227. DIVISION OF ONE FRACTIONAL QUANTITY BY AN OTHER. 5 12 ANALYSIS 1. If * of a bushel of wheat cost of a dollar, what is that per bushel ? To find the cost per bushel, we must divide the price by the quantity (154), that is, we must divide t by But to divide a sumber by a fract non, we multiply it by the denominator, and divide the product by the 3X4 12 numerator (226); hence, we must multiply by 4, as (220), and 12 az is divided by 3, by multiplying the denominator, 5, by 3, as, (121); if of a dollar then is the price of one bushel. Hence, 228. To divide a fraction by a fraction. RULE.-Multiply the numerator of the dividend by the denominator of the divisor for a new numerator, and the denominator of the dividend by the numerator of the divisor, for a new denominator. Note.-In practice, it will be most convenient to invert the divisor, and the proceed as io Art. 224. 5ҳ3° 15 Ans. QUESTIONS FOR PRACTICE. 2. In 74 how many times 5. If g of a yard cost of of + ? a dollar, what is that a yard? 3. In d how many times Ans. Ye=$1.773 Ans. +18=1, 6. If of a piece of cloth 4. At of a dollar a bush- be worth of fe of an eagle, el for oats, how many can I what is the whole piece worth? buy for p of a dollar? Ans. eag. Ans. 1 8 bush. 22? 229. ALTERATION IN THE TERMS OF A FRACTION WITHOUT ALTERING ITS VALUE, ANALYSIS. A fraction is multiplied by multiplying its numerator, and divided by multiplying its denominator (219); hence if we multiply hoth the terms of a fraction at the same time by any number, we both multiply and divide the fraction by the same number, and therefore do not alter its value. Again, a fraction is divided by dividing its numerator, and multiplied by dividing its denominator (219); hence if we divide both the terms of a fraction at the same time by any number, we both divide and multiply the fraction by the same number, and therefore do not alter its value. Hence, 230-233. VULGAR FRACTIONS. 95 230. To enlarge the ternis of 231. To diminish the terms a fraction of a fractionRULE.-Multiply both the RULE.—Divide both the terms of the fraction by the terms of the fraction by such number' which denotes how a number as will divide each many times the terms are to without a remainder. be enlarged. QUESTIONS FOR PRACTICE. 1. What is the expression 1. What is the expression for }, in terms which are 10 for 3, in terms 10 times less? times as large ?-for $, the -for ti, the terms being diterms being increased 9 minished 9 times ? times ? ANALYSIS. 1. If the two terms of a fraction be 8 and 38, what is the greatest number that will divide them both without a remainder ? It is evident that the greatest common divisor of 8 8 ) 38 ( 4 and 38, cannot exceed the smallest of them. We will 32 therefore see if 8, which divides itself, and gives 1 for the quotient, will divide 38; if it will, it is manifestly 6)8(1 the greatest common divisor sought. 'But dividing 38 6 by 8, we obtain a quotient, 4, and a remainder, 6; hence 8 is not a common divisor. Again, it is evident, that the common divisor of 8 and 38 must also divide 6, be. cause 38—4 times 8 plus 6; hence a number which will divide 8 and 6 will also divide 8 and 38; we will therefore see if 6, which divides itself, will divide 8. But dividing 8 by 6, HB bave a quotient 1, and remainder 2; hence 6 is not a common divisor. Again, for the reason above statel, the common divisor of 6 and 8 must also divide the remainder, 2; and by dividing 6 by 2, we find that 2, which divides itself, divides 6 also ; 2 is therefore a divisor of 6 and 8, and it has been shown that a number which will divide 6 and 8, will also divide 8 and 38. Hence 2 is the common divisor of 8 and 38, and it is evidently the greatest common divisor, since it is manifest from the method of obtaining it that 2 will divide by it, and a number will not divide by another greater Hay itself. Therefore, 233. To find the greatest common divisor of two numbers.. RULE.—Divide the greater number by the less, and the divipor by the remainder, and so on, always dividing the last divisor by the last remainder, till nothing remains; then will the last divisor be the common divisor required. mon QUESTIONS FOR PRACTICE. 2. What is the greatest common divisor of 580, 320 common divisor of 24 and 36 ? and 45 ? Ans. 5. No'r E.-When there are 3. What is the greatest than two mimbers, find the common common divisor of 612 and divisor of ewo, then of that divisor 540 ? Ans. 36. and one of the others, and so on to the last. 4. What is the greatest common divisor of 1152 and 6. What is the greatest 1080 ? Ans. 72. common divisor of 918, 1998 5. What is the greatest and 522 ? Ans, 18. 234. THEIR MOST REDUCTION OF FRACTIONS TO SIMPLE EXPRESSION. ANALYSIS. 1. What 18 the most simple expression, or the least termis of 45? The terms of a fraction are diminished, or made more simple, by dk vision (230). Now, if we divide 49 so long as we cau find any number greater than 1 which will divide them both wiihout a remainder, the fracton will evidently be diminished to the Icast terms which are capable of expressing it, since the two terms now contain, no common factor greater than unity. Thus, 2) *2=146, 244=#$, 2)3=5, and 2);=ir. least terms. Or if we find the greatest common divisor of the two teriton, 48 and 272, we may evidently reduce the traction to its lowest terms al auce by dividing the two terins by it. By Art. 233, 've find the greatest common divisor lo be 16, and 16)2482=1, least terms as before. llence, 285. To reduce a fraction to its least terms.. RULE.--Divide both the terins of the fraction by the greatest common divisor, and the quotient will be the fraction in its least, terms. QUESTIONS FOR PRACTICE. 2. What are the least terms 5. Reduce 836 to its least of ? terms. Ans. $. 3. What are the least terms 6. Reduce ## to its least of 188? terms. 4. What are the least terms 7. Reduce %t to ite least of H? terms. Ans. Ans. Ans. 236, 237, 238. VULGAR FRACTIONS. 97 236. COMMON MULTIPLES OF NUMBERS. 1. What number is a coinmon multiplo of 3, 4, 8 and 127 3X4X8X19–1152. Ans. First, 3 times 4 are 12; 12 then is made up of 3 fours, or 4 threes; it is, therefore, divisible by 3 and 4. Again, 8 times 12 are 96; then 96 is divisible hy 8, and as it is made up of 812s, each of which is 'divisible by 3 and 4, 96' is divisible by 3, 4 and 8. Again, 12 times 96 are 1152 ; 1152 then is divisible by 12; and as it is made up of 12 ninety-sixes, each of which is divisible by 3, 4 and 8, 1152 is divisible by 3, 4, 8 and 12; it is therefore a common multiple of these numben $215. Def. 9). 237. 2. What is the least common multiple of 3, 4, 8 and 121 Every number will evidently divide by all its factors: our object then is to find the least number of which each of the numbers, 3, 4, 8 and 12. is a factor. Ranging the uüinbers in a line, and dividing such as are divisible by 4, we separate 4, 8 and 12, cach into two factors, one $)3, 4, 8, 12 of which, 4, is common, and the others, 1, % and 3 respec tively. Now as the products of the divisor, multiplied by 3)3, 1, 2, 3 the quotients, are, severally, divisible by their respective dividends, the products of these products by the other 1, 1, 2, 1 quotients, must also be divisible by the dividends; for *X3X2-24 these products are only the dividends a certain number of times repeated. The continued product then of the divisor, 4. and the quotient 1, 2, 3, (+X1X2X3=24) is divisible by each of the dividends, 4, 8 and 12, and 24 is obviously, the least number which is divisible by 4, 8 al 12, since 12 will not divide by 8, and no number greater than 12, and less than twice 12, or 24, will divide by 12. But the undivided number, 3, must also divide the number sought; we therefore bring it down with the quoticuts, and dividing the numbers by 3, which are divisible by it, we tind ihat 3 is already a factor of 24, and will therefore divide 24. Thus, by dividing such of the given numbers as have a common factor by this factor, we suppress all but one of the common factors of each kind, and the continued product of the divisors, and the numbers in the last line, which include the quotients and undivided numbers, will contain ibe factors of all the given numbers, and may therefore be divided by each of them without a remainder; and since the same number is never taken mone than once as a factor, the product is evidently the least pumber that can be no divided. Hence, 238. To find the least common multiple of troo or more nouiten bers. ROLE.- Arrange the given numbers in a line, and divide by any number that will divide two or more of them without a remainder, setting the quotients and undivided numbers in a line · below. Divide the second line as before, and so on till there are no two numbers remaining, which can be exactly divided by any number greater than unity; then will the continued product of the several divisors, and numbers in the lower line be the multiple required. QUESTIONS FOR PRACTICE. 2. What is the least com 3. What is the least numa mon multiple of 3, 5, 8 and ber which may be divided by 10? 6, 10, 16 and 20, without a re5)3, 5, 8, 10 mainder? Ans. 240. 4. What is the least com2)3; 1, 8, 2 mon multiple of 7, 11 and 13? Ans. 100). 4 1 and 5 X2X3X4=120 Ans, 3, 1, 1. Reduce & of a dollar and of a dollar to a common denominator. If each term of }, the first fraction, be multiplied by 5, the denominator of the second, the of becomes po, and if each term of q, the second, be multiplied by 2, the denominator of the first, s becomes my; then, instead of $ and ea, we have the two equivalent fractions, to and teo (230), which have 10 for a common denominator. 2. Reduce $, f and to a common denominator. Multiplying the terms of $ by 24, the product of 4 and 6, the denominators of the other two fractions, becomes ; again, multiply the terms of x by 18, the product of 3 and 6, the denominators of the first and third fractions, l becomes ; aut lastly, multiplying the terms of by 12, the product of 3 and 4, the denominators of the first and second, becomes 44; then instead of the fractions $, $ and , we have the three equivalent fractions, 46, 4 and 4*, which have 12 for a common denominator. From a careful examination of the above, the reason of the following rule will be manifost. 240. To reduce fractions of different denomiantors to equivatent fractions having a common denominator. ROLE. Multiply all the denominators together for the com mon denominator, and each numerator by all the denominaton except its own for the new numerators, |