Find domain of $(-1)^x$

Question

What is the domain of $(-1)^x$ as a real function?

We have $(-1)^{1/3}=-1$ and $(-1)^{1/2}$ undefined.

I'm confuse.

Answer 1

It's strange as a precalculus question, anyway, $(-1)^x$ is a function defined in the whole complex plane, but the only points on the real axis where it takes real values are the integers, where it takes value $1$ or $-1$ depending on parity.

To calculate it, start from $-1=e^{(2n+1)i\pi}$, so $$ (-1)^x = e^{(2n+1)i\pi x} = \cos((2n+1)\pi x)+i \sin((2n+1)\pi x) $$

For it to be real, you need $$ \sin((2n+1)\pi x) = 0 $$ so $$ (2n+1)x \in \mathbb Z $$ for some $n\in\mathbb Z$, that is, $x$ can be represented as a fraction with odd denominator.

---

Simpler solution.

After a long discussion in the comments, it seems clear that the OP needs an elementary solution that does not involve complex numbers. Exactly as @Arthur says in his comment on the question.

So let's suppose that we "define" $(-1)^{\frac n m}$ ($m$ and $n$ coprime), as the only real number (if any) $r$ that satisfy $$ r^m = (-1)^n. $$ There is no way to define anything similar on irrational numbers anyway.

In this case the function is defined whenever $m$ is odd. If $m$ is even, then $n$ must be odd and the LHS of the equation ($r^m$) is positive where the RHS is negative.

So this function is defined on "some" dense subset of rationals (as said, those whose reduction has odd denominator), and it takes values $-1$ and $1$ in two co-dense subsets of its domain.