I was working on some exponentiation, mostly with rational bases and exponents. And I stuck with something looks so simple:


I know this must be $\sqrt{-2}$, therfore must be imaginary number. However, when I applied some properties I have something unexpected, and I don't know what I did wrong:


I know the above is wrong, but I don't know exactly what is wrong. My initial suspect was from 2nd to 3rd expression, so I checked the property ($x^{mn}=(x^m)^n$), and I realized that it is only true for integer exponents.

I dug a little more, and found the following passage on Wikipedia:

"Care needs to be taken when applying the power identities with negative nth roots. For instance, $−27=(−27)^{((2/3)⋅(3/2))}=((−27)^{2/3})^{3/2}=9^{3/2}=27$ is clearly wrong. The problem here occurs in taking the positive square root rather than the negative one at the last step, but in general the same sorts of problems occur as described for complex numbers in the section § Failure of power and logarithm identities."

But could anyone clarify the explanation here? If we simply follow the order of operation, don't we really get 27?


Note that $4^{\frac {1}{4}}$ has 4 values $(\sqrt {2},-\sqrt {2},i\sqrt {2},-i\sqrt {2})$

When you square a number or an equation then you are increasing solution values.

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The answer is actually much simpler than that. Your first step was correct, you converted the exponent into a square root. Instead of breaking down the exponent into $2*(1/4)$, you should split the root into $\sqrt{2}*\sqrt{-1}$. $\sqrt{-1}$=$i$ so your final answer is $i\sqrt{2}$.

I hope this clears it up.

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  • $\begingroup$ Thank you for comment. But, I do know the answer must be sqrt(2)i. My question is, whats wrong with the "supposedly wrong" derivation? Why am I not allowed to do that? $\endgroup$ – Harry Hong Jul 8 '16 at 1:24
  • $\begingroup$ In your third step, you converted $(-2)^(2*(1/4))$ to $[(-2)^2]^(1/4)$. In doing that, you changed the order of operations, so I think that is what is wrong with the derivation. Instead of it being something equivalent to $-2^(1/2)$, it becomes instead $4^(1/4)$, which is not equivalent to the original question. $\endgroup$ – Lumberjacked216 Jul 8 '16 at 6:20

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