Elements of Algebra

CHAP. XII. OF ALGEBRA. 121 $\frac{n}{1} = \frac{1}{2}$; $\frac{n-1}{2} = -\frac{1}{4}$; $\frac{n-2}{3} = -\frac{3}{8}$; $\frac{n-3}{4} = -\frac{5}{16}$; $\frac{n-4}{5} = -\frac{7}{32}$; $\frac{n-5}{6} = -\frac{9}{64}$. Then, $a^n = a^{\frac{1}{2}} = \sqrt{a}$ and $a^{n-1} = \frac{1}{\sqrt{a}}$; $a^{n-2} = \frac{1}{a\sqrt{a}}$; $a^{n-3} = \frac{1}{a^2\sqrt{a}}, \ldots$ or we might express those powers of $a$ in the following manner: $a^n = \sqrt{a}$; $a^{n-1} = \frac{\sqrt{a}}{a}$; $a^{n-2} = \frac{a^n}{a^2} = \frac{\sqrt{a}}{a^2}$; $a^{n-3} = \frac{a^n}{a^3} = \frac{\sqrt{a}}{a^3}$; $a^{n-4} = \frac{a^n}{a^4} = \frac{\sqrt{a}}{a^4}, \ldots$ 364. This being laid down, the square root of $a+b$ may be expressed in the following manner: $\sqrt{(a+b)} = \sqrt{a} + \frac{1}{2}b\frac{\sqrt{a}}{a} - \frac{1}{2}\cdot\frac{1}{4}b^2\frac{\sqrt{a}}{a^2} + \frac{1}{2}\cdot\frac{1}{4}\cdot\frac{3}{6}b^3\frac{\sqrt{a}}{a^3} - \frac{1}{2}\cdot\frac{1}{4}\cdot\frac{3}{6}\cdot\frac{5}{8}b^4\frac{\sqrt{a}}{a^4}, \ldots$ 365. If $a$ therefore be a square number, we may assign the value of $\sqrt{a}$, and, consequently, the square root of $a+b$ may be expressed by an infinite series, without any radical sign. Let, for example, $a = c^2$, we shall have $\sqrt{a} = c$; then $\sqrt{(c^2+b)} = c + \frac{1}{2}\cdot\frac{b}{c} - \frac{1}{8}\cdot\frac{b^2}{c^3} + \frac{1}{16}\cdot\frac{b^3}{c^5} - \frac{5}{128}\cdot\frac{b^4}{c^7}, \ldots$ We see, therefore, that there is no number, whose square root we may not extract in this manner; since every number may be resolved into two parts, one of which is a square represented by $c^2$. If, for example, the square root of 6 be required, we make $6 = 4+2$, consequently, $c^2 = 4$, $c = 2$, $b = 2$; whence results $\sqrt{6} = 2 + \frac{1}{2} - \frac{1}{16} + \frac{1}{64} - \frac{5}{1024}, \ldots$ If we take only the two leading terms of this series, we shall have $2\frac{1}{2} = \frac{5}{2}$, the square of which, $\frac{25}{4}$, is $\frac{1}{4}$ greater than 6; but if we consider three terms, we have $2\frac{7}{16} = \frac{39}{16}$, the square of which, $\frac{1521}{256}$, is still $\frac{15}{256}$ too small. 366. Since, in this example, $\frac{5}{2}$ approaches very nearly to the true value of $\sqrt{6}$, we shall take for 6 the equivalent quantity $\frac{25}{4} - \frac{1}{4}$; thus $c^2 = \frac{25}{4}$; $c = \frac{5}{2}$, $b = -\frac{1}{4}$; and calculating only the two leading terms, we find $\sqrt{6} = \frac{5}{2} + \frac{1}{2}\cdot\frac{-\frac{1}{4}}{\frac{5}{2}} = \frac{5}{2} - \frac{1}{20} = \frac{49}{20}$; the square of which