E -V: 11 PIE t A VA T I U S T,1~~~~RI1 O.T', l.If ~Ot~r~;~fS RN~,!eSAESRE, G.- "ZIES, PRINTER, 2;1 STAT'DE-STEET IA OF THE Hog RECEIVED IN ExciiAxGE FROM United Stateas Library C t of -Con EXPLANATIO0 N'HE PRINCIlLES OF IRITIMETIC, A NEW PLAN. 2Y COBRNELL:RIORE', E at R O A T E E M ATI CS I' C tIA CI E D GIX A CA E B I 0 HE IS.T E RE 3'COZN IThEVES, PRINTER, 22 STATEISTREET. Bantered according to Act of Congress, in the year 1850, by CORNELL MOREY,:a:. tio Clerk's Office of the District Court of the Northern District of New-York. T E following pages have been prepared amid the labors of the school-room, and with a view to lessen those labors on the part of the teacher, as well as to aid the pupil in obtaining a thorough and practical knowledge of the principles of ARITHMETIC. That something of the kind is much needed, to acceom plish these' ends, every dutiful teacher at once admitso That no improvement can be made upon this production, theauthor does not clairn: but should the system of teaching which it anticipates be approved, the author will soon present to the public, a work upon this branch of Mathematics, which by itself will serve as a text-book, and which perhaps will be more acceptable than the present volume, The author wishes not to force upon the public' an additional work upon this subject, but he only asks for a careful examination of this prodcetion, and of the system of teaching the science which it proposes, leaving itrs introduction into use to the discretion of teacherso An expression of the conclusion arrived at by those who do thus investigate, will be kindly received by C. MOR EY at Macedon Centre, Wayne Co., N. Yi AGusuT 1, 1850O ARITHMETICo A.riITHMETIC comprises both a science and an art. A science, so far as it involves the relations and properties of numbers, as they exist in nature, and an art, so far as these relations and properties are exhibited and applied by means of characterso NOTATION AND NUMERATiONo.obtation consists in writing numbers by figulre-s iNunzeratioln consists in reading the values of fguures vthen written. To represent the absence of every thing, I write that character, (O)called 1aigh1t To represent that 1 have one unit of any kind, I write that character, (1) called one:....... e O. o O. O O.thr.ee.... ee (3) oo... thlree... o.... e....... ir..... O. 4.0 0 0. 0. 0....(4) oourn.. n e-.. a o five......... o....o. (5) o ofive.......r........ si.o.. (1) e o siMX......... * o e'.:seven.......................(7)....sevenri..... o *.... eight. o.oo... o.i.. o...(S) o o o. eigl-h) r..... O.......nine. nno.....r(9) o oinec Now if I represent every different number by a differyent::lharacter; as we have an i7iznite number of niumbers, I should necessarily have an infinzite number of chiat acter'. Hence it becomes necessary to combine these characters so that they may be made to represent every different number. Therefore if I wish to represent that I have ten units oe any kind, i will do it by writing the 1 one place towards the left, and filling the place it formerly occupied with h naught: thus the value of ten is given to I written n thee second place from the right O Ten and one are eleven Ten and two are twelve Ten and three are thirteen Ten and five are fifteen. Ten and nine are nineteen 9 l~ NOTATION' AND NUMERATKIOVI; As twenlty is composed of two tens I will represent it by writing 2 in ten's place 20'; Twenty and two are twenty-two'22 Twenty and five are twenty-five 25 Twenty and nine are twenty-nine - - 2 As thirty is composed of three tens, I will represent it by writing 3 in ten's place - -30 To represent ninety, write 9 in ten's place 9 Ninety and nine are ninety-nine 9. As one hundred is composed of ten tens, I will. represent it by writing I one place farther towards the left, or in the third place from the -right.0.-. 0e0 We mnay observe that in accordance with the system now introcuced, by removing a figure one place towards the le2ft it is made to represent a value ten times as greaLt The next figure at the right is called Units The n.ext is ten times units, called Tens ~ The next is ten times tens, called Hunlrreds c> The next is ten times hundreds, called Thousands Thoe next is ten times thousands, called Tens of T hou0. The next is ten. times tens of thousands, called.:~iun dred ofqf Thousands The next is ten times hundreds of thousands, called Miions c< r The next is ten times millions, called Tens of it illions The next is ten times tens of millions, called flunidreds of "MTiZllionzs c The next is ten times hundreds of nmillions, called Billions c The next is ten times billions, called Tens of Billiotns, The next is ten times tens of billions, called Hundreds Cf Biillions, The next is ten times hundreds of billions, called Trillions cf The next is ten. times trillions, or Tens of Trillions &.;, &e S, &c. 9Now as we do not read farther than Hundreds: is wn NOTATION AND NUMIERATIOND( denomination, and as to represent Hundreds requires three figures, numbers become divided into periods of three figurls each. The first period at the right is called Units, from the Latin word unus, meaning one. The second and third periods have no appropriate names. The fourth period is called Billions. The prefix BiL means two, and the number of the period may be found by adding two to the meaning of the prefix of the name. The fifth period is called Trillions. The prefix Tri, means three. The number of this period may be found by adding two to the meaning of the prefix of the name. The sixth period is called Quadrillions. The prefix Quad. means four. The number of this period may be found by adding two to the meaning of the prefix of the namne. And in general, for all periods at the left of the third, having the name given to find the number, add 2 to the meaning of the prefix of the name. And having the number of the period given, to find the name, take 2 firom the number of the period, aud we have the meaniung of the prefix of the name. 9til th 7th 6th 5th 4th 3d Cd 1st Sept, Sex. Quiit. QOuad. Tri. Bi. Mill. Thou. Units, 222 22 2 222 22.2 222 222 222 222 222 Suppose that I am required to arrange in accordance with this system these numbers, Quadrillions 40 ) In this example the prefix Sex, Sextillions ~ 364 has the greatest value; hence the Trhousands..23' Sextillion's period will be at the Billions - - 489 1 left hand. And Sex. meaning sin, -Units - 376 6 the period will be the eighth. The next prefix in value is Quad., which meaning foul,, the number of the period will be six, and as each period must occupy three places, 1 must write a naught at the left of the 40 to fill the vacant place. There being no Quin-:il!ions given, the seventh period must be filled with naughts. There being no Trillions given, the fifth period must also be filled with naughts. ADDITION. The next prefix in value is Bi., which meaning two, the 489 must be written in the fourth period. There being no Millions given, the third period must be filled with naughts. 2The 23 must be written in the second period, and a naught be placed at the left of it that the period may occupy three places. The 376 being units, must be written in the first period. Therefore, we have for the arrangement, 8th 7th 6th 5th 4th 3d 2d 1st 364 000 040 000 489 000 023 376 The nine figures that have value are called significant iigu res. They are also sometimes called the nine digits. The value a figure represents, when standing alone, is called its simple value. The increased value which a figure represents, when others are written at the right of it, is called the local valoue~ AD D IT IO N Addition consists in uniting several numbers into one, I wish to add several numbers. 3588 As 1 cannot add units of one denomination to 217 units of a different denomination, I have written 4523 the numbers so that units of the same denomination 5460 shall fall under each other. As the values of 13794 figures increase fiom the right hand towards the left, I will commence at the right hand to add. Six units, and three, and seven, and eight units are twenty-four units-equal to four units and two tens. I will write the 4 under the column of units, and add the 2 to the column of tens, that I may have all the tens collected together. Two tens and six, and two, and one, and eight tens are nineteen tens,-equal to nine tens and one hundred. I will write the 9 under the column of tens, and add she 1 to the column of hundreds, that I may have all the SUBTRACTION;' ca hundreds collected together. There are in all seventeen hundreds,-equal to seven hundreds and; one thousand. I will write the 7 under the column of hundreds, and add the I to the column of thousands that I may have all the thousands collected together. There are in all thirteen thousands, which I will write down. The result obtained is called the amount or sum. Hence to add numbers:Write them so that units of the same denomination shall fall under each other. Commence at the right hand and add each column separately. Write down the right hand figure of each amount, arid add the left hand figure or figures to the next column. Thus proceed to the last column, where write down the whole amount. SUBTRACTION Subtractiont consists in- finding- the, dif/irencea between two numbers. From 7854) The greater number is called the I wish to take 32S81 Minuend. 4578 The less numberthe Subtrahend. The difference between them is called the Remainder~. As we cannot take units of one denomination fi'om units of a different denomination, I have written the numbers so that units of the same denomination shall fall under each other. As the values of figures increase from the right hand towards the left, I will commence at the right hand to subtract. One unit taken fiomi four units leaves three units. Eight tens from five tens I cannot take. I will add ten tens to the five tens, making fitteen tens. Eight tens from fifteen tens leave seven tens. Now as- I have added ten tens or one hundred to the Minuend, the Remains der will be too large by one hundred, hence I must take. away from the Minuend one hundred more than I at first intended to; that is, I must take away three hundreds itO MULTIPLICATION. instead of two hundreds. Three hundreds taken froum eight hundreds leave five hundreds. Three thousands taken fiom seven thousands leave four thousands..Hence to take one number from another: Write the less number under the greater, so that units of the same denomination shall fall under each other. Commence at the right hand, and take each figure in the lower line from the one above it. If the value of a figure in the lower line is greater than the one above it, add ten to the upper figure and then subtract: then add one to the next left hand figure of the Subtrahend, and thus proceed. It will appear evident, by examination, that any number may be added to the figure in the Minuend, that will render it of greater value than the figure in the Subtrahecd. But to avoid fractions the number ten is used. 4ULT IPLICATION. 4 ~ I wish to obtain four times 3459. As the values of figures increase from the right hand towards the left, I will commence at the right hand. Four timnes nine units are thirty-six units, equal to six units and three tens. I will write down the six under the units, reserving the three tens. Four times five tens are twenty tens and the three tens reserved being added, give twentythree tens, equal to three tens and two hundreds. I will write the 3 under the tens, reserving the two hundreds. Four times four hundreds are sixteen hundreds, and the two hundreds reserved being added, give eighteen hundreds, equal to eight hundreds and one thousand. I will write the 8 under the hundreds, reserving the one thousand. Four times three thousands are twelve thousands, and the one thousand reserved being added, gives thirteen thousand, which I will write down. Hence the result is 13836. MULTIPLICATiON- Ii This operation is called MULTIPLICATION. Hence Mulb tiplication is taking one number as many times as there are units in another. The number to be multiplied is called the IMuZtiplicand, The number by which we multiply is called the Alulti. plier, and the result obtained is called the Product. Both Multiplier and Multiplicand are called Factors. When the Multiplier does not exceed twelve; —Write the Multiplier under the right of the Multiplicand. Multiply each figure of the Multiplicand by the iMultiplierg commencing at the right hand. Set down the right hand figure of each partial product, and add the left hand figure or figures to the product of the next figure of the Multiplicand. When the iffultiplier exceeds Tzoelveo I wish to multiply four thousand two hundred and fiftysix by three hundred and twenty-five. 42.56 ) First I will take five times the Multipli325 I cand; which gives thisline (21280.) I will 21280 t take two times the Multiplicand which gives 8512 this line (8512,) —but as'the Multiplier is 12768 | tens, I must write the Product one place towards the left. I will next take three times the Multiplicand, which gives this line (127658,)-but as the MAultiplier is hundreds, I must write the product in hundreds' place. Now this line (12768) is three hundred times the Multiplicand, this one (8512) is twenty times, and this one (21280) is five times the Multiplicand. Hence if I add them all together, I shall have three hundred ancl xtwenty-five times the Multiplicand. Therefore when the Miultiplier exceeds twelve,-_ Write the Multiplier under the Multiplicand. Miultiply by each figure of the'Multiplier separately, placing the right hand figure of each partial product directly under its MAlultiplier. Then the sum of the partial products will be the entire producto MULTIPLICATION. WFhen the Multiplier is a. Composite Number. I wish to multiply 4567 by 24. 4567 - First I will multiply it by 8. Now as the 8 multiplier is only one third part as large as it 36536 - should be, the product is only one third of what 3 1 it should be; hence, to make it what it should 109608 J be, I must multiply it by three. Such a number as 24, which is composed of two or more factors, is called a Composite number, and the factors of which it is composed are called the Component parts. Hence when the Multiplier is a Composite number;WYe may first multiply by one of those component parts, and then that product by another, and thus proceed until all the factors have been used. The last product will be the product required. Wh'/en there are Cyphers at the right of the Multiplier. I wish to multiply 3467 by 2000. 8467 -' If I multiply by the 2; as the multiplier 2000 5 will have been only one thousandth part as d934000 large as' it should be, the product will be only one thousandth part as large as it should be,-hence to make it what it should be, multiply by one thousand, which is evidently done by annexing three naughts. Hence mnultiply by the significant figures of the multiplier, and annex to the product as many naughts as are at the right of the multiplier. It is evident that -when there are cyphers at the right hand of both the factors, they may all be omittted until we shall have multiplied the significant figures, and then annex to the product, so obtained, as many naughts as there are At the right hand of both the factors. 'D IV IS IO N 4) 3496 - Suppose thatIwishtodivide 3496 Apples 874- among four individuals, and to find the'number that I can give to each. Suppose that I have them m:rranged in heaps, three of which contain one thousand each, four heaps with one hundred in a heap, nine heaps with ten in a heap, and six single apples. Not having as many heaps of one thousand in a heap as there are individuals, it is evident that I cannot give then one such heap a piece, hence I will cha nge them into heaps,of the next smaller size. One of the heaps with one thousand'in a heap, is equivalent to ten heaps with one hundred in a heap, and three are equal to three times ten or thirty, and four heaps added give thirty four heaps with one hundred i'n a heap. Now, if I go arotund to the four individuals and give themn one heap a piece, it will take four of them, and if I pass around eight times it will take eight times four, or thirty-two heaps, and I shall have two heaps left. Biut as there are not enough heaps left to go around again, I will change the remainder to heaps of the next sinaller size. One of the heaps with one hundred in a healp, is equivalent to ten heaps with ten in a heap, and two such heaps will be equal to two tihmes ten or twenty, and nine heaps added give tvwenty-nine heaps with ten in a heap. Now, if I go around to the four individuals and give them one heap a piece, it will take four of them, and if I pass around seven times, it will take seven times four, or toenty-eight heaps, and I shall have one heap left, which contains ten single apples, and the six apples added give sixteen single apples. Tow, if I giive them one of these a piece, it will take four, and if I give them four apiece it will takefour timesfiour, and there will be none remaining. This operation is called DIVIsIONo Hence Division consists in finding the number of times one number is contained in another. The number to be divided is called the:Dividend 14 3irDVIVS'ION, The number that we divide by is called the Divisor and the resultobtained is called the Quotient. I observe that I first found the number of times the Divisor is contained in as many of the left hand figures of the Dividend as are necessary to contain it, and I had a remainder of two. Then in changing to heaps of the next smaller size, and adding in those next smaller, I obtained 29, the same that I should have obtained by prefixing the remainder to the next figure of the Dividend. Hence the rule. FOR SHORT DIVISION, OR Mhen the.Divisor does not exceed Twelve. Write the Divisor at the left hand of the dividend, and find the number of times that it is contained in as many of' the left hand figures of the dividend as are necessary to contain it, and prefix the remainder (if any) to the next figure of the dividend, then divide, and thus proceed until all the figures of the dividend shall have' been used. TVhen the Divisor exceeds Twelve. 24)6864(286 Suppose that I wish to divide 6864 Apples 48 among 24 individuals, and to find the 206 number that 1 may give to each. Suppose 192 that I have themn arranged in heaps, six 144 of which contain one thousand each, 144 eight heaps with one hundred in a heap, six heaps with ten in a heap, and four single apples. Not having as many heaps of one thousand in a heap, as there are individuals, it is evident that I cannot give them one such heap a piece, hence I will change them into heaps of' the next smaller size. One of the heaps with one thousand in a heap is equivalent to ten heaps with one hundred in a heap, and six such will be equal to six times ten or sixty, and eight heaps added give sixty-eight heaps, with one hundred in a heap. Now, if I go around to the twenty-four individuals, and give them one heap a-piece, it will take twenty-four of DIVISION. 15 them, and if I pass around twice, it will take twice twentyfour or forty-eight heaps, which being taken from the first number, sixty-eight, leaves twenty heaps with one hundred in a heap, remaining. But as there are not enough heaps left to go around again, I will change the remainder to heaps of the next smaller size. One of the heaps with one hundred in a heap is equivalent to ten heaps with ten in a heap, and twenty such are equal to twenty times ten, or two hundred, and six heaps added give two hundred and six heaps with ten in a heap. Now if I go around to the twenty-four individuals, and give them one heap a piece, it will take twenty-four of them, and if I pass around eight times it will take eight times twenty-four or one hundred and ninety-two heaps,'which being taken from the two hundred and six, leaves fourteen heaps with ten in a heap remaining. As there are not enough heaps left to go around again, I will change the remainder to heaps of the next smaller size or to single apples. In one of these heaps there are ten single apples and in fourteen of them there are fourteen times ten or one hundred and forty, and four added give one hundred and forty-four si ngle apples. Now, if I give them six of these a-piece, it will take one hundred and forty-four, and there will be no remainder. HENCE FOR LONG DIVISION, OR, When the.Divisor Exceeds Twelve. Write the Divisor at the left of the dividend, and find the number of times it is contained in as many of the left hand figures of the dividend as-are necessary to contain it: place the result at the right of the dividend, as the left hand figure of the quotient,'multiply the divisor by this quotient figure, and subtract the product from that part of the dividend used: to the remainder annex the next figure of the dividend, then divide again, multiply and subtract as before, and thus proceed until all the figures of the dividend have been used, Id ZDvISION, Wh/en the Divisor is a Composite nublpers )8640 Q 6 )1080o I wish to divide 8640 by 48. 180. First, I will divide by 8. Now, as the dlivisor has been only one-sixth part as large as it should bey the quotient s,,ix times as large as it should be, hence, to make it what it should be I will divide this quotient by 6. Hence, when the Divisor is a composite number-we may divide first by one of the component parts, and that. quotient by another, continuing the operation until all the factors have been used, it is evident from the system of Notation, that to divide by 10 or 100, &e., we have only to strike off from the right' hand of the Dividend as many figures as there are naughts at the right hand of the divisor. Thus, dividing 768 by 10 gives for a quotient 76, and a remainder 8. And 8642 divided by 100 gives 86 for a quotient, and a remainder 423 When there are Naughts at the right of the Divisor. 20) 3457/ Suppose I have 3457 to be divided by 172+17 S 20. The Divisor is composed of two factors, 2 and 10. To divide by 10 otrike off the 70. Then divide the remaining part of the Dividend by 2, I have I for a, remainder, which is evidently of an order of units next above those of the 7. And as the remainder obtained by the second division will always be of the order of tens, compared with the remainder obtained by the first division, we have the following Rule: When there are Naughts *at the right of the Divisor. Strike off the naughts, and strike off as many figures at the right hand of the Dividend, then divide the remaining part of the Dividend by the significant figures of the Divisor. If a Remainder occur, annex to it the figures cut off front the Dividend, and we have the true Remainder. DIVISION. 17 Afier several Divisors have been used, tojind the Remainder in units of the same denomination of the original Dividend. 4 )35678 2 5 ) 8919+21 i 4+4 16 9 )1783+14 EXAMPLE. 5+4+1 20 7) 178341 I I 9+5+4X2-360 7)_198+11 - 28+2 J L Entire rem. 398 I have divided the number 35678 by the factors of 1260, and have found several Remainders. I wish now to find the true Remainder in units of the same denomination of the original Dividend. The remainder 2 is a part of the original dividend, that has not been divided, hence it will forim a part of the true remainder. The second remainder 4 is a part of the second dividend,-but as the second dividend is found by dividing the first dividend by 4, this remainder must evidently be multiplied by 4I The third remainder I is a part of the third dividend,-but as the third dividend is found by dividing by 4 and 5, it is evident that this remainder must be multiplied by 4 and 5 to change it to units of the same denomination of the original dividend. The fourth remainder 2 is a part of the fourth dividend, but as the fourth dividend is found by dividing by 4, 5 and 9, it is evident that this remainder must be multiplied by 4, 5, and 9, to change it to units of the same denomination of the original dividend~ Notice that the last remainder is multiplied by all the divisors before it, except its own, by the 9, 5, and 4, and the same is true of each remainder. Hence the rule. Multiply each remainder by all the divisions before it, except the one immediately preceding, add the several products, thus obtained, together, and their sum will be the true remainder in units of the same denomination of the original dividend. 2 Af. i ~ itREDUCTIO N1, REDUCT I ON. Reduction consists in changing the number of units of one denomination to that of another, without changing their value~ 1 wish to find what number of pence there ~ s. d. are in 12 8 7 20 248 12 2983 As there are twenty shillings in one pound, there will'be twenty times as many shillings as pounds. Hence I will multiply the number of pounds by 20, to find the number of shillings. Twenty times twelve give two hundred and forty, and adding the eight shillings, we have twvo hundred and fcrtyoeight shillings. Now as twelve pence make one shilling, there rwill be twelve times as many pence as shillings. Hence I will now multiply the number of shillings by 12, to find the number of pence. Twelve times two hundred and forty-eight give 2976, and adding the seven pence, we obtain 2983 pence for the result. This operation is called REDUCTION. When, as in this example, units of higher denominations are to be reduced to units of lower denominations, it is called Reduction Descending, to perform which, commence with the highest denomination and multiply by the number of units that it takes of the next lower denomination to make one of that, adding in the units of the next lower denomination, it there be any, and thus proceed until the quantity is reduced to the denomination required. When units of lower denominations are being reduced to units of higher denominations, the operation is called Reduction Ascending, and as it is the reverse of Reduction Descending, we shall, evidently, have the following rule: Divide the given number by the number of units of that lower denomination required to make one of the next higher, and thus continue to the denomination required. IF ADDITION OF COMPOUND NUMBERS. i9 remainders occur, they will be of the same denominatiorn as the dividend fromr which they are obtained. ADDITION OF COMPOUND NUMIBERSo Compound Numbers are such as represent units of different denominations, and do not increase or decrease in a tenfold proportion. A s. d. far. I wish to find the sun of r 5 19 7 2, 6 17 8 1 I 9 12 3 3 -22 9 7 2 As the values of these numbers increase towards the left, I will commence at the right hand to add. Three and one and two farthings give six farthings. Four farthings make one penny-hence, in six farthings there are as many pence as 4 is contained times in 6. It is contained once, and there are two farthings remaining, which I will write under the column of farthings, and I will add the one penny to the column of pence, that I may have all tile pence collected together. One and three and eight and seven pence give nineteen pence. Twelve pence make one shillingshence, in nineteen pence there are as many shillings as 12 is contained times in 19. It is contained once, and there are seven pence remaining, which I will write under the column of pence, and I will add the one shilling to the column of shillings, that I may have all the shillings collected together. One and twelve and seventeen and nineteen shillings give forty-nine shillings. Twenty shillings make one poundhence, in forty-nine shillings there are as many pounds as 20 is contained times in 49. It is contained two times, and there are nine shillings remaining, which I will write under the column of shillings, and I will add the two pounds to the column of pounds, that I may have all the pounds collected together. 'IO ~ISUBTRACTION OF COMPOUND NUMBERSt Two and nine and six and five pounds give twenty-two pounds. Therefore the whole amount is ~22 9s, 7d. 2far, Hence, to Add Compound Numbers:'Write them so that units of the same denomination shall fall under each other. Commence at the right hand, and add the units of each denomination separately, dividing each amount obtained by the number of units of that denomina-tioan required to make one of the next higher. Write down the remainder, (if any,) and add the quotient to the next higher denomination, with which proceed as before. SJBTRACTION OF CO iPOUND NUMBERS. JJ S, dl, Ja.''From 9 5 3 i: wish to take 3 9 8 2 5 i7 9 i Commencing at the right hand two farthings firom three Larthings leave one farthing. Eight pence from five pence I cannot take. I will add twelve pence to the five poece, naking seventeen pence. Eight pence fiom seventeen pence leave nine pence. But, now, as I have added twelve pence or one shilling to the minuend, the remainder Nwill be too large by one shilling; hence, I must take away from the minuend one shilling more than I at first intended to. That is, instead of taking away nine shillings, I must take away ten shillings. But I cannot take ten shillings from seven shillings. I will add twenty shillings to the seven shillings, making twenty-seven shillings. Ten from twenty-seven leaves seventeen shillings. But, now, as I have added twenty shillings or one pound to the minuend, the remainder will be too large by one pound; hence, I must take away from the minuend one pound more than I at first intended to. That is, instead of taking away three pounds, i must take away four pounds. Four pounds fiom nine pounds leave five pounds. IMULTIPLICATION OF COMPOUND NUMBERS. 2 Hence, to Subtract Compound Numbers: Write the less numberr under the greater, with units of the same denomination under each other. Commence at the right hand and take each number in the lower line from the one above it. When the number of units in an4y denomination of the subtrahend exceeds the number in the same denomination of the minuend, add to that denomination of the minuend as many units as make ene of the next higher denomination,, and then subtract;. after which, add one to the units of the next higher denomination of' the subtrahend:, and thus. proceed. MULTIPLICATION OF COMPOUND NUBil ERS11 ~ s. d. I3 wish to multiply 8 13 5 by 7 7 60 13 11 Xs the values of the numbers increa'se from the right towards the left, 1 will commence at the right hand. Sev. en times five pence give thirty-five pence. Twelve pence make one shilling; hence, in thirty-five pence there are as many shillings as 12 is contained times in 35. It ias contained twice, and there are eleven pence remaining, which I will write under the pence, reserving the two shillingso Seven times thirteen shillings give ninety-one shillings, and the two shillings reserved being added give ninety-three shillings. Twenty shillings make one pound; hence, in ninety-three shillings there are as many pounds as 20 is contained times in 93. It is contained four times, and there are thirteen shillings remaininlg, which I will write under the shillings, reserving the four pounds. Seven times eight pounds give fifty-six pounds, and the four pounds reserved being added, give sixty pounds, which I will write down. HIence, to Multiply Compound Numbers: Commence at the right hand and multiply the units.'r 22 DIVISION OF COMPOUND NUMBERS. the lowest denomination by the multiplier. Divide the product thus obtained by the number of units of that denomination required to make one of the next higher. VWrite down the remainder, and add the quotient to the product of the next higher denomination. Divide as before, and thus proceed to the highest denomination, where write down the whole product. VWhen the uITltiplier is a Cozlmposite Number. it is evident that we may use its factors, in the same manner as in simple numbers. W;41/en the.31iltipllier is large, and is not a Comp)osile tNumber. Multiply the price of a unit by 10, to obtain the cost of ten units; then multiply this product by 10, to obtain the cost of one hundred units; then multiply the cost of 100 units by the number of hundreds, the cost of 10 units by the number of tens, and the cost of one unit by the number of units; and the sum of these several products will be the product required. DIVISION 0F COMPOUND NUAtBE RS ~ s. d. 4 )38 9 8 9 12 5 Suppose that I \wish to divide ~38 9s. 3d. amnong four individuals, and to find the sum that may be given to each. If I go around to the four individuals and give them one pound apiece, it will evidently take four of them, and if I pass around nine times, it will take nine times four pounds, or thirty-six pounds, and I shall have two pounds remaining. But as there are not enough pounds left to go around again, I will change this remainder to shillings. There are twenty shillings in one pound, hence there will be twVenty titmes as many shillings as there are poundsa DIVISION OF COMPOUND NUMBERS.c ~.'PTwenty times two give forty, and the nine shillings added give forty-nine shillings. Now, if 1 go around to the four individuals and give them one shilling apiece, it will take four of them, and if I pass around twelve times it will take twelve times four, or forty-eight shillings, and I shall have one shilling left, which is equal to twelve pence, and the eight pence added give twenty pence, of which if I give them five pence apiece there will be none remaining. Hence, I can give to each of the four individuals a9 12s. 5d. Hence, for the Division of Compound Numbers Write the divisor at the left hand of the dividend, and find the number of times it is contained in the units of the highest denomination; write the quotient underneath, and multiply the remainder, (if any,) by the number of units of the next lower denomination required to make one of this higher, and add to this product the given units oL the next lower denomination, (if any,) then divide by the given divisor, and thus proceed to the lowest denomination. It is evident that when the divisor is a composite number, the component parts may be used, as in simple numbers. WVhen the divisor is large, and is not a composite number, represent the work after the manner of Long Division, instead of retaining the whole in the mind-thus: ~ s. d. ~ s. d. 257 )561 2 4 (2 3 s 514 47 20 942 771 i71 12 2056 2056 24 DEFINITIONS OF NUiMBERS. An odd number is a number that cannot'be divided by 2 without a remainder; as 3, 7, 13. An even number is a number that can be divided by 2 without a remainder; as 6, 10, 14.. All numbers are either odd or even. The sum or difference of any two odd numbers, or of any vwo even numbers, is an even number. But the sum or difference of an odd and even -number is an odd number. The sum of any number of even numbers is an even number; hence, if an even number be multiplied by any number whatever, the product will be even. The sum of an even number of odd numbers is an even number, but The sum of anl odd number of odd numbers is an odd number; hence, The product of two or more odd numbers is an odd number. A product is even if one of the factors which compose it is even. An odd number cannot be divided by an even number without a remainder. A number is said to divide another when it is contained inf that numnber without a remainder. A prime number is a number that can be divided only by itself aid a unit; as a7, 1, 19. A factor of a number is one of the numbers which being multiplied together will produce the given numberl Thu: 3 and 7, are the factors of 21. A prime factor is a factor of a number which can b,e divided only by itself and a unit. A conzmosite number is a number that is composed of twt:o or more factors. A priime number cannot be resolved into factors. A11 numbers are either prime or composite. A composite number is resolved into its primene fJctors. when each of the factors intoe which it is resolved is a prime number~ O; FiAC'TORS IN DiVISIOo N5 -'Lhe number 15 is composed of the factors 3 and 5 if we;'.ivide it by 3, we obtain the 5 for a quotient; that is, wv r2ejoct ill the dividend a factor the size of the divisor. Agnain; the number 28 is composed of the factors 4 and,-wif Nwe divide it by 4 we obtain the 7, —that is, we reject in the dividend a -factor the size of the divisor, and-the renmaining fictor is the quotient. And in general, Division may be said to consist in striiking out of the dividend a factor the size of' the divisor. And M{fultiplication may be said to consist in combining,with the multiplicand such factors as occur in the muldiplier. Hence,-it will appear evident, that if we multiply by a n-unmb'Ol, and then dcivide the product by the samee number, 1Te vill cot change the value. Again; if we are to multiply and then to divide the:);r'Oduct by a number eontanaing a factor common to both nmltiplier and divisor, that factor may be omitted. Again; as division consists in striking out of the divi.dend a factor the size o[ the divisor, it is evident that a dividend cannot contain a divisor, unless there is a factor of that divisor in'the dividend. For instance, 15 will contain; as a divisor, for 3 is one of the faletors of 1.5. But 21 w;il'l no contain b as a divisor, for b is not a factor in 21. Again; it will appear evident, that if a divisor is a composite numnbeir, the dividen-d which contai-rs it must be made Hp in parl, at least, of the prime factors of that divisorthat is, in whatever divideatld a divisor is contained, the rime factors of that divisor are contained in the same dividend. Again; it wxill be observed that the quotient is always the faictor which is left after stri'king out of the dividend a factor the size of the divisor. Hence, if we multiply the dividend by atny factor, we increase the value of the quotient by' the same factor-and if we divide thle dividend, we diminish the value of the quotient. If we nlultiply the divisor, we diminish the value of the quotient-for in performing the division, we reject more 2.D,()6 ~ THE LEAIST 1COMMON:MULTIPI,,:factors in the dividend, and hence have less left:fr thae q( uotient. And if we diminish the divisor by any factor, we increase the quotient by the same falctor. if we multiply or divide both divisor and dividend by ihe same number, we shall not affect the. value of tahe q Iuotient. From f he preceding remarks, it is pl'in that the dividend is the product of the divisor and quotient- tat the ditisor nmay e fobund by dividing the dividend by the quotient —~htat if weo multiply together the factors.iven, we obt ainr)'a products —-and that if we divide the, product of an ny ILiumber of factors by tihe product of a part of those fa.ctors, 7( ihall1 obtain the product of ti1e remainin i tactors or qiuotiernt-sand if we divide the: product of any nnumber O"!',hortors by the product- ot all but one oit Theml we saiii obtain that one for a quotient.'in) c3 A 3.,E:~ IE,L Ti0:'"., 1 2 1 5 3X3X2X2X5-1 278 least comm-,n multiple. X. multiple of a number is a nunber that nosy be di; ide hy that number without a remainder. T'hus, 12 is am.i'i, tiole of 4. 8 is a multiplee of 9, Oc A common multiple of two or more numbers is such a:lumber as may be divided by each of them, separaiebif without a remainder, Thus 24 is a common multiple of 4, 3, and 6, THIE LEA ST CO01iMMION MULTITP LE 2E elest comm on multiple of0 several numbers is the Icast number thzat may.5 be divided by each of them without. remtaindero'hus 241' is the least common multiple of T61,:~, and 8o i wisvt tos ufi th o least commton multiple of the number:'t,!. 2 9 ai:d 5. [ I[ f i-du a nu1mber that may be divided by each of these nunebessv witrhout a remntinder, it n-,ust be composed ox the ririer f'actors' of eacth of these numbl-ers. _. J lld the wCes.t -znibUe.r thn, tnav be divided by each of' the; w i-iihout a remainderi, t must be made up of tie primle fattors of these numbers taken- but once each/ Hence if' arl1V two oLr norne of these numbers contain a common prime zaictC., th at factor must be used once and but once. T'here ore for the purpose of eliminating such supertiuo asI fc'tors as may exist in thlese numlber s, I wi ll divide by any' pIrime actor that is common to two or more of themn-re_ serving that factor once in the divisor, and throwing it out of eaci of the numbers in which it occurs.! notice that a factor 3 is comrnmon to three of the numbers, heince I' will divide by 3g. I notice that there is aczoiher factor 3 corntmon to two of the numbers; hence I.,;,ill divide again by 3, There is yet a factor 2, common. to iwo of the numbers, hence I will now divide by 2, There are now no two numbers that contain a Common af toro Hence, I conclude, as I have saved each of'thtcprime factors of these numbers once and buft once, that t}he ploduct of the divisors, quotients and undivided numbers,,viill ib the least common multiple. [-_enc to find the least common multiple of two or more Ylum b rs -s ~ Write the numbers in a line, and divide by any prime) number' that is an exact divisor oc twd or more otf ilem write the quotients and uindivided numbers as a second horizontal line, with which proceed as before —continue the division until no two of the numbers contain a cornmon), factoro Then the product of the divisors, quotients and undivided num:e:ss will be the least common amultiple re, (uired& ,,, ~T-lIIE GREiATEST CO1iMM;ON DITVSOhE:, THIE GREAT'EST COM1I0N BDIMISRW s22 s2;j-) 6(3 46 16 J 616 50 42 20 5 1 2)25 2( 36i 224 23)1 12(4 112 6 1G=2522q- T112:,I A number that is contained in two or more numbers ay-, exact number of times, is called their Commzzona Divisoro And the greatest number that is contained in two or Inore umbers an exact number of times is s called their Greateat Common Divisor. I wish to find the greatest common dlivisor of two numrn bers, as 252 and 616. 1 will first illustrate a principle on.vwhich this demonstration depends. If any ntmbe (as 30) be separated into two parts, (as 16 and 20,) andc each o'c?.he parts and the entire number be divided by tne same divisor, the sum of the partial quotients thus obtained will hje equal to the entire quotient. And if the'e-irle qiotientl and one of the partial quotients are whole numnbers the othler partial quotient must be, for no proper friaction added to a whole number will give a whole nunmuber. A1nd if both the partial quotients are whole numlers, the entire quotient. mnust be, for no two whole numlbers when addelid give a fr-action. I. know that no number lagerl than 252 can be tile greatest common divisor of 25,2 and 616, for no'nunmbe greater than 252 is contained in 252-. Hence. if 252 is a n exact divisor of 616, it is the greatest commolnn divisor of' the two numbers. I find by dividing, that there is a remainder, (112,) hence, 2'.52 is not that greatest common divisor. Now if I can show that the greatest common divisor of the less number and remainder after division, is the greatest comr. nmon divisor of the gr ea.ter and less, I will thlen proceed U!U';a, THE GREATEST COtiMMON DIVISRo D!': kind the greatest common divisor of the less number and remainder, and it will be the greatest common divisor i equired. First, I will show that.the greatest common divisor of the greater and less is -a common divisor of the less and remaindero The greater is always made up of some multiple of th'e less; plus the remainder. It is a divisor of the greater by lhypothesis; it is also a divisor of the less by hypothesis, and it is evidently a divisor of any multiple of the less hence, I have only to show that it is a divisor of the remainder. It must be a divisor of the remainder, for the sunm of the partial quotients is equal to the entire quotient, and the entire quotient and one of the partial quotients are whole numbers: hence, the other partial quotient must be a( whole number, for no proper fraction added to a whole number will give a whole number..Next3 I v ill show that the greatest common divisor of Lhe iess and remainder, is a common divisor of the greater and less. it is a divisor of the less by hypothesis-it is evidently a divi:sor of,ny multiple of the less-and it is a divisor oft the remainder by hypothesis,-henee, it must be a divisor'f the greater; for, the sum of the partial quotients is equd. to fhe entire quotient, and both the partial quotients being whole numbers, the entire quotient must be a whole number, for no two whole numbers added will give a fraction. Now I claim, that the greatest common divisor of the less and remainder, is the greatest common divisor of the greater and less. For suppose the greater and less have a c.ommon divisor, greater than the greatest common divisor o'the less and remainder, —-it could not be contained in the less and remrainder —but I have shown that it nlust be contained in them. Hence, the greaiter and less cannolot have a common divisor grealter than -the greatest common divisor of the less and remainder. Again-suppose thle greatest common divisor of the greater and less to be less than the greatest common divi, sor of the less and remnainder-thern the greater and less B)i" THE GREATEST COMMON DIVISOgo could not contain the greatest common divisor of the 1ie0S1 and remainder,-.-but I have shown that they must contaill it. Hence, I conclude that the greatest common divisor of the greater and less cannot be either greater or less than the greatest common divisor of the less and remaindertherefore it must be the sanme numrber7 Hence, I will now proceed to find the greatest common divisor of the less and remainder, and it will be the greatest common divisor sought. I know that no number greater fthan 112 can be thrat greatest common divisor, for no number greater than 112 is contained in 112; therefore if it'is an exact divisor of 2523 it is the greatest common divisor sought. - find there is a remainder, (28.) Now. consideling: this 1l12 as the less and 28 as the''emainder, I will proceed to find their greatest common divisor. Dividing 112 by 28 I have n o remainder; hence, 28 is the greatest common divisor of 28 and 112; therefore? from what has been shovn, it must be the greatest common divisor of 112 and 252; hence, 28 i*s also the greatest common divisor of 252 and 616. Therefore, to find the greatest common divisor of two numbers Divide the greater.~umber by the' less, and the Iast diviso: by the last remainder, continuing the operation until. odthing remains; and the last divisor will be the greatest common divisor, To find the greatest common divisor of m ore than two, numbers 1?irst find the greatest common divisor of two of them., and then of that common divisor and another of the numn 1bers thus continuing until all the numbers huave been us ed VULGAR P ACTiCONSo 3i. VULGAR FRACT IONS. Introduction to Fractions. 8 7 5 $ -,and 3;- and a-; 3-7; of.; -9 7 4d If I wish to represent that I have a unit of any kind divided into eight equal parts, I will do it by writing the figure 8. If I wish to represent that I have a unit divided into seven equal parts, I will do it by writing the figure 7o If I wish to represent that I have a unit divided into five equal parts, I will do it by writing the figure 5. If I wish to represent that I have a unit divided into three equal parts, J will do it by writing tile figure 80 If of these eight equal parts I wish to represent thit I taKle three, I will do it by writing 3 above the 8 with a line dr:awn between them. If of the 7 equal parts I wish to represent that I take two, I will do it by writing 2 above the 7, with a line between them. If of the five equal parts I wish to relresent that I take four, I will do it by writing 4 above the 5, with a line between them. if of the three equal parts I wish to represent that I take one, I will do it by xwriting I above the 3 with a line between them. The number below the line in each case, shows the number of parts into which the unit is divided. And as these. parts take their name from the number into which the unit is divided, the number below the line becomes the namer of the parts-hence it is called the Denominator. The number above the line, shows the number of the parts that is taken, hence it becomes a numberer, and is therefore called the rnumerator. The whole expression represents a part of a unit, and hence it is called a Fraction. Both Numlierator and Denominator are called ter:s of; the fraction. Vt LGAit Pt1ACTtoNS WThen the numerator is -less than the denominator, the expression represents a parl of a unit —it is then called a proper fraction. As 4 and {, (read five-sevenths and threefifths. ) When the numerator is equal to, or greater than the denominator, the expression denotes something more than a unit, hence it is then called an improper fraction. As - and 8, (read six-sixths and eight-thirds.) W hen we have a fi'action joined to a whole number, the expression is called a mzixed number. As 3t, (read three and five-sevenths.) A fraction of a fraction is called a co7mpound f.raction. As r: of. A fraction having a fraction for either its numerator or denominator, or both, is called a complex fraction. As i. 7, U- -t, (read three-fifths divided by seven, three divided by four-sevenths, and six-elevenths divided by nine-thirteenths.) If I have a unit divided into eight equal parts, and 1 wish to take one-half that unit, it is evident that I must take oneohalf of the number of parts into which the unit is divided. That is, the numerator, which shows the number of parts I take, must be one-half of the denominator, which shows the number of parts into which the unit is divided. If I have a unit divided into twelve equal parts, and I wish to take one-fourth of that unit, it is evident that I must take one-fourth of the number of parts into which the unit is divided; that is, the numerator, which shows the number of parts I take, must be one-fourth of the denominator. If I have a unit divided into twenty equal parts, and I wish to take one-fifth of that unit, it is evident that I must take one-fifth of the number of parts into which the unit is divided; that is, the numerator, which shows the number of parts I take, must be one-fifth of the denominator. And I conclude that, in general, the nzumerator must be VULGARi FRACTIONS.- 33 such a part of the denominator as the valute of the fractio-n is inzended to represent. If I have units divided,into four equal parts each, and I wish to take enough of those parts to make three whole ones, it is evident that I must take three times as many rarts as compose one unit-that is, the numerator must 4e th'ree times the denominator —hence, if the numerator be divided by the denominntor. I shall obtain the value of t he fr'action, (3,) for a quotient. And ill general, tihe value of a fraction is the quotient arising from dividing the numerator by the denominator. In case of division, if we diminish both divisor and dividend bly the same factor, we do not change the value ofthe quotient-and as to finrd the value of a fraction we perforn. division, using the numerator for a dividend and the denorminator for a divisor; it is evident that we may divide both numerator anid denominator of a firaction, by the same nulmber, wvithout changing the value. This operation is called reducing frnactions to their lowest terms. H-tence, to reduce a friaction to its lowest terms. Divide both numerator and denominator by anv numbe r that is an exact divisor of them both, and the rasultingj fraction in the saine manner, tlhus continuing until nor number, greaterl tlhan unit, will divide them both. Or we may find tlhe greatest cornmnn divisor of both numerator and denominator, and then reduce it to its loiwest ternms by one division. To change a /'hole ANmber to an equi'ale0nt I'Frcionri. 5 to Sths. I wish to change'a whole nurmber to an equivw S alent fraction, having a specified denominator;; Cor -- instance, 5 to Sths. 40 4 There are eight eighths in one whole one; th!at is, the, re are eigllht times as many eighths as therie are whole o(nes; leollce, ill five whole ones there are eight times five eighths, orfor!/?eighth,. Hence, to clarJge a whole number to an equivalent firac~ tion having a specified denominator Multiply the whole numnber by the specified denominatorand under tha.t product write the denominator. 34: VULGAR FRACTIONS. 7'o change anl Iiproper Fraction to a VWhole or a M!ixeL Namuber.?j 7)3- I wish to change an improper fiaction to3 7 an equivaleznt whole or mixed nuDmiber, fob lilstance 23 to a whole o, mixed number. There are *-v-en-sevelths in one whole one, and in twenty-tilree sevrilthls there ar e as many whole ones as seven is contained,ti es in twenty-three. Three and two-seventh times. UHence, to change an improper friaction to an equivalent i iole 1or mixed number IDivide the numerator by the denominator. T1 chan2ge a Mized Numnlber to an eaitiva7ent Fractiolo. 4, to 8ths. I wish to change a mixed number to an 4t equivalent flraction; for instance 47 to 8ths. 8 There are eight eighths in one whole one, that is, there are eight timnes as many eighths as whole ones. H-lence, in four whole ones, 7 there are eight times four eighths, or thirtytwo eighths, and the seven eighths added give;39 309 thirty-nine eighths. H1-ence, to change a mixed number to an equivalent fiaction: Multiply the whole number by the denominator of the fraction; to that product add the numerator, and write the result over the denominator. To M3ultiply a Fraclion by a Whole Number. -X5, 37- I wiSh to multiply a fraction by a whole - 2X2, 3) =' nurnber. The value of a friaction is the quotient arising by dividing the numerator by-the denornirat)lr. If the nuinerator be increased by an-y factor, the number of timnes that it will contain the denorminator will be proportionally increased; hence the value of the fraction will be propotrtionally increased. Again lf the denoninator be dimin-ished by any factor, the number of timles that it will be contained in the nunrerator will be proportionally increased; hernce, the value of the fraction will be propoitionally increased. VULGAR VRACTION'o 3 Thofe.'efo0re0 to multiply a fraction by a whole number Multi!ply the nlurnerator, or divide the denonztzator, by the whole number, T'lo Alltiply a Whole Number by a Fraction. 5X 5X3 I wish to multiply a whole number by a -5X;3 4 fiaction; for instance, 5 by -., or by one; iouritl Rait of three. First, I will multiply by the whole a(t three. (5X3.) Now as the Inultiplier has been four times as larg0e as it should be, the produzc is four! times as large as it should be; hence to make it what it should be, I will divide.t by four, represented thus x [-ence, tomultiply a whole number by a fraction Multiply tihe whole number by the numnerator of the fraction, and divide that product by the denominator. o Di:vide ci.F: racltion by a Whole Number. )r)t 3) - I wish to divide a fiaction by a Nwholle i) numlber. The value of a fraction is the -),7 9X4 quotient arising by dividing the' numerator bv the denomninator. If the numerator be diminished by any factor, the number of times that it will contain the denominator wvill be proportionally diminished; hence the value of the firaction will be propeor tionally diminished. sAgain.-If the denominator be increased by any factor, the number of times that it will be contained in the nurnerator -will be proportionally diminished; hence, the' value of' the firaction will be proportionally diminished. Therelore, to divide a friaction by a whole number: D)ivide the numerator, or multiply the denominator by the whole number. To.Divide a }Whole Number by a Fraction. -,3- )5,,5 I wish to divide a whole number by a q5X4 fraction; for instance, 5 by 3, or by one0O' fourth part of three. First, I will divide by 6 VJULGA.R FRPACTIONS. the Wole of threce. ( N.) bow as the divisor has been futJ. times as large as it should be, the quotient(-;) is one-four/It as large as it should be; hence, to make it w\hat it should be, I will multiply it by four, ( 53 ) which is done by mul.. tiplying the numerator by four, as before explained. Hence, to divide a whole number by a firaction' Multiply the whole number by the denomrinator of the fraction and divide that product by the numerator. To Multi2ly one Fraction by another. 3 7, wish to multiplly one friaction by another; X fr insltance, a by, or by one-scventh p)art 3.5 - - " 4X7 of-five. First, I will multiply by the whole st' five, (3 w5,) which is done as before explained, by mnultiplying the numet-ator by five. Now asthe multiplier has boen seven times as larlge as it should be, the product is seven times as large as it should be; hence, to make it vnhat it should be, I will divide it by seven, which is done is before explained, by multiplying the denominalor by seven, (-'-.) I notice that the numerators have been multiplied together for a new numerlator, and the denomi. nators have been multiplied together four a new denomr inateor. IHence, to multiply one 1fraction by another: Multiply the numerators together for a new numerator, nnd the denominators together for a new denominator. If there are mixed numbers, first change them to improper Ifractions. To change Comnpound Fractcions to Simnple ones. *- of - I wish to ireduc'e compound fractions to 5 3X5 simple ones, or to find what of means be7X,8 7-X& tween two fractions; for instance, I wish to obtain -7 of-. First, 1 will find one-seventh of five-eighths, which is done by dividing by seven, and that is done by multiplying the denominator by seven, as before explained. (.5-.) Now I have oneseventh of five-eighths. Three-sevenths will evidently be obtained by multiplying one-seventh by three, and that is ADDfITION OF VULGAR FPRACTIONo $7 done by multiplying the numerator by three, as before explained, (3 5.) I notice that I have multiplied the nunerators together, and the denominators together-the,same as in multiplication of one fraction by another. Therefore, we may conclude, that of occuring between two fractions shows that they are to be multiplied together. To Divide one Fraction by another. 51 ) 3,S - I wish to divide one fraction by another. d3 32X9 For instance, 3 by 5 or by one-ninth part 4X5' 4-X5' of five. First, I will divide by the whole of five, which is done as before explained, by multiplying the denominator by five. (.s'.) Now as the divisor, has been nine times as large as it should be, the quotient is only ione ninth as large as it should be; hence, to make it what it -should be, I will multiply it by nine, which is done as before explained, by multiplying the numerator by nine. (a'. )I notice that in the operation the terms of the divisor have been inverted, and it is then multiplied into the dividend. Hence, to divide one fraction by another ~ Invert the terms of the divisor., and then proceed as in multiplication of one fraction by another. If there are mixed numbers, first change them to improper fractions. ADDITION OF VULGAR FRACTIONS. 7 5 4 3 1 f, 1g,' 5' I wish to add 7 -5, s4 and 3, together. As the fractions have not the same denominator, it is evident that they Cannot be added unless they shall be changed to equivalent fractions having a common denominator. As the numerator of a fraction is always such a part of the denominator as the value of the fraction is intended to represent, it is evidelit that the common denominator must be such a number as will contain each of the given denominators. Such a number as will contain each of the denominators is a common multiple of them. And, that 4 Os 8SUBTRACTION OF FRACTIONS. the number used may be as small as possible,:we will -use for a common denominator, the least common multiple of all the denominators. The least common multiple of 18, 12, 9 and 5, is 180. Hence, 180 is the least common denominator;equired. To obtain the numerators, take such a part of the common denominator as the value cf each fraction represents. For the first numerator, take:7 of 180. One eighteenth of 180 is 10, and seven-eigh. teenths of 180 is 70. Hence, the first equivalent fiaction is 17 —the second is — the third is -— and the fourth is 8. Now, as the fiactions have a common denominator, they may be added by adding their numerators, and writing the result over the common denominator. Hence, to add fractions: Find the least common multiple of all the denominators, for a common denominator-take such parts of that commeon denominator for a numerator, as the values of the given fractions represent-then add the numerators, and:wrrite their sum over the common denominator. Compound fractions must be reduced to simple ones. SUBTRACTION OF FRACTIONS. As subtraction is the reverse of addition, we have the following rule for the subtraction of fractions: Change the given fractions to equivalent firactions having a common denominator, and then subtract their numerators, and write the difference over the common denominator. DECII AL F:FACTIO01N S. 5 8 )5000.6 s 625.6 2 1000.625 I have the expression 8; which shows that 5 is to be divided by. As 5 will not contain 8, I will annex a naught to the 5, which is equivalelt to multiplying it by ten, and then I will divide. DECIMAL FRACTIONS. 39 I obtain 6 for a quotient, and a remainder of 2. Now as the dividend is ten times as large as it should be, the quotient (6) is ten times as large as it should be; hence, to make it what it should be, I will divide it by ten. To continue the division farther, I will annex another naught, equivalent to multiplying the original dividend by 100. I now obtain 2 for a quotient, and a remainder of 4. Now, asthe dividend is one hundred times as large as it should be, the quotient (62) is one hundred times as large as it should be; hence; to make it what it should be, I will divide it by 100. To continue the division still farther, I will annex another naught; equivalent to'multiplying the original dividend by one thousand. I now obtain 5 for a quotient, and I have no remainder. Now as the dividend is one thousand times as large as it should be, the quotient (Q25) is one thousand times as large as it should be; hence, to make it what it should be, I will divide it by 1000. I notice that when I had one figure in the numerator of the quotient, I had for a denominator one naught and a one;-when I had two figures in the numerator, I had twos naughts and a one for a denominator;-when 1 had three figures in the numerator, I had three naughts and a one for a denominator. And I have hero such a set of frace tions as, in general, have for a denominator 1, followed by as many naughts as there are figures in the numerator. MiVathematicians have agreed that to represent such deno. minators, they will write a point at the left of the numerators Thus.6 (six with a point at the left,) has a denominator of 10, and hence is read six-tenths..62 (sixty-two with a. point at the left,) has a denominator of 100, and hence is read sixty-two hundredths..625 (six hundred and twentyfive with a point at the left,) has a denominator of 1000, and hence is read six hundred and twenty-five thousandthso Tenths,-hundredths, —thousandths, —fractional expressions whose values increase (from right to left) in a tenfold proportion, and hence they are called deciaZl fractions, It is plain, that the farther a figure is removed to the right of the decimal point, the less its value. Hence the placing 460 ADDITION AND SUBTRACTION OF' DECIM3AL', of a naught between the decimal point and the first significant figuie, renders the value of the expression only one tenth as great. Again, as a naught annexed has no value itself, and as it does not change the distance of the other figures from the decimal point, it will not affect the value of the expression. As the first figure at the right of the decimal point is of one-tenth as great value as units, it is evident that decimnal fractions may be annexed to whole numbers, as one continuous number, if we observe to place a point between thle tenths and units.o Such expressions are called mixed. numberso ADtDITION AND SUBTRACTION OF DECIMALS. To add or subtract decimal fractions —as they increase' oa decrease as whole numbers, we must write them so that. the decimal points, and units of the same denomination, will fall under each other; and then proceed as in whole ru-mberso MULTIPLICATION OF DECIMALS. -i. wish to multiply one decimal fraction by a nother, and to finld wllat number of figures to point off at the right handl of the product, for decimals. For instance I wish to multiply 2.35 by.7 1.645 From the theory of decimals, before explained, I know theat the 235 may have a denominator of a one and two naughts, (i-n._) and that the 7 may have a denominator of aV one and one naught. (-0)' Here then I have them ex. pressed in the form of vulgar fractions. (.3,X 7" ). But to multiply one vulgar fraction by another, as before explained, we multiply the numerators together for a nest DIVISION OF DECIMALS. 4L numerator, and the denominators together for a new denominator. (1 6 45) Now there will always be as many naughts at the right hand of the denominator of the product, as there are at the right of the denominators of both multiplier and multiplicand,-but, as the number of naughts shows the number of decimal figures, there will always be. as many decimal figures in the product, (if the denominator be removed,) as there are in both multiplier and multiplicand-in this case, three. 1Hence, to multiply one decimal fraction by another Proceed as in whole numbers and point off at the right of the product, as many figures for decimals, as there are in both of the factors. DIVISION OF DECIMALS..7 ).245 I wish to divide one decimal fraction by.35 another, and to find what number of places to point off at the right of the quotient for decimals. For instance I wish to divide.245 by.7 From the theory of decimals, before explained, I know'v that the 245 may have a denominator of a one and three naughts, (T 2T4)-and that the 7 may have a denominator of a one and one naught. (-7-). Here then I have therm expressed in the form of vulgar fractions. (0 —0 — But to' divide one Vulgar fraction by another, as before explained, invert the divisor, and then proceed as in multiplication. 24-5X10, 35;,35. Now the one naught from the denominator of the divisor will cancel one naught fiom the denominator of the dividend. If there had been two naughts in the,denominator of the divisor, they would have-cancelled two naughts from the denominator of the dividend. And there will always be as many naughts left at the right for the denominator of the quotient, as those at the right of the denominator of the dividend, exceed those in the denominator of the divisor. But, as the number of 4~ 4S — ",CANCGELLAT1ON. naughts shows the number of decimal figures, there wili be as many decimal figures in the quotient, (if the denominator be removed,) as those in the denominator of the dividend exceed those in the denominator of the divisor. Hence, to divide one decimal fraction by another: Proceed as in whole numbers, and point off, at the right of the quotient, as many figures for decimals, as those in the dividend exceed those in the divisor. If there are less decimal figures in the dividend than in the divisors. ~annex naughts to the dividend. CANCELLATION. it has already been shown that if we multiply and thenr divide the product by the same number, the value of the original number will not be changed. Hence, it will appear evident that if I wish to multiply 386 by 3 and 4, and then divide it by 3 and 2, that to multiply it by 2- would produce the same result. This operation may be expressed thus: 386 772 Prod. MIultiply 672 by 16, 25 and 8, and divide by 32, 69,; and 20 and what number will result?,;872 OperRation -; 4: 2 1744 Prods CANCELLATION. 43 Multiply 75 by 18, 2&, 36 and 50, and divide by 253 729 84 and 54~,4 I 0 9 i 9 25 25 Result. Multiplyv 68 by:, 18,', and divide by A, - and 36, 100 3 5 4 7 I 22 4 ) 2205 5514 Ans. i wish to multiply 168 by one-fourth part of 3. If I place the factor 3 on the right hand side, I make the dividend too large by a factor 4. Hence, that the quotient may be of proper value, I will increase the divisor by a factor 4. Again, in relation to the 4, if I divide by the whole of 4 it will give a quotient only one-seventh as large as it should be; hence, to make it what it should be, I must increase it by a factor 7, which is done by writing a factor 7 on the dividend side of the line. By a course of reasoning entirely similar, we shall find that in case of fractions, we must write the numerator on the same side of the perpendicular line that it would be if it were a whole num', ber, and the denominator on th/e other side. 44m CANCELLATION. REMiARK. — The active teacher will call the attention of the student to such abbreviations as they may occur, more profitably than it can be done here. And as many oppor. tunities will be offered in the following pages, for practice in this particular, we deem it unnecessary to increase the number of examples. Examples to be solved by Analysis. 1. If 4 yards cost f7, what will 20 yards cost? Ans. $350 NOTE.i-20 yards being 5 times as much as 4 yards, they will cost 5 times as much: hence, to find the cost of 20 yards, multiply the cost of 4 yards by 5. 2. If 28 lbs. butter cost $5 92, what cost 7 lbs. T Ans. $1 48. NOTE.-7 lbs. being 4 of 28 lbs., they will cost only &9 as much: hence, to find the cost of 7 lbs., divide the cost of 28 lbs. by 4. REMARK.-Let the teacher compel the student to give a good reason for each step. APPEAL.-If you are a student, do not pass over a point without knowing fully all the reasons why, and then learn to tell those reasons to others fluently. 3. If 4i tons of hay will keep 3 cattle during the winter, how many tons will it take to keep 25 cattle the same time? Ans. 37,. NOTE.- First find the amount required to keep one, and then multiply by 25. 4. What will 9 yards of cambric cost, if $40 96 be paid for 72 yards! Ans. $5 12. 5. If 12; lbs. of sugar cost'1 34, what will 1 cwt. cost. Ans. $10 72. 6. If a bankrupt pay 12s. 6d. on a pound, how much will A. receive to whom he owes ~1000 I Ans. ~625. 7T What cost 3 hhd. of wine at 10 cts. a pint? Ans. $151 20. 8. If a man performs a journey in 6 days, when the days are 8 hours long, in how many days will he do it whens the days are 12 hours long? Anso 4. CANCELLATION. 45 9. If a pipe empty a cistern in 15 hours, how many pipes will empty the same cistern in 4 of an hour Ans. 20. 10. If 12 pears are worth 21 apples and three apples cost a cent, what will be the price of four-score and four pears. Ans. 49 cents. 11. How long should a person keep $450, to compensate for the use of'150 for 18 months 3 Ans. 6 months. 12. A man bought 18 pipes of wine at 12s. 6d., New Jersey currency, a gallon; how many dollars did it cost M Ans. $3780. 13. If in four months I spend as much as I gain in three, how much do I lay up at the year's end, if I gain every 6 months $428 50? Ans. $214 25. 14. If 2 lbs. of sugar cost 25 cents, and 8 lbs of sugar are worth 5 lbs. of coffee, what will 100 lbs. of coffee cost 3' Ans. V20. 15. H-low many gallons of water must I add to a pipe of wine for which I gave $105, to reduce the first cost to 75 cents a gallon 3 Ans. 14 gallons. 16. If I buy 54 barrels of flour for $297, what will 1I barrels cost at the same rate 3. Ans. W 9. 17. If 14 men can reap 84 acres in 6 days, how many men must be employed to reap 44 acres in 4 days 3 Ans. 11 mten. 18. If by working 9 hours a day, 5 men can hoe IS acres of corn in 4 days, how many acres can be hoed at that rate in 3 days, working 10 hours a day? Ans. 27 acres. 19. If 8 men can build a wall 15 rods long in 10 days, how many men will it take to build a wall 45 rods long in 5 days T Ans. 48 men. 20. A manahaving $100, spent a part of it, and afterward received five times as much as he had spent, and then his money was double what it was at first. How much did he spend? Ans. $25. 21. A merchant bought a number of bales of velvet, each containing 1291-7 yards, at the rate of $7 for 5 yards, and sold them out again at the rate of $11 for 7 yanrds 46: PROPORTION. and gained $200- by the bargain; how many bales were there t Ans. 9 bales. P O PORTION. Suppose thst two yards of cloth cost six dollars, and I wvis.h to i.n;Ld what nine yards of the same cloth will cost. If two yards of cloth cost six dollars, one yard, which is one halif of two yards, will cost one half of six dollars, or three dollars. If one yard cost three dollars, nine yards, which is nine times one yard, will cost nine times three dollars, or twenty-seven dollars. Therefore, if two yards of cloth cost six dollars, nine yards of the same cloth will cost twenty-seven dollars. Now, it is evident that I may always find the cost of any quantity, as I have done in this case, by first obtaining the cost oif a unit, and then multiplying that by the number of units. But as dividing to find the cost of a unit will often give an- awkward fiaction, I will seek some method of solving such problems which shall obviate the necessity for such division. The first number of yards multiplied by the price per yard will give its cost, and the second number of yards mnultiplied by its price per yard will give its cost. But the price per yard being the same in both cases-calling these terms-the first term is multiplied by the same number to make the second, that the third term is multiplied by to make the fourth,-that Is, the same ratio" exists between the first and second terms, that there is between the third and fourth. And whenever we have four terms which have the same ratio between the first and second that there. is between the third and fourth, they are said to be in proportion. And to shbw that such relation exists between the terms, dots are used as you see here. -'Ratio is the quotient arising from dividing one number by anothery PROPORTIO1~. 47 ydso $ yds. 2 6: 9 27 The first and last terms are called Extremes, and the second and third terms are called Means. The first and second taken together are called a couplet, and the third and fourth are called a couplet. The first term in each couplet is called the antecedent, and the second term in each couplet is called the consequent. I will now use the terms of the first couplet as miultiplicands. Now as the second multiplicand is three times the first, if 1 obtain equal products, it is evident that the first multiplicand must be multiplied by a number three times as large as that by which we must multiply the second. If the second multiplicand had been five times the first, then the first must have been multiplied by a number five times as large as that by which we multiply the second. If the second had been nine times the first, then the first must have been multiplied by a number nine times as large as that by which we multiply the second. And whatever ratio exists between the multiplicands must exist between the multipliers, in an inverse order. But friom the nature of Proportion, the same ratio always does exist in the last couplet, that there does in the first; hence, if the second couplet be inverted and multiplied into the first, we shall always obtain two equal products; that is, the product of the means i.s always equal to the product of the extremes. This being the case, if a factor in one of the extremes is wanting, as in the case above, we may multiply together the factors of the means, and divide their product by the product of the given factors of the extremes. Or if a factor in the means is wanting, multiply together the factors of the extremes, and divide their product by the product of the given factors of the means. On the principle that like causes produce like effects, and unlike causes produce corresponding unlike effects, it is evident that if in arranging the terms,the first and third be made to represent the cause of the effects which the second and fourth terms represent, there will be the same ratio between 48 PROPORTION. the first and second terms as between the third and fourthB Hence, tile rule: State the question,'by writing that which represents the cause of one effect in the first term, and its effect in the second term; and that which represents the cause of the other effect in the third telrm- and that which represents the effect itself in the fourth'term; writing some' character to represent the unknown quantity, Then cause the corresponding factors in the first and third terms to be of the same denomination, also the corresponding factors of the second and fourth terms. Then if the unknown factor fall in either extreme, divide the product of the means by the product of the given factors of the extremes. And if the unknown factor fall in either of the means, divide the product of the extremes by the product of the given factors of the means. EXAMPLES IN PROPORTION. 1. If 6 horses consume 34 bushels of oats in a certain time, how many bushels will 18 horses consume in the same time. Ans. 102 bu. C. E. C. E. 6 ~ 34: 18 zx The statemnent reads- If 6 horses consume 34 bushels in a certain time, 18 horses will consume how many bushels I As the unknown factors falls in the extremes, the extremes will form the divisor and the means the dividend, thus e C. E. C. E. 6 34:: 18 x 34 3 Apply the principles of cancelation, as before explained, and we have on the dividend side 34 to be multiplied by 3, which gives 102-hence, if 6 horses consume 34 bushels of oats in a certain time, 18 horses will consumn 1 02 bushels in the same time~ ?IPRPORTION. 49 2. If a tree 20 feet high cast a shadow 30 feet long, how long will be the shadow of a tree 50 feet high T Ans, 75 feet, C. E. C. E. 20: 0.: 0 50: The statement reads-If 20 feet of tree cause 30 feet of shadow, 50 feet of tree will cause how much of shadow'? The extremes are the divisor and the means the dividend, &c. The statement might, perhaps, with equal propriety, have read-if 30 feet of shadow require 20 feet of tree, how many feet of shadow will require 50 feet of tree't The imeans would then have been the divisor and the extremes the dividend-but of course the same result would have been obtained. C. E. C. E. 30: 20:: x: 50 B. What w'ill 20 yards of cloth cost, if 6 yards can be bought for $30? Ans. $100. C. E. C. E. 6 30:: 20: x If 6 vards cost $30, 20 vards will cost how mhuch? 4. Bought IS8 barrels of flour for $72; what will be the cost of 42 barrels at the same rate? Ans. $168. C. E. C. E. 18: 72:: 42 x:, If 18 barrels cost $12, 42 barrels will cost how much? 5. If 8 hats cost $24, what will 110 cost at the same rate? Ans. $330. 6. If 3 barrels of flour cost $24, what will 39 barrels cost? Ans. $312. 7. If 16 pounds of sugar cost $2 56, how much will 13 pounds cost? Ans. $2 08. 8. If a man travel 105 miles in 6 days, how far will he travel in 120 days Ans. 2100 miles. 9. If 6 yards of cloth cost 18 shillings, what will' 11 yards cost e Ans, 33 shillings. 5 50 PROPORTION. 10. If 5 barrels of salt cost $7 25, what will be the cost of 18 barrels. Ans. $26 10. 11. What will 16 tons of hay cost if 6 tons cost $21 Ans. $56. 12. If 6 bushels of corn cost $4 75, what will 75 bushels cost? Ans. $59 37.5. 13. If 7 yards of cloth cost $15 47 what will 12 yards cost. Ans. $26 52. 14. What cost 36 pounds of cloves at 8 cents for 5 oz.. Ans. $9 21`. REMIARK. —In this example the 36 pounds must be reduced to ounces, to agree with the other denomination of weight. 15. If 2 cords of wood cost $11 50, what will 17 cords cost? Ans. $97 75. 16. If $4 75 be paid for 19 lbs. of salmon, how many pounds will $25 50 buy. -Ans. 102. 17, If 7 ounces of silver are sufficient to make 17 teaspoons, how many may be made from 21 pounds. Ans. 816. 18. If 5 and 2 make 10, how much will 9 and 5 be. Ans. 20. 19. If 4- yards cost 89 75, what will 13~2 yards cost I Ans. $29 25. C. E. C. E. 4: 975-::134: x: 9 75: 27: Notice, the 9 must be written On the divisor side, and the 2 (its denominator,) on the dividend side of the perpendicular. All mixed numbers must first be reduced to improper fractions. For suppose that, in this example, we write the 4 on the divisor side, and the 2 on the other side; our divio sor is then only a of 4, but it should be- of 9. 20. If 6- bushels of oats cost $3, what will 91 bushels cost? Ans. $4 26.9. PROPORTION. 51 C. E. C. E. 6 ~ 3:: 91 ~ 9. 137 21. If ~ of an ell-English cost I of a pound, What will 12' yards cost An 5 s. ~5 s. ldo C. E. C. E. 3 1 25 x 5 4 NOTE. If 3 of an ell-English of 5 quarters each cost X-, 2 yards of 4 quarters each cost how much I 22. If 12 — cwt. cost $42~, what will 48- cwt. cost? Ans. $163 50. It wvill appear evident that we may reduce the vulgar, fiactions to equivalent decimals. C. E. C. E. 12.5: 42.25: 48.375: x And now if we will make the decimal places in the two causes equal, the decimal points may be omitted. For suppose we allow them to remain, and we consider the firactions to have denominators expressed122. 422.. 48375 125 A0 00 4225 1000 48375 It will be seen that when the decimal figures are thus made equal, they exactly balance against each other, (as the two one hundreds in this case,) and hence the decimal points may be omitted. 23. If-2 of a yard cost -5 of a pound, what will of a yard cost! Ans. ~1 Is.:24. If 1 of a ship cost ~51, what are 33r of her worth? Ans. 10 1.8s. 6-do. ~d~~J ~V IIVo Yi 52 EPRnOPORTIONS 25. If 97 yards cost $115, what will 1605- ells-Englisa cost 7 A ns. $24. 26. If 41 tons of hay feed 3 cattle during the winter, how many tons wvill keep 25 cattle the same time? Ans. 37-e. 27. W hat cost 1 cwt. of cheese at 24 cents for 34 lbs.? Ans. $6 85 —. 28. If - of a ship is worth $6000, how much is Ts5 of her worth? Ans. $5000. 29. If 5 yvards of cloth are worth $274, how much are 504 yards worth? Ans. $251 25. 30. How much cotton cloth can I buy for ~10 Os. 8d., if I pay 64 pence for. 1 yards? Ans. 496 yards. 31. If 4 men can mow 12 acres of grass in 3 days, how many acres can 8 men mow in 18 dalys Ans. 144. m4 m8 d83 12 dI: dl x If 4 men in 3 days mow 12 acres of grass, S men, in 18 days will mow how many acres? DIRECTION.-ArraDge both antecedents in the same order, and both consequents in the same order. REMARK.-It will be observed that animal strength and time are most conveniently considered as the cause; money and time also are often most conveniently considered as cause. 32. If 4 students spend $96 in 3 months, how much wilte 6 students spend in 10 months, at the sameo rate? Ans. $480..33 If 20 horses eat 18 bushels in 10 days, how many bushels will feed 50 horses 20 days? Ans. 90. 34. If 18 men in 40 days, working 9 hours a day, can build a wall 240 feet long, 8 feet wide, and 6 feet high; in how many days, of 12 hours in a day, can 6 men build a wall 160 feet long, 6 feet high and 4 feet wide? Ans. 30. 35. If 6 lbs. of cotton make 20 yards of cloth, an eli French wide, how many pounds will be required to make a piece of cloth 150 yards long, and an ell.Flemish wide. Ans. 223. 71, PROPORTION. b5 36. How much money will gain $6 in 3 months, if $350 in 9 months gain $15 T Ans. $420. 37. If 45 yards of cloth, 6 quarters wide, will make 10 suits of clothes, how many pieces, each 45 yards in length and 3 quarters wide, will make 500 suits of clothes! Ans. 100. 338. If a cellar 221 feet long, 17%-l feet wide, and 10feet deep, be dug in 2a- days by 6 men working 12u-) hours in a day, how many days of 8- hours each, should 9 men work, to dig another measuring 90 feet long, 34} feet wide, and 12-s- feet deep! Ans. 24 days. 39. A trench of 5 degrees of hardness, 70 yards long, 3 yards wide. and 2 yards deep, can be dug by 336 men, in 5 days of 10 hours each; now, 240 men, in 9 days, can dig how long a trench- of 6 degrees of hardness, 5 yards wide, and 3 deep, if they work 12 hours each day? Ans. 36 yards. 40. If 3 men receive $18 for working 3 days, of 10 hours each, how much should 8 men receive for working 9 days of' 12 hours each? Ans. $172 80. 41. If 22 men can cut 81 cords of wood in 3 days, when the days are 14 hours long, how many men will be required to cut 162 cords in 4 days, when the days are 11 hours long a Ans. 42 men. 42. If a stream of water running into a pond of 190 acres, will raise the pond 10 inches in 12 hours, how much woulld a pond of 50 acres be raised by the same stream, in 10 hours? Ans. 312. 43. If 175 bushels of corn, when corn is 60 cents a bushel, be given for the carriage of 200 barrels of flour 58 miles, —when corn is worth 75 cents a bushel, how much must be given fo6' the carriage of 90 barrels of flour, 200 miles?. Ans. 434v4 bu. 44. A garrison of 900 men have provisions for 4 months; how many must leave, that the provisions may last the remainder 9 months? Ans. 500. 45. If a cistern 74- feet long, 52- feet wide, and 44- feet deep, hold 784 measures of water, how many measures C "fk would it hold if the length were doubled, the width made e tirmes as great, and the depth 4 times as great? Ans. 1881.6. 46. If a of a job may be performed by 12 men laboring!7 hours a day for 15 - days, how many men may be with. drawn and the residue of the job be finished in 15 davs more, if the laborers are employed 7 hours a day? Ans. 4 rneno 47. In what time will $627 50, loaned at 7 per cent, produce as much interest as $Z510, at a3, per cent, will produce in 1 year and 8 months? Ans. 3 -- years. 48. If 1000 men, beseiged in a town with provisions for 5 weeks, allowing each man one pound a day, be reinforced by 500 men more; and supposing that they cannot be relieved until the end of 8 weeks; how many ounces a day tmust each man have, that the provisions may last that time? Ans. 6q ounces. 49. If 6 is 10, how much is 12 Ans. 20. 50. If 12 oxen graze down 16.2;5 acres of grass in 20 days, how many acres of the same pasture will be sufficient for 24 bxen 24 days? Ans. 39. 51. If 8 masters, who have each S apprentices, in 5 weeks of 6 days each, earn $;860 how much will 5 mass ters, who have each 10 apprentices, earn in 8 weeks, each week 5~- days, their daily wages being the same, the masters working as well as the apprentices? Ans. $1075 55.} 52. How many laborers must be employed to finish a piece of work in 15 days, which 5 can do in 24 days? Ans. 8. 53. If when wheat is 83 cents a bushel, the cent loaf weighs 9 ounces, what should it weigh when wheat is $1 24.5 a bushel? Ans. 6 oz. 54. If 9 yards of cloth cost C5 12s., how many yards of cloth can be bought for ~44 16s.? Ans. 72. 55. What will 26 yards of cloth come to, if $6 90 be paid for 13 ells-French? Ans. $9 20. 56. A certain piece of corn may be hoed in 9 days by a man who can hoe 5 rows in the same time that each of two INTER:STd 5I'-oys cah hoe 8; in how many days will the field be hoed, if' they all work together? Ans. 4T days. 57. A can mow an acre of grass in 6 hours, and B in 8 hours; how much will they jointly mow in 10 hours? A ns.,21 58 If A and B can do a piece of work in 7 days, and B alone in 12 days, in how many days can A alone do ~- of it? Ans. I1-. 59. If' 7 oz. 5 pwt. of bread be bought for 4F-d. when corn is 4s. 2d. per bushel, what weight of it may be bought for Is. 2d., when the price per bushel is 5s, 64d.? Ans, 1lb. 4oz. 3- -7pws 60. If the freight of 9 hhds, of suigar, each weighing l2cwt., 20 leagues, cost ~16, what must be paid. for the fieight of 50 tierces, each weighing 21cwt., 300 miles? Ans. ~92 11s. 10Mdo INTEREST interest is anl allowance made, at a given rate by the hundred, for the use of any commodity whatever. For instance, if an individual hires one hundred bushels of grain, and pays six bushels for thle use of it, he pays an interest of six bushels on one hundred, or six hundredths part of the quantity hired. And whatever is the quantity hired, if he pays a quantity equal to six one-hundredths of it for the use of it, he is said to pay an interest of six per cent. If he pays this for the use of that which he hired for one year, he is then said to pay an interest of six per cent per annum. If he pays an interest equal to seven one-hun. dredths part of that which he hires, he is then said to pay an interest of seven per cent per annum, &c. The quantity hired is called the Principal. At 6 per cent per annum, the interest for one year is =.06 of the Principa l...........................nmonthis=.005............... a a4 *e $j jj ev *| v@ @e 6 days is =.001........ 6 NTEREST, I wish to find the interest of $6 75 48, for 2 yrs. 3 moa 18 days, at 6 per cent per annum. If for one year the interest is.06 part of the principal, for two years it will be twice as much or.12 part o' the principal. If for one month the interest is.005 part of the principal, for 3 months it will be three times.005 or.015 part of the principal. If for six days the interest is equal to.001 part of the principal, for eighteen days, which is three times six days, it will be three times.001 or.003 part of the principal. These three results added give.138, as the fractional part of the principal which will be equal to the interest for 2 years 3 months 18 days. Hence, multiply the principal by that and we have the interest. 675.48.138 540384 20'2644 67548 93.21624 Suppose the interest had been required at seven per cent per annum. It is evident that if we divide the inters est at six per cent by six, we shall have the interest at one per cent. Then if I multiply the interest at one per cent by seven, I shall have the interest at seven per cent. Hence, to find the interest on any given sum, for any given time: Multiply.06 by the number of years-then multiply.005 by the number of months-then multiply.001 by the number of times six is contained in the number of days; add the three products, and then multiply the principal by their sum, and the product will be the interest at six per cent per annum. To find the interest at any other rate than six per cent: Divide the interest at six per cent by six, and multiply the quotient by the given rate per cent. Amount is the sum of principal and interest. INTEREST. 57 Interest upon INotes, on which partial payments have been made. The following rule evidently brings to a correct result in computing interest on riotes, when partial payments have been made. Find the amotunt of the principal from the timre it becamQ due until the time of settlement; and then find the amount of' each payment, from the time it was made until settlement, and subtract their sum from the amount of the principal, 1. PROCIaESTER, Jan. Ist, 1848. For value received I promise to pay to John Short, or order, on demand, $1000 with interest from date. CHARLES PATIT. On this note were the following endorsements: April 1st, 1848, received $250. Aug. 1st, 1848, received $.2425. March 16th, 1849, received $116. W l hat was due at the time of settlement, June 25th, 1850 —interest being reckoned at 6 per cent per annum? OPERATION. Principal - - - $1030 Interest for 2 years, 5 mo. and 24 da. - 149 First payment - - $50 Interest for 2 yrs. 2 mo. 24 da. - 3.3 50 Second payment - 22'nterest for 1 yr. 10 mo. 24 da. - 225 65 Third payment -- 116 Interest for 1 yr. 3 mo. 9 da. - 87 Sum of amounts - - - $659 02 Balance remaining due, June 251th, 1850,.489 98 2. BUFFALO, July 7th, 1842. On or before the first day of Jan., 1846, I promise to pay Jarmes Takeall, or order, $500, with interest at 6 per cent, for value received. Wr.O NvwERPAY. r58 INTEREST. On this note were the following endorsements Jan. 7th, 1843, received $25. April 19th, 1843, received $5. Dec. 13th, 1843, received $77. Dec. 25th, 1845, received $200. What was due at the time of settlement? Ans.'297 54.4, The substance of the following rule for computing tlihe interest on notes, is established by law in many of the States, and by the Suprenme Court of the U. S. Raule.-Compute the interest on the given principal, fiom the time the note was given to the time of the first pay rment. If the payment equals or exceeds the interest, add the interest thus found to the principal, and from that amount subtract the payment and use the remainder as a new principal. But if the payment is less than the interest, compute the interest to the date of the second payment, then firom the amount, subtract the sum of the payments then made, unless they are together less than the interest now due; in which case, compute the interest to the time of the third payment, &c. TWhenever the payment (or sum of payments) exceeds the interest due, add the interest to the principal, and fiom the amount subtract the payment (or pavmyents,) and use the remainder for a new principal, with wfhich proceed as before. 1. A bond or note, dated Feb. 1st, 1810, was given for.t)500, interest at 6 pet cent, and there were payments endorsed upon it as follows: May Ist, 1810 - - $40 November 14th, 1810 - - 8 April 1st, 1811 - - - 12 May 1st, 1811 - - - 30 Iow much remains due on said note, September 16th, 1811? Ans. $455 57. BOSTON, Jan. 1st, 1841. 2. For value received, I promise to pay Monroe Bogardus, or order, $850, on demand, with interest at 6 per cent. JONATHAN LOSEITALL. On this note the following payments were endorsed: July 1st 1841, received. - $100 62 Dec. lst, 1841, received - 15 2 COMIP'OUND INTEEREST,;9 Aug. 13th, 1842, received - 175 75 What was due January 1st, 1843. Ans. $650 38. 3. For value received I promise to pay Winm. Cunningham, or order, eight hundred sixty-seven dollars and thirtythree cents, with interest at 6 per cent. JO;0HN BANYRUPT. On this note were the following endorsements: April 16th, 1823, - 136 44 April 16th, 1825, - 319 Jan. 1st, 1826, - - 518 68 WVhat remained due July 1Ith, 1827? Ans.,$215 10.3d COMIPOUND INTEREST. When the amount is found at the end of each period of time, and is used as principal for the next period, the inte. rest is said to be compound. Hence for computing Comr pound Interest, we have the following rule: Find the amount for the first period of time, (one yearl or one half year, as the case may be,) and use it as a new principal, upon which find the amount for the second period of time, and thus proceed through all the different periods required. From the last amount subtract the first principal, and the remainder will be the Compound Interesto EXAMPLES. 1. What is the compound interest of $500 for 3 years, at 7 per cent' Ans. $112 52.1. 2. What is the compound interest of $316 for 3 years, 4 months and 18, days at 6 per cent? Ans. $69 01.7. 3. What will be the compound interest on $500 for one yeai, at 8 per cent, the interest being computed quarterly? Ans. $41 21.6. 4. What is the amount of $1200, at compound interest for 4 years, at 44 per cent, interest being due annually t Ans. $1417 37.7. 80 COMMISS5ION AND INSURANCE. 5. What is the compound interest of ~760 10s, for 4 years, at 4 per cent per annum? Ans. L~128 12s. 3Udo Given, the amount, principal, and timene, tofind the rate. The amount is made up of the p)rincipal and interest added; hence if I subtract the principal from the amount, I shall have left the interest, which is composed of three factors, principal, rate and time, two of which (principal and timne,) I have given, to find the other factor. If I divide the interest, which is the product of principal, rate and time, by the product of principal and time, I shall have the rate per cent. Hence the rule: Subtract the principal fiom the amount, and divide the remainder by the product of principal and time. Given, the amount, principal and rate, to find the tiIme. If I subtract the principal from the amount there will be left the interest, which is composed of the factors principal, rate and time; two of which, (principal and rate,) are given to find the time. Hence, we have to divide the interest, which is the product of principal, rate and time, by the product of principal and rale, and we have the time. Hence, the rule: Subtract the principal from the amount, and divide the remainder by the product of principal and rate. CORMMISSION AND INSURANCE. f;ommission is an allowance of a certain per cent. for transacting business, such as buying and selling goods, &c. Insurance is security against loss, by fire or otherwise, and is obtained by paying a certain per centage on what is called the Premium. Premium is a certain per centage on the value of the property insured. To estimate Commission or Premium: Multiply the given sum by the given per cent, considered as hundredths, STOCKS —LOSS AND GAINo STOCKS. Stock is a general term applied to government funds, and to the capital of incorporated companies. When Stocks sell at their original value, they are said to be at par. When they sell for more than their original value they are said to be above par. When they sell for less than their original value they are said to be below par, or at a discount. LOSS AND GAIN PER CENT. $5.10 4.'25.85 8-5 =.20 Suppose a person to buy cloth at $4 25 per yard, and to sell it $5 10 per yard, I wish to find what fractional part of the first cost is gained. If I subtract the buying price from the selling price, the'emainder will be the gain on the price of one yard, that is, on $4 25. Subtracting, I have a remainder of.85. Now to find what part the.85 is of $4 25. As the numerator of a vulgar fraction is always such a part of the denominator as the value of the fraction is intended to represenet, I have only to vwrite the 85 for a numerator and the 425 for a denominator, (~-?.) Now if I change this vulgar firaction to an equivalent decimal, I shall have the same value represented in hundredths —or the gain per cent. Hence, to find the profit or loss per cent. Make the profit or loss the numerator and the buying price the denominator of a vulgar fiaction, (having both in the same denomination,) which change to an equivalent decimal6 %62 DISCOUNT. To find at what pric goods must be sold lo gain or lose certain per cent. 2o0, I + =_6 If I buy wheat at $1 25 per bushel, 5 )1.25 at what price per bushel must I sell it -- to gain 20- per cent..25 A gain of 20 per cent is equal to T-o4 6 of the cost price; which fraction reduced gives a, hence., I wish to gain $1.50 a, therefore 1 must sell the wheat for 8 of the cost price. A of $1 25 is $1 50. Hence, the sel. ling price must be $1 50 a bushel. Therefore, to find at what price a commodity must be sold to gain or lose at a given per cent: Make a vulgar fraction of the required per cent, which reduce to its lowest terms. Then add it to or subtract it from a unit, as the case is one of gain or loss, and multiply the given price by the resulting fraction. NoTE.-If the selling price be given, to find at what price the commodity must have been bought to gain or lose a certain per cent: Prepare the fraction, as in the rule, and divide the given price by it. DISCOUNT. Discount is an allowance made for the payment of a sum of money before it becomes due. 1.06 1.24 )620.00 (500 620.06 4 00.24 620 1.00 500 1.24 120 Discount. EQUATION OF TIME. 63 Suppose a person to owe $620, to be paid at the expira. tion of four years, without interest, and I wish to find what sum may be paid now, without gain or loss-discounting at 6 per cent per annum. The interest on one dollar for one year is six cents; for four years it is twenty-four cents, and the one dollar being added, gives the amount for four years-one dollar and twenty-four cents. Therefore, one dollar now, will be worth one dollar and twenty-four cents, four years from now; and- for one dollar and twenty-four cents due four years hence, he can afford to take one dollar now. Hence, as many times as one dollar and twenty-four cents is contained in the $620, so many dollars can be taken now. 1'24)620 00(500 Then $500 is the present worth of 620 the $620. And this subtracted fromt the $620 leaves $120, the amount 00 discounted. Hence, to find the present worth of a sum of money due some time hence vithouit interest: Find the amount of one dollar for the given rate and time, by which divide the given sum. NOTE.-TO find the discount-subtract the present worth from the original sum. REMARK.-When Banks discount they reckon the interest upon the given sum in advance, and subtract that, which. is evidently more than they should take, by the interest on the true discount. EQUATION OF TIME. Suppose an individual to owe $800, to be paid at the expiration of four months, and $800 more, due at the expiration of eight months. I wish to find at what one time the whole may be paid without gain or loss to either party. The present worth of the first $800, is $784 31; and the present worth of the second $800, is $769 2$3 The sum of these is $1553 54. EQUATI.ON OF TIME. Now the question is, in what time will $1553 54 amount to $1600, at 6 per cent per annum. Subtracting the $1553 54 from $1600, we have $46 64, the interest to gain on $1553 54. If I divide this interest, which is made up of the factors of principal, rate and time, by the plroduct of the two given factors, principal and rate, I shalli have the tiame 1553 54 06 93 2124 ) 46 46000 (.4984 of a year. 37 28496 9175.040 8389116.4984 - 12 7859240 7456992 5.9408 360 4022480 28.2240 Hence tihe whole should be paid at the expiration of - months 28.2 days. Hence the rule: To find the equaled time of several payments due some time hence without interest: Find the present worth of each payment separately-then find in what time tle sum of' these present worths will amount to the sum of the payments. EXAMPLES. 1. A owes B $197 due at the expiration of 1 year 110 50 due at the expiration orf 1~ yearsy 114 due at the expiration of 2 years. At what time may the whole be paid without gainor loss to eithe r; reckoning at 7 per cent? Ans. 1 years. EQUATION OF TIME. 65 2. What will be the equated time for the payment of the following bills. $103 due in I year; 106 due in 1 year; 112 due in 2 years; rec-koning at 6 per cent. Ans. I year I month 2.99 days. 3. Find the equated time for the payment of the follow. ing: $600 due in 4 years; 400 due in 5 years; 220 due in 2 years; 690 due in 3 years; reckoning at 5 per cent. Ans. 8 years 6 months 28.8 days. 4. What would be the difference in time, in the third example if we had reckolied at 6 per cent'. Ans. 2 months 10 days. The following rule is in common use, to find the equated time of several payments due at different times Multiply each payment by its time, and divide the sum of the products thus obtained by the sum of the payments. REMARKS.-That this rule does not give correct results is evident fiom comparing the answers to the foregoing questions, as obtained by the two different'rules. The error arises in. balancing the interest of a given sum, against the discount of the same sum for the same time and rate. For instance, the equated time for the payment of $800 due in 8 months, and $800 due in 4 months, as found by this rule, is 6 months. That is, the rule seems to suppose the interest on $800 for 2 months equal to the discount of $S00 for the same time. Now, whatever is the rate per cent, the time as *found by this rule, would be the-same. But is evident that for higher rates per cent there will be a greater difference between the interest and discount. Hence, the greater the rate per cent, the greater the error by this rule. 1. What would be the answer to the first example, found by the last rule? Ans. 1 year 6 months 3.6 days. 6* INGLE AND DOUrBLE FELLOWSHIPo &S, What would be the answer to the second example ii Ans. 1 year 2: months 7.6 days. What would be the answer to the third example I Ans. 3 years 7 months 12 days,? 4~ What difference of time would it make in the fourth,,,a-ple Ti Ans. 2a months 23.2 daysSINGLE FELLOWSHIP. I- Fellowship the money invested; is called the Capital,r Iock. Tiah gain is often called the Dividend. 1T1ee individuals commence a co-partnership. A puts'3In f 300; B puts in $400; C puts in $5.00, and they gain 2'4 0 What is each one's share of the gain I find the whole amount invested is $1200; of which A put in -3 oo equal to X part; B put in -~~-, eq:ual to, pr; ~ C put in T%-,0 equal to npart. Hence A should aave; B 3; C -f5 of the $240. eiance, in Fellowship, to find each one's share of the gain or lossMale vulgar fractions, having each partner's share of the capital for the numerators, and the whole capital for tile denominators, and take such parts of the whole gain or Los.s as those fractions represent. DOUBLE FELLOWSHIP. When time is taken into consideration, we have what is erined Compound Fellowship. For this we have the following bule. 1\Mukltiply each man's stock by the time it continues in trade) and uw the product as his stock, and form fractions IN'OLUTION AND EVOLUTION- 670 INVOLUTION. The Root of a number is one of the equal factors into which the number may be resolved. The Power of a number is the product arising by using the number some number of times as a factor. Thus: The 2nd power of 2 is-(22)=2X2=4. " 3rd cc 66 2 is=(23)=2X2x2=8.'" 4th a" " 2 iS=(24)= 16.;" 5th " " 2 is=(2s)=- 32. *" 6th c" " 2 is==(26)= 64. The small figures at the right and a little above the root, are called Indices, or Exponents. They show how many times the root is taken as a factor,-that is, to what power the root is to be raised. If Nwe multiply 2s (the third power of two,) by 24 (the fourth power of two,) we shall be combining a multiplier containing four factors of two, with a multiplicand containing three factors of two; hence, the product will contain seven factors of two, or 27 (2 seventh power.) And in general; to multiply together different powers. of the same number, give the number a new exponent equal to the sum of the exponents in both the factors. Multiply 32 by 34. 36. Multiply 24 by 2'5. 29. Multiply 52 by 57. 59, Multiply 73 by 77. 71,o EVOLUTION, OR EXTRACTING ROOTS. Table of the second and third powers of the nine digits. Roots. 1 2 3 4 5 6 7 8 9 [ Squares.-l 1 4 9'16 25 36 49 64 81 j Cubes. 1 1 8 27 64 125 216 343 512 729t 68 SQUARE ROOT. SQUARE ROOT. The Square Roo.t of a number is such a number as being multiplied by itself will produce the given number. Thus 3 is the square root of 9, or 6 is the square root of 36. I wish to find a means of extracting the square root of a number, as 1296. I have first squared each of the nine digits, and I find that none of the squares occupies more than two figures. But as there are more than two figures in this number, (1296,) I know there will be more than one figure in the root. Anid 1 now wish to find in what places of the square, the square of each figure in the root will be found. 1 2 3 4 5 6 7 8 91 1 4 9 16 25 36 49 64 81 12211 976 2002=4 00 00 9002=81 00 00 202= 4 Off 702= 49 00 12= 1 62= 36 3 6 12.96( 36 Root. 36 9:3X3+3X6 66 )3 96 +3X6+6x6 3 9632+2X3X6+6X6 0 00 For that purpose I have taken the two numbers, 22i and 976, and have decomposed them into the parts 200, 20 and 1; and 900, 70 and 6, and have separated each of those parts. From our system of notation I know that the square of the hundreds' figure, (2,) will occupy the fifth place, (4,) counting from the right; and, if the figure is of sufficient value, its square will occupy both fifth and sixth places, (81.) But it cannot occupy more than those two places, as before shown. That the square of the tens' figure, (2,) will occupy the third place from the right, (4,) and, if the figure is of sufficient value, (7,) its square will occupy the SQUARE R~ OOT. 6 third and fourth places, (49.) That the square of the units' figure, (1,) will occupy the first place at the right, and, if' the figure is of sufficient value, (6,) its square will occupy the first and second places at the right, (36)-but it cannot occupy more than those two places, as beforeshown. Now, as the square of tens cannot reach nearer to the right than the third' place, the square of units will be found in the first two places at the right. And as the square of hundreds cannot reach nearer to the right than the fifth place, the square of tens will be found in the next two places; the square of hundreds in tile next two places, &c. Hence, if in the extraction of square root we point the number into periods of two figures each, the number of periods 0lus forrned will determine the number of figures in the root, and the left hand period, (81,)will contain the square of the left hand figure, (9,) of the root. Hence, having pointed this number, (1296,) I will extract the root of the greatest square in the left harnd period, and it will be the left hand figure of the root. T comrnence at the left hand also for the reason that if I lhave a remainder, I can reduce it to units of a lower denomination, and thus avoid. fractions. The greatest square number in 12 is 9, thee root' of hvnich is 3. Subtracting this square from the left hand jperiod, I have three for a remainder. -Now as the next figure of the root affects the next two places in the square, I will annex to this remainder the next period of the square~ Now that I may obtain the next figure of the r6ot, I will ascertain of what this remainder is composed. For that purpose I have taken tbh number 36 and squared it, representing the operation by appropriate signs. The tens multiplied by tens give that. (3X3.) The tens into the units give that.. (3X:6.) The units into- the tens give that. (3X6.) The units into the units give that. (6X6.) I notice lthat the square of a number consisting of tensl and units, is composed of the square of the tens, (8,) plus: twice the p:oduct of the tens b.y the units, (2x3x6,) pliau the square of the, units, (6X6.) 70 SQUARE ROOT. I have already subtracted the square of the tens, hence this remainder must be composed of twice the product of tens by the units, plus the square of the units: but mostly of twice the product of the tens by the units. Therefore if 1 divide this remainder by twice the tens, I shall; obtain the units-or some number larger. But twice the tens will give tens, and that multiplied by unlits will give tens, hence the units figure of this remainder will form no part of twic'e the tens by the unitstherefore I will reject the units figure, and divide by twice the tens by the units. Twice 3 is 6; and 6 is contained in 39 six times-but as this remainder is also composed of the square of the units, I will write the 6 at the right of the divisor, that it mas be squared when I multiply. Now, multiplying and subtracting, I have no remainder. The square of a decimal fraction always consists of full periods of two figures each: hence, commence at the right hand to point off in decimals, so that if there are not full periods we may annex naughts, and thus give it the form of a complete square. Hence, to extract the Square Root: Point the number into periods of two figures each, commencing at the decimal point and pointing each way. Find the greatest square number in the left hand period, and place its root as the left hand figure of the root. Subtract that square number from the left hand period, and to the retnainder annex the next period for a dividend. Take twice the root already found for a divisor; find the number of titnes the divisor is contained in the dividend, rejecting the right hand fi gre; place the resalt as the next figure of the root, and also at the right of the divisor. Multiply the divisor thus augmented by the last figure of the root found, and subtract the product from the dividend. If there is no remainder or more periods, we have the entire root; but if there is a remainder, annex the. next period, and form a new divisor, by taking twice the root?now found, and thus proceed until all the periods have been usedo EXTRACTION OF CUBE ROOT. 71 EXTRACTION OF CUBE ROOT. 1 2 3 4 5 6 7 8 9 I 8 27 64 125 216 343 512 721'221 976 2002 =4 000 000 9002=729 000 000 202- 4 000 702= 343 000 12= 1 62-= 216 3 6 13,824[24 36 8 3X3+3X6 12) 5 824 3X6+6X 6 48 3 6 96 _-~~~ ~64 3X3X3+3X3X6 +3X3x6+3x6X6 5824 +3X3X6+3x6X6 - +-3x6x6+f-O6X66 0000 32 + 32 X6 +3X62 + 63 I wish to find a means of extracting the cube root of a number-that is of finding a number, which, being used three times as a factor will produce the given number, (13,824.) First, I have cubed each of the nine digits, and I find that no one of the cubes consists of more than three figures. But as there are more than three figures in this number, (13,824,) I know there must be more than one figure in the root. I now wish to find in what places of the cube, the cube of each figure in the root will be found. For that purpose I have taken ihe two numbers 221 and 72X EXTRACTION OF CUBE ROOT. 976, and decomposed them into the parts 200, 20, and Iand 900, 70, and 6, and have cubed those parts. From our system of Notation, I know that the cube of the hundreds figure will occupy the seventh place fiom the right, and if the figure is of sufficient value, its cube may occupy the seventh, eighth, and ninth places from the right. That the cube of the tens figure will occupy the fourth place from the right, and, if the figure is of sufficient value, its cube will occupy the fourth, fifth, and sixth places fiom the right. That the cube of units will occupy the first place at the right, and if the figure is of sufficient value, its cube will occupy the first, second, and third places from the right. Now as the cube of tens cannot reach nearer to the right than the fourth place, —the cube of units will be found in the first three places at the right. And, as the cube of hundreds cannot reach nearer to the right than the seventh place, the cube of tens will be found in the ne xt three places; the cube of hundreds in the next three places, &c.'Hence, if in the extraction of cube root, we point the number off into periods of three figures each, the number of periods thus formed will determine the number of figures in the loot, and the left hand period will contain the cube of the left hand figure of the root. Hence having pointed this number, (13824,) I will extract the root of the greatest cube in the left hand period and it will be the left hand figure of the root. I commence at the left hand, also for the reason, that if 1 have a remainder, I can reduce it to lower denominations and thus avoid fractions. The greatest cube number in 13 is 8, the cube root of which is 2. Subtracting this cube number from the left hand period I have 5 for a remainder. Now as the next figure of the root affects the next three places in the cube, I will annex to this remainder the next period of the cube. Now, that I may obtain the next figure of the root, 1 will ascertain of what this remainder is composed. For that purpose I have taken the number 36 and cubed it,-represeating the operation by appropriate signs. First, I must EXTRACSTION OF CUBE ROOT. 73 tquare it. The tens into the tens give that-(3X3.) The tens into the units, give that-(3X6.) The units int6 t'e. tens give that-(3X6.) The units into the units give that —-(6X6.) I now lhave the square of 36, which being multiplied by: 36 will give its cube. The tens into that line, (3X3+3X6) give that-(3s3x 3+3X3X6.) The tens into that line, (3X6+6x6,) give;hat-(3X3X6+3><6X66.) The units into that line, (3X 3+-3X6.) And the units, into this line, (3X6+6X6,) give that-(3X6x6+6X6 x6.) I notice that the cube of a number consisting of tens and units is composed of the cube of the tens, (33) plus tir ee times the square of the tens by the units, (3X32X6,) plus three times the tens by the square of the units, (3X3X 6:) plus the cube of the units. I have already subtracted the cube of the tens-hence, this remainder mlust be composed of three times the squ are of the tens by the units,-4three times the tens by the square of the units, —and the cube of the units. But a. great part of it is made up of three times the square of the tens by the units. Hence, if I divide this remainder by three times the square of the tens, I shall obtain the units, -or some number larger. But the square of tens gives hundreds,-and that multiplied by units gives hundreds, —-hence the two right hand l;gures.of the dividend can form no part of three times the square of the tens by the units,-therefore reject the two right hand figures, and divide the rest by three times the square of the tens. The square of the tens (2) is 4, which multiplied by gives 12 for a divisor. Twelve is contained in 58 foui tinies,-place the 4 as the next figure of the root. Multiplying the divisor by this fignre, (4,) I obtain three times the square of the tens by the units, (48,) agreeing with the first part of the remainder, which being hundreds, place it in hundreds place. The next part of the remainder is composed of three times the tens into the square of units. Three times the tens (2,) gives 6, which multiplied'by the square of }.nits 7 id4 9t~EXTRACTION OF CtUBE ROO.T (1.6,) gives 96. Three times the tens gives tens, and the square of units being units, three times the tens into the square of units gives tens-hence, place it in tens place. The cube of units being units, place it in units placed Adding, and subtracting from the dividend, I have no red mainder, hence, 24 is the exact root. The cube of a decimal fraction always consists of full periods of three figures each —hence, commence at the left hand to point off in decimals, so that if there is not full periods we may annex naughts% and thus give it the form of a complete cube. Hence, to extract the Cube Root: Point the number into periods of three figures each, commencing at the decimal point and pointing each wayv Find the greatest cube number in the left hand period, and place its root as the left hand figure of the root. Subtract that cube number from the left hand period, and to the remainder annex the next period for a dividend. Take three times the square of the root already found for a divisor. Find the number of times it is contained in the dividend, rejecting the two right hand figures. Place the result as the next figure of the root. Multiply thedivisor by this figure, and place the product under that part of the dividend used. Then take three times the tens, (of the root,) into the square of the units, and place under the first product, one place farther to the right, Then take the cube of the units and place under the last product, one place still farther to the right. Add the three results together, and subtract their sum from the dividend. If we have no remainder or more periods, we have the entire root; but if there is a remainder, annex the next periodi and form a new divisor by taking three times the square of' the root now found, &c., and thus proceed until all the periods have been used, DUODECIMALS. 75 DUODEC JIALS.. Square foot. - I A Prime. 7 3 feet, 4 in. If a square foot be divided into twelve- equal parts, each one foot long and one inch wide, the surfaces so obtained are called primnes. If one of these primes be divided into twelve equal parts, each one inch long and one inch wide, the surfaces so obtained are seconds. If a second be divided into twelve equal parts we obtain thirds, &c. Primes are designated by one stroke, seconds by two strokes, thirds by three strokes, &c.-thus, 3ft. 2' 7" 8"' is read three feet. two primes, seven seconds and eight thirds. It is evident that twelve units of any one of these de, nominations will make one of the next higher. And as their values increase or decrease ina twelve-fold proportion, these numbers are called Duodecimals; and as Iwe shall explain them, they are used in the measurement of, surfaces, but they may be used in the measurement of solids also. Suppose that I have a board four feet and five inches (not primes ) long, and three feet and four inches wide, of which I wish to find the amount of surface. If I give the four feet of length one foot of breadth, I shall obtain tour square feet and five primes. This will extend across the board one foot, but as the whole board 47 ALLIGATION, contairns three and four-twelfths as much surface, I must multiply the 4ft. 5' by First, I will multiply by 3. Three 3 4 twelfths. times five primes are fifteen primes, equal to one foot and three primes. 13 3' I will write the three primes and 1 5' S" reserve the one foot. Three times the four feet are 14 8' S" twelve feet, and the one foot reserved being added gives thirteen feet, which I will write down. Now J will multiply by the four-twelfths. Four times the five would be twenty primes, if the four was a whole number, but as it is twelfths, the product will be twelfths of primes or seconds-equal to one prime and eight seconds. I will write down the eight seconds one place towards the right, and reserve the one prime. Four times four would give sixteen feet if the four was a whole numbe:, but as it is twelfths, the product will be twelfths of squarefeet, or primes, and the one prime being added gives seventeen primes-equal to one foot and five primes. I will write the five primes under the primes and the one foot under the feet. The first partial product represents the surface of the board which is three feet in width, and the second partial product represents the surface of the board which is four inches in width; hence, if I add them [I shall obtain the whole surface of the board. 14ft. 8' 8". Hence, the rule: Multiply one dimension by the number of units in the other, commencing with tle left hand term of the multiplier. Carry one for every twelve, and place the product arising from the successive parts of the multiplier, one place' farther towards the right. Then the sum of the partial products will be the entire product. ALLIGATION. if we mix 4 bushels of oats at 25 cents, with 5 bushels of corn at 40 cents, and 6 bushels of barley at 45 cents-. what will one bushel of the mixture be worth- ALLIGATIONo 7?7 4 bushels of oats at 25 cents' is worth $1 00 5 66 t corn " 0 " " " 2 00 (i; s: barley " 45 " "'" 2 70 15 bushels of mixture is worth $5 70 Hence, one bushel of the mixture is worth one fifteenth part of $5 70, equal to 38 cents. Hence, having the price and quantity of each of several simples given, to find the mean price of the mixture: Multiply each price by the quantity and divide the sum of the products by the sum of the quantities. This process is called Alligation Medial. Hence, Alligation Medial is finding the mean price of a mixture, formed from simples whose quantities and prices are known. When the price or rate of each simple is given, and we wish to find the quantity to be taken of each to make a rmixture of a given rate or quality, the process is called Aligation Alternate. 1 8 3 4 12 8 )54(3 20 1 7 7 54 21 23 -2 2 4 i 2a 3 18 I Suppose a grocer to have sugars —some at 18 cents, some at 20 cents, some at 23 cents, and some at 24 cents a pound, of which he wishes to make a mixture worth 21 cents a pound-how many pounds of each kind must he take? if he takes one pound of the 18 cents sugar, puts it into the mixture and sells it for 21 cents, he will gain three cents. If he takes one pound of the 20 cents sugar, puts it into the mixture and sells it for 21 cents, he will gain one cent. If he takes one pound of the 23 cents sugar, puts it into the mixture and sells it for 21 cents, he will lose two cents, 7* ALLTGAVT oM If he takes one pound of the 24 cents sugar, puts it into the mixture and sells it for 21 cents, he will lose three cents. If he should take four pounds of the 18 cents sugar1, instead of taking but one, he would gain twelve cents. If of the 20 cents sugar he should take seven pounds, he would gain seven cents. If of the 23 cents sugar he should take two pounds, he would lose four cents. We now find that he has gained in all nineteen cents, and that he has lost but four cents-hence, in order that the mixture shall be worth 21 cents a pound, he has yet 15 cents to lose —and he must lose it on the 24 cents sugard if on one pound of the 24 cents sugar he lose three cents, it will take as many pounds, that he may lose 15 cents, as three is contained times in fifteen. It is contained five times. Hence, he may take five pounds of the sugar at 24 cents, —two pounds at 23 cents,-seven pounds at 20cents, -and four pounds at 18 cents, to make a mixture worth 21 cents a pound. Suppose we were required to make a mixture containing 54 pounds. By adding the quantities now obtained, I find we have IS pounds; just one-third of the quantity required-hence,; mInust multiply each quantity already found by 3. Therefore, having the prices or rates of several simples given, td find the quantity to be taken of each to make a mixture of a given rate or quantity Write the prices in a column, with the smallest first, the next larger next, and so:on,-and the mean rate at the left Ilhand. Draw a horizontal line separating the prices less than the mean rate, from those that are greater. Find the difference between the prices and the mean rate, and place opposite the prices themselves. Multiply those differences, omitting one, by numbers taken at pleasure. Then from the sum of the products not on the same side of the horizontal line with the omitted difference, take the -sum -of the products on the same side,-divide the remainder by the omitteddifference, and the quotientttogether with the quan AR1TIIIMETICAL PROGRESSION. 79 tities taken at pleasure, will be the proportionate quantity of each simple, to be taken. If the whole quantity is limited: Divide the quantity to which it is limited by the sum of the quantities fobund by the rule, and multiply each quantity thus found by the quotient-the products will be the quani tities to be taken of each. ARITHMETICAL PROGRESSION. Ist, 2d. 3d; 4th,.5th. 3 7 11 15 19 3 3+4 3+4+4 3+4+4+4 3+4 —4+4+4+ First term-3 19 Last term=19 3 Number of terms5 - C. d,=- 4)16 4=c. d. First term=3 19 Last term-19 8 C. d.=4 - Number of terms=x 4 ) 16 4+1=-No. termds First term-3 3 7 11 15 19 Last term= 19 19 15 11 7 3 No. of terms=5 - -- --- Sum of terms=x 22 22 22 22 22 2 ) 110 Sum=55 A series is a succession-of terms derived from each other according to some fixed and known law. A series of numbers increasing by a common addition, or decreasing by a common subtractions i8 called an Ariths meticaZ Progression, 8~1 ARITHMETICAL PROGRESSION. The common number added or subtracted is called the common diference. The first and last terms are called, extremes. The first term (3) is a term given. The second term (7) is formed by adding the common difference to the first term. (3+4). The third term (11) is formed by adding the common difference to the seconld-term (3+4+4,) -but as the second (7) has the common difference occuring once, the third term will have the common difference occurring twice. The fourth term is found by adding the common differ. ence to.the third term- but as the third term has the common difference occurring twice, the fourth term will have the common difference occurring three times. The fifth- term is formed by adding the common difference to' the fourth term-but as the fourth term has the common difference occurring three times, the fifth term will have the common difference occurring four times. I notice the common difference occurs four times in-the fiftph term, three times in the fourth term, twice in the third term, &c. And in general, the common difference occurs in any term a number of times indicated by the number of terms less one. Hence, to find any term; Multiply the common difference by the number of the term less one, and to the product add the first term, Suppose we have given, the first term, the last term, and the number of terms, to find the common difference. The last term (19) is made up of the common difference (4) taken a number of times indicated by the number of terms less one, added to the first term. Hence, if I subtract the first term from the last, the remainder (16,) will be composed of the common difference, taken a number of times indicated by the number of terms less one. There. fore if I divide this remainder (16) by the number of terms less one, (4,) 1 shall have the common difference. Hence, having the first term, the last term, and the number of terms given, to find the common difference: ARITHMETICAL PROGRESSION. 81 Divide the difference of the extremes by the number of terms less, one. Suppose, that I have given, the first term, the last term, and the common difference, to find the number of terms. The last term (19) is Made up of the common difference (4) taken a number of times indicated by the number of terms less one, added to the first term. Hence, if I subtract the first term from the last, the remainder (16) will be composed of the common difference taken a number of times indicated by the nunmber of terms less one. There. fore, if I divide this remainder (16) by the common difference, (4,) 1 shall have the number of terms less one —and adding 1 I obtain the number of terms. Hence, having the first term, the last term, and the common difference givern,;, find the number of terms: Divide the difference of the extremes by the common di~f erence. and add one to the quotient. Suppose that I have given, the first term, the last term, and the number of terms, to find the sum of the series. If the first and last terms be added together we shall obtain a certain amount. (22.) If we take the second term, counting from the left, which ~ is greater than the first term by the common difference, and add it to, the second term, counting from the right, which is less than the last by the comm-on difference, we shall obtain the saume amount. (22.) Again-if we take thethird term, counting from the left, which is greater than the second by the common dif: fetence, and add it to the, third, counting from the right, which is less than the second by the common difference, we again obtain the same amount: (22.) And, in general., the sum of the extremes is equal to the sum of any two terms equally distant from the extremes. Hence, if I invert the series and add it to itself, I shall obtain a set of' like nuzmbers- the' sum of which may be obtained by multiplying one of therm (22) by the number of them. (5.) GEOtiETRICAL: PRQOGRESSION. But this set of like numbers was formed by adding the original series to itself-hence, it must be twice the origi. nal series-therefore its-sum (110) is twice the sum of the original series,-and hence, divide this sum (110) by 2, and we have the sum of the original series. Hence, having given, the first term, the last term, and the number of terms, to find the sum of the series': Multiply the sum of the extremes by the number of terms, tand divide the product by 2,-or multiply the sum of the extremes by one-half the number of terms, or multi ply one-half the sum of the extremes by the number qf terms. GEOMETRICAL PROGRESS10N. 1st, 2d. 3d. 4th. 5th, 82 8 32 128 512 2 2x4 2X4X4 2X4X4X4 2X4X4x4X4 2 2x4 2x42 2X<43 2X4'4'irst term —2 4 x:: a:; 1 LLast term-=512 2 ) 512 32 —64 No. of terms —5 — = Ratio=l 4x 56=4 Rat.1 First term=2 2 8 32 128 512 Lnst term=.12 8 32 128 512 2018, a tio=4 24 Sum x -~ — - - 3 ) 2046 682 Sum. A series of numbers increasing by a common multiplier or decreasing by a common divisor, is called a Geometrical Progression. The common multiplier or divisor is called the Ratio. The first term is a term given. The second term is found by multiplying the first term by the ratio. The third term is found by multiplying the second term by the ratio; but as the ratio occurs in the second term once as a factor, it will occur in the third term twice as a factor, or raised GEOMETRICAL PROGtESSION. 8 to the second power. The fourth term is found by multiplying the third term by the ratio; but as the ratio occurs in the third term twice as a factor, it will occur in the fburth term three times as a factor, or raised to the third power. The fifth term is found by multiplying the fourth term by the ratio; but as the ratio occurs in the fourth term three times as a factor' it will occur in the -ffth term fovtr times as a factors or raised to the fourth power. We may notice that the ratio occurs raised to thefourth power in the fiJfI termj to the third power in the fourtll term, td the second power in the third term, &c(. And in general, the ratio occurs in any term, raised to a power indicated by the number of the term less one. Hence, to find any term' Raise the ratio to a power indicated by the number of the term less one, and multiply that power by thefirst term1. Suppose that we have given the first term, the last term, and the-number of terms, to find the ratio. The last term is made up of that power of the ratio indicated by the number of terms less one, multiplied by the fil;st term: hence, if I divide the last term by the first, I shall have for a quotient that power of the ratio indicated by the number of terms less one. Therefore if I take that root of the quotient indicated by the number of terms less one, I shall have the ratio. Hence, having given the first term5 the last term and the number of terms, to find the ratio: Divide one extreme by the other, and take that root of the quotient indicated by the number of terms less one. Suppose I wish to find the mean proportional between two numbers, as 4 and 16. As both means of the propor* tion are wanting, we have 4: x:: x 16. But as the product of the rmeans is equal to the product of the extremes, we have the equality, x2=64. Now if x2=64, the root of xe2 must be equal to the toot of 64, (or 8$) Hence, 8 is the mean proportional soughtb ~8~4 xGhANM-UTk1ES AT SIMPLE INTEREST. Therefore the rule to find a mean proportional between two numbers: Multiply them together and take the square root of their product. Suppose we have given; the fiirst term, the last tern and the ratio, to find the sum of the series. If I multiply the original series by its own ratio, I shall obtain a series which will be as many times the original series as there are units in the ratio. Now if I subtract, the first series-from this new series, I shall obtain as many times the original series as there are units in the ratio less one. Subtracting, the terms cancel, except the first and'last. Subtracting the first (2) fiom the last (2048,) I obtain 2046; which, is as many times the first series as there are units in the ratio less one. Hence, diride bv the number of units in the ratio less one, and we have the sumr of the series. Hence, the rule: To find the sum of the series, multiply the last term by the ratio; fror that product subtract the first ter.rm, ant divide.the remainder by the ratio less one, ANNUITIES AT SIMPLE INTEREST., An Annruity is a sum to be paid at stated periods.'When an annuity remains unpaid for some nurnmber of years, it is said to be in arrears, or to be forborne. I wish to show that the book account of an annuity in arrears, at simple interest, constitutes an Arithmezical Series. Suppose the annuity to be $50, to be paid annuall], but that it has Temained unpaid for some years, simple interes, being reckoned, at 6 per cent.'50 At the expiration of the first year, there 50 3 will be charged $50 for annuity. At the end 50 6 of the second year there will be charged for 50 9 annuity $50, and $3 for interest on the first'c. 8c. $50 for the last year. At the end of thethird ANNUITIES AT COMPOUND INTEREST. S5 year there will be charged for annuity $50, and $6 for interest on the last two fifties for the last yea.r. At the end of the fourth year there will be charged for annuity $50, and $9 for interest on -the last thee fifties for the iast year. By examining these charges we shall find that they form an Arithmetical Progression, of which the annuity is the first term, the interest on the annuity for one year is the common difference, and the number of years is the number of terms. ANNUITIES AT COMPOUND INTEREST. 10 00 1 00 100 100 7 7 7 7 7.49 7.49.49.49.0343 ltU 1 107 114.49 1 22,5043 114.49 ) 122.5043 ( 1.07 114 49 8 0143 8 0143 1] wish to show that the book account of an annuity in arrears at compound interest constitutes a Geometricc a Series. Suppose the annuity to be $100,- to be paid annually, but that it has remained unpaid for some years, and compound interest to be reckoned at seven per cent. At the expiration of the first year, there will be charged $100, for annuity. At the end of the second year, there will be charged for annuity $100 and 87 for interest on the 86 ANNUITIES AT COMPOUND INTERESTo first $100 for the last year. At the end of the third year5 there will be charged for annuity $100 and $7 for interest on the first $100 charged, for the last year, and $7 for interest on the second $100 for the last year, and.49 for interest on the $7 charged for the last year. At the end of the fourth year there will be charged $100 for annuity, $7 for interest on the first $100 for the last year,-$7 for interest on the second $100 for the last year,-$7 for interest on the third $100 for the last year —.49 for interest on the first $7 of interest for the last year-.49 for interest on the second $7 for the last year-.49 for interest on the third $7 for the last year-and.0343 for interest on the.49 for the last year. Now if any one of these annual charges, be divided by the one immediately preceding it, the quotient will be the amount of one dollar for one year at the given rate per cent. Hence, the book account of an annuity in arrears at compound interest does form a geometrical series of which the annuity is the first term,-the amount of one dollar for one year is the ratio, —and: the number of years is the number of terms. THEORIES OF ALGEBRA. ADDITION. 4a +6c +7ab -7ab +-12ab — 12tab 5b -4d ~+5ab — 5ab -7ab 7ab 4a+5b. 6c-4d. +12ab. -12ab. -5ab. -5ab. To 4a I wish to add 5b. As the quantities are dissimilar I: can only add them by writing the proper sign between them~ To plus ic I wish to add minus 4d. The 4d being a subtractive quantity, if I annex it to the 6c it will evidenly diminish its value, and the result will be 6c-4d. To plus 7ab I wish to add plus 5ab. As the quantities are similar, and both plus, it is evident that I shall have a plus amount equal to the sum of the two, or +12ab. To minus 6ab I wish to add minus 5ab. As the quantt. ties are similar, and both minus, it is evident that I shall have a minus amount equal to the sum of the two, or — 12ab. To plus 12ab I wish to add minus 7ab. The minus sign before the 7ab shows it to be a subtractive quantity; hence, if I unite it to the 12ab it will take front it, and I will evidently have + 5ab. To minus 12ab I wish to add plus 7ab. The minus Nign before the 12ab shows the 12ab to be a subtractive quantity. Hence, if I unite it to-the 7ab, it will take 12ab friom it. But I cannot take the whole of 12ab from 7ab. Th'fe 12ab is composed of 7ab and 5ab —1 can take 7ab from the 7ab and there will remain 5ab yet to be subtracted -to show that it is to be subtracted, write the minus sign before it. Now as these are all the cases that can occur, we have the rule 88 &SUBTRACTION-U —LTIPLICATI.:Xo When the quantities are dissimilar, write them after 4ach other with their original signs. When the quantities are similar and theirsigns alikeadd their coeficients, annex the literal part, and prefix the sign of the quantities themselves. When the quantities are similar and the signs unhike, take the less coeficient from the greater, to the remainder annex the literal part, and prefix the sign of the greater coeACiento SUBTRACTION. 4b 3c Let 4b represent the sum of all the additive terms in the subtrahend, and 3c 5a-4b+3c the sum of all the subtractive terms. Then from 5a I wish to take 4b minus 3c. Not the whole of 4b, but 4b diminished by the number of units in 3c. First I will take from 5a the whole of 4b, which gives 5a -4b. Now as the subtrahend has been too large by 3c, the remainder is too small by 3c —hence, to make it what it should be, I must add 3c —which gives 5a-4'+3c. I notice that the plus sign of the subtrahend has been changed to minus, and the minus sign has been changed to plus, and then it has been added to the minuend. Hence, for subtraction Change all the-signs in the subtrahend, and then proceed as in Addition. MULTIPLICATION. 7a4bOc4 I wish to multiply one monomial bv 5a3b3c5 another-for instance, 7a4b3C4by 5aab3c 5s Multiplication consists' in uniting with 35a"b6c9 the multiplicand such factors as occur in the multiplier. If with the factor 7 I combine the number 5 as a factor, I shall have for the product 35. As a occurs four times as a factor in the multiplicand, and three times MUiT?IPLCATION. in the multiplier, if I unite them asfactors, it will occur seven times in the product. As b occurs three times as a factor in the multiplicand and three times in the multiplier, if I unite them as factors, it will occur six times in the product- As c occurs four times as a factor in the multi. pilicand and five times in the multiplier, if I unite them as factors it will occur nine times as a factor in the product. Now each letter will'occur as many times as a fictor in the product, as it does in both multiplicand and multiplier taken together. But as the exponents show the number of times each letter occurs as a factor-to find the exponent of any literal factor in the product, add the exponents of that letter in multiplier and multiplicand. Hence, to multiply one monomial by another: Multiply the coefficients together for a new coefficient, write after that all the literal factors, affecting each with an exponent equal to the sum of the exponents in both the actorrso THE SIGNS IN ~MIULTIPLICATION. +5b I — b- b, -b, - -b, - -b, — b +- 3a 3 2 0,-,-I, -3 +15ab I -3b,-2b, -b, 0, + b,+2b,+3b, I wish to find what sign to give the product in multiplication. Plus 5b multiplied by plus 3a will evidently give a positive product. (15ab.) -b multiplied by 3 will evidently give three times as much minus, or -3b. -b multiplied by 2 will evidently give three times as much minus, or -2b' -b taken once will evidently give -b. -b multiplied by naught will give naught. Now while the multiplicands have each been of the same value, and the multipliers have been diminishing by a comdifference of 1, the products bave been increasing by a common difference of b. 8" 0v MULTIPLICATEON o0 POLYH'OMfAL5. And as the multiplicands that yet follow, have the sams value, and the multipliers continue to diminish by a comr mon difference of 1, I conclude that the products must contitnue to increase by a common difference of b —but ift they do, the next one will be plus b, the next plus 2b, the next plus 3b, &c. Hence minus b multiplied by minus 3 must give plus 3b. Therefore we conclude that when the signs of the actors are alike, the product will be plus, and when the signs of the Factors are unlike, the sign of the product will be minusr MUTIPLICATION OF POLYNOMINLS. a-b a-5 c -d d ac —bc-adXbd ab —Dd.I wish to multiply one polynomial by another. Let a represent the sum of all the additive terms in the multiplicand, and b the sum of the subtractive terms. Let c represent the sum of all the additive terms in the multi. plier, and d the sum of all the subtractive terms. Then I wish to multiplya — by c -d —not by the whole of c, but by c diminished by the number of units in d. First, I will multiply by the whole of c. c times a give. a —but before I multiply by c the a should he diminished by b —hence, in multiplying the whole of a, I have multiplied a b which I should not have multiplied, and hence the product is too'large by a times b. Hence, I must sub. tract c times b or ob from it, which gives ac-bc. There~ fore c times a-b is ac -b. But now 1 have multiplied by a quantity too large by the units in d, hence this product (asc — c) is too large by d times a-b —therefore I must obtain d times a-b, and subtract from this product. d times a-b gives ad- d, which being subtracted from ac-dc gives ac-hc-adxb-d. I notice that each term in the multiplicand has been multiplied by each term of the multiplier, hence the rule DIVISION OF MONOMiALS oi Multiply each term in the multiplicand by each term in the multiplier,-affecting each partial product with the plus sign when the signs of the factors are alike, and with the minus sign when the signs of the factors are unlike. DIVISION OF MONOMIALS, b )3a4 b-8c I wish to divide 35a4 b8 c by 7a'bo ____- Division consists in having two terms 5a b5 c given to find a third term, which being multiplied by the first will produce the second. The first is called the divisor, the second the dividend, and the third the quotient. Hence, the dividend may alwaysbe considered as a product of the other two terms. From multiplication of monomials, we know that the co. effcient of the dividend is made up by multipl. ing the coeffil cient of the divisor bly the coefficient of the quotient — hence, if i divide. the coefficient of the dividend by the coefficient of the divisor, I shall obtain the coefficient of the quotient. The exponent of any letter, as a in the dividend, is made up by adding the exponent of that letter in the divisor,) to the exponent of the same letter'in the quotient —hence, if I subtract the exponent of a in the divisor from the exponent of the same letter in the dividend, I shall have the exponent of that letter in the quotient. And in the same manner with b,-if I subtract the exponent of b in the divisor from the exponent of b in the dividend, I shall have the exponent for b in the quotient.. The c must occur in the quotient, for if it does not, to mlultiply the divisor and quotient together will not give the dividend. Hence, the rule for the division of monomials - Divide the coefficient of the dividend by the coefficient of the divisor for a new coefficient-write after that all the literal factors of the dividend, affecting each with an exponent equal to the excess of its exponent in the -dividend over that in the divisor. lRgxar.-Division of monomials cannot be performed, 93 i3!DIVISION OF POLYNOMIALSo when the divisor contains a factor which is not found in the dividend-and when a factor in the divisor has a larger exponent than the exponent oF the same letter in the dividend —and when the coefflcient of the dividend will not contain the coeflicieht of the divisor. DITISION OF POLYNOMIALS, In the product arising fiom the -+ 5 multiplication of one polynomial,a1;~3~; $Iby another, there will always be - +- some irreducible terms —and one,-4I: 3 of those ternms will be the one Nc,4 - formed by multiplying that term in -X ey Ul e term in the multiplier containing -- + + the largest exponent of the same - - < o _I o letter. For, we shall thus obtain a C ~2e s term which will contain a greater i exponent of that letter, than the ex~ + +d4- ponent of that letter in any other C) I I 2 / c I Q term of the product; hence, it will zzl ZZ nuot be like any other term, and 01 ~c es 03 > 2 i ence it will be irreducible. ~ ~ +!t I-+ This is equally true for all proCS 3 GQ < c ducts of polynomials. Hence, if OD m/ we examine any product and select the term which contains the largest.q % i exponent of a letter in that product, we know that such term is t,. — - _ made Lp by multiplying the term an'Z in one of the factors containing the +e~j + 4 r largest exponent of that letter by _ - s the term in the other factor con$-Q P-P.`z, taining the largest exponent of the Z2 Z2;2 - same letter. DIVISION OF POLYNOMIALS. 9 All dividends are products: hence, if I trace along this dividend and find the term containing the largest exponent of some letter, (as a,) 1 know that it is made up by multiplying the term in the divisor containing the largest exponent of a, by the term containing the largest exponent of a in the quotient. Hence, if I divide that term in the di.vidend containing the largest exponent of a, by the term in the divisor containing the largest exponent of a, I shall obtain a term for the quotient containing a larger exponent of that letter than any other term in the quotient. Dividing 10a4 by - 5a2' gives - 2a2 for a term in the. quoi tient. Now, as the dividend: is made up by multiplying the divisor by the qcuotient, if I multiply the divisor by thepart of the quotient found, I shall obtain a part of the div — idend; and subtracting that product from, the whole divideud, I obtain that part of the dividend which is made lup by multiplying the divisor by that part of the quotient not yet ound: hence, this remainder is a product.: But from what was first shown, that term in it which contains the largest exponent of a, is made up by multiplying the term containing the largest exponent of a in the: divisor, by that term containing the, largest exponent of a' in the part of the quotient not yet found. taonce, if I divide that term by the term in the divisor containing the largest exponent of a, I shall obtain that term containing the highest-exponent of a in the part of the quotient not. yet found. Dividing -40a3b by -5a-2 gives 8ab. Multiplying the divisor by the 8ab, and subtracting,. we have no remainder.. Hence, -2a2+8ab is the entire quotient. it has been necessary to compare, the terms in the dvi' — sor containing the largest exponent of some, letter with the term in the dividend containing the largest exponent of the same letter; and then the same term in th.e divisor with the term in the dividend containing the next smaller exponent, &c. flence the. rule for division of polynornials: Arrange both divisor and dividend in reference to some letter, placing the term in each containing the largest exponent of some letter, at the left, the term containing the DIVISION OF POLYNOMIALS~ next smaller exponent next, &c.: thus the exponents of that letter will diminish from the left to the right: then divide the left hand term of the dividend by the left hand term of the divisor, for the left hand term of the quotient: multiply the divisor by that term of the quotient and subtract the product fiom the dividend: if there is a remainder, treat it as a new dividend, with which proceed as before. NLoTE. —When you subtract, be careful to keep the higfhe!t power of the same letter at the left, that you had at the left at first~ ~orn Comstock's Pronetic 11lgazineo AE FONIK GIIRL OV MIJCIGAN. [This Poem is founded on the following fact, oommunicated in a letter from the Hon. Ira Mayhew of Michigan, to Dr. A. Comstock of Philadelphia: "I took up the TREATISE ON PHONOLOGY, and I was unable to lay it down until I had completed its perusal. 1 placed it in the hands of my children, and judge what was my surprise to hear a daughter, not yet six years old, read the firuM chapter of Genesis fluently, in less than one hour from the time she first saw a Phonetic character I" ] A dotur hsm prcpd kqIz wud con, Adornz 5s lek-gurt Stet. Or sumurz six had d'ur hur ccon, Cx copm bs Fconon'z get. Yis; copt bs Tsmpl ov qcr Wu9, In wun cort cpr indxd, And lurnd bs lsturz qix be yui, Bi Fonlk skll, me rxd! Asn stand rxb kt bi 6ildhud'z femn Yx mun hs srur hxd; And 1st an infant's progrss cert Yar livz ov poltri dxd! Se not, az yx hay sad tB mx, Aat " God haz kurst cpr spx6,:And Bebslizd it hcoplesl l!" Nco mcor bis folshud txW. Se not, " Fconstiks ar abstrus, — Abuv bes pxpl'z mind!"A hItl gurl haz lurnd bs yis Ov lsturz trY3 dxsind; A lItl gurl haz lurnd ts rxd B.i sistsm nxt and plenCud not bxs sistsm sirpursxd Acoz ov lorj yxrz ov pen? AkJz not Hsvn'z Gospsl brit, Ov olturn cp(r spx6! For mncosens iz Gospel-lit; Its akcunz vurft tx6! Cal not bis Alfabst bx cond Bi Frxdum'z Stets rxnqpnd! As Gospsl, in it, trumplt-twnd, Gavz nco zunsurtun soind! j UNIVERSITY OF MICHIGAN 1 Illlllllllllllllll lllllllllllL 11 3 9015 04872 4242 r:'imC o1 rec]itatolo, itn J sis d I Cese hft tihe }, ~ e? e'i J g-ve n at'ti co on'l nIcenoentt of the ex)lamna,, itio: snhouid oe coped a iuon the bluci -boarld, and that 1? I(' the stuient sho uld be reiiluired to refer to it wit lle th', -.h, ointini;-''C I'Jet the \exr dlanation be c.nmitteet tt o,?-!i: mlem rs vflai,- mas tic closer the stuacent foll wS j it. i -. hac,.ter w ill toe.is st;,-cess, and thee imoae a ecu-: anI al ci Th eS tu I iabitts il ilhOl)lht wrill hhe haivle e <-X m~-acquired.,1\s the exsamp ple'zr p1ropin-tiolN was on-i.tted a 1.:s pm: i"eop'r piace, we: wVill insert it lherc'!2 ~[- 7 x <,l.:'-:' (1. iM,... D 0!:)',','', i~t -' I b ot _ i - t r X'' It- A I i I;:;j!, a'pli:'I,lace t Ihet i (,' i' 0r o sie, S t I af 0; a l ot3 e p i S e o 1 t S S i n'I o,? 1 di cr V t- tos t ther te wor t ile i' We 1:':t c t -, 1: l to th' o 11 rsl p I t o i 0t1tl S'lO a1 1 t?,ii( /, %C t' 7 f, l-i a Xe'i =:!ve, i 1i X q! iv C'.; a: l,S i u y Ct, i Io c a (I h o -t s he [ 1:. It-ll.. 1 S' 1 n m place c aesicnecl it, 11 a a g' C 1- a" J~~~ ~~THE ATUt OPT J Of~i I'' - v-',f..//% ~ ~ ~ ~ T