The Least Common Multiple. 18. It is often necessary to express fractions as equivalent fractions, having a common denominator; and it is, moreover, convenient to have this denominator as small as possible. Now, there are always an infinite number of numbers which will contain each of the given denominators as a factor, and our problem is therefore to obtain the least of such numbers. DEF.-The least common multiple (L.C.M.) of two or more numbers is the least number which contains each of the given numbers exactly. RULE. Arrange the numbers in a line, putting one of them as a divisor. Strike out the greatest factor common to this divisor, and each of the numbers separately, and place the several quotients in the line below; at the same time bring down every number prime to the divisor. Repeat the operation upon the second line, and so on until we have a line of numbers prime to each other. Multiply the several divisors and the numbers in the lowest line together, and their continued product will be the least common multiple. Ex. 1.-Find the L.C.M. of 12, 16, 36, 45, 60. 19. Reduction of fractions to equivalent fractions having the least common denominator. It is evident that the least common denominator cannot be a less number than the L.C.M., and therefore the following rule needs no explanation: RULE.-Divide the L.C.M. of the given denominators by each denominator in turn, and multiply the corresponding numerator by the quotient. The product thus obtained is the new numerator, and the L.C.M. is the least common denominator. Ex. 1.-Reduce,,, to equivalent fractions, having the least common denominator. 12 3, 12, 8, 10 1, 1, 2, 5 .. The L.C.M. is 12 × 2 × 5 = 120. Dividing 120 by the respective denominators, we get as quotients 40, 10, 15, 12; and hence the given fractions become Ex. 2. Reduce to their least common denominator the following, 18, 43, 38. 52 45 30, 18, 45, 63 2 2, 2, 1, 7 1, 1, 7 Hence the L.C.M. is 45 × 2 × 7 = 630. Dividing 630 by the respective denominators, we get for quotients 21, 35, 14, 10. Hence the required fractions are 11 × 21, 13 X 35, 52 × 14, 29 × 10 X 630 630 630 630 ; or, 231 728, 630 290. 630 NOTE 1.-The operation of dividing the L. C. M. of the denominators is often simplified by using the L.C.M. in its factorial form. Thus, in our present example, the L.C.M. is 45 × 2 ×x 7. Now it is easy to see that the quotient of this by 30 or 3 × 2 × 5 is 3 × 7 or 21, and that the quotient by 18 or 9 x 2 is 5 x 7 or 35, and so on. We thus avoid the process of long division. NOTE 2.It is sometimes necessary, and generally advisable, especially for beginners, to reduce the given fractions to their lowest terms before applying the rule for the least common denominator. Thus, the least common denominator of the fractions, 2, 11, taken as they are, is 60; whereas, if we reduce the second fraction to its lowest terms by striking out the factor 3, common to both numerator and denominator, the fractions become,,, and the least common denominator is 20. If, however, the denominator of any such fraction not in its lowest terms is contained in the L.C.M. of the denominators, when the fractions are all in their lowest terms, it is unnecessary to reduce the fraction to its lowest terms. We strongly recommend the beginner, however, to always commence by reducing the given fractions to their lowest terms. 1. Find the L.C.M. of Ex. IV. (1.) 2, 6, 8, 12. (2.) 4, 9, 10, 14. 45. (4.) 12, 20, 35, 126. (5.) (3.) 15, 21, 40, 39, 65, 52, 140. 2. Reduce to their least common denominator (1.) 2, 3, 15, lb. (2.) †, Ta, Li, H. (3.) 1, †, 11, 132. (4.) 1, 2, 4. 201 (6.) 16, T'', f. 38 (5.), T, 7, 138. 3. A can run round a ring in three minutes, B in four minutes, and C in six minutes, and they start together. In how many minutes will they all be again at the starting point? 4. Arrange the fractions 3, 4, 15, 17, in order of magnitude. 5. Multiply the greatest of the fractions 22, 133, 11 by 339. 6. Divide the least of the fractions, 1, 48, by 6. 7. Reduce to a simple fraction the complex fraction having the greater of the fractions, in the numerator, and the less in the denominator. 8. Which of the fractions, is nearer to ? 10. Arrange in order of magnitude the following: 11. Show that nine times the less of the fractions is eight times the greater. 13 1 4 12. Show that the ratio 18: 7 is a ratio of greater inequality than the ratio 41:16. Addition of Fractions. 20. We have shown (Art. 6) that the numerator and denominator respectively represent the antecedent and consequent of a ratio; and it is evident from the definition of a ratio (Art. 6) that the sum of two ratios having the same consequent is equal to a ratio whose antecedent is the sum of the given antecedents, and whose consequent is unaltered. Hence we have the following rule for addition of fractions— RULE.-Bring the given fractions to their least common denominator, add together the numerators thus obtained, and place under the sum the least common denominator. Ex. 1.-Add together,, T. The least common denominator is easily found to be 42. Dividing this by each of the given denominators, we get as quotients, 14, 2, 3, 7. Ex. 2. Add together 31, 23, 138, 475. The sum of the integral parts of the given fractions = 3 + 2 + 1 + 4 = 10. Hence, 33 + 27 + 111 + 411⁄2 = 10 + 금+급 + ᄒᄒ + fs. The least common denominator is easily found to be 180. Dividing this by each of the given denominators we get as quotients 36, 20, 6, 15. Hence the required sum 21. After the preceding article there will be no difficulty in comprehending the following rule- RULE. Bring the given fractions to their least common denominator, subtract the numerators thus obtained, and under the difference place the least common denominator. Ex. 1.-Subtract from 3. Here the number to be subtracted is the greater. We shall, however, take the less from the greater, and put the negative sign to the remainder, meaning by this that the remainder has yet to be subtracted. .. required result = 2+1=218. We will now give an example involving both addition and subtraction. Ex. 3.-Find the value of 6 78 +41 17. Whenever we have an expression involving both + and signs, the simplest method is to add together all the quantities affected with the plus sign, and likewise those affected with the minus sign; then taking the difference between these two sums, we place the sign of the greater sum before the result. = Thus, taking first the integers, we have 67+ 4 - 1 = 10 - 8 2 (it must be remembered that the sign + is understood before a number which appears without a sign when it stands alone or at the head of an expression). We shall give one more example in order to show how brackets are to be treated. Ex. 4. Find the value of 7 - (21 – 31) + (41 – 14) (21% + 31'). The general rule, which the student will better understand when we come to Algebra, is this— When a minus sign stands before a bracket, it changes all the signs within on removing the bracket; but when a plus sign stands before the bracket, the latter may be removed without changing any of the signs within. |