Elements of Algebra

CHAP. II. ADDITIONS. 519 positively, will be > $\frac{1}{2q^2}$, because $1 - \frac{sH}{q(\sqrt{K} + \frac{p}{q})} > \frac{1}{2}$. So that we shall have the two quantities, $p - q\sqrt{K}$, and $r - s\sqrt{K}$; or rather, making $a = \sqrt{K}$, $p - aq$, and $r - as$: which will be subject to the same conditions as we have supposed in Art. 24, and from which we shall draw similar conclusions: therefore, $\ldots$ (Art. 26), if we had $p^2 - Kq^2 = -H$, then it would be necessary to seek the numbers $r$ and $s$ such, that $ps - qr = -1$, and we should have these two equations, $q\sqrt{K} - p = \frac{H}{q(\sqrt{K} + \frac{p}{q})}$ $s\sqrt{K} - r = \frac{1}{q}(\frac{sH}{q(\sqrt{K} + \frac{p}{q})} - 1)$. As $H < \sqrt{K}$, and $s < q$, it is evident, that $\frac{sH}{q(\sqrt{K} + \frac{p}{q})}$ will be $< 1$; so that the quantity $s\sqrt{K} - r$ will be negative. Now, I say that this quantity, taken positively, will be greater than $q\sqrt{K} - p$; to prove which, it must be demonstrated, that $\frac{1}{q}(1 - \frac{sH}{q(\sqrt{K} + \frac{p}{q})}) > \frac{H}{q(\sqrt{K} + \frac{p}{q})}$, or rather, that $1 > \frac{H(1 + \frac{s}{q})}{\sqrt{K} + \frac{p}{q}}$; that is to say, $\sqrt{K} + \frac{p}{q} > H + \frac{sH}{q}$; but $H < \sqrt{K}(hyp.)$; it is therefore sufficient to prove, that $\frac{p}{q} > \frac{s\sqrt{K}}{q}$, or that $p > s\sqrt{K}$; which is evident, because the quantity $s\sqrt{K} - r$ being negative, we must have $r > s\sqrt{K}$, and much more $p > s\sqrt{K}$, since $p > r$. Thus, the two quantities, $p - q\sqrt{K}$, and $r - s\sqrt{K}$, will have different signs, and the second will be greater than the