University of California • Berkeley The Theodore P. Hill Collection of Early American Mathematics Books Kytdv'cttucmcni, 1 HE design, In making this Com- pilation, IS to collect suitable exercises to be performed by tbe Classes at the Private Lec- tures on Mathematics, given in the Univer- sity. The view, which the Corporation had, of the great advantages, that the Students xnight derive from a judicious work of the kind, produced this attempt to promote their improvement. The parts of the most ap- proved writings, selected for the purpose, are copied, with only such alterations, as appeared to be useful. The Authors of most of the b^arxhes are Dr. Hutton and Mr. Eonny- CASILE ; the Navigation is principally from that of Mr. Nicholson ; and considerable use has been made of a Manuscript^ Mole's Algebra^ and the Works of Emerson. /8oi MATH EM A T I C S, COMPILED FROM THE BEST AUTHORS AND INTENDED TO BE THE TEXT-BOOK O F T H E Coutse of l0rit)ate JLtttmts ON THESE SCIENCES IN THE University at Cambridge, UNDER THE DIRECTION OP SAMUEL WEBBER, ^.m.^.^.s. HOLLIS PKOiESSOR OF MATHEMATICS ANO NATURAL PHILOSOPHY- — «=5=»0<==- IN TWO VOLUMES. — VOL. U Copg Kigfet 0ccurctj. PRINTED FOR THE UNIFERSJTT AT CAMBRIDGE^ . BY THOMAS ^ ANDREWS. ■ » 1801. Contents of the jfirst tlolume. ARITHMETIC. ILXPLANATION of Characters , , ; . 9 Notation . . ,.". , . . .11 Simple Addition .-,.♦•. 14 Subtraction '. , , ♦ • , .18 Multiplication • • • • « . 20 Division ..♦..,•. 26 Reductioti . " , , , , , , » 36 Compound Addition . , , , , ■ , .42 Subtraction , , , ' , . . 46 ^luitiplicatlon . , . , , * 5^ Pivlsion , ' 4 • , ♦ • SZ DuODECIMALJi * • • < ^ • « • 56 Vulgar Fractions .••••. 59 Redaction of Vulgar Fractions . . . . -63 Addition of do, ....... 72 Subtraction of do. . , , . - . '74 Multiplication of do.. . . ... , » . ibid Division of do. ,, • 75 Decimal Fractions •.*•». 76 Addition of Decimals 78 Subtraction of do. ....... ibid Multiplication of do. . . . . • • .79 Division of do. . . . . . , , 82 It*. Juction of do, g ^ . . . , . 85 Federal Vf tJNTENTS. Federal MoneV , ^ . . ^ * r go Circulating Decimals 93 Reduction of Circulating Decimals .... 94 Addition of do. 99 Subtraction of do ^^^ Multiplication of do ^°^*^ Division of do ^^^ Proportion in general ...••• ^^2 Simple Proportion, or Rule of Three . . • 105 Practice . , . . • • • *• ' ^^^ Tare . ^. ,. .. • . • *'^9 by Decimals . - . . . . 193 Equation co^fTE^'YJ. wx Ec^uatlomof Paym^nis . . . . • by Decimals .... CompcHind Interest * * 4 « , . . V - 'by Decimal .... Annuities at Simple Interest Compound do. ..... Value of an Annuity for ever, at Compound Interest . Value pf an. Annuity in Reversion, at Compound Interest Single Position ...•,... Double d^. ........ Permutation, Combination and Composition of Quantities Miscellaneous Questions ...... PajtC 197 202 2C4 207 210 21^ 216 218 220 223 231 LOGARITHMS. Logarithms ...... Computation of Logarithms « . Description and Use of the Table of Logarithms Multiplication by Logarithms jf'ivision by do. ..... (^nvolution by do. .... Efolution by do. ..... 243 246 250 257 258 260 ALGEBRA. Definitions and Notation Addition Subtraction Multiplication Division . . ^ Fractions Involution Evolution . y^irds . , . Infinite Scries 263 270 275 27^ 2B5 • 294 308 312 316 326 Simple \ Vui CONlTENTl. Simple. Eq«atIons • • , 333 Reduction of two, three, or more Simple Equations to one 339 Quadratic Equations •354 Cubic arid Higher Equations . . • • . 370 GEOMETRY. 4}dinitk)ns -. * • 3^^ Prdblems •. •• ...... 39^ Construction, of the Plane Seals (Prob. L.) . . . 420 EXPLANATION of CHARACTERS* SIGNIFIES eqitalitf: as 20 sKilllngS rr i pound, sig- nifiesj that 20 shillings are equal to one pound. -j- Signifies plusy or addition : as, 4 -}- 2 zr 6. — Signifies minus, or subtraction : as, 6-— 2ZZ4. X Ifito^ signifies multiplication : as, 3 X 2 rz 6, -r- By, or ) ( signifies division : as, 6 -^ 2 zr 3, or 2)6(3. Division may alfo be denoted by placing the dividend I overaline, and the divisor under it: thus^iz:6-r- 2Z13. •" : : •• Signifies arithmetical proportion : thus 2 •• 4 : : 6 •• 8; here the meaning is, that 4—211:8 — 6zi2. ; : : : Signifies geometrical proportion : thus 2:41:3:6, which is to be read, as 2 to 4, fo is 3 to 6. ^ Signifies arithmetical progression* -j^ Signifies continual geometrical proportion, or geometrical progression, p*. Signifies therefore, -~i* Signifies /^^ second power, or square, •^^ Signifies /^^ /i??/Vi power, or fw^^. -^l"* Signifies tf«;j power, V^, or — 1^, Signifies /i?)^ /^r/jr^ rr^/, B 10 EXPLANATION 01 CHARACTERS^ 'S/^j or 'P, Signifies t.be cuh^ V ' °^ I "> Signifies any root. m j'' Signifies any j'cot cf any pGiver, Note. The number, or letter, belonging to the above sign^ of powers and roots, is called the i^idcxy or exponent. A line, or vinculum, drawn over several numbers, sig- nifies, that the numbers under it are to be considered jo'irMy : thus, 20 — 7 + 811:5 j but without the vincu- lum, 20 — y-f-Sizai. f\ hzz: a Y, h "^ the product of a and b. ARITHMETIC. »9v^<^^^^^'\*y^'^^« JlVrITHMETIC is the art of computing by numbers, and has five principal or fundamental rules for its opera- tions ; viz. Notation, Addition, Subtraction, Multiplica** jion, and Division. NOTx\TION.* Notation teacheth how to express any proposed number, cither by words or characters. '* As It is absolutely necessary to have a perfect knowledge of our excellent method of notation, in order to understand the reasoning made use of in the following notes, I shall endeavour to explain it in as clear and concise a manner as possible. First, then, it may be observed, that the characters, by which all numbers are expressed, are these ten ; o, i, 2, 3, 4, 5, 6, 7, 8, 9 ; o is called a cypher, and the rest, or rather all of them, are called figures or digits. The names nnd signification of these characters, and the origin or generation of the nutnbers they stand for, are as follow : o nothing ; i dlie, or a single thing called an unit; i-|-i=:2 tv/o ; 2+1^=3 three; 3 + 1=4 four; 44-1=5 five; 5-f| = 6six; 6+1 = 7 feven J 7 -f- 1 = 8 eight ; 8 + 1 = 9 rine ; and 9+i=ten; v/hich ha? no single characler ; and thus ^y the continual addition of one, all numbers are generated. ?. Beside ft ARITHMETIC. ^To read NUMBERS, To the simple value of each figure join the name of its place, beginning at the left hand and reading toward the right. EXAMPLES- Read the following numbers : 37 3<^79^ iiiooGiir loi 70079 1234567890 1 107 3306677 102030405060708090. 3. Beside the simple value of the figures, as above noted, they have, each, a local value, according to the following law : Viz. In a combination of figures, reckoning from right to left, the figure in the first place represents its primitive simple value ; that in the second place, ten times its simple value ; that in the third place, a hundred times its simple value ; and so on ; the value of the figure in each succeeding pUce being ten times the value of i^ in that immediately preceding it. 3. The names of the places are denominated according to their Order. The first is called the place of units ; the second, tens ; the third, hundreds ; the fourth, thousands ; the fifth, ten thou- sands ; the sixth, hundred thousands ; the seventh, millions ; and so on. Thus in the number 3456789 ; 9 in the first place signi- fies only nine ; 8 in the second place signifies eight tens, or eighty; ^ in the third place is seven hundred ; 6 in the fourth place is six thousand ; 5 in the fifth place is fifty thousand ; 4 in the sixth place is four hundred thousand ; and 3 in the seventh place is three millions ; and the v^hole number is read thus, three millions, four hundred and fifty six thousand, seven hundred and eighty nine. 4. A cypher, though it signifies nothing of itself, yet it occupies a place, and, when set on the right hand of other figures, in- creases their value in the same ten-fold proportion ; thus, 5 sig- nifies only five, but 50 is five terxs or fifty, and 500 is five hundred, Sec. $, For NOTATION. IJ To lurite NUMBERS, RULE. Write down the figures in the same order their vakies jire expressed in, beginning at the left hand, and writing toward the right ; remembering to supply those places of the natural order with cyphers, which are omitted in the question., EXAMPLES* 5. For the more easily reading of large numbers, they are di- vided into periods, and half periods, each half period consisting of three figures; the name oftlie first period being units ; of the second, millions ; of the third, billions ; of the fourth, trillions, &c. Also the first part of any period is so many units of it, and the latter part, so many thousands. 7he following Table contains a summary of the tvhole do8rme. fcnods. (iiuadril. Trill. Billions. Millions. Units. V-'V^ Co/^J ^^^>rsJ ^^^\r>J V-^^/^J Half Per. Figures. th. un. th. un. th. un. th. un. cxt cxu Uv^t^/O V.n^(.,vO UvOU-yO V.n^oUv^ \^v^'^.^ Ii:!,4?6 l^'>,or. t^ i. Add up tl^e figures in the row of units, and find how rnany tens are contained in their sum. 3. Set down the remainder, and carry as many units to the next row, as there are tens ; with which proceed as before ; and so on till the whole is finished. Method hers, and carrying for the tens ; both which are evident from the nature of notation : for any other disposition of tlie numbers would entirely alter their value ; and carrying one for every ten, from an inferior line to a superior, is evidently right, since an un^t in the latter case is of the fame value as ten in the former. Beside the method here given, there is another very Ingenious one of proving ad^dition by casting out the nines, thus : Rule r. Add the figures in the uppermost line together, and £nd how many nines are contained in their sum. 2. Reject the nines, and set down the remainder directly even with the £gures in the line. , 3. Do the same with each of the given numbers, and set^alF these excesses of nine togetlier in a row, and find their sum ; then if the excess of nines in this sum, found as before, is equal to the excess of nines in the total sum, the question Is right. EX-AMPLE. 3782 5766 ^755 18303 .2 6 This method depends upon a property of the number 9, which belongs to no other digit whatever, except 3 ; viz. that any num- bcx, divided by 9, will leave the same remainder as the sum of its figures or digits divided by 9 ; which may be thus demonstrated. Demon* l6 ARITHMETIC^ . Method of Proof. 1. Draw a line below the uppermost number, and sup- pose it cut off. 2. Add all the rest together, and set their sum under the number to be proved. 3. Add Demon. Let there be any number, as 3467 ; this separated into its several pans becomes 3000+ 400 -|- 60 -j- 7 ; but 3000=3 XiocGr=3X999-i-i=5X 999+3. In like manner 400=4 X $9+4, and 60=6x9+6. Therefore 34<57=3 X99.9+3+4X 99+4+6x9+6+7=3X999+4X99+6x9 + 3 + 4+6+7. And^ = 3ii999+4X99 + 6X9 3+4+6+2: j^^ 9 9 9 999 + 4X99 + 6x9 is evidently divisible by 9 ; therefore 3467 tiivided by 9 will leave the same remainder as 3+4+6+7 di- vided by 9 ; and the same vv'ill hold for any Other number what- ever. Q^E. D. The same may be demonstrated universally thus : Demon. Let iV= any number whatever, a, h^ r, &c. the digits of which it is composed, and«=r as many cyphers as j, the highest digit, is places from unity. Then N-=ia with «, o's -f-^ with K — i,o's+rwIth« — 2, o^s, &c. by the nature of notation ; =^X«— I, 9's+fl+^X«--2, 9's+^-f rxn— 3, 9*s + r, &c. =5X«— i,9's+^X«— 2,9's+cX«— 3, 9*s, &c. J^a-i^h-^-Cy Zee. hutaxn—i) 9's+^X« — 2, 9's+^x« — 3, 9's,&c. is plain- ly divisible by 9 ; therefore N divided by 9 will leave the same remainder, as ^+^+f, &c. divided by 9. Q;_E. D. In the very same manner, this property may be shown to belong to the number three ; but the preference is usually given to the number 9, on account of its being more convenient in practice. Now from the demonstration here given, the reason of the rule itself is evident ; for the excess of nines in two or more numbers being taken separately, and the excess of nines taken also out of the SIMPLE ADDITION; t^ 3. Add this last found number and the uppermost line together^ and if their sum be the same as that found by the first addition, the sum is right. EXAMPLES. 234S<^ (2) 2234i (3) 3457S 78901 2345^ 78901 2345<5 78901 67890 8752 340 350 78 3750 87 328 17 327 307071 Sum, .99755 Sum; 39087 Sumi 283615 77410 4509 307071 Proofi 99755 Proof. 39087 Proof. 4* Add 8635, 2194, 7421, 5063, 21963 and 1245 to- gether. Ans. 26754. 5. Add the sUm of the former excesses, it is plairi this last excess must be equal to the excess of nines contained in the total sam of all these numbers ; the parts being equal to the whole. This rule was first given by Dr. Wall is, in his Arithmetic, published A. D. 1657, and is a very simple easy method ; thouaii it is liable to this inconvenience, that a wrong operation may some- times appear to be right ; for if we change the places of any two figures in the sum* it will Still be the same ; but then a true sum will always appear to^ be true by this proof ; and to make a false one a^jpear true, there must be at least two errors, and these op- posite to each other ; and if there be more than two errors, they must balance among themselves : but the chance against this particular circurast;mce is so great, that wc may pretty safely Vu€t to this proQf. c i& ARITHMETIC. 5- Add 246034, 298765, 4732I; 58653, 64218, 5376, 9S2 1, and 340 together. Ans. 730528. 6. Add 562163, 21964, 56321, 18536, 4340, 279, and 83 together. Ans. 663686. 7. How many shillings are there in a crown, a guinea, a moidore, and a six and thirty ? Ans. 89. 8. How many days are there in the twelve calendar months .^ Ans. 365. 9. How many days are there from the 19th day of April, 1774, to the 27th day of November, ^775, both days excliw sive ? Ans. 586, SIMPLE SUBTRACTION. Simple Suhtraclion teacheth to take a less number from a greater of the same denomination, and thereby shews the difference or remainder. The less number, or that which is to be subtracted, is called the subtrahend ; the other, the mmiiend ; and the number that is found by the operation, the remamdir or difference. RULE.* I. Place the less number under the greater, so that unit» may stand under units, tens under tens, &c. and draw a line under them, 2. Begin * Demon, i. When all the figures of the les:: number are less than their correspondent figures in the greater, the diiferencc of the figures in the several like places must altogether make the true difference songht ; because as the sum of the parts is equal to the whole, so must the sum of the differences of all the similar parts be equal to the difference of tlie whole* 2. When SIMPLE SUBTE ACTION. »9 2. Begin at the right hand, and iake each figure in tlie Jovver line from the iigure above it, an4 $et down the re- ^nainder. 3. If the lower figure is greater than that ahove it, add ten to the upper figure ; from which figure, so increased, take the lower, and set down the remainder, carrying one to the next lower figure -, with which proceed as before, ;inJ so on till the whole is finished* Mahod of Proop. Add the remainder to the less numher, ar4 if tfi^ suni is equal to the greater, the work is right. EXAUPLES. From 3287625 Take 2343756 From^ 5327467 Take IC08438 From 1231567 Take 345^7^ Reraaind. 943869 Remain. 4319029 Proof 3287625 Proof 5327467 Remain. 888889 Froof ^ 1234567 4. From 2. When any figure of the greater number is less than its cor- respondent figure in the less, the ten, which is added by the rule, is the value of an unit in the next higher place, by the nature of notation ; and the one that is added to the next place of the less number is to diminish the correspondent place of the greater ac- cordingly ; which is only takiing from one place and adding as much to another, whereby the total is never changed, And by this nieans the greater number is resolved into such parts, as are each greater than, or equal to, the similar parts of the less : and tlie difference of the corresponding figures, taken together, will evidently make up the difference of the whole. Q^E. D. The truth of the method of-proof is evident : for the diffei:ence of two numbers, added to the less, is manifeilly c^ual to the (greater. i© ARITHMETIC. 4. From 2637804 take 2376982. Ans. 260822. 5. From 3762162 take 826541. Ans. 2935621* 6. From 78213606 take 278218^0. Ans. 50391716. 7. The Arabian method of notation was first known in England about *the year 1150: how long was it thence to the year 1776 ? Ans. 626 years. 8. Sir Isaac Newton was born in the year 1642, and died in 15727 : how old was he at the time of his decease ? Ans. 85 years. SIMPLE MULTIPLICATION. Simple Multiplication is a compendious method of ad- dition, and teacheth to find the amount of any given numbjsr of one denomination, by repeating it any propos- ed number of times. * The number to be multiplied is called the niultiplicand* The number you multiply by is called the multiplier^ The number found from the operation is called the prodiicl, ^ Both the multiplier and multiplicand are, in gcnera!> called terms or factors, MULTIP^^ICATION AND DIVISION TABL?. i|a|3|4|5|6) 7-1 8 9 | lo | ii | 12 2| 4J 6| 8 1 10 I iz 1 14 1 16 i8 1 20 1 az 1 24 3 1 6[ 9 1 12 1 15 1 18 1 ai 1 24 1 a/ 1 30 i 33 1 36 4 1 8 1 12 j 16 1 20 1 24 1 '-48. 1 32 1 56 1 40 1 44 1 48 5 1 10 1 15 j 20 1 25 1 30 1 IS 1 40 [ 45 f 50 [ J J 1 60 6| 12 1 18 |24 1 30 1 36I 41 I48J 54! 60 1 66 1 7^! 1 7 1 14 1 it 1 28 j ;^.5 f 4i ! 49 1 .S6 j 63 j 70 | 77 | 84; 1 8 1 16 1 24 1 3a i 40 1 48 1 56 1 64 1 li\ 80 1 88 1 961 1 9 i t8 1 27 } 36 1 45 1 54 J 63 1 7^ 1 8H 90I 99 1 xc8 Vic if 20 j 50 j 4Cl4'50 1 60 |'7o 1 80 1 90 1 j;oo j no ( 120 ill \^^ 1 33 U4 I :^S \ 66 | 77 ! ^8 | 99 j lio j 121 | 132J |i2 I 24 ! 3<^ 1 48 1 60 ! 72 1 !?4 i 96 1 Tc8 I 120 1 132 1 144, UhH SIMPLE ?.^ULTIPLICATI0N, 21 Use of the Table in MULTIPLICATION, Find die multiplier in the left-hand column, and the multiplicand in the uppermost line ; and the product is in the common angle of meeting, or against the multiplier, and under the multiplicand. Use of the Table h DiFlSION'. Find the divisor in the left-hand column, and the div^i^ dend in the same line ; then the quotient will be ever the dividend, at the top of the column. R U L E.* I. Place the multiplier under the multiplicand, so that units may stand under units, tens under tens^ &c. and draw a line under them, 2. Begin * Demon, i. When the multiplier is a single digit, it is plain that we find the product ; for by multiplying every figure, that Is, €very part of the multiplicand, we multiply the whole ; and writing dov/n the proid-ucts that are less than ten, or the excess of tens, in the places of the iigures multiplied, and carrying the num- ber of tens to the product of the next place, is only gathering to- gether the similar parts of the respective products, and is, there- fore, the same thing, in effect, as though we wrote down the multiplicand as often as the rriultiplier expresses, and added them together : for the sum of every column ia the product of the ^gures in t|ie place of that column ; and these products, collect- ed together, are evidently equal to the whole required product. 2. If the multiplier is a number made up of more than one digit. After we have Ifound the product of the multiplicand by the first figure of the multiplier, as above, we suppose the multi- plier divided into parts, and find, after the same manner, the })ro- duct of the multiplicand by the second figure of the multiplier ; but as the figure we are multiplying by stands in the place of". tens ,S^^ V 2Z AUITIIMETIC, 2. Begin at the right hand, and multiply the whole mul- tiplicand severally by each figure in the m\iltipiier, setting, down the first figure of every line directly un^er the Eg- ure tens ; the pro4uct roast hz ten times its simple value ; and therC:^ fore the first figare of this product must be placed in the place of tens ; or, which is the same thing, directly under the fig- ure we are multiplying by. And proceeding in this manner sepa- rately with all the figures of the multiplier, it is evident that we shall multiply all the parts of tlie miiltiphcand by all the parts of the multiplier ; or the whole of the multiplicand by the whole of the multiplier ; therefore these several products being added to- gether will be eq^al to the whole required prodoct. Q^E. D, The reason of the method of proof depends upon this propoe sition, " that if two numbers are to be multipUed together, either of them may be made the multiplier, or the multiplicand, and the product will be the same.*' A small attention to the nature of numbers will make this truth evident : for 3X7=^2irr7X3 i and in general 3 X 4 X 5 X 6, &c. =4 XSX6xSi &c. without any regard to the order of tlje tern^ : and this is true of any number of factors v/hatever. The following examples arc subjoiaed to make the reas(>n of she rule appe-ar as ylzisi. as possible. (I) ' (2) 375^5 I3754.3X 5 4567 » ^ 25 zr S"^^ 9628045; = 7. tiiEcstUcinul* 30 = 60X5 8252610* = 6otimcs dp. 7,5 = 500X5 ^^^'17^75 = 500''imes dcr. 35 nr 7000X5 5501740 =4000 times da. 15 = 50000X5 187825 = 37565X5 628x611545 ~ 4567 tuiic* do. Beside die preceding method of proof, th.ere is> another very convenient and easy one by the help of that peculiar property of the number 9, mentioned in addition ; which is performed thus :, RVLE SiMPtt MULTIPLICATION. %% uie you are multiplying by, and carrying for the tens, a.'j in addition^ 3. Add all the lines together^ and their sum is the product. Method Rule i. Cast the nines out of the two factx>rs, as in addition^ and set down the remainder. 2. Multiply the two remainders together, and if the excess of nines in their product is equal to the excess of nines m the total product, the answer is right. EXAMPLE. 42T5 3=:exces3of 9's in the multiplicand. 878 5s=ditto in the multiplier. 33720 29505 33720 3 70G770 6:±ditto ill the product srescess of 9's in 3 X Demonstration OFTHE Rule. Let M and /^T be the number of 9's in the faciors to be multiplied, and a and h "^^vhat remains , then M^a and N^h vnW be the numbers themselves, and their product is Mx iV-f- M-^b^ NXa-^-axh-, but tlie three first of these products are each a precise number of 9's, because one of their factors is so : therefore, these being cast away, there re- mains only <2 X ^ ; and if the 9's are also cast out of this, the excess is the excess of 9's in the total product ; but a and h are the excesses in the fectors themselves, and i^X^ their product j therefore the rule is tree. (^ E. D. This method is Eable to the same incon?enIence with that in addition. Multiplication- may also, very naturally, be proved by division ; for the product being, divided by either of the factors, v.'ili cvi* dendy give the otiier ; but it would have been contrary to good method to have given this rule in the text, becauscT the pupii i*$ supposed, as yet, to be unactjuainted with diyifion. 24 ARITHMETIC. Method of PrOOS. Make the former multiplicand the multiplier, and the multiplier the multiplicand, and proceed as before j and if this product is equal to the former, the product is right. EXAMPLES. (I) (2) Multiply 23456787454 Multiply 32745654473 by 7 by 234 164197512178 Producti 130982617892 . u-*--*- 98236963419 65491308946 Product 7662483146682 3. Multiply 32745(^75474 by 2. Ans. 65491350948- 4. Multiply 84356745674 by 5. Ans. 421783728370. ^. Multiply 3274656461 by 12. Ans. 39295877532. 6. Multiply 273580961 by 23. Ans. 6292362103. 7. Multiply 82164973 by 3027. Ans. 2487 1337327 1. 8. Multiply 8496427 by 874359. Ans. 7428927415293. CONTRJCriONS. I. Whn there are cyphers to the right hand of one cr both the nufnbers to be multiplied* RULE. Proceed as before, neglecting the cyphers, and to the tight hand of the product place as many cyphers as arc in both the number*^ EXAMPLES. SIMPLE MULTIPLICATION. 2$ EXAMPLES, ?. Multiply 1234500 by 7500. 12345 75: 61725 S64I5 9258750000 the iProduct a. Multiply 461200 by 72000. Ans. 33206400000. .3. Multiply 815036000 by 703O0. Ans. 57297030800000. il. Whm the multiplier is the product of two or more tium- hers in the table. RULE.* Multiply continualiy by those parts, instead ^ the whole number at once. £XAMPL£S, I. Multiply 123456789 by 25. 123456789 5 617283945 3086419725 the Product* 2. Multiply * The reason of this method is obvious ; for any number mul- tiplied by the component parts of another number must give the same product, as though it were maltiplied by that number at once ; thus in example the second, 7 times the product of 8? multiplied into the given number, makes 56 times that givea number, as plainly as 7 times 8 jnakes 56, D 26 ARITHMETIC. 2. Multiply 364111 by 56. .Ans. 20390216. 3. Multiply 7128368 by 96. Ans. 68432332B. 4. jMultlply 123456789 by 1440. Ans. 17777777616c. SIMPLE DIVISION. Siifipk Dkision tcachetli to find how often one nu-m- ber is .contained in another of the same denomination, and tlicreby performs the work of many subtractions. Tlie number to be divided Is called the dividend. Tlie number you divide by is called the divisor. The number of times the dividend contains the divisoy is called the quotient. If the dividend contains the divisor any number of times, and some part or parts over, those parts are called the remaindej\ RULE.* I. On the right and left of the dividend, draw a curved line, and write the divisor on the left .hand, and the quotient, as it arises, on the right. 2. Find * According to the rule, we resolve the dividend into parts, and find, by trial, the number of times the divisor is contained in each of those parts ; the only thing then, which remains to be proved, i?, that the several figures of the quotient, taken as one number, according to the order in which they are placed, is the true quotient of the whole dividend by the divisor ; which may be thus demonstrated : Demon. The complete value of the first part of the dividend, is, by the nature of notation, 10, 100, or looo, &c. times the value SIMPLE DIVISION. • 27 2. Find how many times the divisor may be had in as many figures of the dividend, as are just necessary, and write the number in the quotient. 3. Multiply the divisor by the quotient figure, and set the product under that part of the dividend used. 4. Subtract value of which it is taken in the operation ; according as there are i, 2, or 3, 8cc. figures standing before it ; and consequently the true value of tiie quotient figure, belonging to that part of the dividend, is also 10, 100, or 1000, Sec. times its simple value. ' But the true value of the quotient figure, belonging to that part of the dividend, found by the rule, is also 10, ioo» or 1000, &c, times its simple value : for there are as many figures set before it, as the number of remaining figures in the dividend. Therefore this first quotient figure, taken in its com- plete value, from the place it stands in, is the true quotient of the divisor in the complete value of the first part of the divi- dend. For the same reason, all the rest of the figures of the quotient, taken according to their places, are each the true quo- tient of the divisor, in the complete value of the several parts of the dividend, belonging to each ; because, as the first figure on the right hand of each succeeding part of the dividend has a less number of figures, by one standing before it, so ought their quodents to have 5 and so they are actually ordered : conse- quently, taking all the quotient figures in order as they are placed by the rule,, they make one number, which is equal to the sum of the true quotients of all the several parts of the dividend ; and is, therefore,, the true quotient of the whole dividend by the divisor, Q:_ E. D. To leave no obscurity in this demonstration, I sliall illustrate it by an exarnple. liXAMPLB. 28 ARITHMETIC. 4. Subtract the last found product from "that part of the dividend, under wliich it stands, and to the right hand of the remainder bring down the next figure of the EXAMPLE. Divisor 36)85609 Dividendw isl part of the dividend 850QO 36 X 2000=: 72000 20CO the I St quotient,, 1st remainder - 13000 add 6co •Zd part of the dividend 13600 36 X 300= 10800 - - 500 the 2d quotients 2d remainder - 2S00 add 00 3d part of th€ dividend 2800 36 X 70= 2520 - - 70- the 3d quotier% 3d remainder - 280 add 9 4th pai't of the dividend 289 36x8= 288 - - 8 the 4th quotient^ Last remainder ~ 1 2^178 sum of the quotients, • J ', or the anfw t^r. Explanation. It is evident, that the dividend is resolved into these parts, 85000^-600-1-004-9 ;• for the first part of the: dividend is considered only as S^y but yet it is truly 85000 ; and therefore its quotient, instead of 2, is 2000, and the remainder 13000 ; and so of the rest, as may be seen in the operation. When there is no remainder to a division, the quotient is the absolute and perfect ansvyer to the question ; but u'bere there is a remainder, it may be observed, that it goes so much toward another time, as it approaclies to the divisor : thus, if the remain- der be a fpvirth part of the divisor, it will go one fourth of a time more ; if half the divisor, it will go half of a time more ; and so on. SIMPLE DIVISION. ^^ the dividend j which number divide as before 5 and sq on, till the whole is finished. Method on. In order, therefore, to complete the quotient, put the last remainder at the end of it, nbove a small line, and the divisor be<» low it. It is sometimes difficult to find how often the divisor may be had in the numbers of the several steps of the operation ; the best way will be to find how often the first figure of the divisor may be had in the first, gr two first, figures of the dividend, and the answer made less by one or two is generally the figure \yanted : beside, if after subtracting the product of the divisor and quo-, tient from the dividend, the remainder be equal to, or excee4 the divisor, the quotient figure must be increased accordingly. If, when you have brought down a figure to the remainder, it is still less than the divisor, a cypher must be put in the quotient, ^nd another figure brought down, and then proceed as before. The reason of tbe method of proof is plain : for since the quo- tient is the number of times the dividend contains the divisor, the product of the quotient and divisor must evidently be equal to th^ dividend. There are several other methods made use of to prove division ; the best and most useful arc these following. Rule I. Subtract the remainder from the dividend, and divide this number by the quotient, and the quotient found by this divis- ion will be equal to the former divisor, when the work is right. The reason of this r^lc is plain from what has been observed above. Mr. Malcolm, in his Arithmetic, has been drawn into a mistake concerning this method of proof, by making use of particular num- bers, instead of a general demonstration. He. says, the dividend being divided by the integral quotient, the quotient of this du ision will be equal to the former divisor, with the same remainder.— This is true in some particular cases ; but it will not hold, v/hen the 30 ARITHMETIC Method of Proof. Multiply the quotient by the dmsor, and this product, added to the remainder, will be equal to the dividend, V'hen the work is right. (0 (2) 5)^3545728(2705)1451 3<^5)i2345^789(3382g7 10 1095 35 1395 35 ^"=^9$ 45 3oo<^ 45 2920 7 867 5 730 22 137& 20 "^^9^ 2839 2555 284 3. Divide- tht remainder is greater than the quotient, as may be easily demon- strated ; but one instance will be sufficient ; thus 17, divided by 6, gives the integral quotient 2, and remainder 5 ; but 17, di-. vidcd by 2, gives the integral quotient 8, and remainder i. This shews how cautious we ought to be in deducing general rules from, particular examples. Rule II. Add the remainder, and all the products of the several quotient figures, by the divisor, together, according to the order, in which they stand in the work, and tlie sum will be equal to the dividend, when the work is right. The SIMPLE DIVISION. 3t ;;. Divide 3756789275474 by 2. Ans. 1878394637737. 4. Divide 12345678900 by 7. Ans. 1763668414^. 5. Divide The reason of this rule is extremely obvious : for the num- bers, that are to be added, are the products of the divisor by eve- ry figure of the quotient separately, and each possesses, by its place, its complete Value ; therefore, the sum of the parts, together with the remainder, must be equal to the 'whole. Rule III. Subtract the remainder from the dividend, and what remains will be equal to the product of the divisor and quo- tient ; which may be proved by casting out the nines, as was done in multiplication. This rule has been already demonstrated in multiplication. To avoid obscurity, I shall give an example, proved according to all the different methods. EXAMPLE. 57)123456789(1419043 123456789 87* ■ 87 48 364 9933301 1419043) 123456741(87^0"! 348* 1 1352344 1 1352344 48 • 165 9933301 ..87* 123456789 Proof by Mult. 993330I ..786 ..783^ nines. . . . . 378 Proof ly cafi'ing out the .... 348* 4 is the excess of 9's in the quotient. 6 ditto - . - - in the divisor. 309 6 ditto in 4X6, which 261* is also the excess of 9's in (12345674 1 ) the dividend made less by the remainder. 48* 123456789 Proof by Add: iLion. For illustration, we need only refer to the example j except for the proof by addition ; where it may be remarked, that the asterisms shew the numbers to be added, and the dotted Jines 'lieir order. / 3^ ARITHMETIC^ 5. Divide 9876543210 by 8i Aus. 12345679014* 6. Divide 13519153^3^ 9- -^ns. 150886145I.. 7. Divide32i 7684329765 by 17. Ans. i892755488o9f^. 8. Divide 3211473 by 27. Ans. 118943-1^. 9. Divide 1406373 by 108. Ans. 13021^^!-. 10. Divide 293839455936 by 8405. Ans. 34960078-5^'*-^. 11. Divide 4637064283 by 57606. Ans. 80496^ 76 ol* CONTRACTIONS. L To divide hy any nuihher with cyphers amiex^d^ RULE.* Cut ofF the cyphers from the divisor, and the same number of digits from the right hand of the dividend ; then divide, making use of the remaining figures, as usual, and the quotient is the answer ; and what remains, writ-r ten before the figures cut off, is the true remainder. EXAMPLES, 1. Divide 310869017 by 7100. •7i,oo)3io869o,i7{4378444^| the quotient* 284 268 213 497 599 568 310 284. 2617 2. Divide * The reason of this contraction Is easy to cojicclve : for the cutting 9ff the same figures from eacli, is the same as dividing each. of SIMPLE DIVISION. 33 1. Divide 7380964 by 23000. Ans. 320^^-1. 3. Divide 29628754963 by 35000. Ans. 846535- 20965 il. When the divisor is the product of two or more small num- hers in the table, RULE.* Divide continually by those numbers, Instead of the whole divisor at once* EXAMPLES* of them by 10, 100, 1000, &c. and it is evident, that as often as the whole divisor is contained in the whole dividend, so often must any part of the divisor be contained in a like part of the dividend. This method is only to avoid a needless repetition of cyphers, which v/ould happen in the common way, as may be seen by working an example at large. * This follows from contraction the second In multiplication, of which it is only the converse; for the third part of the half of any thing is evidently the same as the sixth part of the whole ; and so of any other number. I have omitted saying any thing, in the rule, about the method of finding the true remainder ; for as the learn-v is fupposed, at present, to be unacquainted with the nature of frac- tions, it would be improper to introduce them In this part of the work, especially as the integral quotient is sufEcIcnt to answer most of the purposes of practical division. However, as the quotient is incomplete without this remainder, and, in some computations, it is necessary it should be known, I shall here shew the manner of fmd- ingit, without any assistance from fractions. Rule. Multiply the quotient by the divisor, and subtract the product from the dividend, and the result will be the true re- mainder. The truth of this is extremely obvious ; for if the product of the divisor and quotient, added to the remainder, -be equal to the dividend, their product taken from the dividend must leave the remainder. Xh'; N 34 ARITHMETIC. EXAMPLES. I. Divide 31046835 by 56^:7X8. 7)31046835(4435262 S)4435262(554407 the quotknS, 28 40 28 40 24 35 21 32 36 . 32 35 ^^ 18 62 * 14 56 43 6 -' 42 15 14 2. Divide •- , .. .. . ■■■ ., — ^- - - The rule which is most commonly made use of is this : Rule. Mukiply the last remainder by the preceding divisor, or last but one, and to the product add the preceding remainder ; multiply this sum by the next preceding divisor, and to the pro- duct add the next preceding remainder ; and so on, till you have gone through ail the divisors and remainders to the first. EXAMPLE. 9)64865 divided by 144. i the last remainder. ' Mult. 4 the preceding divisor. 4)7207 2 4 4)1801 3 Add 3 the second remainder. 450 I 7 Mult. 9 the first divisor. Add 2 the first remainder. Ans. 450 &' 65 To SIMPLE DIVISION. 35 J. Divide 7014596 by 72ZZ8X9. 8)7014596 9)876824 4 97424 8 the quotient. 3. Divide 5130652 by 132. Ans. ^^^68^^, 4. Divide 83016572 by 240. Ans. 345902^-/^. III. 21? perform division more concisely than hy the general rule. RULE.* Multiply the divisor by the quotient figures as before, and subtract each figure of the product as. you produce it, always remembering to carry as many to the next figure as were borrowed before. EXAMPLES. X. Divide 3104675846 by 833. ^33)3io4675S46(3727ioi-B-TT ^^^ quotient. 6056 2257 5915 848 ^546 713 2. Divide To explain tliis rule from the example, we may observe, that every unit of the first quotient may be looked upon as containing 9 of the units in the given dividend ; consequently every unit, that remains, will contain the same ; therefore this remainder must be multiplied by g, in order to find the units it contains of the given dividend. Again, every unit in the next quotient will con- tain 4 of the preceding ones, or 36 of the first, that is, 9 times 4 ; therefore what remains must be multiphed by 36 ; or, which is the same thing, by 9 and 4 continually. Now this is the same as the rule ; for instead of finding the remainders separately, they are re- duced from the bottom upward, step by step, to one another, and the remaining units of the same class taken in as they occur. * The reason of this rule is the same as that of the ^neral rule. 3^ ARITHMETIC* 2. Divide 29137062 by 5317. Ans. 5479|f?4, 3. Divide 62015735 by 7803. Ans. 7947t¥-o-T^ 4. Divide 432756284563574 by 873469. Ans. 495445498|-I^-l|, REDUCTION. TABLES OF COIN, WEIGHT, and MEASURE. MONEY. 4 farthings make i penny 12 pence i shilling ao shillings i pound. fors d denotes pounds shillings pence. J is one farthing, or one quarter of any thing. ~ a half-penny, or a half of any thing. \ 3 farthings, or 3 quarters of any thing. PENCE TABLE. 20 30 40 50 60 70 80 90 100 110 120 is s, I 2 3 4 2 5 5 10 6 8 6 4 2 7 8 9 10 d, 12 24 36 48 60 72 84 96 108 I20 is I 2 3 4 5 6 7 8 9 10 TROY WEIGHT. 24 grains make i penny-weight, marked grs. dwt. 20 dvvt. I ounce, oz. 12 oz. I pound, lb or lb. By. this weight are weighed jewels, gold, silver, corn, bread, and liquors. • APOTHECARIES' REDUCTION. 37 APOTHECARIES' WEIGHT. 20 grains make i scruple, marked gr. (c. or 9. 3 fc. or 9 I dram dr. or 5. 8 dr. I ounce oz. or ^. 12 oz. I pound tb or lb. Apothecaries use this weight In compounding their med- icines ; but they buy and sell their drugs by Avoirdupois weight. Apothecaries' is the same as Troy weight, having only some different divisons. AVOIRDUPOIS WEIGHT. 16 drams make i ounce, marked dr. oz. 16 ounces i pound lb. 28 lb. I quarter qr. 4 quarters i hundred weight cwt. 20 cwt. I ton T. By this weight are weighed all things of a coarse or idrossy nature : such as butter, cheese, flesh, grocery wares, aud all metals, except gold and silver.* DRY lb. * A firkin of butter . is , ^6 A firkin of soap 64 A barrel of pot-ashes ... 200 A barrel of anchovies .... 30 A barrel of candles ....120 A barrel of soap 256 A barrel of butter 224 A fother of lead is 1 9^ cwt. A stone of iron 14 A stone of butcher's meat . . 8 A gallon of train oil ... . 7-%- A faggot of steel 120 A stone of glass 5 A seam of glass is 24 stone, or 120 lb. oz. dr. A peck loaf of bread weighs 17 61 A half peck .... 8 11 A quartern ..... 4 58 make a truss. 56 lbs old hay 1 60 lbs new hay J 36 trusses a load. 4 pecks coal make i bushel. 9 bushels . . . I vat or strike. 36 bushels .... I chaldron. 21 chaldrons . . . i score. 7 lbs wool make 2 cloves .... I clove. 1 stone, 2 stones 38 2 pints make i quart pts.qts. I 8 bushels ARITHMETIC. DRY MEASURE. J\ larked I Marked I quarter qr. 2 quarts i pottle pot. 5 quarters i weytrload wey 2 pottles I gallon gal. 1 4 bushels i coomb co. 2 gallons I peck pe. 5 pecks i bushel water meaf. 4 pecks I busnel bu., 10 coombs i wey 2 bushels i strike str. j 2 weys i last L, Note. — ^The diameter of a Winchester bushel is i8|. inches, and its depth 8 inches. By this measure^ fait, lead, ore, oysters, corn, and other dry goods are meafurcd. ALE AND BEER MEASURE.- Marked 2 pints make i quart pts. qts. 4 quarts i gallon gal, 8 gallons i firkin of Ale fir. 9 gallons I firkin of Beer fir. Marked 2 firkins 1 kilderkin kil, 2 kilderkins i barrel bar. 3 kilderkins i hogshead hhd» 3 barrels i butt butt. Note. — ^The ale gallon contains 282 cubic inches. In London the ale firkin contains 8 gallons, and the beer fir- kin 9 ; other measures being iathe same proportion. WINE 2 stones I tod. 6t tods 1 wey. 2 weys I sack. 2 2 sacks I last. lb. A barrel of pork is .... 220 A barrel of beef 220 A quintal of fish 112 20 things make . . . . i score 12 I dozen. 12 dozen i gross. 144 dozen . . . i greater gross. Further,-^ ^^60 grains rr 1 lb. Troy ; 7000 grains = i lb. A- voirdupois ; therefore the weight of the pound Troy is to that of the pound Avoirdupois, as 5760 to 7000, or as 144 to 175. REDUCTION. 3^ WINE MEASURE. Marked ^ pJnts make i quart pts. qts. 4 quarts i gallon gal. 42 gallons I tierce tier. 63 gallons I hogshead hhd. 84 gallons I puncheon pun. Marked 2 hogsheads i pipe or butt p. or\i, 2 pipes I tun T. 18 gallons I rundlet rund. 31^ gallons I barrel bar. By this measure, brandies, spirits, perry, cider, mead, vinegar, and oil are measured. Note. — -231 solid inches make a gallon, and 10 gallons make an anchor. MEASURE. %\ inches make i nail 4 4 nails 4 quarters x: L O T H Marked nis. qrs. yds. LONG MEASURE. I quarter I yard 3 qrs. 5 qrs. 6 qrs. Markeft I ell Flemish EllFL I ell English EllEng. I ell French EllFr. Marked 3 barley corns make i inch bar.c. in. 1 2 Inches i foot ft. 3 feet I yard yd. 6 feet I fathom fath. 5^- yards i pole pol. 40 poles I furlong fur. 8 furlongs i mile mis. 3 miles I league 1. Marked 6h geographical miles, (?r 69-^ statute miles i de- gree A^g. or ^ 3 60 degrees the circum- ference of the earth. Note — 4 inches make i hand. 5 feet I geometrical pace. 6 points I line. 12 lines I inch. TIME. Marked 60 seconds make i min- ute 60 minutes 24, hours 7 days s.or" m,or I hour h. or ^ I day d. I week w- Marked 4 weeks i month ra, 1 3 months, I day, and 6 hours, or 365 days and 6 hours, i Julian year Y. Note- 40 ARITHMETIC. Note i. The second maybe supposed to be divided into 60 thirds, and these again into 60 fourths, &:c. Note 2. April, June, September, and November, have each 30 days ; each of the other months has 31, except February, which has 28 in common years, and 29 in leap years. CIRCULAR MOTION. 60 seconds make i minute, marked " 60 minutes i degree 0 30 degrees i sign 8* 12 si^ns, or 360° i circle^ Reduction is the method of bringing numbers from one name or denomination ta another, so as still to retain the same value. RULE.* I. When the reduction is from a greater name to a lesu Multiply the highest name or denomination by as many as make one of the next less, adding to the product the parts of the second name ; then multiply this sum by as many as make one of the next less name, adding to the product the parts of the third name j and so on, through all the denominations to the last. II. men * The reason of this rule is exceedingly obvious ; for pounds are brought into shillings by multiplying them by 20 ; shillings in- to pence by multiplying them by 12 ; and pence into farthings by multiplying them by 4 ; and tl.e contrary by division : and this will be true in the reduction of numbers consisun;> of any denoni* .inatioa whatever. REDUCTIOK. 4i ll. When the reduction is from a less name to a greater. Divide the given nmiiber by as many as make one of the next superior denomination ; and this quotient again by as many as make one of the next following ; and so on, through all the denominations to the highest ; and this last quotient, together with the several remainders, will be the answer required. The method of proof is by reversing the question; EXAMPLES. I. In 1465I. 14s. 5d. how many farthings .^ 20 4)1407092 29314 ^2)351773 12 2,0)2931,4 s 35^113 4 Proof 1465I. 14s. 5d. 1407092 the answer. ^2. In 12I. how many farthings? Ans. 11520^ 3. In 6169 pence how tnany pounds ? Ans. 25I. 14s. id. 4. In 35 guineas how many farthings sterling ? Am, 35280. 5. In 420 quarter guineas how many moidores ? Ans. 81 and i8s. 6. In 23 ll. i6s. how many ducats at 4s. 9d. each ? Ans. 976. 7. In 274 marks, each 13s. 4d. nnd 87 nobles, each 6s. 8d. how many pounds ? Ans. 21 il. 13s. 4d. 8. In 1 776 quarter guineas how many six-pences sterling ? Ans. 18648. 9. Reduce 1776 six and thirties to half-crowns sterling. Ans. 25574f. 10. In 50807 moidores how many pieces of coin, each 4s. 6d, ? Ans. 304-842. ^ II. Itt 42 AftlTtiMliTlC. 11. In 2 1 32 10 grains how many lb. ? Ans. 371b. 9grs. 12. In 591b. I3dwts. 5gr. how many grains ? Ans. 34oi57grs, 13. In 8012131 grains how many lb. ? Ans. 13901b. iioz. iSdwts. ipgrs- 14. In 35 ten, lycwt. iqr. 231b. 70Z. I3dr. how many drams ? Ans. 2057ioo5dr. 15. In 37cwt. 2qr. 1 71b. how many pounds Troy, a pound Avoirdupois being equal to 140Z. iidwt. i5Ygrs. Troy ? Ans. 51241b. 50Z. lodwt. ii-|-grse 16. How many barleycorns will reach round the world, supposing it, according to the best calculations, to be 8343 leagues ? Ans. 4755801600. 17. In 17 pieces of cloth, eaeh 27 Flemish ells, how many yards ? Ans. 344yds. iqr. 1 8. How many minutes were there from the birth of Christ to the year 1776, allowing the year to consist of 365d. 5h. 48' 58'' ^ Ans. 934085364' 8'C COMPOUND ADDITION, Compound Addition teacheth to collect several numbers of different denominations into one total. RULE.* I. Place the numbers so that those of the same denom- ination may stand directly under each other, and draw a line below them. 2. Add * The reason of this rule is evident from what has been said in simple addition : &r, in addition of money, as i in the pence is equal COMPOUND ADDITION. 43 1, Add up the figures in the lowest denomination, and find how many ones of the next higher denomination are contained in their sum, 3, Write down the remainder, and carry the ones to the next denomination ; with which proceed as before ; and so on, through all the denominations to the highest, whose sum must be all written down ; and this sum, together with the several remainders, is the total sum required. The method of proof is the same as in simple addition. EXAMPLES. MONEY. £' s. d. C' s. d. I' s. d. 17 13 4 84 75 17 13 Si 4i 175 107 10 13 10 ^3 TO 2 Hi 10 17 3 51 17 B| 89 18 10 8 8 7 20 10 lot 75 12 2i- 3 3 4 17 . 15 4i 3 3 3i 8 8 10 10 ir I i 54 I 4 261 5 8t 45,2 19 2t 3^ 8 I 0 4 176 8 2-1 277 8 4t 54 261 8t 452 19 2t TROY equal to 4 in the farthings ; i in the shillings, to 1 2 in the pence ; and I in the pounds, to 20 in the shillings ; therefore, carrying as directed, is nothing more than providing a method of digesting the money, arising from each column, properly in the scale of de- non^ations ; and this reasoning will hold good in the addition of compound numbers of any denomination whatever. 44 ARITHMETIC. TROY WEIGHT. lb. oz. dwt. ?,^' lb. oz. dwt. gr- lb. OZ. dwt. g^. 17 3 15 11 14 10 13 20 27 10 17 iH 13 2 13 13 13 10 l« 21 17 10 13 13 15 3 14 14 14 10 10 10 13 11 13 I ^^ 10 10 I 2 0 10 I 2 12 1 J? I 4 4 4 4 4 3 3 ^3 14 I 19 2 — — t - — APOTHECARIES* WEIGHT. lb. oz. dr. sc. g- ib. 02. dr. sc. g^- lb. 02. dr. sc. gr. 3 5 7 2 ^7 4 5 6 I 13 5 4 3 I 10 2 7 4 2 18 2 7 5 2 ^7 4 3 2 2 18 I 7 5 I 10 I 6 I 2 7 3 2 I I 17 1 / 5 0 10 3 4 2 £ 4 4 2 I I 4 2 7 3 0 17 2 2 1 2 3 2 IQ 2 6 I I 10 3 I I I I 7 2 2 AVOIRDUPOIS WEIGHT. fwt. (j^r. lb. oz. dr. T. cwt. qr. lb. oz .dr. T. cwt. qr. lb. oz.dr. i:; 2 15 15 15 2 17 3 13 8 7 3 13 2 10 7 7 ^3 '-i 17 13 14 2 ^3 3 14 B 8 2 14 I J 7 6 6 12 2 ^3 14 14 I 16 10 5 4 17 M 6 10 I 17 15 2 13 I 7 2 13 12 7 7 12 I 10 10 I 14 I I 2 2 3 13 10 4 4 10 I 12 ' 7 4 16 I 7 7 5 5 2 12 8 8 — — — J-ONQk CO:»1POUND ADDITION. 45 LONG MEASURE. Mib.fur.pol.yd.ft 37 3 ^4 2 I 28 4 17 3 2 17 4 431 10 5 631 29 2 2 Z 30 4 . in. 5 10 2 7 3 2 Mis. fur.pol. yd.ft. in. 28 2 13 ; I 4 39 I 17 2 2 10 28 I 14 2 2 48 I 17 2 2 7 37 I 29 3 2 20 2 I Ails.fur.po!. yd.ft. 28 3 7 2 30 I 27 6 30 2 2 7 d 20 2 1 5 2 2 7 10 2 in. 7 7 10 2 CLOTH MEASURE. Yd. qr. nl. in. Ell En. qr. nl. in. Ell Fl. qr. nl. in. X20 3 I I 207 2 2 I 200 2 I I •38 2 I 5B 2 2 57 I I 28 2 2 78 I I 28 I I I 38 2 2 21 3 3 2 21 2 28 2 3 20 2 2 3« 3 I 18 3 2 2 3 2 2 "2 WINE MEASURE. T. hhd. gal. qt. pt. T. hhd. gal. qt. pt. T. hhd. g.l. qt. pt. 17 2 10 2 I 27 I 3 I I 37 I 2 I I ID 2 27 2 I 24 ^3 I 27 27 3 £ 8 3 24 2 2 1 3 37 20 2 24 5 2 27 2 10 2 SS I I 20 I 29 2 I 2 I 17 I I 8 2 25 I I 3 39 2 I 3 29 2 I 2 2 35 2 2 37 2 I — - ALE 46 ARITHMETIC. ~ ALE AND BEER MEASURE, hhd. gal. qt. pt. hhd. gal. qt. pt. hlid. gal. qt. pt. 21 2 2 I 27 3 2 I 30 20 3 21 20 3 25 10 » 2 28 29 2 21 21 2 21 13 20 20 ID lO 2 10 17 iS 18 I 3 3 <> J 2 8 4 7 2 2 2 I 17 6 '7 6 I DRY MEASURE. L. qr. ba. pe. gal. , 55231; 32321 22321 12 2 2 2173 562 T I M E. Y. m. w. d. h. m. s. 27 9 2 6 23 25 25 20 7 2 5 20 36 30 18 7 3 4 5 6 7 H I 21 22 23 10 2 4 5 5 8 2 4 3 38 COMPOUND SUBTRACTION. Cjjdp-jiud S;:htracitofi teacheth to find the diiFerence of any two numbers of different denominations. R U L E.* I. Place the less number under the greater, so that those parts, which arc of the same denomination, may stand di- rectly iindsr each other, and draw a line below them. 2. Begin * The reason of this rule will readily appear from what was said ia simple subtraction j for the borrowing depends upon the very same principle, and is only different, as the numbers to be subtracted arc of different denominations. COMPOUND SUBTRACTION. 47 2, Begin at the right hand, and take the number in each denomination of the lower line from the number standing above it, and set down their remainders below them> 3, But if the number below be greater than that above it, increase the upper number by as many as make one of the next higher denommation, and from this sum take the number in the lower line, and set down the remainder as before. 4, Carry the unit borrowed to the next number in the lower line, and subtract as before j and so on, till the whole is finished •, and all the several remainders taken together, as one number, will be the \vhole difference re- quired. The method of proof is the same as in simple sub- traction. EXAMPLES. M 0 N E Y. From Take 176 3. 16 d. 4 6 454 276 s. 14 d. Si £' 274 ^5 s. 14 15 d. 7i Rem. 98 275 16 13 10 4 177 j6 91 188 18 ^\ Proof 454 14 2^ 274 14 Ax ■^4 TROY WEIGHT. lb. From 7 Take 3 oz. dwt. gr. 3 14 II 7 15 20 lb. oz. dwt. gr. 27 2 10 20 20 3 5 21 lb. oz.dwt. gy. 29 3 14 ^ 20 7 15 7 Rem. Proof APOTHECARIES^ 48 AS.ITi^IMETIC. A P O T H E C A R I E S' WEIGHT. lb. oz. dr. sc. gr. lb. oz. dr. sc. gr. lb. oz. dr. sc. gr. From 1147 "^4 2 3 6 I 10 5 i 3 2 19 Take 3 7 i 15 i 8 7 2 12 2251 Rcnii Proof AVOIRDUPOIS WEIGHT. cwt.qr. lb. oz. dr. cwt. qr. lb. oz.dr. cwt. qr. Ib.oz. dr; Frorn 5 1759 2221348 2117613 Take 332117 2011766 13 881 4 Rem. Proof LONG MEASURE. Mis. fur. pol. yd. ft. in. MJs. fur. pol. yd. ft. in. Mis. fur. J5ol. yd. ft.iit. 114 3 17 I a I 70 7 13 I I a 70 3 10 3; :io 7 30 a 10 20 14 2 a 7 17 3 it ^ i 7 Rem. Proof G L O T H M E A S U R E. Yd. From 27 Take 10 qr. nl. 3 3 2 2 Ell En. qr. nl. 127 2 7'^ 3 3 Ell Fl. qr. nl. iri. 270 I I 140 222 Rem. Proof WINE COMPOUND SUBTRACTION. 4^ ■H>. WINE MEASU %E. T. hhd. gal. qt. pt. hhd. gal. qt. pt. hhd. gal. qt. ^rora 2 .3 20 3 I 2 21 2 13 Take i 2 17 3 i 10 27 Rem. Proof ALE AND BEER MEASURE, hhd. fir. gal. qt. pt. hhd. fir. gal. pt. hhd. fir. gal. pt. If'rom 27 2221 29^ 34 27 32 2 Take 10 3 4 3 20 2 4 5 10 3 Rem. Proof DRY MEASURE. L. qr. bu.pe. gal.pot. L. qr. bu. pe. gaL L. qr. bu. pe.gat. From 9 47111 1335 2 I 27 I 2 Take 2 53 7 237 I0 22II Rem. Proof T r M E, m. w. d. h. ' m. w. d. h. ' m. w. d. h. ' From 17251726371 13 171 5 Take 10 18 18 15 2 15 14 17 5 SI Rem. Pooof COMPOUND 50 ARITHMETIC. COMPOUND MULTIPLICATIOR Compound MultipUcatiGti teacheth to find tlie amount of any given number of diiFcrent denominations by repeating: it any proposed number of timesr RULE.* r. Place the multiplier under the lowest denomination ©f the multiplicand. 2. Multiply the number of the lowest denomination by the multiplier, and find how many ones of the next higher denomination are contained in the product.. 3. Write down the excess^ and carry the ones to the product of the next higher denomination, with which pro- ceed as before ; and so on, through all the denominations to the highest, whose product, together v/ith the several excesses, taken as one number, will be the whole amoui^ required. The method of proof is the ssme as in simple multi- plication. EXAMPLES- * The product of a number consisting of several parts, or de- nominations, by any simple number v/hatever, v/ill evidently be expressed by taking the product of that simple number and each part by itself, as so many distinct questions : thus, 25I. 12s. (>&» multiplied by 9 will be 225 1. 1 08s. 54d. = (by taking the shillings from the pence, and the pounds from the shillings, and placing^^ them in the shillings and pounds respectively) 230I. 12s. 66.. which is the same as the rule j and this will be true,, when the multiplicand is any compound number whatev-t;r^ COMPOUND MULTIPLICATION. Cf EXAMPLES OF MONET, ^. ^Ib. of tobacco, at 2s, S^d. per lb. 2S. 8^(1. 9 il. 4s. 4|d. the answer. 2* 3lb. of green tea, at ps. 6d. per lb. Ans. il. 8s. 6d. 3. 51b. of loaf sugar, at is. 3d. per lb. Ans* 61. 3s, 4. pcwt. of cheese, at il. 1 is. 5d. per cwt. Ans. 14I. 2s. pd, ^.12 gallons of brandy, at 9s. 6d. per gallon. Ans. 5I. 14s. CASE I, If the 7nu!i'ipUer exceed 1 2, multiply successively by its component parts, instead of the whole number at once, a§ ill simple multiplication. EXAMPLES, K. i6cwt. of cheese, at il. iSs. 8d. per c\yt, il. 1 8s. M, 4 7 14 8 4 £2^ 18 8 the answer. . ^. 22 yards of broadcloth, at 19s. 4d. per yarcL Ans. 27I. IS. 4dH 3. g6 quarters of rye, at il. 3s. 4d. per quarter, Ans. 112L 4. 120 dozen of candles, at 5s, pd. per doz, Ans, 34I. I OS. 5. 132 yards of Irish cloth, at 2S. 4d. per yard. ■ Ans. 15I. 83, 6. 144 reams of pap$r, at 13s. 4d, per ream. Ans. 96U 7. 1210 5t ARlTHMETfC, 7, 1 2 10 yards of shalloon, at 2s. 2d. per yard. Ans. 131I. IS. 8d. CASE II. If the multipliei' cannot he produced by the midtiplicaiion of small numbersy find the nearest ro it, cither greater or less, which can be so produced ; then multiply by the compo- nent parts as before ; qnd ioj the odd parts, add or sub-? tract according as is required. EXAMPLES. 1. 17 elisof holland, at 7s. 8^d. per ell, 7s. 8id, 4 I JP 10 4 6 3 1 4 £6 II ol the answer, 7. 23 clls of dcv/las, at is. 6~d. per c\L Ans. il. 15s. 5~d^ 3» 46 bushels of wheat, at 4s. 7-^d. per bushel, Ans. lol. us. 9|-d. 4. 59 yards of tabby, at 7s. lod. per yard. Ans. 23I. 2S. 2d. 5. 94 pair of silk stockings, at 12s. 2d. per pair. Ans. 571. 3s. 8d. (5. ii7C'wt. of I\Ialaga raisins, at il. 2S. 3d. per cwt. Ans. 130L 3s. 3d. EXAMPLES OF WEI GMT 3 AND MEASURES, \h. oz.clwt. gr. lb. oz. dr. sc.gr. cwt. qr. lb. oz. mis. fur, pis. yd. 21 I 7 13 243 I 27 I 13 12 24 3 20 2 4 7 12 6 COMPOUND DIVISTOK. 53 Vds. qr. nls. 127 2 2 8 T. hjid. ga!. pt. 29 I 20 3 5 W. qr. bu. pc. 27 I 7 2 M. v.'. d. h. m. 175 3 6 20 59 COMPOUND DIVISION. Compound Divis'ron teacneth to find how often one given number is contained in another of different denomiuation&. RULE.* J. Place the numbers as in simple division. 2. Begin at the left hand, and divide each dcnpminatpn Jjy the divisor, setting the quotients under their respective dividends. 3- But if there he n remainder, after dividing any of the denominations except the least, find how many of the next lower denomination it is equal to, and add it to the num- ber, if any, -which was in this denomination before ; then ^ivide the sum as usual, and so on, till the whole is finished. f he method of proof is the same as in simple division. EXAMPLES * To divide a numl^er consistincr of several denominations, by any simple number whatever, is evidently the same as dividing all the parts or members, of which that number is composed, by the same simple number. And this will be true, when any of the parts are not an exact multiple of the divisor : for by conceiving the number, by which it exceeds that muldplc, to have its proper value by being placed in the next lower denomination, the divi- dend will still be divided into parts, and the true quotient found as J^efore : thus 25I. 12s. 3d. divided by 9, will be the same as 18L 144s. 99d. divided by 9, which is equal to 2h i6s. iid. as b-^ the rule ; and the method of carrying from one denomination to another i? €x^:'>lv the same. [ ARITHMETIC. E3tAMPLES OF MOl^ET^ I. Divide 225I. 2S. 4d. by 2^^ 2)2251. 2s. 4d. 112I. us. 2d. the quotient. 2. Divide 75 Ur 14s. 7^d. by 3. Ans. 250I. lis. 6^A^ 3. Divide 82 iL 17s. p^^d. by 4. Ans. 205U ps. 5^d, 4. Divide 2 81. 2S. i-|-d. by 6. Ans. 4I. 13s. S^^d. 5. Divide 135I. los. 7d. by 9. Ans. 15I. is. 2d. d. Divide 227L los. 5d. by 11. Ans. 20I. 13s. 8d. 7. Divide X332I. us. 8-|-d. by 12. Ans. ml. ii\^* SASE I. Jf the ctiv'isc',' exceed 12, divide continually by It? com- ponent parts, as in simple division. pxAMPLE?, T, What is cTieese per cwt, if i^cwt, cost 30I. i8s, Z^* } 4)301. 1 8s. 8d. 4)7 14 8 £1 18 8 the answer. '2. If 2ocwt. of tobacco comes to 120I. los. what 15 that per cwt. ? Ans. 6\* 6d. 3. Divide 57I. 3s. 7d. by 35. Ans», il. 12s. 8d. 4. Divide 85I. 6s. by 72. Ans. il. 3s. 8^, 5. Divide 3 il. 2s. lo^d. bypp. Ans. 6s. 3-|-d. 6. At 1 81. 1 8s. per cwt. howniuch per lb. ^ Ans. 3s. 4^^ CASE II. ^the divisor cannot he produced by the mult^lication of small numbers^ divide it aft-er the manner of long division. EXAMPLES. COMPOUND Division, ^^ EXAMPLE^. 1. Divide 74!.- 13$. 6d. by 17. 17)74 ^3 <5(4 7 I*- 68 6 20 133 14 12 174 ^7 4 2. Divide 23I. 15s. 7^d. by 37.* AnSa 12s. io-^-g, 3. Divide 315I. 3s. lo-^d. by 365^ Ans. 17s. 3^d., EXAMPLES OF WEIGHTS AND MEASURES^ I* Divide 231b. 70Z. 6dwt. I2gr. by 7* Ans. 31b. 40Z. pdwt. I2giv Ci. Divide 131b. loz, 2dr. logr. by 12. Ans. lib. loz. 2SC. iogi% 3. Divide io6icwt. 2qrs. by 28. Ans. 37cwt* 3qrs. 18IK 4x Divide 3 75mls. 2fur, 7pls. 2yds. ift. 2in, by 39, Ans. pmls. 4fur. 39pls. 2ft. Sin^ 5. Divide 571yds. 2qrs. ml. by 47. Ans. 12yds. 2nly<. 6. Divide 120L. 2qrs. ibu. 2pe. by 74. Ans. iL. 6qrs. ibu. 3peo 7. Divide i2iomo. 2W. 3d. 5h. 20' by iii. Ans, I mo. 2d. loh. 12^ DUODECIMALa _5<5 ARITHMETIC. DUODECIMALS* Duodecimals arc so called because they decrease by twelves, from the place of feet tov/ard the right hand* Inches are sometimes called primes, and are marked thus '5 the next division, after inches, is called parts, or seconds, and Is marked thus '' > the next is thirds, and marked thus ^'^ ; and so on* Duodecimals are commonly Used by workmen and ar-» tlficers in casting up the contents of their work.- MuLTiPLiCArioN of Duodecimals ; or^ Cross Multiplication, RULE. 1. Under the multiplicand write the same names or de- nominations .of the multiplier •, that is, feet under feet, inches under inches, parts under parts, &c. 2. Multiply each term in the multiplicand, beginning at the lowest, by the feet in the multiplier, and write each result under its respective term, observing to carry an unit for every 12, from each lower denomination to its next superior. 3. In the same manner multiply every term in the mul- tiplicand by the inches in the multiplier, and set the re- sult of each term one place removed to the right of those In the multiplicand. 4. Proceed in like manner v.ith the seconds and alt the rest of the denominations, if there be any more ; and the ^m of all the lines will hi the product required. Or ©trODECIMALS. 57 Or the denominations of the particula? products will be as follows : Feet by feet, give feet. Feet by primes, give primes* Feet by seconds, give seconds. &c. Primes by primes, give seconds* Primes by seconds, give thirds. Primes by thirds, give fourths. &c. Seconds by seconds, give fourths. Seconds by thirds, give fifths. Seconds by fourths, give sixths. 3cc. Thirds by thirds, give sixths. Thirds by fourths, give seventhsv Thirds by fifths, give eighths* In general thus : When feet are concerned, the product Is of the same denomination with the term multiplying the feet. When feet are not concerned, the name of the product will be expressed by the sum of the indices of the two fiictors. EXAMPLES. t. Multiply lof. 4' 5'' by 7f. 8' e\ 786 72 611 6 10 x-x 4 5 2 2 (5 79 II 066 Answer- i^j II ^4^ 2. Multiply 4f. 7' by 6L 4' o'\ Ans. 29f. o' ^\ 3. Multiply 5 8 ARITHMETIC* ^ 3c Multiply I4f. 9' by 4L 6\ Alls. 66f. 4' 6''. 4. Multiply 4f. Y 8'^ by pf. 6^ Ans. 44f. o' lo'^ 5. Multiply 7f. 8' &' by lof. 4^ 5''. Ans. 79f. ii'o''6'''6iv. 6. Multiply 39f. 10' 7'' by i8f. 8' 4'^. Ans. 745f. 6' 10'' 2''' 4iv, 7. Multiply 44f. 2' 9'' 2'^' 4iv. by 2f. 10' 3''. Ans. I26f. 2' 10" 8'''^ loiv. 11 v, 8. Multiply 24f. 10'' 8'^ 7'^' 5iv. by pf. 4' 6^'. Ans. 23 3 f. 4' 5'' 9^'^ 6iv. 4V. 6vi^ 9. Required the content of a floor 48f. 6- long, and 24i. 3' broad. Ans. ii76f, i' &'. 10. What is the content of a marble slab, whose length is 5f. 7', and breadth if. 10' ? Ans, I of. 2' lo'''. IT. Required the content of a ceiling, which is 43f. 3' long, and 25f. 6' broad, Ans. Ii02f. 10' 6'^ 12. The length of a room being 2of. its breadth I4f. 6\ and height lof. 4', how many yards of painting are in it, deducting a fire place of 4f. by 4f. 4', and two win-f clows, each 6f. by 6f. 2'?, Ans. 73^ yards, 13. Required the solid content of a wall 53f. 6' long^ 12':. ■ . and 2f. thick, AnSc i3iof. 9', ^'^'•^ 'VULGAR VULGAR FRACTIONS. 59 VULGAR FRACTIONS. Fractions, or broken numbers, are expressions for any assignable parts of an unit ; and are represented by two numbers, placed one above the otber, with a line drawn between them. The figure above the line is called the numerator, ^.nd that below the line, the denominator. The denominator shews how many parts the integer is divided into, and the numerator shews how many of those parts are meant by the fraction. Fractions arc either proper, improper, single, compound, or mixed, 1. A proper fraction is when the numerator is less than the denominator: as ^, ^^ -|-, &c. 2. An improper fraction is when the numerator exceeds the denominator : as y, *-^°, &c. 3. A single fraction is a simple expression for any num- ber of parts of the integer. 4. A compound fraction is the fraction of a fraction : as T of T' i °f h &c. 5. A mixed number is composed of a whole number and a fraction : as 8-^3 i7tt> ^^* Note. — Any whole number may be expressed like a frac- tion by writing i under it : as \. 6. The common measure of two or more numbers 25 that number, which will divide each of them, without :•;. remainder. Thus 3 is the common measure of 12 and T 5 \ and the greatest number, that will do this, is chilled the greatest common measure. 7. A number, which can be measured by two or more niln.b-r ., is called their common multiple ; and if it be the Co ARITHMETIC. least number, which can be so measured, it is called their lea'it common multiple % thus 30, 45, ^o and 75, are multi- ples of 3 and 5 \ but their least common multiple is 15.* PROBLEM I. t To find the greatest common measure of two or more niimhers., RULE.f I. If there be two numbers only, divide the greater by the less, and this divisor by the remainder, and so on ; always dividing the last divisor by the last remainder, till nothing * A prime rMmber is that, which can only be measured by an unit. That number, which is produced by multiplying several numbers together, is called a composite number. . A perfect number is equal to the sum of all its aliquot parts. The following perfect numbers are taken from the Petersburg!! acts, and are all that are known at preseat. 6 496 8128 3355^33^ 8589869056 137438691328 2305843008 139952 1 28 2417851639228158837784576 99035 203 1428297 1 8304488 1 61 28 There are several other numbers, which have received different denominations, but they are principally of use in Algebra, and the higher parts of mathematics. -j- This and the following problem will be found very useful in the doctrine of fractions, and several other parts of Arithmetic. The truth of the rule may be shewn from the first example.— For since 54 measures 308, it also measures 108 + 54, or 162. A gain > Vulgar fractions. Ci tiotliing remains, then will the last divisor be the greatest common measure required. 2. When there are more than two numbers, find tlic greatest common measure of two of them as before ;. and of that common measure and one of the other numbers ; and so on, through all the numbers to the last ; then will the greatest common measure, last found, be the answey, . 3. If I happen to be the co'mmon measure, the given numbers are prime to each otlicr, and found to be incom- mensurable, EXAMPLE*. 1. Required the greatest common measure of 918, 199S and 522. 918)1998(2 So 54 is the greatest common measure 1836' _ of 1998 and 918, 162)918(5 X rry Hence 08(2 08 54)522(9 486 36)54(1. 18)36(2 36 108)162(1 108 54)1' Therefore 18 is the answer required. 2. What Again, since 54 measures 108, and 162, it also measures 5X162-I-108, or 918. In the same manner it will be found to measure 2X9184-162, or 1998, and so on. Therefore 54 measures both 918 and 1998. It is also the greatest common measure ; for suppose there be z greater, then since the greater measures 918 and 1998, it also measures the remainder 162 ; and since it measures 162 and 918, it also measures the remainder 108 j in the same manner it will be found to measure the remainder 54 ; that is, the greater meas- ures 2. What is the greatest common measure of 612 a'rid 540 ^ Ar.s. 36. 3. What is the greatest common measure of 720, 336 and 1736 ? Ans. 8. PROBLEM It. 5n> ^/id the least common multiple of tivo cr more fiutiihers. RULE.* 1. Divide by any number, that will divide two or mord qf the given numbers without a remainder, and set the quotients, together with the undivided numbers, in a Hne beneath. 2. Divide the second line as before, nnd so on, till there are no two numbers that can be divided ; then the con- tinued product of the divisors and quotients will give the muhiple required. EXAMPLES. I. What Is the least common multiple of 3, 5, 8 and 10? 5)3 5 8 10 2)318 2 3 I 4 * 5X2X3X4:=: 120 the answer. 2. What ures the less, which is absurd. Therefore 54 is the greatest com- mon measure. In the very same manner, the demonstration may be applied to 3 or more numbers. * The reason of this rule may also be shewn from the first ex- ample, thus : it is evident, that 3 X 5 X 8 X 10= 1200 may be di- vided by 3, 5, 8, and 10, without a remainder ; but 10 is a mul- tiple of 5, therefore 3x5x8x2, of 240, is also divisible by 3, 5, 8, and 10. Also 8 is a multiple of 2,; therefore 3 X5 X4X 2=120 is also divisible by 3, 5, 8, and lo j and is evidently th^- least number that can be bO dividwd. REDUCTION OF VULGAR FRACTIONS. 6^ 2. "VViiat is the least common multiple Ox- 4 and 6 ? Ans. 12. 3. What is the least number, that 3, 4, 8 and 12 will measure ? Ans. 24. 4. What is the least number that can be diviaed by the nine digits, without a re.-^ainder ? Ans. 25 2c. Reduction of Vulgar Frjctic^'^s. Reduction of Vulgar Fractions is the bringing them oiit.of one form into another, in order to prepare them for ths ooerations of addition, subtfaction, &c. CASE I. To abbreviate or reduce fractions to their lowest terms, RULE.* Divide the terms of the given fraction by any number that will divide them without a remainder, and these quo- tients * That dividing both the tcrm3 of the fraction equally, by any number whatever, will give another fraction equal to the former, is evident. And if those divisions are performed as often as c^n be done, or the common divisor be the greatest possible, the terms of the resulting fraction must be the least possible. Note i. Any number ending with an even number, or a cy- pher, is divisible by 2. 2. Any number ending with 5, or o, is divisible by 5. 3. If the right-hand place of any number be o, the whole is divisible by 10. 4. If the two right-hand figures of any number are divisible by 4, the whole is divisible by 4. '5. If the three right-hand figures of any number are divisible by 8, the whole is divisible by 8. 6. If the sum of the digits constituting any number be divisi- ble by 3, or 9, the whole is divisible by 3, or 9. 7. If 64 ARITHMETIC. tients ngain In the same manner ; and so on, till it appears ihat there is, no number greater than i, which will divide them, and the fraction will be in its lowest terms. Or, Divide both the terms of the fraction by their greatest common measure, and the quotients will be the terms of the fraction required. EXAHPLES. I. Reduce -^44- to its lowest terms. (^) (2) (3L (^) (^) T 4 4 7_^_ 3 O 1 a 6 -—3 ffipnnswrr TT5--~TT^— 6-^ —T^— T-^— T> ^^^ ^^SN\ crt Or thus : 144)240(1 144 96)144(1 48)96(2 ^ 96 Therefore 48 is the greatest common measure, and 4^)^4-5"--- i» ^^^ ^^^^ ^^ before. 2. Reduce 7. All prime numbers, except 2 and 5, have i, 3, 7, or 9, in the place of units ; and all other numbers are composite. 8. When numbers, with the sign of addition or subtraction be- tween them, are to be divided by any number, each of the num- bers must be divided. Thus tX~Jli?==2-f-44-5=:ii. 2 9. But if the numbers have the sign of multiplication between them, only one of them must be divided. Thus i-X- ^ 2'X6 — 3X4Xio_ iX4Xio_ 1x2X10^ 20—20. "" 1x6 ■" 1x2 "" iXi "" I REDUCTION OF VULGAR FRACTIONS. 6| 2. Reduce ^V to its least terms. Ans. ^. 3. Reduce -~-|- to its lowest terms. Ans. -|.. 4. Bring |^ to its lowest terms. Ans. |-i. 5. Reduce y|-^ to its least terms. Ans. ■^. 6. Reduce f-^ to its least terms. Ans. \, 7. Reduce y-f-fl- to its lowest terms. Ans. -|.. 8. Abbreviate j^rr^V^ ^^ much as possible. CASE II. To reduce a mixed number to its equivalent improper fraction. RULE.* Multiply the wbole number by the denominator of the fraction, and add the numerator to the product, then that sum written above the denominator will form the fraction required. EXAMPLES. I. Reduce 27- to Its 27 9 equi 'the valent improper answer. fraction. 243 2 245 9 9 ^^^7X9+^ 9 2, Reduce * All fractions represent a division of the numerator by the de^ nominator, and are taken altogether as proper and adequate ex- pressions for the quotient. Thus the quotient of 2 divided by 5 is f ; from whence the rule is manifest ; for if any number Is mul-* tiplicd and divided by the same number, it pj crideat the quotieot must be the same as the quantity first giyeo^ I ^ AftltllMETJC. 2. Reduce 183^ to its equivalent improper fractiort. Aas. ilf'. 3. Reduce 5147^ to an improper frnction. Ans. ^-^^ 4. Reduce iooj|. to an improper fraction. Ans. -^-jI^. "8 400 CASE III* 5. Reduce 47-|-^-|-^ to an improper fraction. Ans. ^°'^*" ^e reduce an improper fraction to its equivalent whole or tiiix-^ ed number. B-UtE. Divide the numerator by the denominator, and the quo-, tient will be the whole or mixed number required. EXAMPLES. S. Reduce '^■~ to Its equivalent whole or mixed number* i6)98i(6i-V 1(5 5 Or, ^-^ ±:98i-r^i6z=6i-p^ the answer, i. Reduce ^-^ to its equivalent whole or mixed number.- Ans. 7. 3. Reduce ~j-^ to its equivalent whole or' nrlxed num- ber. Ans. ^6^. 4. Reduce ^\\" to its equivalent whole or mixed num- ber. Ans. 183--^. 5. Reduce ^ 7-f J ^ to its equivalent whole or mixed number. Ans. i209yf|. CASK * This rule is plainly the reverse of the former, and has its reasoQ in the nature of common divisionu JIEDUCTION OF VULGAR FRACTIONS. 6'f CASE IV, To reduce a whole number to an equivalent fraction^ having a piven denominator. RULE.* Multiply the whole number by the given denominator^ and place the product over the said denominator, audit v/ill form the fraction required. EXAMPLES. J. Reduce 7 to a fraction, whose denominator shall be 9, 7X9:^:63, and y the answer. And y =z63-^9=r7 the proof. 1. Reduce 13 to a fraction, whose denominator shall Ke 12. Ans. Vr'- 3. Reduce 100 to a fraction, whose denominator shall be 90. Ans. 9 O G '9 o CASE V. To redu;:e a compound fraction to an equivalent single one, RULE.f Multiply all the numerator? together for the numerator, and all the denominators together for the denominator, and they will form the single fraction required. ^ If * Multiplication and division are here equally used, and conse- quently the result is the same as the quantity first pro])osed. f That a compound fraction may be represented by a single one is very evident, since a part of a part must be equal to some part of the whole. The truth of the rule for this reduction may be shewn as follows. I^et the compound fraction to be reduced be y of t. Then y of 4=4-r3=TT? and consequently ^ of 4=AX2=:xt the same as by the rule, and the like will be found to be true in all cases. 68 aAUlTHMETIC. If part of the compound fraction be a whole or mixed number, it must be reduced to a fraction by one of the for- mer cases. When it can Ije done, divide any two terms of the frac» tion by the same number, and use the quotients instead thereof. EXAMPLES. X. Reduce y of ^ of ^ of ^ to a single fraction. 2x3x4x8 ^223== jg the answer. 3x4x5x11 660 55 Or, ^„X.^i<^J<_i=- as before. -2rx>i-X5Xii 2. Reduce -^ of -f- of ■§• to a single fraction. Ans. -• 3. Reduce -^- of f of |- of ^ to a single fraction. Ans. -/i-. 4. Reduce -— gf ^r of ~ of xo to a single fraction. nS. J-^J^ CASE VI, To reduce fractions of different denominators to equivalent fractions^ having a common dctiominator. Multiply each numerator into all the denominators, ex- cept its own, for a new numerator •, and all the denomi- nators continually for the common denominator. EXAMPLES, If the corapound fraction consist of more numbers than 2, the two first may be reduced to one, and that one and the third will be the same as a fraction of two numbers j and so on. * By placing the numbers multiplied properly under one another, It will be seen, that the numerator and denominator of every frac- tion REDUCTION OF VULGAR^fRACTIONS. 69 EXAMPLES. 1. Reduce -{-y y and -^-, to equivalent fractions, having a common denominator. 1X5X7 = 35 the new numerator for 4-. 3X2X7 = 42 do. for .|.• 4X2X5=40 do. for i. 2X5X7 = 70 the common denominator. Therefore the new equivalent fractions are 4---, ~~, and j~y the answer. 2. Reduce ^j yj |- ^i"^<^ |> ^o fractions, having a com- men denominator/ Ans. ^±, ^|^, ^^, -iXf 3. l\.educe y, -| of y, 5y and y*^, to a common denomi- 4. Reduce —, ^- of i-^, y*— and ^ to a common de- nommator. iinb. yToTTj tt^ttj TTTTo"' Td^oT^-' RULE II. ^0 reduce any givsfi fractions to others^ nxjJAch shall have ih^ least common denominator, 1. Find the least common multiple of all the denomina- tors of the given fractions, and it will be the common de- nominator required. 2. Divide the common denominator by the denomina- tor of each fraction, and multiply the quotient by the nu- merator, and the product wiU be the numerator of tlie- fraction required, EXAMPLES. lion are multiplied by the very sarn§ number, and consequently their values are not altered. Xhus in the fust example : X5X7 3 2 I X5X7 5 X2X7 X?X5 X2X5 X2X7 7 In the 2d rule, the common denominator is a multiple of all the denominators, and consequently will divide by any of them j- it is manifest, therefore, that proper parts may be talicn for all th ■ aumeratgrs as required. V^ ARITHMETIC. 5 EXAMPLES. X.' Reduce ~, y and -|-, to fractions, having the least common denominator possible. 6 3 2 3 2 2 I I I I X 1 X I X 2 X 3=6 least common denominator. 6-?-2Xi=3 the first numerator; 6^3x2=4 the second numerator 5 6'^6xs=?.5 the third numerator. Whence tly^ required fractions are |-, ^, I-. 2. Reduce -^ and — to fractions, having the least cona- nr.on denominator. Ans. 4-^, 4-^. 3. -Reduce ~y 4, ^ and ^, to the least common denomi- nator. ^ Ans. ^, -V> TT» ^T. 4. Reduce y, -|-, -|- and ^, to the least common denom- inator. Ans. 4^, --, i-^, Ai.^ 5. Reduce 4-, -^-, |, X, 2-i. and —;-, to equivalent frac- tionsj having the least common denominator. •^"S. -^^-g-, -^-^, ^^, ^^, -^-g., •^. CASE VII. ^o/i;id the value of a fractioji in the hi oivn parts of the integer,, Multiply the numerator by the parts in the next inferior denomination, and divide the product by the denominator; and if any thing remain, multiply it by the next inferior dcnomiiiation, and divide by the denominator as before ; and so on as far as necessary ; and the quotients placed ^f- ter one iuiotlier, in their order, ".vill be the answer required. EXAMPLES. * The numerator of a fraction may be considered as a remain- der, and the denominator as a divisor ; therefore this rule hajS Its reason in the nature of compound division. JREDUCTIO^ OF VULGAR FRxICTlONS. ^i EX^AMPLES. i. Wliat is the value of -f of a shilling ? 12 7)6of8i 2M. Ans= . 4 16 14 i. What IS the value of |- of a pound sterling ? Ans. 7s. 6ti 3. What is the value of -|- of a pound Troy ? Ans. 70Z. 4(lwt. 4. What is the value of -^ of a pound avoirdupois ? Ans. 90Z. 2ydr. 5. What is the value of - of a cwt. ? Ans. 3qrs. 3iK. i*oz. 12-jdr; 6. Wnat is the value of -^ of a mile .^' Ans. ifur. i6pls. 2yd3. ift. 9-,-^in. 7. What is the value of -^ of an ejl English ? Ans. 2qrs. 3^iils. §. What is the vahie of -J- of a tun of v.^ine ? Ans. 3hhd. 3igal. 2qts. 9. What is the valtre of yt ^^ a day ? Ans. i2h. 55' '^3Tr^' CASE viir. Xo rcduci a fraction of one demmhiation to that of a?iciher, rf" taining the same n^alue. RULE.* Make a compound f?acti'i)n of it, and reduce It fo a sin- gle one. ' - EXAMPLES. The reason of this practice is explained in the rule for rc- iJucing compound fractions to single ones. The JLIUrilMrTIC. EXAMPLES. I. Reduce |- of a penny to the fraction of a poTind. i <^^ tV ^^ t^^i^V^^tIt ^^- answer. And--^ of V' of V=:^^r=xd. the proof. 7. Reduce y of a farthing to the fraction of a pound. Ans. -,-^-^-^, 3. Reduce ~-^£. to the fraction of a penny. Ans. •*- . 4. Reduce y of a dwt. to the fraction of a pound Troy. Ans. -j^. 5. Reduce — of a pound avoirdupois to the fraction of a cwt. _ Ans. ~^. 6. Reduce y-Zj-r of a hhd. of wine to the fraction of a pint. Ans. ~p 7. Reduce -,^y of a month to the fraction of a day. Ans. 4f 8. *Reduce 7s. 3d. to the fraction of a pound. Ans. j-^~, 9. Express 6fur. i6pls. in the fraction of a mile. Ans. -j. Addition of Vulgar Fractions. R U L E.f Reduce compound fractions to single ones ; mixed' numbers to improper fractions ; fractions of diiFcreiit iiite- gcrs The rule might have been distributed Into two or three diiferent cases, but the directions here given may very easily be" applied to any question, that can be proposed in those cases, and v/ill be more easily understood by an example or two, than by a multiplicity of words. * Thus 7s. 3d. = 87d. and il. = 24od. .-.^V^srll the answer. f Fractions, before they are reduced to a common denoraina- tor, are entirely dissimilar, and therefore cannot be incorporated with one another ; but when they are reduced to a common de- nominator, and made parts of the same tiding, their sum cr liiii'-r- ADDITION OF VULGAR FRACTIONS. 73 « gers to those of the same j and all of them to a common denominator ; then the sum of the numerators, written over- the common denominator, will be the sum of the fractions required. EXAMPLES. 1. Add 3|-, -f, y of I and 7, together. First3X=vMof^zz^,7=f Then the fractions are */ , |-, 7^ and f ; .% 29X8X10X 1 = 2320 7X8X loX 1= 560 7X8X 8X 1= 448 7X8X 8X10 = 4480 7808 -= 1 2^11 = 1 2f the answer. 8X8X10X 1= 640. 2. Add |-, 7y and |- of -I together.' Ans. 8-|-. q. What is the sum of -|-j 4- of 4 and 9-^ .? Ans. ic-^. 4. What is the sum of -^- of 6|-, y oi ~ and 74 ? Ans. I3i£l. 5. Add ~1. -^s. and ^V ^^ ^ penny together. Ans. mf» or 3s.. Id. 1-1^9. 6. What is the sum of ~ of 15L 3-|-l. \ of \ oi \ of a pound and y of -|- of a shilling .'' Alls. 7L 17s. 54d. 7. Add y of a yard, \ of a foot and -|- of a mile to- gether, Ans. 660yds. 2ft. pin. 8. Add ^ of a week, -^ of a day and \ of an hour to- gether. - Ans. -ad. I4^h. Subtraction ence, may then be as .properly expressed, by the sum or difference of tlie numerators, as the sum or difference of fey two quantities whatever, by the sum or difference of their individuals ; v/hcriCC the reason of the rules, both for addition and subtracfi'On, is manifest, K 74 ARITHMETIC, . Subtraction of Vulgar Fractions. RULE. Prepare the fractions as in addition, and the' difference, of the numeratorSj written above the common denomiaa* tor, will give the difference of the fractions required. EXAMPLES. 1. From y take ^ of -f- ^of f=:, and|= -, 3 •.• YT — i^=TT=T ^^''^- anwt'tr renuireQ. 2. From ^ take f. An.. ^^1 3. From g6~ take 144. Ans. Si^-f. 4. From 14^ take y of 19, Ans. i~. 5. From -ll. take -^^s. Ans. 93. 3d. 6. From j-oz, take ydwt. , Ans.- i idwt. 3gr. 7. From 7 weeks take ^~ days. Ans. 5W. 4d. 7h. 12'. Multiplication of Fulgas. Fractions*. RULE.* Reduce compound fractions to single ones, and mixed numbers to improper fractions ; then the product of the numerators is the numerator ; and the product of the^ de- nominators, the denominator of the product required. EXAMPLES'. * Multiplication by a fraction implies the taking some part or parts of the multiplicand, and, therefore, may be truly expressed by a compound fraction. Thus ^ multiplied by f is the same as ^ of -^ ; and as the directions of the rule agree with the method al- ready given to reduce these fractions to single ones, it is shewn to be right. DIVISION OF VULGAR FRACTIONS. 75-; EXAMPLES. I. Required the continued product of 2^, -g-j i o^ i and 2. 2\-h i of ^ = i^=TV, and 2zz\ i 2. Multiply -V by ^V ^"^- ~V 3. Multiply 4t by h ^^^' '^' 4. Multiply 4- of 7 by f Ans. i^. 5. Multiply I of I by f of 3^. Ans. f ^. 6^ Multiply 4|, -I of -^ and iS-f-, continually together. Ans. 9-^-^, DrrisiON of Vulgar Fractions. . RULE.* Prepare the fractions as in multi-plication ; then invert jhe divisor, and proceed exactly as in multiplication. EXAMPLES. I. Divide 4 of 19 by ~ oi \-. I^IQ 3 ^ ^X>^ 2 •/yXf^--^ — ?r='yi=74 the quotient required. 2. Divide * The reason of the rule may be shewn thus : Suppose It wer.e required to divide | by ~. Now ^-^2 is manifestly \ of |, or — i— ; butjzrf of 2, .*. y of 2, or f must be contained 5 times 4X2 ' . ^ > as often in 4- as 2 Is ; that is ;^ — ^ = the answer ; which is ac- 4x2 cording to the rule ; and will be so in all cases. Note. — A fraction is multiplied by an integer, by dividing the denominator by it, or multiplying the numerator. And divid'jd by an Integer, by dividing the numerator, or multiplying the de- nominator. ^6 1 ARITHMETIC. <• 2. Divide -^ by j. Ans. 4, 3. Divide 9f by -^ of 7. Ans. 2~j-, 4. Divide 3^ by 9^. Ans. ^, 5. Divide I- by 4. Ans. yV 6, Divide 1 of f by | of -|o Ans. yo DECIMAL FRACTIONS. J\. DECIMAL is a fraction, whose denominator Is ai^ unit, or i, with as many cyphers annexed as the numera- tor has places ; and is commonly expressed by writing the rumerator only, with a point before it called the separatrix^ Thus> 0-5 is equal to ^ or 0-25 AVor I G-75 tV^ or 3 13 44 or lA- 24-6 24A^ '02 T^ or T7f •001 ■S T-ons-jo" ^^ TT^J' A finiU decimal is that, which ends at a certain number of places. But an infinite decimal is that, which is under- stood to be indefinitely continued. A repeating decimal has one figure, or several figures, continually repeated, as far as it is found. As "33, &:c. which is a single repetend. And 20*2424, &c. or 20*246246, &c. which are compound repetends. Repeating decimals are al- so called circulates y or circulating decimals, A point is set over a single i DECIMAL FRACTIONS, 77 a slncrle repetcnd, and a nolat over the first and last figures ,of a compound repetend. The first place, next after the dechiial mark, is icth parts, the second is looth parts, the third is loooth parts, arid so on, decreasing toward the right by loths, or in- creasing toward the left by icths, the same as whole ot iiitegral numbers do. As in the following SCALE OF NQTAriGN. r^ ctj to t: ^ CO ■A t/5 O c tn O fis CO o 4-1 a CO O 6 CO 13 CO r/3 -B -o C3 to 'B Ui -t3 o a CO D O CO C .2 en CO c« 3 o •^ O • •-( :3 u ^ s ■u o . '~', U p s ffi H H H t3 rH H H K 8 88 8*88888 8 Cyphers on the right of decimals do not alter their value. For '5 or -^-- is ^ And -50 or -J-^ is {- And -500 or -^°-^- is ^. But cyphers before decimal figures, and after the separ- ating point, diminish the value in a tenfold proportion for ^very cypher. So *5 is A or T But •05 is ■tio or T5- And •005 is •roW 0^ And so ,on. So that, in any mixed or fractional number, if the separ- ating point be moved one, two, three, &c. places to the right hand, every figure will be lo, 10O3 1000, &c. times greater than before. But if the point be moved toward the left hand, then every figure will be dim/mished in the same manner, or the whole quantity will be divided by 10, igoj 1000, &c. Jddition ft AKITHMETIG, Addition of Decimals, RULE. . I. Set the numbers under each other according to the, value of their places, as in whole numbers, or so that the decimal points may stand each directly under the pre- ceding. 2. Then add as in whole numbers, placing the decimal point, in the sum, directly under the other points. EXAMPLES. 7530 l6*20I 3-0142 957-13 6-72819 •03014 85i3'io353 2. What IS the sum of 276, 39*213, 72014*9, 417, 5032 and 2214*298 ? Ans. 79993-411. 3. What is the sum of -014, '9816, -32, '15914, 72913 and -0047 ? Ans. 2-20857. 4. What is the sum of 27*148, 918-73, 14016, 294304, • 7138 and 221*7 ? Ans. 309488*2918. 5. Required the sum of 312*984, 2r39i8, 2700*42, 3*153, 27*2 and 581*06. Ans. 3646*20880 Subtraction of Decimals, RULE. 1. Set the less number under the greater in the sanie manner as in addition. 2. Then subtract as in whole numbers, and place the decimal point in the remainder directly under the other points. EXAMPLES. MULTIPLICATION OF DECIMALS. 79 EXAMPLES. (0 214-8 r 4-90142 209-90858 t. From -9173 subtract -2133. Ans. -7035. 3. From 2*73 subtract 1*9185. Ans. 0*8115. 4. Yv'hat is the difTercnce between 91*713 Jind 407 ? Ans. 315*287. $, What is the difference betvreen 16*37 and 800*135 ? Aus. 783*76^, MuLriPUCAtiGN of Decimals^ RULE.* t. Set down the factors under each Other, and multiply them as in whole numbers. 2. And from the product, toward the right hand, point off as many figures for decimals, as there are decimal places in both the factors. But if there be not so many figures in the product as there' ought to be decimals, prefix the proper number of cyphers to supply the defect. EXAMPLES, * To prove the truth of the rule, let '9776 and '823 be the numbers to be multiplied ; now these are equivalent to, toV/o ^^^ -rVVs- ; whence ^V^^/^Xt''^V=AW/^^o= '8045 648 by the na^ tare of notation, and consisting of as many places, as there are cypliers, that is, of as many places as are in both the Bumbsr^ ; and the same is true of any two numbers whatever. ARITHMETIC EXAMPLES, (1) 91-78 •38 r 917S 73424 ^7534 34-96818: 2. What is the product of 520*3 and •417 ? Ans. 216-9651. 3. What is the product of 51*6 and 21 ? Ans. 1083*6. 4. What is the product of '2^7 and '0431 ? Ans. '0093527, 5. What is the product of '051 and '0091 ? Ans. '0004641. Note. When decimals are to be multiplied by 10, or lOo, or loco, &c. that is, by i with any number of cyphers, It is done by only moving the decimal point so many places fur- ther to the right hand, as there are cyphers in the said mul- tiplier ; subjoining cyphers if there be not so many figures, EXAMPLES. 1. The product of 51*3 and 10 is 513. 2. The product of 2*714 and 100 is 271*4. 3. The product of '9163 and 1000 is 916*3. 4. The product of 21*31 and loooois- 213100. CONTRACTION. When the product ivoiild contahi several more dechnals than are neccessary for the purpose in handy the luork may he much cm-^ tractedy and only the proper number of decimals retained* RULE. 1. Set the unit figure of the multiplier under such deci- mal place o£ the multiplicand as you intend the last of your product MULTIPLICATION OF DECIMALS. SX product shall be, writing the other figures of the multiplier in an inverted Order. 2. Then, in multiplying, reject all the figures in the multiplicand, which are on the right of the figure you are multiplying by -, setting down the products so that their right-hand figures may fall each in a straight line under the preceding ; and carrying to such right-hand figures from the product of the two preceding figures in the multipli- cand thus, viz. I from 5 to 14, 2 from 15 to 24> 3 from 25 to 34, &c. inclusively 5 and the sum of the lines will be the product to the number of decimals required, and will commonly be the nearest unit in the last figure. EXAMPLES. I. Muhlply 27' 14986 by 92*41035, so as to retain only four places of decimals in the product. Contracted. Common way. 27-14986 27-14986 53014-29 92-4103^ 24434874 13 574930 542997 81 44958 108599 2714 986 2715 108599 44 81 542997 2 • H 24434874 2508-9280 2508-9280 1 650510 a. Multiply 480*14936 by 2*72416, retaining four deci- mals in the product. Ans. i3o8'oo35* 3. Multiply 73-8429753 by 4-628754, retaining fiv« decimals in the product. Ans. 341-80097. 4. Multiply 8634-875 by 8437527, retaining only* the integers in the product* An 3. 728569^; to^ ARITHMETIC. Division of Decimals. RULE.* Divide as in whole numbers ; ami to know how many decimals to point off in the quotient, observe the following rules : 1. There must be as many decimals in the dividend, as in both the divisor and quotient -, therefore point off for decimals in the quotient so many figures, as the decimal places in the dividend exceed those in the divisor. 2. If the figures in the quotient are not so many as the rule requires, supply the defect by prefixing cyphers. 3. If the decimal places in the divisor be more than those in the dividend, add cyphers as decimals to the divi- dend, till the number of decimals in the dividend be equal to those in the divisor, and the quotient will be integers till all these decimals are used. And, in case of a remain- der, after all the figures of the dividend are used, and more figures are wanted in the quotient, annex cyphers to the remainder, to continue the division as far as necessary- 4. The first figure of the quotient will possess the same place of integers or decimals, as that figure of the divi- dend, which stands over the units place of the first product. EXAMPLES. I. Divide 3424*6056 by 43*6. 43'^)3424-6o56(78-546 3052 3726 3488 2380 2180 2005 1744 2616 2616 2. Divide * The reason of pointing off as many decimal places in the quotient, as those in the dividend exceed those in the divisor, will easily DIVISION OF DECIMALS. 83 2. Divide 3877875 by '675. Ans. SUS^oo. 3. Divide -0081892 by -347. Ans. -0236, 4. Divide 7-13 by -18. Ans. 39. CONTRACTIONS. I. Jffhe divisor he an integer ivith any number of cyphers at the end ; cut them off, and remove the decimal point in the dividend so many places farther to the left, as there were cyphers cut off, prefixing cyphers, if need be 5 then pro- ceed as before. EXAMPLES. 1. Divide 953 by 210C0. 21*000) 7)-3i766 •04538, &c. Here \ first divide by 3, and tlienby 7, because 3 times 7 is 21. 2. Divide 41020 by 32000. Ans. 1-281875. Note. Hence, if the divisor be i with cyphers^ the quotient will be the same figures with the dividend, having the deci- mal point so many places further to the left, as there arc cyphers in the divisor. EXAMPLES. 217-3 -T-ioo= 2-173. 419 by 10=; 41-9. 5 • 1 6 by 1 000 5; -005 1 5, -2 1 by 1 000 = -0002 1 . JI. When the number of figures in the divisor is great^ the op^ eration may be contra^tedy and the necessary number of deci- tnal places obtained, RULE. 1. Having, by the 4th general rule, found what place of decimals or integers the first figure of the quotient will pos- sess ; easily appear ; for since the number of decimal places in the divi- dend is equal to those in the divisor and quotient, taken together, by the nature of multiplication ; it follows, that the quotient con- tains as many as the dividend exceeds the divisor. 84 ARITHMETIC. sess J consider how many figures of the quotient will scifve the present purpose ; then take the same number of the left-hand figures of the divisor, and as many of the dividend figures as will contain them (less than ten times) ; by these find the first figure of the quotient. 2. And for each following figure, divide the last remain- der by the divisor, wanting one figure to the right more than before, but observing what must be carried to the first product for such omitted figures, as in the contraction of Multiplication ; and continue the operation till the divisor be exhausted. 3. When there are not so many figures in the divisor, as are required to be in the quotient, begin the division with all the figures as usual, and continue it till the number of figures in the divisor and those remaming to be found in the quotient be equal ; after which use the contraction. EXAMPLES. I. Divide 2508*92806505 1 by 92*41035, so as to have four decimals in the quotient.- — In this case, the quotient will contain six figures. Hence Contraction. 92'4io3,5)25o8'928,o6505 1(27*1498 ■• ••••■. 1848207 660721 . 646872 13849 .. 9241 46og . . . 3696 912 ... 80 . . . 74 Cp^imon REDUCTION Of DECIMALS. Common Way. 92*4 1035)2508-92806505 1(27-1498 18482070 ?s 6Co']2i 646872 c6 45 13848 9241 4607 3696 79 73 615 ^35 5800 4140 16605 69315 472901 928280 5544621 2. Divide 721*17562 by 2*257432, so that the quotient may contain three decimals. Ans. 319-467. 3. Divide I2'i69825 by 3^14159, so that the quotient may contain five decimals. Ans. Z'^l^ll' 4. Divide 87*076326 by 9*365407, and let the quotient contain seven decimals. Ans. 9*2976554. Reduction of Decimals, I * CASE I. 7*0 reduce a vulgar fraction to its equivalent decimal, RULE.* Divide the numerator by the denominator, annexing as many cyphers as are necessary j and the quotient vi^ill be the decimal required, EXAMPLES. * Let the given vulgar fraction, whose decimal expression is required, be -rj-. Now since every decimal frartion has 10, ico, 19003 &c. for its denominator ; and, if two fractions be equal, M ARITHMETIC. EXAMPLES, Red nee ^^ to a decimal. 4)5*000000 6)1-250000 i _ '208333, &c. 2. Required the equivalent decimal expressions for 4-^ and f. ^ Ans. '25, -5 and -75. 3. What is the decimal of -^ ? Ans. '375. 4. What is the decimal of ^V ' Ans. '04. 5. What is the decimal of _A_ ? Ans. •015625, 6. Express —-'-^^ decimally. Ans. '071577, &c. CASE IT. 'Tc reduce nin. nhers of different defwminatiom to their equivalent decimal values. R U L E.f 1. Write the given numbers perpendicularly under eacK other for dividends, proceeding orderly from the least to the greatest. 2. Opposite to each dividend, on the left hand, place such a number for a divisor, as will bring it tp the next superior name, and draw a line between them. 3. Begin it will be, as the denominator of one is to its numerator, so is the denominator of the other to its numerator ; therefore 13 : 7 : : 1000, - 7X1000, &C. 70000, &C. o /: .u r &c. : ' 1 =: I r = '53846, the numerator of the decimal required ; and is tKe same as by the rule, f The reason of the rule may be explained from the first ex- ample ; thus, three farthings is ^j- of a penny, which brought to a decimal is '75 ; consequently g^d. may be expressed 9*75d. but 9*75 is -lo-^ of a penny ri-riV^- of a shilling, which brought to a decimal is '8125 ; and therefore 15s. g^d. may be expressed 15-81255. In like manner 15-8 125s. is VAVV of a shillings i-kl-h^l- of a pound ==, by bringing it to a decimal, '790625]. as by the rule. REDUCTION OF DECIMALS. ? " ^. Begin with the highest, and write the q^iotient oi each division, as decimal parts, on the riglit hand of the dividend next below it ; and so on, till they be all used, and the last quotient will be the decimal souglit. EXAMPLES. I, Reduce 15s. p-Jd, to the decimal cf a pound. 4 12 20 3* 975 15-8125 '790625 the decimal required. 2. Reduce 9s, to the decimal of a pound. Ans. '45. 3. Reduce 19s. 5-j-d. to the decimal of a pound. Ans. '972915. 4. Reduce looz. i8dwt. i6gr. to the decimal of a lb. Troy. i\ns. '91 II 1 1, &c. 5. Reduce 2qrs. 141b. to the decimal of a cwt. Ans. '625, &c. 6. Reduce and therefore i must be added ; and when the number of farthings is more than 37, a tV part is greater than i^-, for which 2 must be added ; and thus the rule is shewn to be right. REDUCTION OF DECIMALS, 3^ CASE IV. ^0 Jind the value of any given decimal in Urms cf the integer, RULE. 1. Multiply the decimal by the number of parts in the next less denomination, and cut ofF as many places for a remainder on the right hand, as there are places in the giv- en decimal. 2. Multiply the remainder by the parts in the next infe- rior denominatiouj and cut off for a remainder, as before. 3. Proceed in this manner through all the parts of the integer, and the several denominations, standing on the left- hand, make the answer. EXAMPLES. t. Find the value of '37623 of a pound; 20 7-52460 12 6-29520 4 I -1 8080 Ans. 7s. 6-4CI. 2. What is the value of '625 shiUing ? Ans. 7^d. 3. What is the value of '83229161. ^ Ans. 1 6s. 7^d, 4. What is the value of •6725cwt. ? Ans. 2qrs. 191b. 507. 5. What is the value of "67 of a league i Ans. 2mls. 3pls. lyd. 3in. ib.c« 6. Wfrat is the value of -61 of a tun of wine ? Ans. 2hhc. 27gal. 2qt. ipt. 7. What is the value of -461 of a chaldron of coals } Ans. i6bu. 2peo 8. What is the value of '42857 of a month .? Ans. iw. 4d. 23h. ^^^ 56'''. GASJB- M 90 ARITHMETIC. CASE V. To find the value of a?iy decimal of a pouud h\ inspecthn. RULE. Double the first figure, or place of tenths, for shillings, and if the second be 5, or more than 5-, recko^n another shilling ; then call the figures in the second and third places, after 5, if contained, is deducted, so many farthings y abating i, when they are above 12 ', and 2, when above 37 j and the result is the answer. EXAMPLES. I. Find the value of •785I. by inspection. 14s. = double 7. IS. for 5 in the place of tenths. 8i = 35 farthings. ■- for the excess of 12, abated. 15s. 8^d. the answer. 2. Find the value of '8751. by inspection. Ans. I7s.6d. 3. Value the following decimals by inspection, and find tiieir sum, viz. -9271. + 'SSil' + '203!. + *o<^ih + '021. -j- •oo9h Ans. il. IIS. 5-|d. FEDERAL MONET.'' The denominations of Federal Aicney^ as dctermnicd by an Act of Congress, Aug. 8, 1786, are in a decimal ratio ^ and, therefore, may be properly introduced In this place. A mi 11^ * The coins of federal money arc two of gold, four of silver, and two of copper. The gold coins are called an eagle and half- eagle ; the silver, a dollar, half -dollar^ double dime and dime ; and the copper, a cent diV.d half-cent. The standard for gold and silver is eleven parts fine and on-e pari alloy. The weight of fine gold in FEDERAL MONEY. ' 9 1 A rniil, which Is the lowest money of account, is 'ooi of •a dollar, which is the money unit. A cent is 'O i Or i o mills = i cent, A dime -i marked m. c. A dollar i- lo cents = i dime, d. An eagle ^-o* ^o dimes = i dollar, D. I o dollars = i eagle, E. A number in the eagle is 246*268 grains ; of fine silver in the dollar, 375*64 grains ; of copper in 1 00 cents, 2~\h. avoirdupois. The fine gold in the half-eagle is half the weight of that in the eagle ; the line silver In the half-dollar, half the weight of that in the dollar, &c. The denominations less than a dollar are expressive of their values : thus, w/7/is an abbreviation of milk, a thousand, for 1000 mills are equal to 1 dollar ; cent, of centum, a hundred, for 100 cents are equal to i dollar ; a Jime is the French of tithef the tenth part, for 10 dimes are equiU to i dollar. The mint-price of uncoined gold, 1 1 parts being fine and i part alloy, is 209 dollars, 7 dimes and 7 cents per lb. Troy weight ; amd the miat-price of uncoined silver, 1 1 parts being fine and i part alloy, is 9 dollars, 9 dimes and 2 cents per lb. Troy. In Mr. Pike's " Complete System of Arithmetic," may be seen " Rules for reducing the Federal Coin, and the Currencies of the several United States ; also English, Irish, Canada, Nova- Scotia, Livres Tournois and Spanish milled dollars, each to the par of all the others." It may be su^icient here to observe re- specting the currencies of the several States, that a dollar is equal to 6s. in New-England and Virginia ; 8s. in New- York and North- Carolina ; 7s. 6d. in New- Jersey, Pennsylvania, Delaware and Maryland ; and 4s. 8d. in South-Carohna and Georgia. The English standard for gold is 22 carats of fine gold, and 2 carats of copper, which is the same as 1 1 parts fine and i part al- loy. The English standard for silver is i8oz. 2dvvt. of fine silver, and 1 8dvvt. of copper ; so that the proportion of alloy in their sil- ver is Jess than in their gold. When either gold or silver is finer or coarser 92 ARITHMETIC. A number of dollars, as 754, may be read 754-dollarSj or 75 eagles, 4 dollars -, and decimal parts of a dollar, as •365, may be read 3 dimes, 6 cents, 5 mills, or 36 cents, 5 mills, or 365 mills ; and others in a similar manner, Addition, suhtractiotiy multiplication and division of federal money are performed just as in decimal fractions j and consequently with more ease than in any other kind of rnoney. E X A M P t p S. I. Add 2 dollars, 4 dimes, 6 cents, 4D. 2d., 4d. 9c.> lE. 3D. 5c. 7m., 3c. 9m., iD. 2d. 8c. im., and 2E. /\D,, 7d. 8c, 2m. together. E.D. d. cm. 2*46 4 • 2 • 4 9 ^ 3 • 0 5 7 • 0 3 9 1-281 24-782 m. 8 9 (2) E.D. d. cm, 3 4*123 I - I 7 8 78*001 I • 7 • 3 2 61-789 6-341 m. 0 3 E.D. d.cm, 30-671 3-^23 4-567 - 0 3 70-308 7-17 8-231 46-309 Ans. E.J.^^d.c. From 32-17 Subtract 17-28 70-00 7-81 D. d. cm. 2-652 • 0 7 Remain. 14*88 9 7. Multiply coarser than standard, the variation from standard is estimated by carats and grains of a cajat in gold, and by penny-weights in sil- ver. Alloy is used in gold and silver to harden them. Note. — Carat is not any certain weight or quantity, but -^-2; of any weight or quantity ; and the minters and goldsmiths divide i^ into 4 equal parts, called grains of a carat. CIRCULATING DECIMALS. 93 .7. Multiply 3D. 4d. 5c. im. by iD. 2d. 3c. 2m. D. 3*45^ Note. The figures after ^ "^"^ .ox .on the right hand of, mills ^qoi are decimals of a mill. 1^353 6902 3451 4-251632 =4-251^'^ Ans. D. D. D. 8. Multiply 6*347 by 4*532. Am ». 28'7646o4- D. D. D. g. Multiply 71*012 by 3*703. Ans. 262*95743^- D. . D. 10. Multiply 8o6*222|- by 9. Ans. 7256. D. D. ^ n. Divide 4*251632 by 1*232. j:*232)4*25i632(3'45i Answer. 3696 555^ 4928 6283 6160 1232 1232 D. 12. Divide 20D. by 2000. Ans. o'oi. D. 13. Divide 7256D. by 9. Ans. 8o6'222|. CIRCULATING DECIMALS. It has already been observed, that when an infinite decl- fnal repeats always one figure, it is a single repetend ; and >yhen more than one, a compound repetend ; also that a point is 94 ARITHMETIC. is set over a single repetend, and a point over the first and last figures of a compound repetend. It may be further observed, that when other decimal fig- ures precede a repetend, in any number, it is called a mixed repetend : as '23, or '104123 : otherwise it is a/)//r^, or simple, repetend : as '3 and '123. Similar repetends begin at the .same place : as '3 and *6, or 1*341 and 2*156. Dissimilar repetends begin at different places : as '253 and •4752. Conterminous repetends end at the same place : as '125 and •009, Similar and conterminous repetends begin and end at the same place : as 2*9104 and '0613. ReDUGTION of ClRCVLATING DECIMALS, CASE I. To reduce a simple repetend to its equivalent vulgar fraction* RULE.* I. Make the given decimal the numerator, and let the denominator be a number consisting of as many nines as there are recurring places in the repetend. 2. If * If unity, with cyphers annexed, be divided by 9 ad infinitumy the quotient will be i continually ; i. e. if ^ be reduced to a deci- mal, it will produce the circulate 'i ; and since *! is the decimal equivalent to ^, 'z will = J,- '3 r= |, and so on till •9 = ^ = i. Therefore, every single repetend is equal to a vulgar fraction, whose numerator is the repeating figure, and denomiiator 9. Again, REDUCTION OF CIRCULATING DECIMALS. 9^ 2. If there be integral figures in the circulate, as many cyphers must be annexed to the numerator, as the highest place of the repetend is distant from the decimal point. EXAMPLES. 1. Required the least vulgar fractions equal to '6 and * 1 23. •6=|=i ; and •I23=-^|=^^3V ^ns. 2. Reduce '3 to its equivalent vulgar fraction. Ans. j. 3. Reduce 1*62 to its equivalent vulgar fraction. Alls. --5P5-. 4. Required the least vulgar fraction equal to '769230. CASE 11. 21? reduce a mixed repetetid to its equivalent vulgar fraction. RULE.* I. To as many nines as there are figures in the repetend, annex as many cyphers as there are finite places, for a de- nominator. 2. Multiply Again, -j'-y, or -^^ being reduced to decimals, makes '0101 or, &c. or -001001, &c. ad infinitum =:-oi or '00 1 ; that is, -j^zs •oi',and^-5-^-5^=:*ooi ; consequently -^^p^'oz, -5^V='03, &c. and -5^-5=-oo2, ■5j^='oo3, &c. and the same will hold universally. * In like manner for a mixed circulate ; consider it as divisible ioto its fipite and circulating parts, and the same principle will be seen to run through thcnj also : thus, the mixed circulate '16 is divisible into the finite decimal •!, and the repetend -06 ; but *i=To»and -06 would be=|, provided the circulation began imme- diately after the place of units ; but as it begins after the place of 5v6 ARITHMETIC. 2. Multipl jT the nines in the said denominator by the finite part, and add the repeating decimal to the product, for the numerator. 3. If the repetend begin in some integral place, the finite value of the circulati?ig part must be added to the finite part. r X A M P L E s. 1. What is the viilgar fraction equivalent to '138 ? 9X i3 + 8iri25:r: numerator, and pccii: the de- nominator; .*. •I38n~-|r=:j^^ the answer. 2. What is the least vulgar fraction equivalent to '53 ? Ans.VV 3. What is the least vulgar fraction equal to '5925 ? Ans. ii. 4. What is the least vulgar fraction equal to 'ooS'^py 133.'* 5. What is the finite number equivalent to 31*62 ? Ans. 3 if k CASE III. To mak€ any number of dissimilar repetcnds shnlJar and con-^ termifiouso RULE.* Change them into other repetends, which shall each? consist of as many figures as the least common multiple of the tens, It is § of -s^r:^, and so the vulgar fraction c:'i6 is to4* •^=:/^-|-^=rf|-, and is ,the same as by the rule. * Any given repetend whatever, whether single- compound, pure or mixed, may be transformed into another repetend, that shall contist REDUCTION O.F CIRCULATING DECIMALS. 97 the several numbers of places, found in all the repetends, •>ntains units. EXAMPLES. i. Dissimilar. Made similar and contei^ninous* 9-814 •= 9-81481481 i*_5; =r 1-50000000 87-26 = ■ 87-26(566666 •083 = -08333333 124-09 = 124-09090909 2. Make -3, '27 and '045 similar and conterminous. 3. Make '321, '8262, '05 and '0902 similar and con- terminous. 4. iviake *52i7, 3*643 and 17*123 similar and conter«» niinous. CASE IV. To Jim! whether the decimal fraction^ equal to a given vulgar one, be finite or infinite^ and cf hoiv many places the repe" tend will consist, RULE.* I. Reduce the given fraction to its least terms, and di- vide tlie denominator by 2j 5 or i o, as often as possible. ' 2. If consist of an equal or greater number of figures at pleasure : thus 4 maybe transformed to -44, or '444, or -44, &c. Also •57 = *5757='5757=!l"575 J ^nd so on; which is too evident to need uny further demonstration. * In dividing I 'ooqo, &c. by any prime number whatever, ex- cept 2 or 5, tlie figures in the quotient will begin to repeat over »2ain as soon as the remainder is t. And since 9999, &c. is less than N 9^ ARITHMETIC. 2. If the whole denominator vanish in diviciing by 2, 5 or 10, the decimal will be finite, and will consist of so ma- ny places as you perform divisions. 3. If it do not so vanish, divide pppp, ^cc by the result, till nothing remain, and the number of 9s used will shew the number of places in the repetend ;* wliich will begin after so many places of figures, as there were los, 2s or 5s, used in dividing. EXAMPLES. T. Required to find whether the decimal equal to -"--""- be finite or infinite ; and, if infinite, how many places the repetend will consist of. a a 2 ■rrr^~ 2itV [^ I 4 | 2 ] i ; therefore the decimal is finite, and consists of 4 places. 2. Let than icooo, &c. by i, therefore 9999, &c. divided by any num. ber whatever will leave o. for a remainder, wlien the repeating fig- ures are at their period. Now whatever niunber of repeating figures we have, when the dividend is i, there will be exactly the same number, when the dividend is any other number whatever. For the product of any circulating number, by any other given num- ber, will consist of the same number of repeating figures as before. ^Thus, let '507650765076, &c. be a circulate, whose repeating part is 5076. Now every repetend (5076) being equally multi- plied, must produce the same product. For though these products will consist of more places, yet the overplus in each, being ahke, will be carried to the next, by which means, each product will be equally increased, and consequently every four places will con- tinue alike. And the same will hold for any other number what- ever. Now hence it appears^ that the dividend m^ay be altered at pleasure, and the number of places in the repetend will still be the same : thus -r^=:'9o, and -rV> or ttX3='27, where the number of places in each is alike, and the same will be true in all cases. ADDITION OF CIRCULATING DECIMALS. 99 2. Let -j^ be the fraction proposetf. 3. Let ~ be the fraction proposed. 4. Let ~^^ be the fractioa proposed. 5. Let -g-y'^^: be the fraction proposed. Addition of Circulating Decimals. RULE.* 1. Make the repetends similar and conterminous, and find their sum as in common addition. 2. Divide this sum by as many nines as there are places in the repetend, and the remainder is the repetend of the sum \ which must be set under the ligures added, with cyphers on the left hand, when it has not so many places as the repetends. 3. Carry the quotient of this division to the next col- umn, and proceed with the rest as in finite decimals. EXAMPLES. I. Let 3-6+78-3476-f 735-3+375+-27+i87-4be add- ed together. Dissimilar. Similar and conterminous. 3'6 = y6666666 78-3476 = 78-3476476 735*3 ^ 735*3333333 375* = 375-0000000 •27 = 0-2727272 187-4 = 187-4444444 1 3 80-0648 1 93 the sum. In this question, the sum of the repetends is 2648 191, which, divided by 999999, gives 2 to carry, and the re- mainder is 648193. 2. Let * These rules are both evident from v/hat has been said in re- duction. XQp ARITHMETIC. 2. Let 53pi-357 + 72-384-i87-2i+4-29<^5 + 2i7'849^ -J-42*i7(5+*523 + 58-30048 be added together. Ans. 5974-I037I. 3. A4d 9'8i4-[-i'5 + 87'26-}-'o834-i24'09 together. Ans. 222-75572390, 4. Addi62+i34-o9 + 2-93+97-264.3-769236+99-o8;5 + i'5+*8i4 together. Ans. 501-62651077. Subtraction of Circulating Decimals. RULE. Make the repetends similar and conterminous, and sub- tract as usual ; observing, that, if the repetend of the sub- trahend be greater than the repetend of the minuend, then the right-hand figure of the remainder must be less by uni- ty, than it would be, if the expressions were finite, EXAMPLES. I. From 85*62 take 13*76432. 85-62 = 85-62626 13-76432 = 13-76432 71-8.6193 the difference. 2. From 476-32 take 84*7697. Ans. 391*552/1. 3. From 3*8564 take '0382. Ans. 3*81. Multiplication of Circulating Decimals, RULE. I. Turn both the terms into their equivalent vulgar frac- tions, and find the product of those fractions as usual. 2. Turn DiVISlON OF CIP.CULATING DECIMALS. lOI z. Turn the vulgar fraction, .expressing the product, in- to an equivalent decimal, and it will be the product r*^ ouired. EXAMPLES. £. Multiply '36 by '25. 0^—79 — T"i- ^5 — 90 AX|-|-=:7Vo='0929 the product. 2. Multiply 37*23 by '26. xVns. 9'928. 3. Multiply 8574-3 by 87*5. Ans. 75<='73^'5i^- 4. Multiply 3*973 by 8. Ans. 31*791. 5. Multiply 49640-54 by -70503. Ans. 34998-4199003, 6. Multiply 3-145 by 4*297. Ans. i3*5i<^9533- Division of Circulating Decimals. RULE. 1. Change both the divisor and dividend into their equivalent vulgar fractions, and find their quotient as usual. 2. Turn the vulgar fraction, expressing the quotient, into its equivalent decimal, and it will be the quotient re- quired. EXAMPLES. I, Divide -36 by •25. '36=if-^- ' 9 "— * T •2s=U A^S=Axl?=rlf?=ii?i==i-4229249oii85770750988i the quotient. 2- Divide 3i9'28oo7i 12 by 764*5. Ans. -4176325. 3. Divide 102 ARITHMETIC. 3. Divide 234*6 by "]. Ans. 301-714285. 4. Divide 13*5 169533 by 4-297, Ans 3-145. OF PROPORTION IN GENERAL JN UMBERS are compared together to discover the re- lations they have to each other. There must be tvv^o numbers to form a comparison : the number which is compared, being written first, is called tlie antecedent ; and that to which it is compared, the conse^ quent. Thus of these numbers 2 : 4 : : 3 : 6, 2 and 3 are called the antecedents *, and 4 and 6, the consequents. Numbers are compared to each other two different ways : one comparison considers the difference of tfee two numbers, and is called arithmetical relation^ the difference being some- times named the arithmetical ratio ; and the other considers their quotient, and is termed geometrical relation, and the quo- tient the geometrical ratio. So of these numbers 6 and 3, the difference or arithmetical ratio is 6 — 3 or 3 j and the geometrical ratio is -- or 2. If tw^o or more couplets of numbers have equal ratios, or differences, the equality is named proportion ; and their terms similarly posited, that is, either all the greater, or all the less, taken as antecedents, and the rest as conse- quents, are called proportionals. So the two couplets 2, 4^ and 6, 8, taken thus, 2, 4, 6, 8, or thus, 4, 2, 8, 6, are arithmetical proportionals j and the two couplets 2, 4, and 8, 16, PROPORTION. 103 8, 16, taken thus, 2, 4, 8, 16, or thus, 4, 2, 16, 8, are ge- ometrical proportionals.* Proportion is distinguished into continued and discontinued^ If, of several couplets of proportionals written down in a serieS) the difference or ratio of each consequent and the antecedent of the next following couplet be the same as the common difference or ratio of the couplets, the pro- portion is said to be continued^ and the numbers themselves a series of continued arithmetical or geometrical proportio7Ja!s> So 2, 4, 6, 8, form an arithmetical progression *, for 4 — 2 = 6 — 4=28 — 6 = 2 ; and 2, 4, 8, 16, a geometrical pro- gression -, for 4- = ^= '^ - 2. But if the difference or ratio of the consequent of one couplet and the antecedent of the next couplet be not the same as the common difference or ratio of the couplets, the proportion is said to be discontinued. So 4, 2, 8, 6, are in discontinued arithmetical proportion ; for 4-— 2=8 — 6=2, but 8 — 2 = 6 j also 4, 2, 1 6, 8, are in discontinued geomet'^ rical proportion ; for ^= ^ = 2, but '-/ = 8. Four numbers are directly proportional y when the ratio of the first to the second is the same as that of the third to the fourth. As 2 : 4 : : 3 : 6. Four numbers are said to be reciprocally or inversely proportional, when the first is to the second as the fourth is to the third, and vice versa. Thus, 2, 6, 9 and 3, are reciprocal proportionals ; 2:6:13:9. Three or four numbers are said to be in harmonical pro- portion, v/hen, in the former case, the difference of the first and * In geometrical proportionals a colon Is placed between the terms of each couplet, and a double colon between the couplets ; in arithmetical proportionals a colon may be tmned horizontally between the terms of each couplet, and two colons written be- tween the couplets. Thus the above geometrical proportionals are written thus, 2 : 4 : : 8 : 16, and 4:2:; 16:8.: the arithmet- ical, 2 •• 4 : : 6-8, and 4 •• 2 : : 8 •• 6. 1^4 ARITHMETIC. ancl second is to the difference of the second and third, a5 the first is to the third ; and, in the latter, when the dif- ference of the first and second is to the difference of the third and fourth as the first is to the fourth. Thus, 2, 3 and 6 ; and 3, 4, 6 and 9, are harmonical proportionals j for 3 — 2== I 16^-3 = 3 : : 2 : 6; and 4—^3 = 1 : 9 — 6=3 • • 3 • 9- Of four arithmetical proportionals the sum of the ex- tremes is equal to the sum of the means.* Thus of 2 •• 4 : : 6 .. 8 the sum of the extremes (2-j-8)zz the sum' of the means (4+6)zzio. Therefore, of three arithmet- ical proportionals, the sum of the extrenfes io double the mean. Of four geometrical proportionals the product of the extremes is equal to the product of the"^ means.f Thus, of 2 : 4 : : 8 : 16, the product of the extremes (2X16) is equal to the product of the means (4X8)1:332. Therefore of three geometrical proportionals, the product of the ex- tremes is equal to the square of the mean. Hence it is easily seen, that either extreme of four geo- metrical proportionals is equal to the product of the means divided * Demonstration. Let the four aiithmetical proportionals be ^, B, C, D, viz. A- B '.\ C- D \ then, J—B=C—D and B-\-D being added to both sid^s of the equation, A — B-\-B-\-D zzC — Z)-|-^-f-Z) ; that is, A-\-D the sum of the extremes z=.C-\-B the sum of the means. — And three A, B, C, may he thus expressed, A .. B : : B - C ', therefore yl-^C:=B-^B—2B. Q^ E. 1>. -}■ Demonstratioj^. Lettlie proportion he A : B : : C : Dy A C and let^=2>~^ 5 ^^^^" Az=:Br, and Cz=.Dr ; multiply the for- mer of these equations by /), and the latter by B ; then ADzz BrD, and CBzzzDrB, aftd consequently AD the product of the extremes is equal to BC the product of die means. — And three may be thus expressed, A : B : : B : C, therefore A C=zBx Bz=B\ Q, E. D. PROPORTION. 105 » divided by the other extreme ; and that either mean is equal to the product of the extremes divided by the other mean. SIMPLE PROPORTION, or RULE OF THREE. The Rule of Three is that, by which a number is found, having to a given numbdr the same ratio, which is between two other given numbers. For this reason it is sometimes named the Rule of Proportion, It is called the Rule of Three, because in each of its questions there are given three tiumhers at least. And be- cause of its excellent and extensive use, it is often named the Golden Rule, RULE.* I. Write down the number, which is of the same kind -vtrith the answer of number required. 2. Consider * Demonstration. The following observations, taken col- lectively, form a demonstration of the rule, and of the reductions mentioned in the notes subsequent to it. I. There can be comparison or ratio between two numbers, only when they are considered either abstractly, or as applied to things of the same kind, so that one can, in a proper sense, be contained in the other. Thus there can be no comparison between 2 men and 4 days ; but there may be between 2 and 4, and between 2 days and 4 days, or 2 men and 4 men. Therefore, the 2 of the 3 given nupibers, that are of the same kind, that is, the first and third, when they are stated according to the rul«, are to be com- pared together, and their ratio is equal to that, required between the remaining or second number and the fourth or answer. 2. Though O I06 ARITHMETIC. 2. Consider whether the answer ought to be greater or less than this number : if greater, write the greater of the two remaining numbers on the right of it for the third, and 2. Though numbers of the same kind, being either of the same or of different denominations, have a real ratio, yet this ratio is the same as that of the two numbers taken abstiactly, only when they are of the same denomination. Thus the ratio of il. to 2I, is the same as that of i to 2 =ri- ; is. has a real ratio to 2I. but it is not the ratio of i to 2 ; it is the ratio of is. to 40s. that is, of i to 40 ^=:fo' Therefore, as the first and third numbers have the ratio, that is required between the second and answer, they must, if not of the same denomination, be reduced to it ; and then their ratio is that of the abstract numbers. 3. The product of the extremes of four geometrical propor- tionals is equal to the product of the means ; hence, if the pro- duct of two numbers be equal to the- product of two other num- bers, the four numbers are proportionals j. and if the product of two numbers be divided by a third, the quotient will be a fourth, proportional to those three numbers. Now as the question is re- solvable into this, viz. to find a number of the same kind as the second in tlie statement, and having the same ratio to it, that die greater of the otJier two has to the less, or the less has to the great* er ; and as these two, being of the same denomination, may be con* stdered as abstract numbers ; it plainly follows, that the fourth number or answer is truly found by multiplying the second by one of the other two, and dividing the product by that, which remains. 4. It is very evident, that, if the answer must be greater than the second number, the greater of the other two numbers must be the multiplier, and may occupy the third place ; but, if less, the less number must be the multiplier. 5. The reductibn of the second number Is only performed for convenience in the subsequent multiplication and division, and not to produce an abstract number. The reason of the reduction of the PROPORTION. icy and the other on the left for the first number or term ; but if less, write the less of the two remaining numbers in the third place, and the other in the first. 3. Multiply the quotient, of the remainder after division, and of the product of the second and third terms, when it cannot be divided by the first, is obvious. 6. If the second and third numbers be multiplied together, and the product be divided by the first ; it is evident, that the answer remains the same, whether the number compared with the first be in the second or third place. Thus is the proposed demonstration completed. There are four other methods of operation beside the general one given ^bove, any of which, when applicable, performs the work much more concisely. They are these : 1. Divide the second term by the first, multiply the quotient by the third, and the product will be the answer. 2. Divide the third term by the first, multiply the quotient by the second, and the product will be the answer. 3. Divide the first term by the second, divide the third by the quotient, and the last quotient will be the answer. 4. Divide the first term by the third, divide the second by the quotient, and the last quotient will be the answer. The general rule above given is equivalent to those, which are usually given in the direct and inverse rules of three, and which are here subjoined. The Rule of Three direct teacheth, by having three num- bers given, to find a fourth, that shall have the same proportion to the third, as the second has to the first. RULE. I. State the question ; that is, place the numbers so, that the fkst and third may be the terms of supposition and demand, and the second of the same kind with the answer required. 2. Bring Io8 ARITHMETIC. 3. Multiply the second and third terms together, dlvid 102 18 8i6 .02 1835 73 14688 12852 12 34)143208 ( 5967 232 ■ 160 2,0)49,7 $ j68 Ans. 24I. 17s. 3d. — £-2^ 17 3 The rule is founded on this obvious principle, that the magni- tude or quantity of any effect varies constantly in proportion to the varying part of the cause : thus, the quantity of goods bought is in proportion to the money laid out ; the space gone over by an uniform motion is in proportion to the time, &c. The truth of the PKOPOHTION. 109 Note i. It is sometimes most convenient to multiply and divide as in compound multiplication and division j and sometimes it is expedient to multiply and divide ac- cording to the rules of vulgar or decimal fractions. But ■when the rule, as applied to ordinary inquiries, may be made very evi- dent by attending only to the principles of compound multiplica- tion and division. It is shewn in multiplication of money, that the price of one, multiplied by the quantity, is the price of the .whole ; and in division, tliat the price of the whole, divided by the quantity, is the price of one. Now, in all cases of valuing goods, &c. where one is the first term of the proportion, it is plain, that the answer, found by this rule, will be the same as that found by multiplication of money ; and, where one is the last term of the proportion, it will be the same as that found by division of money. In like manner, if the first term be any number whatever, it is plain, that the product of the second and third terms will be great- er than the true ansv/er required by ns much as the price in the second term exceeds the price of one, or as the-first term exceeds an unit. Consequently this product divided by the first term will give the true answer required, and is the rule. There will sometimes be difficulty in separating the parts of complicated questions, where two or more statings are required, and in preparing the question for stating, or after a proportion is wrought J but as there can be no general directions given for the management of these cases, it must be left to the judgment and experience of the learner. The Rule of Three inverse teacheth, by having three num- bers given, to find a fourth, that shall have the same proportion to the second, as the first has to the third. If more require morcy or less require less, the question belongs to the rule of three direct. But if more require less, or less require ff^crc; it belongs to the rule of three inverse. * ' Note. 1 10 ARITHMETIC. when neither of these modes is adopted, reduce the com- pound terras, each to the lowest denomination mentioned in it, and the first and third to the same denomination ; then will the answer be of the same denomination with the second term. And the answer may afterward be brought to any denomination required. Note 2. When there is a remainder after division, re- duce it to the denomination next below the last quotient, and divide by the same divisor, so shall the quotient be so many of the said next denomination ; proceed thus, as long as there is any remainder, till it is reduced to the low- est denomination, and all the quotients together will be the answer. And when the product of the second and third terms cannot be divided by the first, consider that pro if there be more term$ of each kind ; writing the terms under eacl^ other, which fall on the same side of the middle term. METHOD OF OFERATJQl/, I. By several operations.-rr-TTike the two upper terms and the middle term, in the same order as they stand, for the. first stating of the rule of three ; then take the fourth number, resulting from the first stating, for the middle term, and the two next terms in the general stating, in the same order as they stand, for the extreme terms of the second * The reason of this rule for stating, and of the methods of operation, may be easily shewn from the nature of simple propor- tion ; for every line in this case is a particular stating in that rule. And, therefore, with respect to the second method, it is evident^ that, if all the separate dividends be collected into one dividend, and all the divisor^ into one divisor, their quotient must be tha answer sought. COMPOUND PROPORTION. 127 second stating -, and so en, as far as there are any numbers in the general stating, always making the fourth number, resulting from each simple stating, the second term of the next. So sjiall the last resulting number be t}\^ answer required. 2. By one cperatian. — Multiply together all the terms in the first place, and also all the terms in the third place. Then multiply the latter product by the middle term, and divide the result hj the former product j and the quo^tieut \\dll be the answer required. Note i. It is generally best to work by the latter rheth- od, viz. by one operation. And after the stating, and be- fore the commencement of the operation, if one of the first terms,- and either the middle t^rm, or one of the last terms, can be exactly divided by one and the same number, let them be divided, and the qiiotients used instead of them ; which will much shorten the w^ork. '• Note 2. The first and third terms of each llrre, if of different denominations^ must be redu(!ed to the same de- nomniation. EXAMPLES. I. How many men can complete a trench of 135 yards long in 8 days, provided i6men*can dig 54 yards in 6 davs ? General Stating. 54 8 -yds. or 2 7 ,^ fi^Syds. ore! y > : 16 men ;: -! ^j-' r • days, ar 43 [^ o days, or 33 First Metftod. yds. men. yds. days. men. days. 54>r-27=2 z : 16 :: 5 : 4 : 40 : : 5 : 135-^27=^ 5 3 6 -f- 2^:3 2)80(40 men. 4)120(30 men, the answer. 8 12 Second liS ARITHMETIC, SiicoND Method. 2 1 4 S .,«,, y r 8 • : i6 : r 15 80 16 • 8)240(30 men, the answer as before. 24 o 2. If lool. ill one year gain 5]* interest, what will be tlie interest of 750I. for 7 years ? Ans. 262I. los. 3. What principal will gain 262I. los. in 7 years, at 5L per cent, per aiumm ^ Ans. 75 ol. 4. If a footman travel 130 miles in 3 days, when the days are 12 hours long 5 in howv.tfiany days, of 10 hours- each, luay he travel 360 miles ? Ans. 9I-J- days. 5. If 120 bushels of corn can serve 14 horses 56 days ;. how many days will 94 bushels serve 6 horses ? Ans. I02~ days. 6. If 70Z, 5dwts. of bread be bought at 4-|d. when corn is at 4s. 2d. per bushel, what weight of it may be bought for IS. 2d. when the prica of the bushel is 5s. 6d. ? Ans. lib. 40Z. 3-|4Ttlwts, 7. If the carriage of I3cwt. iqr. for 72 miles be 2I. ics. 6d. what will be the carriage of 7cwt. 3qrs. for 112 miles ? Ans. 2I. 5s.- iid. iT^q^ 8. A wall, to be built to the height of 27 feet, was raised to the height of 9 feet by 12 men in 6 days j hov/ many men must be employed to finish the wall in 4 day.^, at the same rate of working ? Ans. 36 men. 9. If a regiment of soldiers, consisting of 939 men, can eat up 351 quarters of wheat in 7 months j how many soldiers will eat up 1464 quarters in 5 months, at thiit rate ?. Ans. 5433-rVV' 10. If CONJOINED PROPORTION. 1^9 ID. If 248 men, in 5 days of 11 hours each, dig a trench 230 yards long, 3 wide and 2 deep *, in how many days of 9 hours long, will 24 men dig a trench of 420 yards long, 5 wide and 3 deep ? Ans. 288-^^. i CONJOINED PR^OPORTION. Conjoined Proportion is when the coins, weights, or measures, of several countries are compared in the same question ; or it is the joining together of several ratios, and the inferring of the ratio of the first antecedent and the last consequent from the ratios of the several antece- dents and their respective consequents. Note i. The solution of questions, under this rule, may frequently be much shortened by cancelling equal numbers, when in both the columns, or in the first column and third term, and abbreviating those that are commens- trable. Note 2, The proof is by so many statements in the single rule of three, as the nature of the question requires, CASE I. Jf^hen it is required to find how many of the last kind of coin, iveight, or measure, menfmied ifi the question^ are equal tj a given number of the first, RULE. 1. Multiply continually together the antecedents for the first term, and the consequents for the second, and make the given number the third. 2. Then find the fourth term, or proportional, which will be the answer required, EXAMPLES. R 130 ARITHMETIC. EXAMPLES. , " J. If lolb.at Boston make plb. at Amsterdam ; polb. at Amsterdam^ iislb. at Thoulouse j how many pounds at Thoulotise are equal to 5olb. at Boston ? Ant. Cons. 10 : 9 ■90 : 112 900 ; ioq8 : : 50 ;. 50 ) 5 0400(56 the answer. 4500 5400 5400 Or by abbreviation. 10 : 9:: 50 1 0 : I : : 50 i : i : : 5 90 : 112 10 : 112* 10 : 112 2 : 112 :: I s6 ^6 the answer, 2. If 20 braces at Leghorn be equal to 10 vares at Lis- bon ; 40 vares at Lisbon to 80 braces at Lucca j how ma- ny braces at Lucca are equal to 100 braces at Leghorn ? Ans. 100 braces, CASE It. W7:en it is required to fitid hoiv many of t,he first hind of coift, ^weighty or measure^ nienticned in the question^ are equal to a given number of the last. RULE. Proceed as in the first case, only make the product of the consequents the first term, and that of the antecedents, the second. EXAMPLES. * In performing this example, the first abbreviation is obtained by dividing 90 and 9 by their common measure 9 ; the second by dividing 10 and 50 by their common measure 10 ; the third by dividing 10 and 5 by their common measure 5 ; and the fourth, or answer, by dividing 2 and ;i2 by their common measure 2. BARTER, 131 EXAMPLES. I. It* looib. in America make 951b. Fiemlsh ; and 191b. Flemish, 251b. at Bolognia ; how many pcund^ iu Apaerir ca are equal to 5olb. at Bolognia ? Cons. Ant. 95 : ICO 25 : 19 . 475 190 237^ : 1900 : : 50 50 )9jooo(4oIb. the answer. 9500 o Or by abbrevlatioru 95 ; 100 5 : 100 5:4 >^5 » 19 ;: 50 25 : I :: 50 i : i ;: 50 i : 4 :; 10 : 4 Ans. 4olb. 2. If 251b. at Boston be 22lb. at Nuremburg -, 881b. at Nuremburg, 921b. at Hamburg ; 461b. at Hamburg, 491b. at Lyons ; how .many pounds at Boston are equal to 981b. at Lyons .'' Ans. looJb. 3. If 6 braces at Leghorn make 3 ells English ; 5 ells English, 9 braces at Venice ; how many braces at Leghorn will make 45 braces at Venice ? Ans. 50 braces. I BARTER. Barter is the exchanging of one commodity for anoth- er, and directs traders so to proportion their goods, tjiat neither party may sustain loss. RULE. 132 • ARITHMETIC. RULE.* Find the value of that commodity, whose quantity is giv- en ; then find what quantity of the other,' at the rate pro- posed, you may have for the same money, and it gives the answer required. EXAMPLES. I. How many dozen of candles at 5s. 2d. per dozen, must be given in barter for 3cwt. 2qrs. of tallow at 37s, 4d, per cwt. ? qr. 6. d. 4 : 37 4 : 12 cwt. qr. : : 3 2 ; 4 448 14 1792 448 M 4)6272 12)1568 2,0)13,0 8 61. los. s. d. doz. 52:1 12 8d. 1. s. d, : : 6 10 8 20 62 130 12 62)1568(25 124 328 310 18 12 62)216(3 186 Ans. 25 doz. and 3. — » 30 2, How This rule is evidently, only, an application of the rule pf three. LOSS AND GAIN. ' ^33 2. How much tea at 9s. per ib. can I have in barter, for 4 cwt. 2qrs. of chocolate at 4s. per lb. ? Ans. 2cvvt. 3. How many reams of paper at 2s. p-^-d. per ream must be given, in bartet", for 37 pieces of Irish cloth at il. 12s. 4d. per piece ? Ans. 428I4. 4. A delivered 3 hogsheads of brandy at 6s. 8d. per gal. to B, for 1 26 yards of cloth j what was the cloth per yard ? Ans. 10s. 5. A and B barter •, A hath 4icwt. of hops at 30s. per cwt. for which B gives him 20I. in money, and the rest in prunes at 5d. per pound ; what quantity of prunes must A receive ? Ans. lycwt. 3qrs. 41b. 6. A has a quantity of pepper, weight net i6oolb. at 1 7d. per lb. which he barters with B for two sorts of goods, the one at ^d. the other at 8d. per lb. and to have ~ in money, and of each sort of goods an equal quantity : how many pounds of each must he receive, and how much in money ? Ans. i394-|-|-lb. and 37I. 15s. 6yd. LOSS AND GAIN. Loss AND Gain is a rule that discovers what is got -or lost in the buying or selling of goods 5 and instructs mer- chants and traders to raise or lower the price of their goods, so as to gain or lose a certain sum per cent. &;c. Questions 'm this rule are performed by the Rule of Three. 134 ARITHMETIC. EXAMPLES. r. How must I sell tea per pound, that cost mc 13s. ^d» to gain at the rate of 25 per cent. ? 1. 1. s. d. ICO ; 125 : : 13 5 ? 12 161 805 -522 161 1,00)201,25 ■II « » 12)201 25 i6s. pd. AV the answer. Or thus : 4)135. 5d. 3 4 1 6s. 9^d. the same as before. 2. At 33. 6d. profit In the pound, how much per cent. ? Ans. 17I. I OS. 3. Bought goods at 4-^d. per lb. and sold them at the rate of 2I. 7s. 4d. per cwt. what was the gain per cent. ? Ans. 12I. 135. I id. 4. Bought cloth at 7s. 6d. per yard, which not proving so good as I expected, I am resolved to lose 1 7^ per cent. 1-y it : how must I sell it per yard ? Ans. 6s. 2-^d, 5. If I buy 28 pieces of stuffs at 4I. per piece, and sell 1 3 of the pieces at 61. and 8 at 5I. per piece ; at what rate per piece must I sell the rest to gain 20 per cent, by the whole ? Ans. 2I. 6s. 10^. 6. Bought 40 gallons of brandy at 3s. per gallon, but by accident 6 gallons of it are lost ; at what rate must I sell the^emainder per gallon, and gaiii upon the whole prime cosjp^t the rate of roper cent. ? ' Ans. 3s. io|-d. 7. Sold FELLOWSHIP. 135 7. Sold a repeating watch for 175 dollars, and by so do- ing lost 17 per cent, whereas I ought in deahng to have cleared 20 per cent, how much was it sold for under the just value ? Ans. 23I. 8s. o-^d. FELLOWSHIP. FelloW'ship Is a general rule, by which merchants, &c. trading in company, with a joint stock, determine each per- son's particular share of the gain or loss in proportion to his share in the joint stock. By this rule a bankrupt*s estate may be divided among his creditors 5 as also legacies adjusted, when there is a de- ficiency of assets or effects. SINGLE FELLOWSHIP. Single Fellowship is when different stocks are employed for any certain equal time. HULE.* As the whole stock is to the whole gain or loss, so is each man's particular stock to his particular share of the gain or loss. METHOD OF PHOOF* Add all the shares together, and the sum will be equal to the gain or loss, when the question is right. EXAMPLES, * That the gain or loss, in this rule, is in proportion to their stocks is evident : for, as the times the stocks are in trade are equal, if I put in \ of the whole stock, I ought to have t of ;he whole gain ; if my part of the whole stock be f , my share of the whole gain oi* loss ought to be 3- also. And, generally, if I put in ^ of the stock, I ought, to have ^ part of the whole gain or loss; ^at is, the same ratio, that the whole stock }ias to the whole gain or loss, must each person's particular stock have tc his particular gain or loss. S^di ARITHMETIC. EXAMPLES. I. Two psrsons trade together ; A put into stock. 130!, and B 220L and they gained .500L. what ia each person's share thereof ? i30-f-22o=:35o 350 : 500 : : 130 ; n<=> 15000 500 « 35,o)65oo,o(ia5l, 300 200 20, 35)50.0(^5- 150 10 12 35)i2o(3d. IS 4 35)6o(iq, 25 350 : 500 : : 220 220 10000 1000 35,0)11000,0(3141. 15a 10 20 185I. 14s. 3id. ||=A*s share. 314I. 5s. 8id.|f=B's share. - 500I. OS. od. the Proof. 35)8o(2 Ans. IcA. 40I. and 60L 4. Three persons make a joint stock- ; A put in 184!* I OS. B 96I. 15s. and C 76I. 5s. they trad-c and gain 22 oh I2s. what is each person's share of the gain ? Ans. A 1 13I. 16s. 4ff, B 59I. 143. Vtt. C 47'- 's. -'^. ^. Three merchatits A, B and C, freiglit a ship with 340 tuns of wine ; A loaded no tuns, B 97, and C the r^st. \i\ a Gtorm the seamen were obliged to throw 85 tuns overboard j how much must each sustain of the loss ? A 27i, B 24^, and C 33i. 6. A ship v/orth 860I. being entirely lost, of which ^ belonged to A, -^ to B, and the rest to C ; what loss will each sustain, supposing. 500I. of her to be insiired .f* Ans. A 45!. B 90I. and C 225L 7/ A bankrupt is indebted to A 275I. 14s. to B 304!. 7s. to C 152I. and to D 104I. 6s. His estate is worth only 675I. 15s. how must it be divided ? Ans. A 222I. 15s. 2d. B 245I. 1 8s. i-^-d. C I22h 'kSs* 2{-d. and D 84]. 5s. 5d. 8. A and B, venturing equal sums of money, clear by joint trade 154I. By agreement A v/as to have 8 per cent, because he spent his time in the execution of the project, and B was to have only 5 per cent. 3 v/hat was A allowed for his trouble ? . Ans. 35I. 105. 97-Vd. DOUBLE FELLOWSHIP. Double Fellowship is when different or equal stocks arc employed for different times* RULr, 138 ARITHMETIC. U U L E.* Multiply each man's stock into the tin^e of iis continu- ance, tlien say, As the total sum cf all the products is to the wliole gain or less. So is each man's particular product to his particular share of the gain or loss. EXAMPLES. I. A and B hold a piece of ground in cojumcn, for which they are to pay 36I. A put in 23 oxen for 27 days, and B,2i oxen for 35 days 5 what part of the rent cu^ht: each man to pay ? 23x27=621 1356 : 36 ; : 6:?i : 21x35=735 <^2i 135^ 3^ 216 >» ,« 356)22356(161. 135^^ 8796 8136 • 660 20 1356)13200(93. 12204 996 12 I356)ii952(8d» 10848. 1104 4 I356)44i6(3q. 4068 348 135^ * Mr. Malcolm, Mr. Ward, and several other authors, have given an analytical investigation of this rule : but the most general aad FELLOWSRir. 139, 356 : 36 : : 735 73? 180 108 252 ^56)26460(191. 135^ 1 2900 12204 . 20 1356)13920(105. 1356 360 12 16I. 9c.. Sja.y,VV=A'se 19I. I OS. 3d. 4ff|-B's s »35%32o(3(l. 4068 36I. OS. od. the Proof. 252 . 4 1008 % Three graziers hired a piece of land for 6cl. los. A put in 5 sheep for 4^- months, B put in 8 for 5 months, and C put in 9 for 6-^ months : how much must each pay of the rent ? Ans. A nl. 5s. B 20I. and C 29I. 5s. 3. Two merchants enter into partnership for 1 8 months ; A put into stock at first 200I. and at the end of 8 months he put in lool. more ; B put in at first 550!. and at the end of 4 months took out 140I.. Now at the expiAttion of the time and elegant method, perhaps, is that, v/hicli Dr. Kutton has giv- en in his Arithmetic, viz. When the times are equal, the shares of the gain or loss are evi- dently as the stocks, as in Single Fellowship ; and when the stocks are equal the shares are as tlie times '; -where Fore, when nei- ther are equal, the shares must be as their products. 14® ARI'THMETie. time they find they have 'gained 526I. wIiul ib catjh man'tj just share ? Ans. A 192L i(;s. od. -~-y ., B 3331.0S. iiid.-'^; 4. A, with a capita] of loool. began trade. January 1, 1776, and meeting with success in business he took in B as -a partner, with a capital of 1500I. on the first of March following. Three months after that they -admit C as a third partner, who brought into stock aSooL'and after trading together till the first of the qext year, they find, there has been gained since A's commencing business, 1776L los. flow must this be divided among the partners ? Ans. A 45 7I. 93. /^~d. B 57 iL 163. 8^d. C 747I. 3s. 11^4. ALLIGATION. Alligation teaches how to mix several simples of dif- ferent qualities, so that the composition may be of a mid- dle quality ; and is commonly distinguished into two prin- cipal cases, called Alji^i^tion nicdial and AH'igation alternate* ALLIGATION MEDL4L. Alligation medial is the method of finding the rate of the compound, from having the rates and quantities of the sev- eral simples giycn. * RULE.* Multiply each quantity by its rate" ; then divide the sum of the products by the sum of the quantities, or the whole conipositian, * The truth of this rule is too evident to netd a demonstration. Note. If an ounce or any other qiiMctity of pure gold be re- duced into 24 equal parts, these parts are cuiicd carats ; but gold ALLIGAT10>: MEDIAL. t^% composition, and the quotient will be the rate of the com- pound required. EXAMPLES. I. Suppose 15 bushels of wheat at 5s. per bushel, and 12 bushels of rye at 3s. (k\. per bushel were mixed togeth- er : how must die compcund be sold per bushel without loss or gain? - 00 ^5 4; 15 12 12 300 60 5^4 -7 9'^o (JOO 27)i404(52d.=43. 4d. the answer 135. 54 54 2. A composition being made of 51b. of tea at 7s. per pound, 91b. at 8s. 6d. per pound, and i4-^lb. at 53. lod. per pound, what is a pound of it worth ? Ans. 6s. io~d. .3. Mixed 4 gallons of wine at 4s. lod. per gallon, with 7 gallons at ^s. 3d. per gallon, and p-^ gallons at 5s. 8d. j)er gallon 5 what is a gallon of this cpmpofition worth ? Ans. 5s. 4-^d. 4. A goldsmith melts 81b. si^^' ^^ g^^^ bullion of 14 or-ats fine, with i2lb. S^-oz. of 18 carats fine : how many- carats fine is this mixture ? Ans. 1 64-^4- carats. J O o is often mixed with some baser metal, which is called the alloy, and the mixture is said to be of so many carats fine, according to the proportion of pure gold contained in it : thus, if 22 carats of pure gold and 2 of alloy be mixed together, it ii said to be 22 ca- rats fine. ■If any one of the simples be of little or no value with respect to the rest, its rate is supyosed to be nothing, as water mixed witli wine^ and alloy with gold or silver. 14^ ARITHIvIETlC. 5. A icuncr melts lolb. of gold of 20 carp.ts fine with loib. of 18 carats fine ; how much alloy must he put to it to make it 22 carats fine ? Ans. It is not fine enough by 3^ carats, so that no al- io v must be put to it, but more gold. ALLIGATION ALTERNATE. Alligation alternate is the method of nnding \vhat quan- tity of any number of simples, whose rates are given, will compose a mixture of a given rate ; so that it is the re- verse of alligation medial, and may be proved by it. 1. Write the rates of the simples in a column under each other. 2. Connect * Demonstration. By connecting the less rate to the greater, and placing the differences between them and the mean rate alternately, the quantities resulting are such, that there is pre- cisely as much gained by one quantity as is lost by the other, and therefore the gain and loss upon the whole are equal, and are ex- actly the proposed rate : and the same will be true of any other- two simples, managed according to the rule. In like manner, let the number of simples be what it may, and with how many soever each is linked, since it is always a lisss with a greater than the mean price, there will be an equal balance of loss and gain between every two, and consequently an equal bal- ance on the whole. Q^ E. D. It is obvious from the- rule, that questions of this sort admit of a great variety of answers ; for, having found one answer, we may find as many more as we please, by only multiplying or di- viding each of tlie quantities found by 2, 3, or 4, &c. the reason of which is evident ; for, if two quantities of two simples make a balance of loss and gain, with respect to the mean price, so must also ALLIGATION ALTERNATE. M3 •1. Connect or link with a continued line the rate of each simple, which is less than that of the compound, with one or any number of those, that are greater than the com- pound •, and each greater rate with one or any number cf the less, 3. Write the difference between the mixture rate and that of each of the simples opposite the rates, with which they are respectively linked. 4. Then if only one difference stand, against any rate, it will be the quantity belonging to that rate ; but if there be several, their sum will be the quantity. EXAMPLES. 1. A merchant would mix v/ines at 14s., 19s. 15s. and 22s. per gallon, so that the mixture may be worth i8s. the gallon : what quantity of each must be taken ? 18- H 4 at 14s. 1 at 15s. 19 3 at 195. * of- '»'» The two last rules can want no demonstration, as they evident- ly result from, the first, the reason of which has been already explained. i INVOLUTION. 147 a. A grocer would mix teas at 1 2s. ros. and 6s. with lo\h. at 4s. per pound ; how much of each sort must he take to make the composition worth 8s. per lb ? Ans. 2olb. at 4s. 10 at 6s. 10 at los. and 20 at 12s. 3. How much gold of 15, of 17 and of 22 carats fine, must be mixed with 50Z. of 18 carats fine, so that the composition may be 20 carats fine ? Axis. 50Z. of 15 carats fine, 5 of 17, and 25 of 22. INVOLUTION. A Power is a number produced by multiplying any giv- en number continually by itself a certain number of times. Any number is itself called the Jirst power ; if it be multiplied by itself, the product is called the second ponver^ or the square ; if this be multiplied by the first power again, the product is called the third power ^ or the cube ; and if this be multiplied by the first power again, the product is called the fourth power ^ or biquadrate ; and so on 5 that is, the power is denominated from the rmmber, which ex- ceeds the multiplications by i. Thus, 3 is the first power of 3. 3X311: 9 is the second power of 3. 3X3X3:::i:27 is the third power of 3. 3X3X3X311:81 is the fourth power of 3. &c. &c. And in this manner is calculatecf the following table of powers. TABLE 148 ARITHMETIC. TABLE of the first twelve PoiVERS of the 9 DIGITS, ON GO VO VO ON 1 CO VO ON VO ON N 00 IT- N VO '^ 0 CO ON cc Ti- 0 00 CO CO VO oc '+• CO ON to 0 CO CO CO 1 CO to 0 M CO 00 1 ^2- IN "os c^ VO «o vo N VO 00 CO CO ^ 1 CO 0 ON tr> CO ON ON 00 to CO VO CO VO ON l>. CO VO t-» ? CO CO 0 0 00 VO ON CO Th »o CO N CO 5 ! 0 00 VO rt- 1 CO VO 1 ^ t- 1 CO 10 1 0 ON c^ 00 CO Ti- VO M CO ON 0 CO VO VO CO VO VO VO c< M SO VO VO CO ON ON VO VO ON »^ VO M VO o\ VO r- 8 VO VO 1 VO to 0 1-- ON !>• C< VO CO VO CO CO CI 00 VO r^ M tTk lO VO lO *-* CO lO VO 10 00 to p» VO 0 ON CO to ca CO «o c» VO to VO ON to 00 N CO 00 to N VO % ^ SO ^ VO 10 0 ^ - 00 CO VO VO CO lO VO VO !>. to 00 t t CO On vo CO Ov CO CO ON 00 VO to VO CO 00 VO ON ON to 1 CO »o c^ '^ t 00 VO i N so 00 •0 0 00 VO - 1 - 1 - - 1 - 1 - - - M - 1 - - s s s & & & S ! -S ■! ?o -J- VO 1 •5 00 ON S ^ \ i Pi HI Note i. INVOLUTION, 14^ Note i. The number, which exceeds the multipHca- dons by i, is called the index, or exponenty of the power \ so the index of the first power is i, that of the second power is 2, and that of the third is 3, &c. Note 2. Powers are commonly denoted by writing their indices above the first power : so the second power of 3 may be denoted thus 3*, the third power thus 3^, the fourth power thus 3"*, &c. and the sixth power of 503 thus 503*^. Livolutioji is the finding of powers j to do which 'we have evidently the following RULE. Multiply the g>ven number, or first power, continually by itself, till the number of multiplications be i less than the index of the power to be found, and the last product will be the power required.* Note. "Whence, because fractions are multiplied by taking the products of their numerators and of their de- nominators, they will be involved by raising each of their terms to the power required. And if a mixed number be proposed, * Note. The raising of powers will be sometimes shortened by working according to this observation, viz. whatever two or more powers are multiplied together, their product is the power, whose index is the sum of the indices of the factors ; or if a power be multiplied by itself, the product, will be the power, whose index is double of that, which is multiplied : so if I would find the sixth power, I might multiply the given number twice by it- self for the third power, then the third power into itself v/ould give the sixth power ; or if I would find tlie seventh ])cv. :r, I might first find the third and fourth, and their praduci: v/onii be the seventh ; pr lastly, if I would find the eighth povvcr, I might first find the second, then the second into itsr^f -would be tlse fourth, and this into itself would be the eighth. 150 ARITHMETIC. proposed, either reduce it to an improper fraction, or re- duce the vulgar fraction to a decimal, and proceed by the rule. EXAMPLES. 1. What Is the second power of 45 ? Ans. 2025. 2. What is the square of '027 ? Ans. '000729. 3. What is the third power of 3*5 ? Ans. 42*875. 4. What is the fifth power of .029 ? Ans. '000000020511149. 5. What is the sixth power of 5*03 ? Ans. 16196*005304479729. 6. What is the second power of ~ f Ans. --t EVOLUTION. The Root of a^y given number, or power, is such a number as, being multiplied by itself a certain number of times, will produce the power ; and it is denominated the jftrst, second, third, fourthy ^c. root, respectively, as the number of multiplications made of it to produce the givers power is o, i, 2, 3, &c. that is, the name of the root is taken from the number, which exceeds the multiplications by I, like the name of the power in involution. Note i. The index of the root, like that of the power in involution, is i more than the number of multiplications, necessary to produce the power or given number. ' Note 2. Roots are sometimes denoted by writing y/ before the power, with the index of the root against it : 3 so the third root of 50 is -y/ 50, and the second root of it is ^ 50, the index 2 being omitted, which index is always understood, when a root is named or written without one. But if the power be expressed by several numbers with the sign + or -— , &c. between them, then a line is drawn from EVOLUTION. 151 from the top of the sign of the root, or radical sign, over all the parts of it : so the third root of 47 — 15 is 3 >/47 — ^5* ^^^ sometimes roots are designed like powers, with the reciprocal of the index of the root above the given number. So the 2d root of 3 is 3"^; the 2d root of 50 is 50"=^ ; and the third root of it is 5o"3' j also the third root of 47 — 15 is 47 — 15I . And this method of notation has justly prevailed in the modern algebra ; because such roots, being considered as fractional powers, need no other di- rections for any operations to be made with them, than those for integral powers. Note 3. A number is called a complete power of any kind, when its root of the same kind can be accurately ex- tracted ; but if not, the number is called an Imperfect pow- er, and its root a surd or irrational number : so 4 is a com- plete power of the second kind, its root being 2 ; but an imperfect power of the third kind, its root being a surd number, Evolution is the finding of the roots of numbers either accurately, or in decimals, to any proposed extent. The power is first to be prepared for extraction, or evo- lution, by dividing it from the place of units, to the left in integers, and to the right in decimal fractions, in- to periods containing each as many places of figures, as are denominated by the index of the root, if the power con- tain a complete number of such periods : if it do not, t|[ie defect will be either on the' right, or left, or both ; if the defect be on the right, it may be {supplied by annexing cyphers, and after this, whole periods of cyphers may be annexed to continue the extraction, if necessa- ry *, but if there be a defect on the left, such defective peripd must remain unaltered, and is accounted the first period of the given number, just the same as if it were complete. , Now 1^2 ARITHMETIC. Now this division may be conveniently made by writing a point over the place of units, and also over the last fig- ure of every period on both sides of it ; that is, over every second figure, if it be the second root 5 over every third, if it be the third root, &c. Thus, to point tliis number 2io'^^^g6'i2'j2S > for the second root, it will be 21035896*127350 j but for the third root . 2 103 5 896* 12 73 50 ; and for the fourth 2 103 5 896" 1273 5000. Note. The root will contain just as many places of figures, as there are periods or points in the given power ; and they will be integers, or decimals respectively, as the periods are so, from which they are found, or to which they correspond ; that is, there will be as many integral or deci- mal figures in the root, as there are periods of integers or decimals in the given number. To EXTRACT THE SqUARE RoOT» I. Having distinguished the given number into periods, find a square number by the table or trial, either equal to, or * In order to shew the reason of the rule, it will be proper to premise the following Lemma. Th^e product of any two numbers can have at most but as many places of figures, as are in both the factors,' and at least but one less. Demonstration. Take two numbers consisting of any num- ber of places, but let them be the least possible of those places, viz. unity with cyphers, as 1000 and 100 ; then their product will be I with as many cyphers annexed as are in both the numbers, viz. EVOLUTWN. 153 or the ncpct less than, the first period, a^d put the root of it to the right hand o£ the given number, after the manner of a quotient figure in division, and it virill be the first fig- ure of the root required. ■ 1. Subtract the assumed square from the first period, and. to the remainder bring dovrn the next period for a dividend. 3. Place- viz, looooo ; but icocoo has one place less than 1000 and-too together have : and since 1000 and 100 were taken the least pos- sible, the product of any other tv/o numbers, of the same number of plaqes, will be greater than i 00000 ;. consequently the product of any two numbers can have, at least, but one place less than both, the factors. ... Again, take two numbers of any number of places, that shall be the greatest of those places possible, as 999 and 99. Now 999 X 99 ^s ^^55 ^ban 999 x 100 ; but 999 X 100 (==99900) con- tains only as many places of figures, as are in 999 and 99; there- fore 999 X 99 or the product of any other two numbers, consisting of the same number of places, cannot have more places of figure^^^ than are in both its factors. Corollary i. A square number cannot have more praces of figures, than doable the places of the root, and, at least, but one less. ., '; .u Cor. z. 'A cube number cannot have more places of figures than triple the places of the root, and, at least, but two less. The truth of the rule may be shev/n algebraically thus ; Let iVrr the number, whose square root is to be found. Now, it appears from the leinma, -that there will be always g^ many places of figures in the root, as there are points or periods in the given number, and therefore the figures of these places may be represented by letters. • ' u 154 ARITHMETIC. 3. Phcc the double of the root, already found, on the left hand of the dividend for a divisor, 4. Consider what figure must be annexed to the divisor, so that if the result be r^ultiplied by it, the product inaf be equal to, or the next less than, the dividend, and it will be the second figure of the root. 5. Subtract the said product from the dividend, and to the remainder bring dovim the next period for a new divi- dend. 6. Find «t- ■ ■■ - • ' ^ - Suppose N to consist of two periods, and let the figures in the root be represented by a and b. Then a-^-b zza^ '\-2ah-\'h'=N=::: given number j and to find the root of N is the same as finding the root of a^ •^zab-^-b'^^ the Tijethod of doing which is as follows : ist divisor fl}^* + 2^^-f ^^ (a -f*=: root. zd divisor 2a'{-b)2ah-\rh^ tal-^-b' Again, suppose N to cbnsist Of 3 periods, and let the figures of the root be represented by a, b and c. Then a-{-b-\-c =:^* + 2^3-f 3* -f-2^ 3X36^X4 = T5552 3X36 X4'=: "1728 ^ add .43 1572544 subtrahend. 2. What is the cube root of 1092727 ? Ans. 103. 3^ What is the cube root of 27054036008 ? Ans. 3C02. 4. What is the cube root of '0001357 ? Ans. '05138, Sec, 5. What is the cube root of ~-t^ •'' ^"s. -i-. 6. What h the cube root of -j ? Ans. '873, &c. Rule for extracting the Cube Root by ApFRoxiMArioy* ^ I. Find by trial a cube near to the glren number, and call it the supposed ciibt\ 2. Then * That this rule converges extremely fast may be easily slicv/n thus ; ' ' Let EVOLUTION. l^i ^. TliSn, twice tL^ supposed cube acklecl. to the given number, is to twice the given nuir/oer added to the sap- posed cube, as the root of the supposed cube is to the too* required nearly. Or as the first sum is to the diff'-rence of the given and supposed cube, so h the supposed toot to the dificrence of the roots nearly. 3. By talcing the cub6 of the roOt thus found for the supposed cCibeV aftd repeating the operation, the root Vv'ill be had to a still greateii- degrse of exactness. example:. i. It is required to find the cube root of 9800344^, Let Let A^=: given number, a'= supposed cube, and x:=z cor- rection. Then 2a' -{-N : iN-^a'^ :: a : a-{-x by the rule, and con- sequently 2a' +Nxa-\-xz=2N-{-a^ Xa, or 2a' +a'-^x^ X a-^x z=2N-{-a'^ Xa, Or 2a'^ + 2a'x-4-a*-f 4a^Ar -{- Ca^x"- -f ^x"^ + .v'*=2aiV+a^ and by transposing the terms, and dividing by 2a A^=a'-f-3a*x-{-3a.v*-fx'-f-Ar' -{-■—, V/hich by neglecting the terms x'^ -f ..., as being very small, becomes Nzza'' ■\-'ia'^x-\' 2a 3'T;v''+^:'= the known cube of a-\-x. Q. E. L W 1 61 ARITHMfiTICi Let 125000000 rz supposed cube, whose root I& 50^; Then 125000000 2 250000000 98003449 348003449 98003449 2 J96006898 125000000 321006898 : 500 500 [or root nearly, 348003449)160503449000(461 zi corrected root, 1 3920 1 3 796 2130206940 2088020694 421862460 348003449 73859011 2. Required the cube root of 2 103 5 '8. Here we soon find that the root lies between 20 and 30^ and then between 27 and 2"8. Therefore 27 being taken, its cube is 19683 the assumed cube. Then 19683 2 39366 21035-8 As 6040 1 '8 21035-8 2 42071*6 -^[9683 : 61754.6 ■ 27 4322822 1235092 1667374-2 27 27'6ozi'7 60301*8) X, . *-'L. c i !A.)N. K»- 6o40i*§)i667374*2(27"6o47 the root nearly 459338 422813 36525 36241 242 42 Again for a second operation, the' cube of this root Is 2io35'3 18645 I55^23i -and the process by the latter meth- od is thus : 2io35'3i8645, &c. 42070*637290 21035*8 21035-8 21035-3 1 864.5, 8^. As 63iQ$*43729 : difF. '481355 :: 27*6047 : the diS'.zr •000210834 27*604910834 zz: the root required. 3. What is the cube root of 157464 ? Ans. 54. 4. What is the cube root of |- ? Ans. '763, &c. 5. What is the cube root of 117 ? Aiis. 4-89097. To EXTRACT THE RoOTS OF PoWERS IN GENERAL. KULE.* I. Prepare the given number for extraclicn, by pointing off from the units place as the root required directs. 2. Find * This rule will be sufficiently obvious from tlie work In the following example : Extrac }6^ ARITHMETIC. 2. Find the first figure of the root by trial; and subtract Its power from the given number. 3. To the remainder bring down the first figure in the next period, and call it the di-vide7id, 4, Involve Extract the cabe root of a^ -\-6a^ — 40^^4-9^^ — ^4* a^ -^Ga^ — 40^?^ -\-Q6a — 64(^*4- 2a — 4 ^a^)6a^ [J^2a a«4.6«^4-i2a^ + 8a5=a^ + 2a ^^ 4-2^ X3=3«*'^ + 1 2^^ + 12^*) — I2dr^ — 48/2^+96^ — 64 ( — /^ a^ -\' 6a^ — 40^' + 9^^ — 64 = ^ " 4- 2a — 4 When the index of the power, whose root is to be extracted^ is a coiTipositc number, the following rule will be serviceable : Take any tv;o or more indices, whose product is the given in- dex, and exiract out of the given number a root answering to one of these indices ; and then out of this root extract a root answer- inor to another of the indices, and so on to the last. Thus, the fourth root = square root of the square root. The sixth root == square root of the cube root, Sec. The proof of all roots is by involution. The following theorems may sometimes be found useful in ex- tracting the root of a valvar fraction : \/ — zzY zz ^_ — =: 1- ; "^ ' " b ^b b ^^ or, universally, ^1 ^Jl ah i ^l in 7F\ "EVOLt^TlON-. l6^ 4* Involve the rcx)t to the next inferior power to that, which is given, and multiply it by the number denoting the given pdvv'er for a divisor. 5. Find how many times the divisor may be had In the dividend, and the quotient will be another figure of the root. 6. Involve the whole root to the given power, and sub- tract it from the given number as before. 7. Bring down the first figure of the next period to the remainder for a new dividend, to which find a new divisor^ pnd so on, till the whole be finished. EXAMPLES. I, "What is the cube root of 531^7376 ? 53i5737^(37<^ 3*X3:^27)26i dividend. o-^oT 2 50^53=37 I I 3* X3i:::4i 07)25043 second dividend. 53157376 2. What is the biquadrate root of 19987 173376 ? Ans. 376 3. Extract the sursoUd^ or fifth root, of 307682821106 715625. Ans. 3145. 4. Extract 4. Extract the square cul>e(i> or §ixth root, of 43572838 1009267809889764416. Ans. 27534, 5. Find the seventh jroot of 344B771 74673075 13 18249 ^153794673, Ans. 32017. 6. Find the eighth root of 11210162813204762362464 97942460481. Ans. 13527. to extract any root whatever by Approximation. RULE. 1. Assume the root nearly, and raise it to the same power with the ^iven number, which call the assumed power, 2. Then, as the sum of the assumed power multiplied by the index more i and the given number muUiplied by the index less i, is to the sum of the given number multi- plied by the index more i and the assumed power mul- tiplied by the index less i, so is the assumed root to the required root. Or, as half the first sum is to the difference between the given and assumed powers, so is the assumed root to the difference between the true and assumed roots ; which difference, added or subtracted, gives the true root nearly. And the operation may be repeated as often as we please, by using always the last found root for the assumed root, and its power as aforesaid for the assumed power. EXAMPLES. I. Required the fifth root of 21035*8. Here EVOLUTION. 167 Here It appears tKat the fifth root is between 7*3 and 7*4. 7*3 being taken, its fifth power is 20730*71593. Hence then 21035-8 rr given number. 20730*716 = assumed power. 305-0^4 = difference. 5 = index. 20730-716 2x035*8 5 + 1=6 3 .2 5 i-_4 6-2=3 4^2=2 62192-148 42071-6 42071-6 1042637 io4i63-748=i tlie first sum. : 305-084 : ; 7-3 : -0213605 73 2135588 io4263'7)2227'ii32(-o2i36o4 = difference. 208527 7'3 141 84^ 7*321360 = root, 10426" true to the last figure. 3758 3128 630 626 4 2. What is the third root of 2 ? Ans. 1-25992 1. 3. What is the sixth root of 21035*8 ? Ans. 5'254037. 4. What is the seventh root of 21035*8 ? Ans. 4'M5392' c What i6S iEiTKMtTlC* 5. What is tlic riintli root of 21035-8 ? Ans. 3'02223i> ARITHMETICAL PROGRESSION. Any rank of numbers increasing by a common excess, or c^ecreniring by a common difference, are said to be in j^rithnctical Progression ; such are the numbers i, 2, 3, 4, ,5, &c. 7, 5, 3, I •, and '8, '6, 'zi, '2. When the numbers increase they fofm an asce?iding scries ; but v/hcn they de- crease, they form a descending series. The numbers, which forrii the series, are called the terms cf the progression; Any three of the five following terms being given, the other two may be readily found. 1. Tiie first term, 7 commonly called the 2. The hst termt,- 3 extremes* 3. The number of terms. 4- The comm.on difference. 5- The sum of all the terms, PROBLITM I. *Thc first iernty the last term^ and the number of terms lacing given, to find the siwi of all the terms. RULE.* Multiply the sum of the extremes by the number of terms, and Iralf the product will be the answer. EXAAIPLIZS. * Suppose another series of the same kind with the given one be placed under. it in" an inverse order ; then ulll the sum of every two corLesponding terms be the same as that of the first and last ; consequently, ARITHMETICAL PROGRESSION, l6^ EXAMPLES. I. The first term of an arithmetical progression is 2, the last term 53, and the number of terms 18 ; required the yum of the series. 53 55 18 440 55 2)990 495 Or, 53 + 2x18 , ^•^-i =495 "^^ answer. 2. The first term is i, the last term 21, and the numbct of terms 1 1 , required the sum of the series. Ans. 121. 3. How many strokes do the clocks of Venice, which go to 24 o'clock, strike in the compass of a day ? Ans. 300. 4. If consequently, any one of those sums, multiplied by the number of terms, must give the whole sum of the two series, and half that sum will evidently be the sum of the given series : thus, Let I, 2, 3, 4, 5, 6, 7, be the given series ; and 7, 6, 5, 4, 3, 2, i, the same inverted ; then 8 + 8 + 84.8 + 8 + 84-8=8x7=56and 1+3+4+5 + 6+7=1?= 28. Q: E. D. \']0 ARITHMETIC, 4. It 100 Stones be placed in a right line, exactly a yard asunder, and tlic fu'st a. yard from. a basket, what length of ground will that man go, who gathers them up shigly, re- turning with them one by one to the basket ? , Ans. 5 miles and 1300 yards. PROBLEM II. T/jd first ierWi the last terniy av.cl the tiumher of terms being giveiiy to find the commsn. difference, RULE.* , Divide the difference of the extremes by the number of terms less i, and the quotient will be the common differ- ence sought. EXAMPLES. I. The extremes are 2 and 53, and the number of terraa is 1 3 ; required the common difference. 53 2 17)51(3 17 51 Or, 53—2 Kl ,, ^^ — zz - — zr 1, the answer. i8~i 17 ^ 2. If * The difference ofthe first and last terms evidently shews the Increase of the first term by all the subsequent additions, till it becomes equal to the last ; and as the number of those additions is evidently one less than the number of terms, and the increase by every addition equal, it is plain, that the total increase, divided by the number of additions, must give the difference at every one separately ; whence the rule is manifest. ARITHMETICAL rROq|lESSION. 37I 2. If the extremes be 3 and 19, and the number of 1 erms 9, it is required to find the common difference, und the sum of the whole series. Ans. The diiFerence is 2, and the sum Is 99. 3. A man is to travel from London to a certain place in 12 days, and to go but 3 miles the fir^t day, increasing every day by an equal excess, so that the last day's journey may be 58 miles ; required the daily increase, and the dis- tance of the place from London. Ans. Daily Increase 5, distance 366 miles. PROBLEM III. Given the first tenn^ the last term^ and the common difference^ t^ jind the number cf terms. RULE.* # - ■ Divide the difference of the extremes by the common difference, and the quotient, increased by i, is the num- ber of terms required. EXAMPLES. * By tlie last problem, the diifercnce of the extremes, divided by the number of terms less i, gives the common difference ; consequently the same, divided by the common difference, must 'uve the number of terms less I ; hence this quotient, augmented by I, must bo the answer to the question. In any arithmetical progression, the sum of any two of its terms Is equal to the sum of any other two terms, taken at an equal distance on con^iraiy sides of the former ; or the double of any one term is equal to the sum of any two terms, taken at an equal distance from it on each side. The sum of any roimber of terms («) of tlie arithmetical series of odd numbers i, 3, c, 7, 9, Sec. is equal to the square {«^) of that number. That 172 ARITHMETIC. EXAMPLES. I. The extremes are 2 and 53, and the common diiFcr- ence 3 ; what is the number of terms ? 53 3)51 I 18 Or, 3J. J- 1 = 18 the answer. 2. If That is, if 1,3, 5, 7, 9, &c. be the numbers, Then will i, 2^, 3*, 4*, 5^, &c. be the sums of i, 2, 3, &c^ of those terms. For, o-f-i or the sum of i term = i* or i I -}- 3 or the sum of 2 terms = 2 * or 4 4-J-5 or the sum of 3 terms = 3* or 9 94-7 or the sum of 4 terms = 4^ or 16, &c. Whence it is plain, that, let n be any number whatsoever, the £um of n terms v/iil be n^. The following table contains a summary of the whole doctrine of arithmetical progression. Cases of Jujthmetjcal Progression. Case;Giv. | Req. r" / adn Solution. n — I xd-^-a. 2 Case ^«; ARITir METKAL PROGRESSION. 1/3 2. I£ the extremes be 3 and 19, and the common ilifTer- ,eiice 2, what is the number of terms ? Ans. 9. 3- A Casei Giv. 1 Req. Solution. 2. 3* 4- f ^+- ;• 2^ 1 ads <] 1 / ^ — • — ^ — \ s/ 2a-— d -4- ^ds -— 2a— d ^d ^ 2a-^d\ -f- 8^j- — d^ I % /z/x < n L i+a X I— a 2s — i-{-a 2S_ 5- 6. 2 X s — an n — I X n ans { 2s a. n aln < ^ I— a n — i' J fl + /X« 2 Case 1/4 ARITHAIETIC. 3. A man, going a journey, travelled the first day 5: miles, the last day 35 miles, and increased his journey every day by 3 miles j how many days did he travel ? Ans. 1 1 days. GEOMETRICAL V€ase Giv. Req. Solution. j ' a l—n^i X ^- 7- dnl < f nxl-—n^ix£ 2 a. s dxn — I n 2 8. r-i/ .^ . M 1 I' a s dxn — I n 2 ' ^±V 2l^d\'—^ds 2 9- dls ^ \ n r 1 " 2d 2S — L 10. d n 2 X nl—s n — IX» r^ 7 = least term. 1 ' ; = number of terms. Here, ^ L,r 4- asl < r s — a n Ljs — a — L,s — / Case GEOMETRICAL PROGRESSION. ' 179 5. Divide the dividend by the divisor, and the quotient will be the term sought. Note. When the first term of the series is equal to the ratio, the indices must begin vi'ith an unit, and the in- dices added must make the entire index of the term re- quired ; Case Giv. I Req. ans < an} i 8. »/ <^ rns -^ Solution. a a l^s-l^-'zz.ay^s^^ L\^^ — I. >.«— -I /— /-f XJ. r — I Case i8o LITHMETIC. quired ; and the product of ,tlie different terms, found' a^j before, will give the term required, EXAMPLES. I. The first term of a geometrical series is 2, the number of terms 13, and the ratio 2 j required the last term. 1, 2, 3, 4, 5, indices. 2, 4, 8, 16, 32, leading terms. Then 4+4+3+2 rr index to the 13th term. And 16X 16X8X413:8192 the answer. In this example the indices must begin with i, and such of them be chosen, as will make up the entire index to the term required. 2. Required Cas Giv. I Req. rls ij 2, 3, 4, 5, 6, indices. ' 3, 6, 12, 24, 48, 96, 192, leading 'terms. Then 6-^^ zz index to the 12th term. And 192 X96ZZ 18432 zz dividend. The number of terms multiplied is 2, and 2 — iini is the power to which the term 3 is to be raised ; but the first power of 3 is 3, and therefore 18432 -7- 3 zz 6 144 the 1 2th term required. 3. The first term of a geometrical series is i, the ratio 2; and the number of terms 23 5 required the last term. Ans. 4194304. QUESTIONS TO BE SOLVED BY THE TWO PRECEDING PROBLEMS. 1. A person, being asked to dispose of a fme horse, said he would sell him on condition of having one farthing for the first nail in his shoes, 2 farthings for the second, one penny for the third, and so on, doubling the price of every nail to 32, the number of nails in his four shoes : what would the horse be sold for at that rate ? Ans. 4473924I. 5s. 3|d. 2. A young man, skilled in numbers, agreed with a farmer to work for him eleven years without any other re- ward than the produce of one wheat com for the first year, and that produce to be sowed the second year, and so on from year to year till the end of the time, allowing the in- crease to be in a tenfold proportion : what qirantity of wheat is due for such service, and to what docs it amount at a dollar per bushel ? Ans. 226056^ bushels, allowing 7680 wheat corns to be a pint -, and the amount is 226056— dollars. 3. What 182 ARITHMETIC. 3. What debt will be discharged in a year, or twelve months, by paying il. the first month, 2L the second, 4I. the third, and so on, each succeeding payment being double the last J and what will the last payment be ? Ans. The debt is 4095I. and the last paym,^t 2C48I SIMPLE INTEREST/ Interest is the premium, allowed for the loan of money* The sum lent is called the principal. The sum of the principal and interest is called the amount. Interest is allowed at so much per cent, per atmum^ which premium per cent, per annum y or interest of lool. for a year, is called the rate of interest. Interest is of two sorts, simple and compound. Simple interest is that, which is allowed only for the principal lent. Note. Commission, Brokerage, Insurance, Stocks,* and, in general, whatever is at a certain rate, or sum per cent, are calculated like Simple Interest. RULE.f 1. Multiply the principal by the rate, and divide the product by 100 ; and the quotient is the answer for one year. 2. Multiply * Stock is a general name for public funds, and capitals of trad- ing companies, the shares of which are transferable from one per- son to another. f The rule is evidently an application of Simple Proportion and Practice. SIMPLE INTEREST. 1S3 2» Multiply the interest for one year by the given time, . and the product is the answer for that time. 3. If there be parts of a year, as months or days, work for the months by the aliquot parts of a year, and for the days by Simple Proportion. EXAMPLES. T. What is the interest of 450I. for a year, at 5 per cent, per annum ? 45©1. 5 1,00)22*50 20 lo'oo Ans. 22I. and ■/-^=To-=^*5=^ OS. 2. What IS the interest of 720I. for 3 years, at 5 per cent, per annum ? 720L 36 5 3 36-00 108I. Ans, 3. What Is the interest of 170I. for i|^ year, at 5 per cent, per annum ? 170L 2)81. IDS. interest for i year. 5 4 5 8*50 X2l. 15s. Answer. 20 lO'OO 4. What 184 ARITHMETIC. 4. What is the interest of loyl. for 117 days, at 4-| pet cent, per annum ? 107I. 5 I 7 3'2 $17 3'2 4l II 7 35: II 6 2-4 428 53 10 26 15* 31x104-7=1x7 20 55 18 >■ 3.2 10 559 I 6 0 35 II 6 2-4 365)594 13 o r4(il. I2S. 7tAi^. ^ '^5 3<^S ^^^ answer. . 229 7'8o 20 4 )4593 3-20 365 4. q. ^ — - 2-4=2f=:fd. 943 « 213 T _ 8 ,:; 32556 If 5'. What is the interest of 32I. 5s. 8(1. for 7 years, at 4-^ per cent, per annum ? Ans. 9I. 12s. i-r|-5-d. 6. What is the interest of 319I. 6d. for 5^ years, at 3-| per cent, per annum ? Ans. 681. 15s. pd. 2~-q. 7. What is the interest of 607*500. for 5 years, at 6 per cent, per annum ? Ans. 182*250. 8. What is the interest of 213I. from Feb. 12, to June %, 1796, it being leap year, at 3-^ per cent, per annii-m ? Ans. 2I. 6s. 6d. 3iVrTq* SIMPLE \ SIMPLE INTEREST BT DECIMALS. 185 I SIMPLE INTEREST BY DECIMALS. RULE.* Multiply continually the principal, ratio and time, and it will give the interest required. Ratio is the simple interest of il. for i year, at the rate per cent. ?.greed on *, thus the ratio f 3 per cent, is '03. •035- •04. •045. I 5 •05- 5~ 055. 16 '06. 13 r If- at <|4t EXAMPLES. * The following theorems will shew all the possible cases of simple interest, where /> = principal, / = time, r = ratio, aad a = amount. I. ptr^p-a, II. ^[ittzt. rp III. ^^^. IV. fZf =,, Z « lS(S ARITHMETIC. EXAMPLES. I. What is the interest of 945I. los. for 3 years, at 5 per cent, per annum ^ 945*5 47*275 141-825 20 16*500 12 6-poo Ans. 141L 1 6s. 6d. 2. What is the interest of 796I. 15s. for 5 years, at 4 per cent, per annum ? ^ Ans. 179I. 5s. 4^d. 3. What is the interest of 537I. 15s. from November II, 1764, to June 5, 1765, at 3|- per cent. ? Ans. I il. ~d. COMMISSION. Commission is an allowance of so much per cent, to a factor or correspondent abroad, for buying and selling goods for his employer. EXAMPLES. BROKERAGE. 1S7 EXAMPLES. I. What comes the commission of 5 col. 13s. 6d, to at 2~ per cent. ? 500I. 13s. 6d. 3t 1502 250 0 6 6 9 17-52 20 7 3 10-47 12 5-^7 4 2*68 Ans. 17I. I OS. 5-|-d. 2. My correspondent writes me word, that he has bought goods on my account to the value of 754I. i6s. What does his commission come to at 2-^ per cent. ? Ans. 1 81. 17s. 4-jd. 3. What must I allow my correspondent for disbursing on my account 529I. i8s. 5d. at 2-^ per cent. ? Ans. 111. 1 8s. 5~d. BROKERAGE. Brokerage is an allowance of so much per cent, to a person, called a Broker, for assisting merchants or factors in procuring or disposing of goods. EXAMPLES. J 88 ^ ARITHMETIC. EXAMPLES. V What is the brokerage of 610I. at 5s. or ^ per cent. ? 5s. is ~ 610I. 1*52 10 ^ 20 10-50 12 6'oo Ans. il. I OS. 6d^ 2. If I allov/ my broker 3-^ per cent, what may he de- mand, when he sells goods to the value of, 8 7 61. 5s. lod. ? Ans. 32I. 17s. 2^d» 3. What is the brokerage of 879I. i8s, at |- per cent. ? Ans. 3I. 5s. ii^d, INSURANCE. Insi/rance is a premium of so much per cent, given to certain persons and ofEces for a security of making good the loss of ships, houses, merchandize, &c. v^hich may happen from storms, fire, &c. EXAMPLES, DISCOUNT. 189 EXAMPLES. I. What is the insurance of 874!. 13s. 6d. at n^ P^^ cent. ? 874I. 13s. 6d. 12 10496 2 0 874 13 6 437 6 9 iiB-oS 2 3 20 1-62 12 7*47 4 1*88 Ans. 1 1 81. is. 7^d. 2. What is the insurance of 900I. at io-|- per cent. ? Ans. 96I. 15s. 3. What is the insurance of 1200I. at 7-|- per cent. ? Ans. 91U I OS. DISCOUNT. Discount is an allowance made for the payment of any sum of money before it becomes due j and is the differ- ence between that sum due some time hence, and its pres- 190 ARITHMETIC. The present ivorth of any sum, or debt, due some time hence is such a sum, as, if put to interest, would in that time and at the rate per cent, for which the discount is to be made, amount to the sum or debt then due. RULE.* ,1. As the amount of lool. for the given rate and time is to I col. so is the given sum or debt to the present worth. 2. Subtract the present worth from the given sum, and the remainder is the discount required. Or, As the amount of lool. for the given rate and time is to the interest of looI. for that time, so is the given sum or debt to the discount required. EXAMPLES. * That an allowance oUgbt to be made for paying money be- fore it becomes due, which is supposed^ bear no interest till af- ter it is due, is very reasonable ; for, if I keej? the money in my own Jiands till the debt becomes due, if is plain I may make an advantage of it by putting it out to interest for that time ; but if I pay it before it is due, it is giving that benefit to another ; there- fore we have only to enquire what discount ought to be allowed. And here some debtors will be ready to say, that since by not paying the money till it becomes due, they may employ it at in- terest, therefore by paying it before due, they shall lose that inter- est, and for that reason all such interest ought to be discount- ed : but that is false, for they cannot* be said to lose that interest till the time the debt becomes due arrives ; whereas we are to consider what would properly be lost at present, by paying the debt before it becomes due ; and this can, in point of equity or justice. y DISCOUNT. 191 EXAMPLES. 1. What Is the discount of 573I. 15s. due 3 years hence> at 4~ per cent. ? 4I. I OS. 3 13 10 100 I. s. 1. s. 13 10 : : 573 15 : 20 20 113 10 ; 20 2270 270 I 1475 270 803250 22950 (2,0) 227,0)309825,0 ( 136,4 828 1472 68 4 1 105 197 12 227)2364(10 94 4 Ans. 681. 376(1 4S. io|:d. 149 2. What jastice, be no other than such a sum, as, being put out to interest till the debt becomes due, would amount to the interest of the debt for the same time. — It is beside plain, that the advantage arising from discharging a debt, due some time hence, by a pres- ent ZgZ AKlTHMEXiC. 2. What is the present worth of 150I. payable in ^ of a year, discount being at 5 per cent. ? Ans. 148I. 2S. ii^d. 3. Bought a quantity of goods for 150I. ready money, and sold them again for 200I. payable at ~:of a year hence 5 what was the gain in ready money, supposing dis- count to be made at 5 per cent. ? Ans. 42h 15s. 5d. 4. What is the present worth of 120L payable as fol- lows, viz. 50I. at 3 months, 50I. at 5 months, and the rest at 8 months, discount being at 6 per cent. ? Ans. 117L 5s. 5-^-d. DISCOUNT ent payment, according to the principles we have mentioned, is exactly the same as employing the whole sum at interest till the time the debt becomes due arrives ; for if the discount allowed for present payment be put out to interest for that time, its amount will be the same as the interest of the whole debt for the same time : thus, the discount of 105I. due one year hence, reckoning interest at 5 per cent, will be 5I. and 5I. put out to interest at ^ per cent, for one year, will amount to 5I. 5s. which is exactly equal to t;ie interest of 105I. for one year at 5 per cent. The truth of the rule for working is evident from the nature of simple interest : for since the debt may be considered as the amount of some principal (called here the present worth) at a certain rate per cent, and for the given time, that amount must be in the same proportion, either to its principal or interest, as the amount of any other sum, at the same rate, and for the same time, is to its principal or interest. DISCOUNT BY DECIMALS. ^93 DISCOUNT BY DECIMALS. RULE.* As the amount of il. for the given time is to il. so is the interest of the debt for the said ^ time to the discount required. Subtract the discount from the principal, and the re- mainder will be the present worth. EXAMPLES. I. What is the discount of 573I. 15s. due 3 years hence, at 4^ per cent, per annum ? •04^ ^ ■ ■ * Let m represent any debt, and n the time of payment ; then will the following tables exhibit all the variety, that can happea with respect to present worth and discount. Of the Present Worth of Money paid before it is due AT Simple Interest. The present worth of any sum m. Rate per cent. For n years. ] n months. n days. r per cent. loom 1200W 36500W «r-|-ioo «r-i-i200 3 per cent. it)om 3«-j-Ioo 400W «-i-4QO 36500^2 3« 4- 36500 4 per cent. 1 25m 300m «-|-30o 9125W «4-9i25 5 per cent. 20m 240m «-f.240 7300;n « + 7300 «4.20 A A Of IQ.1 ARITHMETIC. •c45X3+izrri35zr amount of il. for the given time. And 573-75X'045X3 =:: 77-45625 zr' interest of the debt for the given time. IT35 : I : : 77*45^^5 • i-J35)77-45625(63'243 6S10 935^ 9080 2762 2270 4925 4540 3850 34^5 68-243 = 681. 43. lo^d. Ans. 445 1. What Of Discounts to be allowed t6k paying of Money KEFORE IT FALLS DUE AT SiMPLE INTEREST. ^ The discount of any sum m. Rate per cent. For n years. n months. n days. r per cent. mnr mnr mnr nr-\-\ CO nr-\- I2C0 ;;r -1.365 00 3 per cent. 7,mn mn 3/K.7 3/2-}- 100 tt-f 4C0 3/^4-36500 4 per cent. mn mn mn «4-30O ^ + 9135 5 per cent. inn r,in mn n-\-20 «4-240 /i-l-7300 EQUATION CF PAYMENTS. I«^5 2. What is the discount of 725I. i6s. for 5 months, at 3|- per cent, per annum ? Ans. iil. los. 3^d. 3. What ready money will discliarge a debt of 13771. j^s. 4(1. due 2 years^ 3 quarters and 25 days hence, dis- counting at 4-|- per cent, per annum ? Ans. 1226I. Ss. S-d. EQUATION OF PAYMENTS. Eqitation of Payments is the finding a time fo pay at once Several debts, due at different times, so that no ioss shall be sustained by either party. RUtE.* Multiply each payment by the time, at Vvhich it is due j then divide the sum of the products by the sum of tlic payments, and the quotient will be thp time required. EXAMPLES. * This rule is founded iipoq a supposition, that tlie Pian of tlic interests of the several debts, which are payable before the equated time, from tlicir terms to that time, ought to be ccjiisl to the sum of the interests of the debts payable after the equated time, from that time to tlieir terms. Among others, that defend this prin- ciple, Mr. CocKCR endeavours to prove it to be right by this ar- gument : that what is gained by keeping some of the debts after tiiey are due, is lost by paying others before they are due : but tliis cannot be the case ; for though by keeping a debt unpaid af- t^r it is due there is gain.ed the interest of it for that time, yet by paying a debt before it is due tlie payer does not lose the inrcrc :: for that time, tiut the discount orily, Vvhich is less than tiie inter- est, and therefore the rule is not: true. Although 1^6 ARITHMETIC. EXAMPLES. I. A owes B 190I. to be paid as follows, viz. 50I. in 6, months, 60I. in 7 months, and Sol. in 10 months j what is the equated time to pay the whole ^ 50X 6 = 300 60X 7 = 420 80X10=800 50 + ^0 + 80=190)1520(8 1520 Ans. 8 months. 2. A Althoiigh this rule be not accurately true, yet in most questions, that occur in business, the frror is so trifling, that it will be much used. That the rule is universally agreeable to the supposition may ^iie thus demonstrated. f J = first debt payable, and the distance of its term of I payment t. J^et ^ -O = last debt payable, and the distance of its term T, I X = distance of the equated time. (^ r = rate of interest of il. for one year. fThe distance of the time f and k is = X — L The distance of the time T and x is = T—x. Now ttie interest of J for the time x — t is x — /X^'" > and the interest of JD for the time T—x is T—xX-Dr ; therefore x — t X^r = 7' — .Y >^ /)r by the supposition ; and frpm this equation^ X is found =""2)+T' which is the rule. And t^e same might be shewn of any number of payments. The true rule is given in equation of payments by decimak. EQUATION OF PAYMENTS BY DECIMALS. T^'j 2. A owes B 52I. 7s. 6d. to be paid in ^-f nionths, Sol. I OS. to be paid in ^t nionths, and 76I. 2s. 6d. to be paid in 5 monihs ; what is the equated time to pay tlie whole ? Ans. 4 months, 8 days. 3. A owes B 240I. to be paid in 6 months, but in one month and a half pays him 60I. and in 4— months after that Sol. more ; how much longer than 6 months should P in equity defer fhe rest ? Ans. 3-— months. 4. A debt is to be paid as follows, viz. ^ at 2 months, I- at 3 months, -J- at 4 months, |- at 5 months, and the rest at 7 months 5 what is the equated time to pay the whole ? Ans. 4 months and 18 days. EQUATION OF PAYMENTS BY DECIMALS. STiw debts being due at different times, to find the equated time to pa^ the nvhole. RULE.* I. To the sum of both payments add the continual product of the first payment, the rate, or interest of il. for one year, and the time between the payments, and call this the first number. 2. Multiply * No rule in arithmetic has been the occasion of so many dis- putes, as that of Equation of Payments. Almost every writer upon this subject has endeavoured to shew the fallacy of the meth- ods made use of by other authors, and to substitute a new one in their stead. But the only true rule seems to be that of Mr. Malcolm, or one similar to It in its essential principles, de- rived from tlie consideration of interest and discount. The I9S AS.ITHMETIC. 2. Multiply twice the first payment by the rate, and call this the second number. 3. Divide The rule, given above, is the same as Mr. Malcolm's, cxcci^fe that it is not encumbered with the time before any payment is due, that being no necessary part of the operation. Demonstration of the Rule. Suppose a sum of money to be due imraedrately, and another sum at the expiration of a certain given time forward, and it is proposed to find a time to pay the whole at once, so that neither party shall sustain loss. Nov/, it is plain, that the equated time must fall between those of tlie two payments ; and that what is got by keeping the first debt after 4t is due, should be equal to what is lost by paying the second debt before it is due. But the gain, arising from the keeping of a sum of money after It is 'due, is evidently equal to the interest of the debt for tliat tune. And the loss, which is sustained by the paying of a sum of money before it is due, is evidently equal to the discount of the debt for that time. Therefore, it is obvious, tl^at the dcbtot rpust retain the sunit immediately due, or, the first payment, till its interest shall be equal to the discount of the second sum for the time it is paid be- fore due ; because, in that case, the gain and loss will be equal> and consequently neither party can be the loser. Now, to find such a time, let a = first payment, h •=. second,^ and / rr time between the payments ; r =r rate, or interest of iL for one year, and y rr equated time after the first payment. Then arx = interest of a for x time, J htr — Irx J. i r 1 r I and =: discount of L for tiie time I -^-tr—rx Bat. E<^ATION OF PAYMENTS BY DECIMALS. I99 3. Divide the first number by the second, and call the cjuotient the third number. 4. Call But arx z:z — ■ by the question, from which equation ps i-{-tr — rx ^ h found = ititff;: + '•±1±'!!l\ ' _ if Let — ■ — ■ be put equal to «, and — = in. 2ar ar Then it is e-vldent that n, or its equal «^j" is greater than «^ — m\^) and therefore x will have two affirmative values, the quantities «-|-»^ — m\ and n — n^ — m\ being both positive. 3ut only one of those values will answer the conditions of the question ; and, in all cases of this problem, x will be = » -*^ « - — jn\ Por suppose the contrary, and let .\* zz n -|-«* — ni\ . Then tsz::L — n — n ^ — wp= t — u 1 ^ | — « * — mV" =1 e — 2/« + «M* — «* — m\ =«^4-r — 2tn\ — «^ — m\ Nov/, since a-\-h-^atr X — =«> acd It X — = ???> we ^hall liavc from the first of tliesc equations /*— 2/«nr — ht-^at X — and consequently / — x = ?;^ — ^/— ^/ X — '_;,'-^/X— I Bir Too ARITHMETIC. 4. Call tlie bquare of the third number the fourth num- ber. 5. Divide the product of the second payment, and time between the payments, by the product of the first pay- ment and the rate, and call tlie quotient the lifth number. 6. From the fourth number take the fifth, and call the Square root of the difference the sixth number. 7. Then the difference of the third and sixth numbers is the equated time, after the first payment is due. EXAMPLES. I. There is lool. payable one year hence, and 105I. payable 3 years hence ; what is the equated time, allow- ing simple interest at 5 per cent, per annum ? 100 ^i^ But ti'^ — k X ] is evidently greater than n^ — l?i — at >< — ar \ ar and therefore n* — l>t X — I — n"- — k—atX- — ar , or its equal t — X, must be a negative quantity ; and consequently x will be greater than t, that is, the equated time will fall beyond the sec- ond payment, which is absurd. The value of x, therefore, can- , a-\-hA-atr . a-^-b-^-atr not be = -X — ? + -J — ! 2ar , but must in all cases bc=''J±±fl:-1±i±^\ -±\ , which is the same as 2ar 2ar \ ar | the rule. From this it appears, that the double sign made use of by Mr. Malcolm, and every author since, who has given his method, cannot obtain, an-d that there is no ambiguity i^ ^e problem. In like manner it might be shewn, that the directions, usually given for finding the equated time when there are more than two payments, EQUATION OF TAYMENTS BY DECIMALS. 201 lOO 100 •05 1 5*oo 200 2 •05 lO'OO lO'OOS • 100 105 lumber. [0)215= 1st ri 2i-s = 3d nun^ber^ 21-5 1075 215 430 462*25 = 4th number. 105 2 V I St payment X rate = 5)210 v/ 42 = 5th number. 462-25 21-5 42 (2o'5 = 6th number. ... I = equated time ftom the firist 420*25(20*5 payment, and .*. 2 years 4 = whole equated time. 405)2025 2025 2. Suppose payments, will nol agree with the hypothesis, but this may be easily seen by working an example at large, and examining the truth o£ the conclusion. / The Be / . 202 ARITHMETIC. 2. Suppose 400I. are to be paid at the end of 2 years, and 2 1 col. at the end of 8 years ; what is the. equated time for one payment, reckoning 5 per cent, simple inter- est ? Ans. 7 years. 3. Suppose 300I, are to be paid at one year's end, and 300I. nfore at the end of i^ year ; it is^ required to find the time to pay it at one payment, 5 per cent, simple interest being allowed. Ans. 1*248637 year. COMPOUND INTEREST. Compound Interest is that, which arises from the principal and interest taken together, as it becomes due, at th- end of each stated time of payment. I RULE.'^' I. Find the amount of the given principal, for the time of the first payment by simple interest. 2. Consider The equated time for any number of payments may be readily found when the question is proposed in numbers, but it would not be easy to give algebraic theorems for those cases, on account of the variation of the debts and times, and the difficulty of finding between which of the payments the equated time would happen. Supposing r to be the amount of il. for one year, and the othw /oo-.^r^-f"^ er letters as before, then / — s will be a general theorem log. r for the equated time of any two payments, reckoning compound interest, and is found in the same manner as the fQ|mer. * The reason of this rule is evident from the definition, and the principles of simple interest. COMPOUND IN-y^REST. 203 2. Consider this amount as the principal for the second payment, whose amount calculate as before, and so on through all the payments to the last, still accounting the last amount as the principal for the next payment. EXAMPLES. r. What is the amount of 320I. los. for 4 years, at 5 per cent, per annum, compound interest ? ^)32ol. , los. 0 6d. I St year's principal. ist year's interest. ^)336 16 10 16 6 4 2d year's principal. 2d year's interest. /o)353 17 7 13 4 3d yeafs principal. 3d year's interest. Tc)37i 18 0 II 4i 0 4th year's principal. 4th year's interest. 389 II 4-^ whole amount, or the answer re^ i^ulred. 2. What is the compound interest of 760I. los. forborn 4 years at 4 per cent. ? Ans. 129I. 3s. 6^d. 3. What is the compound interest of 410I. forborn for 2^- years, at 4— per cent, per annum ; the interest payable half-yearly ? Ans. 48I. 4s. ii-|d. 4. Find the several amounts of 50I. payable yearly, half-yearly and quarterly, being forborn 5 years, at 5 per cent, per annum, compound interest. Ans. 63I. 1 6s. 3-^d. 64I. and 64I. is. p^d. COMPOUND 204 ARITHMETIC. COMPOUND INTEREST BY DECIMALS, RULE.* I. Find the amount of iL for one year at the given rate per cent, * 2. Involve * Demonstration. Let r = amount of il. for one year, and p = principal or given sum ; then since r is the amount of il. for one year, r* will be its amount for two years, r' for 3 j^ears, and so on ; for, when the rate and time are the same, all principal sums are necessarily as their amounts ; and consequently as r is the principal for the second year, it will be as i : r : : r : r^-=. amount for the second year, or principal for the third j and again, as i : r :: r* : r'r= amount for the third year, or prin- cipal for the fourth, and so on to any number of years. And if the number of years be denoted by t, the amount of il. for t years will be K Hence it will appear, that .the amount of any other principal sum p for t years is pr^ ; -for as i : r' i : /» : pr\ the sarai^ as in the rule. If the rste of interest be determined to any other time than a year, as f, -4-, &c. the rule is the same, and then t will represent that stated time. r r = amount of il. for one year, at the given rate j per cent. p zz principal, or sum put out to interest. Let { . .- \ ' t =r mtertst- t r= time. r,i = amount for the time i. Then the following theorems will exhibit the solutions of all /i:e cases in compound interest, I. p/zzm. II. pr^ — pzz'i^ III. --=:/, IV. — I = f\ The COMPOUND INTEREST BY DECIMALS. 205 2. Involve the amount thus fcund to such a power, as is denoted by the number of years. 3. Mukiply this power by the principal, on given sum, and the product i^ill be the amount required. ^ ■ 4. Subtract The mbsfj convenient way of giving the theorem for the /;W, as well as for all the other cases, will be by logarithms, as follows : I. tX^og. r-\-Iog. p:=:Iog.m. II. log.m — ty.log.rzzlog. p, III. % m-log.p _^ j^^ h'm—log.p _j^^^ log. r t "^ If the compound interest, or amount of any sum, be required for the parts of a year, it may be determined as foUov^s : I. When the time is any aliquot part cf a year, RULE. 1. Find the amount of il. for one year, as before, and that root of it, which is denoted by the aliquot part, will be the amount sought* 2. Multiply the amount thus found by the principal, and it wi'I be the amount of the given sum required. II. When the time is not an aliquot part of a year. RULE. 1. Reduce the tirtie into days, and the ^^^\.h 1001 ol i_,c amount of il. for one year is the amount for one day. 2. Raise this amount to that power, v/hose index is equal i-c die number of days, and it will be the amount of il. for the giv en time. 3. Multiply this amount by the principal, and it will be li.f amount of the given sum required. To avoid extracting very high roots, the same may be done by logarithms thus : divide the logarli'im of the rate, or amount oi il. for one year, b,y the denominator of the given aliquot part, and the quotient will be the logarithm of the root sough^. 206 ARITHMETIC. 4. Subtract the principal from the amount, and the re- mainder will be the interest. ' EXAMPLES. I. What is the compound interest of 500I. for 4 years, at 5 per cent, per annum ? 1*05 zi amount of il. for one year at 5 1*05 per cent. 1050 1-1025 11025 22050 110250 11025 I '2 1 550625 11:4th power of 1*05. 500IZ principal. 607*753 12500ZI amount. 500 107*753125 IT 107I. 15s. o^d. iin interest required. 2. What is the amount of 760I. los. for 4 years, at 4 per cent. ? Ans. 889I. 13s. 6^d. 3. What is the amount of 72 il. for 21 years, at 4 per cent, per annum ? Ans. 1642I. 19s. lod. 4. What is the amount of 217I. forborn 2-^ years, at 5 per cent, per annum, supposing the interest payable quar- terly ? Ans. 242I. 13 s. 4yd. ANNUITIES. ANNUITIES. 207 ANNUITIES. An Annuity. is a sum 01 money payable every year, for a certain number of years, ox for ever. When the debtor keeps the annuity in his own hands, beyond the time of payment, if is said to be in arrears. The, suni of all the annuities for 'the time they have been forborn, together with the interest due upon each, is called the amount. If an annuity be to be bought ofF, or paid all at once, at the beginning of the first year, the price, which ought to be given for it, is called t\\Q^ present nvorth. 21? Jind the Amount of an Annuity at Simple Interest. RtJLE.* I. Find the sum of the natural series of numbers i, 2, 3, &c. to the number of years less one. 2. Multiply * Demonstration. Whatever, the tipie is, there Is due up- on the first year's annuity, as many years* interest aS the whole number of years less one ; and gradually one less upon every suc- ceeding year to the last but one ; upon which there is due only one year's interest, and none upon the last ; therefore in the whole there is due as. many years' interest of the annuity, as the sum of the series i, 2, 3, 4, &c. to the number -of years less one. Consequently one yearns interest, multiplied' by this sum, must be the whole Interest due ; to which If all the annuities be added, the sum is plainly the amount. Q^ E. D. Let r be the ratio, n the annuity, t the time, and a the amount. Then will the folio v/ing theorems give the solutions of all the dliTerent cases. T t'^rn — trn . ^_ la — 2/« 1. \-tnzza. II III. 2©8 ARITHMETIC. 2. Multiply this sum by one years interest of the an- nuity, and the product will be the whole interest due upon the annuity.^ 3. To .this product add the product of the annuity and time, and the sum will be the amount sought. .. ■. . Note. When the annuity is to be paid half-yearly or quarterly ; then take, in the former case,^ -f the ratio, half the annuity, and twice the number of year& j and, in the latter case, ■^- the ratio, -J the annuity, and 4 times the number of years, and proceed as before. EXAMPLES. I. What is the amount of an annuity of 50I. for 7 years, allowing simple interest at 5 per cent. ? 1 + 2+3+4+5+6=21=^:3X7 2I. 10s. zi I year's interest of 50!. 3 10 7 52 10 350 o =r 50I. X 7 402I. 10s. = amount required. 2. I£ 2a III. tZ—=n, IV. ~ +- t^r — tr-{-2t rn 4 2 ~~ * In the last theorem J= , and in theorem first, if a sum rn cannot be found equal to tfie amount, the problem is impossible in whole years. Note. Some writers look upon this method of finding the amount of an annuity as a species of compound interest ; the annui^ ty itself, they say, being properly the simple interest, and the cap- ital, whence it arises, the principal. ANNUITIES. 209 2. If a pension of 6ocI. per annum be forborn 5 ycr.rs, what will it amount to, allowing 4 per cent, simple in- terest ? Ans. 3240I. 3. What will an.^nnuity of 250I. amount to in 7 years, to be paid by half-yearly payments, at 6 per cent, per an- num, simple interest ? Ans. 209 il. 5s. 'To find the present Worth cf an Annuity at Simple Interest, RULE.* Find the present worth of each year by itself, discount- ing from the tirrie it becomes due, and the sum of all these v/ill be the present worth required. EXAMPLES. * Tlie reason of this rule is manifest from the nature of discount, for all the annuities may be considered separately, as so many sin- gle and independent debts, due after i, 2, 3, &c. years ; so that the present worth of each being found, their sum must be the present worth of the whole. The estimation, however, of annuities at simple interest \i high- ly unreasonable and absurd. One instance only will ho. sufficient to shew the truth of this assertion. The price of an annuity of 50I. to continue 40 years, discounting at 5 per cent, will, by ei- ther of the rules, amount to a sum, of which one year's interest only exceeds the annuity. Would it not therefore be highly ri- ' diculous to give, for an annuity to eontinue only 4O years, a suraj which would yield a greater yearly interest for ever ? It is most equitable to allow compound interest. Let p = present worth, and the other letters as before. Then { nx 1 1 — ,&c. to '■■ z=:p. i-fr i-f2r i-4-3r i-{-tr /---^4--^+— ^.&-to ' 14-r i4-2r 14-3^ i-{-ir C c The aiO ,, ARITHMETIC. EXAMPLES. 1. What is the present worth of an annuity of looi. td continue 5 year^, at 6 per cent, pet annurfi> simple interest ? ic6 I 100 :: 100 : 94-3396= present worth for i year. 112 : 100 : : 100 ; 89*2857= 2d year. 118 : 100 :: 100 : 84*7457=: 3d year. 124 : 100 : : 100 : 80*6451= 4th year. 130 : ICO : : 100 : 76*9230= 5th year. 425 "939 1 = 425I. 1 8s. 9-Jd. = present worth of the annuity required. 2. What is the present worth of an annuity or pension of 5001. to continue 4 years, at 5 per cent, per annum, simple interest ? Ans. 1782!. 58. 7d. To find the Ammmt of an Avjin'ity at Compound Interest, RULE.* I. Make i the first term of a geometrical progression^ and the amount of il. for one year, at the given rate per cent, the ratio. 2. Carry The other two theorems for the time and rate cannot be given in general terms. * Demonstration. It is pdain, that upon the first year's an- nuity, there will be due as many years' compound interest, as the given number of years less one, and gradually one year's interest less ANNUITIES. 211 2. Carry the series to as many terms as the nunoiber of years, and find its sum. 3. Multiply the sum thus found by the given annuity, and the product will be jthe amount sought. EXAMPLES. less upon every succeeding year to that preceding the last, which has but one year's interest, and the last bears no interest. Letr, therefore, = rate, or amount of il. for i year ; then the series of amounts of il. annuity, for several years, from the first to the last, is I, r, r*, r', &c. to r^-'. And the sum of this, according to fj I xhe rule in geometrical progression, will be , = amotmt of •ll. annuity for / years. And all annuities arc proportional to their amouots, therefore i : - — - : : si : -^~. x » = r — I r — I amount of any given annuity «. Q^ E. D. Let r = rate, or amount of il. for one year, and the other letters as before, then y(nz=a, and =zn, r — I r- — 1 And from these equations all the cases relating to annuities, or pensions in arrears, may be conveniently exhibited in logarithmic terms, thus : •> I. Log. n-\-Log» r^ — I — Log. r — i -zzLog, a* II. Log. a — Log. r^ — i -\-Log. r — izzLog. n. III. itS-^r-a^-n-Lo!..,,^^^ I V. r'- 21 + f- - I = o. 212 ARITHMETIC. EXAMPLES, I. What is the amount of aji annuity of 40I. to con- thiuc 5 years, allowing 5 per cent, compound inferest ? i-f i-o5+ro5| +i*o5|+i'o5| = 5*525^3125 5-52563125 40 221 •02525 20 0-505 ■" ^2 6'o6 Ans. 22 il. 6cl 2. If 5oiv yearly rent, or. annuity, I>e forborn 7 years, what will it amount to, at 4 per cent, per annum, com- pound interest ? • ' Ans. 395!, 'Z'o find ihc present Value of Annuities at Ccmpound Iniei'esU' RULE.* I. Divide the anmiity by the ratio, or the amount of iL for one year, and the quotient will be the present worth of I year's annuity. 2. Divide * The reason of this . rule is evident from tlie nature of the question, and v/hat was said upon the same subject in the purchase ing of annuities at simple interest. Let p = present worth of the annuity, and the other letters as r^ I before, then as the amount = X'?? and as the present wortli r — I or AXNUIT 213 2. Div'iuc the annuity by the square of the r:ulo, and ^he quotient will be the present worth cf the annuity for two years. 3. Find, in like manner, the present worth of each year by itself, and the sum of all these will be the value of the annuity sought. EXAMPLES. or principal of this, according to the principles of compound In- terest, Is the amount divided by r, therefore r" — I , r'-l-T — / «X -7- r -py and/ X -r^ =«. r'-j-l — r^ r — I And from these theorems all the cases, where the purchase of annuities is concerned, may be exhibited in logarithmic terms, as follows : I, Z^^. n-^Lcg. I — — Log, r — i~Lcg.p, II. Log. p-^- Log. r — I — Log. I — — "TzI^ng. n. III. — £ 1 — JX.^.JL^zz.t. IV. r-'^ — ---fi X/-/ + — =ro. Log.r P p Let t express the number of half years or quarters, n the half year's or quarter's payment, and r the sum of one pound and \ or i" year's interest, then all the preceding rules are applicable to half-yearly and quarterly payments, the same as to whole years. The amount of an annully mciy also le found for years and parts of a year, thus : 1. Find the amount for the whole years as before. 2. Find the Interest of that amount for the given parts of a year. 3. Add this interest to the former account, and it will give the whole amount required. ne 214 ARITHMETIC. EXAMPLES. I. What is the present worth of] an annuity of 40I. to continue 5 years, discounting at 5 per cent, per annum, CPinpound interest ? ratio = i'o5)40'ooooo(38'095=rpresentworthforiyear. ratio) = i'io25)40'ooooo(36'28i= do. for 2 years. ratio| = ri57525)4o-pooop(34'556=: do. for 3 years. ratio| = i'2i55o6)4o'ooooo(3a-S99=: do. for4years. — 7-5 ratio! 1= I '2 762 18)40-00000(3 1 "342 =2 do. for 5 years. 173*^73 = 73^' 3S- 5t^- = whole present worth of the annuity required. 2. What is the present worth uf an annuity of 2x1. I OS. p^d. to continue 7 years, at 6 per cent, per annum, compound interest ? Ans. 120L 5s. 3. What is 70I. per annum, to continue ^g years,, worth in present money, at the rate of 5 per ceat. per "annum ? Ans. 1321*30211. To The present ivorth of an annuity for years and parts of a year may be found, thus ■ : 1. Find the present wortli for the whole years as before. 2. Find the present worth of this present worth, discounting for the given parts of a year, and it will be the whole present worth required. ilN3S[UITIES. 215 "To find the present Worth of a Freehold Estate ^ or an Atmui" ty to continue for every at Compound Interest, RULE.* As the rate per cent, is to lool. so Is the yearly rent to the value required. EXAMPLES. * The reason of this rule Is obvious : for since a year's interest of the price, which is given for it, is the annuity, there can neither more nor less be made of that price than of the annuity, whether It be employed at simple or compound interest. The same thing may be shewn thus : the present worth of an annuity to continue forever is }- ~~-\- J- — , &c. ad infinitum, r r^ r* r+ as has been shewn before ; but the sum of this series, by the rules of geometrical progression. Is .. ; therefore r — i : i i: n z r — I —— , which is the rule. r — I The following theorems shew all the varieties of this rule. I. — L-=r/>. II. r — I %pz=.n. III. ^ + i=:r, or -=r— -i. r— I p p The price of a freehold estate, or annuity to continue for ever, at simple interest, would be expressed by — , , -{ i-j-r 14-2/ + — h — ■• J &c. ad infuutum : but ihe sum of this sfe- ries is Infinite, or greater than any assignable number, \\\\ich. sufficiently shews the absurdity of using simple interest in these. 2i6 ARiTHMETlG. EXAMPLES I. An esratc brings in yearly 79I. 4s. what would it ?elt for, allowing the purchaser a— per cent, compound interest fcr his money ? 4*5 : 100 : : 79*2 loo 4'5)792o'o{i';6ol. the answer^ 45 342 315 270 270 2. "What is the price of a perpetual annuity of 40L dis- counting at 5 per cent, compound interest f Ana. 2ooL . 3. What is a freehold estate of 75I. a year worth, all- lowing the buyer 6 per cent, compound interest for his money ? Ans. 125CI.' To find the present TForih cf an Annuityy or Frcehcld Estate^ in Reversion, at Compound Interest, RULE.* I. Find the present worth of the annuity, as if it wer^ to be entered on immediately. 2. Find ** This rule is sufficiently evident without a demonstration. Those, who wish to be acquainted with the manner of comput- ing the values of annuities upon lives, may consult the writings of Ivlr. Demoivre, Mr. Simpson, and Dr. Price, all of whom have handled this subject in a very skilful and masterly manner. Dr, ANNUITIES. 217 " 2. Find the present worth of the last present worth, discounting for the time between the purchase and com* mencement of the annuity, and it will be the answer re- quired. EXAMPLES. 1. The reversion of a freehold estate of 79I. 4s. per an- num, to commence 7 years hence, is to be sold •, what is k worth in ready money, allowing tho purchaser 4- per cetit. for his money ? 4*5 : 100 : : 79*2 100 4''5)792o*o(i)6o= present worth, 45 if entered on im- • — - mediately. 315 270 ^270 and i*o45'i == 1*360862)1 760*000(1 293*297 = 1293I. 5s. ii-^d. = present worth of 1760I. for 7 years, or the whole present worth required. 2. Which is most advantageous, a term of 15 years in an estate of lool. per annum, or the reversion of such au estate forever, after the expiration of the said 15 years, computing Dr. Price's Treatise upon Annuities and Reversionary Pay- ments is an excellent porformance, and will be found a very val- uable acquisition to those, whose inclinations lead them tq studies of this nature. 2 1 S ATIITHMETIC. computing at the rate of 5 per cent, per annum, convpound interest ? Ans. The first term of 15 years is better than the ric- version for ever afterward, by 75I. i8s. 7^^' 3. Suppose I would add 5 years to a running lease of 15 years to come, the improved rent being 186I. 7s. 6d. per annum 5 what ought I to pay down for this favour, discounting at 4 per cent, per annum, compound interest .? Ans. 460I. 14s. l|;dr POSITION. Position is a method of performing such questions, a^ cannot be resolved by the common direct rules, and is ol two kinds, called si/jgle and double, SINGLE POSITION. ^ Sif^gk Position teaches to resolve those questions, whose' results are propoi onal to their suppositions. , RULE.* 1. Take any number and perform the same operations with it, as are described to be performed in the question. 2. Then say, as the result of the operation is to the po- sition, so is the result in the question to the number re- quired. EXAMPLES. * Such questions properly belong to this rule, as require the multiplication or division of the number sought by any proposed number ; or when it is to be increased or diminished by itself, or any parts of itself, a certain proposed number of times. For in thi? POSITION. 219 EXAMPLES. I. A's age is double that of B, and B*s is triple that of" C, and the sum of all their ages is 140 : what is each person's age ? Suppose A's age to be 60, Then will B's = V=30> AndCs = V° = 10. 100 sum. As TOO : 60 : : 140 : :^^-?i^ =84= A's age. Consequently \-* =42 = B*s. And Y =i4 = Cs. 140 Proof. 2. A certain sum of money is to be divided between 4 |)ersons, in such a manner, that the first shall have y of it j the second -^ ; the third y *, and the fourth the remainder, which is 28I. : what is the sum ? Ans. 112I. 3. A person, after spending — uxxd -^ of his money, had 60I. left : what had he at first ? Ans. 11 4J. 4. What number is that, which being increased by ~, -^ and -J of itself, the sum shall be 1 25 ? Ans. 60, 5. A this case the reason of the rule is obvious ; it being then evident, diat the results are proportional to the supposidons. Thus ex : X : : na : a — : .V : : n a n a - +— ,&c. n — m I X ; : j — ,&c. : 5,&soon. n — m Note, i may be macie a constant supposition in all questions, and in mi>st cases it is better tliao any other number. 2iO ARITHMETIC. 5. A person bouglit a chaise, horse and harness, for 60I. ; the horse came to twice the price of the harness, and the chaise to twice the price of the horse and harness ; what did he give for each ? Ans. 13I. 6s. 8d. for the horse, 6L 13s. 4d. for the har- ness, and 40I. for the chaise. 6. A vessel has three cocksj A, B and C ; A can fill it in I hour, B in 2, and C in 3 : in what time will they all fill it together ? Ans. -^ hour. DOUBLE POSITION. Double Position teaches to resolve questions by making two suppositions of false numbers. RULE.* I. Take any two convenient numbers, and proceed ^yith each according to the conditions of the question. 2. Find * The rule is founded on this supposition, that the first errgr, is to the second, as the difference between the true and first sup- posed number is to the difference between the true and second supposed number : when that is not the case, the exact answer to the question cannot be found by this rule. That the rule is true, according to the supposition, may be thus demonstrated. Let A and B be any twp nurnbers, produced from a and b by similar operations ; it is required to find the number, from wh^ich JV" is produced by a like operation. Put X = number required, and let N — Azzr, and N—B:=:s, Then, according to the supposition, on which the rule Is found- ed, r IS : : x— •« : x — b, whence, by multiplying means and extremes. POSITION. 221 2. Find how much the results are difFerent from the re- sult in the question. 3. Multiply each of the errors by the contrary supposi- tion, and find the sum and difference of the products. 4. If the errors be alike, divide the dIfFevence of the products by the difference of the errors, and the quotient will be the answer. 5. If the errors be unlike, divide the sum of the prod- nets by the sum of the errors, and the quotient will be the answer. Note. The errors are said to be allkey when they are both too great or both too little ; and U7jlikey when one is too great and the other too little. EXAMPLES, 1. A lady bought tabby at 4s. a yard, and Persian at 2s. a yard ; the whole number of yards she bought was 8, and the whole priuc u;uc. . Kow inauj j-ardo kad olic o£ cach sort f Suppose extremes, rx — rhzzsx — sa ; and, by transposition, rx — siKz:zr!/ — sa ; aod, by division, xz= = number sought. r — s Again, if r and s be both negative, ws shall have — r : — s ; : X — a : x — If, and therefore — rx-\-rh=z — sx-^-sa ; and rx— ^.v=r3 — sa ; whence x == ■ as before. r — J- In like manner, if r or j be negative, we shall have sxrrr — ^ — , by working as before, which is the rule. Note. It will be often advantageous to make 1 and o the suppositions. 522 ^ ARITHMETIC, Suppose 4 yards of tabby, value i6s. Then she must have 4 yards of Persian, value 8 Sum of their values 24 So that the first error is + 4 Again, suppose she had 3 yards of tabby at 12s. Then she must have 5 yards of Persian at i o Sum of their values 22 So that the second error is -|~ ^ Then 4 — 2n:2zz: difference of the errors. Also 4X.2z:zSzz product of the first supposition and second error. And 3X41:^12^1 product of the second supposition by the first error. And 12 — 8zz4zr their difference. Whence 4-4- 2ZZ2Zi: yards of tabby, ? , A J o ~/c J r r» • ^ > uiQ answer. And 8 — 2 = 0= yards of Persian, 3 2. Two persons, A and B, have both the same income j A saves y of his yearly ; but B, by spending 50I. per an- num more than A, at the end of 4 years finds himself locl. in debt :^ what is their income, and what do they spend per annum ? Ans. Their income is 125I. per annum; A spends lool. and B 150I. per annum. 3. Two persons, A and B, lay out equal sums of money in trade ; A gains 126I. and B loses 87I. and A's money is now double that of B : vi^hat did each lay out ? Ans. 3ooh 4. A labourer was hired for 40 days, upon this condi- tion, that he should receive 2od. for every day he wrought, and forfeit lod. for every day he was idle j now he receiv- ed rERMUTATlON AND COMBINATION. 223 cd at last 2I. IS. 8d. : how many days did he work, and how many was he idle ? Ans. He wrought 30 days, and was idle 10. 5, A gentleman has two horses of considerable- value, and a saddle worth 50I. ; now, if the saddle be put on the back of the first horse, it will make his value double that of the second ; but if it be put on the back of the sec- ond, it will make his value triple that of the first : what is the value of each horse ? Ans. One 30I. and the other 40L 6. There is a fish, whose head is 9 inches long, and his tail is as long as his head and half as long as his body, and his body is as long as his tail and his head '. what is the whole length of the fish ? Ans. 3 feet* PERMUTATION and COMBINATIOR The Permutation of ^tantittes is the shewing how many different ways the order or position of any given number of things may be changed. This is also called Variation^ Aliernatttm^ or Changes ; and the only thing to be regarded here is the order they stand in ; for no two parcels are to have all their quanti- ties placed in the same sitaation. The Combination of ^lantiiies is the shewing how often a less number of things can be taken out of a greater, and combined together, without considering their places, or the order they stand in. This is sometimes called Election, or Choite ; and here every parcel must be different from ail the rest, and no two are to have precisely the same quantities, or things. The Compcsiiion of ^mniitles is the taking a given num- b^er of quantities out of as many equal rows of different quantities, ^24 ARITHMETIC. quantities, one out of each row, and combining them to-^ gether. Here no regard is had to their places ; and it differs from combination only, as that admits of but one row, or set, of things. Cotnbhiations of the same form are those, in which there is the same number of quantities, and the same repeti- tions : thus, ahccy bbady deef Sec. are of the same form ; but Mcy abbhy aaccy &c. are of different forms. PROBLEM I. u^o find the number of permutations ^ or changes y that can be made of any given number of things^ all different from each other. RULE.* Multiply all the terms of the natural series of numbers^ from I up to the given nnmher, r^oiT-tinually together, and the last product will be the answer required. EXAMPLES. * The reason of the rule may be shewn thus : any one thing a is capable only of one position, as a. Any two things^ a and 3, are only capable of two vaiiations j as abi la ; whose number is expressed by 1X2. If there be 3 things, ^, b and r, then any two of them, leaving out the third, will have 1x2 variations ; and consequently, when the third is taken in, there will be 1x2x3 variations. In the same manner, when there are 4 things, every 3, leaving out the fourth, will have 1X2X3 variations. Then, the fourth being taken in, there will be 1X2X3X4 variations. And so 0% as far as you pleaje. PERMUTATION AND COMBINATION. 22^ . EXAMPLES. I. How many changes may be made with these three letters, abc ? I 2 Or 1X2X32=6 the answer. 6 The changcoi ahc acb hac lea cab cla 2. How many changes may be rung On 6 bells ? Ans. 720.' 3. For how many days can 7 persons be placed in a dif- ferent position at dinner ? Ans. 5040 days. 4. How many changes riiay be rung on 12 bells, and how long would they be in ringing, supposing 10 changes to be rung in i minute^ and the yea): tp^ consist of 365 days, 5 hours and 49 minutes ? v^.^ Ans. 479001600 changes, and 9iy. 26d. 22h. 41m. ^. How many changes may be made of the words in the following verse ? 7i/ tibi sunt dotes, virgo^ quot sydera coeh, Aris. 40320. PROBLEM E t 226 ARITHMETIC. PROBLEM II. \Afiy nuniher of different things being given y to find honv rtian^ changes c^n be made out of thcm^ by taking any given num* her at a ti?ne» RULE* Take a series of numbers, beginnlnp; at the number of things given, and decreasing by i till the number o^, terms be equal to the number of things to be taken at a time, and the product of all the term.s will be the answer re- quired. EXAMPLES. * This rule, expressed in terms, is as follows : myim — i X f^i — 2 Xm — 3, &c. to n terras ; where m = number of things given, and n = quantities to be taken at a time. In order to demonstrate the rule, It will be necessary to pre- mise the following .L E M M A. The number of change^ of m things, taken ri at a time, is equal to m changes of m — i things, taken n — -i at a time. • Demonstration. Let any 5 quantities, abcde^ be given. First, leave out the a, and let ir -^ number of all the variations of every two, be. Id, &c. that can be taken out of the 4 remain- ing quantities, bcde. Now let ^ be put in the first place of each of them, abc^ ahd^ Sec. and the nuMber of changes will' st?ll remain the same ; that is» •0 =: number of variations of every 3 out of the 5, abcde, when a 13 first. In like manner, If b, c, d, e, be successively left out, the "num- ber of variations of all the twos will also = v ; and b, r, d, e^ being respectively put in the first place, to make 3 quantities out ©f 5, there will still be tf variations as before. But these are all the variations, that can happen of 3 things out «f 5, when a, by c, d^e, are successively |^ut first ; and therefore the PERMUTATION AND COMBINATION. 2^7 EXAMPLES. 1. How many changes may be made out of the 3 letters abcy by taking 2 at a time 3 2 Or 3X2e:6 the answer. 6 The changes. ab ha ac ca he ch 2. How many words can be made with 5 letters of the alphabet, it being admitted that a number of consonants may make a word ? Ans. 5100480. PROBLEM tlje sum of all these is the sum of all the changes of 3 things out ofS- . But the sum of these is so many times ^', as is the number of things ; that is, 51;, or wv, =; all the changes of 3 things out of 5, And the same way of reasoning may be applied to any num- bers whatever. Demonstration of the Rule. Let any 7 things, alcdef^j be given, and let 3 be the number of quantities to be taken. Then m=7, and «=3. Now it is evident, that the number of changes, that can be made by taking l by i out of 5 things, will be 5, which let rr-u. Then, by the lemma, when mz=:() and «=2, the number of changes will =:m'y=6X5 ; which let =:i? a second time. Again by lemma,, when w=:7 and n=:3, the number of changes =;nTJ=:7X6X5 ; that is, mvz=.m%m — i Xw — 2, continued to 3, or n terms. And the same may be shev/n for any other numbers. ?l^8i ARITHMETIC. * PROBLEM III. Any fiumher of things being ^iven^ whereof there are several given things of one sort^ several of another ^ ^c, to find haiu many changes can be made out of them all. RULE.* I. Take the scries i, 2, 3, 4, &c. up to the number of things given, and find the product of all the terms. 2. Take * This rule is expressed in terms thus : ^ — T — I 5 vvhere m = number i X 2 X 3,&c.to/ X I X 2 X 3,&c. to ^, &c. of things given, p =: number of things of the first sort, q = num- ber of things of the second sort, &c. The DEMONSTRATION may be shewn as follows : Any two quantities, ^, b, both different, admit of 2 changes ; but if the quantities be the same, or ab become aa, there will be but one alternation ; which may be expressed by -Jil_ =1. 1X2 Any three quantities, akf all different from each other, afford 6 variations ; but if the quantities be all alike, or c^r become aaa^ then the 6 variations will be reduced to i j which may be ex- pressed by — — — _lA~i. Again, if two of the quantities only 1X2x3 be alike, or abc become aac, then the six variations will be re- duced to these 3, aacy caa and aca j which may be expressed by J2ii2<32l4 ^ ,,. 1X2 And by reasoning in the same manner, it will appear, that the number of changes, which can be made of the quantities alhccc, is equal to 60 ; which may be expressed by ^X^X3X 4X5x6 1X2x1X2x3 ;=6o ; and so of any other Quantities whatever. 230 ARITHMETIC. 2. Hovy many different numbers can be made of the fciiowing figures, 1220005555 ? Ans. 12600. 3. What is the variety in the succession of the follow- ing musical notes, fa, fa, fa, sol, sol, la, mi, fa ? Ans. 3360. PROBLEM IV. fTo find the changes of any g'rjen number of things, tahen a given number at a time ; in *which there are several given things of one sorty several of another , ^c» RULE.* 1. Find all the different forms of combination of all the given things, taken as many at a time as in the question. 2. Find the number of changes in any form, and mul- tiply it by the number of combinations in that form. 3. Do the same for every distinct form ; and the sum of all the products will give the- whole number of changes required. Note. To find the different forms of combination proceed thus : 1. Place the things so, that the greatest indices may be first, and the rest in order. 2. Begin with the first letter, and join it to the second, third, fourth, &c. to the last. 3. Then take the second letter, and join it to the third, fourth, &c. to the last ; and so on through the whole, always remembering to reject such combinations as have occurred before j and this will give the combinations of all the tv/os. 4. Join * The reason of this rule is plain from what has been shewn before, and the nature of the problem. PERMUTATION AND COMBINATION. 23 1 4. Join the first letter to every one of the twos follow- ing it, and the second, third, &c. as before i and it will give the combinations of all the threes. 5. Proceed in the same manner to get the combinations of all the fours, &c. and you will at last get all the sev- eral forms of combination, and the number in each form. EXAMPLES. I. How many changes may be made of every 4 letters, that can be taken out of these 6, aaabbc ? No. of Forms. No. of com- No. of changes in forms. binations. each form. X2X3X4=24 isfc a^li, a'^c 2 -l — =4. X2X3 =6 2d a'h 1: 1.1X2X1X2=4 r 1x2x3x4=24 i ~=i2. LIX2 =2 3d a^hc, b'^ae 2 4X2= 8 6X1= 6 12X2=24 38 = the number of changes required. 2. How many changes can be made of every 8 letters out of these 10, aaaahhccde ? Ans. 22260. 3. How many different numbers can be made out of i unit, 2 twos, 3 threes, 4 fours, and 5 fives, taken 5 at a time ? Ans. 21 1 1. PROBLEM ^3^ ARITHMETIC. PROBLEM V, fn* find the mfmher of cGmbinatioin of any given nuinler of thitigSy all different from one a?ioiher^ tahen any given num' her at -a time. I. Take the series ij 2, 3, 4, &c. up to the number to be taken at a time, and find the product of all the terms. 2. Take * This rule, expressed algebraically, is — X X X ^Lni. , Sec. to s^ .terms 1 where m is the number of given quanti- 4 tics, and n those to be takeij at a time. . Demonstkation of the Rule, i . Let the number of things to be taken at a time be 2, and the things to be combined =;n. Now, when m, or the number of things to be combined, is on- ly two, as a and by it is evident, that there can be only one com- bination, as ab ; but if m be increased by i, or the letters to be combined be 3, as abc, then it is plain, that the number of com- binations will be increased by 2, since with each of the former letters, a and ^, the new letter c may be joined. It is evident, therefore, that the whole number of combinations, in this case, will be truly expressed by i -|- 2. Again, if m be increased by one letter more, or the whole number of letters be four, as abed ', then it will appear, that the whole number of combinations must be increased by 3, since with each of the preceding letters the new letter d may be com- bined. The combinations! therefore, In this case, will be truly expressed by i -f- 2 -|- 3. PERMUTATION AND COMBINATION. 233 2. Take a series of as many terms, decreasing by i, from the given number, out of which the election is to be made, and find the product of all the terms. 3. Divide the last product by the former, and the quo- tient will be the number sought. EXAMPLES. In the same manner, it may be shewn, that the whole number of combinations of 2, in 5 things, will be 14-24-3+4; of 2, in 6 things, 1 4-24-3-^-44-5 ; and of 2, in 7, i 4-24-3+4+5 4-6, &c. Whence, universally, the number of combinations of m things, taken 2 by 2, is =14-24-34-44-5-1-6, &c. to m — i terms. Bat the sum of this series is ::z!!L-^!!tII^ ' ; which is the same as I 2 the rule. 2. Let now the number of quantities In each combination be supposed to be three. Then it is plain, that when mzz'^y or the things to be combined are ahci there can be only one combination ; but if m be increased by I, or the things to be combined be 4, as abcd^ then will the number of combinations be increased by 3 ; since 3 is the num- ber of combinations of 2 in all the preceding letters ahcj and with each two of these the new letter d may be combined. The number of combinations, therefore, in this case, is 14-3. Again, if m be increased by one more, or the number of let- ters be supposed 5 ;^hen the former number of combinatiojis will be increased by 6 ; that is, by all the combinations of 2 in the 4 preceding letters, abed ; since, as before, with each two of these the new letter e may be combined. The number of combinations, therefore, in this case, is 1 + 3 + 6. Whence, universally, the number of combinations of m things', taken 3 by 3, is i + 3 4-6-f.io, 6cc. to m — 2 terms. But F F 234- ARITHMETIC. EXAMPLES. I. How ms^ny combinations can Be -matlc of 6 lefteris^ out of 10? 1X2K3X4X5X 6(=: the number to be taken at a time)= 720 10x9x8 X 7 X6X5(= same number from 10)3=151200 720)151200(210 t^e answer-. 1440 720 720 ^. How many combinations can tie Tha& cr£ 2 letfer^j- out of 24 letters of the alphabet I An Si 276.^ 3.- A general, who hacT often been successful in \^ar,- was asked by his King, what reward he should confer upon' him for his servieoa y the general only desired a farthihg for every file, of 10 men in a file, which he could make^ with a body of 100 men : what is the amount in pound* sterling ? Ans. 1 303 15723 50I. 9s. 2d. PROELEJ-f- Burthe smn of this series is = HyJILJ vJ^^Ill y which 13- I 2 3 the same as the rule. And the same thing wii! hold, let the number of things, to be taken at a time, be what it may ; therefore the number of com* binations of m tilings, taken n at a time, will == — X— ^ X~ ■ 123, y,^^^, &c. to n terms. Q^ E. P. 4 PERMUTATION AND COMBINATION. 235 PROBLEM VI. 21? find the numher cf combinations of any given number of things, by taking any given number at a time ; in ivhith ^fy^ are several things of one scrty several cf another , C3*c. RULE. 1. Find, by trial, the number of difFerent forms, which .the things, to be taken at a time, will admit of, and the num- ber of combinations in each. 2. Add together all the combinations, thus found, and the suln will be the number required. EXAMPLES. I. Let the things proposed be aaabhc ; it is required to find the number of combinations that can be made of every -three of these quantities. Forms. Combinatioas. 1 a'^by a*Cf b^a^ b^c 4 abc I 6 = number of combina- .tions required. 2. Let qaahbbcc be proposed j it Is required to find the number of combinations of these quantities, taken 4 at a time. Ans. 10. 3. How many combinations are there in aaaabbccde, 8 being taken at a time ? Ans. 13. 4. How many combinations are there in aaaaabhbhhccccdd ddecfcfffg, \ o being taken at a time ? Ans. 2819. PROBLEM 226 ARITHMETIC* PROBLEM VII. 31? jfind the compositions of any numhery vi an equal number of setSy the things themselves being all different. RULE.* Multiply the number of things in every set continually together, and the product will be the answer required. EXAMPLES. * Demonstration. Suppose there are only two sets ; then it is plain, that every quantity of one set, being combined with every quantity of the other, will make all the compositions of tv/o things, in these two sets ; and the number of these composi- tions is evidently the product of the number of quantities in one set by that in the other. Again, suppose there are three sets ; then the composition of two, in any two of the sets, being combined with every quantity of the third, will make all the compositions of 3 in the 3 sets. That is, the compositions of 2, in any two of the sets, being mul- tiplied by the number of quantities in tlie remaining set, will pro- duce the compositions of 3 in the 3 sets ; which is evidently the continual product of all the 3 numbers in the 3 sets. And the same manner of reasoning will hold, let the number of sets be what it will. (^E. I>. The doctrine of permutations, combinations, &c. is of very ex- tensive use in different parts of the mathematics ; particularly in the calculation of annuities and chances. The subject might have been pursued to a much greater length ; but what has been done already will be found sufficient for most of the purposes to which things of this nature are applicable. MISCELLANEOUS QUESTIONS. 237 EXAMPLES. I. Suppose there are 4 companies, in each of which there are 9 men ; it is required to fmd how many ways 4 men may be chosen, one out of each company ? 9 9 81 9 729 9 6561 Or, 9X9X9X9 = 6561 the answer- 2. Suppose there are 4 companies, in one of which there are 6 men, in another 8, and in each of the other two 9 ; what are the choices, by a composition of 4 men, one out of each company ? Ans. 3888. 3. How many changes are there in throwing 5 dice ? Ans. 7776^ ■•*«*« LOGARITHMS. ^a**^©©®^-'^^*-^®®®*^©^ JuOGARITHMS are numbers 60 contrive!, and adapt- ed to other numbers, that the sums and differences of the former shall correspond to, and shew, the products and quotients of the latter. Or, logarithms are the numerical exponents of ratios ; or a series of numbers in arithmetical progression, an- swering to another series of numbers in geometrical progression. 'hus C o, I, 2, 3, 4, 5, 6, indices, or logarithm*. ^ I, 2, 4, 8, 16, 32, 64, geometric progression. Qj. ^^y ^3 2, 3, 4, 5, 6, indices, or logar, C ^> 3j 9j 27, 8 1, 243, 729, geometric progress. Q^ ^o> i> 2, 3, 4, 5, ind. orlog. (^ I, 10, 100, 1000, loooo, icocooj gcom. prog. Where it is evident, that the same indices serve ecpi Jlv t<'r any geometric series ; and consequently there may be an endless variety of systems of logarithms to the sam-j cnmmon numbers, by oidv cliangiiig the seCoivd term 2, 3, oi 244 LOGARITHMS. or lo, Sec. of the - geometrical series of whole numbers i and by interpolation the whole system of numbers may be made to enter the geometric series, and receive their pro- portional logarithms, whether integers or decimals. It is also apparent from the nature of these scries, that if any two indices be added together, their sum will be the index of that number, which is equal to the product of the two terms in the geometric progression, to which those indices belong. Thus, the indices 2 and 3, being added together, make 5 ; and the numbers 4 and 8, or the terms corresponding to those indices, being multiplied to- gether, make 32, which is the number answering to the index 5. In like manner, if any one index be subtracted from an- other, the difference will be the index of that number, which is equal to the quotient of the two terms, to which those indices belong. Thus, the index 6 minus the index 4rz2 ', and the terms corresponding to those indices are 64 and 16, whose quotient r:;4 j which is the number an- swering to the index 2« For the same reason, if the logarithm of any number be multiplied by the index of its power, the product will be equal to the logarithm of that power. Thus, the index or logarithm of 4, in the above series, is 2 ; and if this number be multiplied by 3, the product will be zz6 ; which is the logarithm of 64, or the third power of 4. And if the logarithm of any number be divided by the index of its root, the quotient will be equal to the logar- ithm of that root. Thus, the index or logarithm of 64 is 6 ; and if this number be divided by 2, the quotient will be 1=3 ; v/hich is the logarithm cf 8, or the square root Di 64. The logaritlims most convenient for practice are such, 3S are adapted to a geometrical series, increasing in a ten^? fold proportion, as in the last of the above forms ; and are those, NATURfi OF LOGARITHMS. 245 tjiose, which are to be found at present, in mo-st of the^ common tables of logarithms. The distinguishing mark of this system of logarithms is, that the index or logarithm of 10 is I j that of 100 is 2 ; tliat of 1000 is 3, &c. And, in decimals, the logar- ithm of 'I is -—I 5 that of 'pi is — 2-5 that of 'ooi is . — 3, &c. the logarithm of i being o in every system. Whence it follows, that the logarithm of any number between i and 10 must be o and some fractional parts ; and that of a number between 10 and 100, i and some fractional parts ; and so on, for any other number what- ever. And since the integral part of a logarithm, thus readily found, shews the highest place of the corresponding num- ber, it is called the hia'ex, or characteristic, and is common- ly omitted in the tables ; being left to be supplied by the person, who uses them, as occasion requires. , Another definition of logarithms is, that the logarithm of any number is the index of that power of some other number, which is equal to the given number. So if there be JVzzr", then n is the log. of N \ where n rriay be either positive or negative, or nothing, and the root r any num- ber whatever, according to the different systems of log- arithms. When n\s zro, then AT is zri, whatever t\\Q value p£ r is ; which shews, that the logarithm of i is always o, in every system of logarithms. When n is :=i, then AT is rzr ; so that the radix r is always that number, whose logarithm is i in every system. When the radix r is zr2'7 1828 1828459, Sec. the in- dices n are the hyperbolic or Napier's logarithm of the numbers N ; so that n is always the hyperbolic logarithm of the number A^ or 2*7 18, &c.| . But when the radix r Is zzio, then the index n becomes lj:e common or Briggs' logarithm of the number N ; so that 24^ LOGARITHMS, that the common logarithm of any number lo" or N is ft the index of that power of lo, which is equal to the said number. Thus, loo, being the second power of lo, will have 2 for its logarithm j and looo, being the third power of lo, will have 3 for its logarithm : hence also, if 50 be ---jqI'6 9 8 9 7^ then is i "69897 the common logarithm of 50. And, in general, the following decuple series of terms, viz. 10*, loS 10% 10% 10*, lo""', 10"*, 10""*, IQ-S or loooo, 1000, 100, 10, I, 'I, '01, '001, '0001, have 4, 3,. 2, I, o, — i, —2, — 3, —4, for their logarithms, respectively. And from this scale of numbers and logarithms, the same properties easily follow, as before mentioned. PROBLEM. To compute the logarithm to any of the natural numherSy i, 2, 3, 4, 5, ^c, RULE. Let b be the number, whose logarithm is required to be found ; and a the number next less than ^, so that b — a'zz I, the logarithm of a being known ; and let s denote the sum of the two numbers ^r-f-^. Then 1. Divide the constant decimal '8685889638, &c. by /, and reserve the quotient ; divide the reserved quotient by the square of s, and reserve this quotient 5 divide this last quotient also by the square of j, and again reserve the quo- tient ; and thus proceed, continually dividing the last quotient by the square of j", as long as division can be made. 2. Then write these quotients orderly under one anoth- er, the first uppermost, and divide them respectively by the COMPUTATION OF LOGARITHMS. 247 the odd numbers, i, 3, 5, 7, 9, &c. as long as division can be made ; that is, divide the first reserved quotient by i, the second by 3, the third by 5, the fourth by 7, and so on. 3. Add all these last quotients together, and the sum will be the logarithm of b-^a ; therefore to this logarithm add also the given logarithm of the said next less number ^, so will the last sum be the logarithm of the number h proposed; That is, log. of b Is log. ^ + ~ X : iH — i-] — ^H 5 s 31 ^S^ ']S -}- &c. where n denotes the constant given decimal •8685889638, &c. EXAMPLES. Example t. Let it be required to find the logarithm of the number 2. Here the given number b Is 2, and the next less number a Is I, whose logarithm Is o ; also the sum 2+izr3zrj", and its square ^^1^9. Then the operation will be as fol- low* : 3)-868588964 i)'289529654(-289529654 9)-289529654 3) 32i69962( 10723321 9) 32169962 5) 357444o( 714888 _ 9) 3574440 7) 397i^o( 56737 9) 397160 9) 44i29( 4903 9) 44129 ,11) 4903( 446 9) 4903 13) 545( 42 9) 545 15} 6i( 4 9) 61 • Log. of f -3 01029995 Add log. I 0 Log. of 2 *3 00000000 01029995 Example 248 LOGARITHMS* Example 2. To compute the logaritlim of the number a* Here h'zzi,, the next less number aiZ2y and the sum c-|-^rz5:i:j-, whose square s^ is 25, to divide by which, always multiply by '04. Then the operation is as follows i 5)-868588964 25) 6948712 277948 25) ^5) nii8 445 18 II i)-i737i77^3(-i737i7703 3) 69487i2( 2316237 277948( iiii8( 445( I8( 5) 7) 9) 5559^ 1588 5^ 2 Log. of -|- '176091260 Log. of 2 add '301029995 Log. of 3 sought '477121255 Then, because the sum of the logarithms of numbers gives the logarithm of their product, and the difrerence of the logarithms gives the logarithm of the quotient of the numbers, from the above two logarithms, and the logar-* ithm of 10, which is i, we may raise a great many log* arithms, as in the following examples : Example 3. because 2 X 2 zz 4, therefore To logarithm 2 •3010299954- Add logarithm 2 '301029995-1- B( Sum is logarithm 4*602059991-^ Example 4. Because 2 X 3 1:1 6, therefore To logarithm a '30102999^ Add logarithm 3 -4771212^5 Sum is logarithm 6*77815 1250. Ez>5Ml»LE computation of logarithms. 249 Example 5. Because 2^ zr 8, therefore Logarithm 2 •3010299954- Muhiplied by 3 3 Gives Logarithm 8 -903039987. ExATvlPLE 6. Because 3^ z;i 9, therefore Logarithm 3 •477121254-;;^- Multiplied by 2 ^ Gives logarithm 9 -954242509. Example 7. Because "-^ n: 5, therefore Prom logarithm 10 1*000000000 Take logaritlmi 2 -301029995-^- Reniains logarithm 5 •698970004J Example 8. Because 3X4:= 12, therefore To logarithm 3 -477121255 Add logarithm 4 602059991 Sum is logarithm 12 1*079101246. And thus, computing by this general ruk, the logar- ithms to the other prime numbers 7, II, 13, 17, 19, 23, $ic, and then using composition and division, \rc may ea- sily 2^0- LOGARITHMS. sily find as many logarithms as we please, or may speedily- examine any logarithm in the table.* Deschiption and Use of the TABLE of LOGARITHMS. Integral numbers are supposed to form a geometrical se- ries, increasing from unity toward the left •, but decimals are supposed to form a like series, decreasing from unity toward the" right, and the indices of their logarithms are negative. Thus, -f-i is the logarithm of lo, but — i is the logarithm of -p^-, or 'i ; and -{-2 is the logarithm of loo, but — 2 is the logarithm of ~-^, or 'oi 5 and so on. Hence it appears in general, that all numbers, which consist of the same figures^ whether they be integral, or fractional, or m.ixed, will have the decimal parts of their logarithms the same^ 'difFerihg only in the index, which will be more or less, and positive or negative, according to tlie place of the 'first figure of the- liiimber. Thus, the logarithm ot 26^1 being 3*4234097, the logarithm of —-y <^i" xoo> or ,;^^, &c. part of it^ will be as fol I0W&: Numbers. Logarithms. 2651 3-4234097 265*1 2-4234097 26*51 I •4234697 2*651 0-4234097 •2651 — 1-4234097 •0265 1 2-423405)7 '00 2 65 1 — 3*4^34097 Hence * Many other ingenious methods of finding the logarithms of numbers, and peculiar artifices for constructing an endre table of them, may be seen in Dr. Hutton*s Introducikn to his Tables^ and Baron M4SEres* Scrlptons Logarithmicu DESCRIPTION AND USE OV THE TABLE. 2$ I Hence it appears, that the irtdex, or characteristic, of ;iny logarithm is always less by i than the number of inter gcr figures, which the natural number consists of ; or it is £qual to the distance of the first or left hand figure from the place of units, or first place of integers, whether on the left, or on the right of it : and this index is constant- ly to be placed on the left of the deciinal part of the logarithm. When there are integers in the given number, the index is always affirmative"-, but when there are no integers, the index is negative, and is to be marked by a. short line drawn before, or above, it. Thus, a number having • i, 2> 3i 4) 5j &c. integer places, the index of its logaritiim Is Oj I, 2, 3, 4, &c. or I less than the number of those places. Aad a decimal fraction, having its first figure in the ist, 2d, 3l^4th, &c. place of decimals, has always — i, — 2, — 3,*^^^— 4, &c. for the index of its logarithm. It may also be observed, that though the indices of fractional quantities be negative, yet the decimal parts of their logarithms are always afhrmative. I. To riND, IN THE Table, the Logarithm to any Number.* I. If the number do net exceed iqo.ooo, the decimal part of the logarithm is found, by inspection in the table, standing against the given number, in this manner, viz. in most tables, the first four figures of the given number are -* The Tables, considered as the best, are diose of Gardinrs. in 4t6. first published ia the year 1742 ; of Dr. Hutton, In 8vo. first printed in 1785 ; of Taylor, in large 410. pviLlishcd in 1792 ; and in France, dioss of Cai.li-t, the second edition published In 1 795. 252 LOGARITHMS. are In the first column of the page, and the fifth figure in the uppermost line of it ; then in the angle of meeting are the last four figures of the logarithm, and the first three figures of the same at the beginning of the same line ; to which is to be prefixed the proper index. So the logarithm of 34*092 is 1*5326525, that is, the decimal 5326525, found in the table, with the index i pre- fixed, because the given number contain? two integers. 1. But if the given nwnocr contain more than Jive figures^ take out the logarithm of the first five figures by inspec- tion in the table as before, as also the next greater log- arithm, subtracting one logarithm from the other, and al- so one of their corresponding numbers from the other. Then say, , As the diiterence between the two numbers *^ Is to tlie difference of their logarithms, 80 is the remaining part of the given number To the proportional part of the logarithm. "Which part being added to the less logarithm, before tiken cut, the whole logarithm sought is obtained very nearly. EXAMPLE. To find the logarithm of the number 34*09264. - The log. of 3409200, as before, is 5326525, and log. of '3409300 is 5326652, the diff. 100 and 127 Theuj as too : 127 :: 64 : 81, the proportional part. This added to 5326525, the first logarithm, elves, with the index, i •532660.6 for the logarithm of 34-09264. Or, in the best tables, tlie proportional part may often be taken out by inspection, by means of the small tables pf proportional parts, placed in the margin. DESCRIPTION AND USE OF THE TABLE. 25:3 If the number consist both of integers ?.nd frnctlonr,, or be entirely fractional, find the decimal part of the logar- ithm, as if all its figures were integral ; then this, the proper characteristic being prefixed, will give the logarithm required. And if the given number be a proper fraction, subtract the logarltlmi of the denominator from tlie logarithm of the numerator, and the remainder will be the logarithm nought ; which, being that of a decimal fraction, must al- ways have a negative index. But if it be a mixed number, reduce it to an improper fraction, and find the difR:rcnce of the logarithms of the jiumciator and denominator, in the same manner as before. exa:.iples. I. To find the logarithm of |-|. Logarithm of 37 1*5682017 Logarithm of 94 ^'913^^19 DifF. log. of A{- —1.-^-0738 YV"here the index i is negative. 2. To find the logarithm of i y4-j. First, i7fi = VV- Then, Logarithm of 405 2*6074556 Logarithm of 23 1*3617278 IP iff. log. of I7{-1 1*2457272 II. Tq fi>,'d, in the Table, the natural Nltmeer to ANY Logarithm. This is to be found by the reverse method to the former, aamely, by searching for the proposed logarithm among those in the table, and taking out the corresponding num- ber 254 LOGARITHMS. ber by inspection, in whkh the proper number of integers is to be pointed oiF, viz. i more than the unit.j of the af- firmative index. For, in findings the number answering. tp any given logarithm, the index always sliews how far the first ligm-e must be removed from the place of units, to the left or in integers, when the index is afTirmative j but to the right or in decimals, when it is negative. EXAMPLES. ' So, the number to the logarithm 1*5326525 is 34'092. And the number of the logarithm — 1*5326525 is 34092. But if the logarithm cannot be exactly found in the tahlcy take out the next greater and the next less, subtracting one of these logarithms from the other, an^ also one of their natural numbers from the other, and the less logar- ithm from the logarithm proposed. Then say. As the first difference, or that of the tabular logarithms, Is to the difference of their natural numbers. So is the difference of the given logarithm and the last tabular logarithm To their corresponding numeral difference. Which being annexed to the least natural number above taken, the natural number corresponding to the proposed logarithm is obtained. EXAMPLE. Find the natural number answering to the given logar- ithm 1*5326606. Here the next greater and next less tabular logarithms, with their corresponding numbers, 3cc. are as below : Next greater 5326652 its num. 3409300 ; giv. log. 5326606 Next less 5326525 its num. 3409200; next less 5326525 Differences 127 100 81 Then, MULTIPLICATION BY LOGARITHMS- ^^c; Then, as 127 : 100 : : 81 : 64 nearly, the numeral difference. Therefore 34*09264 is the number sought, two integers being marked off, because the index of the given logarithm is I. Had the index been negative, thus, -1—1*^326606, its corresponding number would have been '3409264, wholly decimal. . • Or, the proportional numeral difference may be found, in the best tables, by inspection of the small tables of pro- portional parts, placed in the margin. MuLriPLICATlON RT LOGARITHMS. R U L E. Take out the logarithms of the factors from the table, then add them together, and their sum will be the logar- ithm of the product required. Then, by means of the ta- ble, take out the natural number answering to the sum, for the product sought. Note i. In every operation, what is carried from the decimal part of a logarithm to its index is aiFirmative ; and is therefore to be added to the index, when it is affirma- tive ; but subtracted, when it is negative. Note 2. When the . indices have- like signs, that Is, both -j- or both — ^, they are to be added, and the sum has the common sign ; but when they have unlike signs, that is, one -{- and the other — , their difference, with the sign of the greater, is to be taken for the index of the $um. EXAMPLES, 25<5 LOGARITHMS. EXAMPLES. I. To multiply 23*14 by 5*062. Numbers. I.ogaritlims. 23*14 1-3643634 5*062 0*7043221 Product 117*1347 2*0686855 f 2. To multiply 2*581926 by 3*457291. Numbers. Logarithms. 2*581926 0-4119438 3*457291 0-5387359 Product 8-92647 0*9506797^ 3. To multiply 3*902, 597* 16 and '0314728 all t<3H gether. Numbers. Logarithms. 3*902 0-5912873 597-16 2*7760907 •0314728 — 2*4979353 , I* Product 73'33533 ^'8653133 Here the — 2 cancels the 2, and the i, to be carried from the decimals, is set-down. 4. To multiply 3*586, together. Numbers. 2*1046, 0-8372 Logarithms. and 0*0294 all 3-586 2*1046 0*8372 0*0294 0-5546103 0323 1696 — 1-9228292 —2-4683473 Product 0*1857618 — 1*2689564 Kcre DIVISION BT LOGARITHMS. ft^j Here the 2, to be carried, cancels the -—2, and there re- tnains the -^i to be set down. Dins ION ST Logarithms. RULE. From the logarithm of the dividend subtract the logar-- athm of the divisor, and the number answering to the re« xnainder will be the quotient required. Note. If t be to' brc -carried to tJie'Jndex of the sub- trahend, apply it according to the sign of the index ; then change the sign of the index to — -, if it be -j~> or to -f-j if it be — ; and proceed according to the second note un«* - ALGEBRA, fi!afmS^S&SMl^l^f>»' DEFINITIONS and NOTATION. 1. jl\.LGEBRA is the art of computing by symbols^ It is sometimes also called Analysis •, and is a general kind of arithmetic, or universal way of computation. 2. In Algebra, the given, or known, quantities are usualTy denoted by the first letters of the alphabet, as a, b, c, dy &c. and the unknown, or required quantities, by the last let- ters, as X, y, z. Note. The signs, or characters, explained at the be- ginning of Arithmetic, have the same signification in Al- gebra. And a point h sometin^s used for X : thus 3. Those quantities, before which the sign -[- ^s placed, are called positive, or affirmative ; and those, before which the sign — is placed, negative. And it is to be observed, that the sign of a negative quantity is never omitted, nor the sign of an affirmative one. ^64 ALGEBRA. one, except it be a single quantity, or the first in n series of quantities, then the sign -j- is frequently omitted : thus tif signifies the same as +j, and the series is multiplied by the simple quantity c j so that if a were lO^ h 6, and c 4, then would a-^-hYsC be 10+6X4} or 16 in* to 4, which Is 64 J and rt+i^Xc-f-i expresses the product DEt^lNITlONS AND NOTATION. 26^ lof the compound quantities a-{-b and c-^d multiplied to- gether. 7. When we would express, that one quantity, as a, is greater than another, as b, we write ^C~^> or c; > 3 •, and if we would express, that a is less than ^, we write a^h or ^ <3 ^. " 8. When we would express the di|Ference between two quantities, as a and ^, while it isunknown which is the greater of the two, wc write them thus, t? £o ^, which de- notes the difference of a and ^. 9. Powers of the same quantities or factors are the prod- ucts of their multiplication : thus oXt?, or af2y denotes the ■square^ or second poiuer^'oi the quantity represented by ar \ ^jX«X«, ox aaay expresses the cube, on third power ; 2^nd >flX«X^X«> or aciaay denotes the biquadrate^ or fourth power of ay Sec. And it is to be observed, that'the quantity a is the root of all these powers. Suppose azzi^y then will aaz:zaXa:iZ 5X5^:25=1 the square of .5,^ ^1^^—^7X^^X^1=5X5X5 = J25ZZ the cube of 5 5 and aaaazHaXaXaXaZZ^X ^X S X5ZZ625ZI the fourth power of 5. I o. Powers are likewise represented by placing above the root, to the right hand, a figure expressing the number of fafctors, that produce them. Thus, instead of aa, we writer* ; instead of aaa, we write a^ -, instead of aaaa^ we write a"^, &:c. IT. These figures, which express the number of factors, that produce powers, are called their indices, or exponents : thus, 2 is the index or exponent oi a' \ 3 is that of x^ ; 4 is that of x"^, &c. But the exponent of the first power, though generally omitted, is unity, or i ; thus a signifies the same as «, namely, K K 266 ALGEBRA. namely, the -first power of a ; aXa^ the same as aX(^^ or «" + % that is, a'y and a'Xd is the same as a'Xn, or^?*'*^^ or fl^ 12. In expressing powers of compouiKl quantities,, we -usually draw a line over the given quantity, and at the end of the line place the exponent of the power. Thus, a J^h\ denotes the square or second power of a'\-h^ consid- ered as one quantity ; a'\'}\ the third power \ a-j-I^l ths fourth power, &c. And it may be observed, that the quantity a-\-h, called the first power of /3-j~^> ^^ ^^^ ^°^^ ^^ ^^^ these powers. Let <3iZ4 and ^r:2, then will ^-f-3 become 4+2, or 6 .5 and «+^[ zz:4-j-2| zz6*rz6X6zz36, the square of 6; also fl+^l =:4+2| i=6^i=6X6X6zr2i6, the cube of 6. 1 3. The divi-slon of algebraic quantities is very frequent- ly expressed by writing down the divisor under the divi- dend with a line between them, in the manner of a vulgar a fraction : thus, — represents the quantity arising by • di- viding a by r ; so that if a be 144 and c 4, then will _- be , or 36. And ■ — - — denotes the quantity ans* ■V 4 * a — c ing by dividing ^z^-^ by a — c ; suppose ^^12, ^rz:6 and .1, -n ^+^ u ^^ + ^ '^^< czzg, then will become or — rro. a — c 12 — 9. 3 J ^ •4- ^ 14. These literal expressions, namely, — and -, arc called algebraic fractions ; whereof the upper parts are call- ed DEFINITIONS AND NOTATION. 2^ ed the numerators, and the lower the denominators : thus, a is the numerator of the fraction — , and c is its denomi- c nator j a^c? is the numerator of , and a — c is its a — c denominator. 15. Quantities, to which the radical sign is applied, arc called radical quantikesy or surds ; whereof those consisting of one term only, as ^^ and y/ ^ a', are called simple surds ; and those consisting of several terms, as ^ ab 4- cd and ^rt"* — b^-{-bcy compound surds. i6. When any quantity is to be taken more than once, the number is to be prefixed, which shews how many times it is to be taken, and the number so prefixed Is called the numeral coejjicient : thus, 2a signifies twice a, or a tak- en twice, and the numeral coefficient is 2 ; 3a,'* signifies, that the quantity ^^ is multiplied by 3, and the numeral coeiiicient is 3 ; also Sv^A^^-j-^" denotes, that the quan- tity ^>^^-{-a^ is multiplied by 5, or taken 5 times. When no number is prefixed, an unit or 1 is always un- derstood to be the coefficient : thus, i is the coefficient of n or of .V ; for a signifies the same as la, and x the same as iXi since any quantity, multiplied by unity, is still the same. Moreover, if a and d be given quantities, and a;^ and v required ones *, then ax"" denotes, that x" is to be taken a rimes, or as many times as theje are units, in a ; and dy shews, that y is to be taken d times ; so that the coefficient of <7.v' is ay and that of dy is d : suppose jZZfS and dzz^, then 268 ALGEBRA. 3fAr then will ax^rz6x*, nnd dyzr:4y. Again, -at, or — , de-^ notes the half of the quantity Xj and the coefficient of ^^ is -f- ; so likewise -^x, or — , signifies ^ of x, and the co^ efficient of -^x is -!-• 4 4 1 7. iii^^ quantities arc those, that are represented by the same letters under the same powers, or which differ only in their coefficients : thus, 3^2, 5^ and a are like quanti- ties, and the same is to be understood of the radicals t^ x"" •{■ a^ and ']\/ x^ -^-a^ . But unlike quantities are those, which are expressed by different letters, or by the same letters under different powers: thus laby a^h^ lahcy ^ah^ y ^^ , J, y^ and z^ are all unlike quantities. 1 8. The douHe or amhiguous sign +_ signifies plus or mi-- nus the quantity, which immediately follows it, and being placed between two quantities, it denotes their sum, or dif-. ference. Thus, -j^ + V^ " ^ shews, that the quan- 4 ^^ tity \/^ — — ■ ^ is to be added to, or subtracted from, ^.7. 4 19, A general exponent is one, that is denoted by a letter instead of a figure : thus, the quantity x^ has a general exponent, viz. w, which universally denotes the ?«th pow- er of the root x. Suppose mzz2y then will x"':zzx^ ; if m :rz3, then will x^^znx^ j if ni=:4y then will x'"zz:x^, &c. In like manner, a — 1?\ expresses the ///th power of n- — k 2 J. This root, viz. a — A, is called a residual root, be- cause its value is no more than the residue, remainder, or difference,' of its terms a and k It is likewise call- . ed DEFINITIONS AND NOTATION. 2% ed a binomial, as well as a-^b^ because it is composed of two parts, connected together by the sign —- w 2f. A fraction, which expresses the root of a quantity, is also called an ifidtXy or exponent ; the numerator shews the power, and th^ denomihiitor the root : thus ^r* signifies the same as -y/ ^ 5 itnd a-^-ab"^ the same as ^a-^-ab 9 likewise n\ denotes the square of the cube root of the quantity a. Suppose ^11:64, then will a^'z=L6^z:zd^^ ziz 16 •, for the cube root of 64 is 4, anti the square of 4 is 16. Again, a + bf expresses the fifth power of the biquad- ratic root of a^b» Suppose orzp and ^iry, then will <2 + /i'^ii:9+7i^zi:i6|'^~2^zz32 -, for the biquadratic root of 16 is 2, and the fifth power of 2 is 32. Also, qH signifies the wth root of ^. If //IZ4, then will ^Turr/'^ ; if /'/ = 5, then will aH'iz.di, &c. Moreover, a-\~hY denotes the /;7th power of the ;nh root of a-\-h, \i ^ = 3 and ;/=2, then will a^hY •::^a-\-b\ , namely, the cube of the square root of the quantity a-\-b \ and as «" equals ^/^S or y/a^ so a-^-bY = Y/^^^^^, namely, the «th root of the wth pov/er of a-{-b. So that the wth power of the n'Oii root, or the «th root of the f;2th power, of a quantity are the very same in 'effect, though differently expressed. 22. An exponential quantity is a power, whose exponent is a variable quantity, as a;''. Suppose .v=2, then will ^*=2*=4 5 if ^^'=3, then wilU^*' = 3^=27. ADDITION. ALGEBRA. ADDITION. Addition, in Algebra, is connecting the quantities to- gether by their proper signs^ and uniting in simple term« such as are similar. In addition there are three cases. CASE I. TVhen like quaiitliics have like signs. R U L E.* Add the coefficients together, to their sum join the common letters, and prefix the common sign when nece^;-- sary. EXAMPLES. * The reasons, on which these operations are founded, will readily appear fi-om a little reflection on the nature of the quanti- ties to be added, or collected together. For with regard to the first example, where the quantities are 3^ and 5^, whatever a represents in one terra, it will represent the same thing in the other ; so that 3 times any tiling, and 5 times the same thing, col- lected together, must needs make 8 times thartfiing. As, if a denote a shilling, then ^a is 3 shillings, and $a is 5 shillings, and their sum is 8 shillings. In like manner — zab and — 7^^, or — 2 times any thing and — 7 times the same thing, make — 9 times that thing. As to the second case, in which the quantities are like, but the signs unlike ; the reason of its operation will easily appear by re- flecting, that addition means only the uniting of quantities togeth- er by means of the arithmetical operations denoted by their signs j^ and — , or of addition and subtraction ; which being of con- trary or opposite natures, one coefficient must be subtracted from the other, to obtain the incorporated or iinitcd mass. As Add Sum Sa AI5DITI0N- < . ^ '271 EXAMPLE 3. I. 2. 3- 4- 3^ —7^ — b —2b 7«x— y ^ax—2y 6a X — 2y Sa — 10b i6.vjy = ^y — 2,y ax—^ y 26a X — loy 5- As to the third case, where the quantities are unlike, it is plain, that such quantities cannot be united Into one, or otherwise added than by means of their signs. Thus, for example, If a be supposed to represent a crown, and b a shilling ; then the sum of a and b can be neither 2a nor 2b, that Is, neither 2 crowns nor 2 shillings, but only i crown plus 1 shilling, that Is, a-{'L In this rule, the word addition is not very properly used, being much too scanty to express the operation here performed. The business of this operation is to incorporate into one mass, or alge- braic expression, different algebraic quantities, as far as an actual incorporation or union is possible ; and to retain the algebraic marks for doing it in cases, where an union is not possible. When we have several quantities, some affirmative and others neg- ative, and the relation of these quantities can be discovered, in whole or in part ; such incorporation of tv/o or more quantities into one is plainly effected by the foregoing rules. It may seem a paradox, that what is called addition in algebra should sometimes mean addition, and sometimes subtraction. But the paradox wholly arises from the scantiness of the name, given to the algebraic process, or from employing an old term In a new and more enlarged sense. Instead of addition, call it incorpora-^ Hon, union, or striking a balance, and the parado^f vanishes. 272 5- 7^ — 6b /\a — 3^ la — 8^ a — '' b 3<3r .2b ALGEBRA. 6. 3^i — Pcy tx- — 2xy 7* CASE II. When like quantities have iinlihe signs » RULE. Subtract the less coefficient from the greater, to the re- mainder prefix the sign of the greater, and annex their common letters or quantities. EXAMPLES. V ^' ^- 3-* 4- S'_ To +6a — 7^ ■ +2C — cd -hSs/a'+b^ Add — 2a +6b — 2c +2cd — 8^^^^+^ Sum +4^ b * +2ed 5\/a^+b' * In example 3, the coefficients of the two quantities, viz. 4-2C and — 2c, are equal to each other, therefore tliey destroy one another, and so their sum makes o, or #, which is frequently used, in algebra, to signify a vacant place. APDITION. » 273 6. To +2* 1- —6b 8. 9- -yd 10. Add — 6a +7* + b — 4c -{.cd -3^/a^+i* f3f^ —2cd Sum —4^ +5v/«'+i* Note. When many like quantities are to he added together ^ ^whereof some are affirmative and others negative ; reduce them first to two terms, by adding all the affirmative quan- tities together, and all the negative ones j and then add the two terms according to the rule. Thus, 11. Add 4^ *-|-7ri*— 3^*4-12^'' — Srt'+^*.— 5^* to- gether. First, 4o*-j-7a**{-i2fl*-|-a*zr24^?% the sum of the af- firmative quantities. And — 3a* — 8a* — 50*= — i6a^y the sum of the nega- tive. Then 24^*^1 6a* =8/3% the sum of the whole. 12. Add 5a»*— "4aAf*-|-i0fl;c* — 8fl;v*-— j5a;r* together. First, ^ax *-|- 1 oax * = i ^ax^y And — 4fl/)f ^ — ^ax"- — ^ax"" = — l ^ax^ j Therefore the sum of these quantities is -{-15^;^*— -i8a ^3- 14- , — 6^ ax — 2y '\-2ax^ + 2\/aX + y + ^«* -— 6^ ax ""^Jy — 3^^* + iov/'«;v +Sy +3^^* Li case 2-74 ALGfefiRA. CASE 111. When the quantities are unlike* *RUL£. Setthem dovm- in a line, with their signs arid coef- ficients prefixed. EXAMPLE. To 3^ ''^Sxy '\-2^a^—h'- Add zb -|-5 y — 9 Sum 3fl-}-2^ — ^xy-\-^y'\' 2^/^* — b' — -9 OTHER EXAMPLES IN ADDITION. I. 3«» -pP -j-pP + XV V +7%/"* -Sd -2^ 7«' — 12 b^ + 10 — 6 Suit I 15^^- -z^'+Sx/"^ +6^*— S'^+'O'? -2/+4* 2. * Here the first column is composed of like quantities, which are added together by case i. The terms — 9^' and -f"9^' de- stroy one another ; and the sum of — 12^' and 4-5^' is — 7^', by case 2, Tlie sum of ■^'^^ah and — ^^ ab is +3-v/-f'lo* ^ — 10+^* — X — 8a; ^jy — S+v/^j' lo — a — x^ — y SUBTRACTION. RULE/ Change each -j- into — , and each -— into 4** ^^ ^^^ subtrahend, or suppose them to be thus changed j ther^ proceed as in addition, and the sum will be the true re- liiainder. EXA?.r?LES. * This rule is founded on the consideration, that addition and subtraction are o^osite to each other in dieir nature and operation, as are the signs -f- and — , by which they are expressed and rep- resented. And since to unite a negative with a positive quantity of the same kind has the effect of diminishing it, or subducting an equal positive quantity from it ;. therefore to subtract a positive, which is the opposite of uniting or adding, is to add the equal negative quantity. Tn like manner, ta subtract a negative quan- tity Is the same in effect, as to add or unite an equal positive quantity. So that, by changing the sign of a quantity from -f- to — , or from — to -{-, its nature is changed fronv a subductiv+i 2y/^}'-{-2-J-^7 6a;' — io-l~4^ — ;;* r^i-^ ^fy^-xyytiri-^-jT^^-^/^-^ M)c. 19. 20. 21. 3^>— 20 3A:' — 8X^+^ iV';);*+2«^.vj+io MULTIPLICATION. In multiplication of algebraic quantities there is one general rule for the signs ; namely, when the signs of the factors are both affirmative or both negative, the product is affirmative j but if one of the factors be affirmative and the other negative, then the product is negative.* CASE * That like signs make +, and unlike signs — , in the prod- uct, may be shewn thus : i. When l^S ALGEBRA. CASE I. When both the Jactors ar& jimple quantities. ^ITLE. Multiply the coefficients of the two terms together, to the product annex all the letters of the terms, and prefix the proper sign. EXAMPLES, I. 2 3. 4. 5. Multiply a — 33 ^ab — ^cd — a by •« br ••, — '2c 3 '• — 4A? h Product ab -\-6bc izab -{*20cdx — c 6. Multiply 1. IV/jen -{-« is to he multipled by -{-b ; it implies, that -f-^r is to be taken as many times, as there are units in b ; and since the sum of any number of affirmative terms is affirmative, it follows, that 4-^X 4-^ makes +ab. 2. W/jen ttvo quantities are to he multiplied together ; the result will be exactly the same, In whatever order they are placed ; for a times b is the same as b times a ; and therefore, when — a is to be multiplied by -f 3, or '\-h by — a^ it is the same thing as tak- ing — a as many times as there are units in -f^ 5 ^"d since the sum of any number of negative terms is negative, it follows, that — a% -{-b, or -f ^x — b, makes or produces — ab, 3. IVben — a is to be multiplied hy — b ; here — a is to be sub- tracted as often as there are units in b ; but subtracting negatives is the same as adding affirmatives, by the demonstration of the rule for subtraction 5 consequently the quotient is b times a, or -i-ab. Otherwise. Multiply- by 6. JMUtTIPLICATlON. ' — I2ah 8. Scd —4^ Product —6bc *-^2Qcdx Multiply by 10. II. jxyz ^^6ax Product ^^i^yt 279 XAf' Note i . To tnultiply any power by another of the same root / add the exponent of the multiplier to that of the multipli- cand, and the sum will be the exponent of their product. Thus the product of a^ multiplied into a^ is «^*^, or a^. That of at" into x is a-"*'. That of .r" into ;v* is :v"+*. That of x"^ into a?" is ;v'''+". And that of cf^"-' into /"'' is cf^'''^^-\ or o^*"'^^ Again, the product of ^+^1 multiplied into ^-f-A? is And that of x-^-yX into a^+jI is ^+)'l • This Otherwise. Since a — a=:o, therefore a — a X — h is also =0, beause o multipKed by any quantity b still o ; and since the first term of the product, or « X — h, =iab, by the second case ; there- fore the last term of the product, or — «X — b, must be -^-ab, to- make the sura =:c, or • — ^3 4-^=0 ; that is, — aX— ^=+^^« 28o ALGEBRA. This rule is equally applicable, when the exponents of any roots of the same quantity are fractional. Thus, the product of ^z* multiplied into a' is «*X«*±i In like manner, a;"^X^^Xa;^ z: a;^"*'^'*'^ zz x^ :zzx' Hence it appears, that, if a surd square root be multi- pheH ifito*itself, the prdcHlct will be rational ; and if a surd cube root be multiplied into itself, and that product into the same root, the product is rational. And, in gen- eral, when the sum of the numerators of the exponents is divisible by the common denominatorj \^ithOut a re- mainder, the product will be rational. X. ' 5_i_? 111. 8 Thus, a^Xa^ZZa^'^'^ZZa ^ ZZa^ZZ:a\ 8 Here the quantity a^ is reduced to a^, by actually di- viding 8, the numerator of the exponent, by its denomi- nator 4 ; and the sum of the exponents, considered mere- ly as vulgar fractions, is ■^'{'^zz^zz:2. When the sum of the numerators and the denominator of the exponents adjuit of a common divisor greater than unity, then the exponent of the product may always be reduced, like a vulgar fraction, to lower terms, retaining still the same value. Thus, x^Xx'^zzx^-ixK Compound surds of the same quantity are multiplied in the same manner as simple ones. J. -X a , I Thus, a + x\* X a + xl"" = a+xl^ = c+«| = a+x ; MULTIPLICATION. 28 1 So likewise \/ a'^a X V^«+^ =>/a+^ =:«-i"2 *• And i^a^x Xx/*^"!"^ =\/j+a? =i:a-|-A; *. And ^a'\'X X\/^+^='2+^- These examples shew the grounds, on which the prod- ucts of surds become rational. Note 2. Different quantities under the sarnie radical siign are multiplied together like rational quantities, only the product, if it do not become rational, must stand un- der the same radical sign. Thus, >/7X\/3=v/7X3=v/2i. ^"icx y.s/ly z=z^i^cx^ And y/4iX\/2^^=>/8^^' =8c'^*!"- It may not be improper to observe, that unequal surds have sometimes a rational product. As ^^Xv/2 =:y/6^=r8. ^I2ab y,y/ -^ab Z=:y/'},6a'b''z=.6aL V^'y y^^^y'^s/^'y'= Multiply by 6. — 2.V* 7- -8.V* — yx Product CASE MULTIPLICATION. ZSj CASE III. U^hen both the factors arc coinpouud quantities. RULE. Multiply each term of the multiplicand by each term of the multiplier ; then add all the products together, and the sum will be the product required. "^ EXAMPLES^ I. 2. Multiply by ^+3 a\-b .+b a'+ab a^'J^2ab+b'*- a'+ab —ah—h' Product a" * — ^* 3.1 * In the first example, we multiply a-^-h, the multiplicand, into ^7, the first term of the multiplier, and the product is a'^ J^ah ; then we multiply the multiplicand into ^, the second term of the multiplier, and the product is ah^^b'^. The sum of tlicse two products is dE* + 2«3-f-3*, as above, and is the square of a-\-h. In the first example, the like terms of the product, viz. ah and aby together make zab j but in the second, example, the terms -\-ab and -^ah^ having contrary signs, destroy each Other, and the product is <«^' — b"^ , the difFerence of the squares of ^ and 3. Hence it appears,, that the sura and difference o^ two quantities, multiplied together, produce the diiference oi their squares. And by the next following example you mav ob. ser^, that the square of the difference of two quantities, as ci and^j is equal to a* — 2a^-|-i^, the sum of their sq\?arer. mimiT* twice their product. 284 AL— 3 I4^>' — 83; Froduct I 4A:_); — 2 1 AT — 8j-j- ^ ^ 8. Multiply DIVISION. 2S5 8. Multiply x^-\'ioxy'\-y — 6x^y — 60^';''— 42^;; -J-4X-* -\-/\oxy +23 Product X ^ +4^' ^y—'6oxly ^^\ix^ — 2a?/)+2 8. 9. Multiply x^ '{'X^ y-^-Xy"- -{-y^ by a: — y. Ans. X* — y* . 1 o. Multiply x^+xy-^-y"" by a: ' — .v;- +;' * . 11. Multiply 3;^' — 2;c;+5 by p:^+2.vj?— 3. Ans. 3^; "* -j-.v ^ j — 4A; * — /\x *;' ^ + 1 6;^; — 1 5 12. Multiply 2«* — 3^a:-1"4^* ^Y S^^ — ^^^ — 2^''' Ans. lofl^ — 2']a^X'{-24a*x^ — iS^jat^ — Zx*. DIVISION. Division in Algebra, as well as in Arithmetic, is the converse of multiplication, and is performed by beginning at tiie left hand, and dividing all the parts of the dividend by the divisor, when it can be done ; or by setting them down like a vulgar fraction, the dividend over the divisor, and then reducing the fraction to its lowest terms. In division the rule for the signs is the same as in multi- plication, viz. if the signs of the divisor and dividend be alike. 286 ALGEBRA. alike, that is, both -{- or both — , then the sign of the quotient must be -J- ; but if they be unlike, the sign of the quotient must be — .* CASE I. When the divisor and dividend are both simple quaniities. RULE, 1. Place the dividend above a line, and the divisor un- der it, in the form of a vulgar fraction. 2. Expunge those letters, that are common to the divi- dend and divisor, and divide the coefficients of all the terms by any number, that will divide them without a re- mainder, and the result will be the quotient required. EXAMPLES. I. 2. 3- Divide I 8 a;* — I2fl^ abc by 9a; =Z2X 3 — /1<7^ bed Quotient ^ 9^ I2ab 3 ~" abc a bcd—d 4. Divid( * Because the divisor, multiplied by the quotient, must produce tlie dividend. Therefore, 1. When loth the terms are -f- ; the quotient must be -{-, be- cause -|- in the divisor x -f- in the quotient produces + in the dividend. 2. When the terms are both — ; the quotient is also -f-, be- cause — in the divisor X + in the quotient produces — in the dividend. 3. When DIYISIOSt. ^^7 4. 5- 6. Divide ab "^^^5 by 2b 3/? .5 a Jabcx^ sax' Quotient— 7 zi-=:i-^.—— —— ^^ a/- 2 3a 'jabcx^ Jbcx'' 5 7. Divide i6x* by 8/if. Ans. 2x 8. Divide 12a* x* by 3^';^. Ans. 4«. 9. Divide— 15^^* by 3^)?. Ans. 5j^. •10. Divide —I 8^a:*j? by — *3axz Ans. ■^-^. 4Z It may not be amiss to observe, that when any quanti- ty is divided by itself, the quotient will be unity, or i j because any thing contains itself once : thus x—x gives i, and ^ 2ab divided by y/ 2ab\ gives i. Note i. 51? divide a?iy power by another of the same root ; subtract the exponent of the divisor from that of the dividend, and the remainder will be the exponent of the quotient. Thus, 3, When one term Is + and the other — ; the quotient must be — , becaase + in the divisor X — in the quotient produces — ' in the dividend ; or — in the divisor X + iii the quotient gives — in the dividend. So that the rule is general ; like signs give +, and unlike signs give — » in the quotient. t88 ALGEBRA, Thus, the quotient of ^^ divided by ^^ is a^""^, or a^c That of ;v"by A;is.V'-\ That of ^" by at" is x'''^ That of ;v"^-^" by a;" is x'\ And that of a;" by x*' is a;""''. But it is to be observed, that when the expoilent of the divisor is greater than that of the dividend, the quotient ■will have a negative exponent. Thus, the quotient of x^ divided by x'^ is x^""^, or x"^. And that of ax^ by ^ ^ is ax"^. And these quotients, viz. x"^ and ax'~^y are respective- ly equal to — - and — j ; for x^ being actually divided by X X ' X^ 1 ax^ a «^ gives -y =~7 *, and aX divided by x^ gives •— ^zir-^ X X XX' as above. aX^ a In like manner, dx" divided by f;v *" gives — — zz - — . And the quotient of a* -^x* \ divided by «' -}"^' * I ^^ Moreover, aV divided by <2* gives a"^ ir^^=j. -rl ^ -'-- a-^xV divided by ^-j-AcI^ gives ^-[-^vl^ = «+Arp = ..+AiT And ah^x * I " divided by ^^ -j- ;r * T gives ah-^-x * | " . 3CH0LIUM. PIVISlONi ^89 SCHOLItTM. When fractional expo?ients of the powers of the scjne rovt have hot the same denominator, they may be brought to a common denominator, like vulgar fractions, and then their numerators may be added, or subtracted, as before. Thus> the quotient of /7<:-|-^P divided by y/^-f-;?!'*^ is ac +^h ^=:ac'{'X\ 4 ±=:z ac*^x\ . Note 2. Surd quantities under the same radical sign are divided, one by the other, like rational quantities, only the quotient, if it do not become rational, must stand un- der the same radical sign. Thus, the quotient of y/^i divided by ^3 is y/y. That of ^ab by ^a is ^L 3 3 — 3 That of ^i6c by ^2c is -y/8, That of v^^f by ^^H'' is i. Andthat of i2a'';^y|" by s^V;'^!'' is 4;^;;')" CASE II. When the divisor is a simple quantity and the dividend a com- pound quantity. RULE. Divide every term of the dividend by the divisor, as in the first case. EXAMPLES- N N 2g6 ^ ALGEBRA. EXAMPLES. t. 30i5'2^+3MS^4~^ quotient, tient. 3. Divide 3A;*-— i5-j-6»v-^3^ by 3^^ Ans. ^ l-2-f: 4. Divide 3a^r-{- 1 2ahx — 9^ * ^ ^7 3^^* Ans. f-j-'4/v — 3^» 5. Divide ioa*x-^i$x*'^^x hf ^x. Ans. 2a!*-^3Ar-^i. CASE III* IFhen the divisor and dividend are both compaund quantities, * . RULE. 1. Range the terms according to the powers of some letter in both of them, placing the highest power of it first, and the rest in order. 2. Divide the -first term of the dividend by the first term of the divisor, and place the resuft in the quotient. 3. Multiply the whole divisor by the quotient term, and subtract the product from the dividend. 4. To the remainder bring down as- many terms of the dividend as are requisite for the next operation ; call the sum a dividual^ and divide as before j and so on, as in Arithmetic. EXAMPLES. DIVISION* 291 EXAMTLES. I. Let it be required to divide <7 ^ — 3a * Af— 3«;v * -j-.v ^ — 4a^X-^2^^^ ^^^^ dividual. ——4a * X- — ^ax * -j- ax^ -{-N^ second dividual. -j- ax' -{-x^ 2. Divide * The process may be explained thus ; First, a^ divided by a gives a* for the first term of the quo^ tient, by which v\'e multiply the whole divisor, viz. a-\^x, and the product is «- -^a^x, which, being taken from the two first terms of the dividend, leaves — 4^*^ ; to this remainder we bring down — ^ax^, the next term of the dividend, and the sum is — ^^x — $ax^, the first dividual ; now dividing — 4«^.v, the first term of this dividual, by a, the first term of the divisor, there comes out — ^ax, a negative quantity, which we also put in the quotient ; and the whole divisor being multiplied by it, the prod- uct is —-^.a'^x — 4flx^, which being taken from the first dividual, the remainder is -^-ax^ ; to which we bring down x^ , the last term of the dividend, and the sum is +^7a;^ -}-*.% the second di- vidual ; and -^-ax^, the first term of the second dividual, divid- ed by a, the first term of the divisor, gives x^ for the last term of the quotient ; by which we multiply the whole divisor, and the product is -}-«>;* -|-x', which being taken from the second di- vidual leaves nothing ; and the quotient required is n^ — ^^ax^x"^. 2^2 ALGEBRA, H 2. Divide a '^'\'a*X'-^a^x^ -r^.'ja^x^ — 6;v* by o*— *;a?', a*^x')a''\'a'x—aK->c'—'ja'x^—6?('{a^'^a*x—6x^ a^ -fl3^ i ' * • '-'6a\x^—6x^ —6a'x^—6x' * * 3. Divide c-—^ hy ^a — -^/L -d. Divide * Here a, the f^rst term of the dividend, being divided by jy^a, the first term of the divisor, gives ^/« for the first term of the quotient. Torazza^f and ^a =z a'^x and the difference of the exponents is i— i, or t j therefore a^ divided by a"^ gives a^ '^=:a^ zzy/a, as above. Or it muy be considered thus : ask what quantity being muItipHed by \^a will give a, and the answer is \/a ; then the divisor being multiplied by /^a | the fToduct is a — ^i/ab 5 but there being no term in the dividend, that DIVISION. 293 4. Divide «^ — 3«'-^4-4«<:*— #^*"by a'— 2^1:+^ . .» /.I ac — c ^ * ^2^^+c * )« ' -r 3^ ' ^+4^^ ' — 2 or — ---— ' . ^ adX-^-anx d-{-n , . a^hx-i^cX^-^-aX^ . Here the quotient —-,- — = • is reduced to , adx-YanX ah'\-cx-\-x'' ,,..,. ■ ^ ;• > by dividing e^^ery term of its numerator a^d denominator by ax, 6. And a-^-ab'^d^ divided by « ' -— ^^-J-^ * ^ * gives Here the quotient cannot be reduced to lower terms, be- cause the factor a is not to be found in the term d^ . But it is to^be observed, tlwt though a fraction cannot be reduced to lower terms by a simple divisor, yet it may sometimes be so reduced by a compound one j as Vviill ap- pear in the reduction of fractions. ^ 7. Divide c^-j-^^ by ^z+a:. Ans. «*— ^ft:-}-;c*. 8. Divide a? — 3«*}'"[-3^j'* — y"^^ by a — y* Ans. a^ — 2^>'-f7*' 9. Divide 6;!^^ — -96 by 3A?— 6. Ans. 2a;^+4a?'4"^''^4"^^' 10. Divide a^ — -^a^X-^lOa^^x'^ — 10^ 'a; ^+5'^^'^'^ — ^^ ^7 c*— 2^^:-}-;;*. Ans. a}"^l^a^ x-\-y*^^ "^^^^ FRACTIONS. Algebraic Fractions. have the same names and rules of operation, as fractions in Arithmetic. PROBLEH Iractio^s. $^J PROBLEM I. ?i find the greatest common measure of the terms of a fractiotti RULE. 1* Range the j^uantities according to die ^diaieasions of some letter, as is shewn in division. 2. Divide the greater term by the less, and the last di- vIs|oi% by the. last remainder, and so on till nothing re- main ; then the divisor last used vjrill be the common measure required. ^ Note. All the letters or figures, which are common to each term of any divisor, must be rejected before such di- visor is used in the operation. EXAlklPtES* I. ^o find the greatest common measure of — r-; — r". or C'\-x)ca * -\-a * !x{a * Therefore the greatest common measure is r-j-^r. a. To find the greatest common measure of — —2bx''—2b'-x)x * + 2hX'\-b'' or ^+^. )x''\-2bx+b\x'\-h bx+i^' Therefore X'\'b is the greatest common measure. 3. To ^p^ ALGEBRA. 3. To find the greatest common measure of . Ans. x-^t. 4. To find the greatest comtiion measure of * Ans. x'J^br I>ROBLEM Hi ^0 reduce a fraction to its lonvest terms* RULE. 1. Find the greatest common measure, as in the last problem. 2. Divide both the terms of the fraction by the com- mon measure thus found, and it will be reduced to its lowest terms. EXAMPLES. Cx -I- ii^ \. Reduce - — . .- to its lowest terms. 4 ca^'^a'^x cX'{'X^)ca^+a*X or ^4"^ )ca^ -^-a^ SC{a*' Therefore r-j-itf is the greatest common measure 5 2. Having and c4-ii) -~~--r-( -r is the fraction required I nP-ACTlONS. 297 5j5 l-^X . . . , ^. Having -5^ T-- -T" given, it is required to reduce :t to its least terms. or fc-{-b )x''+2hx-\'b*{^X'^h x^ -^ bx bx-^-b^ p. """^ Therefore X'\-b is the greatest common measure, i^nd ;4-.5) '^-^- -— (^- ;- is the fraction required. 2. Reduce -vT-TT-T to its lowest terms. Ans. — ' — . yf^ x^-> — y* I 4. Reduce — — ^ to its lowest terms. Ans. . ^A ^4. c. Reduce -j r r to its lowest terms. Ans. — ^ — , IPROBLEM III. ^'o reduce a mixed quantity to an Improper fraction. RULE. Multiply the integer by the denominator of the fraction, and to the product add the numerator ; then the dcno ni* nator O o -ZpS ALGEBRi*.. nator being placed under this sum will give the inipropet fraction required. EXAMPLES. 1. Reduce ^-j ^o an improper fraction. 5 3x7+5 21+5 26 . o-i-zz^ — ^-i-i=z= — -^-^=— the answer, 7 7 7 7 2. Reduce a—— to an improper j^^?iction.* c b aXc — b ac — b - fli^— ■— =z=: zn the answer-. c c c 3. Reduce ;v-j — to an improper fraction, '* ath- — =:• rr — ! — > the answer. 4. Reduce 8~ to an improper fraction. Ans. y*» 5. Reduce i to an improper fraction. Ans. ^ T> 1 ^X + X* ^ . r • * ^;^f> 0. Reduce a: to an improper fraction, j^ttf ITn^ 7,(1 • p ^U- 7. Reduce io-| to an improper fraction. — -~— — ~ PROBLEM IV. 21? reduce an improper fraction to a nuhole or mixed quantity'. RULE. Divide the numerator by the denominator for the inte- gral part 5 and place the remainder, if any, over the de- nominator FRACTIONS. 299 UDminator for the fractional part ; the two joined together will be the mixed quantity required. EXAMPLES. |. To reduce y to a mixed quantity. y z=z 1 7-^5 r:3y the answer required. ^. Reduce f^"^^- to a whole or mixed quantity. --^ — rzzax-f-a^ '^xzzn-f'--' answer. 3. |leduce — 7 — to a whole or mixed quantity. ai> — a , zz ab-r^a * -f-^=^ 7- answer. V ^ 4. Reduce --^-i-^! — to a whole or mixed quantity. ay-{-2y* ^y-Y'^f •^'"\-y—y^{:^^ answer. 5. Let -^ — be reduced to a whole or mixed quan- tify. Ans. 3^- — — (J. Let — ^ — be reduced to a M'hole or mixed quantity Ans. ^-f-.v-f- •;. Let ^ be reduced to a whole or mixed quantity Ans. ^''■\* the numerators. cXbXc =12 c' I? J hyicy^d zz=z hcd the common denominator. ah , (7 acd h' d 1 ly, the fractioiiS n-^uircd. Thereiore -7-,— and --~: ----, ---- and -7-7 respectivc- d bed bed bed 3. Reduce - ^KACTIONS. 3©! 3. Ecducc— and -r- to equivalent fractions, having n fiommon denominator, Ans. ' and — . ac ac 4. Reduce -2. and --^ to fractions, having a comntq* be denominator. Ans. 7— and - — r — . <^<; ^<; e. Reduce ~, — and dy to fractions, having a common ^ 2a y denominator. . ^ ^^^^ Ans. ■ -2 — , --. — and -7 — -. 0^<: oac oac 6. Reduce ~, — and a4 — , to fractions, having ^ 4 3^ ^ommon deno^linator. Ans. -^^ , and X_Z_, 12^ 12-^ 12J! PROBLEM VI. To ^^^ fractional qutintities together. RULE. * X, Reduce the fractions to a common denominator.* 2. Add * In the addition of mixed quantities, it is best to bring the fractional parts only to a common denominator, and to affix their sum to the sum of the integers, interposing the proper sign. ^QZ ALGEBRA. 2. Add all the numerators together, and under the sum write the common denominator, and it will give the sun^ pf the fractions required. EXAMPLES. X. Having — and ~ given, to find their sum." Here '^•^ ^ > the numerator^. And 2X3^^*6 the common denominator, ?X 2X Cx Therefore --- +-— zz=^ is the sum required. 006 • 2. Having -j-y -j- and -- given, to find their 6Un>y aXdXf=: adf ^ Pere cX^by^f z=z chf > the numerator^. eXhXd := ebd J And bXdXf = ^^ the common denominator. Therefore -rir+'-^^T+'Tir^^ ij/^ — > ^^^ sum ^df hdf hdj bdf required. 3. Let a —— and ^-|- - — be added together. * ^ ; ^ , > the numerators. And ^X*^ = ^^ the common denominator. Therefore a — ^ — la_l — ^ — ^ — Li_L — 3i-+.+^=.-^;^+.+-^=:.+ 2c23x — yy:^ ^H 7 ^^ sum required. 4. Add Fractions, 303 4. Add t and 1 togetKet. Ans. iil±i!f , 26 5 10^ c. Add -—, -T- and — toerether. Ans. »v4-*— » or -J k. ^ 2' 3 4 ^ ' 12 12 6. Add ^-^^ and ~ together. Ans. — ?^^^^""I-3 . 3 7^ ii 7. Add ATH to ^flr**- — —-• Ans. 4;cH ,3 ^ ^ 4 I I ox — 17 2 PROBLEM VII, ^<7 subtract one fractional quantity from another^ RULfi. 1. Reduce the fractions to a common denominator, as in addition.* 2. Subtract one numerator from the other, and under their difference write the common denominator, and it will give the difference of the fractions required. EXAMPLES. I. To find the difference of — and — . 3 " Here 2X\ v^ ^ > the numerators. ixY, 3 = 6x^ And 3X1 1 = 33 the common denominator. Therefore * The same rule may be observed for mixed quantities in sub- traction as in addition. 304 ALGEBRA.. Therefore — ' zn-^ is the difiference reqiiirccl. 33 33 SS %, To find the difference of and -^ — i-. 3^ . 5c Here Pc—a Xs^ =z ^cx—s^c / the nu-merat« ■ ors. 2a — 4;vX3^ ZZ 6ab — I2hx And 3^X5^ zz i^bc the common denominatori ?.v Til jtcy — Sac Sab—- 1 zhx ^^j^Scx-rSac — 6ab •{• i ih IS be iSbc l^bc ^^ the difference required. 3. Required the difFerchce of sy and-^. . Ans. ■^~. o 8 4. Required the difference of ^ and — . Ans. ~-. 1 '9 ^3 5. Subtract -7 from —7— » Ans. — ^-7-7 '* ^J ^ _ , 2x4-7 r 3^ + ^ 6, Take- — ^ from ^ ^ > Ans. ^4y+^^'-^Q^^--35^ 40^ . 7. Take ;tf— — ^- from 3^+y. . , ex — bxArah Ans. 2«+ ^-_ PR0BLET4 JTRACTIONS, 305 PROBLEM Vlil. To multiply fractional quantities tcgethtr, R U L E.* Multiply the numerators together for a new numerator, and the denominators for a new denominator ; and it will give the product required. EXAMPLES. 1. Required to find tlie product of — and ~. Here ii^ii^r: — rr"^ the product reqUifcd. 6X9 54 27 a A-X I Ox Z. Required the product of — , — and — - — . jlere - — =^^ — "^^-^ — the product requirea. 2x5X21 2io 21 3. Required the product of — and --^ . Here — —— f- = •—■ — • the oroduct rcquued, ex a^c ^'+^^ 4. Required * I. When the numerator of one fraction, and the denomin^t* or of the other, can be divided by some quantity, . which is com- men to both, the quotients may be used instead cf them. 2. When a fraction is to be multiplied by an integer, the prod- uct is found by multiplying the numerator by it ; and if the inte- ger be the same with the deuomioator, the numerator may be tak- en for the product. 3. When a fraction is to be multiplied by any quantity, it fs the same thing, whether the numerator be multiplied by it, or the denominator divided by it. r p loC . ALGtBRA. 4. Required the ' product of ~ and 4^. Ans. -2^ , 5. Rec^uiied the product of t-^-— and — . . ao-4-L\s Ans. — i — -4 ^; Required the prbdiict of ^~t — ^^'^^^ '• , - 7. Required the product of x, — •^— and ' j. a a -Yo Ans. PROBLEM IX. 21? divide one fractional quantity by another^ RULE,* Multiply the denominator of the divisor by the numer- ator of the dividend for a new numerator, and the nu- merator * I. If the fractions to be divided have a common dcnomi*». nator, take the numerator of the dividend for a new numerator, and the numerator of the divisor for the denominator. 2. When a fraction is to be divided by any quantity, it is the same thing, whether the numerator be divided by it, or the de- nominator multiplied by It. 3. When the two numerators, or the two denominators, can be divided by some common quantity, that quantity may be thrown out of e^h, and the quotitnW used instead oi' the factions P^rst- proposed* merator of the divisor by the denominator of the dividend for a nevir denominator. Or, invert the terms of the divisor, and then multiply Ijy it,- exactly as in multiplication. |LXAMPL£% I. Divide ii by ^. 3 9 Herie iLx~ = ^'=T= iv is the quotient required. 3 2.V 6x r^. .J 2a . AC 2, Divide --- by -V. „ 2a d 2ad ad ... ,. ^ . , Here -— X — :=— 7— zz — -- is the quotient required. 3, Divide .- ■■ ■ ■ by — \ — . 2x — 2b ' $x-\-a ^•\-a Kx-X-a Ca;*4.6«.v+^* , . . , 4. Divide by -4-- -T-;; — rX ^:=^ ■ = -r ^ — is inc quotient r9quircd. 5. Divide -ii- by c;^. Ans. ^-. I €» Divide -~- by —. Aiis. —^ . ' ' * "^ 3 4^^ 7. Divide 3^8 a:^gecra. by — X — I ^ 2 Ans. ^ , '"-^* bT^' + ^^ Ans. ^+^* 8. Divide INVOLUTION. Involution is the continual multiplication of a qu iT or ax. And 5if ILfiEBRA. Apd the mth power of ^*-}-;tf*| ^'" is ^'-j-^'l ' \ 0? o^-J"'^'MS namely, the ?;th power of the cube root qf '"' Note. All the odd powers raised from a negatire ro4^^ are negative, and all the even powers are positive,' Thus, the second power of -—(2 is — « X*^~^=^='4~'^ * » ty the rule for the signs in multiplication. The third power of — --/2 is -|"^* X — ^=— ^^% The fourth power is -—a^ X— '^^^4"^'** ^^he fifth power of -— ^ is -^a^ X — ^^ = ~^-^*5 ^^, EXAMPLES FOR PRACTICE. I. Required the cube of -^^x'y^. Ans. ''S^2X^y\ 2. Require^ the bjquadrate of 3*" Ans. 3. Requited the 5th piower of «— r-;v. Ans. a^ — ^a^x-{-ioa^x'' — ^loa^x^ '\'^d^^'-T'9c\ Sir ISAAC NEWTON's Rule Jfor raising a binomial or residual quantity to any power ivhdtever* I. To Jind the terms ivithout the cocjfflcients. The index; ©f the first, or leading quantity, begins with that of the givcA * This rule, expressed in general terms, is as follows ; '2 '23 iNVdtUTldN; §t! glVeri power, ami decreases continually oy i, in every term to the last •, and in the following quantity the indices of the terms are o» I, 2, 3, 4, Sec. 2. To find the unkx or coefficients. The first is always 1, iand the second Is the index of the power ; and in general, if the coefficient of any term be multiplied by the index of the leading quantity^ and the product be divided by the number of terms to that place, it will give the coefficient of the term next following. Note. The whole number of terms will bd onfe itiorcr than the inc^ex of the given power ; and, when both terms of the root are -f-, all the terms of the power will be -f- ; but if the second term be -^, then all the odd terms will be -f > and the even terms — . EXAMPLES. % . Lit a-\-K be involved to the fifth power. The terms without the coefficients will be and the coefficients will be 5X4 foX3 .^0X2 5X1 or I, 5, 10, 10, 5, I ; And therefore the 5th power is a ^4"5^ V-}- 1 o^ ^A' ' -j- 1 o« ' :v ^ + 5(2.v ^-^-x ^. f 2. Lei \ 2 23 b\ Sec. Note. The sum of the coefficieats, in every power, is equal to the number 2, raided to that power. Thus, 14-1=2, for the first power ; 1 -{-2 + 1=4=: 2^, for the square; i+3-f3-|-i=r lzz2 ', for tho cube, or third power 5 and so qn. '^12 AlGEBllA. •z. Let >:• — .7 be involved to the slxtli powei** The terms without the coefficients will be a;^, X^ch X^^a^ x'a'f fic^i"^, xa^y a^ ^ and the coellicients Vv-i]l be . 6X5 1^X4 20x3 15x2 6XJ 2 3 4 5 6 or I, 6, 15, 20, 15, 6, I ; And therefore the 6th power of a,' — -a is 3. Find the 4th power of x — a. Alls. X * -^4.^ "^ ^ •4-6rv "" a ^ *-^4J^^ ^ -^-a ^ . 4. Find the 7th power of Ar+/7. Ans. A'^-j-7.v^^^+ 2 lAJ^rt ^4-3 s-v'^.z ^ -f-3 5^' ' '^''^+ 2 ^ -v * ^r EVOLUTION. Evolution is the reverse of involution, and teaches to find the roots of any given powers. CASE I. To find the rovts of simple quantities, RULE.* Extract the root of the coefficient for the numerical part, and divide the indices of the letters by the index of the power, and it will give the root required. EXAMPLES. * Any even root of an affirraative quantity may be either -f- <»* — : thus, the square root of 4-^* is either 4-^, or —a; for And feVOLUTION. J 13 EXAMPLES. t. The square root of ^x^ zz-^x^zr^X, 2. The cube root of Sx^ =z 2X ^ = 2x. ± A 3. The square root of 3^ ^ a; ^ = « - a; - ^3 = ax ^ y/i'. i £ 4* The cube root of — 125^^^?*^= — S^^x^'zz — 5^A?*. 5. The biquadrate root of i6a^;v^ = 2«'^A^'^=2^iV*. CASE II. ^ofind the square root of a cotnpound quantity. liULE. 1. Range the quantities according to the dimensions of some letter, and set the root qf the first term in the quo- tients 2. Subtract the square of the root, thus found, from the first term, and bring down the tWo next terms to the remainder for a dividend. 3. Divide the dividend by double the root, and set the result in the quotient. 4. Multiply And an odd root of any quantity will have the same sign as the quantity itself : thus, the cube root of -j-^^ is -f-, Ans. a:* — ►^-H:- 5. Required EVOLUTION. 31^, i; Required the square root of a'^tc', CASE III. *^o find the roots of. powers in general . RULE. 1. Find the root of the first term, and place it in the quotient. 2. Subtract the power, and bring down the second term for a dividend. 3. Involve the root, last found, to the next inferior pow- er, and multiply it by the index of the given power for a divisor. 4. Divide the dividend by the divisor, and the quotient will be the next term of the root. 5. Involve the whole root, and subtract and divide as before 5 and so on> till the whole be finished. EXAMPLES. I. Required the square root of ^2''- — 2a^ x-\-'^a^ x^ — 2 a' — .'7A--|-v "' la'^) 2a^X ^4 2a^ X-^-a'" x"" 2a')2a'x' a ^ — 2fl 5 A'4-3^ * X ' 2aX ^ +,v ^^ * 2. Extract 3^6 ALGEBRA.' 2. Extract the cube root of ;v*^-j-6;!tf ^— 40;^ ^ -|-96;tf!-«— 6/|^ fc^-{'6x^ — 40A? 2 +96,v — 64(5? * + 2x — 4 3x^)6, ,5 a:^+6^^+i2;c^+8a;^ 3A;'*) I2A?* iV^-J-6/v^ — 40.v3 +96;v — 64 I z Required the square root of a*rj-2^3!^-|-2^^-|-^*+2^^ ^ . Ans. «3-f~^4~^« 4. Required the ciibe root of a;^ — 6!<;^-\-i^x^ — 2o:v^-{- JiJa;^ — 6x+i, Ans. a;^ — -2a;+i, , 5. Required the biquadrate root of i6a^ — g6a^X-\--2i6 \(i^x^ — '2i6ax'^'\-^ix'^, Ans. 2«— 3^", SURDS. Surds are such quantities as have no exact root, being usually expressed by fractional indices, or by means of the radical sign y/. X Thus, 1^, or Y^2, which denotes the square root of 2. And' 3^, or \/'3^j signifies the cube root of the square of 3 ; where the numerator shews the power, to which the quantity is to be raised, and the denominator its root. PROBLEM SURDS. 117 PROBLEM I. To reduce a rational quantity to. the form of a surd^ ' RULE. Raise the quantity to a power equivalent to that,, denote cd by the index of the surd ; then over this new quantity place the radical sign, and it will be of the form required. EXAMPLES. 1. To reduce 3 to the form of the square root. First 3X3=3 ^==9 j then ^/p is the answer. 2. To reduce 2a;* to the form of the cube root. First, 2A?'X2A;'X2Ar'=2;^.*| zzZx^ -, Then s/^^^i or 8;c^l , is the answer. 3. Reduce 5 to the form of the cube root. Ans. 125LS or -v/125. 4. Reduce ~Xj to the form of the square root. Ans. \/\x^y''\ e. Reduce 2 to the form of the 5th root. , Ans. 3 2 p. PROBLEM II. 21? reduce quantities of different indices to ether equivalent oneSy that shall have a common index* RULE. I. Divide the indices of the quantities by the given in- dex, and the quotients will be the new indices for those quantities. 2. Over 31 8 ALGEBRA. 2. Over the said quantities with their new indices place the given index, and they will make the equivalent quantities required. Note. A comnion index may also be found by reducing the indices of the quantities to a common denominator, and involving each of them to the power, denoted by it^ numerator. EXAMPLES. I X T. Reduce 15 ^ and 9^ to equivalent quantities, having the common index ■-, •^-4-t~tXt=^=t the first index. ^-7r~^=:^XY^i^==^y the second index, >■ i Therefore 15 ''I and 9' » are the quantities required. 2. Reduce a^ and x'^ to the same common index 4-, •f-^|=:f Xf=|- the first index. ■^--yrr-X-f^::^ the second index. — ^ TjT Therefore a^\^ and ^"^i are the quantities required, 3. Reduce 3"" and 2^ to the common index -J. Ans. 2']^ and 4*^, 4. Reduce «* and 3*^ to the common index |-. _^ Ans. ^[^ and Fi^. 1. — 5. Reduce a" and y" to the same radical sign. mn mn Ans. y/aP" and ^'b"". PROBLEM er*. Surds. 1 1^ PROBLEM IdU \ reduce surds to their most simple terms* RULE.* frind the greatest power contained in the given surd, and set its root before the remaining quantities, witli the proper radical sign between them. EXAMPLES. t. To reduce s/^% to its most simple terms* v/48=v^i6X3=:v/i6Xv/3==4Xv/3=4\/3 the answer. 3 2. Required to reduce v^ i68 to its tnost simple termsi 3 v/io8=v^27X4=v/27Xn/4=3Xv/4=3\/4 the answer. 3. Reduce \/i2$ to its most simple terms. 4. Reduce y^^^V to its most simple terms. Ans. •^v^<5i 5. Reduce \/243 to its most simple terrtis. Ans. 3y/9i 6. Reduce v^-|-f to its most simple terms. • Ans. 1^1. 7. Reduce ^<^Zci^x to its most simple terms. Ans. "jasZ-zx, PROBLEM * When the given surd contains no exact pov/er, it is akeady in its most simple terms. 2'^<^ Algebra. PROBLEM IV. 7*0 add surd qtiafitkies together. RULE. i. Reduce suqh quantities, as have unlike indices, td other equivalent ones, having a common index. 2. Reduce the fractions to a common denominator, and the quantities to their most simple terms. 3. Then, if the surd part be the same in all of them, annex it to the sum of the rational parts with the sign of multiplication, and it will give the total sum required. But if the surd part be not the same in all the quanti-* ties, they can only be added by the signs -■[- and •— . feXAMPLES. i. It is required to add \/27 and \/48 together- First, ^zv^^/pXa—sv's i And y'48 = y^i6X3~4'/3 3 Then, 3'/3+4//3=3+4X >/3=:7v'3= sum re- quired. 2. It is required to add ^^500 and 'v/ioS together* 3 3 3 First, \/5ooiZ'/i25X4=^5'/4 ; 3 3 3 And \/io8=y'27X4=^3'/4 » Then, 5>/44-3>/4=5 4-3Xv^4=8v'4=r sum te-* quired. 3. Required th^ sum of y'72 and ,^128. Ans. i4v'2. 4. Required the sum of a/ 2'] and >v/i47' Ans. ioy^3. ll 5. Required the sum of >/j and v'-^i* x ^ ^ Ans.. \i.^^, 6. Required SURDS. 321 3 5 -3 6. Required the sum of y/40 and v/13^. Ans. S\/S* 3 3 3 7. Required the sum of y/^ and v^jV* ^^^' i\/^' PROBLEM V. To subtract, or find the difference of, surd quantities. RULE. Prepare tKe quantities as for addition, and tKe differ- ence of the rational parts, annexed to the common surd, will give the difference of the surds required. But if the quantities have no common surd, they can only be subtracted by means of the sign — . EXAMPLES. 1, Required to End the difference of v^448 and y/xI2. First, v/443=z^64X7 = 8v/7 5 And v^ii2=:-v/i6X7=4\/7 5 Then 8^/7 — 4^/7 — 4 v/T ^he difference required. Z X 2. Required to find the difference of 192^ and 24^. First, 192^ z=64X3l' = 4X3 And 24^= 8X3P = 2X3' 5 I T I Then, 4X3^ — 2X3^=^X3^ ^^ difference re- quired. , I 3. Required the difference of 2^/50 and v^i8. Ans. 7v/2. 4. Required the difference of 3 20 ^ and 40 ^ . X Ans. 2X5"^. R R 5. Required 322 ALGEBRA. \ 5. Required the difference of -y/j- and -\/^. Ans. ^v/15. 3 3 ' 6. Required the difference of y/j and \/-jV* 3 Ans. tVv^J^S. 7: Find the difference of \/ Soa^x and \/20rt^A;^ Ans. /^*-— 2^/;? Xv^S^* PROBLEM VI. jH? multiply surd quantities together, RULE. I.' Reduce the surds to the same index. 2. iNlultiply the rational quantities together, and the surds together. 3. Then the latter product, annexed to the former, will give the whole product required ; which must be reduced to its most simple terms. EXA:jvJLrLE3. I. Required to find the product of 3v/8 and 2^/6. Here, 3X 2X>v/8X^6=:6v^8X6=:6^48=:6Vi6X3 = 6X 4 Xv^3=24>y/3, the product required. 3 3 2. Required to find the product of ^'^^ and ^V'-l-. Here,. iX|v'|x'./f=iXv'^=|X^H=|X|X V 15:=:^ \/ i5=-|-\/i5, 'the product required. 3. Required the product of 5 V' 8 and 3 V 5. Ans. 30^10. 3 3 , 4. Required the product of -^V6 and yV 18. Ans. V4. 5. Required • SURDS. 323 5. Required the product of jy/-|- and ^\/-~» 6. Required the product of ^/iS and 5\/4- 3 Ans. iOy/9. 7. Required the product of a ^ and « ^ . Ans. oH^ or a, PROBLEM VII. To dividd one surd quantity by anothew RULE. 3. Reduce the surds to the same index. 2. Then take the quotient of the rational quantities, and to it annex the quotient o£ the surds, and it will give the whole quotient required \ which mvist be reduced to its most simple terins. EXAMPLES. T. It is required to divide 8^108 by 2^6. 8-f-2Xv'io8-4-6=4^i8=z=4-/9.X2=4X3'/2Z=:i2y^2 the quotient required. 2. It is required to divide 8-^/512 by 4y^?,. Ill 1 8-^4:^:2, and 5i2^-f-2^ z:256~~:4X4^ ; Therefore 2X4X 4^ zzB X 4^ 1:18^/4, is the quo,tient re- quired. 3. Let 6y/ioo be divided by 3y^2. Ans. 10^/2. 3 ^3 3 4. Let 4y/ioco be divided by 2^/4. Ans. 10^/2. 5. Let •fv'-rly be divided by ^y/^. Ans. ^v/3. 6. Let 324 ALCEBRA. 353 6. Let fv/i- be divided by Iv'i-. Ans. |f ^3, 7. Let f-/^, or yo*, be divided by -^a^, Ans. -^a^. PROBLEM yiii. Jo ifivolve^ or raise^ surd quantities to any fowsr^ HULE. Multiply the index of the quantity by the index of the power, to which it is to be raised, and annex the result to the power of the rational parts, and it will give the power rec^uired. EXAMPLES. 1. It is required to find the square of ~a^y First,. ffzz^Xf =1-5 And ^§-1 -a'^ z::=u-^'z=,a'[^ ; Therefore -a^i =- -|- • a""! ^ = ^V'^*, the square xc^ quired. 2. It is required to find the cube of \Vjy --3 ' • iirsr, -yi — _XyXT-^T4T * Therefore i^vj =z\\-^-fV=\i^l-^^^\-, the cube required. 3. Required the square of 3V'3. Ans. pv^p. 4. Required the cube of 2* , or s/z* Ans. 2y 2. 5. Required SURDS. 325 V 5. Required the 4th power of ^V6. Aug. yg-. ^. It is required to find the wtji power of a'\ PROBLEM IX. To extract the roots of surd quantities^ RULE.* , Divide the index of the ^iven quantity by the index of the root to be extracted ; then annex the result t;o the roo^ of the rational part, and it will give the root required. EXAMPLES. 3' I, It is required to find the square root of 9^/3. First, Vpzis ; And3^l^z::r3^-^'=3^; Therefore ^^2 I ^^3 X3^ is the square root required. 2. It * The square root of a binomial or residual surd, A-^-B, ox 4— 'By may be found thus : take V-^^ — B^z=:D ; Then ^jT^-^t±^J^j±i:^^ 2 2 A-^^D ,A^D And V A^B=J-^ J "2 "2 Thus, the square root of 8+2-v/7=i + v^7 ; And the square root of 3 — ^/%z=:.yj 2 — i ; Jut for the cube, or any higher root, no general rule is giveni 326 ALGEBRA. 2. It is required to find the cube root of ■g-V^2, First, \<^=i ; And "i/7F= z^'^^zz 2^ ; Therefore ~^2\'^ z=:^X^^ is the cube root required. 3. Required the square root of lo^ Ans. icv/io.. / 4. Required the cube root of ~a^. Ans. fa.l 5. Required the 4th root bf 3a;''» Ans. s'^X.a?^. INFINITE SERIES. . An infinite series is formed from a fraction, having a compound denominator, or by extracting the root of a surd quantity ; and is such aSj being continued, would rua -on infinitely, in the manner of some decimal' fractions. But by obtaining a few of the first terms, the law of the progression will be manifest, so that the series may be continued without the continuance of the operation, by which the first terms are found. - PROBLEM I. 21? reduce fractional quajitities to infinite series* RULE. Divide the numerator by the denominator ; and the op- eration, continued as far as may be thought necessary, will give tliC series required. EXAMPLES, JNFIKTE SERIES. 327 EXAMPLES. I. Reduce to an infinite series. I X :—x) I (i-f-x 'i-A;--t-^^-t-A:^-1 -,«( I X i' +x -^-x—x' — +x^ +x^ '— «' +**—«'' +x\ &c. '^ answer. Here it i§ easy to see how tHe succeeding terms of the quotient may be obtained without any further division. This law of the series being discovered, the series may be continued to any required extent by the appHcation of it. I ^ . . . 2. Reduce to an infinite series. 1 4-^ -i_ ~ i—x+x'—x^+x"^-^, &c. the answer. Here the exponent of x also increases continually by i from the second term of the quotient j but the signs of the terms are alternately -j- and — . 3. Reduce 32S ALGEBRA. •2. Reduce — -*- to an infinite seriesw ,^,) c (i^-- 1:+ fil_f:^+,&c.*=-^, and cA — , is the answer. * a ex a ex ex^ '7~7 ex" a ex* , tx^ + 7t+-r ex^ tx'^ cx^ .4 + - i. &<:• 4. Reduce * Here we divide c by ^, the first term of the divisor, and the quotient is — , by which we multiply a-^-x, the whole divi- sor, and the product is — + — or c-j , which being subtract- a a a ex cd from the dividend r, there remains ; this remainder, be- ts cx ing divided by 5, the first term of the divisor, gives — for the 2d term INFINITE SERIES. 3^9 4. Reduce — to an infinite series. ~ a — ;v 5. Reduce , I ■ ■ • ' ■ ■ - ■ ... ■ ■- term of the quotient, by which we also maltiply a-{-x, the divi- aix cx^ ex cx^ sor, and the product is — — r 7-, or — ■ , , which, being taken from , leaves H . a a* The rest of the quotient is found in the same manner ; and four terms being obtained, as above, the law of continuation be- comes obvious 5 but a few of the first terms of the series are gen- erally near enough the truth for most purposes. And in order to have a true series, the greatest term of the di- visor, and of the dividend, if it consist of more than one term, must always stand first. Thus in the last example ; if a: be greater than ^, then x must be the first term of the divisor, and the quotient will be zz: *+^ ' € ac , a^c ^^ . « t . f -r , , — r H J 37- 4-,&c. the true series : but \ix be \tyi X X- x^ x^ than or -J ; two terms are deficient by -r^-2 ; three terms exceed it by tV ; four terms are deficient by ^^ > five terms will exceed the truth by -^-V* ^c. So that each succeeding term of the serieiJ brings the quotient continually nearer and ftearer to the truth bjr ttit half of its last preceding difference ; and consequently the series will approximate to the truth nearer than any assigned num- ber or quantity whatever ; and it will converge so raiich the swift- er, as the divisor is greater than the dividend. But the second series perpetually diverges from the truth ; for the first term of the quotient exceeds the truth by i — f, or 4 » two terms thereof are deficient by f ; three terms exceed it by 4 » four terms are deficient by ^-/ ; five terms ejCceed the truth by V, &c. which shew the absurdity of this series. For the sanae ncason, x must be less than unity in the second example ; if at were there equal to unity, th^n the quotient would be alternately I, and nothing, instead of \ ; and it is evident, that x is less than \inity in the first example, otherwise the quotient would not have been afnimative ; for if ,v be greater than unity, then i — x, the divisor, is negative, and unlike signs in division give negative quo- tients. Tiom the whole of which it appears, that tlie greatest term of the divisor must always stand first. INFINITE SERIES. 53? PROBLEM IK f 0 reduce a compound surd to an infinite series, RULE.* Extract the root ps in Arithmetic, and the operation, continued as far as may be thought necessary, will give the series required. EXAMPLES. }, Required the square root o^ a*-\-X^ in an infinite series. ^ ' 2a 'Sa' "^ i6a' , OtL. ^+r) x' '^ V* — x' &c. 2, Required *^ This rule is chiefly of use in extracting tiie square ropt ; the operation being too tedious, when it is applied to the higher powers. f Here the square root of the first term, a"^, is a, the first term of the root, which, being squared and taken from the given surd tf*4-*:^, leaves ** ) tliis remainder, divided by 2r?, twice tlic 33 2 ALGEBRA. 2. Requirednhe square root of a*— ^* in an infinite series. Ans. O' 3. Convert -i/ i-}-! into an infinite series. 4. Let V^— a;* be converted into an infinite series. i J, 1 Z Ans ^_— — ^— i^~-^ &c. 2 8 16 128' 5. Let y/i — *v^ be converted into an infinite series. Ans. i—j — ^— g-, &c, SIMPLE ihe first term of the root, gives — for the second term of the 2fl root, which, added to 2a, gives 2^-| for the first com- X* pound divisor, which being multiplied by — > and the product «* 2a x^ . . x^ J taken from the first remainder w*, there remains 1 4«^ 4a this remainder, divided by 2«, gives — —-- for the third term of die root, which must be added to the double of (z-^ , the tv/o 2a first terms of the root, for the next compound divisor. And by proceeding thus, the scries may be continued as far as is desired. Note. In order to have a true series, the greatest term of the prpposed surd must be always placed first. SIMPLE EQUATIONS. 333 SIMPLE EQUATIONS. An Equation is when two equal quantities, difFerently expressed, are compared together by means of the sign = placed between them. ^ Thus, 1 2 — '5=7 ii an equation, expressing the et^uaKty of the quantities 12— ?-5 and 7. A simple equation is that, which contains only One un«s known quantity, in its simple form, or not raised to any power. Thus, a;— fl-^^zr^ is a simple equation, containing on-e \j the unknown quantity x, "Reduction of equations is the method of finding the valu^ of the unknown quantity. It consists in ordering the equation so, that the unknown quantity may stand alone on one side of the equation without a coefficient, and all the rest, or the known quantities, on the other side. RULE I,* Any quantity may be transposed from one side of the equ^,tion to the other, by changing its sign. Thus, if flf-|-3=7, then will iv::=7 — 33=4. And, if ;vr— 4+6~z8, then will ;tf=:8+4— 6z=6. Also, if ii—^a-^fhzzc-^dy then will ;vZZr— -^-{-^ — h* And, in like manner, if 4^---8z;3:v-|-2o, then will 43? f--3iv=;:2o-j-8, or ;v:;:;28. RULE * These are founded on the general princ^le of performing equal operations on equal quantities, when it is evident, that the results must still be equal ; whether by equal additions, or sub-, tractions, or multiplications, or divisions, or roots, or powers. 334 AlGEBRi. RULE 2. If the unknown term be multiplied by any. quantlty^^ that quantity may be taken away by dividing all the other terms of the equation by it. Thus, if aX=ab — a, then wiU ATzzr^ — I. And if 2 a: -|- 4=1 6, then will a:-1-2Zz8, and x=zz^-rr-2 rr6. In like manner, if ax-^2baz=:2c*i then will x-^\bz:z '- — , and A^= —2^. a a RULE 3. If the unknown term be divided by any quantity, that quantity may be taken away by multiplying all the other terms of the equation by it. Thus, If - =5-l"3> ^^" will A;::i:io-|-5==:i^. And, if - zzh'X'C — dy the^ will XZZab'Ar^'^ — ^d. a ■ ■ In like manner, if •2Zz6-f-4> ihen will 2a;-i— 5=q 3 ^8+12^ and 2^=184-12+6=36, or A:=;y=i8, . RULE 4. The unknown quantity in any equation may be made free from surds by transposing the rest of the terms ac- cording to the rule, and then involving each side to such a power, as is denoted by the index of the said surd. Thus, if \/ X — 2=6, then will Y/Arl=6+2=:r8, and ;f=8'=i:64. And, if v'4«-i" i^= 12, then will 4;f-J~i6=: 144, ami 128 4!tf=i:i44 — i6i:ri28, or ^= — 32. 4 In 15IMPLE EQUATIONS. 33I In like manner, if \/ 2A;+3+4=:8, then will \/ 2x-^^ sr8 — 4=4> And 2;v4-3=4'=^4> ^n^l 2;v=64 — 3 = 61, or x=z^-J r=3oi-, RUtE j^i If that side of the equation, which contains the iin* known quantity, be a complete power, it may be reduced by extracting the root of the said power from both sides of the equation. .• - - Thus, if a;*-}-6x4-9=:25, then will »+ 31=^/25= 5 j or Ar = 5 — 3 = 2. And, if 3;*" — 9=21 + 3, then will 3;^* = 21 + 3+9 = 33, and;f* = y = ii, or;v=\/ii. 2a;' In like manner, if — *-l-ic=S2o, then will 2A;*-4-30 3 S=:6o, and /v*+-i5 = 3o, or a;* = 30 — 15= i5,or Ar=:\/i54 RULE 6*. Any analogy, or proportion, may be convCftdd into an equation, by making the product of the two mean terms equal to that of the two extremes. Thus, if 3.V t 16 :: 5 ! 10, then will 3JvX 10= 16X5, and 30^:= 80, or ^ = -~= 2j. 2X 2CX And, if — : a i: h : d then will =r^^, and 2cx:afi' 3 3 3(7^ 2ahf or xi=— ^. 2C A? 2 4.V 9C In like manner, if 12 — x : — :: 4 : i, then will 12— 2 4.V X=:—=z 2.V, and 2;v+.t =: 1 2, or ;* =i y ~ 4... RULB 33^ /lg£bra. RULE 7. if any quantity be found on. both sides of the equation with the same sign, it may be taken away from them both ; and if every term in an equation be muhiplied or divided by the same quantity, it may be struck out of them all. Thus, if 4X^a^b'-\'a, then will 4X=:l>y and ;^zz— . 4 And, if 3«;>;-|-5j^zr8«r, then will 3A?-j*53— :8r, and 8.: — a JV= ^. 3 In like manner, If ——* -1=: Y —*|-> then will 2.vr:! 16, and ;v=r8. Miscellaneous Examples. I, Given 5A?— i5zza^r+6 5 to find the value of x> First, 55?— ;:2Ai;i=6+i5 Then 3a:zz:2I And ;v=:Yri7.. a. Given 40— 6a?-— i6::z:iao— i4Jf ; to find A'. First, 14A;— 6;v=zi20—- 404-16 Then 8^=^96 And, therefore, A;rzy = i2. 3. Let $ax — 2^irz2dx^c be given 5 to find ^% First, 5/i;v— 2f/;vrrr-}-3^ Or 5^ — 2d X^=<^+3^ And, therefore, ^r: 2^ 4. Let SIMPLE EQUATIONS, 33^ 4. Let 3^* — ioA:zr8A;4-^'* be gvveiv; to find;;. First, 2^ — I03=:8-|-a; An4 then 3 a; — ;)czz:8-f-io Therefore ix^z 1 8, and xz=: V^ —9'' 5. Given 6ax^ — i2^^A;*=3^/v^4"-6^i?tf-* *, to find x* First, dividing the whole by 3^/^'*, \vb shall have. 2X — ^^hzzX-{-2 ^ And then 2x — Xznl-f^h Whence a;iz:2-{-4^' X ' X X 6. Let -I rz: 10 be given ; to find ^* 234 2X 2X • First, ^ + — = 20 3*4 And then 3X' — 2x-\ r:6o » ^ And i2;c — 8?^4"<^^'~240 Therefore ioArz=:240 And a;:=:^V-:24. AT— 3 y- ^+^9 \» , ♦7. Given z=2o— ; to find ^ 2*3 2 2X First, X — 3-] = 40 — X — 19 And then 3A? — 9-}-2Arzz:i2o — 3.V — 57 And therefore 3;f+2;f-^3/scz::^i2o — 57-j"9 That is, 8^=72, orAr=:yi=9. 8. Let y/y^-[-5zir:7 be given ; to find X» First, -^i-A?z=r7 — ^Z=2 And then yAm2^ZZ4 And 2;;= 1 2, or;v=yr=5. p. Let T T , ,^ ALGfEBRA. la 9. Let i(^s/ " * H"^'^ * ~ — "" "^ ^^ g^^'^^"^ » ^0 find >r. First, Si^a' ^^x' -^a^-^-x'^iia^ And then a;^« * -J-a; * r=^2 * — a;'^ And X^ X^' -|-.v' 1=^^— jc'fizrj^— 2a*iv''4..v^ Or fl * ;v * +;^ ' nzrt ^ — 2^ ^A; * -^-a; * "Whence « ' ;c ' + 2a ' ;c * ^a^ Or 3^^;;^=.^* ' ' «^ And consequently a;*=: , And ;v=:^-— - zz^ to find a;. " ^ ' aU Ans. x = bc-^-ac-^ab* SC *C "V 6. Given — 4--^^ ---=t ; to find a;. Ans. x=:^, 2*3 4 * ' ^ 7. Given y/ia^/vrra+v^Af -, to find x. Ans, ^=4. 8. Given SIMPLE EQUATIONS. 33^ 8. Given v/«'+.v' =:3^4"^^i » to find .Y. Ans. Kzz.s/ T"* ^a 9. Given »v-|-az:Y/^/'^-f ''^V^^^'i'^'* ' to find :v:. Ans. ;viz — •— =20+4j, Or ir^'— 115— 20=95^ And y=^-s> Whence xZ^-^ zzd. I. Given \ ""-^ , > ; to find x and v. From the first equation A'z=:r2-—j;, And from the second, KZ:zh'\-^y Therefore a — 'yzzdo-^yy or 2)j=;a — -^^ ^ — -I' And ccnsequentlj^, ^= — — , And ii'zia — v=:^ 3. Given ^ +t=' l^ 3 2 rt-|:-^ > \ to find it: and "^^ From the iirst equation a:zzi4— ~, And from the second, *Yrr:24- 3j' 2V "V Therefore, 1 4 ~z=z:2 4 — -^j .^ 2 And 42 — 2yzz']2' 9y Or 34— 4J 11:144— 9j;; Whence j;;':=ii44 — 84i:::6c, Andy=:^=i2, 2y 4. Given SIMPLE EQIJATI0N6. 34 I 4. Given ij^-j-,vz^34, an4 4v-|-A'Zi:i6 ; to find x and r. Ans. a;-z:8, and jzra. J. Given - +^ =J'-. =^'"1 7 -h j = T-^ -. to ^"'1 ft; and J. Ads. :>':=:-, and jrzzvv Y 6. Given *v-4-;z:;:/, and .v*-- — -j'* :z:zd *, to find »v and y. Ans. A'^ 5 and v^::: . 2s '. IS RULE 2. 1. Consider which of the unknown quantities you would iirst exterminate, and let it$ value be found in that equa- tion, where it is least ixxvolved. 2. Substitute the value thus found for its equal in the Other equation, and there will arise a new equation v/icji Qiily one unknown q[uantity, whpse value may be found as before. ex;a?v1ples, 1. Given \ '■ "r^^~- -^7 I fQ £j^jj ^ ^^ From the first equation A-nz 1 7 — 2r, And this value, substituted for a' in the second, gives 17 ^J^-XS ^,V~2, Or 51 — 6;—;— 2, or 5 1 — -jy^iz y That is, 7j'=5i — 2=^=49 ; Whence y^""-^—^, and a:—!;- — 2j-— 17 — 14-^3. 2. Given < "n'—^H ( ^q r^^^^ ^ ^^^ lx—y~ 2>S From the first equation >:^ 13 — y. And this value, being substituted for pc in the sccaiid, Gives i3«— j--y=:3^ or 13 — 2;'=:3 ; That is,: 2jj;= 13 — 3= 10, Or ;y= V=S> and ;v=:i3—)':^i 3—5=^ 8. 3. G:vca 34- -ALGEBRA. 5. Given j , z __ r h to nnd ;v and y. yhe first analogy turned inXQ an equatioa Is l;x:r:ayy qt x=z y, And this value of x^ substituted in the second. Gives 2' +>''■='■' «■ -^+:»''='^> Or «'/+i'/=r^S or^*= -^, And therefore, y: cb a'+b' y and x: ca '+^ 4. Given - — J-7_)=:?99j and -^^ — |-7.v = 5Lj to find Pf. end j,'. A ns. a; = 7, and j;^!: 1 4. c. Given 12= -^ +8, and —^ 4 r 8 = ^ 2 ' 4 5 3 ly — X 4-273 to find X and y. Ans. k = 60, and y'=:=^^o. 6. Given ^ : (^ : : »v : Vj and at ^ — -j^^zrij to find a; and ^v Ans. da^^ :.v, and ^^^ '—h- RULE 3. Let the given equations be multiplied or divided by such numbers or (|Uiintities, as will make the term, which con- tains one of the unknown quantities, to be the same in both equations j and tlien by adding or subtracting the equations, according as is required, there will arise a new- equation \vith only one unknown quantity, as before. EXAMPLES, SIMPLE EQTTATIONS. 343 EXAMPLES. 1. Given \ 3-^+5;' = 407 ,^ ^^^^ ^ ^^j First, multiply the second equation by 3, And we shall have 3^+6)' = 42 ; Then, from this last equation subtract the first. And it will give 6y — 5)=42 — 40, or;' =2, Atid therefore, x= 14 — 2y=: 14 — 4= 10. 2. Given s ^^"T^-^'"^ ^ f ; to find x and y. ^2^+5jy=i6 3 •" Let the first equation be raultiplied by 2, and the sec- ond by ^, And we shall have loAT — • 6y^ 18 IOA*-j-25^= ^^ » And if the former of these be subtracted from the latter. It will give 31;' = 62, or y=:~^:=2, 9"4"3v And consequently, x:=: ^> by the fipst equation, 5 ' ^ Another Method. Multiply the first equation by ^, and the second by 3, And we shall have 25^ — 15^ = 4^ 6a:-}-i5;; = 48 ; Now,, let these equations be added together, And it will give 3i,v=93, or .v = -|-J-= 3» f6— 2v And consequently, yz=L y by the second equation, 16—6 Or >•=; =: y =r2, as before. MlSCELLANEOUi 3^-f ' ALGEBRA. AfjSCELLANEOUS EX/IMPLES, 1. Given • -|- 8^'=:3i, and — j-ioATir: 19a ; to 3 4 find .V and J. Ans. a'= 19 and ^ = 3. 2x — -J -Sy-f-x- 2. Given [-i4=i3> and \ri6=.i(^ ; ta 2 3 find i: and )•• Ans. a- = 5 andjj;=2. 3. Given - — -: 1 =3, and '■ yz=: 11.; to 63 2 find .V and y. Ans. a^=:6 and ;;rr8. 4. Given ^A;-f-'^7=:^> nnd dx-\'ey—f\ to find a; and j?. ce lif af- — dc Ans.- a;= ~ and v=r -. at' — do ae-'^db PROBLEM II. Td exterminate three unhmivn quantities y or to reduce the three simple equations containing them to one. RULE. 1. Let .V, y and s, be the three unknown quantities to be exterminated. 2. Find the value of K from each of the three given equations. 3. Compare the first value of .r v/ith the second, and an equation vi^ill arise involving only y and %» 4. In like manner, compare the first value of x with the thirds and another equation will arise involving only v and z. 5. Find the values of y and % from these two equations, according to the former rules, and a:, y and 2 will be ex* terminated as required. Note. Any number of unknown quantities may be ex- terminated in nearly tjie same manner, but there are often much SIMPLE EQUATIONS. 345 much shorter methods for performing the operation, which will be best learnt from practice. EXAMPLES. I. Given \ x-\r y + 2^=29 ^ I ^ » ^ k. to find;f,vanda, ;^ y z l'^ 3 4 From the first equation xrzip — -j— -z. From the second x=:62 — 2y — 32. 2y z From the third Xz=^2o . 3 2 Whence 29 — y — 2:^62— 2jy' — 32, 2y z And 29— V— ^zz=2o ; 3 2 But, from the first of these equations, )'Zz62''*— 2p— 2« 3Z And from the second yr227— — ; 2 QZ Therefore 33 — 22:=27 — — , or2r:i2. And ;;rr62— -29 — 2zzr62 — 29 — 24^9, And Arz=29 — •;' — zz:29— 12 — 9==:8. i. Given -+^ + ^ = 62 ~ -f- — -j ^^47 Yi to find A", ^ and z. X , y , z .7+7 + ^=33 U u Fir;Jt, % 34^ ALGEBRA. ■ First, the given equations, cleared of fractions> become I2.V-f- Oj-\- 62 ~z 1 40 3 2cAr-|- 1 5y--{- 1 2s n: 2 S 20 3 c,v-|-24v-f- 202 11: 45 60 Then, if the secorid of these equations be subtracted from double the first, and three times the third from five times the second, v/e shall have 4;c-[- J— 156 io;<-j-3v^^42o And again, if the second of these be subtracted from three times the first, it will give 1 2X — I c.v=465 — 4 2 o, or x zz '^y =z 24 ; Therefore jzrz It; 6 — 4X=z6y 1488— 8r— i2;c And 2= '■ zri2o. 0 3. Given x-\-y^z-=zi,ii x-\-y — 211:25, ^^"^^^ AT-*—)' — r mp *, to find K, y and 2. Ans. xzii:2o, jzzzS and2iizz3» 4. Given x-^-yZZiay ^-}-2Zr:^, and ^y-j-szz^ ; to find x^ y and z. 5. Given < .7.Y-[-f)-|-y2;=:;/ > ; to find AT, )? and z* -^ Cgllec'Tion of ^UESTIONSy producing Sim* FLE E.^UATIOyS, I. To find two such numbers, that their sum shall be 40, and their difference 16. Let A? denote the less of the two numbers requirsd. Then will x-^j6= the greater, And ;)f-j^.v-^ 1 6=1=40 by the question ; That is, 2.vnz40 — 16^2:24, Or X:=:'y zzzizzz less nutober, And .\'-J-i6z:=:i2-^i6zi2oz=z greater number required. 2. What SIMPLE EQIJATI0N3. 347 2. What number is that, whase j part exceeds its -^ part by 16 ? ' , Let X equal number required^ Then will its 4 P^rt be — , and its 4 part — $ ^3 ^ 4 a: a; And therefore zzii6 by the question, 3 4 That is,*Y — '^2-485 or ^x-- — 3A'ZI 192 ; 4 "V/hcnce a\::z: 192 the number required. 3. Divide 10001 between A, B and C, so that A shall Jjave 72I. more than B, and C lool. more than A. Let A:m' 3 bhnre of the given sum. Then will a;-}-72z:;A's sliare, And a;-j- i , iZzC's share, And the sum of all their shares ^-}-Ar-F|-72-|-Ar-j^i72, Or 3^-{-244ZZiooo by the question ; That is, 3A,'= 1 000-— 244=756, Or xzz'-^=2S2\. == B's share, And ;v-J-72=252+72Zr324l. = A's share. And ;v-f- 1 7211:25 2+ 1 72=424). :;==: C's share. 25 2I. 324I. 424I I cool, the proof. 4. A prize of loool. is to be divided between two per- sons, whose shares therein are in the proportion of 7 to 9 ; required the share of each. Let *r equal first person^s share, Then will 1000 — <«^equal second person's share, And X I looQ — .V :: 7 : 9, by the question. That 348 ALGEBRl. That is, p^^zriooo-— ;vX7=:7ooo— ^7« Or 16.VZZ7000, 7000 Whence xzp — 7-~437l. los. —first share. And 1000 — ;vzi: 1 000— 43 7I. los. z=z 562!. los. 26. share, 5. The paving of a square at 2s. a yard cost as much as the inclosing of it at 5s. a yard ; required the side of the square. Let X equal side of the square sought. Then 4^v= yards of inclcsure, And ^-^ in yards of pavement ; Whence 4A;X5rz:2oA; equal price of inclosing, And .v^XanaA^* equal price of paving. But 2X^ z=2oxhY the question, Therefore, x^zzii ox, and a- = i o = length of the side required. 6. A labourer engaged to serve for 40 days upon tiiese conditions, that for every day he worked he should receive 2od. but for every day he played, or was absent, he was to forfeit 8-d. ; now at the end of the time he had to re- ceive il, IIS. 8d. The question is to find how many days he worked, and how many he was idle. , Let X be the number of days he worked. Then will 40 — x be the number of days he was idle ; Also *Y X20 = 2oa; = sum earned. And 4c — X X 8 = 3 20 — 2x = sum forfeited. Whence 2oa: — 320 — 8A:=:38od. (-zii. lis- 8d.) by the question, that is, 20;; — 32c-j-8/v= 380, Or 28;c= 380-1-320 = 700, And Ar=z VV ~ ^5^^ number of days he worked. And 40-- ;v = 4o — 25=15— number of days he was^ idle. 7. Out SIMPLE Ec^JATIONS. 349 7. Out of a cask cf wine, \vhich had leaked away y, 21 gallons were drawn ; and then, being gauged, it ap- peared to be half full : how much did it hold ? Let it be supposed to have held x gallons, Then it v/ould have Jeaked — gallons, 3 And consequently there had been taken away 2ir-j gal. .V X But 21-f-— =r — , by the (question;. That is, 63+;cr=::^, Or i26-\-2x—i,x ; Hence 3a;-^2a;= 126, Or ;v := 126 :i= number of gallons required. 8. What fraction is that, to the numerator of which if I be added, the value will be y ; but if i be added to the 4enominator, its value will be ^ ? X Let the *fraction be represented by — , y X-\-l __ ^ Then will y ^'' Or 3Af-f-3 —J, And 4^:zzj5-j-i, Hence 4.V — 3^ — 3-3;-}- 1— -y, That is, a: — 3= i. Or A?=r4, and ;=3x4.3z= 124-3=^55 So that -j^ ^zz fraction required. 9. A market woman bought a certain number of eggs,^ at 2 a penny, and as many at three a penny, and sold them™ all again at the rate of 5 for 2d. and, by so doing, lost 4d. What number of eggs hacf she ? Let 35* ALGEBRA. Let Ar=: number of eggs of each sort. X Then will — == price of the first sort, X And — rr price of the second sort. But 5 : % x; 2x (the whole number of eggs) : — ^ Thcrefors — price of both sorts together, at 5 for ad, A? ^ 4X And 1 = 4 by the question ; 235 ^ That IS, AT-i = 8 ; 3 5 Or 3A;-f-2^ = 24, Or I 5a;-}- 10^ — 24^;= 120 ; Whence Ariz: 120= number of eggs of each sort required, 10. A can do a piece of work alone in ten days, and B in thirteen •, if both be set about it together, in what time will it be finished ? Let the time sought be denoted by x, X Then i o days : i .work : : x days : 7- , X And 13 days : i work : : x days : — , ; Hence — :rz part done by A m ;v days 5 10 And — = part done by B in re days. X X Consequently, j rz i 5 ^ 10 13 That SIMPLE ECUJATIONS. 35 i That IS, \-x=±i2f or i3a:*j-io;;= 13d ; And therefore 23^= 130, or ;tf = '-^ = 5-|^ days, the time required. 1 1. If one agent A alone can produce an effect e in the time a, and another agent B alone in the time if 5 in what time will they both together produce the same effect ? Let the time sought be denoted by x. Then a : e : : » : — =1 part of the effect produced by Av And h : e :-. X : -y-zr: part of the effect produced by B, ex ex Whence \--r=zehy the question j X X Or -+-=:, That IS, x-f-yzi^ia ; Or bx'^-ax-^iha ; ba And consequently, y — y~p--= time required. QUESTIONS FOR PRACTICE. 1. What two numbers are those, whose difference is 7, and sum 33 ? Ans. 13 and 20. 2. To divide the number 75 into two such parts, that three times the greater may exceed seven times the less by 15. Ans. 54 and 21. 3. In a mixture of wine and cider, ~ of the whole plus 25 gallons was wine, -|. part minus 5 gallons was cider ; how many gallons were there of each ? Ans. 85 of wine and 35 of cider. 4. A 352 ALGEBRA. 4. A bill of 120I.. was paid in guineas and moidores. and the number of pieces of both sorts used was just 100 j how many, were there of each ? Ans. 50 of eac}i. ij. Two travellers set out at the same time froni London and York, whose distance is 150 milps i oht of then^i goes $ miles a d^tf, and the other 7 ; in v/hat time will they meet ? Ans. ic days. 6. At a certain election 37^ persons voted, and the can- didate chosen had a majority of 91 ; how many voted for each ? Ans. 233 for one and 142 for the other. 7. There is a fish, wliose tail weighs pib. his head wifighs as much as his tail and half his body, and his body weighs as much as his head and his tail ; what is the whole weight of the fish ? Ans. 721b, 8. What number is that, from which if 5 be subtracted, •J of the remainder will be 40 ? Ans. 6^> 9. A post is one fourth in the mud, one third in the wa- ter, and I o feet above the water 5 what is its whole length ? Ans. 24 feeto 10. After paying av/ay cn^ fourth and one fifth of my money, I found 66 guineas left in my bag •, what was in it at first ? Ans. 120 guineas. 11. A's age is double that of B, and E's is triple that of C, and the sum of all their ages is 140 ; what is the age of each ? Ans. A's = 84, B's =2 42 and C's = 14. 12. Two persons, A and B, lay out equal sums of money in trade 5 A gains 126I. and B loses 87I. and A's money is now double that of E ; v/hat did each lay cut .? Ans. 300!. 13. A person bought uch parts, that if the first be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, the sum, diiFerence, product, and .quotient, shall' be all equal to each other. Ans. The parts are 18, 22, 10, and 40, respecllvely. 22. The hour anld minute hands of a clock are exactly- together at 12 o'clock J when ^re they next together ? Ans. I hour, ^yV niin. 23. There is an island 73 miles; in cireumferenee, and 3 footmen all start together to travel the same way about it j A goes 5 miles, a. day, B 8, and C lo ; when will, they all come together again ? Ans. 7 3. days. 24. If A can do a piece of work alone in 10 days, and A and B together in 7 days, in what time c;in B do it alone } Ans. 23ydays. 25. If three agents, A, B and C, can produce the ef- fects £j, b, c^ in the times e^f, g^ respectively ; in what time would they jointly produce the eiFect d ? Ans. ■~^,--, — tnne. 26. If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C in 10 days j how many days will it take each person to per- form the same work alone ? Ans. A i4j-J-days, B i7|-f, and C 23/^. QUADRATIC EQTJATIONS; A SIMPLE, QUAD . ' "at, which involves the square of the unknown quantity only. An QUADRATIC EQjTATIONS. ' 35^ An AFFEC^TED QIJADRATIC EQUATION is that, wllich In- vclves the square. of the unknown quantity, together with the product, that arises from multiplying it hy some known quantity. Thus, n>:':^b is a simple quadratic equation, And ax''-\-lxzzc is an affected quadratic equation. The rule for a simple quadratic equation has been given already. All affected quadratic equations fail under the three folld'^^ing forms. 1. X^-\-aX:=zh 2. r/^ riX-=:h 3. X' — J/.v=: — b. The rule for finding the value of /V, in each of these equations, is as follows : Ji u L E,* I. Transpose all the terms, that involve t\\Q unknown quantity, to one side of the equation, and the known terms to the other side, and let them be ranged according to (heir dimensions. 2. When * The square root of any quantity may be elt-her -f- or — , and therefore all quadratic equations admit of two solutions. Thus, the square root of ■\-n'^ is -j-«> or — n ; for either -f-'^X ^n, or — nx — n is equal to -\-n'^. So in the first form, where ff 4" — ^s found ^-y/ ^4" — >the root maybeeithcr -f v' ^~l~ 2 A 4 or — ^/ b-^- , since either of them being multiplied by itself 4 will produce b-{- — . An' ^^^■'-"-.^■'■•■■""*: expressed by v^•nti^igthe 4 uocertain sign +^ before ^^ i-f- — ; thus a'= + y' b-^ ~~ — f 2 In 35<5 ALCEBRA. 2. When the square of the unknown quantity Kas any coefficient prefixed to it, let all the rest of the terms be divided by that coefficient. 3. Add la the first form, where ;v=: + -/ d-^-t £., the first valu^ 42 z of X, vis. ;c=-|-v^ ^4- — — — Is always affirmative ; fof 4 2 ^ince J-5 IS greater than — , the greatest square must necessa> 4 • 4/ rily have the greatest square root ; \^ ^-j- _. will, there^i 4 2 ' fore, always be greater than -v/ — > or its equal — ; and cons^ quently -t-\/ <^-| — will always be affirmative* 2 2 ■ ' ■ The second value, viz. xzz — t^ ^-j- -i, will always 42 be negative, because it is composed of two negative terras. Therefore, when x * ■xaxzz.h^ we shall haye x= -J- */ h-\- — — , 4 -% 5- for the ajHvmative value of ic^ and a;:zz — jJ b-\- i i for the negative value of x. Jn the second form, where x:=:'^ji/ b-^- ^ -{- JL, the first 4-2 value, viz. xzn-^- \f 3 -J- — -j- J5. is always affirmative, since ■ 4 2 it is composed of two affirmative terms. The second value, viz. ;)fiz: — y^ ^-(- ^ 4" ~ > ^^^^ always be negative ; for sincQ 4 2 ■• ■■,■•■-• <2UADRATIC E<5UATI0NS. 357 3. Add the square of half the coefficient of the second term to both fides of the equation, and that side, which involves the unknown quantity, will then be a complete square. 4. Extract ^-f — is greater than fL, the square root of^4-fL (-y/3-|-f_) 4 4 4 4 will l?e greater than V' — » or its equal — j and consequently ^— ./ ^-f — -f — is always a negative quantity. Theiefore, 4 2 yhen x*-T-f=-|-|/ ^-f- f_ -j forthi? ' jj^rmatlve value of y, and xrr — y' ^4 4 for the nef^a^ ^ive value of .v^ .a' , . ^7 la the third form, where 5:= -y/-- — h -| , both the values 4 ^ pf ^' will be positive, supposing — is greater than h. For the 4 first value, viz. xz=.'\'a/- 3-{- — , is evidently afHrmative, 4 ^ being composed of two affirmative terms. The second value, viz. x= — -/ h A , is also affirmative ; for since — is 4 2 4 greater than 1^ therefore */ — or — Is greater than */" ii and consequently — y' — — ^ 4 'wIII always be an affirmative quantity. Therefore, when «*— ^xn: — h^ we shall have 35? iLLGEBRA. 4. Extract the square root from both sides of the equa- tion, and the value of the unknown quaxitity wiii be dcr termined, as required. Note i. The square root of one side of the equation is always equal to the unknown quantity, with half the co*. efEcient of the second term subjoined to it. Note 2. i^ 11 equations, v/herein there are two terms involving the unknown quantity, and the index of one is just double that of the o^iher, are sbived like quadratics by completing the square. _«_ Thus, x^'-\'^':X^-:zzb, or x'"-\-axrzzl;, are the same as quadratics, and the value of the unknov/n quantity may be determined accordingly. EXAM^LE$, 1 /7 have xzz-^-a/ — — 5 -f- for the affirmative valiic of x, an4 43 • '■* a - — y'^ — If 4" — for the negative value of :;, But in this third form, if l> be greater than — , the solution of 4 the proposed question will be impossible. For, since the square of any quantity (whether that quantity be affirmative or negative) is always affin-matlve, the square root of a negative quantity is im- possible, and cannot be assigned. But if b be greater than—, 4 then — — ii is a aiegative quantity ; and conseauently ^ — — d 4 4 is impossible, or only imaginary, when — is less than b ; and -V therefore in that case x=z — J^a/ ^ is also impossible or 24 imaginary. EXAMPLES. {. Given .v^-|-4.Vi=:i4o ; to find X. First, A,''-{-4A;+4m4o-f-4~i44 by completing the square ; Then ^x ' -|-4A? + 4^\/ 1 44 try extracting the root ; Or X'\-2—i2y And therefore a;— 12 — 2=io. 2. Given a;*— 6»V"|-8=8o ; to find x. First, x^ — 5Ars=8o — 8 = 72 by transposition ; Then .v* — 6Ar+9zr 72-t-9=8i by completing the square 5 And X — 3= x/ 8 1 1=9 by extracting the root ; Therefore ;ci=:9-j-3Z=: 1 2» 3. Given 2.v*-j-8.';— .20=70 •, to find^. First, 2A?*-j-8.vr=:7c-}-20=:90 by transposition^ Then x ^ +n^ ^ 45 ^Y dividing by 2j And .v^-f-4^+4 = 49 by completing the square j Whence .Y-(-2 = \/ ^gzz:']- by extract;ing the root. And consequently x::r 7 — 2=115. .. 4. Given 3>;* — 3A:-}-6=z5y 5 to find X. Here a;* — Ar-j-,2Z?: i|- by dividing by 3, - And x'' — X=i^ — 2 by transposition ; Also A?* — X'\--^=:i^ — 2+^=;j-j by completing the" square. And X — -zz^—'n^hj evolution ; Therefore Arnf -f 4=4-. 5. Given — -J-20y=42-|- ; to find x, r^* a: .Here -~ ,----—4:52. _2p|-=22-|- by tcanspositiP-a.; ■- » 2^ And ;v' — ~z=44i. by dividing by ~ ; 2*V ^' "Whence ;v*— ~-j^|=44| + ^=441 ^Y -Complet- ing the square. And 2^a: Algebra, * And.v— |r:v^44-i=6|. Therefore >:=±6^ + l-zzi, 6, Given j,v'4"^.vr::f4-i — T ==— H T t>v completinff the square ; lution ; And/v-^ = v/ — h-X=\/ • 1 — by evo- Therefore fc+v/^dbfL^i. 7. Given /zA?*— ^A'-4-^=i ; to find ic. Here, ax^'^-^lxz^zd — c by transposition^' And ;v ^ «•= by division ; a a ^ 3*^2 ^-^^ ^' Also »v*- -;\;-| -zz ^ by completing the square ; b Jd^c T^ And ;v zz+v/ 1 by evolution \ Therefore Arrr h v/ j . 2fl"- a * 4a* 8. Given ^*+2flr/v*=r^ ; to find x* Here, /if * +2^;v * -f a * =^+a * by completing the square. And Pc*'^a=zy/b-\-a^ by evolution ; Wlience x^zzz^b-^a — j. And consequently ;v=:y/ : -y/^4"^ — a* 9. Given QJJADRATIC EQTJATIONS. 3^! ^. Given jat"— i^^* — c:=i — d % to find ^. n First, ^A?"— ^x * nrc— ^ by transposition, b !L c—d And A?"— — a;*zz:— — by division .; hi. 5» c— ^ ^» , . , Also a;" a:* -f "i = + 2 by completing the square. And ;c* — ~ =: s/ 1- — r by evolution 5 2-=z2^6 by completing the square j Also A^-|-4 1=^/25 6= 16 by evolution ; And therefore xzniO — 41=12= less number, and 12-J-8 1=20:::= greater. 2. To divide the number 60 into two such parts, that their product may be 864. Let X zz greater part, Then QUADRATIC »E(^ATIONS. 363 Then will 60 — Xzn less, And k)<6o — x:=6ox — x'z:z^6^ by the quesfi^n, That is, X' — 60XZZ — 864; Whence iv' — 6ox-\-90o= — 3644-9^'=^3^ ^Y complet- ing the square ; Also X — 30ZZY/30— 6 by extracting the root ; And therefore .\rz:6+3c— 3'v'5z=: greater part, And 60— ;czz6 — 36^124= less. 3. Given the sum of two numbers =:io(//), and the sum of their squares =58 (^) 5 to find those numbers. Let x=: greater of those numbers, Then will a — xzz less *, And AT^+rt— A'j ZZ2X'']'a'^ — 2nxz=.h by the (question, Or ;v* + ax=.— by division, • * 2 2 ^ h a"- h—a^ Or »Y*— .7.v=r T— zz. — : — by transposition 3 22 2 "^ Whence x ' — ax A zz: A rz: • by com^ 4 2*4 4 pleting the square. Also X = That is, x"" +4x1:^320 *, Whence iv* 4-4a;-|- 4^:320-1- 4=1: 3 24 by completing the square ; And Ar-}"«<^^^\/324=i8 by evolution ; Consequently ;czzi8 — >2=:i6 =: number of oxen re- quired. . 6. What two numbers are those, whose sum, product and difference of . their squares are all equal to each other ? Let X r= greater number, And y zz less j Then ^ADHATIC E(^ATI0N5. 3^5 Then 5^T'^^{ . i by the question, X And izz ~— 3:^— V, Qr ArnrtTf-i from the second equation. Also ^f+i-f-jzir v+i Xj ^ron; the first equatjon. Or 2jv-j-i::^;'^+j', That is, ;;' — ■;.•=:; i. Whence ;?* — y^-^zzi^ by completing the square 5 Also y — y=:v/i|===v/^= — by evolution j Consequently ;t= [- y=z -*— •, A A I — v/5+3 And xzzz^'f-iz:^^ . 7. There are four numbers in arithrnetical progression, whereof the product of the two extremes is 45, and that of the nieans 77; what are the numbers ? Let X = less extreme. And y = common difference ; Then x, x-^-y, X'{'2y, X-j-^y ^'^^^ be the four numbers, And 3 ^^^+3;'=^* +3^^=45 ? by the J I v^*- I 2 I > * \ question. ix+yXx-\;2yZ=x'+2>^y+V =11 J Whence 2y^ Z^'}'] — 451=332, and j)*:ryzzi6 by sub- traction and division. Or y=:\/ 1611=4 by evolution, Therefore x'' '\- i^xyzzx"" ■^iix'iz.^^ by the first equation, Also x^ -\-\2X-\-i^6z=.^'<^'\-2,6zz'6i bycompleting the square. And x-^-bzzzs/ ^iz=z^]yy the extracting of roots. Consequently x-=.g-~^6:=zii And the numbers are 3, 7, 11 and z.y 8. To 5 \66 ALGEBRA. 8. To find three numbers in geometrical progression, whose sum shall be 14, and the sum of their squares 84. Let X, y and z be the numbers sought ; Then xzzny^ by the nature of proportion, A"d [itlP^'ts^] by the question, But x-{-zzzi4 — y by the second equation, Aiid x"" '■^2Xz-{-z^ zi:ig6-^2oy-j'y^ by squaring both sides, Or X^-{-z^"{'2y'' = 196 — 28j-f-;'* by putting 2y' for its equal 2^2 ; That is, X * -\-z ' -{ry * = 1 96— ^28^ by subtraction. Or 196 — 28j'=o4rby equality ; 196 — 84 Hence ^'=x ;:; — —14 by transposition and division. Again, Afzzrv ' i:z 1 6, or ;v— — by the first equation, z And x-\~j-\'Z=: [-4-J-2:= 14 by the 2d equation. Or i6-[-42:'4"2:'' == 14-J or s^ — -icz = — 16., Whence z^ — ioz-{-25=;: 25 — 16=9 by completing the square ; Ar>d 2—5 = ^/9 = 3, or% = 3-f5=r8i Consequently ;v= 14 — y — 2=14 — 4 — 8=2, and the Ti umbers are 2, 4, 8. 9. The sum .(j-) and the product (/>) of any two num- bers being given j to find the sum of the squares, cubes, biquadrates, &c. of those numbers. Let the tviro numbers be denoted by x and y ; Then will j '^"^•^^^]- by the question, ' ^ But x-^-yl z:rx^-\~2xy-\~y^z=s^ by involution^ And a;* -|* 2^^+^' — 2,tj— j* — 2/> by subtraction. That is, x^-{-y'^zzs'^ — zpzusnm of the squares. Again QUADRATIC EQUATI0>;S, 367 Again, :c'"f^^ xx+yzzs^ — 2/»XJ by multiplication. Or X ' -\->:y X .v -j-y -\-y^=:s'^ — 2 j/, Or x^-{'Sj>-\-y^=:s^ — 2sj> by substituting s/> for its equal xyXx-^y ; And therefore a?'-|-j>'=:j' — 3jr/r? sum of the cubes. In like manner, .■v'+j* x^+j— j' — 3Jf/'XJ'by multiplication. Or x'^-\-xyXx''+y'' +y*=s'^—3s^p. Or iv*4-/x J^ — zjf'^-y'^zzs'*- — p^p by substituting / x J^— 2^ for its equal xyX^^-\-y^ ; And consequently, x"^ +y'^—s'^'- ^s'p—pxs^-^zp z^s"^ — 4^* /4-2/*^=: sum of the biquadrate.s, or fourth powers ; and so on, for any power whatever. 10. The sum (a) and the sum of the squares (l^) of four numbers in geometrical progression being given j to find those numbers. Let X and y denote the two means, a; ' v^ Then will — and — be the two extremes, by the nature y PC of proportion. Also, let the sum of the two means zn/, and their product =: p. And then will the sum of the two extremes :=za — j- by the question. And their product rz /> by the nature of proportion. Cx- ^y^ -s^-2p -1 Hence < a;-* ^* ___ :* > by the last problem, Cx- ^y^ -s^-2p -1 And x^-fj;*-j r + -^ — J* + ^— j| —^pzzh by the quest- ion, y ^' Again, 1- -^ zza — s by the question, y X Or 3<58 AtGEBRAiT Or x'-j-y^ =ivy X a—s =rp x a-s: But x^-^-y^zzis^ — pp by the last problem, And therefore pxa—s=s^-—^sp by equalityi Or pa—ps -|- 3/j =/ J -\-2pszzs'^. a-\-2s 4.' TOience /^4-a— j| — 4/^::if>-f.a— f! ? -- =z5 by sub- stitution, Or j'4-— j= —11- by reduction. a 2 And j-rr -^/ 1 [ — — by completing the square, and extracting the root. And from this vtiiue of / all the rest of the quantities />, X and y may be readily' determined. QUESTIONS . FOR PRACTICE. I. What two numbers are those, v/hose sum is 20, and their product 36 ? Ans. 2 and 18. ^ 2. To divide the number 6ci into two such parts, that their product may be to the sum of their squares in the ratio of 2 to 5. Ans. 20 and 4a. ^ 3. The difference of two numbers is 3, and the differ- ence of their cubes is 117; what are those numbers ? Ans. 2 and 5. 4. A company at a tavern had 81. 15s. to pay for their reckoning ; but, before the bill was settled, two of them sneaked off, and then tUose, who remained, had los. a piece more to pay than before *, how many v/ere there in the company ? Ans. 7. ^LGEBHA. 15. To find four numbers in geometrical progrossion, whpse sum is 15, ami the $um of their squares 83/. Alls. I, 2, 4 and 8. 1(5. Given x' value of X. _1- ^ "^ zz — ; to find the a Ans. .vrz 1^^ + v/^— ■ '4 * CUBIC AND HIGHER EQUATIONS- A CUBIC EQUATION, or equation of the third degree or power, is one, that contains the third power of the un- known quantity : as .v^ — ax^'\'^x'zzc. A biquadrat'tc^ or double quadratic, is an equation, that contains the fourth power of the unknown quantity : as 9i * — ax ^ -j-^^ * — cxzzd. An equation of the fifth power, or degree, is one, that contains the fifth power of the unknown quantity : as ^^ "■^ax "* -^bx ^ — cx ' -J-^.v zzf . An equation df the sixth powery or degree, is one, that contains the sixth power of the unknown quantity : as ^* — r^^ ^ '\-bx ^ cx ^ -i-dx * ex ZZf. And so on, for all other higher powers. Where it is to be noted, however, that all the powers, or terms, in the equation are supposed to be freed from surds, or fractional exponents. There are various particular rules for the resolution of cubic and higher equations ; but they may be all easily re- solved by the following rule of Double Position. RULE. CUBIC AND HIGHER EQUATIONS. ^ft RULE.* 1. Find by trial two numbers, as near the true root as pos- sible, and substitute them separately in the given equation^ instead of tlie unknown- quantity j marking the errors, which arise £ron> each Qf them. 2. Multiply the difference of the two numbers, found by trial, hf the l^a-st etror, and divide the product- by the difference of tlie errors, when they are alike, but by their sum, when they are unlike. Or say, as the. difference or sum of the errors is to the difference of the two numbers, so is the least error to the correction of its supposed number, 3. Add the quotient last found to the nyijiber belonging to the least error, when that number is too little, but sub- tract it, when too great ; and the result will give the true root nearly, 4. Take this root and the nearest of the two former, or any other, that may be found nearer ; and, by proceeding in like manner as above, a root will be had still nearer dian before ; and so on, to any degree of exactness requir-* ed. Each new operation commonly doubles the number of true figures in the root. Note i. It is best to employ always two assumed num- bers, that shall differ from each other only by unity in the last figure ow the right hand , because then the differ- ence, or multiplier, is only i. EXAMPLES. * This rule may be uced for solving, the questions of Double Position, as well as, that given in the Aritlimetic, and is prefera- ble for the present purpose. Its truth is easily deduced from the same supposition. For, by the supposition, r : j : : rv — a : x — h, therefore, bt" division, r — ^ \ s : : h — a : x — b j which is ths riilc ■*7^-' &LGEBIIA. EXAMPLES. I. To find the root of the cubic equation flf^-f^v^-l"''^ ssrioo, or the value of x in it. Here it is soon found, that x lies between 4 and 5. Assume, therefore, these two numbers, and the operation ■will be as follows : St supposition. 4 16 64 X X' Zd suppositioi, 5 125 84 sums '55 — 16 errors + 55 The sum of which is 71. Then, as 71 : I : : 16 : '225. Hence :cz=:/[' 22^ nearly. Again, suppose 4*2 and 4*3, and repeat the work a2 follows : 1st supposition. 7d supposition, 4-2 X 4*3 17-64 X' i8'49 74-088 X^' 79*507 95-928 -4-072 sums errors 102-297 + 2-297 The sum of which is 6*369. As 6*369 : '1 : : 2*297 : 0-036 ■Jfhis taken from 4*300 X^eaves x nearly zzn 4-264 Agaiuj CUBIC AND HIGHER EQUATIONS. 373 Again, suppose 4*^64, and 4*265, and work as follows : I St supposition. 2d supposition, 4*264 ►V 4*265 18*181696 .V* . 18*190225 77-526752 ^^ 77-581310 99*972448 sums 100*036535 — -o'o275<;2 errors -1-0*036535 The sum of which is '064087. THen, as '064087 : 'ooi : : 4*264 ; 0*0004299 To this adding 4*264 We have ^ very nearly ==4*2644299 ' 'I 2. To find the root of the equation ;tf^— i5A;*-j-63;tfi:: '50, or the value of x in it. Here it soon appears, that 9C is very little above i. Suppose, therefore, 1*0 and i*i, and work as follows : 1*0 X VI 63*0 6ja; 69*3 tp^i5 • — 15A;* 18*15 I a;' i'33i 49 sums 52*481 '—I errors ■4-2*481 3*481 sum of the errors. A$ 3*481 : 'I ; : i : '029 correct. 1*00 Hence x==. 1*029 nearly. 374 ALGEBRA. Agnin, suppose the two numbers 1*03 and 1*02, sn4 ^ork as fallows : 1*03 ic rot 64-89 — 15-9135: 1-092727 63^' -i5.v = sums errors •01 :: n from learly zz 64-26 •— 15-6060 1*061208 50-069227 49.715208 -{-•069227 •284792 ^•284792 A? -354019 • This take •069227 : •0019555 Leaves at r 1-02804 Note 2. Every equatieij. has as many roots as it con- tains dimensions, or as there are units in the index of its highest power.. That is, a simple equation has only one value or root ; bat a quadratic equation has two values or roots y a cubic equation has three roots j a biquadratic equation has four roots, and so on. And when one of the roots of an equation has been found by a|>proximQtion, as before, the rest may be found as foUov/s -.-•*^Takc for a dividend the given equation, with the known term- transposed, its sign being changed, to the unknown side ; of the equatLoji ? and for a divisor take K minus the root just found. "rDividje the said dividend by the divisor, and the quotient will, be the equation depressed a degree lower than the given one. Find a root of this hew equation by approximation, a§ before, and it will be a secotid root of the original equa- tion. ■ Then, by means of this root> depress the second equation eUfilC AND HiaiJER E(^UTI0N'3. 37^ equation one degree lovv-er, and thence find a -third root, and so on, till the equation be reduced to' a quadratic j then the two roots of this being found, by the method of completing the square, they will make up the re^nainder of the roots. Thus, in the foregoing equation, having found one root to be 1*02804, connect it by jniuiis with a; for a divisor, and take the given equation with the known term transposed for a dividend : thus, ^'— i'o28o4).v'~i5jtf* + 65.v— 5c(.x*-i3-97T9dv-|-48'<55€27i:ro. Then the two roots of this quadratic equation, or «* — 13*97 196^;=: — 48*63627, by completing the square, are 6*57653 and 7*39543, which are also the other two roots of the given cubic equation. So that all the three roots of that equation, viz. a;^— i5.v*4"^3^— 5^* are 1*02804 and 6*57653 and 7*39543 Sum 15*00000 And the sum of all the roots is found to be 15, being equal to the coeiTicient of the second term of the equation, which the sum of the roots always ought to be, when they are right. Note 3. It is also a particular advantage of the fore- going rule, that it is not necessary to prepare the equa- tion, as for other rules, by reducing it to the usual final form and state of equations. Because the rule may be ap- plied at once to an unreduced equation, though it be ever so much embarrassed by surd and compound xjuantities. As in the following example : 3. Let it be required to find the root .v of the equation ^i44A;'—A;^ + 2pf4-^i96;c'—.v' 4-241' =114, or the value of x in it. By 37<5 ALGEDRA. By a few trials it is soon found, that the value of x is but little above 7. Suppose therefore first, that xzz^^ and then that A? =8. 'irst, when x: ==7. Second, when x::zS< 47-906 V^i44a;* — A;'-f-2o|'^ 46*476 113*290 114-000 V/196A:' — at' +24!" 69*283 the sums of these 115*759 the true number 1 14*000 — 0*710 +^*759 As 2*469 the two errors "f"i*759 : 1 ;: 0*710 : 0*2 nearly. 7*o lvrr7-2 nearly. Suppose again »r:7'2, and, because it turns out too great, suppose also xzzyi. 3pose X — .7* 2. Suppose ivr::7*r. 47*990 v/144^'— .V^+2of 47*973 66*402 the sums of these the true number the errors •123 :: 'I : -024 the ( 7*ioo 65*904 114*392 114*000 113*877 114*000 4.0*392 ©•123 —0-123 •515 ' Dorrection. Therefore :i=zyi2^ nearly, the root required. Note CUBIC AND HIGHER EQUATIONS, 377 l^OTE 4. The same rule also, among other more dlffi- €ult forms of equations, succeeds very well in what are called exponential equations^ or those, which have an un- known quantity for the exponent of the power ; as in the following example. 4. To find the value of x m the exponential equation ft;*zzioo. For the more easy resolution of this kind of equations, it is convenient to take the logarithms of them, and then compute the terms by means of a table of logarithm.s. Thus, the logarithms of the two sides of the present equation are, ^X log. of xzni^ the log. of 100. Then by a few trials it is soon perceived, that the value of x is somewhere between the two numbers 3 and, 4, and indeed nearly in the middle between them, but rather nearer the latter than the former. By taking therefore first xzzzy^y and then a:z=:3'6, and working with the logarithms, the operation ^ill be as follows : First, suppose ►Yzz 3*5. Logarithm of 3*5 zz 0*5440680 Then 3*5 X log- 3*5 = 1-904238 The true number 2*000000 Error, too little, — -095762 Second, suppose x — : Logarithm of 3-6 iz Then 3'6X log. 3-6 z=l The true number 3 '6. 0*5563025 2*002689 2*000000 Error, too great. -J--0C2689 — '095762 4-'oo2689 •098451 sum of the errors. Then, Z z , As 3/8 ALGEBRA. As '098451 : •! : : '002689 : 0*00273 Which correction^ taken from 3*60000 Leaves 3'59727ZiAr nearly. On trial, this is found to be very little too smalL Take therefore again A'=i:3'59727, and next A:zr3'597285. and repeat the operation as follows : First, suppose x zr 3*59727. Logarithm of 3*59727 is 0*5559731 3*59727 X log. of 3-59727= 1-9999854 The true number 2*0000000 Error, too little, — 0*0000146 Second, suppose Am 3*59728. Logarithm of 3*59728 is j^o 5559743 3-59728 X log. of'3-5972811: 1-9999953 The true number 2*0000000 Error, too little, —0*0000047 — 0*0000146 -—0*0000047 0*0000099 difFerence of the errors. Then, As -OC00099 • 'ooooi :: -0000047 : 0*00000474747 Which correction, added to 3*59728000000 Gives nearly the value of x zz 3*59728474747 5. To find the value of x m the equation .v ^ -}- 1 c»v ' -}- 5a;iz26oo. /ins. /vzz 1 1-0067 3. 6. To find the value of >: in the equation x"^ — 2X=:^. • Ans. 2*004551. 7. To CUBIC AND HIGHER EQUATIONS. 379 7. To find the value of x in the equation a;^-{-2.v* — 23^^=70. Ans. ^=5*1349. 8. To find the value of s in the equation a;^ — 17^;* + 54^ = 350- ^"S. ^z:iii4-95407. 9. To find the value of a m the equation Ar'' — 3^*""" 75A'=ioooo. Ans. Afzz 10*2615. 10. To find the value of x in the equation 2x^ — i6x^ ^4o;v* — 3ca;zz — i. Ans. A:=r284724. 11. To find the value of ►v in the equation ;c^-|-2:v'' + 3;v^-|-4:v'-j- 5*^=5432 1- ^ns. ;vzi8-4i4455. 12. To find the value of X in the equation a;^ZI ^23456789. ' Ans. -;vii:8'6400268. s^J5^= END OF ALGEBRA. geometry; -=s.a®®^j<^S^gS5>!'©e®« DEFINITIONS, I. J\. POINT is that, which has position, but not magnitude. 2. A /hie is length, without breadth or thickness. 3. A surf ace f or superficies, is an extension, or a figure, of two dimen- sions, length and breadth, but with- out thickness. 4. A bocly, or solid, is a figure of three dimensions, namely, length, breadth and thickness. \[ Hence surfaces are the extremities of solids ; lines the extremities of surfaces 5 and points the extremities of lines. 5. Lines '"^ A Tutor teaches Simson^s Edition of Euclid's ElE" MwNT^ of Geometry in Harvard College, ;S2 GEOMETRY. 5. LixiGs are either right, or curved, or mixed of these two. "t6. a right liney or straight line, lies all I in the same direction between its extrem- ities, and is the shortest distance between two points. 7. A curve continually changes its direction between its extrenie points. 8. Lines are either parallel, oblique, perpendicular, or tangential. 9. Parallel lines are always at the same distance, and never meet, though ever so far produced. I o. Oblique right lines change their distance, and would meet, if pro- duced, on the - side of the least distance. II. One line is perpendicular to another, when it inclines not more on one side than on the other. 12. One line is tangential^ or a tangent, to another, when it touches It without cutting, if both be pro- duced. !l3« An tJEFINITKJNS. 3S3 13. All angle is the IncUnation, or opening, of two lines, having dilFerent directions, and meeting in a point. 14. Angles are right or oblique, acute or obtuse. 15. A right angle is that, which is made by one line perpendicular to an- other. Or when the angles on each side are equal to. one another, they are right angles. 1 6. An olllqtie angle is that, which is made by two oblique lines, and is either less or greater than a right angle. 1 7. An acute tingle is less than a right angle. 18. An oltuse angle is greater than a right angle. 1 9. Superficies are either plane or curved. 20. A ^lane superficies ^ or a plane ^ is that, with which a right line may, every way, coincide. But if not, it is curved, 21. Plane figures are bounded either by right lines or curves. 22. Plane figures, bounded by right lines, have names according to the number of their sides, or angles ; for they have as many sides as angles ; the least number being three. 23. A 3«4 GEOMETRY. 23. A figure of three sides and angles is called a trian^ gle. And it receives particular denominations from the re- lations of its sides and angles. 24. An equilateral irmtigle is that, whose three sides are equal. /\ A / \ 25. An isosceles triangle is that, which / \ has two sides equal. / \ LJ-: 26. A scalene triangle is that, whose three sides are all unequal. 27. A right-angled triangle is that, which has one right angle. / // / 28. Other triangles are oblique-angled, and are either ob- tuse-angled or acute- angled. 2p. An DEFINITIONS. 38^ 29. An ohtuse-angled triangle ha* one obtuse angle. 30. An acute-angJed triangle has all its three angles acute. 31. A figure of four sides and angles is called a quad' tangle^ or a quadrilateral, 32. A parallelogram is a quadrilateral, which has both pair of its opposite sides parallel. And it takes the fol- lowing particular names. . 33. A rectangle is a parallelogram, having all its angles right. 34. A square is an equilateral rectan- gle, having all its sides equal, and all its angles right. 35. A rhomboid is an oblique-angled parallelogram. LJ 36. A 7-homhusis an equilateral rhom- boid, having all its sides equal, but its angles oblique. Ala 37. A 3^^ GEOMETRY, 37. A trapezium is a quadrilateral, which has not both pair of its opposite sides parallel. 38. A trapezoid has only one pair of Opposite sides parallel. 39. A diagonal is a right line, joining any two opposite angles of a quadrilateral. 40. Plane figures, having more than four sides, are, in general, called polygons ; and they receive other particular names, according to the number of their sides or angles. 41. A pentagon is a polygon of five sides ; a hexagon has six sides ; a heptagon, seven ; an octagon, eight ; a non- agon, nine ; a decagon, ten ; an undecagon, eleven ; and a dodecagon, twelve. 42. A regular polygon has all its sides and all its angles equal. — If they be not both equal, the polygon is ir- regular. 43. An equilateral triangle is also a regular figure of three sides, and the square is one of four ; the former be- ing also called a trigon, and the latter a tetragon. Pentagon. Hexagon. DEFINITIONS. Heptagon. Octagon. 587 Nonagon, V. Undecagoiu. Decagon. Dodecagon. 44. A circle is a plane figure bounded by a curve line, called the circumference .^ which is every w^here equidistant from a certain point within, called the centre. Note. The circumference itself is often called a circle. 45. The 38S GEOMETRY. 45. The racl'nts of a circle is a right line, drawn from the centre to the circumference. 46. The diameter of a circle is a right line, drawn through the centre, and terminating in the circumference on both sides. 47. An arc of a circle is any part of the circumference. 48. A chord is a right line, joining the extremities of an arc. x 49. A segment is any part of a / circle, bounded by an arc and its j chord. '* 50. A DEFINITIOKS. 3«9 50. A semkircle is half the cir- cle, or a segment cut off by a di- ameter. \ 51. A sector Is any part of a circle, bounded by an arc, and two radiij drawn to its extremi- ties. 52. A quadrant y or quarter of a circle, is a sector, having a quar- ter of the circumference for its arc, and its two radii are perpen- dicular to each other. 53. The height y ox attitude ^ of a figure is a perpendicular let fall from an angle, or its vertex, to the opposite side, called the base. 54. In a right-angled triangle, the side opposite to the riglit angle is called the hypotenuse ; and the other two the legSy or sides y or sometimes the base and perpendicular. ^5. When an angle is denoted by three letters, of which one stands at the angular point, and the other two on the two_sides, that, which stands at the angular point, is read in the middle. 390 t>pOMETRY. 56. Th'j clrcuinfercncc of every circle is supposed to be divided iato 360 equ«l parts, called degrees ; and each de- gree iiiio 60 mh:::ieSf each minute into 60 seconds, and so on. Hence ^ semicircle contain? 180 degrees, and a quad- rant 90 degrees. « 57. The measure of a right- lined angle is an arc of any cir- ,• cle, contained between the two / lines, v/hich form that angle, the \ angular point being the centre \ \ and it is estimated by the num- ber of degrees, contained in that arc. Hence a Tight angle is an angle, of 90 degrees. 58. Identical fgures 7XXQ. snc\ as have all the sides and all the angles of one respectively equal to all the sides and ail the angles of the other, each to each ; so that, if one figure were applied to, or laid upon, the other, ail the sides of it would exactly fail upon and cover all the sides of the other 5 the tv.ro becoming coincident. 59. An angle in a segment is that, which is contained by two lines, drawn .from any point in the arc of the^ segment to the ex- tremities of the arc. 60. A right-lined figure is inscribed in a cf'cley or ^e circle circutnscribes it, when all the angular points of the figure are in the circumference of the circle. 61. A right-lined figure c'lrcnmscrihes a cir- cloy or the circle is inscribed in it, when all the sides of the figure* touch the circumference ©f the circle. 62. 0ns bEHMlTtbNS. %pf \/ 62. One righf' lined figure is inscribed In ar.- hihery or the latter circurnscribes the firmer^ \/ "^ when all the angular points of the former arc j)laccd in the sides of the latter. 63. Similar Jigures zre tho^Cy that have all the angles of one equal to all the angles of the other, each to each, and the sides about the equal angles proportional. 64. The perimeier of a fgiire is the sum of all its sides, taken together. 65. A proposition is Something, which Is either propojJed to be done, or to be demonstrated, and 19 either a probldni or a theorem. 66. A problem Is something proposed to be done. 67. A theorem is something proposed to be demonstrated. 68. A lemjia is something, which is premised, 6x pre- viously demonstrated, in order to render what follows more e^y. 69. A corolktry is a consequent truth, gained immediate- ly from some preceding truth, or demonstration. 70. A scholium is a remark, or observation, made upon something preceding it. PROBLEMS. 39^ - tersecting in m. Draw the line B w, and / it will bisect the angle, as required. Note. By this operation the arc A C is bisected ; and in a similar manner may any given arc of a circle be bisected. PROBLEM PROBLEMS, 393 PROBLEM III* To divide a right angle ABC into three equal parts^ From the centre B, with any radius, describe the arc AC. From the centre A, with the same radius, cross the arc AC in n ; and with the centre C, and the same radius, cut the arc AC in w. Then through the points m and n draw B m and B n^ and they will trisect the angle, as required. PROBLEM IV, To draiv a Line parallel to a given Line A B. Case i. — 'IVhcn the parallel Line is to be at a given Dis* tance C. ""■!> B From any two points m and «, in the line AB, with a rad- ius equal to C, describe the arcs r and o. Draw C D to touch these arcs, without cut- ting • them, and it will be the parallel required. Case l.-'—When the parallel Line is to pass through a given Point C. From any point m^ in the ^ line AB, with the radius /tzC, ^1 "~ — " \ E describe the arc Qn. From / / the centre C, with the same A--, i ■— 4> radius, describe the arc mr. Take the arc Qn in the compasses, and apply it from >« to r. Through C and r draw DE, the parallel rL-quircd. Bub Note 394 GEOMETRT. Note. In practice^ parallel lines ate niore eatily drawn with a Parallel Rule. PROBLETvI V. To eyed a Ferpendiadar from a giv^ii' Point- ' A in a given Line B C ' ' Case i. — When the Po'mt u mar the mlcIc^J of the Line. On each side of the point A, take -..^r ,.. any two equal distances Am, An. From ^' i ' ^ tlie centres m and «,. with any radius- ! greater than Am or A«, describe two \ area, intersecting in r. Through A and | r draw the line Ar, and it will he the, perpendicular jcecj^uircd. •Bu CjS£ 2. — When the Pvint is near the end of the Lina With the centre A, and any radius, describe the arc m n j, From the point »;, v/ith the same radius, turn the compasses twice over on the arc, at n and s. A- gain, with the centres n and /, describe arcs intersecting in r. Then draw Ar, and it will be the perpendicular required. 2i Another Method. From any point w, as a centre, with the radius or distance m A, describe an arc cuttliig the given line in ;; and A. Through u -.xn^ m draw a right line .cutting the arc in r. Lastly, draw A r, and It wilt be the pcrpendiculcir ic- .^^ ' q^ih-ed. Another PROBLEMS. 09 f Another Method. From .nny plane scale of equal parts, set oft Am equal to 4 parts. ! With centre A, and radius of ^ 1 three parts, describe an arc. And <''T: "v with centre w, and radius of 5 'I * parts cross it at n. Draw A;; for 3"" j^j]^ ~ — - the perpendicvilar required. Or any other numbers in the same proportion, as 3, 4, 5, will answer the s^me purpose. PROBLEM VI. From a g'weti Point A, cut of a given Line BC, to let fall a Pei-pendicular» Cas^ l.-^lVhm the point is iiearl^^ opposite the middle of the Line, With the centre A, and any rad- -A, lus, describe an arc cutting B C in 1 VI and //. With the centres m and j «, and the same, or any other rad- t?^v___ j l> ^^ ius, describe arcs intersecting in r. ^^^""••••••1 "'ST Draw A D r for the perpendicular T required, "J * "■'-»'■' Case %~Whcn the Point is marly opposite the end cf the Line, From A draw any line A w to a meet B C, in any point m. Bi- sect Am zt ;;, and with the cen- ^< tre «, and radius An or mn, de- ^/i •scribe an arc, cutting BC in D. Draw AD, the perpendicular re- ig / // quired. '-*'^' - 39(5 GEOMETRY. Another Method, From B or any point in B C, as a centre, describe an arc through the point A. From any- other centre tn in B C, describe another arc through A, cutting the former arc again in «. Through A and « draw the line A D « ; and AD will be the perpendicular required. .»^...> "::.A. C: i jax D • •* • n // Note. Perpendiculars may be more readily raised and let fall, in Practice, by means of a square or other fit in« strument. PROBLEM VII. To divide a given Line A B into afty proposed number of- equal Parts, From A draw any line A C at random, and from B drav/ B D parallel to it. On each of these lines, beginning at A and B, set off as many equal parts, of any length, as A B is to be divided into. Join the opposite, points of division by the lines A 5, i 4, 2 3, &c. and they will divide A B as required. PROBLEM TROBLEMS^ 597 PROBLEM VIII. ^0 divide a given Line A B in the same proportion^ as another Line C D is divided. f. -?.. r^^n JV \ \ From A draw any line AE equal to C D, and upon it trans- fer the divisions of the line C D. Join B E, and parallel to it draw the lines i i, 2 2, 3 3, ^ A ^ Sec. and they will divide A B a§ required. PROBLEM IX. Jit a given Point A, in a given line A D, /^ malce an equal to a given Angle C. With the cqntre C, and any radius, describe an arc mn. — With centre A, and the same radius, describe the arc- rj-.— • Take the distance mn in the com- passes, and apply it from r to j. Then a line, drawn through A & J-, will make the angle A equal to the angle C, as required. Angl'^ PROBLEM X. At a given Point. A, in a given l^ine A B, to male an Angk of any proposed number of degrees^ With the centre A, and radius equal to 60 degrees, taken from a scale of chords, describe an arc cutting A B in m. Then take in the compasses the pro- posed number of degrees from the same scale of chords, and an f> apply 398 GEOMETRY. apply th&in from m to ??. Througli the point ;; draw A «, and it will make the angle A of thf number of de- grees proposed. Or the angle rriT^y he made with any divided arc, or in« strument, by applying the centre to tlie point A, and iti radius along A B -, tlien make a mark « at the proposed number of degrees, tlirough ^yhicIl draw the. line A r^., as before. Note. Angles of more than oo degrees are usually- taken off at twice. PRO B L E M X I^ ^0 t?:easurg a given Angle A. Describe the arc inn with the chord of 60 degrees, as in the last Problem. Take the arc ni n in the compasseS;, and that extent, applied to the chords, will shew the degrees in the given angle. PROBLEM XIF, To find the Centre of a Cirtk. Draw any chord A B ; and bisect it perpendicularly with C D, which will be a diameter. Bisect C D in the point o, trv% which will be the cen,- J\ PROBLEM PROBLEMS. 3f^ PROBLEM XllU ^9 dascrlhe the Clrcinnfc/cncc of a Circle thi-ough thne given Feints A, B, C. From the middle point B draw chords to the other two poiii^. liisect these chords perpendicu- larly by lines meeting in o, which will be the centre. Then 'from the centre o, at the distaiice o A, ©r oB, or oC, describe thcr circle* Note. In the sanie miniier may the centi'e of an atC «f A circle be found. PROBLEM XIV, Through a given Point A to draw a Tangent to a given Cireti. Case i. — When A is in the Ciramfcrence of the Circle* From the given point A, draw Ac to the centre of the circle. Then through A draw B C perpendicular to Ao^ and it will be \},\^ tangent re- fill ired. Cab% 4trO GEOMETRT. CaS£ 2 — J^hen A is out of the Circumference. From the given point A draw Ao to the centre, which bisect in the point m. With the centre w, and radius m A or 771 o, describe ail arc, cut- ting the given circle in n. Through the points A and n draw the tangent B C. TS^^ PROBLEM XV, ^0 find a third ProportioJtal to two given Lines AB, AC. Place the two given lines, AB, A C, making any angle at A, and join BC. In AB take AD equal to AC, and draw DE par- allel to B C. So shall AE be the third proportional to AB and AC. That is, AB : AC :: AC : A. A- .a: A^. ^B PROBLEM XVl. To find a fourth Proportional to three given Lines AB, AC, AD. Place two of them, A B, AC, A- so as to make any an^lc at A, and '^' join B C. Place AD on A B, and draw D E parallel to BC. So shall A £ be the fourth proportional re- quired. That is, A^ ; AC ;; AD : AE. JV -D -B JB PROBLEM TROELEMS. m PROBLEM XVII, ^0 find a mean Proporirjua! betivccji tivo given Lines K B, B C Join A B and B C in one A — straight line AC, and bisect ■^'~~ it in the pvoint o. With the centre o, and radius o A or o C, descrijbe a semicircle. Erect the perpendicular BD, »nd it will be the mean pro- j\ '— portional required. That is, AB ; BD : : BD : BC. B -C b ii PROBLEM XVIIT. To divide a Line A B in Extreme and Mean Ratis. Raise B C perpendicular to A B, and equal to half A B. Join A C. With centre C, and radius C B, cross AC in D. Lastly, with cen- tre A, and radius A D, crois A B in E, which will divide the line A B in extreme and mean ratio, name- ly, so that the whole line is to the greater part, as the greater part is to the less part. That is, AB : AE Ail : Ljj. 13 Ccc PROBLEM 4'->»' GEOMETRT. PROBLEM XIXo 21? inscribe an isosceles triangle in a given circle, that shall have each of the angles at the base double the angle at the vertex. Draw any diameter A B of the givers circle ; and divide the radius CB, in the point D, in extreme and mean ratio, by the last problem. From the point B apply the chords BE, BF, each equal to C D 'y then join A E, A F, E F, and A E F will be the triangle required. PROBLEM XX« To male an enu'ilateral Triangle on a given Line A.b. From the centred A and B, with the radius A B, describe arcs intersecting in C. Draw A C and B C^ and it is dene. Note. An isosceles trian;^Ie may be made in the same manner, by taking for the radius the given length of one ef the equal sides. PROBLEM PROBLEMS. t»ROBLEM XXI. 403 ^0 niah a 'Triangle nvith three given Lines A B, AC, B C» "With the centre A and radius A C, describe an arc. "With the centre B and radius B C, describe another arc cutting the former in C. Draw AC and B C, and ABC is the triangle required. c PROBLEM XXII. To make a Square upon a given Line A B. Draw BC perpendicular and -equal to AB. From A and C, with the radius AB, describe arcs Intersecting in D. Draw A D and C D, and it is done. 3> J-^ Another Way, On the centres A and B, with the radius A B, describe arcs crossing at o. Bisect A o in «. With centre o, and rad- ius ony, cross the two arcs in C and D. Then draw AC, B D, C D, i> / "^ IX j.-^ B PROBLEM / 4C| GEOMETRY. PROBLEM XXIII. Q describe a Hect angles or a Paralleiogra?ny of a given Length and BreaJik. Place B C perpendicular to A B. "W ith centre A, and rad- ius B C, describe an arc. With centre C, and radius A B, de- scribe another arc cutting the former in D. Draw A D and CD, and it is done. Note. In the same manner is described any oblique parallelogram, except in drawing B C so as to make the given oblique angle with A B, instead of a right one. PROBLEM XXIV, 21? mahe a regular Pent/igon en a given Line A B, Make B /;; perpendicular and equal to half AB. Draw A/;/, and produce it till in n be equal to B ;;/. With centres A and y B, and radius B «, describe • \ arcs intersecting in o, which '\ \ will be the centre of the cir- \\ ^^ cumscribing circle. Then with "- -; - -- the centre o, and the same rad- ius, describe the circle -, ancl about the circumference of it spply A B the proper number of times. Another ._J: JJ PROBLEMS. 4of Another Method. Make B in perpendicular and equal to A B. Bisect A B in « ; then with the centra «, and radius « w, cross A B produced in O. With the centres A and B, and radius Ao, describe arcs intersect- ing in D, the opposite angle of the pentagon. Lastly, v/ith centre D, and radius A B, cross those arcs ngain In C and E, the other two angles of the figure. Then draw the lines from angle to angle, to complete the figure. A thh*d Method, nearly true, On the centres A and B, with the radius A B, * de- scribe two circles intersecting in m and rt. With the same radius, and the centre w, de- scribe r Ao B S, and draw m n cut- ting it in o. Draw r o C and S o E, which will give two angles of the ^ pentagon. Lastly, with radiu* AB, and centres C and Ej describe arcs i itersecting in^D, the other angle of the pen- tagon nearly. PROBLEM 4<^i GEOMLT!^ PROBLEM XXV* To viahe a Hexazofi en a f^lven Line With the radius A B, and the centres A and B, de- scribe arcs intersecthig in o. With the same radius, and centre o, describe a circle, which will circumscribe the hexagon. Then apply the line A B six times round the circumference, marking out the angular points, which connect with right lines. AB. PROBLEM XXVI. fTo mah an Octagon on a ^'^iven Line A B. Erect A F and B E per- pendicular to A B. Pro- duce A B both ways, and bisect the angles in A F and /2 B E with the lines A H and B C, each equal to AB. Draw CD and H G parallel to A F or BE, and each equal to A B. With radius A B, and centres G and D, cross AF and BE, in F and E. Then join G F, F E, ED, and it is done •'V / ..'•\. PROBLEM 4*^7 PROBLEM?. PROBLEM XX YP- ffo ihalc a?r; regulai- Polygjfi on a given Line A B. Draw A o and B o, male- ^g the angles A and 15 eacli equul to half the angle of the pulygpn. With the centre o, and radius o A, describe a circle. Then ap- ply the line A B continually- round the circumference the proper number of times, and it is done. Note. Tho. angle of any polygon, of which the angles o A B and q B A are each one half, is found thus : divide the whole 360 degrees by the number of sides, and the quotient will be the angle at the centre o ; then subtract that from 180 degrees, and the remainder will be the an- gle of the polygon, and is double of o A B, or of o B A. And thus you will find the following table, containing the degrees in the angle c at the centre, and the angle of th& polygon> for all the regular figures from 3 to 12 sides. No. of siacs. Nanxe of ihc I'olyoon. Angie 0 at il;f centre. An"le of tl»c Pulyi'oii. Ant^lc 0 A B j or 0 P. A. ' i 3 Trigon 120° 60° 30° i 4 Tetragon 90 90 45 5 Pentagon 72 108 54 6 Hexagon 60 120 60 I ' 7 Heptagon 5 4 1284- 6a\ ! 8 j Octagon 45 135 6/1- ; 9 Nonagon 40 140 70 j 10 Decagon 36 V 144 r- , II , Undecagon S^tV i4;tV 73tV 1 12 Dodecagon 30 150 7=; luoniXM 4oS GEOMETRY* PROBLEM XXVIII. Jji a given Circle to inscribe any regular Polygon / ^r, to divlis the Clrcamfcrcnce into any number of equal Parts, [See the last Figure.] At the cenj.re o make an angle equal to the angle at the. centre of the polygon, as contained in the third columrr of the above table of polygons. Then the distance A D. ■will be one side of the polygon j which, being carried round the circumference the proper number of times, will complete the figure. Or, the arc A B will be one of th& equal parts of the circumference. Another Method, nearly txxxt^ Draw the diameter A B which divide into as many equal parts as the figure has sides. With the radius A B, and centres A and B, describe arcs crosc^ing at n ; whence draw n C through the second. division on the di- ameter ; so shall A C be a side of tiie polygon nearly. yi ..^*' Another PROBLEMS. 409 Another Method, still nearei% Divide the diameter A B into as many equal parts as the figure has sides, as before. From the centre o rJiise the perpendicular ow, which produce till m n be three-fourths of the radius o in, T'rom n draw n C through the second division of the diameter, and the line AC will be the side of the polygon still nearer than before ; or the arc A C, one of the equal parts, into which the divided. ; ,211: circumference is k? be PROBLEM XXJX, About a given Circle to circumscrihe any Polygon* Find the points my «, /, Sec. as in the last problem, to which draw radii ;«o, rjo, &c. to the centre of the circle. Then through these points my ;;, &c. and perpendicular to these radii, draw the sides of the polygon. Dod PROBLEiA* 4tQ GEOMETRY. PROBLEM XXX. To find the Ce?itre of a given Polygon^ or the Centre of its in^ scribed or circumscribed Circle, Bisect any two sides with the per- pendiculars mOi no ; and their inter- section will be the centre. Then, with the centre o, and the distance only describe the inscribed circle j or with the distance to one of the an- gles, as A, describe the circumscrib- ing circle. Note. This method %yni also circumscribe a circle- about any given oblique triangle. PROBLEM XXXI. In any given Triangle to inscribe a Circle* Bisect any two of the angles with the lines Ac, Bo, and o will be the centre of the circle. Then, with the centre o, and radius the nearest distance to any one of the «ides, describe the circle. PROBLEM XXXII . About any given Triangle to circwnscribe a Circle, C Bisect any two of the sides A B, B C, with the perpendiculars wo, no. With the centre o, and distance to any one of the angles, describe the circle. PROBLEM PROBLEMS. 411 PROBLEM XXXIII, /«, or ahuty a given Square to describe a Circle* Draw the two diagonals of the square, and their intersection o will be the cen- tre of both the circles. Then, with that centre, and the nearest distance to one side for radius, describe the inner circle ; and with the distance to one angle for iradius, describe the outer circle. ^ PROBLEM XXXIV, ItiyOr ahout^a given Circle to describe a Square, or ati Octagon. Draw two diameters A B, CD, perpendicular to each other. Then connect their extremities, and they will give the inscribed square A C, ^ B D. Also through their extremi- ties draw tangents parallel to them, and they will form the outer S is one twelfth of the circumference, or 30 degrees, an|l gives the side pf the dodecagon. Note. If tangents to the circle be drawn through ai' the angular points of any inscribed figure, they will form the $idcs of a like gircumscribing figure. PROBLEM XXXVI. In a given Circle io inscribe a Pentagon^ or a Decagon, Draw the two diame- ters AP, w;/, perpendic- ular to each other, and bisect the radius on at q. With the centre q^ and radius q A, describe the arc A r ; and with the centre A, and radius Ar, describe the arc rB. Then is A B one £fth of the circumference \ PROBLEMS. 413 circumference ; and A B, carried round five times, will ^?0J the pentagon. Also the arc A B, bisected in S, wiU give A S> the tenth part of the circumference, or the side of the dcc:igon. Another Method, Inscribe the isosceles triangle ABC, -^- having each of the angles ABC, A C B, double the angle B A C. Then bisect the two arcs A D B, A E C, in the points D, E ; and draw the chords AD, DB, AE, EC ; so shall ADB QE be the inscribed pentagon re- quired. And the decagon is thence obtained as before. Note. Tangents, being drawn through the angular points, will form the circumscribing pentagon or decagon. P ROLL EM XXXVII. STo divide the Circumferc7ice of a given Circle into tivdve equal Parts y each being 30 D egret's. Or to iiiscribe a Dodecagon by another Method, Draw two diameters i 7 and and 4 10 perpendicular to each other. Then, with the radius of the circle, and the four extremi- ties I, 4, 7, 10, as centres, de- scribe arcs through the centre of the circle j and they will cut the circumference in the points re- quired, dividing it into 12 equal parts at the points marked 'Svith the numbers. PROBLEM 4^4 PROBLEM XXXVIII. To divide a given Circle into any proposed Numher of Parts hy equal LineSy so thai those Parts shall he piutually eqtmly both in Area a:;d Perimeter. Divide the diameter Ah into the proposed number of equal parts at the points ^, by r, &c. Thenon Afl, hb, Ar, &c. as diam- eters, describe semicircles on one side of the diameter A B ; and on Vtd, Br, B^, &c. describe semicircles on the other side of the diameter. So shall the corresponding joining semi- circles divide the given circle in the manner proposed. And in like manner we may proceed, when the spaces are to be in any given proportion. As to the perimeters, they are always equal, whatever may be the proportion o( the spaces. PROBLEM XXXIX. On a given Line AV> to describe the Segment of a Circl^^ capable of containing a given Angle, Draw AC and BC, making the angles BAG and ABC each equal to the given angle. Draw A D perpendicular to AC, and BD per- pendicular to B C. With centre D, and radius D A, or D B, de- scribe the segment AEB. Then any angle', as E, made in that seg- ment, will be equal to the given angle. PROBLEM PE.OBLEMS. 4^5 PROBLEM XL, To cut off a Segment from a given Circle^ that shall contain a given Angle C. Draw any tan- gent AB to the given circle ; and a chord A D, to make the angle DAB equal to the given angle C ; then D E A will be the segment required, any angle E made in it being equal to the given angle C. 2\. PROBLEM XLI. *To make a Triatigls similar to a given Triangle ABC. Let ^ ^ be one side of the re- quired triangle. Make the angle a equal to the angle A, and the an- gle b equal to the angle B \ then the triangle ab c will be similar to ABC, as proposed. Oj Note. U ab be equal to A B, the triangles will also be equals as well as simijar. PROBLEM 4i6 CEOMETRV. PROBLEM XLII. *To make a Figure similar to any other given Figure ABCDE. A draw diagonals From any angl to the other angles. Take A^ a side of the figure required. Then draw h c parallel to B C, and c d to CD, and de to D E, &c. Otherwise, )i B h Make the angles at a^h^e^ Sec. Tespcctively equ il to the ailgles at A, B, E, and the lines will intersect in the angles of the figure re- quired. PROBLEM XLIII. 2"o make a Triangle equal to a given Trapezium A BCD. Draw the diagonal DB, and CE parallel to it, meeting AB produced in E. Join D E •, so shall the triangle ADE be equal to the trapezium AB CD. PROBLEM PROBLEMS. 4^7 PROBLEM XLIV* 'To tnahe a Triangle equal to the Figure ABCDEA^ Draw the diagonals DA, DB, and the lines EF, CG, parallel to them, meeting the base AB, both ways produced, in F and G. Join D F, D G ; and DFG •will be the triangle required. Note. Nearly In the same manner may a triangle be jnade equal to any right-lined figure whatever. PR0I3LE*M 3^LV, To male a Rectangkf or a ParaUeiogramy equal to a given Triangle ABC. Bisect the base A B in m> Through C draw Qno parallel to A B. Through m and B draw ;;/ n and B o parallel to each other, and either perpen- dicular to AB, or making any angle with it. And the rectan- gle or parallelogram junoVt will be equal to the triangle, as required. Eec PBOBLEM 41^ GEOMETRY. PROBLEM XLVr, 4 To 77iah a Square equal to a given Rectangle ABGl). r Produce one side A B, till B E be equal to the other side B C. Bisect A E in o ; on; which as a centre, with radius / Ac, describe a semicircle, and [_ produce B C to meet it at F. ^ On B F make the square B F G H, and it will be equal, to the rectangle A B G D, as required* R -a PROBLEM XLVIK To make a Square equal to two given Squares P and Q. Set two sides AB, B C, of the given squares perpendicular fto each other. Join their ex- tremities AC; so shall ths square R, constructed on A C, be equal to the two P and Q^ taken together. Note. Circles, or any other similaY figures, are added in the same manner. For if A B and B C be the diame- ters of two circles, A C will be the diameter of a circle equal to the other two. And if AB and B C be the like sides of any two similar figures, then A C will be the like side of another similar figure equal to the two for- mer, and upon which the third figure may be constructed, "by Problem rSiu PROBLEM PROBLEMS. |>ROBLEM XLVIII. 419 Tb male a Square equal to the Difference cf iivQ given Squares P, R. On the side A C of t\\Q greater square, as a diapieter, describe a semic'iTcle ; in which apply A B the side of the less square. Join BC, and it will be the side of a square equal to the difFerencc between i^c two P and R, as required. PROBLEM XLIX. ^0 male a Square equal to the sum of any numher cf Squares tahfi together. Draw two indefinite lines A my A «, perpendicular to ^ach other at the point A. On one of these set off* AB the side of one of the given squares, and on the other A C the side of another of them. Join BG, and it will be the- side of a square equal to the two together; Then take AD equal to BC, and AE equal to .!> li the side of the third given square. So shall D E side of a square equal to the sum of the three squares. An^ so on continually, always setting sides of the given squares on the line An, and th of the successive sums on the otliev line A w. be the given more e sides KOTF. s,v 420 GEOMETRY, Note. And ttius any number of any kind of figurcf piay be added together. PROBLEM L, To construct the lines of the Plane Scale, The divisions on the Plane Scale are of two kinds *, on^ kind having relation merely to right lines, and the other to the circle and its properties. The former are c;illed Linesy or Scales^ of Equal Farts, and are either siraj)le or diagGnal, By the Lines of the Plane Scale we here mean the follow- ing Lines, most of which commonly, and all of them sometimes, are drawn on a Plane Scale. 1. A Line or Scale of Equal Parts, marked f. P, 2. . . • » < • Chords . . 4 . Cho. 3 Rhumbs .... Rhu. 4. . . . • . • Sines , . . . , Sin. ^ ". tangents / * ^ . . Tan. 6. ..... . Secants . , . . Sec. 7. ••.... . Semiiangents . . . S. T. '8 . Longitude .... Lon. 9 Latitudes .... Lat. 10 Hours .... Ho. II » . Inclination f Meridians Ll. Mer. PROBLEMS. 411 I. To construct plane diogonal Scales. Draw any liije, as A B, of any convenient length. Di- vide it into II equal parts.* Complete these into rectan- gles ol* a convenient height, by drawing parallel and per- pendicular lines. Divide the altitude into lo equal parts, if it be. for a decimal sc?le for common numbers, or into 12 equal parts, if it be for feet and inches •, and through these points of division draw as many parallel lines, the whole lencrth of the scrde. Then divide the length of the first divlbic, AC h: ; to equal parts, both abover and be.- low •, and ..,.!,, .r , .-■ ;ioints of division by diagonal lines, and the scale is hnished, after being numbered as you please. u -t PLANE SCALES. fOR rwo Figures. U-LJJ_U_:_UJ L.—^.. J-^ Ci Only 4 parts are here drav/n for want cf room. 42^ GEOMETR"£. Of the preceding three forms of scales for two figure^ the first is a decimal scale, for taking off common num- bers consisting of two figures. The other two are duo-» decimal scales^ and serve fqr feet and inches. In order to construct the other lines, 4cscrtbe a 'circum- ference with any convenient radius, and draw the diame- ters A B, D E, at right angles to each other ; continue B A at pleasure toward F 5 through D draw D G parallel toBFj and draw the chords BD, BE, AD, AE. Cir- cumscrihe the circle with the square HMN, whose side^ HM, MN, shall be parallel to AB, ED, 2, To cctistri^'ct the Line of Chords* Divide the arc AD into 90 equal parts ; mark the iotl\ divisions with the figures 10, 20, 30, 40, 50, 60, 70, 80, po ; on D, as a centre, with the compasses, transfer the several divisions of the quadrantal arc, to the chord AD, which marked with the figures corresponding, will bq 31 line of phord?. Note. In the construction of this and the fallowing scales, only the primary divisions are drawn ; the inter* mediate ones are omitted, th^^t the figure may not appeav too much crowded. 3. To construct the L'lns of ' Rhumbs^ Divide the arc BE into 8 equal parts, which mark ^vith the figures i, 2, 3, 4, 5, 6, 7, 8 ; and divide each of those problems; Jtn g to 6\» 'v "i^ CO X4 .2 cr 1 :\i 424 GEOMEITs-T. those parts into quarters ; on B, as a centre, transfer ttie divisions of the arc to the chord B E, which, marked with the corresponding figures, will be a line of rhumbs. 4. To const ruct the Line of Sines:^'' Through each of the divisions of the arc AD drav/ right lines parallel to the radius A C ; and C D will be divided into a line of sines, which are to be numbered from C to D for the right sines ; and from D to C for the versed sines. The versed sines may be continued to 180 degrees, by laying the divisions of the radius CD from C to E. 5. 7*5 construct the Line of Tangents, ^' A rule on C, and the several divisions of the arc A T), will intersect the line D G, which will become a line of tangents, and is to be figured from D to G with 10, ao, 30, 40, &c. 6. To construct the Line of Secant. The distances from the centre C to the divisions on the line of tangents, being transferred .to the line CF from the centre * For De^nltions of Sines, Tangents and Secants, see PlXns Trigonometry ; and for that of Rhambsj-see N-avigation. PROBLEMS. 42S. centre C, will give the divisions of the line of secants > which must be numbered from A toward F with lo, 20j -3 0, &C. 7« To construct the Lifte of Semitangenfs^ or the Tattgents of half the Arcs, A riile on E, and the several divisions of the arc A D, -vlil intersect the' rndius C A, in the divisions of the semi •or half tangent<5 j mark these with the corresponding fig- ures of the arc A D. U iiC fcermtangents on the plane scales are generally con- tinued as far as the length of the rule, on which they arc Inid, will adrjir ; the divisions beyond 90° are found by dividing tive arc A E like the arc AD, then laying a rule by E and these divisions of the arc AE, the divisions of the semitiingents above '90 degrees v.'lli be obtained on the fine C A continued. 8. To construct the Line (f LorgituJr, Dlvitk AH into 60 equal parts ; through each of these divisions parallels to the radius AC, will intersect the arc AE in as many points ; from E, as a centre, the di- visions jjf the arc EA, being transferred to the chord EA, w-iil give the divisions of the line of longitude. The points thus found on tlie quadrantal arc, taken 'from A to E, belong to the. sines of the equally increas- ing sexagenary parts of the radius j and fho:3e s.rcs, reck- oned Fpf 4l6 ^ GEOMETRY. oncd from E, belong to the cosines of those sexagenary- parts. 9. To construct the Line of Lafiiudes^, A rule on A, and the several divisions of the sines on. CD, will intersect the arc BD. in as many points ; on B, as. a centre, transfer the intersections of the arc B D, to the right line B D ; number the divisions from, B to D v^^ith- 10, 20, 30, &c. to 90 ; and BD will be a line o( latitudes, 10. To construct the Line of Hcurs^ Bisect the quadrantal arcs B D, BE, in /r, h ; divide- the quadrantal arc a b into 6 eq^ual parts, which gives 1 5 de- grees for each hour ; and each of these into 4 others, which win give the quarters. A rule on C, and the several divisions of the arc ab^ will intersect the line MN in the hour, &c. points, which are to. be marked as in the figure* II. To construct the Line of Inclination of Aferidiaiis, ' Bisect the arc EA in c ; divide the quadrantal arc be in^ to 90 equal parts ; lay a rule on C and the several divisions of the arc be, and the intersections of the line HM wili be the divisions of a line of inclinati-on of meridians* ssse^ses END OF FOZUME FIRST. m' V.I