Solution to $(-1)^{2x}=1$

Question

So I encounter this problem when studying quantum mechanics:

(-1)^2x=1, find the set of values of x

Obviously, x ∈ {Z}, as 2x is even for all integral x.

Just to be rigorous, I wrote down the steps and come out with something like this:

(-1)^2x=1

[(-1)²]^x=1

1^x=1, so x can be any real numbers.

Where is my mistake?

Welcome to Mathematics Stack Exchange. Non-integer powers of negative numbers are not well-defined — J. W. Tanner, Nov 27 '19 at 18:41
WA says this here $$c_1\in \mathbb{Z}\land \left(x=1-\frac{i (2 i \pi c_1+i \pi )}{\pi }\lor x=-\frac{i (2 i \pi c_1+i \pi )}{\pi }\right)$$ — Dr. Sonnhard Graubner, Nov 27 '19 at 18:44
$(-1)^{x} = 1$ if x is a rational number in lowest terms with even numerator. — Doug M, Nov 27 '19 at 18:44
@J.W.Tanner But if there's nothing wrong with my steps why does it give wrong answer? — TheLearner, Nov 27 '19 at 18:44
Cf. this question; the "general rule" $(a^m)^n=a^{m×n}$ does not always work when $ m$ and $ n$ are not integers. — J. W. Tanner, Nov 27 '19 at 18:45
@The99sLearner there is something wrong with your steps: you cannot blindly apply the same basic exponent rules to non-integer powers of negative numbers. — Nap D. Lover, Nov 27 '19 at 18:50

score 2 · Accepted Answer · answered Nov 27 '19 at 20:04

2

Your mistake is that non-integer powers of negative numbers are not uniquely defined,

and the "general rule" $(a^m)^n=a^{m\times n}$ does not always work when $m$ and $n$ are not integers.

answered Nov 27 '19 at 20:04

J. W. Tanner

1 Answers1