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We have nth roots that can be rewritten as fractional powers:

$$\sqrt[n] x = x^\frac 1 n$$

I was looking around on Wikipedia and some other online material, but I couldn't find any definitive set of numbers that the index, $n$, belongs to.

Wikipedia says this in its nth root article:

The nth root of a number x, where n is a positive integer...

It then goes on to include this a few lines down:


So one part states that $n$ must be positive integral, but another leads toward non-positive integral. However, I couldn't really pull anything from the Exponentiation article that was referenced.

Is it permitted for the index to be fractional, such that:

$$\sqrt[\frac 1 n] x = x^n$$

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2 Answers 2

up vote 3 down vote accepted

There's no mathematical reason for left-hand side of your equation to be invalid. It's just that the right-hand side is simpler and more conventional, just as one would prefer to write $a/b$ rather than $1/(b/a)$.

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That's all I was after, thanks. It would be pretty silly to use it, but then again, I can think of some non-serious situations where it can come in handy as long as it's still valid. Programming isn't the only thing you can obfuscate :) Oh, and congrats on your first 1k. –  chris Dec 5 '12 at 12:06

In general we can define $x^{a/b}$ to be a number $r$ such that $r^b = x^a$. This agrees with the usual definition of $x^a$ (take $b=1$) and $\sqrt[b]x$ (take $a=1$). When $x$ is positive, there will be exactly one such $r$, also positive.

We can even define $x^z$ where $z$ is irrational by observing that if $z_0, z_1, z_2\ldots$ is a sequence of rational numbers that converges to $z$, then $x^{z_0}, x^{z_1}, x^{z_2}\ldots$ is also a convergent sequence, and its limit depends only on $z$, and not on the particular sequence $z_0, z_1, z_2\ldots$ that we use to approximate it. Then we can define $x^z$ to be this limit.

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