fraction power

Let $m$ be an integer and $n$ a positive factor of $m$.  If $x$ is a positive real number, we may write the identical equation

$${(x^{\frac{m}{n}})}^n=x^{\frac{m}{n}\cdot n}=x^m$$

and therefore the definition of $n^{\mathrm{th}}$ root (http://planetmath.org/NthRoot) gives the

$\sqrt[n]{x^m}=x^{\frac{m}{n}}.$(1)

Here, the exponent $\frac{m}{n}$ is an integer.  For enabling the validity of (1) for the cases where $n$ does not divide $m$ we must set the following

Definition.  Let $\frac{m}{n}$  be a fractional number, i.e. an integer $m$ not divisible by the integer $n$, which latter we assume to be positive.  For any positive real number $x$ we define the  fraction power  $x^{\frac{m}{n}}$ as the $n^{\mathrm{th}}$

$x^{\frac{m}{n}}:=\sqrt[n]{x^m}.$(2)

Remarks

1. The existence of the in the righthand side of (2) is provedhere (http://planetmath.org/existenceofnthroot).

2. The defining equation (2) is independent on the form of the exponent $\frac{m}{n}$:  If  $\frac{k}{l}=\frac{m}{n}$,  then we have  ${(\sqrt[n]{x^m})}^{ln}={[{(\sqrt[n]{x^m})}^n]}^l=x^{lm}=x^{kn}={[{(\sqrt[l]{x^k})}^l]}^n={(\sqrt[l]{x^k})}^{ln}$,  and because the mapping $y\mapsto y^{ln}$  is injective in ${ℝ}_{+}$, the positive numbers $\sqrt[l]{x^k}$ and $\sqrt[n]{x^m}$ must be equal.

3. The fraction power function  $x\mapsto x^{\frac{m}{n}}$  is a special case of power function.

4. The presumption that $x$ is positive signifies that one can not identify all $n^{\mathrm{th}}$ roots (http://planetmath.org/NthRoot) $\sqrt[n]{x}$ and the powers $x^{\frac{1}{n}}$.  For example, $\sqrt[3]{-8}$ equals $-2$ and $\frac{2}{6}=\frac{1}{3}$,  but one must not

   $${(-8)}^{\frac{1}{3}}={(-8)}^{\frac{2}{6}}=\sqrt[6]{{(-8)}^2}=\sqrt[6]{64}=\mathrm{\hspace{0.33em}2}.$$

   The point is that ${(-8)}^{\frac{1}{3}}$ is not defined in $ℝ$.  Here we have  $l=6$  and the mapping $y\mapsto y^{ln}$  is not injective in  ${ℝ}_{-}\cup {ℝ}_{+}$. — Nevertheless, some people and books may use also for negative $x$ the equality  $\sqrt[n]{x}=x^{\frac{1}{n}}$  and more generally  $\sqrt[n]{x^m}=x^{\frac{m}{n}}$  where one then insists that  $\gcd (m,n)=1.$

5. According to the preceding item, for the negative values of $x$ the derivative of odd roots (http://planetmath.org/NthRoot), e.g. $\sqrt[3]{x}$, ought to be calculated as follows:

   $$\frac{d\sqrt[3]{x}}{dx}=\frac{d(-\sqrt[3]{-x})}{dx}=-\frac{d{(-x)}^{\frac{1}{3}}}{dx}=-\frac{1}{3}\cdot {(-x)}^{-\frac{2}{3}}(-1)=\frac{1}{3\sqrt[3]{{(-x)}^2}}=\frac{1}{3\sqrt[3]{x^2}}$$

   The result is similar as $\frac{d\sqrt[3]{x}}{dx}$ for positive $x$’s, although the root functions are not special cases of the power function.