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I know this: $\sqrt{x}^2 = |x|$, but

$\sqrt{(-1)^2} = \sqrt{(-1)^2}$

$(-1)^\frac{2}{2} = \sqrt{-1 * -1}$

$(-1)^1 = \sqrt{1}$

$-1 = 1^2$, then

$-1 = 1$

What step is wrong?

amWhy
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ESCM
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  • $\sqrt x \neq |x|$, do you mean $\sqrt{x^2} = |x|$? – Andrew Li Apr 12 '18 at 18:35
  • Yes sorry, edited – ESCM Apr 12 '18 at 18:36
  • At the very first step. Consider amending to $(\sqrt{|x|})^2 = |x|$. – Robert Soupe Apr 12 '18 at 18:40
  • You have misunderstood the actual identity $\sqrt {x^2}=\sqrt{(x^2)}=\lvert x\rvert$ (notice where the overline stops), which holds for all $x\in\Bbb R$, for the made-up thing $(\sqrt x)^2=\lvert x\rvert$, which does not hold because $\sqrt x$, whenever it is defined, is made so that it satisfies $(\sqrt x)^2=x$. –  Apr 12 '18 at 19:27

1 Answers1

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The square root function has a branch cut discontinuity on the negative real axis. That means you can't do this: $\sqrt{(-1)(-1)}=\sqrt{-1}\,\sqrt{-1}$, because you'd be approaching that discontinuity from two different directions, and expecting them to be equal. For your derivation, $((-1)^2)^{1/2}\not=((-1)^{1/2})^2$, a step implicit in your derivation.

Adrian Keister
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    This doesn't really have to do with the branch cut. After all, you can choose the branch cut to lie on a different axis. The key issue is that the property $\sqrt{ab} = \sqrt{a} \sqrt{b}$ doesn't hold anymore. –  Apr 12 '18 at 18:39
  • @user296602: And the reason you can't do that if both $a$ and $b$ are negative is because you're crossing the default branch cut. If someone were to specify a different branch cut, that'd be fine, but the algebra above would have to look different. – Adrian Keister Apr 12 '18 at 18:41
  • See @Alexander McFarlane's answer here: https://math.stackexchange.com/q/1602274 for a more detailed explanation. – Adrian Keister Apr 12 '18 at 20:33