Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 209
Transcription
172
ELEMENTS
SECT. III.
The decimal fraction which expresses the value of $2/3$, is also continued ad infinitum; for we have
$3)2.0000000$
$0.6666666 = 2/3$.
Which is also evident from what we have just said, because $2/3$ is the double of $1/3$.
528. If $1/4$ be the fraction proposed, we have
$4)1.0000000$
$0.2500000 = 1/4$.
So that $1/4$ is equal to $0.2500000$, or to $0.25$: which is evidently true, since $2/10 + 5/100 = 25/100 = 1/4$.
In like manner, we should have for the fraction $3/4$,
$4)3.0000000$
$0.7500000 = 3/4$.
So that $3/4 = 0.75$: and in fact
$7/10 + 5/100 = 75/100 = 3/4$.
The fraction $5/4$ is changed into a decimal fraction, by making
$4)5.0000000$
$1.2500000 = 5/4$.
Now, $1 + 25/100 = 5/4$.
529. In the same manner, $1/5$ will be found equal to $0.2$; $2/5 = 0.4$; $3/5 = 0.6$; $4/5 = 0.8$; $5/5 = 1$; $6/5 = 1.2$, \ldots
When the denominator is $6$, we find $1/6 = 0.1666666, \ldots$ which is equal to $0.666666 - 0.5$: but $0.666666 = 2/3$, and $0.5 = 1/2$, wherefore $0.1666666 = 2/3 - 1/2$; or $4/6 - 3/6 = 1/6$.
We find, also, $2/6 = 0.333333, \ldots = 1/3$; but $3/6$ becomes $0.5000000 = 1/2$; also, $5/6 = 0.833333 = 0.333333 + 0.5$, that is to say, $1/3 + 1/2$; or $2/6 + 3/6 = 5/6$.
530. When the denominator is $7$, the decimal fractions become more complicated. For example, we find $1/7 = 0.142857$; however it must be observed that these six figures are continually repeated. To be convinced, therefore, that this decimal fraction precisely expresses the value of $1/7$, we may transform it into a geometrical progression, whose first term is $142857/1000000$, the ratio being $1/1000000$; and consequently, the sum $= \frac{\frac{142857}{1000000}}{1 - \frac{1}{1000000}} = \frac{142857}{999999}$ (by multiplying both terms by $1000000$) $= 1/7$. [See Art. 520.]
531. We may prove, in a manner still more easy, that the decimal fraction, which we have found, is exactly equal to $1/7$; for, by substituting for its value the letter $s$, we have
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