Elements of Algebra

CHAP. I. ADDITIONS. 471 remainder $-23$, which gives either the quotient $-9$, and the remainder $9$, or the quotient $-10$, and the remainder $-14$, and so on. In this way, we obtain $\frac{1103}{887} = 2+\frac{1}{-1+\frac{1}{-3+\frac{1}{-9+, \&c.}}}$ in which we see that all the denominators are negative. 7. We may also make each negative denominator positive by changing the sign of the numerator; but we must then also change the sign of the succeeding numerator; for it is evident that $\{\mu + \frac{1}{-\nu+\frac{1}{\pi+, \&c.}}\} = \{\mu - \frac{1}{\nu-\frac{1}{\pi+, \&c.}}\}$ Then we may also, if we choose, remove all the signs $-$ in the continued fraction, and reduce it to another, in which all the terms shall be positive; for we have, in general, $\{\mu + \frac{1}{-\nu+, \&c.}\} = \{\mu - 1 + \frac{1}{1+\frac{1}{\nu-1+, \&c.}}\}$ as we may easily be convinced of by reducing those two quantities to vulgar fractions*. We may also, by similar means, introduce negative terms instead of positive; for we have $\mu + \frac{1}{\nu+, \&c.} = \mu + 1 - \frac{1}{1+\frac{1}{\nu-1+, \&c.}}$ whence we see, that, by such transformations, we may always simplify a continued fraction, and reduce it to fewer terms: which will take place, whenever there are denominators equal to unity, positive, or negative. In general, it is evident, that, in order to have the continued fraction approximating as nearly as possible to the * Thus, the mixed number, $1+\frac{1}{\nu-1} = \frac{\nu}{\nu-1}$; therefore $\frac{1}{1+\frac{1}{\nu-1}} = \frac{\nu-1}{\nu}$; and, consequently, $\{\mu - 1 + \frac{1}{1+\frac{1}{\nu-1}}\} = \mu - 1 + \frac{\nu-1}{\nu} = \mu - \frac{1}{\nu}$. B.