By straightforward evaluation, $$(0^{-1})^{-1}=(NaN)^{-1}=NaN$$ where $\frac{1}{0}$ is taken to equal $NaN$ (not a number), or undefined or indeterminate. However, the laws of exponents state that $$(x^m)^n=x^{mn}$$ so the original equation should theoretically be evaluated as: $$(0^{-1})^{-1}=0^{(-1)(-1)}=0^1=0$$ Is one more correct than the other, or is this just one of the discrepancies when working with $\frac{1}{0}$?
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$(x^{m})^n=x^{mn}$ is not true for any $x$ and any $m,n$. Either $m,n\in\Bbb N,m,n\geq 1$ or $x$ has to be positive.
This is a classical mistake. Abusing this rule you can also prove that $$-1 = (-1)^{1}=(-1)^{2/2}=\sqrt{(-1)^2} = \sqrt{1}=1$$
Surb
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Ah, that makes sense! I've seen the -1 = 1 concept before, but now I understand the reasoning against it. Thanks for your help. – Dodo Aug 20 '15 at 10:07