Elements of Algebra

CHAP. XII. OF ALGEBRA. 173 $s = 0 \cdot 142857142857142857, \ldots$ $10s = 1 \cdot 42857142857142857, \ldots$ $100s = 14 \cdot 2857142857142857, \ldots$ $1000s = 142 \cdot 857142857142857, \ldots$ $10000s = 1428 \cdot 57142857142857, \ldots$ $100000s = 14285 \cdot 7142857142857, \ldots$ $1000000s = 142857 \cdot 142857142857, \ldots$ Subtract $s = 0 \cdot 142857142857, \ldots$ $999999s = 142857.$ And, dividing by 999999, we have $s = \frac{142857}{999999} = \frac{1}{7}$. Wherefore the decimal fraction, which was represented by $s$, is $=\frac{1}{7}$. 532. In the same manner, $\frac{2}{7}$ may be transformed into a decimal fraction, which will be $0 \cdot 28571428, \ldots$ and this enables us to find more easily the value of the decimal fraction which we have represented by $s$; because $0 \cdot 28571428, \ldots$ must be the double of it, and, consequently, $= 2s$. Now we have seen that $100s = 14 \cdot 28571428571, \ldots$ So that subtracting $2s = 0 \cdot 28571428571, \ldots$ there remains $98s = 14$ wherefore $s = \frac{14}{98} = \frac{1}{7}$. We also find $\frac{3}{7} = 0 \cdot 42857142857, \ldots$ which, according to our supposition, must be equal to $3s$; and we have found that $10s = 1 \cdot 42857142857, \ldots$ So that subtracting $3s = 0 \cdot 42857142857, \ldots$ we have $7s = 1$, wherefore $s = \frac{1}{7}$. 533. When a proposed fraction, therefore, has the denominator 7, the decimal fraction is infinite, and 6 figures are continually repeated; the reason of which is easy to perceive, namely, that when we continue the division, a remainder must return, sooner or later, which we have had already. Now, in this division, 6 different numbers only can form the remainder, namely, 1, 2, 3, 4, 5, 6; so that, at least, after the sixth division, the same figures must return; but when the denominator is such as to lead to a division without remainder, these cases do not happen. 534. Suppose now that 8 is the denominator of the fraction proposed: we shall find the following decimal fractions: