Elements of Algebra

564 ADDITIONS. CHAP. VII. Since we have $t'''^2 - A u'''^2 = 1$, $t^{iv}^2 - A u^{iv}^2 = 1$, and $\theta^2 - A v^2 = 1$, we shall have $\theta^2 - t'''^2 = A(v^2 - u'''^2)$, and $t^{iv}^2 - \theta^2 = A(u^{iv}^2 - v^2)$; whence we see that, if $\theta > t'''$ and $t^{iv} > \theta$, we shall also have $v > u'''$, and $u^{iv} > v$. Farther, we shall also have these other values of $t$ and $u$; namely, $t = \theta t^{iv} - A v u^{iv}$, $u = \theta u^{iv} - v t^{iv}$, which will satisfy the same equation, $t^2 - A u^2 = 1$; for, by substitution, we shall have $(\theta t^{iv} - A v u^{iv})^2 - A(\theta u^{iv} - v t^{iv})^2 = (\theta^2 - A v^2) \times (t^{iv}^2 - A u^{iv}^2) = 1$, an identical equation, because $\theta^2 - A v^2 = 1$, and $t^{iv}^2 - A u^{iv}^2 = 1$ (hyp.). Now, these two last equations give $\theta - v \sqrt{A} = \frac{1}{\theta + v \sqrt{A}}$, and $t^{iv} - u^{iv} \sqrt{A} = \frac{1}{t^{iv} + u^{iv} \sqrt{A}}$; hence, substituting instead of $\theta$, in the expression, $u = \theta u^{iv} - v t^{iv}$, the quantity $v \sqrt{A} + \frac{1}{\theta + v \sqrt{A}}$; and, instead of $t^{iv}$, the quantity $u^{iv} \sqrt{A} + \frac{1}{t^{iv} + u^{iv} \sqrt{A}}$, we shall have $u = \frac{u^{iv}}{\theta + v \sqrt{A}} - \frac{v}{t^{iv} + u^{iv} \sqrt{A}}$. In the same manner, if we consider the quantity $t''' u^{iv} - u''' t^{iv}$, it may likewise, on account of $t^2 - A u^2 = 1$, be put into the form, $\frac{u^{iv}}{t''' + u''' \sqrt{A}} + \frac{u'''}{t^{iv} + u^{iv} \sqrt{A}}$. Now, it is easy to perceive, that the preceding quantity must be less than this, because $\theta > t'''$, and $v > u'''$; therefore, we shall have a value of $u$, which will be less than the quantity $t''' u^{iv} - u''' t^{iv}$; but this quantity is equal to $v$; for $t''' = \frac{(T+v\sqrt{A})^3 + (T-v\sqrt{A})^3}{2}$, $t^{iv} = \frac{(T+v\sqrt{A})^4 + (T-v\sqrt{A})^4}{2}$, $u''' = \frac{(T+v\sqrt{A})^3 - (T-v\sqrt{A})^3}{2\sqrt{A}}$, $u^{iv} = \frac{(T+v\sqrt{A})^4 - (T-v\sqrt{A})^4}{2\sqrt{A}}$, whence,