Elements of Algebra

580 ADDITIONS. CHAP. VIII. seen (Art. 3) that, in general, in continued fractions, we may render the denominators positive or negative, according as we take the approximate values less, or greater, than the real ones; but the method of Problem I. (Art. 23, et seq.), absolutely requires the approximate values $μ$, $μ'$, $μ''$, &c. to be all taken less than the real ones. 87. Now, since the fraction $p/q$ is equal to a continued fraction, whose terms are $μ$, $μ'$, $μ''$, &c. it is evident, from Art. 4, that $μ$ will be the quotient of $p$ divided by $q$, that $μ'$ will be that of $q$ divided by the remainder, $μ''$, that of this remainder divided by the second remainder, and so on; so that calling $r$, $s$, $t$, &c. the remainders in question, we shall have, from the nature of division, $p = μq + r$, $q = μ'r + s$, $r = μ''s + t$, &c. where the last remainder must be $= 0$, and the one before the last $= 1$, because $p$ and $q$ are numbers prime to each other. Thus, $μ$ will be the approximate integer value of $p/q$, $μ'$ that of $q/r$, $μ''$ that of $r/s$, &c. these values being all taken less than the real ones, except the last $μ^z$, which will be exactly equal to the corresponding fraction; because the following remainder is supposed to be nothing. Now, as the numbers $μ$, $μ'$, $μ''$, &c. $μ^z$, are the same for the continued fraction, which expresses the value of $p/q$, and for that which expresses the value of $√A$, we may take, as far as the term $m^z$, $p/q = √A$, that is to say, $p^2 - Aq^2 = 0$. Thus, we shall first seek the approximate, deficient value of $p/q$; that is to say, of $√A$, and that will be the value of $μ$; then we shall substitute in $p^2 - Aq^2 = 0$, instead of $p$, its value $μq + r$, which will give $(\mu^2 - A)q^2 + 2μqr + r^2 = 0$, and we shall again seek the approximate, deficient value of $q/r$; that is, of the positive root of the equation, $(\mu^2 - A) \times (q/r)^2 + 2μq/r + 1 = 0$, and we shall have the value of $μ'$. Still continuing to substitute $μ'r + s$, instead of $q$, in the