Leonhard Euler · 1770 (1828 English ed.) · Internet Archive
Elements of Algebra
All 627 leaves · leaf 95
Transcription
58 ELEMENTS SECT. I.
ner, $\frac{1}{\sqrt[3]{a}}$ is equal to $a^{-\frac{1}{3}}$. Again, if the quantity $\frac{a^2}{\sqrt[4]{a^3}}$ 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 $a^{2-\frac{3}{4}}$, or to $a^{\frac{5}{4}}$, or lastly, to $a\sqrt[4]{a}$. 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 $\sqrt{a}$ being the same as $a^{\frac{1}{2}}$, and $\frac{1}{2}$ being transformable into the fractions, $\frac{2}{4}$, $\frac{3}{6}$, $\frac{4}{8}$, \ldots it is evident that $\sqrt{a}$ is equal to $\sqrt[4]{a^2}$, or to $\sqrt[6]{a^3}$, or to $\sqrt[8]{a^4}$, and so on. In the same manner, $\sqrt[3]{a}$, which is equal to $a^{\frac{1}{3}}$, will be equal to $\sqrt[6]{a^2}$, or to $\sqrt[9]{a^3}$, or to $\sqrt[12]{a^4}$. Hence also we see that the number $a$, or $a^1$, might be represented by the following radical expressions:
$\sqrt{a^2}$, $\sqrt[3]{a^3}$, $\sqrt[4]{a^4}$, $\sqrt[5]{a^5}$, \ldots
205. This property is of great use in multiplication and division; for if we have, for example, to multiply $\sqrt{a}$ by $\sqrt[3]{a}$, we write $\sqrt[6]{a^3}$ for $\sqrt{a}$, and $\sqrt[6]{a^2}$ instead of $\sqrt[3]{a}$; so that in this manner we obtain the same radical sign for both, and the multiplication being now performed, gives the product $\sqrt[6]{a^5}$.
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 $\frac{1}{2}+\frac{1}{3}$ is $\frac{5}{6}$, and consequently the product required is $a^{\frac{5}{6}}$, or $\sqrt[6]{a^5}$.
On the contrary, if it were required to divide $\sqrt{a}$, or $a^{\frac{1}{2}}$, by $\sqrt[3]{a}$, or $a^{\frac{1}{3}}$, we should have for the quotient $a^{\frac{1}{2}-\frac{1}{3}}$, or $a^{\frac{3-2}{6}}$, that is to say, $a^{\frac{1}{6}}$, or $\sqrt[6]{a}$.
QUESTIONS FOR PRACTICE RESPECTING SURDS.
1. Reduce 6 to the form of $\sqrt{5}$.
Ans. $\sqrt{36}$.
2. Reduce $a+b$ to the form of $\sqrt{bc}$.
Ans. $\sqrt{(aa+2ab+bb)}$.
3. Reduce $\frac{a}{b\sqrt{c}}$ to the form of $\sqrt{d}$.
Ans. $\sqrt{\frac{aa}{bbc}}$.
4. Reduce $a^2$ and $b^{\frac{3}{2}}$ to the common index $\frac{1}{4}$.
Ans. $a^{\frac{8}{4}}$, and $b^{\frac{6}{4}}$.
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