Take $\sqrt[n]{m}$ where $n$ is a positive non-integer number. Is this possible?

Question

The title pretty much says it all. I am trying to figure out how to take the $n$-th root of a number $m$ where $n$ is a fractional number. Unfortunately all I could turn up on Google was how to take a square root of a fraction, which is quite different. Is such an operation even possible?

Answer 1

The usual definition of $$ m^{\frac{p}{q}} $$ is that $$ m^{\frac{p}{q}} = \left( m^p \right)^ {\frac{1}{q}} $$ i.e., that you raise $m$ to the $p$ power -- that's easy -- and then take the $q$th root of the result.

Since you're asking for the $n$th root rather than the $n$th power, we have an inverse in there:

$$ \sqrt[n]{m} = \left( m^q \right)^ {\frac{1}{p}}. $$

As a practical matter, an alternative definition is that $$ m^{\frac{p}{q}} = \exp(\frac{p}{q} \log (m)) $$ where $\exp$ and $\log$ are the natural exponential and logarithm functions, which can be defined (via calculus) without reference to anything more than integer powers (just in case you were worried about this being circular).

Answer 2

Suppose $m>0$.

If $n\in\mathbb Q$, write $n=p/q$, where $p,q\in\mathbb Z$ and $q>0$. Define $\sqrt[n]{m}$ as $\sqrt[p]{m^q}$.

If $n\in\mathbb R \text{ or } \mathbb C$, $\sqrt[n]{m}$ is also well defined. (see wikipedia)