Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 505
Transcription
468
ADDITIONS.
CHAP. I.
tity lies. There is, however, an essential difference between these two methods of taking the approximate values, with respect to the continued fraction which results from it: for if we always take the approximate values *less* than the true ones, the denominators $β$, $γ$, $δ$, &c. will be all positive; whereas they will be all negative, if we take all the approximate values *greater* than the true ones; and they will be partly positive and partly negative, if the approximate values are taken sometimes too small, and sometimes too great.
In fact, if $α$ be less than $a$, $a - α$ will be a positive quantity; wherefore $b$ will be positive, and $β$ will be so likewise: on the contrary, $a - α$ will be negative, if $α$ be greater than $a$; then $b$ will be negative, and $β$ will be so likewise. In the same manner, if $β$ be less than $b$, $b - β$ will always be a positive quantity; therefore $c$ will be positive also, and consequently, also $γ$; but if $β$ be greater than $b$, $b - β$ will be a negative quantity; so that $c$, and consequently also $γ$, will be negative, and so on.
Farther, when negative quantities are considered, I understand by *less* quantities those which, taken positively, would be *greater*. We shall have occasion, however, sometimes to compare quantities simply in respect of their absolute magnitude; but I shall then take care to premise, that we must pay no attention to the signs.
It must be remarked, also, that if, among the quantities $b$, $c$, $d$, &c. one is found equal to an integer number, then the continued fraction will be terminated; because we shall be able to preserve that quantity in it: for example, if $c$ be an integer number, the continued fraction, which gives the value of $a$, will be
$a = \alpha + \frac{1}{\beta + \frac{1}{c}}$.
It is evident, indeed, that we must take $γ = c$, which gives $d = \frac{1}{c - \gamma} = \frac{1}{0} = \infty$; and, consequently, $d = \infty$; so that we shall have
$a = \alpha + \frac{1}{\beta + \frac{1}{\gamma + \frac{1}{\infty}}}$,
the following terms vanishing in comparison with the infinite
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