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Simple question: Is the following assertion valid?

$$ (x^2)^{\frac{n}{2}} = |x|^n $$ for $x \in \mathbb{R}$ and integer $n>0$. Whether yes or no, please help me understand why or why not.

RickyBoy
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    $(x^2)^{\frac{n}{2}} = |x^2|^{\frac{n}{2}} = |x|^{2\frac{n}{2}} = |x|^n$ – Klaus Aug 05 '22 at 17:12
  • Thank you @Klaus, you clearly support the assertion! – RickyBoy Aug 05 '22 at 17:13
  • @Klaus. Is it the round brackets that validate it? – RickyBoy Aug 05 '22 at 17:17
  • What do you mean by that? – Klaus Aug 05 '22 at 17:23
  • @Klaus Perhaps the following is a more precise way of illustrating the exact point of the question. Suppose $x=-1$ and $n=1$. Now $x^{2\times\frac{1}{2}}=x^{1}=-1$ whereas the assertion in the question results in $|x|^1=1$. – RickyBoy Aug 05 '22 at 17:30
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    You are using $(x^a)^b = x^{ab}$ for negative numbers, which is not allowed. So yes, the brackets make sure you evaluate $x^a$ first and then continue. – Klaus Aug 05 '22 at 17:34
  • That's clear, much appreciated. – RickyBoy Aug 05 '22 at 17:37
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    This is correct. For further understanding you can also look at my answer here https://math.stackexchange.com/a/3356598/399263 which explains how to deal with negative $x$ – zwim Aug 05 '22 at 18:27
  • Let n=1. Then you have $\sqrt{x^2}=\pm x$, ignoring convention of taking only positive solution. – herb steinberg Aug 05 '22 at 22:05
  • @Klaus Your engagement is appreciated. The way this topic is described in Wikipedia (not that what is there is guaranteed to be correct) appears to read that $(x^a)^b = x^{ab}$ is permissible even for negative $x$ when $a$, $b$ are integers, but that $x^r$ is defined only for positive $x$ when $r$ is rational. What are your views? – RickyBoy Aug 06 '22 at 07:45
  • @zwim, your views as well? – RickyBoy Aug 06 '22 at 07:45
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    @RickyBoy Yes, if $a$ and $b$ are integers, then $(x^a)^b = x^{ab}$ is fine even for negative $x$. The difference between integer powers and non-integer powers is that integer powers are defined via multiplication while non-integer powers are defined either by roots (e.g. $\sqrt{\cdot}$ is power $\frac{1}{2}$) or via the exponential function. Multiplication of negative numbers is fine, taking roots of negative numbers can be a problem if you are not careful. – Klaus Aug 06 '22 at 09:58
  • @Klaus. OK thanks for that. But perhaps you can clarify for me... In my question, $n$ is integer. You have above advised that my example for $n=1$ is not allowed for $x=-1$. So I am a bit confused. It is leading me to believe that it is only because of the brackets surrounding $(x^2)$ in the question that is allowing us to write $x^2=|x|^2$ so we may assert $(x^2)^{\frac{n}{2}} = |x|^n > 0$ for all $x \in \mathbb{R}$. Would you agree with that? – RickyBoy Aug 06 '22 at 10:13
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    @RickyBoy $n = 1$ is an integer, but $\frac{n}{2} = \frac{1}{2}$ is not. $x^2 = |x|^2$ is true for all $x \in \mathbb{R}$. – Klaus Aug 06 '22 at 10:48
  • @Klaus. You are indeed correct and of course it is obvious! There is a lot to be said for enlisting the help of someone who hasn't stared at a problem for too long. Haha. Again, your engagement is very much appreciated. Thank you. (I cannot "accept' the answer because it is in a comment, but let it be known this is the accepted answer, with thanks.) – RickyBoy Aug 06 '22 at 11:12

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