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I am sure that this has been discussed many times and the answer may be trivial. But I was confronted with the following issue and could not give a satisfactory answer: $$1=(-1^{6})^{\frac{1}{2}}=-1^{6\cdot\frac{1}{2}} = -1^{\frac{1}{2}\cdot 6}=(-1^{\frac{1}{2}})^6 = (\sqrt{-1})^6.$$

On the left handside of the chain, we have $1$, and on the right handside, we have the square root of a negative number, which is not defined with real numbers. Also moving towards complex numbers does not seem to help, as $$(\sqrt{-1})^6 = (\sqrt{i^2})^6 = i^6 = (i^2)^3 = (-1)^3 = -1.$$ Where is the mistake or the step that is not allowed?


marked as duplicate by mrf, The Dead Legend, Especially Lime, Shailesh, mlc May 10 '17 at 0:33

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  • 1
    $\begingroup$ Left-hand side, not "left handside" $\endgroup$ – Omnomnomnom May 8 '17 at 15:09
  • 5
    $\begingroup$ $(x^{a})^b \ne x^{ab}$, in general unless $x\ge0$ $\endgroup$ – Mark Viola May 8 '17 at 15:11
  • $\begingroup$ Dr. MV, is that a general convention? $\endgroup$ – Florian May 8 '17 at 15:37

Whomever presented the "proof" was not rigorous in their argument and skipped a step:

$$1 = (-1)^3 = ((-1)^6)^{\frac {1}{2}}= ... $$

That skipped step is false.


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