Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 538
Transcription
CHAP. II.
ADDITIONS.
501
must be an integer, positive number, we may prove, that the quantity $\frac{aq^{iv}-p^{iv}}{p''' - aq'''}$ will necessarily be positive, and greater than unity, and that $\mu'''$ can only be the integer number immediately below the same quantity; and so on.
It follows, 1st, that the quantities $p - aq$, $p' - aq'$, $p'' - aq''$, \ldots will successively have different signs; that is, alternately positive and negative, and will form a series continually increasing. 2dly, that if we denote by the sign $\angle$ the integer number, which is immediately less than the value of the quantity placed after that sign, we shall have, for the determination of the numbers $\mu, \mu', \mu'', \ldots$
$\mu \angle \frac{aq''-p''}{p'-aq'}$
$\mu' \angle \frac{aq'''-p'''}{p''-aq''}$
$\mu'' \angle \frac{aq^{iv}-p^{iv}}{p'''-aq'''}$.
Now, we have already seen, that the series $q, q', q'', \ldots$ must terminate in 0; and that then the preceding term will be 1, and the term corresponding to 0 in the other series $p, p', p'', \ldots$ will be $= \pm 1$ at pleasure.
For example, let us suppose that $q^{iv} = 0$, we shall then have $q''' = 1$, and $p^{iv} = 1$; therefore
$p''' - aq''' = p''' - a$, and
$p^{iv} - aq^{iv} = 1$;
therefore $p''' - a$ must be a negative quantity, and less than 1, abstracting from the sign; that is, $a - p'''$ must be $\succ 0$, and $\angle 1$; so that $p'''$ must be the integer number immediately below $a$; we shall therefore know the values of these four terms,
$p^{iv} = 1 \qquad q^{iv} = 0$
$p''' \angle a \qquad q''' = 1$
by means of which, going back through the former formulæ, we may find all the preceding terms. We shall first have the value of $\mu''$, then we shall have $p''$ and $q''$, by the formulæ,
$p'' = \mu'' p''' + p^{iv}$, and
$q'' = \mu'' q''' + q^{iv}$;
from which we shall get $\mu'$, and then $p'$ and $q'$; and so of the rest.
In general, let $q^\rho = 0$, then we shall have $q^{\rho-1} = 1$, and $p^\rho = 1$; and shall prove, as before, that $p^{\rho-1}$ can only be the
This text was produced by machine and has not been corrected by hand. Where it matters, read the scan. Confidence is the model's own, not an external check.