Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:

For the extension of powers and roots to indices that are not positive integers, see exponentiation.

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:

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

share|improve this question
add comment

2 Answers

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 $1/(b/a)$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.