the or's with +, and consequents with — ; Sin (or cos) times cos (or sin) is SorC. CorS, a dissyl. [30] Set mol Sir's two SorCHaM.CdrSHaD, (10, 11) Ca mol Cot's mol two CorSHaM.CorSHaD... (12, 13) Sine (or Cos) Half Suil/ x Cos (or Sin) Half Diff. in (10, 11.) M and D are the suM and Diff. of a and t on the left. Pron. CH and SH as in Cheshire. [31] CorS is SU'D or two CHa.SH. (16) v.SUD [20]. Cos or Sin is SUD (Cos Half, Sin Half) or two (Cos Half. Sin Half.) Vou should frequently write out these expressions (5, 6,) (10, 11,) &c. from memory, talking to yourself in these mnemonic syllables. Let the sides of our field be a, b, c : by [26] ' sq. b, &c' b 2 = a 2 +c 2 - 2ac Cos B, and by [31] Cos B = Cos 2 |B- Sin 2 ±B; < Cos is SuD CHaSH.' but 0=1- Cos 2 ±B - Sin 2 £ B, by [25] ' DUQ SiCo.' .-. Cos B = 2 Cos 2 1 B - 1, by subtraction. P. Hence b 2 = a* + c 2 - 2ac (2 Cos 2 | B - 1 ), £r = a 2 + c 2 + 2ac - 4>ac Cos 2 i B, a. o 2 = (a + c) 2 - 4ac Cos 2 £ 5, by [14] ; then transposing, and dividing by 4, 2 4 2 2 by [14, c]. A. here (a + c) 2 is the ' sq.b,' and b 2 is the * sq.a' of [14, c]. D 74 FIRST MNEMONICAL LESSONS. [§ 3G. Again by addition instead of subtraction above, CosJ? = l-2Sin 2 AJ?. Q. Hence b 2 = a 2 + c 2 - 2ac (1 - 2 Sin 2 £ J3), i 2 = a 2 + c 2 - 2ac + 4ac Sin 2 £ B = (a - c) 2 + 4ac Sin 2 1 B, b. .. ,, r, b 2 -(a-c) 2 b-(a-c) b + (a-c) . _,, _. «cSin 2 i£ = -V_^L = _A_ '. L ',by[14,c] b-a+c b + a-c ~ = . , by 1111- "• 2 2 ' * L J Next by multiplication of equals by equals in (A) and /T>\ ^ ~. o , ti a + b + c a-b+c a + b-c -a + b + c a 2 c 2 .Cos 2 ±BSin 2 ±B = ^—.— -=— .— — • «— ■ a + b + c a + b + c 2a -a + b + c L e t s = — — — , the semiperimeter of our triangle (abc); then s — a ■ b = 2 2 2 a + & + c 2b a-b + c ~2 "2 ~ 2 a+i + c 2c a + i-c 6 ~'' - 2 2.2' whence by substitution, c 2 a 2 Cos 2 ±B. Sin 2 i B = s.(s-b).(s-c).(s- a), and, extracting roots of equals, ca Cos £ 2? sin |B=± „/T (*- 6).(*-c) . (*- a) = ca.-^ » b y [31] j C. for Sin B = ' 2.CHa.SH,' [31]. Now c Sin5, [H, 22], is the perpendicular on the side a of the triangle ; and \ ac Sin B is the area, giving Area of a = ± Js . (s-a) . (s -b). (*-c), < ' § 37.] FIRST MNEMONICAL LESSONS. 75 In our second question, s = 1 (600 + 560 + 720), s - a = § (- 600 + 560 + 720), s - b = i (600 - 560 + 720), * - c =£ (600 + 560 - 720) : the field contains 163458-12 square feet. The double sign in the expression for the area is explained thus. Two points A and B with a third point C determine a triangle, whose area is of one sign when C is on one side of the line AB, of either sign, when C is in that line, and of the con- trary sign, when C is on the contrary side of AB. To remember (C) and its demonstration, say : RodP sleb.s.slec.sla, le or 1 for less or - ; sleb's a dissyl. Is Sine Bang half ca, vid. [18i [32] Is CHaSHca, is Area. Write 'sqb' in quaCH and quaSHaBang, By ' CorS is SUD'...and ^^vr d a b angle, we have by [22], CD standing for the length of the perpendicular, CD = b Sin A = a Sin B, and if A be obtuse, so that D is without the triangle, we have (32) [23, L], CD = b Sin (w-A) = a Sin B, i.e. by [23, L], b sin A = a Sin B, in either case, and all cases, b a Sin B Sin A ' from division by Sin A . Sin B. By placing the triangle on CB instead of AB for base, we prove the same about b, B, and c, C. Hence o:Sin B = a:Sin A = c:Sin C, E. §37.] FIRST MNEMONICAL LESSONS. 77 It follows that - = a Sin£ = ^ — ~i > by multiplying by Sin B: a in e. , , b , SinB , wherefore - ± 1 = w . — 3 ± 1 a Sin A or since ± 1 = =fc a ± Sin v4 " a ~ Sin J ' &±a Sin £± Sin A a Sin A by addition of fractions : b + a Sin 5 + Sin A i.e. = a Sin A b — a a Sin B-SinA Sin A whence by division, b + a Sin B + Sin A b-a ' SinB-SinA' Now Sin E + Sin A = Sin ±(B + A). Cos £ (£ - A), Sin £ - Sin A = Cos £ (B + A) . Sin | (S - 4), by [30], ' Sa mol Sib's.' SinB + SinA _ Sin±(B + A).Cos±(B-A) whence gin £ _ gin A ~ Cos i (# + ^) . Sin ±(B-A) Sin± (B + A) Sin^(B-A) ~Cosi(B+A) : Cos j>(B- A) tan ± (B + A) tan ± (B-A) Sin£+S in^ _ b_+a, tan \(B+A) 1-e ' Sin J e-SinJ~6-a"tani(,B-^)' and consequently Sin C + SinB _ c + b _ tan ^(C+£ ) Sin C- Sin B ~ c -6 ~ tan ± (C- 5) ' Sin ^ + Sin C _ a + c _ tan ^ (^4 + C) Sin A - Sin C~7r^~c~tan± (A -Cy are also true for the sides and angles of any triangle [22, K]. (F). 78 FIRST MNBMONICAL LESSONS. [§ 38. You may repeat (E) and (F) thus : [34] Side to Sinop as side to Sino'p (E). 1352 SubyD (Sines or Sides) is taf Sum by taf Diff (F). Side to the Sine of opposite angle, as Side to Sine of opposite angle. Sum by D'\S. of Sines (or of Sides opposite) = tan of hal/ Sum by tan of half Diff. of angles opposite those sides. If we divide equals by equals [30], Sin (a ± t) _ Sin a . Cos t ± Cos a . Sin t Cos (a d= t) Cos a . Cos t =f Sin a . Sin t ' or, dividing numerator and denominator by the number (cos a x cos t), which cannot change the value of the frac- tion on the right, e. [22, K]. (G). I have put to for t (which makes no difference, since both are equally general), for the sake of saying more smoothly ; (pron. to as o) : [_36~] tan (a molto)'s ta mol tw by DorS (tin ta . ta>). to, tco, for tana, tan ; and we have by the last equation nx ap = Oa xnx AP. This must be true, however great n the number of sides of the inscribed polygon be supposed. If we suppose n = a million, we shall have two polygons immeasurably near in figure and area to the two circles. We may sup- pose n inconceivably greater than any assignable number. even to infinity. When n is infinite, the polygon coincides with the circle; for a circle is a regular polygon the num- ber of whose sides is infinite. It follows that, (f = Oa), § 39.] FIRST MNEMONICAL LESSONS. 81 A. Circumf. apq... = Oa x circumf. APQB .. = 2irr, -^. circumf. apq... = Oa x --- . circumf. APQB..., 360 l l 3b0 or a degree of the circumf. apqr =Oa x a degree of the scale circle. If Oa' be the radius of any other circle, it is proved in the same way that its degree is Oa' times the scale- degree; or its minute, or second, &c. ; i. e. Similar arcs of two circles (i. e. arcs on which stand equal angles at the centre) are proportional to the two radii. 39. A triangle has six parts or elements, three sides and three angles. Of these if any three be given, the re- maining three can be found, except the case in which the three angles are given. It is evident that any number of triangles abc, a x &,c,,... may have the same three angles, if a and a x are parallel, as also b and &,, c and c 1} &c. When the triangle is right-angled at C, we require only two more data to determine the three remaining parts. The dif- ferent cases that can occur are : 1. Given c and a, 2. Given a and b, Required b, A, B. Required c, A, B. 3. Given C and A, 4. Given a and A, Required b, a, B. Required c, b, B. 5. Given a and B, Required b, c, A. 1 . Since c 2 = a 2 + b\ [7], c 2 - a 2 = b\ and b = ± Jc^i 2 Sin^=^; B = j-J t (Prop. D.) or Cos B = ~ [23]. From the tables, A is found by its sine, and B is its com- plement, for A + B + C = 7r or two right angles, and C = — , or one right angle. 2. c = sja 2 + b 2 ; tan A = ^ [22] ' tan is vi (op. ad.) tan B = - , or B = i tt - A. a' 2 3. b = cCosA [22] a = c Sin A, Cos B = a:c. T>5 82 PIRST MXEMONICAL LESSONS. [§ 39. 4. c = a-.Sin A, by division, from a = c Sin A ; b=a cot A, by multiplication, from b-.a - cot J = l:tanJ; [24]. 5. ft = « tan jB by mult, from b:a = tan Z> c = a:Cos B = a sec 5 ; [24], from c Cos B = a. The case of c and b given, differs in nothing, more than the exchange of two letters a and b, from the first case. When the triangle is oblique-angled, the cases are 1. Given abc, 2. Given ab C, 3. Given ABc, Sought ABC. Sought ABc. Sought abC. 4. Given ABa y 5. Given aM, Sought bcC. Sought £Cc. 1. Any of the formulae (C, 34) or (cABD, 36) are suf- ficient. The most useful in practice are (ABD, 36), the last of which f tasquaf A..' is used whenever the angle A does not approach 2ir or 180°; and the first one when it does. Theoretically, all these formulae are equally effective ; i. e. supposing that no decimals are neglected in the computa- tions. 2. Two sides and the contained angle are given. Since in every a, by Prop. D, A + B + C = tt, A + B is known : and, since ^ (A + B) = ^ it - A C, b y[W «»t(^j>- g:8::{g -gE^.Tp»gi = cot£C, by [24]; then i r-^-i a + ° cot| C , by 1351 r = t-~ r- , whence 3 L J a-b tan £ (J - B) tan f ( A - 1?) _ o - b _ cot ^ C ~ a + b' for equal numbers have equal reciprocals : thus, _ 10 ,12 5 = — , and - = — - : 2 5 10 .-. tan 5 (A-B) = . cot ^ C, by multiplication by cot | C ; this is a given number, since a and b are known, and cot | C § 39.] FIRST MNEMONICAL LESSONS. 83 is given in the tables. We thus have ^(A-B) and \(A + B), and consequently both A and B by [28] ' HaS(ab)..' (9, 36). The side c is found by [26, B]. 3. Two angles and the side between them are given. C = ,r - A - B (by Prop. D) ; .-. Sin C = Sin (A + B), and a = c.SmA:Sm(A + B) [34], b = c . SinI?:Sin {A + B). 4. Two angles and a side opposite one of them is given. tA _ . Sin£ SinC Sin (J + 5) C = ,-(A + B), b = a- s . ml , c = a^ 2 = a-^ sJnJ ^. 5. Two sides and an angle opposite one of them are given. Sin B Sin A . „. „ _,. . 7 — - — = gives Sin B = Sm A . b:a. b a This gives the sine of B, but as B and v-B have the same sine, it remains doubtful which of these supplementary arcs, the acute, or the obtuse, is the one sought. We may sometimes decide it thus : If b < a, B < A; for cut off from C on a a portion = b, (CB'=CA): then ACB' is isosceles, and CAB=CB'A, which is >CBA; for CB'A = CBA + CAB, by Prop. D • i.e. In any triangle, the greater of two sides is oppo- A site tlie greater angle. If then A is acute, the obtuse must be rejected, for no obtuse angle B is < A. If A is obtuse, the obtuse B must be rejected; for there cannot be two ob- tuse angles in a triangle, by Prop. D. If b>a, B> A. Then A must be acute, and both values of B, the acute and the obtuse, may ----7T\ be suitable, or there may be two tri- ^^^~ /' | x^ angles (AB l C, ABC) which have the r^ s^ » " sides a, b, and the angle A. B being found, C is it = A - B, and c = ~ — - A a, ° ' ' Sin .4 or c may be expressed in terms of the data abA, thus: draw the J_ CD: then c = AD±DB, according as D falls within the a or without it . (BD) 2 + (DC) 2 = a 2 ; .-. £D=± Ja'-(DC)'=±Ja*-(bSmAy, or, asAD=bCosA c = bCosA± Ja 2 -b 2 Sin 8 A, 84 FIRST MNEMONICAL LESSONS. [§ 40. taking two like signs if D falls within the triangle, and two unlike if it falls without. The two values of c answer to the two values of B. You are already in possession of the knowledge requisite for geometry in its original etymological meaning, the measurement of land; for every plane figure can be divided into triangles, of which you can now discuss and solve all possible cases. Whole provinces have often been surveyed and mapped by measuring a single line, and taking angles with a proper instrument, such as the theodolite. Let AB be a base known by exact measurement. If the an- gles PAB and PBA are observed, which can be easily done if P be visible, whatever be its distance, from the points A and B, the triangle APB is known, as in case 3 of oblique triangles. Thus the lengths AP and PB and the angle between them are known by calculation, and the position of P can be fixed on the map or plan. If the point Q can be observed from any two angles of the triangle APB, Q is found in like manner with P: or if Q be visible only from P, not from A or B, and if PQ. is a line known either by measurement, or as may easily happen, from calculation of some triangle of which it is one side, the angle BPQ can be observed, and QB is found as in Case 2, as well as the angles PQB and PBQ. 40. You can write the expression for the area of a tri- angle, if either ab and c are given, or ab and C, as in \_32~\, for it is * Sine Cang half ba ; or putting a for area, a = 4 ab Sin C = 5 ca Sin B = -kbc Sin A. You can write it also in terms of the data of Case 3, ABc, by putting for a in ^ ac Sin B its value in terms of ABc, giving , c Sin A , . Sin A Sin B A -*-Sm{A + B)- CbmB -* C Sin(A + B)' In fact the area can be expressed in terms of any data which are sufficient to determine the triangle. The three perpendiculars from the angles on the sides ale are given with the sides and area, thus : ap x = bp 2 = 2a, ap l = cp z = 2a, wherefore p, = 2a:«; p 3 = 2^:b; p 3 = 2A:c;also a:b = p 2 :p, ; a:c = p a :/), ; from dividing the first equation by bp u and the second by § 40.J FIRST MNEMONICAL LESSONS. 85 We can substitute for a in the values of p lt p 2 , and p a any expression equivalent to a, and thus obtain various for- mula? for the perpendiculars. Thus by [[32], 2 t p l = - „Js . (s - a) . (s - b) . (s — c), or squaring, p*=j.S'(s-a).(s-b).(s-c), or , a + b + c -a + b + c a-b + c a + b- c jPl_4 ' 2a '^Ta ' 2 * 2 ; then dividing both sides by a 2 , f\_. a + b + c —a + b + c a — b + c a + b — c ^ a 2 ~ ' 2a ' 2a 2a 2a ' for remembering that . means times, the denominators on the right make 2a x 2a x 2a x 2a = 16V x a 2 . But the frac- — is not altered by dividing both numerator and tion 2a denominator by a ; therefore be b c . b c , b c „, 2 1+-+- -1+-+- 1--+- 1+ ]h . a a a a a a a a « 2 2 2 2 or multiply both sides by 4, ¥-(i + i + i)'.(-i + » + !).(i-! + i).(ui-ft a" \ a a) \ a aj \ a a) \ a a) be. If we now put for - and - the equivalent fractions just now found, *— and '— , Pi p 3 ^„( 1+ a + a\( r -i + a + &V(i-& + &Vfi+ e - & V « \ Pa Ps/ \ Pa Pa/ \ P2 ]hJ \ Pa Pa/ whence, extracting the square roots of these equals, 2pi — =±V(l+pi:p 2 +p l ip 3 ).{-l+p i :p 2 +p l :p 3 ).{l-p l :p 2 +p 1 :p 3 ).{l+p i :p r -p i :p 3 ), then taking the reciprocals of these equals, and multiplying by 2p„ by which we can find the side a, if p,, p 2 > an d p 3 are given. 86 FIRST IfNKMONICAL LESSONS. [§ 41. This is the solution of the problem, Given the three perpendiculars from the angles on the op- posite sides, to find the three sides. A more elegant exhibition of this solution is obtained by simply dividing the last equation but two by p*; i.e. by pi • P\ ■ Pi 'Pi> S ivm S ft f ter multiplication again by p, 2 * =p ,.(i + i + iy(_i + i + i).(i_i + i).(i + i_i) ; « \Pi P2 PJ \ Pi Pa PJ \Pi P2 PJ \Pi Pa PsJ and thence you get 4:i 2 and 4:c s , by (17) going round (1231...). The squared reciprocals of - - and - being thus known, and - are found, and also a b and c, in terms of the 2 2 2 three perpendiculars. LESSON XI. 41. Jane: — I have wished to find the triangle whose three perpendiculars are 6, 8, and 9, on the sides a, b, and c. We have by the last equation you gave us, ;hKs4 + §)(-S + H)(5 - H)(s + H) = ,— — x 20735. (lay f =8 *(s + H)( _ H + s)(H + 5)(H-s) (W x20m - != fl *(H + D(-H + £KH4)(K-5) lW j»«*w § 41.] FIRST MNEMONICAL LESSONS. S7 20735 is (144) 2 nearly: so that |=TTT = 6 nearly; ^=| nearly, and - = 4 nearly ; or the sides of the triangle are nearly a = 12, 6 = 9, c = 8, at least within the thousandth part of unity; a being, more exactly, 12-00029, 6=9-00022, and c — 8-00019. I confess that to me these arithmetical computations are not less troublesome than the algebraic reasoning, and require as great an expenditure of time and thought. Uncle Ben. ; — It is evident that no triangle can have its perpendicular on b equal to c, except a right-angled triangle. The triangle just found is nearly right angled at A, as appears from 12 2 = nearly 9 2 + 8 2 . Richard : — This is all very interesting about the sides and the perpendiculars. You have introduced us to fere, bic, and biCang Ql8]: can the three bisectors of the sides and angles be handled in the same manner as the perpen- diculars ? Uncle Pen. : — You may readily find either the bisector of a side or that of an angle in terms of the three sides. Let Ce — h 3 be ' bic,' the bisector of c. By [26, B], «' =V +(i<0" " - h 3 c Cos BeC, 6 2 =/,, 2 +GH 2 - - h 3 c Cos AeC ; but AeC and BeC are supplements, therefore by (23, L) a 2 + b 2 = 2h 3 2 + 2 . (£ c)*, or dividing by 2, h 3 2 +(±cy=±(a 2 +b 2 ), or V = * (a 2 + b 2 ) - a c) 2 ; h? = k (b 2 + c 2 ) - (l a) 2 ; h 2 = ±{c 2 +a 2 )-{ib) 2 - (B) if /«! and h 2 be the bisectors of a and b. The square of the line drawn from any angle of a triangle to bisect the opposite side, together with the square of half that side, is equal to half the sum of the squares of the other two sides. [37] DUQ {bic (half c)}'a half DUQ {ab}. v. bic [18J ; v. DUQ [13]. The two squares of bic and Q c) are half the two squares of a and b. 88 FIRST MNEMONICAL LESSONS. [§ 41. Let CF be the bisector of B ^ the angle C in the triangle ABC. Draw AB' parallel to CF, to meet CB produced in B'. The line BB\ cutting the parallels CF, B'A, makes FCB = CB'A = § C, ' ext. = int. op/ ; and AC cutting the same parallels makes FCA = CAB'^^C, 'alter, ins =' \_l~\- Therefore ACB' is an isosceles triangle, and Cf the perpendicular on B'A bisects B'A by QlS], ' perc is bic:' hence B'A = 2/A = 2 AC. Cos CAF = 2b . Cos | C. XT , r _ CF BC a Now by [6], -^- = BB , = ^n, > or > h y the preceding, CF = B'A.~ = ^ r Cos±C; C. a + b a+b z which gives the bisector CF in terms of ab and C. By squaring equals we obtain (CT)2 = (^ C08 '* C: and ^ M (3? ' A) ab Cos 2 5 C = (.? — c) . s, whence follows by substitution, 4- fib (CF)*=-^ FVn .(s-c).s. C. v ' (a + 6)- v ' This gives us the square of the bisector CF in terms of a, b and c ; * being g (« + 6 + c). If 4,j A- 2 , £ 3 be the bisec- tors of A, B, and C, we thus obtain k i = ^Jbc.s.(s-a); k^-^-Jca .s .{s - b) ; IntheA^iZ?C, if e = -4Fj Bf = c-e: and we have £34], rt c — e Sin CFB Sin£C SinCivl Sin^C § 42.] FIRST MNEMONICAL LESSONS. 89 whence, as Sin CFB = Sin CFA, [23, L], a c — e . 7= — -, A. b e by division of equals by equals.' The following theorem is expressed in this equation (A). The bisector of the vertical angle (C) of any triangle (ABC) divides the base (c) into segments {e and (c — e)} which have the same ratio with that of the two sides (b and a) adjacent to those segments. 42. By help of this property we can make a useful transformation of the formula for the bisector. From D. T = , and - = , follow b e a c — e a „ c — e , b e r + l = + 1, and- + l = + 1, or b e a c — e a b c—e e , b a e c—e t + 7= + -j an d - + - = + , or bo e e aac—ec—e „. a + b c , b +a c D'. — -j— = - , and = . e a c—e , T , N (a + b) + c (a + b) — c Now s.(s-c) = K ^ .^ -I 2 (a + Vf-c 1 [14, , , N2 ~, (a + by K ' 4 (a + b) 2 or / t - 3 2 = ab .(l - — ^L- I = ab - c" . {a + b) 2 } (a + b) a The multiplication of the equal members of the two equations D' gives (a + b) 2 c 2 ab _ e(c-e) ab ~ e(c-ey ° r {a + b) 2 c" ' whence by multiplying equals by c 2 , comes, C "'(a~+b) 2 = e ( C ~ e )> glVmg * 3 Q = ab - e {c - e) ; or * 8 = ^6 - e (c - e). E. 90 FIRST MNEMONICAL LESSONS. [§ 42. Def. The number 2ab:(a + b), obtained by dividing twice ike product of any two numbers a andb by their sum, is called the harmonic mean of those two numbers. The equations (C) and (E), with (A), may be remem- bered thus : (E) BiCang is RooFDuP (db, segs), vid. biCang [181. [38] (C) Is C6sHaCang of HarM legs ; in CosIIa pron. sh as in cash ; of stands for times. (A) And vi (ab) is the vi (segs). vid. vi. [5]. [39] HarM (db) is two db by S (db). S for Sum of. As either S or M is used forSuM, so either D or F may stand for DiF, or difference. DuP is two or duo Products of the indicated pairs of num- bers ; RooF is square Root of DiF ; biCang, i.e. k 3 = the square Root of the Difference of the duo (two) Products {ab and the segments, e and (c — e), of c} ; k 3 also is equal to the Cos of Half Cang of the Harmo- nic il/ean of the leg?, a and b. N.B. between two numbers, of always mean times, as g x ^, is h of f , &c. The fourth line is the above de- finition. In equations (C) and (E) are expressed the proposition following : The bisector of any angle of a triangle, included within it, is equal to the harmonic mean of the sides containing the angle multiplied by the cosine of half the angle; and the square of that bisector is equal to the difference of the rect- angles of those two sides and of the two segments of the third side made by the bisector. If it is granted^ (and it shall presently be proved) that a circle can be drawn through three given points, the property (E) can be established very simply thus. Let the bisector CF be produced to meet at B' the circle through AB and C; join AB' ; then i. BCF = ACB' ; /: CBF= CB'A, because these are angles on the same arc AC [a, 1Q~|. Hence the remaining pair are equal, or z CFB = t CAB' ; for the three angles of the tri- angles CFB and CAB' have equal sums, and make up in each two right angles : Prop. D. We have in these triangles, which are similai because they have angles of the same magnitudes, [34], CF CB Sin CBF ~ Sin CFB ' AC CB' Sin CB'A Sin Ii'AC' § 42.] FIRST MNEMONICAL LESSONS. 91 whence by division, Sin CB'A Sin CBF CF Sin B' AC CB ' AC ~ Sin CFB ' CB' or CF CB AC~ CB" by reason of the equal angles and sines : Hence CF .CB'= AC .CB, by multiplying by AC . CB' ; or CF(CF + FB') = ba y (CF) 2 +CF.FB'=ba, (CF) 2 =ab - CF. FB'= ab- AF . FB = ab-e(c-c); for CF.FB'=AF.FB, by [20]. In reply to Richard's enquiry about the handling of bisectors, I shall only say at present that the sides a b and c can be found by equations (B), when h l} h 2 , and h 3 are given, and from equations (C') when k l} k 2 , and k 3 are given; but the solution of these problems, especially of the latter one, is too difficult for you to attempt, until you are more dexterous algebraists. Yet you may bear these ques- tions in mind, as matters of future interest and enterprise. Jane : — The device of adding unity to* both sides, by which (D') is proved, was employed in the demonstration of [352 ' SubyD Sines &c/ and is a very pleasing con- trivance. Uncle Pen. : — It is both a simple and a fertile expedient. Let abc and tfi£>,c, be triangles having the same angles AB and C, or similar triangles. From «:Sin A = &:Sini? E34fj, comes by dividing by 6:Sin^4, a:b = Sin ^iSin B ; and in the same way a^fr, = Sin A:8in B. We have then a _ o, b _ b l c c, b ~ b t ' c~ c x ' a ~ «, ' Add to each side of each ± 1, and you have as in (37), a±b a,^bi b±c_b x ±c x c±a c,±a, . b b x ' c c, ' a «i ' F. 92 FIRST MNEMONICAL LESSONS. [§ 42. Divide G with the upper sign by G with the lower, member by member; a b a, + b, b + c b.+ H. are simitar, liKe m. ' v preceding figure. \ / e tCB'B=CAF, \\/ / a — b a 1 — b 1 ' b — c b l — c 1 > c Multiplying in (F) by bb lf &c, ab x = ajj, bc x = 6,c, ca, = c x a ; F'. and then dividing by «,&,, &c, a:ai = b:b 1} b:b i = c:c 1 , c:c l = a:a } . F". All the equations FGHF'F" are true of the sides of two similar triangles. Thus, let CF be any line drawn from C to meet in B' the circle through AB and C ; join AB' ; jfe- — - draw Cf making z BCf= B'CA : then the /7p> triangles CBf and CAB' are similar, like a£_ CBF^ and C^' in the Join Z?Z?', and you have both on the arc CB [19], and tfCA (=ACB'+FCf), equal to the zFCB=(BCf + FCf); wherefore the triangles ACf and BCB' are also similar. It follows that fB.CB'=AB'.CB, and Af.CB'=BB'. AC, by the reasoning in (F) and (F'), (i.e. ca x = c x a). Hence we have CB'. (fB + Af)=CB.AB'+CA. BB', or CB'.AB=CB.AB'+CA.BB'. (I) As ACBB' are any four points of the circle, we have (Euclid vi. Prop. 16, Cor. 4) in the equation (I) this propo- sition : The rectangle of the diagonals (AB, CB') of a quadri- lateral in a circle, is equal to the sum of the two rectangles of the opposite sides (AB'. CB and AC . BB^. [40] In quad, inscri. two opps. are v, (c) (31). And pro digs is DuPo'ppo.si. (I). i.e. in a f/wWrilateral inscribed in a circle, two opposite angle.* are tt, or together make two right angles : and the product of the diagonals is the duo V'roducts, or double ;>roduct, of the o/);>osite rides. §43.] FIRST MNEMONICAL LESSON'S. 93 43. I promised to prove that a circle can be drawn through three given points. Let them be A B and C, and let d and e be the middle points of the lines CB and AB: let the perpendicu- lars to those lines at d and e meet in the point //. The triangles Bhd and did hav- ing two sides in the one equal to two sides in the other, and the contained angles equal (34, B), have their remaining sides hB and hC equal: and in the same way, in the triangles Bhe and Ahe, the sides hA and hB are equal. The point h is therefore equidistant from the three given points : a circle having its radius = hB, and its centre at h, will pass through them all. The perpendicular on AC from h bisects AC by \_\7, a]: hence it appears that the three perpendicidars to the sides of any triangle at their middle points meet in a point, viz. the centre of the circumscribing circle. To find therefore the centre of the circle which passes through three given points, it is merely necessary to join one to the remaining two, and to raise perpendiculars at the mid points of the joining lines ; these perpendiculars meet in the centre required. Let CD be the diameter of the circum- scribed circle, and Cf the perpendicular from C on the side AB or c. Join AD: then CD A and CBf are similar triangles, being right angled at J. and/ £19]> and having equal angles at D and B, and con- sequently the other pair of acute angles at C equal. Hence (F', 42), CfxCD = CB. CA, and CD=CB. CA:Cf. The same thing is proved by the consideration that the angles at D and B have equal sines. Thus we see that the rectangle under any two sides (CB, CA) of a triangle is equal to that under the diameter of the circumscribing circle and the perpendicular let fall on the third side from the opposite angle. (J). The area is \AB .Cf, or Qf=-^, putting a for area, whence by substitution, CD=CB.CA.AB:2a, r CD = abc:2 a = 2R, 94 FIRST MNEMONICAL LESSONS. [§ 43. if we put R for the radius of the circle. Hence R = abc-AA, or m = abc:± Js (s-a).(s-b). (s - c). The radius of the circumscribed circle is found by dividing the product of the three sides by four limes the area of the tri- angle. (K). Let CD the bisector of the angle C of the triangle ABC, meet in D the bisector of A ; and let Df and De be perpendiculars from D on AC and CB. By [22], DC Sin 1 ACB = Df = De, and DA Sin £ (7^.£ = Z>/= Dg, .-. Dg=-Df=De, and since the sides of the triangle are perpendicular to these lines, these sides by [17, b] are tangents of the circle which passes through gf and e, having its centre at D. Thus it is evident that a circle can be inscribed in any triangle ABC. Because De = Dg, if we now draw the line DB, we see that it is the bisector of the angle CBA ; for [22] DBe and DBg have equal sines. Thus the three bisectors of the angles of a triangle meet in a point, which is the centre of the inscribed circle. By Prop. B. (20), AC x Df+BC xDe + ABx Dg= twice the area of the triangle ABC, i.e. putting r for De the radius, r (a + b + c) = 2a, or, s being ^ (a + b + c), r=A-.s. (Z). The radius of the circle circumscribed about a triangle is equal to the product of the three sides divided by four times the area : and the radius of the inscribed circle is equal to the area divided by the semiperimcler. (J) Dim. out circ is ba to perc. ba is one syll. [41] (Z) In. rad. of semp. is Ar'e, Are is one syl. ; of is x. (K) Four out. is bac by Are. bac a monosyl. .DJawictiT of out (i.e. circumscribed) circle is the quotient or ratio ba : perc. vitl. [11! | ; to is here equivalent to by. The inscribed mcfius of (i.e. limes, vid. [SoC.]) scnnjienmcicx — Arc (French for area) : four out (radii ) - bac by Area. ; or Ali = 6ac:A. § 43.] FIRST MNEMONICAL LESSONS. 95 If DA be the bisector of the angle at A below the base AB, the perpendiculars Df, Dg, and De are still equal as before: and DB being then joined is proved as before to be the bisector of the angle at B, below the base. Hence a circle whose centre is D and radius De will touch the sides of the triangle below the base in ef and g. We have also (Prop. B, 20), A C x Df+ BBxDe- AB xDg = twice the a ABC, or r,(a + 6-c) = 2A, and exactly in the same manner, r 2 (a — b + c) = 2 a, r 3 (- a + b + c) = 2 a ; putting r l r 2 and r 3 for the radii of the cirles touching the sides below the bases c b and a. We have also, as already proved, r (a + b+c) = 2a, whence by multiplication of equals, a 4 being aaaa, 2s = a + b + c, &c, rr i r 2 r 3 .8s(s — a). (s — b).(s-c) — 8 a 4 , or dividing both sides by 8 a 2 , ^32^], rr x r,r z = a 2 , and ± s jrr ) r i r z = a. It is usual to call the three circles whose radii are r^^ the escribed circles. You may add, if you think it will aid your memory, to the last mnemonic this line more, (in for inscribed, e for escribed, rad.) [jll'3 RooP. in. e. rads is Are ; vid. RooP [32.] which expresses the elegant theorem, that the square root of the product of the radii of the itiscribed and three escribed circles of the area of the triangle. Example. If the sides are 6, 8, and 9, a = 23-525252, R = 4-59081, r = 2-045673. What are r„ r 2 , and r 8 ? Find them, and verify C 41 '3- If c be 8, we obtain from (B, 41), (D r , E, 42), h 3 = 6-5192024, e = 4-8, # 3 = 6-2161081. 9G FIRST MNEM0X1CAL LESSONS. [§ 44. LESSON XII. 44. Def. Similar figures are such as have the same angles in the same order. Let ABC and A ,B,C, be similar triangles, whose sides are ale and a x b x c x . The areas of these triangles (40) are ic 2 Sin A Sin B: Sin (A +'B), and | c 2 Sin A Sin jB:Sin (A +B), of which numbers the quotient or ratio = c 2 :c 2 . By this, since c 2 :c 2 = a 2 :a 2 = b 2 :b 2 , (F", 42) we see that the areas of two similar triangles are proportional to the squares of their like sides, i. e. sides opposite the same angles. Let now CBD and C l B l D l be similar triangles : ACDB and A&D^, will be similar quadrilaterals, A CD and AiC\D, being equal, as sums of the same triangles, and ABD also = A 1 B l D 1 . If we put CB:C l B l = N, we have (F", 42), N^DB.D.B = DC.D.C, , and also N=AB:A 1 B 1 *=AC:A 1 C 1 . We have proved that (a CDB):(a C.D.B,) = N*, i.e. (a CDB) = N* (a CAA), and (aADB)=N 2 (aA 1 D 1 B 1 ); whence, by addition, (a CDB+a ADB) = A T2 (a C.D.B, + a AA#,), or fi^CD^ = N2 =( AB f : ( A > B iY = (BDy+BM 2 = &e. If the figures be made similar pentagons by the addi- tion of similar triangles DBE and D l B l E l , it can be proved that N=DB:D l B l = BE:B l E l , and (a DBE) = N* (a D x B x E& whence § 44.] FIRST MNEM0N1CAL LESSONS. 97 fig.(ACDB + DBE ) m N * = (BEr , BiEiT = (ABy:{A l B i y = &c. In this way can be demonstrated, for any number of sides, the truth of the following proposition. Sirnilar polygons have their corresponding sides in a con- slant ratio, i. e. the ratio of every such pair is the same ; and the areas of the figures are to each other as the squares of the corresponding sides. In szmils, quot. Ares [42] Is vi(lik.si. squares) pron. villiksy : vid. vi. [5'.] i.e. in two similar figures, the quote of the areas is the quote of the squares of like sides. Corresponding or like sides lie between like pairs of angles. This includes the former part of the proposition : for if the ratio of the squared sides is constant, that of the sides is constant. Let a b c d..f be a regular polygon of n sides inscribed in our scale-circle; which implies that the n chords ab, be... fa are all equal and that each of the n angles subtended by them at the centre is 2iv.n, or the « th part of four right angles. If Op be perpendicular to af it bisects the angle aOp [18], and the chord af, which is evidently twice the sine of a Op; or ab = bc= ... =fa 2 Sin (|.^ = 2 Sin (*■:»). Let OA, OB...OE, OF, be taken on Oa, Ob, &c, each equal sec ir-.n, and let AB, BC,...EF, FA be joined, making a polygon of n sides. In the isosceles triangle OAF, Op the bisector of AOF is perpendicular on AF [18], which is therefore parallel to af; and we have by [6], Oq:Op = OA-.Oa, or since Oa = 1, Oq = Op.OA = Cos (7r:«). Sec (ir:«) = 1 ; [24] wherefore q is the extremity of the radius, and the line AF perpendicular to Oq is a tangent. In the same way it can be proved that all the n sides AB, BC, CD. . . are tan- E 98 FIRST MNEMONICAL LESSONS. [§ 44. gents parallel to the chords ah, be, cd...; they are also all equal (B, 34); and the / ABC = l abc, &c. ; wherefore ABC...F is a regular polygon circumscribed about the circle and similar to the one inscribed. The side AF is manifestly = 2 tan (tt-ji) ; and the peri- meter of the circumscribed polygon is 2w tan (tt:?i), while that of the inscribed is 2n sin (tt:w). Wherefore Perimeter of inner polygon _ 2w Sin (vr:n) . . Perimeter of outer polygon 2« tan (•«•:«) ^ ' '* When n is exceedingly great, ir-.n is exceedingly small, and its cosine is immeasurably near unity, in -which case the two terms of the fraction Sin (7r:H) : tan (tt:h) differ in value by a quantity immeasurably minute, and either of these perimeters may without appreciable error be taken for the other, or for that of the circle which lies between them. It follows that if we can find correctly the sine or the tangent of an arc ir:n as small as we please, we can find the perimeter of our circle as nearly as we please, and thence by proportion that of any other circle whose radius is given. To find the value of ir, the length of our scale- semicircle, is no more practicable, and no less, than to find the length of the diagonal of the square whose side is unity, i.e. the square root of 2. The value of ^2 is bobodatusdipta, or 1-4142135623731, that of ir is tafaloud- sutuknoint, or 3-141592653589793, at least within a ten- thousand-billionth; these decimals have been extended to hundreds of places, but it is idle to seek for the termi- nation of these infinite series of figures. There is nothing difficult, except the laboriousness of the arithmetical work, in finding the perimeter of a polygon of a million sides We know [31], Sin0 = 2Sin^ Cos hj, and [22], tan^ = ^; Sin 6 • tan ±6 = 2 Sin 2 1 0. Also [25], l=Cos 3 |0 + Sin 2 A0, [31], Cos0 = Cos 2 £0-Sin 9 i0, l-Cos0 = 2Sin 2 £0; §45.] FIRST MNEMONICAL LESSONS. Hence Sin0tan<|0= 1 - Cos 0, t . i - Cos e tan ^ = " Sin0 • tansy 1-Cos 2 1 - Cos _ = t — ~ — ^: ri4. cl. 1+Cos0' l."»-j» the numerator and denominator being both divided by (1-Cos0). By formula b, if we know the cosine of the angle at the centre subtended by the side of any regular polygon, we can find the sine of its half; and knowing the sine of that angle we can find the tangent of its half by equation c'. This half-angle is the angle subtended at the centre by the polygon whose sides are in number double of the first ; the half of this half-angle is that at the centre under a side of the polygon having four times as many sides as the first : and thus we can proceed bisecting angles and doubling the number of sides and finding the perimeters, 2ra Sin (inn), and 2n tan {ttvi), as far as we please. 45. Thus to find the perimeter and area of the inscribed and circumscribed squares. Let p and P be the perimeters ; we have » = 4, and p = 8 Sin (ir: 4) P = 8 tan (tt:4). In equation (b) (44), is the known angle at the centre, which is 7r:2, the fourth part of the circumference ; and g is found by Sin 2 ( 7 r:4) = i(l-Cos0, or Sin'.(7r:4) = N /|"=l: N /2= JS / : 5. tt 8 4.2 Hence p = __=_ _ = 4/0. ^2 J2 * V ~ By equation c', we have tan (tt:4) = (1 - 0):1 = 1, there- fore P = 8 x 1 = 8. E2 100 FIRST MNEMONICAL LESSONS. [§ 45. The area of the inscribed polygon (vid. last figure) is 11 x (area of triangle aOf), or ^32], £ wOa . O/Sin aOf= £ n Sin (2tt:«), and that of the circumscribed is a »i . OF . Sin A OF = i n Sec 2 (tt: n) . Sin (2w.n). J Sin (£«•:») Area of inner polygon 2 _^ Hence Area of outer polygon » ^ ^ sin ^.^ 1 Sec 2 (w:m) Cos 2 (ir:w). Whereby we see that these areas approach without limit to equality with each other, and consequently to the area of the circle, as n increases without limit until Cos 2 (tt:m) = Cos 2 .0=1, These formulae give us, since Cos^yi-Sin^j-, Area of inscribed square = _ Sin - = 2 square units. Area of circumscribed square = iSec 2 ^Sin^ = 2:Cos 2 ^ = 4. 2 4 2 4 These are the perimeters and areas when the radius is unity. When r is the radius, we have, considering the tri- angle made by any side of the polygon of n sides and the radii through its extremities, in the' two circles whose radii are 1 and r, if s, s' be those sides, s' s - = - , or s = sr, r 1 whence it follows, us' - r.ns, or perimeter for rad. r = r x perimeter for rad. unity. In the same way the areas of the similar elementary tri- angles of the similar polygons in these two circles arc as § 45.] FIRST MNEMONICAL LESSONS. 101 r 2 -.l 2 , [[42], and area of polygon for radius r = r 2 x area for radius 1 ; i. e. nr 2 . 2ir area of inscribed = -— Sin — , 2 n r. . Ml nr * C 2 7r O' 2-7T area of circumscribed = — oec 2 - . bin — . 2 n n Required the sides, perimeters, and areas of the inscribed and circumscribed regular octagons, the radius being r. The inscribed side is 2r Sin (tt:8), and Sin (tt:8), by (44, b) is 1-Cos \ _ J 1 f 5 , = -3826834. 2 V 2 The circumscribed side is 2r tan (tt:8), and tan(7r:8), by (44, c'), is ~j^ = J ^-1=^2-1 = '4142136. The perimeters are therefore l6r x 0-3826834, and l6r x 0-4142136, at least within a millionth of unity. For the areas we have Sin(27r:8) = x / : 5 = "7 71068; 1 1 SeC ■ ^ = Coi^S) = ^l-Sin»( W :8) = 1 =T0823922. 7l-('3826834) 2 The area inscribed is 4r 2 x 0-7071068. The area circumscribed = 4r 2 x (1-0823922) 2 x 07071068. We can now find Sin(-n-:l6) and tan(7r:l6), from Sin(7r:8); for Cos(7r:8) = 7l-Sin 2 (7r:8), in the formula? (b) and (c'); and thus we determine the perimeters and areas of the outer and inner polygons of sixteen sides, and finally of 102 FIRST MNEMONICAL LESSONS. [§ 45. polygons that approach inconceivably near to each other and to the circle. Jane : — I see that in Hutton's tables, the Sine and tan- gent are set down as equal for arcs below 18' ; thus, sine 3' = -0008727 = tan 3'. Now 6' is —r, — of the circumference : according to Hutton, 3600 ... the semi-sides and perimeters of the inscribed and circum- scribed polygons of 3600 sides are sensibly alike. Then either perimeter is equal to that of the circle, at least to our perception equal. We should have 2tr for the product of 2 x 3600 x -0008727 ; but this is 2 x 3-14172, and 2tt = 2 x 3-141 593 ; how is this ? Uncle Pen. : — You have a right to say that the sine and tangent of 3' or 7r:3600, and therefore the sides of the poly- gons, are sensibly equal ; for they differ certainly by less than one ten millionth of the radius, a difference which in itself is an error of little importance; but when you infer that the perimeters are also equal, you multiply that error by 3600, and it then becomes of consequence. You may however see plainly, from what we have been doing, that you can calculate tt, and solve the famous problem of squar- ing the circle, as nearly as you please by arithmetical opera- tions. To square the circle, is to find a square whose area is equal to that of the circle. The area of the inscribed polygon of n sides, when n is unlimitedly great, is the area of the circle, and this is n times the area of the elementary triangle of the polygon. This element is the product of the perpendicular into half the base, or Op x Ap = Cos (tt:?i) . Sin (win) for radius unity, and = r s . Cos (tt:>i) . Sin (7r-.11) for radius r, ^42]. The perimeter of the polygon is 2nr Sin (tt:m), so that the area is perimeter times - Cos (jr-.n), which, when n is great beyond limit, making the angle 7t:m small beyond all limit, (that is, nothing,) becomes = perimeter x -, or (since the perimeter of the circle is 2irr) (38 A) = irr. That is, § 46.] FIRST MNEMONICAL LESSONS. 103 C. The area of a circle is the product of the radius and semiperimeter, and—"w x (rad.) 2 . It is easily seen that the area of any polygon, regular or not, circumscribed about a circle, is the product of the radius and semiperimeter of the polygon. To square the circle whose radius is r, we have to solve the equation s s = irr 2 , giving s = r Jtt, the length of the side of the sought square, which is found by extraction of the root, with more or less exactness, as you take more or fewer of the decimal places in ir. The important equa- tions (b and c) may be easily remembered,- thus, putting for J - Cos e its value ver. (Def.'5, 32) ; [43] Taf. Sin is ver.; (c) vid. taf. [35.] Two quasif is ver. (b) [43 J Rim's two . ir . rad ; Are's r' . -it . rad. (A. 38.) C. Sif means sine of hal/, as taf is tan of half: (tan of half 0) times sin 6 = ver ; and 2 (squared sin of hal/ 6) = ver 6. You can saiely omit after taf, sin, sif, and ver. The rim or circumf. = 2irr; Area = rirr = irr 2 = rad . tt . rad ; pron. r j>y rad. I should have remarked, just now, that you will be hereafter introduced to more expeditious methods of calcu- lating the value of tt, so that it would be a profitless labour to attempt this by the method pointed out above. There are too methods of determining the sines, cosines, &c, of arcs, somewhat more compendious and generally applicable than the one I have described ; of which you may see some stated in the introduction to the tables. My object is an- swered for the present by giving you the information you now possess. From the known sine and cosine of 90° or of any other arc, the sines, cosines, &c, of all arcs can be found, by [27], [29], and [43], to any required extent of decimal places. LESSON XIII. 46. You know that if a be any number, a-a = a s = a 1+1 : a-a-a = a s = a 1+,+1 = a 2 a = a w , 104 FIRST MXEMONICAL LESSONS. [§ 4G. and whatever whole numbers b and c may be, a b -a c = {a'a-a"-(b factors)} (a-a~a---c factors) = a bH ; or (a to 6 th ) x (a to c ,h ) = a to (b + c) ,h (power). Further a*:a = a 3 =a*~ l ; a*:a s =a i =a*~ i ; a*:a 3 =a l = a*~ 3 =a I a*:a* = «° = « 4-4 = 1 ; for a 4 :a 4 is certainly = 1, so that we may affirm that a l :a A = a , if by a" we mean unity. . ,. aaaa 1 . r aaaa 1 a 4 :a 5 = = - ; a 4 : a 6 = — — = -,; aaaaa a aaaaaa a hence we may affirm that a*:a b — a 4-5 = a~\ and that a 4 :« 6 = a*~ 6 = a~ s , if we mean by a~ l and tf" 8 merely the reciprocals of a and a 2 . In the same way, we can always affirm a b :a c = a*-*, whether b > c, or b = c, or 6 < c ; if we understand that a i_6 = a is unity, and a~ p is the reci- procal of a r . We are to remember then that a~ p , (a to less p,) — reciprocal of a r , (cip a to p). £44] (a to le p') is cip (a to p'). We have proved that a b+c means a b xa c , and a'"* means a b :a c , for all integer values of b and c. What if they be fractions ? Suppose b = § = c : can we maintain intelligibly that a-, a' 2 = a- + * = a l = a} Certainly, J a x Ja = a; so that we may affirm a 1 . a'- = a, if we define a- to mean the square root of a; and a'-.ar.a 3 = d A 3 —a, if « 3 be ^a, the cube root of a. If then 6 = J = c, can we say, a*, a 3 - rt < + 3 = a 3 ? Unquestionably a 3 , a 3 = (a 3 ) 2 ; for (the cube root of a) times itself must be the square of (the cube root of a). We are then safe in putting a*' for a*.fl , if we define that « 3 = the square of a = (a ) 3 , and it is plain that a r .a r = a r r = a r , if it be understood that 2 J[ rt r = (« r ) 2 = the square of the r th root of a. Can we then say, A_rt 3 .a 3 = a 3 + 3 + 3 = a 3 = a 2 ? We know well that HJa 1 . ^/a J . ^/« 2 = a 2 ; for the same reason that § 46.] FIRST MNEMONICAL LESSONS. 105 ^2.^/2.1/2 = 2. We can then affirm the preceding with safety, if it be laid down that a 3 = the cube root of a 2 . Are then the cube root of a 2 , and the square of a 3 the cube root of a, the same number? To consider this, let us denote the cube root of a by y, and write the identity A y = a 3 ; thence, cubing equals, it follows that y 3 = a, and, squaring equals, The cube root of a 2 (or y 6 ) is clearly y 2 , for y 2 .y 2 .y 2 =y 6 ; and the square of a 3 , by A, is also y 2 . i. e. the square of a 3 , which is a 3 , is also the cube root of a 2 . ^45] e to vi(two three), v. vi. [5] pron. tovvy. is Croot of sq.e, Croot for cube root. is squared Croot e : sq. e = e 2 . i.e. e*, e to vi. (2, 3), e to the (2:3) ,h power, is the Cube root of squared e, and is the squared Cube root of e. Croot e is cube root of e. You will digest this little nicety best with this mnemonic, which will prompt the meaning of e». Generally, let y denote the n ih root of a. From y = a n , comes, taking n th powers of equals, y" = «, and then taking m lh powers of equals, y n .y n .y\..(m factors) = a m , or ^-+»+...(m umeo _ y™ = «, Now since y m .y m .y m ...(n factors) = y m+m+ - in{imet:) = y'" n } it is clear that the n lh root of « m (or y m ") is y m ; and the m ih power of a" is y m , by the first equation. Therefore the m th power of a", the ?i lh root of #, is also the ra lh root of a m , the ?n ih power of a ; and a n is either this root or this power, whichever you please. You can easily generalize e :! of the last mnemonic E5 106 FIRST MNEMONICA.L LESSONS. [§ 47. into e n . It is not necessary to read e" at length, as the m {h power of the « th root of e, or the w tb root of the ?« th power ; for we may call it the (m by w) th power at once, and read it e to (in by »)*, or e to (w« by n). A little attention will pre- vent us from confounding to in the reading of a power with to in proportion. We can distinguish the reading of e:p and e p , as e to p and e to j) lh (power understood). The (in:n) lh power of a number is always given: thus 2 5 the (2:3) th power of 2, is obtained either by squaring 2, and then taking the cube root of 4, or by finding the cube root of 2, and then squaring that root. 47. No power of a is altered in arithmetical value by changing merely the form of the index. Then a" = a* = c$ = a*"; for a*™ is the m* power of a*" , i. e. of a"*, the »i th root of a- ; but the m th power of the m th root of a number is the number. You may feel, and you ought to feel satisfied, without trying to conceive exactly how it is, that whatever real number a' or a c represents, a"' } or a mc must represent the same; and you have a right to affirm, when once the interpretation of a fractional index is fixed upon, that the (mc) lh root of the (?w6) ,h power of a is the c ,h root of the 6 th power of a ; or that the m th power of the (b:mc) th power of a is the m ih root of the (mb:cy h power of a. Thus we can dispense with the word root altogether, if we please: it is perfectly correct to call 12 the (s) th power of 144, or 10 the half-power of 100. We may write 10=100 ' 5 , and 12 = 144 ' 5 ; which assert that 10 is the 5 th power of the 10 th root of 100, and that 12 is the 10 th root of the 5 lh power of 144. We are satisfied, that for a b . a c we can put a 1 **, when b and c are equal fractions ; is the same allowed when they are unequal? Thus, is a'-. a* = a'- + * = a? '? It is certain that a', a* = a*, a*- Let «* the Cth root of a be denoted by y: then y 3 .y* is a € .a s , the product of the third and fourth powers of cr : but //.,'/ -}f' ', i- e. a 6 , a* =g 7 ; .-. aK a* = y 1 and is - a , the 7th power of y, or of a". § 47-] FIRST MNEMONICAL LESSONS. 107 Further, a 2 :a* = a 2 ~* = a'^ ; for In the same way it can be proved that the product of any powers of a number, positive, or negative, whole or fractional, is the number raised to the (sum of all the in- dices)* power. Thus, -^—. = 2 3 . 2 2 . 2-\ 2 6 = 2 3 " 2 . 2 s " e 2 2 .2* = 2 (3-2 + i-|) = 2 l-| = 2 ! =2 3 } or twice the square root of 2, divided by the sixth root of 2 5 is the cube root of 4. We can thus prove, of. xf. x c . x d . . . . = x (5 or 10^=71000 = 31-6227766; therefore 31 is some power of 10 less than 1*5, and 32 is a power of the same greater than 1*5. In Hutton's tables of logarithms, you find opposite the numbers 31 and 32, the numbers 1-4913617, and 1.5051500, that is, 31 = io 1-4913617 , 31-6227766 = 10 1 ' 5 , 32 = io 1 " 50515 . t„ , ,, 14913617 , „, rt .« If then you seek the th power of 10, you will obtain 31, or a number which differs from 31 by a quantity of no sensible value; i.e. 31 is the ten millionth root of 1000 000; the zeros being in number 14913617, nearly 15 millions. The number 1*4913617 = log 31, or is the logarithm of 31 to the base 10. ' Richard: — There is a sum for you, Jane; you shine in evolution. How many miles would those zeros occupy ? Jane : — I fancy that Mr Hutton did not find out by in- volution and evolution what power of 10 31 is ; but I should greatly like to know how he discovered it. This is one of the most interesting subjects we have yet seen opened. Uncle Pen. : — You will be introduced to this secret in due time. For the present, you may safely take it for 110 FIRST MNEMONICAL LESSONS. [§ 48. granted that Hutton is correct, and be satisfied with a lesson about the use of his tables, leaving the full theory of their construction till you are more advanced. And I can assure you that it is a rich remuneration for all the previous study required, to be able to understand the beautiful doctrine of logarithms. LESSON XIV. 48. Every number is some power of every other. 1 1 1 8 4 2 1 2 4 8, are in order equal to 2 -3 2 -2 2 -1 2° 2 1 2 2 2 3 , all powers of 2 ; 1 1 1 27 9 3 1 3 9 27, are the foll ? . powers of 3, 3-3 3 - 2 3-1 3° 3 1 3 2 3 3 ; 1 1 1 1000 100 10 1 10 100 1000, are powers of 10, viz. io~ 3 io- 2 io- 1 10° 10 1 10 2 10 s . The numbers between 4 and 8 are powers of 2 greater than the square and less than the cube; and so on of the rest. A list of numbers, 1, 2, 3, 4 &c, having under each the index of the equivalent power of 2, would be a table of logarithms to the base 2 : and such a table can be con- structed to any base. The base commonly employed is 10, and it is usual to write 2 = com. log 100, 3 = com. log 1000, Sec. ; or the base is sometimes written as a subindex, thus : 2 = log )0 100, 3 = log 3 27, &c, read 2 is log 100 to base 10 ; 3 is log 27 to base 3 ; but it is usual enough to omit all indication of the base when there is no risk of being misunderstood. Whatever the base may be, is always log 1 ; for, (46), 3 =2° = a =10°=l ; also if the base be greater than unity, all proper fractions will have a negative, and all numbers above unity a posi- § 48.] FIRST MNEMONICAL LESSONS. Ill tive, logarithm. The contrary would be true if the base were a proper fraction. The chief use of logarithms is to save the labour of mul- tiplication and division, and to facilitate evolution and in- volution. Let n and m be two numbers, suppose each of six or eight places, and let I and /, be their common logarithms. We have thus n = 10', m = 10'', and - = 10~'i [44], w»i = 10'. lO'i = 10'+'' by [46], and n:m = 10'. 10"' 1 = 10'~'i ; or log (nm) =1 + l { = log n + log m, and log (jiwi) = /—/, = log n — log m. To find the product or quotient of n and m, we find their logarithms / and Zj standing opposite n and m ; then their product nm is seen standing opposite the logarithm (I + /,), and their quotient n:m opposite I — /,. Thus, in- stead of a tedious multiplication, we have only an easy addition, and, for a long division, a simple subtraction to perform. All products of two quantities {t and a), are ob- tained by the rule log t + log a = log (ta), a. which includes all quotients, for t:a =t . - , and log - = — log a ; [44], so that log t + log - = log f t . - ) , or log t — log a = log (t:a). a'. All involution and evolution is effected by this rule, for all values of/), loga p =ploga. b. To prove this, let p = q:r, where r may be unity, if p is integer. Then if I be the logarithm of a ; we have / = log a, and a = 10'=10 ,osa l l r 1, and therefore log PQ > 0. He tabulates this logarithm, which is 10 too great, for log Sin 8 = log PQ - log 10 10 = log PQ - 10. And every circular logarithm, log sin 0, log cos 8, &c. is too great by log 10 10 or 10, in his tables. We have then always to subtract this 10, which is added for typogra- phical convenience and correctness, from the tabulated value, in order to obtain the accurate logarithm of any circular function of 8. An example or two will suffice for illustration. § 50. J FIRST MNEMONICAL LESSONS. 115 To find log cos 6°. 8'. 42". Call this log cos (N+i)', t being 07' = 42". We read log cos 6°. 8' = 9*9975069 = log cos if', and log cos 6°. 9' = 9*9974933 = log cos (N+iy, whence log cos 6°. 9' -log cos 6°. 8' = - 136 = A log cos N' = log cos (N+ 1)' - log cos N'. As the log cos 6 decreases at about this rate, — 136 for each added minute, in this part of the table, we obtain the decrements due to 07' by proportion, as before. We want to know log Cos (A + ^') / - log Cos N' ; and we have it by log Cos (N + i)'- log Cos N' : - 136 :: 07' : 1', or log Cos (N + i)'- log Cos N'= (A log Cos N) x i = -136x07 = -95'2. Adding this decrement — 95 to log Cos N', we obtain, by deducting 10, log Cos (N + i)'= log Cos 6°. 8'. 42" = 9-9974974 -10 = - T-9974974. You see that this is exactly the process by which we find the logarithm of any given number. The odd seconds are our i, and are to be expressed as a fraction or decimal of a minute. To find the arc by the logarithm of its sine or cositie, Sfc. ; as to find that whose log cosine = 9*9974974, or more correctly, whose log Cos is T'9974974. We want i in the arc (2V + i)' ; knowing N from log Cos N'= log Cos 6°. 8'= 9*9975069, by table ; log Cos (N+ t)'= 9*9974974, as given, log Cos (A r + i)'- log Cos N' = - 95 ; also A log Cos N'= - 136, by the table, 1 16 FIRST MNEMONICAL LESSONS. [§ 50. therefore by the equation, log Cos (N + i)'- log Cos N'= (A log Cos N) i f , i'= ~wr = 0-698'= 42" nearly, whence 6 n . 8'. 42" is the arc required. If you digest this well in the shape following, you will have a sufficient general knowledge of the use of the tables. For more detailed information, particularly about the method of finding i when the tabular difference A log N 1} varies greatly from minute to minute, you can consult the introduction to the tables. You want to fi', fi for find. Dilogs or i. Di = Diff. and Di {loSCiVOloW} is (tab. diff.) i: [49] first take i for point i. «4- S. L14],lo= log. Neg. is 16. of frac. pro. Pow'.ten next sub num. is -dex. (A, 49) sub = below. You always want to Jind, after the first inspection, either the Dit of logs log (N+i)- log N, when N+i is given, or the decimal i, when log(iV + i) is given : and both these are found equally from the equation log (N + i) - log N= O&ular diff.) x i : (line third.) loS(Ni) = log Sum (N+i). Di(loS loN) = the Diff. (log Sam -log X). You may first take i for 0-i, point i, as in our first example. The num- bers 0-1, 5-9, &c, are best read, point 1, 5 point 9, &c. JVe^rative is the log of a fraction proper. The power of ten 7iext sub (= below) the num- ber is marked by the index of its log. It shoidd have been remarked, that, as there are no negative logarithms in the table, in order to find the num- ber answering to such, as to -3"5(J78921, it must be written under the equivalent form 4-4321079, fo r > (2), - 3-5678921 = -4321079 - 4-0000000. Note also that the numbers whose logs, are in the table are positive numbers : of the logs, of negative numbers you have some curious knowledge to acquire hereafter. § 51.] FIRST MNEMONICAL LESSONS. 117 LESSON XV. 5]. You can now comprehend and remember the solutions of innumerable questions of the greatest practi- cal importance and scientific interest. Ex. 1. A triangular field has its sides a = 1050, b = 600, and c = 500 yards : required the side of an equilateral triangle of equal area. The angle of an equilateral triangle is 1 x 180° = 60°; and if s be the sought side we must have by [32], ! z 2 Sin 60° = Js.(s-a).(s-b).(s-c), or, dividing both by | Sin 60, *-snrSo('-('-»M'-*)-C'-<0>». and, extracting square roots of equals, z = (2:Sm60°) h {s.(s-a) . (s-b) .(s-c)V*. 1:2 . (e$ = e 2 ; £45] generalized, or e* = \f J~e.) If a perpendicular is let fall from an angle on the oppo- site side of the equilateral, it bisects that side, by [18]; and [22], z Cos 60° = | z, whence Cos 60° = £, Cos 2 60 + Sin 2 60=1, [25], and Sin 60-Jl -^ = | J 3 ; wherefore * = (4:j3)Hs.(s-a).(s-b).(s-c)}l 2 = "J . ** (s - a)* (s - 6)* (* - c% by [47, d], and [48], log 2 = log 2 - \ log S + \ log s + i log (s - a) + i log (s - b) + 1 log (s - c). s = ±(a + b + c)= 1075, s - a = 25, (s-b) = 475, s - c = 575. 118 FIRST MNEMONICAL LESSONS. [§ 51. log 2 = 0-3010300 1 log 1075 = 0-7578521 1 log 25 = 0-3494850 1 log 475 = 0-6691734 1 log 575 = 0-6899169 2-7674574 -1 log 3 = -0-1192803 logs= 2-6481771 = log 444-8126, and z = 444-8126 yards, the side of the equilateral required. Ex. 2. CP is a perpendicular cliff, surmounted by a beacon P, seen across a creek from A. Wishing to know the height PC, and the distance AC, 1 measure AB = 500 yards level with C, and I take the angles PAC = 5°. 7'. 30", CAB = 45°. 48'. 7", and CBA = 94°. 2'. 9". Required PC and A C. The angle ACB = 180°- (139°. 50'. 16") = 40°. 9'. 44", by Prop. D. By [34], AC= AB . Sin 94°. 2'. 9":Sin 40°. 9'. 44", log AC = log AB + log Sin 94 . 2'. 9"- log Sin 40°. 9'. 44", or, since Sin f| + e\ = Cos (- 0) = Cos 6, [23, G], log AC = log AB + log Cos 4°. 2'. 9"- log Sin 40°. 9'. 44", log AC = 2-6989700 + 1-9989216 - 1-8095286 = 2-SS83630. .-. AC = 773-3267 yards, by the table of logs. Next CP = AC tan 5°. 7'. 30", log CP = log A C + log tan 5°. 7'. 30" = 2-8883630 + 2-9527310 = 1-8410940 = log 69-3576, .\ CP = 69-3576 yards. § 51.] FIRST MNEMONICAL LESSONS. 119 Ex. 3. A and B are two light- houses, neither visible from the other. AG and CB are two level roads making the angle ACB = 61°. 32', AG = 4 miles, BC= 8'75 miles. The distance AB is required to be found by logarithms. c 2 = a 3 + b 2 - 2ab Cos G, [26, B], gives the value of c or AB; but this is not adapted for logarithms, which are of use only in finding products and quotients. You can obtain a 2 6 2 , but not a 2 +b\ by the logarithms of a and b. A convenient formula is obtained from the preceding, thus, by [31] and [25], c 2 =a 2 +b 2 -2ab (Cos 2 \G- Sin 2 \ C) = a 2 + b 2 -2ab (2 Cos 2 ± C-l), or c 2 = a 2 +b 2 + 2ab - 4>ab Cos 2 1 C, = {a+ b) 2 - 4ab Cos 2 1 G, by [14], < a+b y{'-^vbj c ^^ cx ^ for the same reason that 8— 3 = 8{l-f}. Now I say that 4ab, a 2 -2ab + b 2 = - c, or (a -b) 2 = - c, [14], which is absurd ; for the negative quantity - c cannot be the square either of (a - b) or of (6 - a), (v. 23). Therefore 4a6:(« + b) 2 is not an improper fraction, and consequently Cos 2 | G. 4>ab:(a + b) 2 is a proper and still smaller fraction, the value of which is known. We may call it Sin 2 7, and we can find by the tables the arc y whose (sine) 2 is this fraction. We have now the equation c 2 = (a + b) 2 {1 - Sin 2 7 } = (a + b) 2 Cos 2 7 , [25] ; .♦. c = (a + b) Cos y, (a) and log c = log (a + b) + log Cos y - log 10 10 . In our problem ab = 35, a + b = 1275; i (7= 30°. 46', 120 FIRST MNEMONICAL LESSONS. [§ 51. Cos I C 2 Jab:(a + b) = Cos 30 n . 46' x 2 7^5: 12-75 = Sin y, (b) log Sin y = log Cos 30°. 46' + log 2 + A log 35 - log 1275 = 9-9341234 + '3010300 + -7720340 11-0071874- 1-1055102 = 9-90l6772 = log Sin 52°. 52'. 58", .-. 7 = 52°. 52'- 58", log Cos 7 - 10 =7-7806407 log 12-75 = 1*1055102 •8861 509 = log 7 , 69398; .-. c = AB = 7-69398 miles. This is a much readier mode of finding c, especially when a and b are large numbers, than the formula [26, BJ, and it must be remembered. Putting Si.7 and C0.7 for Sin 7 and Cos 7, say : (pron. 7 like g) [50] The Side c' is C67 Sum (db), (a) Where Si 7 mean (db)'s CHaCang. IlarM (db). (b) vid. mean (ab) [8] ; HarM (ab) v. [39] ; C#a, v. [31], Cang [18]. To determine the side c in terms of a, b, and C by logarithms, we use the equation above (a), in which y is given by S'my .\/ab= cosh C.2ab:(a + b), the equation (b) above, ■with the multiplier \/ab introduced on both sides, for mnemonical con- venience. Vid. (15). If we had wished to obtain the angles CAB and CBA in terms of abC, and not the side c, we have them by ^S'] and case 2, (39) : log tan 1{A-B) = log (a - b) - log (a + b) + log cot ^ C = log 4-75 - log 12-75 + log cot 30°. 46' = 0-6766936 - 1-1055102 + 10-2252412 = 9-7964246 = log tan 32°. 2'. 15"; 81.] FIRST MNEMONICAL LESSONS. 121 .-. ±(A-B) = 32\2'. 15", i(A+B)*=59°.W.O"; for £ (A + B + C) = 90°, (Prop. D). Therefore A = 91°. 1 6'. 15" and B = 27°. 1 1'. 45", [28]. Richard: — The right side of (b) as written last, is 'CosHaCang of HarM.legs,' or biCang [38]; so that I could make a shorter mnemonic still, by saying Where Siy mean(ab)'s biCang, instead of the second line of [50]. Uncle Pen. : — You are right : and the readiness with which you have observed this, proves both the power of these mnemonics, as instruments of rapid thought, and the attention you have paid to them. Here then you might make a little theorem, if it were worth while : the bisector of any angle of a triangle is never less than the mean proportional of the containing sides. This is evi- dent from the consideration that no sine can be greater than unity. If Sin 7 = 1, Cos 7 = 0, and the base c = 0, by (a) ; this is the case of C = 0, and gives no triangle at all. Our theorem then is : the bisector of an angle of a triangle is always less than the mean proportional of the sides. Scores of such theorems can easily be made. But it is a great and common mistake, to lose about triangles and circles that time and labour, which would be more profitably employed in mastering the magical secrets of the more advanced analysis. Ex. 4. B, A and C are three lights ashore, whose mutual position is known ; M is a sand-bank, whose exact position is to be determined by observations thereat made of the two angles, BM A = 0, and AMC = (p. Let c = BA, b = AC, A, wangle BAC, all numbers already supposed known. Let AD be the unknown diameter through A of the circle passing through B, A, and M, and AE the unknown diameter of that through A, C and M. Because by [19], AMD and EMA are both right angles, DME is a straight line to which AM is perpendicular. 122 FIRST MNEMONICAL LESSONS. [§ 51. tBDA = tBMA = Q by [19], CAE = CME=, no matter whether obtuse or acute, the whole figure is the sum of the four triangles having their common vertex at s y the intersection of the diagonals, and is = |Sin<£ (As.Ds+Js.Bs + Cs.JDs + Cs. Bs) = ±Sm

= Sin (tt — , (x 2 y 2 ), be the other two. Let x 2 and y, through q and p meet in r, and x x meet y 2 in s: join Os. The triangle Oqp = pqs + Oqs + Ops, if all the co-ordinates are positive, as in the first figure, and = pqs + Ops — Oqs, if x 2 is negative, as in the second. The triangle pqs is half the parallelogram sprq (D 34) ; the triangles Ops and Oqs are halves of the parallel- ograms RpsQ and sqmP, being on the same bases ps, qs, with them, and between the same parallels, i. e. having the same altitudes (Prop. B, 20). Therefore a Oqp = ^ {prqs + sqmP + RpsQ) in one figure, and = ^ {prqs + sqmP - RpsQ) in the other : the first is | {ORrrn - OQsP) ; the second is 5 (prmP- RpsQ), or i (ORrrn + OQsP). From (D 34) and [32], it follows that the area of a parallelogram whose sides a and b include the angle to is ab.Sinw. (E.) If then YOX = 10, we see that as Om = x l} OR=i/ 2 , OQ=y t , PO = x 2 , a Oqp = \x x y 2 Sin ia-\y l x 2 Sin o> = ^ Sin to. (x l y 2 -y l x s \ 126 FIRST MNEMONICAL LESSONS. [§ 52. in either figure; for — y,x 2 = a positive area in the second, x 2 being negative and y l positive, [3~\. \_5\~\ Sin to (xy le yx) at un two', pron zy le wix at unto. Is twice Are (Or' un two), un for unity. i.e. Sino>x (x l y 2 — y^i), xy at 1, 2, fess yx at 1, 2, is twice the area included within the points Origin, (*j»yi) and (x 2 ,y 2 ): to being the angle between the co-ordinate axes. This is true for every pair [» u «/,), The area may have either sign, as it is considered to be space from Oq to Op, in the positive direction from OX to- wards OY, or space from Op to Oq, in the opposite direction from OY towards OX; and these two spaces, if the sign is to be taken into account at all, must have opposite signs. Look now at any triangle 12 3; we see that, if Ol 2 is the a (012) &c, a 123 = 012- (013 + 032), if we consider 012, 032, and 013, to be spaces described by angular motion from OX towards OY. Then, if 1, 2, and 3, be (x i y^{x i y 3 ) ) an d 0*s#a)j *(W3)=iSmw{x l y 2 -y l x 2 -(x 3 y a -y 3 x.J-(x 1 y 3 -y l x 3 )}by[5l2,or a (123) = | Sin a. {x, . {y 2 -y 3 )+x 2 . (y 3 -y l )+x 3 . (y-y 2 ) }• This then is the area of any triangle, whose angular points are(x = x i ,y = y l ), {x = x 2 ,y=y 3 ), (x=-x 3 , y=y 3 ) ; provided that in this expression you take care to put for x,, &c. the proper numbers with the proper signs. Can you remember this ? Richard: — Why it is exactly ) +(- r 23/3-#2* 3 ) +(* 3 y* -yax 4 ) + (xyy- t/ 4 .*i)} Sinw, and this certainly true, whatever be the points x, ?/, &c, or the co-ordinate axes employed, if in this expression the proper values and signs be put for the co-ordinates. The same thing is true of the area of any polygon 123456... n made by joining n points in thai order, Area of polygon = 012 + 023 + 034 + ...+ 0(n -l)n + Onl, where 0(n-\)n = ^ Sin w(x n -iy„-y n -iX„), in which for x„ &c. 128 FIRST MNEMONICAL LESSONS. [§ 53. are to put then* proper values with their proper signs. All this is true, even of polygons whose sides drawn in that order cross each other (not produced) ; but you had better not trouble yourself with such figures at present. or x y az + aE > v. (9), be the equation of any line PQ, and A,x — B,y = c>, x y ~ *> that of a line P,Q U the two lines subtending supplemental angles at 0, and let the co-ordinate angle POQ = w. It is required to find from these equations the length (disregard- ing the signs) of the perpendiculars OL and OL, on these lines from the origin. By Prop. (B 20) and [32 J, we know that OL . QP = Sin a. . OP . OQ = 2 (Area OPQ), OL, . Q l P l = Sin (tt - w) OP, . OQ l = 2 (Area OP,Q,) ; whence by division of equals, OP. OQ OL = Sin to QP and OL, = Sin (tt - a,) - P * _° Ql ; i. e., [26 B], nT e . OP. OQ OL = Sin « . r ; ± (OP* +OQ 2 -20P. OQ Cos . »)* and n7 a ., , OP,.OQ, \)L, = om(7r — i») -, k ' * {OP, 3 + OQ? - 20P, . OQ, Cos (tt - a,)}:* Observe, that in [2(i B] a and c represent two numbers of the same sign, and consequently OP, and OQ, must be considered as of the same sign; to find Q,P„ we have no account to take of their signs, but of their lengths only, which are by [5'J C,:B, and C,:A,. Then § 53.] FIRST MNEMONICAL LESSONS. 129 OL- G . C C Sma 'B'A (C 2 C 2 Q c c n \*' and OL x - Sin (tt - ») . — . — ^/C, 2 ^ C 2 2C, c, r , «>} J The numerator and denominator of this latter fraction being L i. ,.,.,, B t A t . . /B 2 A 2 \t both multiplied by ~ or its equal l— -jy*~ ) ' we ° tain, [23], ± C, Sin , , 0L= *" C^ + JS, 2 )* (A 2 + 5 2 )** The denominators of OL and OL, are exactly of the same form, if you take into account that B i is negative, while B is positive : so that if ax + by = c be any line, re- ferred to these axes, the perpendicular on it from the origin = ± c Sin w:(a 2 + b 2 -2ab Cos aft. For example, the perpendicular on y + 3x — 4 is * 4 Sin ft.:(l 2 + 9 - 6 Cos «)*, and the perpendicular on y — 3x = 4 is ±4Sinft):(l 2 + 9 + 6CoSft))i F5 130 FIRST MNEMONICAL LESSONS. [§ 53. We can in general take our choice of signs, as in ± JH ; and may consider the perpendicular positive, so long as we contradict no previous supposition thereby. Let us for shortness call the perpendicular on any line from the origin the Lor of that line, distance of Line from Origin. And let us call the expression (a 2 + b 2 — Slab Cos w)-', bas(abw), (pron. abu;): it represents, when a and b are numbers of the same sign, the base of the triangle [[26 B]] whose sides are a and b containing the angle to. If a and b in bas^abw) are numbers of unlike signs, this geometrical ex- planation fails ; but, from what precedes, you can see that (a 2 + b 2 + 2ab Cos w)' is the base of the triangle whose sides are + a and + b, con- taining the angle (tt - w) : in fact it is {a 2 + b 2 - 2ab Cos (tt - »)}i. Let us call A and B, in Ax + By - C = 0, the fics, i. e. the coefficient? of x and y : then bas(fics a>) will stand for the expression (A 2 + B 2 - 2AB Cos 0)2, whether the Jics have like or unlike signs, i.e. whether -2AB represents a negative or positive number. You may say []523 Lor's Con. Sin « by b;is(fics w) : Is Con. cipoth fics, if right is «. ci = cip, v. [44], poth vid. [12.J i.e. the Lor = (the constant term of the equation) x sina>, divided by has (fics od) explained above ; and = when u> is !»0° Constant x the reci- procal of the poth (21) of the coef/i'cients. You will find the value "of these abbreviations as we proceed. The two lines y + 3x = ± 4, have Lors of the same length, and the pair y — 3x = ± 4, have equal Lors also. If the equation of the first line were written V + -' T -is 4 4:3 the Lor would be Sin«,:^ + ^-^Co8») , or is^jg, when— 90. You see readily that this is the same value as before, p. 12.0. § 54.] FIRST MNEMONICAL LESSONS. 131 54. Let ax + by ± c be the equation of any line referred to any axes OX OY, and let I be the Lor of this line, or perpendicular on it from the origin. It is evident that max + mby = mc is the same line still ; for every value of x and y that satisfies the first will satisfy the second equa- tion. Let m = he, then mc = I ; and max + mby = I, or a x x + b^y = I, (if ma = c, &c), is the equation to the same line. What is the geometrical meaning' of «, and b Y considering their values only and not their signs ? Suppose PQ to be the line ; when x = 0, we have b l y = l, or b^hy^OLiOP, i. e. 6, - Sin OPQ ; and when y = 0, we have a lX = l; a, = l-.x = L-.OQ = Sin OQP. The equation then is SinOQP. * + Sin 0PQ.y = l; 1. whence it appears that if the absolute term of the equation to a line is the Lor or X from the origin, the coefficients of x and y, considering their numerical values apart from their signs, are the sines of the angles which the line makes with OX and with OY. And you may easily prove also, by writing the equation in the form x y . mrOQP + /:Sin OPQ =1 ' as m L9 ^' that if the coefficients are numerically the sines of these angles, its absolute term is exactly the perpendicular from the origin. Let then ax + by -1=0 be a line PQ whose Lor is /. We know at once what the numbers a and b are. The equation is not true for a point (x^i) not in the line; ax x + by t — l is not = 0, but ax x +by l — l = \, or ax x + by l — l-\ = 0, is the truth, X being some number positive or negative. Yet the equation ax + by -I- \ = represents some line pq of which (.tj yi) is a point, because its equation is true at 132 FIRST MNEMONICAL LESSONS. [§ 54. (x, yd ; and by f_5] we know that this line is parallel to ax + by - I = 0, both lines being parallel to ax + by = 0. Hence both lines make the same angles with OX and OY, and the coefficients a and b of the variables in each line are as just pointed out, numerically the sines of these angles • therefore ax + by = I + \ is a line whose Lor is l + \: i.e. 0L t = 0L+\, and \=0L-0L = LL i . Thus we find that ax, + by i — 1 = \ = LL lt and this is exactly the perpendicular distance between the lines PQ and pq : in other words, ax x + by x - I = the perpen- dicular from (x x #,) [2~] upon the line PQ, which has for its equation ax + by - Z = 0. Thus we have proved that if in the equation L. ax + by - 1 = = u, u standing for (ax + by- 1), I is the perpendicular from the origin, a and b being the sines of the angle between the line and the axes OX and OY, the values of u for different co-ordinates (x x y t ) (.r 2 y 3 ) &c, which values we may call u,, u 2 , &c, u, being (ax, + by x - V) &c, are exactly the lengths of the perpen- diculars, from those points respectively, on the line u=0, This affirmation u = is true of all points in the line u = 0, i.e. the perpendiculars from any point of this line upon this line =0. If a line be given by any equation y + cx-b = = u, in which b is or is not the Lor or perpendicular from the origin, we obtain u = of the form L, by multiplying u by lib, I being the Lor; so that (l:b)u = u, as at the beginning of this article, when m=l:c [_53^\ vals'. vi LorCo'n is per lin. upon. i.e. if m = = U{Sev) be the equation (a#+ Jy-0 = 0) of any given line ; val », the value of that »{oei>) (made by putting any (*,jr,) to V h ^ c ?'.°" ..T nates,) x the quotient L( the line from the point term. t Lor:Con, (t»i LorCon),is the perpendicular up oint (J'iyi). Con means the constant or absolu § 54.] FIRST MNEMONICAL LESSONS. 133 Jane : — What is this ahev? Richard: — It is Greek for nothing; you are to pro- nounce owden ; » is ou. Uncle Pen. : — Find now the distance of the point (x x -2, y x = 7) from the line 3y - 5x + 4 = = u, referred to axes inclined at the angle w = 30°. Richard: — By \J>3~\ the perpendicular is (3;/, - 5x x + 4) . (/:4), if / be the Lor ; and by [52], I Sin 30 4 ~ ± Js 2 +5 2 + 30 Cos 30° ' What now is Sin 30? Jane: — Half the angle of an equilateral is 30°; and besides, Sin 30 must be Cos 60 by £23, G] • and this we have seen (51, Ex. 1 ) to be \. Hence by [25], Cos 30 =+ 7^ . The expression for the perpendicular is (3y x - 5x x + 4) (21-10 + 4). (9 + 25 + 15^3)' 1 2 sj '34 + 15 J3 7'5 ±734 + 1573' Uncle Pen. : — Find now the distance of the point (x x = - 3, y x = 5) from the line 8x + 7y = 0, referred to right axes. Richard: — This will be simpler, I fancy: but there is no Con ! and of course no Lor, the line passing through the origin. Then vi. Lor Con is - . I give it up. Uncle Pen. : — Proceed boldly to find Lor by [52]. Richard : — Am I then to say 0-- ° .nd»— ' > ±78+7* ° ±78'+7 2 134 FIRST MNEMONICAL LESSONS. [§ 54. Uncle Pen. : — You can safely say, putting I and c for Lor and Con, Z = — , and of any line Sx+'Jy = c; and this is true for all values posi- tive and negative of c, each different c giving of course a different line and Lor. It is true also when c = ; for we know that the value of I vanishes when c = 0, i. e. when the line passes through the origin. Here then is a case of - , arising from the unlimited diminution of both numerator and denominator of a fraction, the value of the fraction con- tinuing still unchanged. The perpendicular required is certainly 8x-3 + 7*5 _ II and this you will find to be the Lor of the line, 8x + 7y = -8x3 + 7x5, which passes through (-3, 5), and is parallel to the given line through the origin. This Lor and the perpendicular are equal, being opposite sides of a rectangle, of which the two parallel lines form the other sides. Generally the perpendicular from Xtf-L on ax + by — c = is t» / ? \ Sin m P. (ax y + by x - c) ■ j . \ a > + b s -2ab Cos »)* The abbreviation bas is founded on c sq.b &c.,' £2(T], and always has the same import, as to signs of the alge- braical symbols, whatever shape it may assume after sub- stitution of arithmetical values. The following observation will presently be useful. If we have the given lines u = = ax + by + c, v = = a,x + b x y + c„ rv = = a a x + b. 2 y + c 2 , z = = a 3 x, &c, and /, m, n, be known numbers; then u + Iv + mw + nz = 0, is also a given line; for it is (a + la x + ma a + na 3 ) x + (b + lb t + mb a + nb 3 ) y + (c + Ic l + mc 3 + tiCj) = 0. 55.] FIRST MNEMONICAL LESSONS. 135 LESSON XVII. 55. The geometrical meaning of e in y - h = ex, when y has the coefficient unity, and is on the opposite side of the equation from ex, is very im- portant. Take [5] the paral- lel line y = ex, or Op. If pq be the ordinate at p, pq:Oq = e, or by [34f], -^7 x SmpO i=e = Sin^ , altel ,i ns . = ; D] . Let the angle between any line Ax + By- C = 0, and OX, on the ?/>er side towards the right, be called ft ; then is YOp = oo-ft, pOq being the /3 of Op: Sin/3 _ _ Sin pOq ( a ) Sin (co -/3) - e ~Sm(YOX-pOq)> or by [27], Sin ft = e (Sin a, Cos ft - Cos o> Sin ft), whence, transposing, Sin ft + e Cos w Sin ft = e Sin w Cos /3, and by division by cos/3, [22.] tan ft + e Cos w tan /3 = e Sin &> ; then dividing by 1 + e cos u, 0>) t»/3- T ^ j ^:tlu.l.,bjrpsl (b') tan ft = e, if w = £ tt, the case of right axes. All lines parallel to y = ex have the same /3, as ft = pOq = PQO = EQ l X; and Sin/3 and Sin(w -ft) are the same for all, consequently e has the same meaning and value, Sin/3:Sin(w -ft), in them all. You may pronounce ft like b, and say, 136 FIRST MNEMONICAL LESSONS. [§ 55. [54] (a) Si/3 to SiD(o>/3), or ta/3 in RAx, pron. D -/3), which is /an/3 in i?ight ^4.res, is e, in y = ex : /3 being always upper angle on the right between the Line and the ^4xis of x. For all axes, ta/3 = e sinw by 5um (wwity and ecos u>). It is very easy now to Jind the angle hetween any two lines y = ex + b and y = e x x + b l3 the axes and their angle w being given. Let them be PQ and PQ^ meeting each other in P in the positive angle, and OX in Q and Q, ; draw the figure, and suppose PQ t X>PQX. Then by Prop. D, PQ,X = QPQ l + PQX, PQ 1 X-PQX = ft 1 -ft=QPQ l the angle between the lines. By M ■■W-fl. ^ , or by [54, bj e Sin o> / \ . r» r>r» 1 + e, Cos w 1 + e Cos w , m (c) tan QPQ l = i-— - = tan (/3, - ft). v ' e. Sin o> e Sin to v ' 1 h ! , . 1 + e, Cos w 1 + e Cos w When to = 5 ttJ this is simply, [25], (d) tanQPQ^-^i; = tan (/?, - ft). You may effect for yourselves the reduction of (c) to (c') tan QPQt - (?.-«) Sin* = ^ ' 1 + (e + e t ) Cos w + ee { w ' ™ This gives the angle between two given lines for any axes right or oblique : the distance between two points you know for the case of right axes by [12]. Try for your- selves to prove that when w is oblique, the distance r between the point (/, a) and (x, y) is always r = {(x - iy + (a-yy-2. (x - I) . (a - y) Cos »}*, e. or = the base of the a whose sides, when both of one sign, are (x - I) and (a - y), and the vertical angle w, not (tt - w). Call the angle {ft,- ft), LiL Difts, Diff of ft's, angle between Line and Line, and say : § 56.] FIRST MNEMONICAL LESSONS. 137 [55~] (e) bas[Di(xl) w D(ay)] joins xy to la, D = Dift'. xy a dissyl. (c) LiL Di/3's you know by 'tan(a mol &>'), £36]. 5 = 90. The second line, together with Prop. D [36] and [54 b] will always put you in posses- sion of tan (/3i -/3). Cos to changes sign with (x-l).(a - y). Draw this. 56. If y = ex + b, and y — e x x + 6, , are lines at right angles to each other, /3—/3 x = ^tt, or else j3 x — /3 = g t, whence either /?i = 5*-+/3, or Pi=fi— 5T. In the first case, , „ , . Sin (.8 + h tt) n ^.„ Cos (- /3) Cos ft - 1 • - in the other case, r> , n . > — Sin (A 7T — /3) 1 . n tanA = tanG3-|,) = - Cos( ^_^ = - tan> = -cot/3 ; so that, in either case, tan /3, = - cot /3, and cot /3, = - tan /?, (a) • • 1: r^ i-T e Sin w 1 + e, Cos w n N giving, by [54, b], j - - ^ - = - - g> ^ - , (b) or, mul. both sides by (l + e Cos w) (e, Sin o>), ce, Sin 2 w — — 1 — e Cos w — e, Cos w — ee x Cos 2 w, whence by transposition and [25], e, (e + Cos &i) = - (1 + e Cos «), and then by division, _ 1 + e Cos 10 . . * 1= ~ e + Cosft) ; {C) which becomes e x = , for right co-ordinates. To remember (a) (b) (c), say, thinking of Q54, b], E 5 (T] gil' s ]e cota./3 is perlin's ta/3. gil is given line ; a given line's negative cota /3 (le — - or ncg.) is the tan (3 of the /i/;e perpendicular to it. Thus fi x and e x are found by /3 and e. Let now the sides OB = «, and OA = 6, of any triangle, be taken for the positive axes of y and x. If BP and AR 138 FIRST MNEMONICAL LESSONS. [§ 57. be perpendiculars from B and A, on OA and OB, and t BOA = », = 1, is BP, [5'] [22]; a Cos to and ,~ % + 7=1, is ^K. b Los w o When x and y are the same in both these, by subtraction 1 1 \ /I 1\ « • — T = 0, is true, or, b Cos w/ ' \a Cos o> 6/ multiplying both sides of this equation by ah Cos w, the line y . (b Cos w — a) + x (b — a Cos w) = 0, (p) contains the intersection of AR and BP. The equation to BA is [5'J y x , a - + r = 1, or « = - T x + a : the perpendicular on this line from is of the form u = e,x, # = - _~ a ;r+ coTftT ^ by ^ and &** b ^ ( ebein s ~ a '- h ) which you can easily prove to be none other than the line (p), through the intersection of AR and BP. We have proved that the three perpendiculars from the angles of any triangle let fall on the opposite sides meet in a point. 57. Jane: — You have shewn us how to find 'perc bic, and biCang' in any given triangle, and the equations of these lines can be formed by [9], if we have two points known in each. Suppose that a triangle is given, can you find from its three angular points the equations of these three lines ? We can find the sides and angles. Uncle Pen. : — Let us try what we can make of the problem: the co-ordinates of any three points being given. determining a triangle, to find thereby the equations to the perpendicular from C on c, the bisector of c from C, and the bisector of the angle C in the triangle. We can write down the equations of the sides, having two points in each, by [9] whatever be the axes and origin chosen: we can find by [52] the Lor of each line, and reduce the equations of the § 57.] FIRST MNEMONICAL LESSONS. 139 three sides to a = 0, b = 0, c = 0, of the form (54, L). What meaning do you see, Jane, in this symbol a ? Jane : — I see three constant numbers, two of them sines, and one of them a Lor : I see my friends x and y amusing themselves as usual: so long as a = is true, they drag each other along the side a of the triangle ; and when a not = 0, a is always the perpendicular on that side a from the point (xy) exhibited. Uncle Pen. : — If I now say, without taking the trouble to form these equations actually, let a — cph = = 7 be the equation of O, (Art. 30), the bisector of the angle C, what do you gather, Richard, from this assertion, a — assignable, as of lines through G, i. e. an infinite number, and thus we shall have the equation of the line Ge before us, if we can determine the proper is found to be unity, and a — b = = 7 is the equation of the line Ce, being true at two points of it, C and e. 140 FIRST MNEMONICAL LESSONS. [§ 58. We seek next the equation to Cd the perpendicular on c, and it is evident that a— 0,b = 0=P C is this equation/ if

), meet in a point. By the same argument P h = when P c = P a = 0, and the three perpendi- eulars on the sides meet in a point, as already proved above. Also (^ = 0, if both Q b and Q a = 0, or the three bisectors of § 58.] FIRST MNEMONICAL LESSONS. 141 the sides meet in a point. In this article and (43) you have proof of this proposition : In any triangle, the three perpendiculars at the centres of the sides meet in a point, as do also the three bisectors of the angles, the three bisectors of the sides from the opposite angles, and the three perpendiculars on the sides from the angles. Let BPB', APA', CPC be lines drawn from the angles of a triangle ABC through any point P without or within the triangle to meet the sides in A'B'C. Draw two figures. By [34T\ we have in the triangle AC'P, &,c, AB'= SinAPB'. PA.S'm AB'B BC'= Sin BPC. PB:SmBC'C CA'= Sin CPA'PCiSln CA'A ; AC'=SmAPC'.PA:SmAC'C BA'=S'mBPA'.PB : SmBA'A CB'= SmCPB'.PC.SinCB'B whence it follows that A B' . BC .CA' = AC. CB' . BA', if you put for Sin APC its equal Sin CPA', for Sin 4 CC its equal [23 L~] Sin BC'C, and so on. Further we have [34], Sin A CC = AC. Sin A: CC Sin BA'A = BA'. Sin B-.AA' SinCB' B=CB'. Sin C:BB' Sin ABB' = AB' . Sin A-.BB' Sin BCC = BC. Sin B: CC Sin C A A' =C 'A'. Sin C.A A' - whence it follows, of the six left members, Sin ABB'. SinBCC. SmCAA'=SmACC SinCBB'. SinBAA'. To see this, it is merely necessary to perform the equivalent multiplications of the right members of the equations: and it is a beautiful and easy deduction from [34] and [23 L~]. The reasoning and results are unaltered in their truth if you put, for P A'B'C, P^^Bfi,. This is the general pro- perty of three lines drawn from any point P through the angles of a triangle ; and is worth remembering, [592 P---Ang cuts a' in dot A', &c, AB'.BC.CA"s AC.CB'.BA'; dot alterna' pron. abblcca, ackibba. Then for lines put oppo. sines : i.e. If the line through P and A cuts a in doited A', through P and B cuts b in B', &c. ; you dot ihe alternate letters in the assertion AB'. BC. CA' = AC. CB'.BA' ; then you may put for the 6 lines AB', &c, the sines of the 6 angles at ABC opposite to them ; and it is still true. 142 FIRST MNEMOXICAL LESSONS. [§ 59. If through any point P lines be drawn through the angles ABC of a triangle to meet the opposite sides in A'B'C, then of the six segments AB', B'C, CA', A'B, BC, CA, the product of the first, third and fifth is equal to that of the second, fourth and sixth, and the like holds of the sines of the six angles which stand over the segments at the vertices of the triangle. What is proved above concerning a point and a triangle can be readily established by a similar argument of a point and any polygon of an odd number of sides. 59. It frequently happens that a line is required of which we know only expressly one point, as when we were looking for the equations to ' perc. bic. and biCang,' in Art. (56). Suppose that we were in search of the equation of the line which passes through the given point (x = e, y = i), and through the intersection of two given lines ax + by — c = = v, and a x x + b x y — c x = = u. We can write down what we know about the first point thus: let x - e = (p(y-i) be the line required : for this is one of the innumerable lines passing through (e, /), the equation being true at that point, whatever may be. What value of

Gr c,6 — co, c,a — ca x a A b - ab l b t a — 6a, dividing both sides of the equation by (ajj-ab,) *-c y-i (c,6 - cb t ) - e (a x b - ab,) - c t a + ca x - i (a t b - ab t ) ' § 59.] FIRST MNEMONICAL LESSONS. 143 This line we can now draw by the method of Art. (9), even though the point (X, Y) should be ten thousand miles off. We might have begun the investigation thus : let ax + by - c = (« ,.r + b x y — c^, or v =

is unknown. By subtraction of equals from equals, we get rid of this unknown quantity, and obtain y-Y=e(x-X), the equation required. And you can easily prove from this equation that it gives a line parallel to y = ex, and con- taining the point (X, F). Let soughl (pron. saul) stand for sought line ; and say (pron. . {y - i) is a sought /ine through the point (e, i) ; 0-v - (pu is a sought line through the point (v, v) the intersection of v = and u = 0, y- (f> = ex is a sought line parallel to y = ex. 144 FIRST MNEMONICAL LESSONS. [§ 60. LESSON XVIII. 60. Jane: — You have not yet shewn us how to write down the equation of the circle through three given points i x iy\\ (^V/s)' an d (#3^3)- I am somewhat curious to see how it will look, and I expect it to be no less entertaining when understood than the equation of a line through two points, so easily acquired in \jf\. Uncle Pen. : — You will learn this most agreeably when I have given you a lesson on permutations and combinations; but first we must talk a little about progressions. Do you remember what you have met with on this latter subject in your arithmetic ? Jane : — Indeed I do not ; I never understood them. Richard : — And I never tried, for I do not love the look of them ; do give us a trial with some mnemonical aids : I will be very attentive. Uncle Pen. : — Well then, here is an arithmetical series, a, a + d, a + 2d, a + 3d,...a + (n-S)d, a+(n-2)d, « + («-!>/; the number of its terms is n, and each is made from the next preceding term by the addition of a certain number d, called the common difference. The sum of the first and last term is 2a + (w — 1) . d, that of the second and last but one is 2a + d + (n - 2)d = 2a + (w - ] ) . d, that of the third and last but two is also = 2a + (n — l)d, and so on. Let us sup- pose then n is an even number, say n = 2.6 = 12 : then will there be 6 pairs of terms, first and last, second and last but one, &c. each pair = 2a + (n — 1) .d, or if Ave call the last term s, each pair = a + 2, the sum of the first and last. The sum of all the terms is 6 . (a + z) = - (a + z). Next suppose II odd, say n = 2.6 + 1 = 13. In this case there will be a middle term; call it M. It stands between M-d and M+d, which make one of the 6 pairs each equal to a + z. M is half the sum of this pair, or = ^(a + z). The whole series consists of 6 equal pairs + M ; i. e. 6 . (« + ~) + 1 (a + =) = ^ (a + z) = I (a + z). § 60.] FIRST MNEMONICAL LESSONS. 145 Thus whether n be 12 or 13, if S stand for the sum of the series, we have S = \ n . (a + z). If now you suppose n to be either equal 2m, any even number, or 2m + 1, any odd one, and put m for 6 in all the preceding reasoning, you have a demonstration of the theorem, (a) #=!(« + *); and (b) s = a + (n-l).d, is evident of itself, and may be written a - z = d - dn, by transposition. [60] If abc... forth is Ari.Se : Ult and a is penult, and b ; Sum Ari.'s half n.Sum(az), and (d le dn) is Dif.(az). dn a dissyl. i.e. if a, b, c, , and so forth, be the terms of an arithmetic series, the (ultimate term +o) = the (penultimate +b), and so on. The sum of the arithmetic series is always in times the sum (a + z), and (d-dn) = the Diff. (a-z.) By the equations (a) and (b) we can determine any two of the five quantities S, d, a, z, n, if the other three are given. This gives rise to ten different problems, for the two unknowns may be any of the ten pairs, S and d, S and a, S and z, S and n, d and a, d and z, d and «, a and z, a and n, z and n. If S and n are unknown, and a z and d given, we obtain first n from (b) by transposition and division, n = — - ; then putting this for n in (a), o d-a+z What is the number of terms, and the sum of the arith- metic progression whose first term is 1, last term 10, and common difference l£? These formulae give 1-1 + 10 3 + 18 i « r / n = - = — — -7, and S=%(l + I0)-S8|: the series is 1, 2-5, 4, 5-5, 7, 8*5, 10, whose sum is 38±. What is the number of terms and sum of the series whose first term is 10, last term 1, and common difference - 1^? G 146 FIRST MNEMONICAL LESSONS. [§ GO. The answer is as before n = 7, S = 38% ; for neither n nor 5 is altered in value, if you exchange z and a, and make d negative. The series is the former one read backwards. This is the solution of the fourth of the 10 problems above mentioned : and you will find it a profitable exercise to prove that the solutions of the other nine are in order from the first, as follows: 1. 2S=n(a+z), d = (a-z):(l-n). 2. 2S=n.{2z-d(n-l)}, a = z-d.(n-\). 3. 2S=n.{2a+d(n-l)}, z = a+d.(n-l). 5. d = (2/S-2nzy.(n-n*), a = (2S-nz):n. 6. d=(2an-2S):(n-?i 2 ), z = (2S-na)-.n. 7- d=(a 2 -2 2 ):0 + z-2S), n = 2S : (a+z). 8. a~S:n + ±(d-dn), [28] z=S:?i-±(d-dn). 9- n = [2z+d±{(2z + d)*-8Sd}*l :2d, a = z+d.(l-n). 10. n =rd-2a±{(d-2a) a +8Sd}H 2d, s=a-d.(l-n\ The reason why the ninth and tenth problems require the solution of a quadratic equation is, that (a) with (b) is of the second degree in the unknown quantities, containing their product, an and zn. If (a) had happened to contain both n and d, the seventh would have been also a quadratic problem, because the product dn occurs in (b). In the ninth, n must be of course an integer ; therefore the data z d and S must be such that {(2z + df - 8Sd\ shall be a square number M 2 , and that 2d shall divide without remainder either the sum or the difference of 2z + d and M, or else both. When it so divides both, and both quotients are positive, there are two values of n, and consequently two of «, found by putting for n its values successively in the equation a = z + d(l - n). What is the first term and the number of terms of the series whose sum is 67, its last term 18, and its common difference 2^? Here (2z +dy~ 8Sd= (36+2-25)*- 8x67 x 2-25 = 257'0625, which is not any square M 3 , and there is consequently no number « assignable ; that is, no such arithmetic series ex- §61.] FIRST MNEMONICAL LESSONS. 147 ists. But if we put 67^ for S in the question, the data are found to be congruous : we have (15"75) 2 = M 2 , and 36+2-25± N /(38-25) 2 ~ 8x67-5x2-25 54 22"5 no m = ^~ v '- = — or — — - = 1 2 or 5. 2x2-25 4'5 4-5 Thus there are two values of n ; and a = 18 + 2-25 x (l-12) = -6f, or a = 18 + 2-25 .(l -b) = 9. There are in fact two series having 2 = 18, d = 2'25, and £ = 67-5, viz. 9, Hi, 131, 15f, 18 ; and -6f, -4|, -2J, 0, 2$, 4|, 6£, 9, Hi 13^, 15f, 18; the first of 5, the second of 12 terms. If the three numbers - , — , T , are an arithmetical pro- a m b r gression, they must be of the form d, — , — + d, for s ' J m m m some value positive or negative of d. The middle term is evidently half the sum of the two extreme terms, i. e. m~ 2 \a + b)~ z '\ab + ab) ~ 2ab ' You have learned already by definition (42) that 2ab-.(b+a) is the harmonic mean between a and b; but this mean is m; hence if -, — , -r form an arithmetic progression, a, m, 6, form an harmonic progression. The reciprocals of the arith- metical series taken in the same order form an harmonic series; and generally, if abed... to n terms, are a progression of the former, - j — -.... the reciprocals of the same n terms, are said to be a progression of the latter kind. Thus §, 2f , 5, are an arithmetic, and 2, ~ v \, are an harmonic pro- gression. You may add to the preceding mnemonic this line [Q0 r ] cips Ari.Se. are Harmo.Se. cips for reciprocals of. 61. A geometrical progression of t terms is seen in the following series: a, ae, ae 2 , ae 3 ae' -3 , aV~ 2 , tf-e' -1 , in which e is the ratio, and every term after the first is e G2 148 FIRST MNEMONICAL LESSONS. [§ 01. times the preceding one, a and e being any positive or negative numbers. If we put z for the last term, and S for the sum of the / terms, we have S-a + ae + ae 2 + ae 3 + + ae'~ 3 + ae'- 2 + z, and eS = ae + ae 2 + ae 3 4- + ae'~ 3 + ae*' 2 + z + ze, whence S-eS = a-ze, and S = (a - ze):(l - e). We have here, as in the last Article, five quantities, S, a, z, t, e, viz. the sum of the series, its first and last terms, the number of terms and the ratio, and two equations about them: (G) z = ae'~\ S=y^. Ten problems may be proposed, by considering in turn every pair of these five quantities as unknowns, viz. the pairs zS, eS, aS, az; ae, ze; ta, tz, te, IS. The first is already solved; the second gives, [47, d], e = (sw) 53 , 8 = 0'^ - (Fy.ljF - a'- T ). The third and fourth are easy. The fifth and sixth you may attempt, but you will arrive at an equation beyond your present power to solve. The remaining four are readily managed by the application of [4S b~\. Thus, when t and e are the unknowns, S-a (S-a\- 1 S-a {S-a\> eas 5W Z = a \s^z) > S=-z* = a \S=-*)> whence \o g (S-a)-)og(S-z)+log2 = \oga + tAog(S-a)--tlog(S-z) t from which t is readily found after transposition and division. When e is a positive proper fraction and / is a very great number, e' _1 is very small ; if e = *l and t— 1 = 100000, z = ae'~ i = a millionth part of a, and s diminishes, as /, the number of terms, increases. When I is infinite, z = 0, and ze = ; i.e. S = = a + ae + ae 3 + ac 3 + ..., which differs from it only in the mutual position of 4 and 5, and thus you will have 60" pairs each alike, save as to 4 and 5. If now you put 5 for 4 throughout, your six elements are 333556, and the number of different per- . , ,„ , . . • n 1.2.3.4.5.6 mutations is half what it was : it is exactly — — — - — - — — . By repeating this argument for any value of p, you prove: // p elements contain m a's, e b's, i c's, the different per- mutations of those p elements are in number 1.2-3 Q-2).(p-l).p 1.2. 3...(w* - 1) . m . 1 . 2 . 8...(e -1). e. 1 . 2 . 3...(i -l).t" The permutations of aaaaabbbbcccdddcf amount to 1.2. 3. 4. 5. 6.7. 8. f).10.1 1.12. 13. 14. 15. 16. 17 1.2.3.4.5.1.2.3.4.1.2.3.1.2.3 Let fags stand for /actor digit*: then 1.2.3 is :; faff?, § 63.~\ FIRST MNEMONICAL LESSONS. 151 1.2.3.4.5 is 5 fags, &c. : and you may say the above proposition thus : [62] If p elems. have m a's, e b's, i c's, The perms, in p's are p fags by (m fags, e fags. i fags). 63. There are many permutations of the same combination. Thus abed, aedb are the same combination, but different per- mutations of it. Suppose six symbols, 123456; with each of them you can combine in turn every other, as 12, 13, 14, 15, 16; 23, 24, 25, 26, 21 ; 34, 35, 36, 31, 32 ; and thus can be completed six fives; but every pair is twice written, as 12 and 21, so that the exact number of combinations two together that can be made out of 6 elements is ^ . 6 . 5. In the same way you can make with n elements — — duads, or combinations of two. Again, each of the 15 duads made with 6 symbols can be combined with the remaining 4, as from 12 are made the four, 123, 124, 125, 126, and thus you can complete 15 such fours; but 123 will be thrice written, being made once from 12, once from 23, and once from 31. The correct number of combinations of threes made out of 6 is one third of those 6-5.4 15 fours, or — : — ^ -. In the same way can be proved that with n elements can be made triplets in number «.(»-!).(» -2) 1.2.3 With every different triplet can be combined each of the remaining n — 3 symbols, thus forming n.(n-l).(n-2) , quadruplets or combinations of four ; but each of these will be made four times over, each time by adding a different fourth single symbol ; and this consideration reduces the number of different 4-plets to ».(»-!). (n- 2). (»- 3) 1.2.3.4 152 FIRST MNEMONICAL LESSONS. [§ 64. the fourth part of the preceding. By carrying on this mode of reasoning, it is easy to prove that The number of non-repeating combinations of d together, that can be formed with n symbols, is n . (n - 1) . (re - 2) ...{»- (d-l)} l.2.3...(d-l).d We may call the quantity 9.8.7, three nine-backs, and 9 '8. 7 -6. 5, five nine-backs; re. (re — 1) . (re — 2) is three re-backs, and the numerator just written is d re-backs. By non-repeating combinations are meant such as contain no repeated letter : abcde is a non-repeating, and aabcd is a repeating combination. You may add to your stock of mnemonics the following: (com. for combinations, repea. for repeating). \_GS~\ Comb, non-repea.'s of re in d's, Are d re-backs by d fags. fags vid. [62J. The number of repealing combinations qfn things taken d together, is (the proof follows below) re . (re + 1) . (re + 2) ...{» + (d-l)} 1.2.3...(d-l).d And you may join this mnemonic to the preceding : (repe. combs, for repeating combinations). [64] The repe. combs, of re in d's-, Are d w-iips by d fags. Three 5-ups is 5-67; 4 n-ups = n •(n + \)'(n + 2) -(n + 3). Among the repeating combinations of 7 in 4's are counted 2222, 2776, 4441, &c, as well as all the non- repeating combinations of 7 in 4's. The proof of this propo- sition [64] is not so easy as that of the preceding, and you will perhaps have to meditate it somewhat longer than those. Now, I say, that the theorem is true for any values N and 1), of re and d, if it be true for all values of n and d both less than D. We are seeking the number of repeating D-plets that can be made with N symbols: and our hypothesis shall be that for values of re and d both less than D, [64] is true. 64. m and c Let D be divided into any two positive numbers, 7, so that 7/1 not > A 7 , and suppose all the non-repeat- § 64.] FIRST MNEMONICAL LESSONS. 153 ing D-plets to be written out that are possible with (N + D - 1) things. Their number, by [63}, is D (N+D-l )-backs by D fags, or (N+D-l)(N + D-2)...{N + D-(D-l)}(N+D-D) l.S.3...(D-.l).D which is D iV-ups by D fags also, the same fraction with that over [_6±]-> w hen n = N and (/ = />. We may suppose our N + D — 1 symbols to be letters in alphabetical order. Let A represent any D-plet of those just written out, which contains m letters and no more of the first N: A will con- tain e of the remaining D — 1 letters, because D = m + e; and the same m letters of A will form part of D-plets A, A ls A 2 , A 3 , &c. in number equal to the non-repeating e-plets that can be made of those D - 1 remaining letters. That is, this r«-plet, which we may call a, will appear (D-l)(D-2)(D-3)...{D-l-(e-2)}{D-l-(e~l)} 1.2.3...(e-2).(e-l).e times, by [63~], or, since m = D-e, m + e-l = D-l, m.(m + l).(m + 2)...{m + (e-2)}{m + (e-l)} fi 1.2.8...(e-l).e * which is e ?«-ups by e fags, the exact number of repeating combinations of m symbols e together, as, by hypothesis, we know beforehand ; m and e being both less than D, whatever numbers they may be. If we now erase all the letters, that lie alphabetically beyond the first 2V, from these D-plets, A, A 1} A 2 &c, we can replace the erased e-plets by the repeating combinations e together of the m letters in a, and we shall have in place of J., A u &c, all the Z)-plets that can be made by adding to a e repeated letters out of its in symbols. When D is greater than N, our restriction, m not > N, makes it necessary that e not < (D — N). As e will have every value less than D, and not <{D-N), we shall, after thus collecting together all the D-plets of those above supposed written that contain m of the first N letters, for all values of m not > N and < D, and substituting repeated letters out of those m for the letters beyond the first N, we shall, I say, thus have before us every D-plet possible with N things, in which there is any D - 1 repeated letters, or any D - 2 repeated G5 154 FIRST MNEMONICAL LESSONS. [§ 65. letters, or any e repeated letters, whatever e may be. We shall have also, among the system supposed written out, all the non-repeating Z)-plets possible out of N letters. But Ave have made no change in the number of the D-plets by these erasures and substitutions, which is still N.(N+l).(N+2)...{N+(D-2)}(N + D-l) m 1.2.S...(D-1).D and this proves the truth of [64], for n = N, d = D, if only it be known true for n = m and d = e both less than I). Now we see easily that when n = 2 and d = 2, [64] is true ; for 2 2-ups by 2 fags = 23:1 "2 = 3, and the repeating combinations of 2 in twos are an, ab, bb, exactly three. Therefore [64] is true, by the preceding argument, for n = N, d = 3; because we do know it to be true for n = c Z and d = 2, which are less than 3: consequently it is true for n = N, and d=4>; for we can prove it true for all values of n and d less than 4 : and thus proceeding, we can esta- blish its truth for every value of d. In all this N may be any number we choose. The above argument will be at first perplexing. I advise you to read it aloud putting small numbers for N and D; say N = 6, D = 3, and again N=6, D = 4, all through. When D is 3, m and e must be either 1 and 2, or 2 and 1 ; when D is 4, m and e may be 1 and 3, 3 and 1, or 2 and 2 ; and when m = 1, a is of course a single letter. The repeating combinations of 5 in threes are, in addition to the 10 non-repeating, aaa aab aac and aae abb ace add aee bbc bbd bbe bec bdd bee ccd cce edd cee dde dee bbb ccc ddd eee, 25 more, making in all 5-6'7-l'S'S. Those of 5 in sixes are 210 in number, aaaaaa, aaaaab, aaaaac, aaaaad, aaaaae, aaaabb, &c. 65. If you take all the non-repeating combinations of six in threes, and write down the six permutations of each one, you have what are called the non-repeating variations of six in threes, which are in number 6 - 5'4, viz. that of the combinations [63] multiplied by the number of the permutations of each one, [62]. The no)i-rcpeati?ig variations of n symbols taken d toge- ther, are [n . (n - 1 ) . (n - 2) . . . {n - (d - 1 )}]. [65] Non-repe. vara, of n in d's Are d n-backs. § 66.~\ FIRST MNEMONICAL LESSONS. 155 If to the non-repeating variations of six in twos you add the six repetitions, 11, 22, 33, 44, 55, 66, you obtain the repeating variations of six in twos : the number of which is by [65^ 6.5 + 6 = 6. (6-1) + 6 = 6 2 . And the repeating variations of n in twos are in the same way n . (ti - 1) + n = n 2 . By adding to each of. these n 2 duads in turn the n symbols, n . n 2 triplets are formed, which are the repeating variations of n in threes. Every symbol of the n can be added to each of these, giving n . ?i 3 = n 4 4-plets, the repeating varia- tions of n in fours ; and so on. The repeating variations of n in p\s are in number n p . [66~] The repe. vars. of n in p's Are n to p ,h . pth power How many different whole numbers of seven figures can be made with the digits 1, 2 and 3? The answer is the number of repeating variations of three in sevens, which is=3 7 = 21S7. LESSON XX. 66. The product (l+r,)(l+r 2 ) = 1 + (ft + r 2 ) + r x r 2 = A ; (1 + r,) . (1 + r 2 ).(l + r 8 ) = (1 + r 3 ).A = 1 + (r, + r 2 + r 3 ) + (r x r 3 + r t r 3 + r,r 2 ) + r x r 2 r 3 = B ; (1 + rJ.B = 1 + (r, + r 2 + r 3 + r 4 ) + (r-3'2:l'2'3'4, quadruplets r^r 3 r A r 5 &c., i.e. all the non-re- peating combinations of five in fours. And as ™.(m-l)...{m-(e-<2>)}{m-(e-\)) 1.2. ..(e- 1). e m.(m-l)..,{m-(e-2)\ + 1.2...(e-l) (m-c+l)[m.(m-l) ..{?«-(e-2)n 1.2...(e-l).e e[jn.(m-\)...{m-(e-2)}2 + ~~ 1.2...(e-l).e (m + l).m.(m-l)...{m-(e-2)} 1 .2...(e-l).e = e (m + 1) -backs by e fags, it is evident by [6S~] that the number of non-repeating comb illations of m in e's + that of those of m in (e - l)'s = that of those of (m + 1) in e's. We had m = 4 above, and again, 9-8.7.6 9-8.7 = 10.9-8.7 1.2.3.4 1.2. % 1.2.3.4 ' By forming thus the product (1 + r 6 ) C= D, (1 + r,) D = E, &c, until (1 + r u ) is introduced, it can be proved that (l+rOCl + TaXl +r.)... (l + fr-0 (t + r.) = 1 + P, + P 2 + P 3 + . . . + P, + P, +1 + . . . + P„_, + P„ ; in which P ; is the sum of the non-repeating combinations of the n letters r, r 2 &c, in i's, whatever i may be from e = 1 to i — n. All this is true whatever be the numbers r, r 2 &c. ; let us then suppose them all equal, r l = r a = ... =r n ; the sub- indices may be erased now, and the last written equation will be n.(n-l) , n. (n-l).(n-2) , n.(n-l).(ii-2)...{n-(»-l)} H 1.2.3...i «. ( — l).(n-2)...(»-t) r , +1 1 .2.3.4... •.(» + !) § 66.] FIRST MNEMONICAL LESSONS. 157 K..(*-l).(tt-2)...{ll-(»l-2)} , 1.2.3...(ra-l) W .( W -l).(»-2)...{ W -(n-l)} rn . 1.2.3...A ' for every pair rr will be now r 2 , and the number of these pairs is — — by [632; every triplet rrr will be r 3 , „ , . , , n. (n-l).(n-2) of which there are — J — — -— , &c. Thus, if for n we put in succession the values n = 1, n = 2, n = 3, n = 4, n = 5, we obtain from the last equation 9 1 (1 + r) 1 = 1 + r ; (1 + r) 2 = 1 + 2r + -^ r 2 as in [14], (l + r) 3 = l+3r + ~ r 2 +^^* r 3 =l+3r + 3r 2 +r 3 , 4X , 4.3 , 4.3.2 , 4.3.2.1 , ( 1+r ) = 1 + 4r+ iT2 r+ TT2T3 r+ T^374 r = 1 + 4>r + 6r 2 + 4r 3 + r 4 , (l+r) 5 =l +5r + 10r 2 + 10r 3 + 5r 4 + r 5 . . If ?w = 2ra+l, any odd number, there will be an even number of terms, namely, 2m + 2, and no middle term ; if n — 2m, there will be an odd number of terms, namely, 2w + 1, and therefore a middle term; in the former case every coefficient will occur twice, as in the last equation 1, 5, and 10, each occur twice, for the first term, unity, or 1 r°, may be called the coefficient of the zero power of r ; in the latter case every coefficient will occur twice, except that of the middle term in which r m appears. The reason of this is, that one n-backs by one fag — n = (n — 1 ) n-backs by (n — 1 ) fags ; two n-backs by two fags = (n — 2) n-backs by (n — 2) fags, as you easily convince yourself. We have now proved the celebrated Binomial Theorem of Newton, at least for whole and positive values of the index n ; and this theorem is expressed thus, A. , n.(n-l) . w.(w-l).(n-2) , „ the &c, denoting that the terms are supposed to be con- 158 FIRST MNEMONICAL LESSONS. [§ 6G. tinued to the right for ever, the powers of r constantly rising by one, and the «' lh power of r, r\ being always mul- tiplied by (i n-backs by i fags). Jane: — Are we to conceive the terms carried on for ever, when n = 1 or 2 or 3 or 4 or 5 ? How can this be true when we see, in the examples you have just written, that they do not so go on ? Uncle Pen. : — You are to distinguish between the alge- braic form and the arithmetical value of the series. The form is the same for innumerable values of n ; for the truth of the equation depends no more on any particular value of n than on the value of r. The number of terms must be always the same if the form is the same. Now we can prove that however great an integer n may be, the number of terms is at least n + 1 ; i.e. number of terms is greater than any number, however great; it is therefore, no finite number, but infinite. It happens that, when n is a positive whole number = n', the (n + 2) ,u term and all the succeed- ing ones become zeros, by reason of the factor (n — n') in the numerator ; when n = 1 the third term = by reason of (w-1); when n = 2, the fourth becomes zero by reason of (w - 2), &c. But if you put for n on the left side of the equation, a value which is not a positive whole number, and the same value for n on the right, none of the terms can vanish, because (n — i), whatever positive integer i may be, cannot be zero for such a value of n. Thus put n = | ; ,, v7s , rrr -75(75-1) , 75(75- 1 ) (75 - 2) - „ (l+r) 75 =l+75r+ y — - r+ -i o g L r 3 +&c. is what Newton's theorem becomes, and no term will ever become zero. The question now arises, is this infinite series on the right really equal to (1 +r)' 7i ? We have proved the truth of this equation only for the case of n whole and posi- tive. It is just possible, that there may be, in the expanded value of (l + ry, besides the infinite series of the form above written, a certain term T which has a value for « = |, and which vanishes for every whole and positive value of ;/. If so, the series on the right will be either greater or less than (1 +»')*, and its real value will be some unknown number II, different from (1 +ry. For the present you must take it, on my assurance, that this theorem of Newton is per- fectly general, and true for all values of r, whole or frac- tional, positive or negative, possible or impossible. I shall § 66.~\ FIRST MNEMONICAL LESSONS. 159 content myself with giving you a clear notion of the shape of the successive terms when n is negative or fractional. To prove a negative, e.g. the non-existence of this supposed term T, is a very difficult task; and I strongly suspect that some of the proofs which I have seen in writers on Algebra of this point, are no proofs at all. When n is the fraction the fifth term of the series is ©• ».(«-!). (n-2).(«-3) 4 _ (n:e)(n:e-l)(n:e-2)(n:e-3) A 1.2.3.4 1.2.3.4 _ (n:e)(n:e-l)(n:e-2)(n-3e) 4 _ (n:e)(n:e-l)(n-2e)(n-3e) 4 1.2.3.4.e 1.2.3.4.e 2 _n.(n-e)(n-2e)(n-3e) r 4 _ 1.2.3.4 ~" "e*' because the numerator and denominator are both multiplied by the same quantity e.e.e.e, which makes no change in the value of the term. A similar transformation of every term being made, we have _ , v i r n.(n-e) r 2 n.(n-e).(n-2e) r 3 B. (l+r) e =l + n-+ ; n J .- 2 + — v , ,, J .— 3 v ' e 1.2 e 2 1.2.3 e 3 n . (n - e) . (n - 2e) . (n - 3e) r* + _ 1.2.3.4 ~~ *e 4+ " which differs from the expansion of (1 + r) n in these two points — first, that it proceeds by the powers of - instead of those of r, and secondly, that every digit 12 3 4 &c. in the numerator of any term, is multiplied by e. When e is nega- tive, the minus signs in all the numerators will of course become plus signs, for - (- e) is + e ; and if r be not also negative, every odd term, first, third, &c. will be negative in (B), because r:(- e)=— r:e. If both r and e are negative while n is positive, every term of the series (B) will be positive, and the signs in the numerators will all be positive. But whatever be the signs of r n and e, you can always obtain the correct expansion of (1 + r) n:e , if you put for those three symbols their proper signs and values in the last written equation. To develope or expand (1 + r)°, you first write out r o + r > + r *+ r 3 + r 4 + ...&c. (r°-l), 160 FIRST MNEMONICAL LESSONS. [§. 67. then multiply r* by (i n-backs by i fags). If you have to develope (1 + r) e , you first write out r° r 1 r e r s r* -5 + -r+ — + -J + -j + &C, and then multiply r { by an expression differing from (i n- backs by i fags) in this, that the subtracted digits, 12 3 &c. in the numerator, are multiplied by e. To remember (A) and (B), say [67] Siiton(un r)? write fr» on r; v . [61 H.] frbon. Theu r to i you multiply (A). By (i n-backs by i fags): vid. [62], [68]. If n has den.e, (B). Put r vi(re); top dits wed e. v i vid. [6]. Do you want the expansion of Sum to power n of wnity and r ? Write &c. : r to i is r* (to power i). There is no fear of your mistaking this for the ratio r-.i. Den. for denominator. If the index has e for its den., you write frtf on x'i(re) for fr» on r. vi(re)=r:e ; and the top digits (of the numerators) 1, 2, 3, &c, wed e, i.e. are multiplied by e, as in (B). 67« Examples of expansions when n is fractional are, (1 ± rf l - 1 =f r + r 2 =?r 3 + r 4 =f &c, as in [6l H] v. [44] (1 - rf 1 - (1 - r)- 3 = 1 + 3r + 67* + 10r 3 + 15r 4 + ... 0-) ! =l-ir-I^-A r .. &c .... All these you may prove from (B) by making the requi- site substitutions for n and e, and then simplifying. You will now find no difficulty, except the mere length of arithmetical operations, in expanding (a + b)', whatever numbers a b n and e may be. This is called a binomial quantity; (a + b + c) m is the w* th power of the trinomial (a + b + c) ; a binomial or a trinomial is a quantity consist- ing of two or of three terms. Since a (1 + b:a) = (a + 6), (a + b) m = a m { 1 + b:a) m , by [47, c], you have only to put {b:a) for r in (A), to expand (l + b:a) m , and then to multiply every term by a m . For example, (a + by = a 2 (1 + b:a) 2 = a 2 (l + 2b:a + b*:a 3 ) = a' + 2ba + b s , I 67.] FIRST MNEMONICAL LESSONS. 161 A 4 »a-5- & c. = a 3 + 3a 2 b + 3ab 2 + b 3 , (a + by = a 4 (l + - J = a* + 4 anc * m the last line, n+0 = n, n — l + l = n, n — 2 + 2 ~n, &c. You may exercise yourself to prove the following expansion of the cube root of 31 ; 41* _,_ 4 16 6561 '- 3 + 0-14815- 0-007SI (Sl)»-(27 + 4)*=:27*[l+~}*-8{l+^-- 1594323 129140163 + 0-0006 - 0-00006 + ... = 314138 ; a result correct to the last decimal ; and this correctness might be increased by taking a greater number of terms of the series. In this expansion the terms diminish rapidly in value, and therefore a small number of them will give a tolerably accurate result, the rest of the infinite series being so small that it may be neglected. Such a series is said to be converging ; but not every infinite series is of this kind: many are diverging, and of no use for arithmetical calcula- tion. On this subject there are very many curious and interesting things known to mathematicians, and many more yet to be discovered. 162 FIRST MNEMONICAL LESSONS. [.§ 68. LESSON XXI. 68. We have now cleared our way fairly into the most attractive fields of geometry, and are about to commence in earnest the study of the circle and those other curves of the same family, whose beautiful properties were for so many centuries the delight of the Indian, the Egyptian, and the Grecian sages of the olden time. You were curious, my dear Jane, to see how the equation to the circle through three points (x,j/,) (x 2 y„) (x 3 y 3 ) would look, and to compare it with that to the line through two given points. This last has a secret to reveal to us of immense importance, if we arrange its terms in the fashion following : x y x \ 2 - x y 2 \ l + .r^lo- x lt y l 2 + x 2 y l l -x 2 y l 1 = 0, (a) which is merely the equation (8) of Art. (16), the units being introduced for symmetry only. Looking at the subindices, you see every permutation of 012, and observe that every two terms, whose subindices differ by the ex- change of a single pair, are of opposite signs ; as the first and second, differing by the exchange of 1 and 2, the first and last, differing by that of and 2, as also the second and. fifth. The effect of this arrangement is, that if you suppose any two subindices to coincide in value, the whole expres- sion is reduced to pairs of terms that destroy each other. Suppose and 1 to coincide, in other words, let x = x l and y» — y\\ then the first and fourth terms destroy each other, as do the second and third, and the fifth and sixth ; which proves that the line represented passes through the point (x = x,, y =_?/,). The same thing happens if and 2, or if 1 and 2 be supposed to coincide in value. This last suppo- sition is of course only admissible by defining (x y ) as one of the two given points, and either (a^y,) or (x 2 y 2 ) to be the co-ordinates of the current variable point. Further, no two terms which have the same sign can be made alike by the exchange of a single pair of subindices. Thus x y l 1 a requires two exchanges, 102 and 120, before it can be made the same with the third term, first that of and 1, then that of and 2. We see then that in this equa- tion the whole of the 1.2.3 permutations of 012 occur P>2], and that the signs are determined by this law, that § 69.] FIRST MNEMONICAL LESSONS. 163 two terms which can be made to agree in their subindices by one exchange of one pair, performed in one term, have opposite signs; while two terms, which cannot be made so to agree without performing two such exchanges, have the same sign. If we put x 2 for every x the expression becomes *Q a .yiU-Xa S gah + Xi'y2lo-x l 'y l a + a: a '!t l l - or 2 y x \ = 0, (b) which vanishes like the preceding if x = x, and y =y i} or if x = x 2 and y a =y 2 , and for the same reason, being then reduced to three pairs of terms, each pair equal zero. But this is no equation to a right line, for it is of the form Ax 2 +By +C = 0, or x 2 = (By + C):A, where ABC are constant numbers ; and gives x = ±J(By + C):A, yielding two values of x for every value of y. Thus if y = 0, which is true only in the axis of x, (11), x = ± J 'cTa, so that this locus has two points in the axis of x, equidistant from the origin. It is, however, not a circle, (28), because the coefficients of x 2 and y 2 are not alike. It is in fact, A 2 x 2 + 0.y 2 + By+C = 0, zero being the coefficient of y 2 . Look now at the expression (Vo + *o*) y^2- {y 2 + *o) M + {y 2 + <0 i/ 2 i o - (y 2 + *i") ^ . + (j//+<> < yoi 1 -(#/ + <) y.i = o: 00 the arrangement of subindices is exactly as before. If you suppose x = Xi and y =y,, the first and fourth term, the second and third, the fifth and sixth are self-destroying pairs, as in (a) and (b) : wherefore (.r^/i) is a point of this locus, the equation being satisfied at that point ; and if we put x = x 3 , y =y 2} we see in the same way that (x 2 y 3 ) is also a point of it. This equation is exactly 1 - 1/2) y" + Oi - y 2 ) * 2 + (y* + ** - .y. 2 - *?) y + (v? + **) y* - (y* + •*•/) y , = o, (c') 164 FIRST MNEMONICAL LESSONS. [§ 69. of the form H (29), and is therefore, when the axes are rectangular, the equation of a circle (28) which passes through (^i^)) and (x 2 y 2 ), and has its centre in the axis of x, with radius = i f ,y,' + * a '-y,'-*,' 1 (V + *,').y.-Cy.' + *»'). y. yi 2 1 (yi-y*T yx-y* )' This becomes if V\=3i and *i" = *a*> an indeterminate quantity, having any one value as much as any other. This shews that if (^ij/,) be (3, 4), (e.g.) and (x 2 y 2 ) be (-3, 4), an infinite number of circles can be drawn to have their centres in the axis of x, and to pass through the two points : and it is obvious, if you draw the figure, that every point in that axis is equidistant from them. We have seen that the equation of the line through any two points (x l y l ) (x 2 y 2 ) exhibits all the permutations of 012, with a certain law of signs : this makes it probable that the equation of the circle through any three points (x x y } ) (•^2.^2) (^23/2) will exhibit all the permutations of 0123, arranged by the same law of signs. At least it is worth while to examine this. 69. Let the co-ordinates be rectangular : we know that in the circle if and or must have the same multiplier. Let us form our equation thus, — + (yo a +x s ).y l .x 2 .l 3 -(y 2 +x 2 ).y l .x 3 .l 2 + (y 2 +x s ).y 2 .x 3 .l x - (yo + x 2 ).y 2 .x x .\ 3 +{y 2 + x 2 ).y 3 x x A 2 -(y 2 + xi).y 3 .x 2 .l x - (jti'+xfl-ifrXv 1 + (y * + x x 2 ).y 2 .x • 1 3 - (yx s + ^).y 3 .x . i , + (y i 2 + x 2 ).y,.x 2 A -(y 2 + x; 2 ).y .x 3 .l 3 +(y 2 +x 2 ).y .x 3 .l 2 + O2 2 + x.{).y T x . 1 , - ( y 2 + x 2 2 ).y 3 .x x . 1 + (y. 2 + *,■)#„.*,. 1 , - (yf + x 2 ).y .x 3 . 1 , + (iff + xi).y r .r 3 . 1 „ - (y.r + x 3 2 ).y x .x . 1 , - (yf + x 2 ).y .x v 1 2 + (yf + xf).y -x 2 . i , - (y 3 * + *,").#.** I o + W +x 3 2 ).y x .x A a - ( 7/ 3 a + x 3 2 ).y 2 .x .l , +(j/ 3 a +x 3 *).y 2 .x x . 1 . (d). § 70.] FIRST MNEMONICAL LESSONS. 165 70. The law of the signs in (a) may be thus stated : the permutations made by going round any triplet, or as it is said by cyclically permuting it, have all one sign : those made by going round any duad have alternately + and —. Applying this law generally to all odd and even multiplets, we have the rule: if in a multiplet of n symbols abcdefghi... you consider any continuous portion, and carry the first letter (or last) of that portion to the place below the last (or ahove the first) letter of it, you either change the sign of the multiplet abed..., or not, according as the portion considered has an even or an odd numher of symbols. By this rule the signs in the preceding equation are de- termined ; the first six terms, as also the next six, &c. are merely the arrangement (a), if you consider the last three subindices. The first, seventh, thirteenth and nineteenth terms have alternate signs, being cyclical permutations of the quadruplet 0123. The subindices of y 2 and x 2 are to be counted as one and the same. It follows from the law of the signs that if for x and y you put any of the three pairs x iVi) J-VAm £>£sj tne equation is reduced to a system of self- destroying pairs of terms. Take the third term, + (y 2 + x 2 )y 2 .x 3 .\ i , and let (x y ) be the point (x 3 y 3 ). The term is now identical in value but not in sign with (x 3 a +y 3 )y 2 .x .l 1} which stands (the 23 permutation) in the equation with a negative sign, for by our law, 0231 and 3201 have contrary signs, as the first can be transformed into the second by one exchange of a pair, 3 and 0. Or they have contrary signs, if you like, because the first be- comes the second by first going one step round 023, thus — 302, which changes no sign, and next going round 02, thus — 320, which changes the sign. In fact you may easily see that any term of the (4-3'2 # l =) 24, [62], if you give to any pair of co-ordinates the values of any other pair, (as if you supposed x 3 = x 2 , y 3 =y 2 ) will become identical in value, but not in sign, with another term. Thus (y 2 2 +x 2 2 ).y l .x 3 . 1 and (y 3 2 + x 3 ).y 1 .x 2 .l are of opposite signs, and destroy each other, if (x 2 y 2 ) and (x 3 y 3 ) be the same point. It is enough, however, if the coincidence of (x tj ) with any of the other three points reduces the expression to zero, in order that the equation be true at those points, or that the locus may contain them. As this reduction to zero does thus happen, we are sure that, whatever this locus may be, it passes through (#,y,), {x 2 y 2 ) } and (x 3 y t ), and this too, whatever 166 FIRST MNEMONICAL LESSONS. [§ 70. co-ordinate angle we employ. As the above equation is of the form (y* + x *)A + By +Cx +D = O, it represents a circle, if our axes are rectangular, (28) which passes through the three points. The coefficient A is all in the six first terms of (d), and by (51 B) this is exactly 2(a123) the double area of the triangle made by (•*•,#,), ( x 3ya)> (#3^3)' The coefficient B is + W+ X t*) • (9*-yJ + (#»"+<) • (y»-#)+ ( y?+*i) • (# -^)}> and D is - {(y* +*i 2 ) • (^tfa-ya-**) + (#'+*«•) ■ (y 3 *i -yi * 3 ) + (y 3 2 + x 3 *).(y i x 2 -y ;i x l )}. These are all known numbers, so that, as in (28), the centre and the radius can be instantly found, of the first the co-or- dinates, and the length of the second. We may write the equation (d) thus, {Sin w = 1 in (51 B)} (a 123) (y* + .O - (a 230) (y* + x*) + ( a 301 ) (y> + x 2 B ) -(a 01 2) (j,/ + = 0, a form of little value, except for its symmetry, and for prac- tice in algehraic symbols. Or it may be more briefly sym- bolized thus, ^)(yo s + **)-yi-**-h = o, in which 2 is the token of a summation -of terms, the terms being all alike, except in their subindices and signs, and all deduced from + {(y 2 +x 2 ).yi.x 2 .l 3 } by permutations of the subindices, with their signs determined by the rules above given. The symbol (±), bracketted, may denote that the signs have to be properly affixed, after the permutations are written out. To remember the rule for affixing them, you may perhaps find it sufficient aid to say by rote, £68] Round ev. or o. with pin or no For signs you go. pin is h 1 1 — . i.e. if you cyclically permute any multiplet, go round it, you write clown the successive permutations with alternate signs, or no (i.e. without change) as that multiplet is even or odd ; pin for plus mimis. Thus + 123456-162543 + 162345 - 312654, are four of the 1-2 3-4-5'6 permutations of six things, with congruously § 70.] FIRST MNEMONICAL LESSONS. 167 determined signs. To find the sign of the second, I take the following steps to reduce the first to that shape, + 123456 + 1(62345) + 162(534) - 1625(43). As I have gone a step round an odd multiplet, a 5-plet, in the second of these last, I have changed no sign, nor in the next by a step round a triplet, but the next step round a duad changes the sign. The sign of 162345 requires the steps - 12(6345) + 1(62)345, the sign of 123456 being twice changed in its transformation; or two operations on — 162543 would have sufficed, thus, + 1625(34) + 162(345). If we had before us all the permutations of six things, there would be no study at all required after a few steps : you can see that in equation (d) all depends in the correct writing of the first six terms, the rest being three sixes made from those first by goittg round quadruplets — of course with alternate signs. I may state here, although somewhat too early perhaps for strict method, that a curve of the second degree can always be made to pass through any five points. Its equa- tion is The number of terms is 6 Jags [7)2], and if you understand what I have just said about the signs, + or — , of the permu- tations of six symbols, you can write out the equation your- self; the only difficulty being the length of an operation almost merely mechanical. It matters not what origin or axes be chosen ; the curve thus given passes infallibly through every point. Suppose e. g. x = x 6 , y = y 5 , the term is destroyed by the term and the whole equation = 0, as it ought to be, by similar pairs of internecine terms. What the nature of the curve is, whether a circle or otherwise, will be a delightful enquiry that will ere long occupy us. You can easily see that, referred to right axes, 2H«+^)i, = o is the equation of the circle which has its centre at the origin, and passes through the point (x lt y t ): that 168 FIRST MNEMONICAL LESSONS.' [§ 70. is that of the circle (29 G) which touches the axis of x at the origin, and passes through (r,.?/,) ; and that SBK' + rO^O touches at the origin the axis of y, and passes also through (x, , y { ) ; as also that is the equation of the line through the origin and through the point (x l5 y^). What curves are represented by the above equations when the co-ordinates are not rectangular, is reserved to be examined hereafter. The equation C contains the terms following : +y 2 .y i x l .x 2 2 .y 3 .x i A 5 -y 2 .y x .x, 2 .y 3 .x i A 5 +yi i -y^ 2 .x 2 .y z .x i A 5 -y l 2 .y 2 x 2 .x 2 .y .x A A b +y l 2 .y 2 x 2 .x 3 2 .y i .x A & -y 2 .y 2 x 2 .x 2 .y i .x s A , which are exactly the following, + y* Q/i*i • *2-y 3 • *4 • 1 5 ) - y x 0/, 2 . ^-y 3 -x t .i 5 ) + x o {yi-y^i -y* ■ ** • i *) - y (jti'.y&t • * 3 S . «« • U) + x o (y*-y* x * • *3- y* ■U)-i (jr*. jua • x *- y* ■ *«)• Let A stand for 2 (±)j/,a-,. x 2 2 .y 3 .x 4 . 1 5 , the sum of 120 terms made by permutation of subindices only, and let B, C, &c. each stand for such a sum ; then the equation is Ay 2 - Byx + Cx 2 - Dy + Ex - F = 0. It is easy to write out completely any of the coefficients : thus D is obtained from y 1 2 .y 2 x 2 .x 3 2 .x i A s by adding to it 119 terms differing only in subindices. Writing subindices only, — D is -{ 12345-12354+ 12453-12435 + 12534-12543 - 13452 + 13425-13524 + 13542- 13245 + 13254 + 14523 - 14532 + 14235 - 14253 + 14352 - 14325 - 15234 + 15243 - 15342 + 15324 - 15423 + 15432 + 23451-23415, &c.} The same series of permutations of 12345 will give all the six coefficients from their six first terms. Something further on this mode of writing out explicitly loci, plane or solid, of any class or order, may be seen in a Memoir by the Author On Linear Constructions, in the ninth volume N.S. of the Memoirs of the Philosophical and Literary Society of Manchester. I JUVENILE CONVEKSATIONS. On §§ 2, 3. William : — I see no difficulty in adding or subtracting negative quantities. Putting on a decrement is taking off something positive; and taking off a decrement is an augmentation, evidently, for it must be the reverse of taking off a positive quantity. It is easy to see that 5+(-3) is 5-(+3), the same as 5-3=2, and 5-(-3) is 5 + 3 = 8, vid. [11.] Between 5 + (- 3) and 5 - (- 3) there is a difference of 6 ; and while a + (-b) is = a-b, a-(-b) is =a + b. But I am puzzled about these co-ordinates. You say that every pair of numbers determines a point; shew me then the point belonging to the numbers 2 and 500. Richard :— Do you want the point (2,500) or the point (500,2) ? Just now, you read for me the points here written, (3, 4) (3, -4) (-3,4) (-3,-4), (4,3) (4,-3) (-4,3) (-4,-3); read them again. William : — The upper line is — the point whose x is 3 and whose y is 4, the point whose x is 3 and whose y is minus 4, the point whose x is minus 3 and whose y is 4, and that whose x is minus 3 and whose y is minus 4. The lower line is— the point whose x is 4 and y = 3, that whose * is 4 and y = — 3, and so on : that is the lesson you have taught me. But points should have no magnitude— they cannot be one bigger than another: shew me the points (3, -4) and (2, 500), that I may compare them. Jane : — I think William does not quite see the use of x and y. Two numbers can determine a point only by measurement ; and, before we can measure, we must know from what point and in what directions we are to measure. When you ask Richard for the point (2, 500), you should give him your origin and axes : there is no sense in co-ordinates without these. You must have settled the point O, from which to measure x — two units, (i.e. two inches, on our supposition) along a fixed line OX, and y= 500 along, or rather in the direction of, the fixed line OY. William : — Well, then, I draw two lines, and mark them XOX' and YOY' : you now take two inches along OX, and 500 inches along OY: pray where is the point (2, 500) ? You have found a brace of points, one on each axis ! Richard: — Look again, you philosopher, at my first figure, (vid. p. 3.) I shewed you three ways of coming at the point q, (3,2): first, your H 170 JUVENILE CONVERSATIONS present method, to take Op = 3 and Or = 2, inches, and then to draw through p and r parallels to OY and OX, — these will meet in the point (3, 2) : secondly, to take on the axis of x Op = 3, and at p to raise an ordinate pq parallel to OY, two inches in length ; this brings you to the point (# = 3, y = 2) or (3,2): thirdly, to measure y=two inches on the axis of y, to r, then from r to draw a parallel to OX, three inches mea- sured on this parallel brings you to the same point q. The second of these methods is the one to bear in mind, when you think of the point (3, 2). Now tell me how you would find (3, -4), and (2, 500). William: — I first march three inches on OX to p, and thence four inches on an ordinate in the negative direction parallel to 01", this gives me (3, —4) : two inches along OX, and then 500 inches along a positive ordinate, bring me to (2, 500). Jane : You see it now, and in the same manner you can find your way to the point ( -, — ) whose x is half an inch positive, and whose y is five-sevenths negative. Did you tell William what we mean by the points (■*■!, «/i) (-*i,yi) (*i,-yi) C-*i»-yi)? Richard: By x and y in general we understand the co-ordinates of a variable point, some point or other, no matter for the moment what or where : it is enough for us that * and y are the two distances of a point from two given lines, its distance from each line being measured in a direction parallel to the other line. By x x and y, we understand the co-ordinates of a fixed point, which we know of and can find. Perhaps x x and y x are two numbers of which we have thought, or which we have down in a list along with other numbers, marked x Q , y 2 , x 3 , y 3 , &c. ; or they are numbers which we can find, when we please, by measurement, of the distances of the given visible point from OX and OY. The points above written are, the point whose x is the known length (.rat 1) and whose y is the known number (y at 1), that whose x is the given negative length (x at 1) and whose y, &e. vid. p. 5. The same thing is meant by the point (x = m, y= n), or the point (m, »). We conceive m and n to be known numbers. Suppose now you had before you the given points (m, n) and {-g, h), m, n, g, and h being positive numbers; where would you look for the points (— m, — tt) and (g, — h ) ? William : — I should expect (m, n) to be somewhere between the lines OX and OY, as both the co-ordinates are positive ; but {-g,h) is found by proceeding g inches along OX', and the point must be within the angle X'OY. As the co-ordinates of (-m, -») and (g, -h) have signs all contrary to those, the former of these is within the angle A"01", and the latter within XOY'. Jane:— True: and you can see that the eight points first written by Jtichard form the angles of two distinct parallelograms, whether our axes are oblique or rectangular. Have you read, William, the points (.»',,»/,) (*2> y*)j &c., in (3) ? p. 5. There is nothing like reading aloud. ON THE OPENING LESSONS. 171 William : — The first is that whose x is f and whose y is one ; the second that whose x is f and whose y is J, and so on ; the fifth (x s , y & ) has its ^ = | and its y = \: all easy enough. But how is this, about the point (0,0) at the end of (3) ? Richard: — Do you not see, from (a'), that if x =0, y being \ of it, is nothing also ? How much is § x ? On § 4. William : I am ashamed to say that I stuck fast for a few moments at the pronoun it, in the second line of (4). It is nonsense to suppose that OO' can meet the origin a second time. Why must it meet the curve again ? Richard: It would be a difficult thing to draw a line from O, a point in the supposed curve, which should nowhere meet the curve again; but there is no must in the argument : any line you please that cuts the curve in any point O' suffices for the reasoning. William : Why did your uncle not set down the lengths in numbers of the co-ordinates of O' instead of saying let them be m and n ? Richard: — I dare say he never measured them : they are considered as known numbers, for the point O' is given and before you. We under- stand m to be the number of inches in Ob, and n that of those in bO'. William: — How am I to "find a series of points in the locus (a')" ? What is a locus ? Jane: — A locus is simply a line, either straight or curved, formed by a series of points arranged in close succession according to a certain law. This law is expressed by an equation, and this equation is always an as- sertion about x and y, which is true of the co-ordinates of every point in the locus, and true of no other points. We speak of the line 2y = bx, y = ex, &c : which means, more at length, the line whose law is this or that equation. In the first of these two the law is, that double the y of every point is 5 times the x of it ; that of the second is, that y is e times the x at every point of the locus. The point (20, 51) is not in the first line, be- cause it is not true that 2x51=5x20; but since 2 x 50 = 5 x 20, is truly affirmed, (20, 50) is a point of it. If e happens to be the number 2iJ, (20, 51) is a point of the line y-ex, for 51 = 2^x20, is true; but as 50 = 2^5 x 20, is false, (20, 50) is not in this locus. For a reply to your first question, William, you have only to glance at the process by which a series of points is found in § 3. You may make the same measurements along OX" and O'Y", that are there made along 172 JUVENILE CONVERSATIONS OX and OY. Your points so found will have different positions from the former. William: — I see plainly that my point (jg, y D ) will be near O', while your point (jjj, 15) will be near O. Why is there but one series of points traced in the figure ? There are two supposed curves mentioned. Jane : — You can imagine a second supposed series on the opposite side of OO' ; the argument does not require that it should be drawn. Richard:— Uncle says I was right in calling the argument in § 4 a re- ductio ad absurdum. You see, William, that (a') cannot give a curve when referred to OX and OY, without giving also a different curve, when referred to O'X" and OY". But from the law (a) by which both curves are traced out, it follows that any point of my curve, as (j? = Ot, y = st), is also a point of yours, viz. the point (.r=0'u, y = sv); so that the curves are not different. The hypothesis which leads to this contradiction must be false ; and there is no way of escaping the absurdity, but by sup. posing that both series of points lie in the right line OO". On §§ 5, 6. Jane: — Cousin Henry shewed me this morning a pretty proof that the product of two negative numbers is positive. We know that (5-5)x(2-2)=0x0 = 0: i.e. 5x (2-2)-5x(2-2)mustbe = 0, i.e. 0-5x2-5x(-2) =0, i.e. - 10 -ox (-2) =0, is true, which cannot be, unless — 5x-2=+10. In like manner, (m — m) times (»— n) = 0x0 = 0, which requires of necessity that (-»») times (-n) = + wm. Divide both these equals by the product — nxn, and you get — m:n — m:—n, as in p. it. Is not this charming ? I feel quite scientific. Richard :— Do you not think, Jane, that Article (f>) might somehow have been dispensed with ? After proving in (4) that (ij = }») is a right line, it seems easy to grant, that y = - §*, or, what is the same, if we mul- tiply both sides by - 1, -y = §*, is a right line also. Jane :— It is one thing to be convinced of a truth, and another thing to frame a valid proof of it. When you have invented a shorter demonstra- tion than this of (6), I will try to give you my opinion upon it. William:— 1 see that the Roman co-ordinates refer to the left band figure. How do you know that y = $x is a right line? Tin different from those in (4). Ah! 1 see my answer in what precedes here in (ti). ON THE OPENING LESSONS. 173 Jane: — You observe that the question now is, whether (y=— %x) gives points in a right line or no. Whatever this locus may be, two points in it, besides O, can be found ; p and m are so found. The question is : are O, p, and wina right line ? Now they form a group exactly similar to O, p, m, in the left-hand figure ; and as these are proved in (4) to lie in a right line, O, p, tn, are in one also. On §§ 7, 8. Richard: — William began to read (7) thus : "We find a point in the locus y which is equal to e,r." I laughed, and so did he ; but you must bring in the which in the sentence below — " If qp which is equal to q x p\, which is equal to 9 2 p 2 )" &C. Uncle says the ivhich ought to have been written here; but that the ellipsis is very common in algebraic reasoning, and seldom occasions any ambiguity. William : — I see that (y = ex) is supposed to be the line q, q u q 2 pass- ing through the origin O, which letter you have forgotten to insert, and that p, p u » 2 is a line parallel to it, which does not pass through O, but cuts OY and OX'. Jane : — There are of course innumerable lines parallel to q, q x , q 2 , •■■ some cutting OY and OX', others cutting OY' and OX. The lines y = ex + l, y = ex + 2, y = ex + b, if b is positive, all cut the former pair, while y = ex — 1, y = ex~2, y = ex-b all cut the latter; if e has in all one value. William .-—Where is your figure for § 8 ? Jane : — Draw it yourself : you may choose any point you like for (x u y { ) in any of the four angles about O ; this point and O determine a given line. If x { and y { are both of one sign, as (2, 3), (-6, -40), the line is drawn in the angles A'OFand X'OY' ; if they have different signs, as (-2, 3), (6, -40), the line is drawn in the angles XOY' and X'OY. The number of lines through O is infinite; for between any two, however little they may diverge from each other, you can always imagine a line drawn, which diverges still less from either. It is something worth know- ing, if it can be proved, that any of those lines through O may be repre- sented by the equation y=ex. Richard:— It appears to me that this is proved in (6); for it there appears, that for every different value of e, positive or negative, that you can name or think of, there is a distinct corresponding line passing through O. Jane : — True ; but is it there demonstrated, that no line can pass through O besides those which correspond to the positive and negative values which e may receive ? This negative assertion remains to be proved, and is established in (8). Choose any line, so it be a given line through O. If it is given, a point of it is known besides O. Give me this point, i.e. give me its co-ordinates in numbers, and I will write down the equation of a line, as it is done in (8), and make you confess that it is your line, and that it has an equation of the form asserted. H3 174 JUVENILE CONVERSATIONS William :— It is evident, looking at the second proposition in (8), that a line which does not pass through O must meet YOY' somewhere, at some distance b, positive or negative. But is this proposition true of every line, of lines through O also ? Richard ;— Of course, if b may have any value : for lines through O, b = 0. On §§ 9, 10, 11. William :— I can make nothing of your equation (c'), p. 12. How do you read it ? Jane : Did you never handle a fraction whose denominator was a fraction ? If my master had taught you arithmetic ! Look at the middle of the next page— you see y divided by the fraction f, and the negative quotient x divided by the fraction §. You should always read an argu- ment through, and all the illustrations, before you despair. I read (c') thus : " the negative quotient * by the fraction (b by e), +y by b, = 1 ; " or, if you like, "y by b, minus x by (b by e), = 1." "When y = 0, which is true of every point in XOX', this becomes -4 = 1, or since = — , (5), n -n fr 1 ' e whence, as the denominator and numerator must be equal, we have b x = — . e Richard :— The example at the end of the Lesson, (p. 15), became much clearer to me, after I had drawn the figure, and examined the case of y = 2x+l and y = -2x + l, which are I+_ =1 and y + 7 = 1 - I took ra = 3 and obtained y, =(*» + 1 = 7, y., = — 6+ 1 =-5 ; whence #, +7^ = 7 -5 = 2, or PB-PCi%.PA, QA being the parallel to OX through (0, 1), or Q. Now from PB - PC = PA + PA) just proved, comes PR = PA + PA + PC, by addition of + PC to the equals; and hence we get PB-PA=AP+PC, or AB^AC, by subtraction of PA from the equals. ON THE OPENING LESSONS. 175 Thus I prove that QA always bisects the base of the triangle QBC, whatever ordinate BAPC may be. Jane :— Well demonstrated, Richard; but you speak as if you had found all that out yourself. I will now shew William something more, the discovery of which he will of course ascribe entirely to me. The line through (0, 6) parallel to OX, spoken of, p. 15, has for' its equation y = b, or y- 6 = 0; for at every point of it, the ordinate is of the same length b ; and every line parallel to OX can be represented by the same equation, for some fixed value or other of b ; thus y = — b, or y + 6 = is the equidistant parallel on the other side of OX, through the point (0, —b). Every parallel to OY has an equation of the form x = b; thus a- -4 = 0, and # + 4 = 0, cut OX and OX' at equal distances from O, and are both parallel to OX, because x never changes in either line, however y may vary. The equation of the ordinate drawn in your figure through jPis x — 3=0; and every ordinate can be represented by x — n = Q for some value positive or negative of n. The property you have just proved may be stated thus: the line y — b = 0, (QA), bisects that portion of the line x — n = (BAPC), which lies between the lines y-ex — b = and y + ex — 6 = 0; whatever be the numbers e or n. This is true of course when 6 = 0, in which case the bisecting line QA is y-0 or OX. The two lines are now y = ex and y= — ex, and we know something worth remembering about the four lines y=0, x = 0, y—ex=0, y + ex = 0, which all meet in O. Richard : — There could hardly be a simpler combination ; there is only one constant in the four equations, the number e, used with opposite signs. Jane : — Uncle Penyngton remarked that simple combinations are the most instructive, and lead, when their meaning is completely mastered, to the most extensive and entertaining views. The thing to be remembered is, that the first of these four lines bisects every parallel to the second drawn from the third to the fourth ; and that the second bisects every paral- lel to the first, which is drawn from the third to the fourth. Thus let y =m in both the latter, which will be the case where each meets the line y — m = Q; the equations become rn = ex x , m=—ex 2 , whence by subtraction, m - m = ea\ — (- ex 2 ), which is = ex x + ex 2 , or since -f- e = 0, = x x + x 2 , by div. of equals by e, or, the x of the point on {y-ex = 0) whose y is m, differs from the x of the point on (y + ex = 0), whose y is m, only in sign. In other words, OY bisects the parallel to OX, which is drawn between those two lines. This is true whatever m may be, and is not affected by the value of e ; 176 JUVENILE CONVERSATIONS neither does it depend on the angle XOY between x = and y - 0, any more than on that between y — ex = 0, and y + ex = 0. I am told that these four lines form an harmonic pencil, and that such a pencil has many curious properties which we shall be delighted to learn. On §§ 12, 13. William : — How can anybody be expected to remember such a long piece of reasoning as this in the 12th Article? I am utterly confused in it. Richard : — So it was at first with both Jane and myself; but it is plea- sant to find how rapidly such difficulties melt away, after a little examina- tion. I think it all through at a glance, when I look at the figure in (13). First of all 1 know that if y.x = e be any line through the origin, no matter what be the angle between the axes OX and OY, containing the points (x lf y,), {x 2 , y. 2 ), {x 3 , y s ), &c, 3^1 = ^2 = ^3 = ^4 =&( , X{ ^2 ^3 ^4 because each of these fractions is one number e. From the first pair, multiplying the equals by — , comes This is equation B' (p. 17), if OY and OX are my two axes, and OP the line through the origin, and it is equation b', if OY and OP are my axes, and OX' the line through the origin. The two ordinates y x and y t are the same, whether OX or OP be the axis of x, except that in the first case they are both positive, qp and QP, in the direction 01', and in the latter both negative, pq and PQ, in the direction YO. This makes no difference in the truth of the last equation, since -yi'.-y-i = y\'-y 2 - Thus I have proved the third line of [0] ; for the para/l. cutters are the two ordinates y x and y 2 , and the quotient of these is that of the two abscissa?, whether these be Op and OP, or Oq and OQ. The whole argument of (12) is in the equations B', b', and the equation lib ; the last deduced from B and b by division, the former pair by multiplication. The second line of [G] may be considered to assert the equality of the right members of B' and b', as well as to express (Bb). On § 14. William :— He was puzzled at ' turned face downwards ?' If the tri- angle ABC is turned over in its place without disturbing A, where can C fall but on C, or D but on D' ? multiplication of a line by a line is not so easy a notion — how can you multiply but by a number? Jane: — Vou see that line times line appears in the equations: this ON THE OPENING LESSONS. 177 cannot be the same thing with number limes line, which is just a longer line. We must accept the definition : and it is easy to conceive an inch square to be made by the repetition of an inch line through a height equal to its length, and to call such a square pile of inch lines, an inch times an inch. A quarter-inch repeated to a height equal to half an inch, is £ x i - \ x a square unit ; and so of any two lines. On § 16. Richard :— The numbers x x , x 2 , y u y 2 , eq. (B) are lengths measured from the two visible points (x t , y^) (x 2 , y 2 ) to our axes OX and OY. Suppose now that we had chosen other axes, inclined at any particular angle : these numbers would be different, some or all of them ; but the position of the points is fixed, and therefore the position of the line through them ; while the equation of the line alters for every different pair of axes. Is it not odd, and confusing to imagine, that the same visible line should be represented and determined by so many different equations ? Jane : — Just as odd, as that the same point in our plane should be represented and determined by innumerable pairs of numbers, i.e. by a different pair for every different pair of axes. But we use only one pair OX, OY, in one argument ; we know always what x and y stand for, and how they are measured. Don't you think that this is the chief beauty of equation (8), that it applies to any axes we may choose ? Is it not worth knowing, that, when they are chosen, whatever be the distances » ti x.,, y x , y 2 , that equation is sure to be true of every point (x , y n ) in a line with (j?i, y,) and (x 2> y 2 ) ? Thus, if j?j =0 = y t , it takes the shape y — ex = 0, or rather Ax + By = 0, — the same thing, as to the form of the equation. On § 21. William: — How can you say that p 2 m 2 =?m 2 q 2 + q 2 p 2 ? Is not the distance from q 2 to p 2 negative, q 2 p 2 being a negative ordinate ? Richard: — You may take it thus: +p 2 m 2 = + p 2 q 2 + q 2 m 2 ; we are measuring from p 2 to m 2 in the direction OY, which is positive: and if you state it thus : —m 2 p 2 = - m 2 q 2 — q 2 p 2 , putting negative signs because we measure now from m 2 to p 2 , it comes to the same thing ; for multiply- ing both sides of this equation by —1, you obtain the former equation. It is useful to remember that you may change the signs of every term on both sides of any equation, without altering the truth of it ; for this is merely multiplying equals by — 1. If one side of the equation, as it fre- quently happens, is zero, you may change the sign of every term on the other side. Thus 5-2 = 3, _5 + 2 = - 3, 5-2-3 = 0, -5 + 2 + 3 = 0, are all equally true. It is evident that x - 1 = 0, for + and - are the same quantity, at least in arithmetic value. I thought it an odd and 178 JUVENILE CONVERSATIONS mysterious remark of Uncle Penyngton, which fell from him the other day, that if ever I mounted into the higher regions of analysis, 1 might learn to make important distinctions between positive and negative zero ! William : — I am not satisfied about the two equations y 2 = - q 2 p 2 , and 1/3 = — <73/>3- It is plain to be seen that y 2 is negative, and so is y 3 , and that y 2 and y 3 are nothing else than the two lines q 2 p 2 and q 3 p 3 ; why then do you not write - y 2 = — q 2 p 2 , and -y 3 = - q 3 p 3 ? Jane (after a pause):— This is very acute of you; and it has cost me some thinking to find a sufficient answer to your objection. But I have it now: in the equations with which (21) begins, the co-ordinates are of course numbers of inches , I is the number I, a is the number a, and Oq { P\Qu Pflz, &c, are put for the numbers of inches in those lines. We must read, let p 2 be the point whose X is the positive number (of inches in) Oq 2 , and whose y is the number (of inches in) /j^ taken negatively. If we were to write, as you propose, — V2 = — q 2 p 2 , — y 3 = — qiPs, it would follow immediately that y 2 = q 2 p 2 , and y 3 = q 3 p 3 , making y 2 and y 3 to be positive ordinates, which they are not. On § 21, 23. " William : — You ask (p. 35) for the distance between (2, 3) and (•3 — -4) referred to right axes. Would the distance be dift'ereut if the axes were oblique ? Richard: — Of course not: referred here agrees with points. But (# — b) 2 + (y — af is not the squared distance between (,r, y) and (6, a), if these co-ordinates are measured parallel to oblique axes. I do not exactly see why not, at this moment; but I do see the proof of the rule (p. 30) laid down in [12]. The argument leaves the choice of axes to me. Provided that they are at right angles, and the same axes all through the reasoning, it matters not where they are. This is to me the great beauty, as Jane says, of equation D' and the rule under it, that R is sure to come out the right distance between (,r„ y,) and (x , y ), whatever be the right axes OX and Ol", from which co-ordinates are measured. Jane : — The reason of this is, that R depends not exactly on the distances » v . . r i!l, is true in the case of u> negative also. Richard: — I see that in (34) the segments of the side AS arc mea- sured from D to its extremities, whether D is in AB or AB produced. It is pleasant to have Euclid n. 12 and 13, in so small a compass in [26 A]. 14 DAY USE RETURN TO DESK FROM WHICH BORROWED LOAN DEPT. ARY (WAY 3 1962 MAR 1 3 \m 3 1,1) 21A-50m-8/57 (C8481sl0)476B 4k ±. General Library University of California Berkeley f U C BERKELEY LIBRARIES lilt IV £70568 Q/- \<5 THE UNIVERSITY OF CALIFORNIA LIBRARY \