### Part I. Section I. Chapter 19. “Of the Method of representing Irrational Numbers by Fractional Exponents.”

195 We have shown 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, namely 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 $$a$$. Whence we see that $$a^{\frac{1}{2}}$$ is equal to $$\surd a$$; which new method of representing the square root demands particular attention.

197 We have also shown, 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 $$\sqrt[3]{\vphantom{1}}a$$.

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 $$\sqrt[4]{\vphantom{1}}a$$; the fifth root of $$a$$ will be $$a^{\frac{1}{5}}$$, which is consequently equivalent to $$\sqrt[5]{\vphantom{1}}a$$; and the same observation may be extended to all roots of a higher degree.

200 We may therefore entirely reject the radical signs at present made use of, and employ in their stead the fractional exponents which we have just explained: but as we have been long accustomed to those signs, and meet with them in most books of Algebra, it might be wrong to banish them entirely from calculation; there is, however, sufficient reason also to employ, as is now frequently done, the other method of notation, because it manifestly corresponds with the nature of the thing. In fact, we see immediately that $$a^{\frac{1}{2}}$$ is the square root of $$a$$, because we know that the square of $$a^{\frac{1}{2}}$$, that is to say, $$a^{\frac{1}{2}}$$ multiplied by $$a^{\frac{1}{2}}$$, is equal to $$a^1$$, or $$a$$.

201 What has been now said is sufficient to show how we are to understand all other fractional exponents that may occur. If we have, for example, $$a^{\frac{4}{3}}$$, this means, that we must first take the fourth power of $$a$$, and then extract its cube, or third root; so that $$a^{\frac{4}{3}}$$ is the same as the common expression $$\sqrt[3]{\vphantom{1}}a^4$$ Hence, to find the value of $$a^{\frac{3}{4}}$$, we must first take the cube, or the third power of $$a$$, which is $$a^3$$, and then extract the fourth root of that power; so that $$a^{\frac{3}{4}}$$ is the same as $$\sqrt[4]{\vphantom{1}}a^3$$, and $$a^{\frac{4}{5}}$$ is equal to $$\sqrt[5]{\vphantom{1}}a^4$$, etc.

202 When the fraction which represents the exponent exceeds unity, we may express the value of the given quantity in another way: for instance, suppose it to be $$a^{\frac{5}{2}}$$; this quantity is equivalent to $$a^{2\frac{1}{2}}$$, which is the product of $$a^2$$ by $$a^{\frac{1}{2}}$$: now $$a^{\frac{1}{2}}$$ being equal to $$\surd a$$, it is evident that $$a^{\frac{5}{2}}$$ is equal to $$a^2 \surd a^5$$; also $$a^{\frac{10}{3}}$$, or $$a^{3\frac{1}{3}}$$, is equal to $$a^3 \sqrt[3]{\vphantom{1}}a$$; and $$a^{\frac{15}{4}}$$, that is, $$a^{3\frac{3}{4}}$$, expresses $$a^3 \sqrt[4]{\vphantom{1}} a^3$$. These examples are sufficient to illustrate the great utility of fractional exponents.

203 Their use extends also to fractional numbers: for if there be given $$\frac{1}{\surd a}$$, we know that this quantity is equal to $$\frac{1}{a^{\frac{1}{2}}}$$; and we have seen already that a fraction of the form $$\frac{1}{a^n}$$ may be expressed by $$a^{-n}$$; so that instead of $$\frac{1}{\surd a}$$ we may use the expresssion $$a^{-\frac{1}{2}}$$; and, in the same manner, $$\frac{1}{\sqrt[3]{\vphantom{1}} a}$$ is equal to $$a^{-\frac{1}{3}}$$. Again, if the quantity $$\frac{a^2}{\sqrt[4]{\vphantom{1}} a^3}$$ be proposed; let it be transformed into this, $$\frac{a^2}{a^{\frac{3}{4}}}$$, which is the product of $$a^2$$ by $$a^{-\frac{3}{4}}$$; now this product is equivalent to $$a^{\frac{5}{4}}$$, or to $$a^{1\frac{1}{4}}$$, or lastly, to $$a\sqrt[4]{\vphantom{1}}a$$. Practice will render similar reductions easy.

204 We shall observe, in the last place, that each root may be represented in a variety of ways; for $$\surd a$$ being the same as $$a^{\frac{1}{2}}$$, and ½ being transformable into the fractions, ²⁄₄, ³⁄₆, ⁴⁄₈, ⁵⁄₁₀, ⁶⁄₁₂, etc. it is evident that $$\surd a$$ is equal to $$\sqrt[4]{\vphantom{1}}a^2$$ or to $$\sqrt[6]{\vphantom{1}}a^3$$, or to $$\sqrt[8]{\vphantom{1}}a^4$$, and so on. In the same manner, $$\sqrt[3]{\vphantom{1}}a$$, which is equal to $$a^{\frac{1}{3}}$$, will be equal to $$\sqrt[6]{\vphantom{1}}a^2$$, or to $$\sqrt[9]{\vphantom{1}}a^3$$, or to $$\sqrt[12]{\vphantom{1}}a^4$$. Hence also we see that the number $$a$$, or $$a^1$$, might be represented by the following radical expressions:

$\sqrt[2]{\vphantom{1}}a^2, \; \sqrt[3]{\vphantom{1}}a^3, \; \sqrt[4]{\vphantom{1}}a^4, \; \sqrt[5]{\vphantom{1}}a^5, \; \textrm{etc.}$

205 This property is of great use in multiplication and division; for if we have, for example, to multiply $$\sqrt[2]{\vphantom{1}}a$$ by $$\sqrt[3]{\vphantom{1}}a$$, we write $$\sqrt[6]{\vphantom{1}}a^3$$ for $$\sqrt[2]{\vphantom{1}}a$$ and $$\sqrt[6]{\vphantom{1}}a^2$$ instead of $$\sqrt[3]{\vphantom{1}}a$$; so that in this manner we obtain the same radical sign for both, and the multiplication being now performed, gives the product $$\sqrt[6]{\vphantom{1}}a^5$$. The same result is also deduced from $$a^{\frac{1}{2}+\frac{1}{3}}$$, which is the product of $$a^{\frac{1}{2}}$$ multiplied by $$a^{\frac{1}{3}}$$; for $$\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$$, and consequently the product required is $$a^{\frac{5}{6}}$$, or $$\sqrt[6]{\vphantom{1}}a^5$$.

On the contrary, if it were required to divide $$\sqrt[2]{\vphantom{1}}a$$, or $$a^{\frac{1}{2}}$$, by $$\sqrt[3]{\vphantom{1}}a$$, or $$a^{\frac{1}{3}}$$, we should have for the quotient $$a^{\frac{1}{2}-\frac{1}{3}}$$, or $$a^{\frac{3}{6}-\frac{2}{6}}$$, that is to say, $$a^{\frac{1}{6}}$$, or $$\sqrt[6]{\vphantom{1}}a$$.

#### Editions

1. Leonhard Euler. Elements of Algebra. Translated by Rev. John Hewlett. Third Edition. Longmans, Hurst, Rees, Orme, and Co. London. 1822.
2. Leonhard Euler. Vollständige Anleitung zur Algebra. Mit den Zusätzen von Joseph Louis Lagrange. Herausgegeben von Heinrich Weber. B. G. Teubner. Leipzig and Berlin. 1911. Leonhardi Euleri Opera omnia. Series prima. Opera mathematica. Volumen primum.