# Elementary question power absolute values

If | | denotes the absolute value then is it always true $|x^{s}|=|x|^{s}$ whenever s is any positive real number? I know this holds for any positive integer. What about positive reals?

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I think you should analyse how $x^s$ is defined in your book. – AD. Nov 11 '10 at 6:03

Assuming you're talking about real $x$, the only time you would be concerned with whether the equality holds is if $x$ is negative. (Otherwise, $|x|=x$ and $|x^s|=x^s$, so there's nothing to show.) Now, how do you define $x^s$ if $x$ is negative and $s$ is irrational?

One answer is given in more generality by user3296. You can use complex numbers, in which case it works out well. However, if you only want to work with real values, then trying to take irrational powers of negative numbers is best avoided. To see this, think first about how $x^s$ can be defined when $x$ is positive. One way is by taking rational numbers $p/q$ approaching $s$, and taking a limit of $x^{p/q}=\sqrt[q]{x}^p$. But this definition won't work if $x$ is negative, because for some choices of $p/q$ the result will be imaginary, for others negative, and for others positive. Another way is using the definition $x^s=e^{s\ln{x}}$, where $\ln$ is the natural logarithm defined on the positive reals. To extend this definition to negative $x$, you need a way of defining a logarithm of negative numbers. This actually works (although there is not a unique choice), but the result is that you end up with complex numbers rather than real numbers. For example, you can define $\log$ on $(-\infty,0)$ by $\log(x)=\ln(-x)+\pi i$, so that $x^s=e^{s(\ln(-x)+\pi i)}=e^{s\pi i}(-x)^s$. But this is not real when $s$ is not an integer.

As long as you have no problem using complex exponentials (and logarithms, at least implicitly), then the answer is yes, as user3296 says.

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Yes, even when $x \in \mathbb{C}$. To see this, write $x = |x| e^{i \theta}$, in which case $$|x^s| = ||x|^s e^{i s \theta}| = |x|^s |e^{i s \theta}| = |x|^s.$$

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What is $s$ here? – AD. Nov 11 '10 at 5:55

This is a fairly standard argument proceeding along the lines of first extending the relation from $\mathbb{Z}$ to $\mathbb{Q}$ and finally to $\mathbb{R}$ (of course only for the values where the relation makes sense).

For the first extension use the fact that every rational $r \in \mathbb{Q}$ can be written as $r = p/q$ with $p, q \in \mathbb{Z}$. For the second, use the continuity of the norm and density of $\mathbb{Q}$ in $\mathbb{R}$.

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