17 1 2 17X4 2 ratio (12 9. To divide a fraction by a whole number, we may either multiply the denominator by the number, or we may divide the numerator by it. Ex.-Divide *} by 4. We may proceed thus, (1.) 13 = 4 1 2:4 1 (2.) 11 = 4 As to the first method The ratio of 12:17 will evidently be diminished 4 times if we divide its antecedent 12 by 4. We thus get the • 4):17 or 3:17; and it therefore follows that ** • 4 = As to the second method The ratio of 12:17 can also be diminished 4 times by increasing its consequent or divisor 4 times, so that we thus get the ratio 12:(17 x 4) or 12:68. It therefore follows that 14 = 4 It may be remarked, as in Art. 8, that the two results, 2 and t, have exactly the same value, for the latter can be obtained from the former by multiplying each of its terms by 4 (see Art. 7). And again, in actual practice, we usually take the first method when the numerator contains the divisor as a factor, but not otherwise. Thus (1.) 19 = 6 (2.) 1} = 5 :51 (3.) 11 + 8 = -13 Here it is convenient to divide the numerator and denominator by the common factor 4 (Art. 7). We then have is : 8 1 2 - 4 ,,, 10. To reduce a mixed number to an improper fraction. Looking at our definition of a mixed number (Art. 6), the following rule is evident: Multiply the integral part by the denominator of the fractional part, and add in the numerator ; this gives the required numerator, and the denominator of the fractional part is the required denominator. 186 13 X 5 17X8 5 5 9 Ex.-Reduce 53, 74 to improper fractions. (1.) 53 5 X 9+2 - 4. (2.) 7 = 7x8+ 3 11. To reduce a complex fraction to its equivalent simple fraction. Before stating a rule, let us take an example. 35 Suppose we have to reduce to an equivalent simple fraction. 31 3 X 5+1 S Now, by the last Art., 5% 5X9+2 Again, the ratio į : will not be altered in value if we multiply both its terms by the same quantity. Let us multiply them by 9 and it becomes 4 x 9:x 9. Now, by Art. 8, 46 x 9 = = 16X9 and 4 x 9 7 꾸 = 47. The ratio then becomes 1849 : 47. We will again multiply the terms of this ratio by the same quantity, viz. by 5, and we get the ratio 16X9 x 5:47 x 5. Now, by Art. 8, 16X9 x 5 16x9 = 16x9 = 16 x 9. Hence the ratio 46 : 4 is equivalent to the ratio 16 x 9 : 47 x 5, and hence the fraction 3 16X9 Now 16 and 9 are called the extreme terms of the complex fraction 3 and 5 and 47 are called its mean terms. 5 99 5 5 + 5 1 4 7 X 5 16 We arrive then at the following rule :RULE.—Bring the numerator and denominator to the form of simple fractions, then multiply together the extreme terms for a new numerator, and the mean terms for a new denominator. 7} 335 36 Thus, 21 7 X 5 + 1 5 9 = 36 X 9 5 X 19 9 12. To reduce a compound fraction to its equivalent simple fraction. Let it be required to find the simple fraction equivalent to the compound fraction of . 5 x 3 5 X 3 7 7 X 4 3 X 5 Now i of « is the ratio 3:4, where the unit of this ratio is . It is therefore, from the definition of ratio, equal to 3 times this unit divided by 4. Now 3 times (Art. 8), And :. 3 times ; 4 = 5 X 3 4 = 5*_3 = 3**5. 4 X 7 Hence we arrive at the result of And in the same way we might show that Ž of of of 11 7 X 4 X 3 X 11. Hence the rule : RULE.—Multiply together the several numerators for a new numerator, and the several denominators for a new denominator. Ex. 3} of 21 of 6 = 3 X 8 + 1 of 2 X 2+1 of f of of 8 X 9 X 5 X 1 8 2 2 5 X 5 X 6 Ex. II. 1. Reduce the following to improper fractions 3},4%, 1001'0, 351, 1, 1134. 2. Reduce the integer 19 to sixths, tenths, thirteenths, eighteenths, nineteenths, and twentieths. 3. Bring the following fractions to integers, and reduce them respectively to fourths, sixths, eighths, tenths, twelfths, and fourteenths 3®, 1 }, 84, 43, 4, ?. 4. Multiply the following fractions each by 10, 11, 12 4, 16, I's, 13, 7, 31. 5. By how much does 8 times the fraction y exceed the quotient of 686 by 3 ? 6. Divide the following fractions each by 6, 7, 18– 37, 14, 16, 315, 103, 1101: 7. Diminish the following ratios respectively 6, 7, 8-fold 12:5, 375 : 4, 97:2}. ligits is divisible by 3. 0. is also divisible by 3. med by the last three figures is digits is divisible by 9. he digits in the odd places (that uth, &c.) is equal to the sum of or the one exceeds the other by formed by its last two figures is of its digits is a multiple of 3. le by 37, when it is composed of three times, or any multiple of 1444, &c. is three figures repeated in the 11, 13. 3023 are divisible by 7, 11, and Etten 023023. four figures repeated in the same 137. are both divisible by 73 and 137; 02760276. on then be reduced to its lowest out factors determined by in lowest terms. uch divisible by 4, for the numformed by the last two figures of nee, dividing numerator and de each i Hence 79 2 79 2 + 4 = 198 5 6 1 66 187; |