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This is a very short argument that I have often seen about the definition of $|x|=\sqrt{x^2}$ for $x\in \mathbb{R}$. I'm not quite sure what is the flaw in this argument below, can someone please illustrate at what line this goes wrong and reference what "restriction" was broken from the original index law.

Claim: $\sqrt{x^2} = x$ where $x\in\mathbb{R}$.

Proof: Note that $$\begin{align*} \sqrt{x^2} &= (x^2)^{\frac{1}{2}} \\ &= x^{\frac{2}{2}} \\ &= x^{1}=x.\end{align*}$$

What is wrong with this?

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  • $\begingroup$ In general, $(a^b)^c\ne a^{bc}$. $\endgroup$ – Hagen von Eitzen Sep 4 at 6:22
  • $\begingroup$ Note that fractional powers are not nicely behaving for negative numbers. $\endgroup$ – Matti P. Sep 4 at 6:23
  • $\begingroup$ $\sqrt{(-1)^2}=\sqrt1=1\neq-1$. $\endgroup$ – trisct Sep 4 at 6:25
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First the counterexample to this is $x = -1$. Now to the error in your reasoning is the following $(a^b)^c = a^{bc}$ is only true if $a \geq 0$.

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  • $\begingroup$ @Q the Platypus Thanks that makes sense $\endgroup$ – user523384 Sep 4 at 6:35
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    $\begingroup$ @Q the Platypus Sorry just a follow up question, about the "only true" part $((-2)^2)^3=64=(-2)^6$ but $-2$ is not positive. Do you mean it is only true for non-integer but rational powers when $a\ge 0$? $\endgroup$ – user523384 Sep 4 at 8:24
  • $\begingroup$ if you take the sixth root of 64, you get 2 not -2 because it's nearly always assumed you want the positive version. $\endgroup$ – Roddy MacPhee Sep 4 at 12:10

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