Elements of Algebra

CHAP. XI. ADDITIONS. 499 Now, as we must have $pq' - qp' = \pm 1$, we shall also have, by substituting the preceding values of $p$ and $q$, and effacing what is destroyed, $p'q'' - q'p'' = \mp 1$. Again, substituting in this equation the values of $p'$ and $q'$, there will result $p''q''' - q''p''' = \pm 1$, and so on; so that we shall have, generally, $pq' - qp' = \pm 1$ $p'q'' - q'p'' = \mp 1$ $p''q''' - q''p''' = \pm 1$ $p'''q^{iv} - q'''p^{iv} = \mp 1, \ldots$ So that, if $q'''$, for example, were $= 0$, we should have $-q''p''' = \pm 1$; also, $q'' = 1$, and $p''' = \mp 1$: but if $q^{iv}$ were $= 0$, we should have $-q'''p^{iv} = \mp 1$; therefore $q''' = 1$, and $p^{iv} = \pm 1$; so that, in general, if $q^\varrho = 0$, we shall have $q^{\varrho-1} = 1$; and then $p^\varrho = \pm 1$, if $\varrho$ is even, and $p^\varrho = \mp 1$, if $\varrho$ is odd. Now, as we do not previously know whether the upper, or the under sign is to take place, we must successively suppose $p^\varrho = 1$, and $= -1$: but I say that one of these cases may at all times be reduced to the other; and, for this purpose, it is evidently sufficient to prove, that we can always make the $\varrho$ of the term $q^\varrho$, which must be nothing, either even, or odd, at pleasure. For example, let us suppose that $q^{iv} = 0$, we shall then have $q''' = 1$, and $q'' > 1$, that is, $q'' = 2$, or $> 2$, because the numbers $q, q', q'', \ldots$ naturally form a decreasing series; therefore, since $q'' = \mu'''q''' + q^{iv}$; we shall have $q'' = \mu'''$, so that $\mu''' = 2$ or $> 2$; thus, if we choose, we may diminish $\mu'''$ by unity, without that number being reduced to nothing, and then $q^{iv}$, which was $0$, will become $1$, and $q^v=0$; for putting $\mu''' - 1$, instead of $\mu'''$, we shall have $q'' = (\mu''' - 1)q''' + q^{iv}$; but $q'' = \mu'''$, $q''' = 1$; wherefore, $q^{iv} = 1$; then having $q''' = \mu^{iv}q^{iv} + q^v$, that is, $1 = \mu^{iv} + q^v$, we shall necessarily have $\mu^{iv} = 1$, and $q^v = 0$. Hence we may conclude, in general, that if $q^\varrho = 0$, we shall have $q^{\varrho-1} = 1$, and $p^\varrho = \pm 1$, the doubtful sign being arbitrary. Now, if we substitute the values of $p$ and $q$, given by the preceding formulæ, in $p - aq$, those of $p'$ and $q'$, in $p'-aq'$, and so of others, we shall have $p - aq = \mu (p' - aq') + p'' - aq''$ $p' - aq' = \mu' (p'' - aq'') + p''' - aq'''$ $p'' - aq'' = \mu'' (p''' - aq''') + p^{iv} - aq^{iv}$ $p''' - aq''' = \mu''' (p^{iv} - aq^{iv}) + p^v - aq^v, \ldots$ whence we find KK 2