Finding -1 to irrational powers I've been curious about $A=(-1)^r$ when $r$ takes different values. I know that if $r$ was an even number, $A$ will become $1$ and if it was odd, it will become $-1$. Now I want to know what will $A$ be if $r$ were an irrational number like $\sqrt{2}$.Here's what i have done: According to Euler's identity we have $e^{\pi i}=-1$ and so $e^{\pi i\sqrt{2}}=(-1)^\sqrt{2}$ and $e^{i(\sqrt{2}\pi)}=(-1)^\sqrt{2}$. We also know that $e^{i\theta}=\sin\theta+i\cos{\theta}$. According to what i just showed, $(-1)^\sqrt{2}=e^{i(\sqrt{2}\pi)}=\sin{\pi\sqrt{2}}+i\cos{\pi\sqrt{2}}$ and from this we can find the exact value of $(-1)^\sqrt{2}$. Is what I have done correct? Is $(-1)^\sqrt{2}$ even a number?
 A: I would prefer not to consider $(-1)^r$ as definable in any sense. Because for me, $(-1)^{p/q}$ is $\bigl(\sqrt[q]{-1}\bigr)^p$ or $\sqrt[q]{(-1)^p}$, and I get:


*

*If $p,q$ are both odd, then both expressions are equal to $-1$, since $(-1)^q=-1$ therefore $\sqrt[q]{-1}=-1$.

*If $p$ is even, then both are clearly equal to $1$.

*If $q$ is even, then neither is well defined for me (it can be, but in a different way).


The problem here that the numbers $p/q$ with $q$ odd are dense in $\mathbb R_+$, therefore this exponential is quite badly discontinuous.

You can as well think that $(-1)^{p/q}$ is any (or all) of the roots of the polynomial $X^q=(-1)^p$, getting a similar thing as in the previous case, just with all the complex roots considered as well.

(All of these considerations strongly mimic the problem of $0^0$, which is considered to be $1$ by many people in combinatorics, however, we all know that it is not as simple as that.)
In the end, you can define $(-1)^r$ to be whatever you want, as long as you define it properly. However, as a reviewer of a paper, I would very likely reject such notation.
