Elements of Algebra

CHAP. XVI. OF ALGEBRA. 293 $\frac{q}{p}$ must express the root $x$ with tolerable exactness; that is to say, we have $\frac{q}{p} = x$ nearly. We shall have also $\frac{r}{q} = x$*, and the multiplication of the two values will give $\frac{r}{p} = x^2$. Farther as $\frac{s}{r} = x$, we shall also have $\frac{s}{p} = x^3$; then, since $\frac{t}{s} = x$, we shall have $\frac{t}{p} = x^4$, and so on. 793. For the better explanation of this method, we shall begin with an equation of the second degree, $x^2 = x + 1$, and shall suppose that in the above series we have found the terms $p, q, r, s, t, \ldots$. Now, as $\frac{q}{p} = x$, and $\frac{r}{p} = x^2$, we shall have the equation $\frac{r}{p} = \frac{q}{p} + 1$, or $q + p = r$. And as we find, in the same manner, that $s = r + q$, and $t = s + r$; we conclude that each term of our series is the sum of the two preceding terms; so that having the first two terms, we can easily continue the series to any length. With regard to the first two terms, they may be taken at pleasure: if we therefore suppose them to be 0, 1, our series will be 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, $\ldots$ and such, that if we divide any term by that which immediately precedes it, we shall have a value of $x$ so much nearer the true one, according as we have chosen a term more distant. The error, indeed, is very great at first, but it diminishes as we advance. The series of those values of $x$, in the order in which they are always approximating towards the true one, is as follows: $x = \frac{1}{0}, \frac{1}{1}, \frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, \frac{13}{8}, \frac{21}{13}, \frac{34}{21}, \frac{55}{34}, \frac{89}{55}, \frac{144}{89}, \ldots$ If, for example, we make $x = \frac{21}{13}$, we have $\frac{441}{169} = \frac{21}{13} + 1 = \frac{34}{13}$, in which the error is only $\frac{1}{169}$. Any of the succeeding terms will render it still less. 794. Let us also consider the equation $x^2 = 2x + 1$; and since, in all cases, $x = \frac{q}{p}$, and $x^2 = \frac{r}{p}$, we shall have * It must only be understood here that $\frac{r}{q}$ is nearly equal to $x$.