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. Or the following rule may be used: Expunge the decimal point of the divisor, and move that of the dividend as many places to the right as there were places in the divisor, using ciphers if necessary. Then proceed as in common division, making one decimal place in the quotient for every decimal place of the final dividend which is used. Thus 17·314 divided by 61·2 is 173·14 divided by 612, and the decimal point must precede the first figure of the quotient. But 17·314 divided by 6617·5 is 173·14 by 66175; and since three decimal places of 173·14000 ... must be used before a quotient figure can be found, that quotient figure is the third decimal place, or the quotient is ·002..... EXAMPLES. 3·1 ·00062 ----- = 1240, ------ = ·00096875 ·0025 ·64 EXERCISES. 15·006 × 15·006 - ·004 × ·004 Shew that ----------------------------- = 15·002, 15·01 and that ·01 × ·01 × ·01 + 2·9 × 2·9 × 2·9 --------------------------------- = 2·9 × 2·9 - 2·9 × ·01 + ·01 × ·01 2·91 1 1 365 What are -------, ---------, and ------, as far as 6 places 3·14159 2·7182818 ·18349 of decimals?--_Answer_, ·318310, ·367879, and 1989·209221. Calculate 10 terms of each of the following series, as far as 5 places of decimals. 1 1 1 1 1 + --- + ----- + --------- + ------------- + &c. = 1·71824. 2 2 × 3 2 × 3 × 4 2 × 3 × 4 × 5 1 1 1 1 1 + --- + --- + --- + --- + &c. = 2·92895. 2 3 4 5