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Is $|x^n|=|x|^n$

for any rational $n$ and for any real number $x$?

If the above is true, what is the proof?

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    $\begingroup$ If one feels it requires proof (I don't) the result is clear if $x\ge 0$. If $x\lt 0$, then $x=-|x|$, and therefore $|x^n|=|(-1)^n |x|^n|=(|(-1)^n|)|x|^n=|x|^n$. $\endgroup$ – André Nicolas Sep 26 '14 at 21:29
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Definition of Absolute Value \[ |x|=\left\{ \begin{array}{cc} x & : x\ge 0 \\ -x & : x<0 \\ \end{array} \right. \] Multiplicative Identity of Absolute Value \[ |xy|=|x|\cdot|y| \] So if $n\ge 0$, we have $$ |x^n|=|x\cdots x|=|x|\cdots |x|=|x|^{1+\dots +1}=|x|^n $$ However, if $n\lt 0$ and $x\neq 0$, we have $$ |x^n|=\left|\frac{1}{x\cdots x}\right|=\frac{1}{|x\cdots x|}=\frac{1}{|x|\cdots |x|}=\frac{1}{|x|^{-n}}=|x|^n $$

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  • $\begingroup$ This doesn't seem correct. Let x = -1 and n = 1. Then we have (sqrt(x^2))^1 = x^(2*1/2) which expands to 1 = -1. $\endgroup$ – Milo Brandt Sep 26 '14 at 21:52
  • $\begingroup$ @Meelo, please see my edit. $\endgroup$ – k170 Sep 27 '14 at 0:25
  • $\begingroup$ how is abs(x)......abs(x) = (abs(x))^-n ? - for the case n<0 how do you get from the 4th term to the 5th term $\endgroup$ – zebra1729 Oct 4 '14 at 12:29
  • $\begingroup$ @zebra1729, if $n\lt 0$, then $n$ is negative and $-n$ is a positive number. So the absolute value of $x$ is multiplied $-n$ times. It's a bit tricky because it's counter intuitive to think in negative values. As an exercise, let $n=-3$ and try to follow every step while keeping the negative sign in front of the $3$. $\endgroup$ – k170 Oct 4 '14 at 12:54

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