230. To enlarge the terms of 231. To diminish the terms a fraction of a fraction“RULE.-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 ļ, in terms which are 10 for 1%, in terms 10 times less? times as large ?—for $, the —for 2 f, the terms being diterms being increased 9 minished 9 times ? times ? 232. OF THE GREATEST COMMON DIVISOR OF TWO NUMBERS. ANALYSIS. 1. If the two terms of a fraction be 8 and 39, what is the greatest num. ber that will divide them boil without a remainder ? It is evident that the greatest common divisor of 8 8 ) 38 ( 4 and 33, 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)3(1 t'ie greatest common divisor sought. But dividing 38 6 hy 8, we obtain a quotient, 4, and a remainder, 6; hence 8 is not a common divisor. Again, it is evident, that 2)613 the common divisor of 3 and 38 must also divide 6, be cause 38=1' times 8 plus 6; hence a number which will divide 8 and 6 will also divide 8 and 38 ; we will there. fore see if 6, which divides itself, will divide 8. But dividing 3 by 6, wo have a quotient 1, and remainler 2; hence 6 is not a common divisor. Again, for the reason above stated, the common divisor of 6 and 8 must also divide the remainder,”; and by dividing 6 by 2, we tind that 2, which ativides itsell, 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 3, will also divide 8 and 38. Hence 2 is the common divisor of 3 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 than itself. Therefore, 233. To find the greatest common divisor of two numbers. RulE.—Divide the greater number by the less, and the divisor 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. QUESTIONS FOR PRACTICE. 2. What is the greatest common divisor of 580, 320 common divisor of 24 and 36 ? | and 45 ? Ans. 5. Ans. 12. NOTE.-When there are more 3. What is the greatest | than two numbers, find the common common divisor of 612 and divisor of two, 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. of 48 ANALYSIS. 1. What is the most simple expression, or the least terms 272 ? The terms of a fraction are diminished, or.made more simple, by di. vision (230). Now, if we divide so long as we can find any number greater than 1 which will divide them both without a remainder, the fraction will evidently be diminished to the least terms which are capable of expressing it, since the two terms now contain no common factor greater than unity. Thus, 2.4=146, 2)24=1,2)=, and 2)344=it least terms. Or if we find the greatest common divisor of the two terms, 48 and 272, we may evidently reduce the fraction to its lowest terms ai ance by dividing the two terms by it. By Art. 233, we find the greatest common divisor to be 16, and 16).2192 4-8-=-3, least terms as before. Hence, 235. To reduce a fraction to its least terms. RULE.—Divide both the terms 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 56 to its least of 48,? terms. 3. What are the least terms 6. Reduce 244 to its least of 199? terms. 4. What are the least terms 7. Reduce 14to its least of 17 ? terms. Ans. it Ans. Ans. Ans. . Ans. 236. COMMON MULTIPLES OF NUMBERS. 1. What number is a common multiple of 3, 4, 8 and 12? 3X4X8X12=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 by 8, and as it is made up of 8 12s, 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 hy 3, 4, 8 and 12; it is therefore a common multiple of these numbers (215. Déf. 9). 237. 2. What is the least common multiple of 3, 4, 8 and 12? 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 numbers in a line, and dividing such as are divisible by 4, we separate 4, 8 and 12, each into two factors, one 4)3, 4, 8, 12 of which, 4, is common, and the others, 1, 2 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 4X3X2_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, (4X1X2X3–24) is divisible by each of the dividends, 4, 8 and 12, and 24 is obviously the least number which is divisible by 4, 8 and 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 quotients, and dividing the numbers by 3, which are divisible by it, we find that 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 the 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 more than once as a factor, the product is evidently the least number that can be so divided. Hence, 238. To find the least common multiple of two or more numbers. | Rule.-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. 9 QUESTIONS FOR PRACTICE. 2. What is the least com 3. What is the least num. 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). 3, 1,4 1 and 5X2X3X4=120 Ans. NALYSIS. 1. Reduce 3 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 } becomes to, and if each term of 2, the second, be multiplied by 2, the denominator of the first, becomes ; then, instead of } and we have the two equivalent fractions, to and f (230), which have 10 for a common denominator. 2. Reduce }, i and to a common denominator. Multiplying the terms of f by 24, the product of 4 and 6, the denominators of the other two fractions, becomes 4; again, multiply the terms of & by 18, the product of 3 and 6, the denominators of the first and third fractions, & becomes 59; and lastly, multiplying the terms of by 12, the product of 3 and 4, the denominators of the first and second, a becomes 4; then instead of the fractions $, f and , we have the three equivalent fractions, 44, 43 and 44, which have 12 for a common denominator. From a careful examination of the above, the reason of the following rule will be manifest. 240 To reduce fractions of different denominators to equivalent fractions having a common denominator. RULE.-Multiply all the denominators together for the common denominator, and each numerator by all the denominators except its own for the new numerators. 1 common EXAMPLES 3. Reduce and to a 5. Change &, & and the to. common denominator. fractions having a 을 Ans. 189, 279 and . 3x4=12 com. denom 6. Express and of a Then and dollar in parts of a dollar of 4. Reduce , and to the same magnitude. a common denominator. Ans. 16 and 45. Ans. Ja, ik and 1. Ans. 241. TO REDUCE FRACTIONS TO THEIR LEAST COM MON DENOMINATOR. ANALYSIS. 1. Reduce }, 4, , and 11 to their least common denominator. The common denominator found by the foregoing rule is a common multiple of the denominators of the given fractions, but not always the least common multiple, and consequently not always the least common denomi. nator. The least common multiple of the denominators, 3, 4, 8 and 12 is 24 (238), which may be divided into thirds Courths, eighths and twelfths ; for the new numerators we must therefore take such parts of 24 as are denoted by the given fractions; and this is done by dividing 24 by each of the denominators (438, 236, 2433, and 44=?), and multiplying the quotients by the respective numerators, (8X1, 6X3=18, 3Xő-15, and 2x1122), and the new numerators (8, 18, 15 and 22) written over 24, the common denominator, give it and for the new fractions, having the least possible common denominator. Hence, 242. To reduce fractions of different denominators to equivalent fractions having the least common denominators. RULE.--Reduce the several fractions to their least terms (235). Find the least common multiple of all the denominators for a common denominator. Divide the common denominator by the denominators of the several fractions, and multiply the quotients by the respective numerators, and the products will be the new numerators required. |