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Suppose $x$ and $r$ are real numbers. Is $|x^r|=|x|^r$? If so, how do you prove it at the lowest level? (That is, using definitions and theorems available at K-12 level. If this is not sufficient then please feel free to rise to the appropriate level.)

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Note that $x^r$ is not defined for all real $x$ and $r$ (e.g. negative $x$ and $0<r<1$). –  Dirk Sep 23 '11 at 11:55
Yes, this is true, but note that $x^r$ itself may be complex (not real). –  GEdgar Sep 23 '11 at 12:15
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2 Answers

up vote 2 down vote accepted

Sorry, I don't know what K-12 level means. But evaluating $x^r$ for real $r$ and negative $x$ really only makes sense in rather few cases, as long as you stick to real numbers. Is does make sense if $r$ is a positive integer, and in this case the proof should be obvious.

It does also make sense if $r=p/q$ with $p$ a positive integer and $q$ a positive odd integer. In that case $x^r$ is the $q$-th root of $x$ to the power of $p$ and again, the proof of the equation is easy.
If $r$ is either of the form $p$ or $p/q$, $p$ a negative integer and $q$ a positive odd integer, the proof is also easy.

Edit: Now for the general case, where we understand complex numbers but only allow real values for $x$ and $r$.

If we do not restrict ourselves to real numbers, then $x^r$ is in general not unique, but it is of the form $e^{r\ln x}$, where there are in general infinitely many choices for $\ln x$. However, if $r$ is real, then $|e^{r\ln x}|$ is actually unique, and is equal to $e^{r\cdot\mbox{Re}(\ln x)}$. Now, the real part of $\ln x$ is $\ln|x|$. This immediately shows that $|x^r|=|x|^r$.

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"The proof is obvious/easy" - that may be, but isn't the detail what the OP requested ? –  The Chaz 2.0 Sep 23 '11 at 12:22
I don't really think so. The point is that in these cases the proof is really easy, while if you see the general problem, it certainly looks more complicated. Also, see the new part that I added to my answer. –  Stefan Geschke Sep 23 '11 at 12:46
@Stefan: K-12 refers to school (in America), in the sense of from Kindergarten to 12th grade. –  Zhen Lin Sep 23 '11 at 12:50
The addendum is helpful. –  The Chaz 2.0 Sep 23 '11 at 13:12
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Let's restrict first to real $x$:

There is a pesky special case to consider first: If $x=0$, then only $r>0$ makes mathematical sense, in which case: $$ |x^r| = |0^r| = |0| = |0|^r = |x|^r.$$

Next assume $x \neq 0$, and rational $r = p/q$ where $p$ is a whole number and $q$ is a positive odd whole number, so that $x^r$ is guaranteed to be a real number. Recall, for all real numbers $x$, it is true that $|x| = \sqrt{x^2}$.

$$ |x^r| = \sqrt{ (x^r)^2 } = \sqrt{ x^{r\cdot 2} } = \sqrt{ x^{2r} } = \sqrt{ (x^2)^r } = (\sqrt{x^2})^r, $$ where the last equality only works because $x^2 > 0$. Then by the same trick as above, $$ (\sqrt{x^2})^r = |x|^r. $$

The same proof works when $x >0$ and $r$ is an arbitrary real number. All that we require is that $x^r$ is real.

Hope this helps!

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Thank you to both Stefan Geschke and Shaun Ault. –  Sony Sep 23 '11 at 21:46
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