Where is the mistake? $$-1=(-1)^{2/2}=\left(\left (-1\right)\displaystyle ^2\right)^{1/2}=1^{1/2}=\sqrt 1=1$$
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Thank you for the response! Not really I see it here, though. – Math Student Nov 29 '20 at 12:22
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1The problem is that the formula $x^{\alpha \beta}=(x^{\alpha})^{\beta}$ is valid for $\alpha,\beta \in \Bbb Q$ only if $x>0$. And here you used exactly this property for $\alpha = 2, \beta =1/2$ and $x=-1$ therefore your equality is wrong. – Surb Nov 29 '20 at 12:29
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1@Surb $1/2\in Q$ – aarbee Nov 29 '20 at 12:41
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Thank you for the responses! @Surb, but why $x^{\alpha\beta}=(x^\alpha)^\beta$ only if $x>0$? – Math Student Nov 29 '20 at 12:44
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Cf. this question – J. W. Tanner Nov 29 '20 at 13:26
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@aarbee yes, this is what I said, isn't it? – Surb Nov 29 '20 at 15:25
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Each $=$ but the second is clearly right. That the second $=$ is wrong proves $\color{blue}{x^{a/b}=(x^a)^{1/b}}$ admits counterexamples with $x<0$. We use $x>0$ in the manipulation$$(x^{a/b})^b=x^a\implies\color{blue}{x^{a/b}=(x^a)^{1/b}},$$which is our only reason to expect the blue part to be true.
J.G.
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