Elements of arithmetic - Augustus De Morgan

Thus,

1	is less	1	by	3	which is not	1
10	than	7	by	70	so much as	10
14		1		2		1
100		7		700		100
142		1		6		1
1000		7		7000		1000
1428		1		4		1
10000		7		70000		10000
14285		1		5		1
100000		7		700000		100000
142857		1		1		1
1000000		7		7000000		1000000
&c.		&c.		&c.		&c.

In the first column is a series of decimal fractions, which come nearer and nearer to ¹/₇, as the third column shews. Therefore, though we cannot find a decimal fraction which is exactly ¹/₇, we can find one which differs from it as little as we please.

This may also be illustrated thus: It is required to reduce ¹/₇ to a decimal fraction without the error of say a millionth of a unit; multiply the numerator and denominator of ¹/₇ by a million, and then divide both by 7; we have then

1	=	1000000	=	1428571¹/₇
7	=	7000000	=	1000000

If we reject the fraction ¹/₇ in the numerator, what we reject is really the 7th part of the millionth part of a unit; or less than the millionth part of a unit. Therefore ¹⁴²⁸⁵⁷/₁₀₀₀₀₀₀ is the fraction required.

EXERCISES.

Make similar tables		3	,	17	, and	1	.
with these fractions		91		143		247

The recurring	3	is	329670,329670, &c.
quotient of	91
	17		118881,118881, &c.
	143
	1		404858299595141700,4048582 &c.
	247

130. The reason for the recurrence of the figures of the quotient in the same order is as follows: If 1000, &c. be divided by the number 247, the remainder at each step of the division is less than 247, being either 0, or one of the first 246 numbers. If, then, the remainder never become nothing, by carrying the division far enough, one remainder will occur a second time. If possible, let the first 246 remainders be all different, that is, let them be 1, 2, 3, &c., up to 246, variously distributed. As the 247th remainder cannot be so great as 247, it must be one of these which have preceded. From the step where the remainder becomes the same as a former remainder, it is evident that former figures of the quotient must be repeated in the same order.