Elements of Algebra

56 ELEMENTS SECT. I. CHAP. XIX. *Of the Method of representing Irrational Numbers by Fractional Exponents.* 195. We have shewn in the preceding chapter, that the square of any power is found by doubling the exponent of that power; or that, in general, the square, or the second power, of $a^n$, is $a^{2n}$; and the converse also follows, viz. that the square root of the power $a^{2n}$ is $a^n$, which is found by taking half the exponent of that power, or dividing it by 2. 196. Thus, the square root of $a^2$ is $a^1$, or $a$; that of $a^4$ is $a^2$; that of $a^6$ is $a^3$; and so on: and, as this is general, the square root of $a^3$ must necessarily be $a^{\frac{3}{2}}$, and that of $a^5$ must be $a^{\frac{5}{2}}$; consequently, we shall in the same manner have $a^{\frac{1}{2}}$ for the square root of $a^1$. Whence we see that $a^{\frac{1}{2}}$ is equal to $\sqrt{a}$; which new method of representing the square root demands particular attention. 197. We have also shewn, that, to find the cube of a power, as $a^n$, we must multiply its exponent by 3, and consequently that cube is $a^{3n}$. Hence, conversely, when it is required to find the third, or cube root, of the power $a^{3n}$, we have only to divide that exponent by 3, and may therefore with certainty conclude, that the root required is $a^n$: consequently $a^1$, or $a$, is the cube root of $a^3$; $a^2$ is the cube root of $a^6$; $a^3$ of $a^9$; and so on. 198. There is nothing to prevent us from applying the same reasoning to those cases, in which the exponent is not divisible by 3, or from concluding that the cube root of $a^2$ is $a^{\frac{2}{3}}$, and that the cube root of $a^4$ is $a^{\frac{4}{3}}$, or $a^{1\frac{1}{3}}$; consequently, the third, or cube root of $a$, or $a^1$, must be $a^{\frac{1}{3}}$: whence also, it appears, that $a^{\frac{1}{3}}$ is the same as $\sqrt[3]{a}$. 199. It is the same with roots of a higher degree: thus, the fourth root of $a$ will be $a^{\frac{1}{4}}$, which expression has the same value as $\sqrt[4]{a}$; the fifth root of $a$ will be $a^{\frac{1}{5}}$, which is consequently equivalent to $\sqrt[5]{a}$; and the same observation may be extended to all roots of a higher degree.