Reduce the following fractions to decimal fractions:

1,36,297, and1 .
800125064128
Answer, ·00125, ·0288, 4·640625,
and·0078125.

Find decimals of 6 places very near to the following fractions:

27,156,22,194,2637,1,1, and3.
4933370001399072908466277
Answer,·551020, 4·727272, ·000594, 14·923076, ·266175,
·000343, ·002145, and ·010830.

149. From (121) it appears, that if two fractions have the same denominator, the first may be divided by the second by dividing the numerator of the first by the numerator of the second. Suppose it required to divide 17·762 by 6·25. These fractions (138), when reduced to a common denominator, are 17·762 and 6·250, or ¹⁷⁷⁶²/₁₀₀₀ and ⁶²⁵⁰/₁₀₀₀. Their quotient is therefore ¹⁷⁷⁶²/₆₂₅₀, which must now be reduced to a decimal fraction by the last rule. The process at full length is as follows: Leave out the cipher in the denominator, and annex ciphers to the numerator, or, which will do as well, to the remainders, when it becomes necessary, and divide as in (145).

Here four ciphers have been annexed to the numerator, and one has been taken from the denominator. Make five decimal places in the quotient, which then becomes 2·84192, and this is the quotient of 17·762 divided by 6·25.

150. The rule for division of one decimal by another is as follows: Equalise the number of decimal places in the dividend and divisor, by annexing ciphers to that which has fewest places. Then, further, annex as many ciphers to the dividend[18] as it is required to have decimal places, throw away the decimal point, and operate as in common division. Make the required number of decimal places in the quotient.

Thus, to divide 6·7173 by ·014 to three decimal places, I first write 6·7173 and ·0140, with four places in each. Having to provide for three decimal places, I should annex three ciphers to 6·7173; but, observing that the divisor ·0140 has one cipher, I strike that one out and annex two ciphers to 6·7173. Throwing away the decimal points, then divide 6717300 by 014 or 14 in the usual way, which gives the quotient 479807 and the remainder 2. Hence 479·807 is the answer.

The common rule is: Let the quotient contain as many decimal places as there are decimal places in the dividend more than in the divisor. But this rule becomes inoperative except when there are more decimals in the dividend than in the divisor, and a number of ciphers must be annexed to the former. The rule in the text amounts to the same thing, and provides for an assigned number of decimal places. But the student is recommended to make himself familiar with the rule of the characteristic [given in the Appendix], and also to accustom himself to reason out the place of the decimal point. Thus, it should be visible, that 26·119 ÷ 7·2436 has one figure before the decimal point, and that 26·119 ÷ 724·36 has one cipher after it, preceding all significant figures.