<s>[INST] Why does the square root of a square involve the plus-minus sign? If $\sqrt{x^2}$ can be simplified as follows:
$\sqrt{x^2} = (x^2)^\frac{1}{2} = x^{\frac{2}{1}\times\frac{1}{2}} =x^\frac{2}{2} = x^1 = x$
Then why would $\sqrt{x^2} = \pm x$?
 [/INST] Because $\sqrt{x^2}$ can't be simplified "as follows." To be more precise: the notation $\sqrt{x^2}$ naturally designates a number $y$ such that $y^2=x^2$. Among real numbers there are two such $y$, namely, $y=-x$ and $y=x$. Your mistake is in extending the rule $(x^a)^b=x^{ab}$, valid for $a$ and $b$ integers, to allow $a$ and $b$ to be fractions. 
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