(e.) When the numerator of the unit of the compound fraction is not a multiple of the denominator of the other fraction, the preceding solution will give a fraction in the numerator of the result. In all such cases the following soluIndeed, applying, as it does, the principles of cancellation, it really includes the preceding. 15. Reduce of to a simple fraction. tion should be adopted. First Solution. Since of any fraction must equal 3 times as many parts, each as large as before, & of may be found by multiplying the denominator of by 8, and the numerator by 3. This gives the following written work. Second Solution. the denominator, and expressed by making 3 6 of of written work as before. may be expressed by making 8 a factor of must be 3 times this result, which may be a factor of the numerator. This gives the same NOTE. In writing the work of such examples as the above, the fraction on which the operation is to be performed should always be written first. 16. Reduce of 1 of 3 to a simple fraction. 8 First Solution. 31, or, being the number on which the operation is to be performed, should be written first. of this must equal 15 times as many parts, each as large, and may be expressed by making 15 a factor of the numerator, and 28 a factor of the denominator. of this must equal 8 times as many parts, each as large, and hence may be expressed by making 8 a factor of the numerator, and 9 a factor of the denominator. This would give the following work. Second Solution. 31, or, is the number on which the operation is to be performed, and should therefore be written first. of may be expressed by making 28 a factor of the denominator, and of must be 15 times this result, which may be expressed by making 15 a factor of the numerator. of this result may be expressed by making 9 a factor of the denominator, and §, or 8 times the last result, by making 8 a factor of the numerator. This would give the same written work as before. NOTE.- It will be seen that both of the above solutions give the same numerical process, viz., to make all the numerators of the compound fraction factors of the new numerator, and all the denominators factors of the new denominator, and then cancel and reduce. Reduce each of the following fractions to simple ones. 142. Multiplication of Fractions. 1. What is the product of X? Solution. multiplied by = of, which, found by multiplying the numerator of by 3 and the denominator by 8, gives the following written work. (a.) By reading the sign X as times, we have = times of, which, found by multiplying the numerator of by 4 and the denominator by 9, gives the following written work. (b.) The following form of explanation may be adopted when a very thorough analysis is required. 11. What is the product of 98 Solution. First write as the number on which the operation is to be performed; we then have multiplied by 1 equals, and multiplied by 's will equal g's of this result, which may be expressed by making 98 a factor of the denominator. If multiplying by gives this result, multiplying by must give 81 times this result, which may be expressed by making 81 a factor of the numerator. Hence, NOTE. It will be seen that all the forms of solution give similar forms of written work, the numerators of the several fractions being in all cases factors of the numerator of the product, and the denominators factors of the denominator. * For models of written work, see 171st page, solution to 8th example. 143. Reduction of a Vulgar Fraction to a Decimal Form Reduce to a decimal fraction. Solution., or of 1, = of 7 of 70 tenths 8 tenths, with a remainder of 6 tenths. But 6 tenths 60 hundredths, and of 60 hundredths = 7 hundredths, with a remainder of 4 hundredths. But 4 hundredths 40 thousandths, and of 40 thousandths = 5 thousandths. Hence, } = .8 + .07 + .005 = .875 Solution. 2, or of 1,= 28 of 25 = z's of 250 tenths = 8 tenths, with a remainder of 26 tenths. But 26 tenths = 260 hundredths, and of 260 hundredths 28) 25.00 2.60 ; &c. Hence the following written work. .892857 Ans., or dropping the 28, = or of a millionth, we shall have the approximate value of 2.892857 .080 .0240 .00160 .000200 .000004 Proof. .892857 1000000 = 28. NOTE. = 892857 times 1000000 = 6259000 of Compare the above solutions with 87, (c.) (b.) The fractions may also be reduced by the following solutions: , or units, 10 times as many tenths = 25 X 10 tenths, 25 X 10 X 10 = hun 28 25 × 10 × 10 × 10 × 10 x 10 x 10 28 7 millionths = 6250000 millionths.8928574 25 28 25 × 10 × 10 x 10 x 10 x 10 x 10 14 ៗ 6250000 6250000 = of .000001.892857+ (c.) It is obvious that equivalent decimal form, for introducing into the numerator cannot be reduced to an exactly |