What's wrong with this argument about fractional exponents?

Question

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?

Left-hand side, not "left handside" – Ben Grossmann May 08 '17 at 15:09 — Ben Grossmann, May 08 '17 at 15:09
$(x^{a})^b \ne x^{ab}$, in general unless $x\ge0$ – Mark Viola May 08 '17 at 15:11 — Mark Viola, May 08 '17 at 15:11
Dr. MV, is that a general convention? – Florian Biermann May 08 '17 at 15:37 — Florian Biermann, May 08 '17 at 15:37

Andy Walls · Accepted Answer · 2017-05-09T11:14:30.513

0

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.

edited May 09 '17 at 11:14

answered May 09 '17 at 10:12

Andy Walls

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What's wrong with this argument about fractional exponents?

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