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While working on some field extensions I came across this "proof" that $i = 1$, but I am not sure what is wrong with it:

We know $ i = \sqrt{-1} = (-1)^{1/2}$. Now we have the following equation: $$\begin{align} i \sqrt[4]{-2} =(-1)^{1/2}(-2)^{1/4} &= (-1)^{1/2}(-1)^{1/4}(2)^{1/4} \\ &= (-1)^{3/4}(2)^{1/4}\\ &= (-1)^{1/4}(2)^{1/4} \\ &= \sqrt[4]{-2} \end{align}$$

Dividing both sides by $\sqrt[4]{-2}$ would yield $i = 1$.

Juan Sebastian Lozano
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    You assume $(ab)^c=a^cb^c$, which is not true for negative bases with non-integer exponents. – Arthur Apr 11 '17 at 05:56
  • That is extremely important to know, I'm not quite sure how I didn't know that. – Juan Sebastian Lozano Apr 11 '17 at 05:57
  • I am quite sure that every time you've seen rules for powers like this in text books, for non-integer exponents, they've said $a,b>0$ on the side. However, they don't usually call attention to it. – Arthur Apr 11 '17 at 05:59

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