Elements of Algebra

522 ADDITIONS. CHAP. II. $p^{\text{vii}} = p^{\text{i}}$, and that the difference between the two indices, 1 and 7, is even; whence it follows, that all the succeeding terms will likewise be the same as the preceding; so that we shall have $q^{\text{vii}} = 4$, $q^{\text{viii}} = -3$, $q^{\text{ix}} = 7, \ldots p^{\text{vii}} = -7$, $p^{\text{viii}} = 10, \ldots$ so that, if we choose, we may continue the above series to infinity, only by repeating the same terms. Secondly, let us take the radical $\sqrt{79}$ with a negative sign, and the calculation will be as follows: $q^{\text{ix}} = -6 \times 2 + 5 = -7$ $q^{\text{viii}} = 9 \times 1 - 4 = 5$ $q^{\text{vii}} = -7 \times 1 + 3 = -4$ $q^{\text{vi}} = 10 \times 1 - 7 = 3$ $q^{\text{v}} = -3 \times 5 + 8 = -7$ $q^{\text{iv}} = 5 \times 3 - 7 = 8$ $q^{\text{iii}} = -6 \times 1 - 1 = -7$ $q^{\text{ii}} = 13 \times 1 - 14 = -1$ $q^{\text{i}} = 9 \times 5 - 59 = -14$ $q^0 = -59$ $\ldots, \ldots$ $p^{\text{ix}} = 5$ $p^{\text{viii}} = -6$ $p^{\text{vii}} = 9$ $p^{\text{vi}} = -7$ $p^{\text{v}} = 10$ $p^{\text{iv}} = -3$ $p^{\text{iii}} = 5$ $p^{\text{ii}} = -6$ $p^{\text{i}} = 13$ $p^0 = 9$ $\ldots$ $p^{\text{x}} = \frac{49-79}{-6} = 5$ $p^{\text{ix}} = \frac{25-79}{9} = -6$ $p^{\text{viii}} = \frac{16-79}{-7} = 9$ $p^{\text{vii}} = \frac{9-79}{10} = -7$ $p^{\text{vi}} = \frac{49-79}{-3} = 10$ $p^{\text{v}} = \frac{64-79}{5} = -3$ $p^{\text{iv}} = \frac{49-79}{-6} = 5$ $p^{\text{iii}} = \frac{1-79}{13} = -6$ $p^{\text{ii}} = \frac{196-79}{9} = 13$ $p^{\text{i}} = \frac{59-\sqrt{79}}{9} = 5$ $\mu^{\text{x}} = \frac{7+\sqrt{79}}{5} = 3$ $\mu^{\text{ix}} = \frac{-5-\sqrt{79}}{-6} = 2$ $\mu^{\text{viii}} = \frac{4+\sqrt{79}}{9} = 1$ $\mu^{\text{vii}} = \frac{-3-\sqrt{79}}{-7} = 1$ $\mu^{\text{vi}} = \frac{7+\sqrt{79}}{10} = 1$ $\mu^{\text{v}} = \frac{-8-\sqrt{79}}{-3} = 5$ $\mu^{\text{iv}} = \frac{7+\sqrt{79}}{5} = 3$ $\mu^{\text{iii}} = \frac{1-\sqrt{79}}{-6} = 1$ $\mu^{\text{ii}} = \frac{14+\sqrt{79}}{13} = 1$ $\mu^{\text{i}} = \frac{7+\sqrt{79}}{9} = 1$ We may stop here, since we have found $q^{\text{ix}} = q^{\text{iii}}$, and $p^{\text{ix}} = p^{\text{iii}}$, the difference of the indices 9 and 3 being even; for, by continuing the series, we should only find the same terms that we have found already.