Background
Let $f(x) = x^n$ with $n\in\mathbb Z$ and $x \in \mathbb R$.
For $f$ to have an inverse, $n$ must be odd.
Question
Can we change the domain of $x$ such that $x^n$ has an inverse for all integers $n$ except $0$?
Background
Let $f(x) = x^n$ with $n\in\mathbb Z$ and $x \in \mathbb R$.
For $f$ to have an inverse, $n$ must be odd.
Question
Can we change the domain of $x$ such that $x^n$ has an inverse for all integers $n$ except $0$?
For every integer $n\neq0$, the function $f_n:(0,\infty)\to\mathbb R$ given by $f_n(x)=x^n$ is monotone, and so it is one-to-one, and so it is invertible.
The proof that each $f_n$ is monotone is simple: if $n>0$, then $a>b>0\implies a^n>b^n$; this can be easily seen from the identity $$ a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+b^{n-1}) \, . $$ If $n<0$, then $-n>0$, so $a>b>0\implies a^{-n}>b^{-n}$; rearranging this final inequality gives us the desired result.