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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 $0$?

Alec
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    Of course, and it is commonly done as follows: take $(0,\infty)$ as a (smaller) domain. –  Dec 22 '21 at 21:10
  • @StinkingBishop - I'm ashamed. Obviously the function would be strictly increasing or decreasing on this domain. – Alec Dec 22 '21 at 21:20
  • @Alec Not necessarily. Here is a silly example: define $f(x)=x^n$ on nonnegative algebraic numbers and negative transcendental numbers. Then for even $n$, $f$ is not increasing or decreasing on its whole domain, but it's nevertheless a bijection from its domain to $\Bbb R^+$. – Jean-Claude Arbaut Dec 23 '21 at 16:46

1 Answers1

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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.

Joe
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  • The counter-domain has to be $(0, \infty)$ for this to work (just a small detail, but I guess I’m picky…) – Gauss Dec 22 '21 at 21:18
  • @Gauss: That's true if you include the codomain as part of the definition of a function. Personally, I prefer to define a function as a set of ordered pairs $f$ such that if $(a,b)\in f$ and $(a,c)\in f$, then $b=c$. In this approach, the inverse of $f$ is defined as the set $$f^{-1}={(b,a):(a,b)\in f} , .$$ – Joe Dec 22 '21 at 21:22
  • @Gauss: Under this set-theoretic definition of function (which can be found in Michael Spivak's Calculus, for instance), the functions $f:\mathbb R_{\ge0}\to\mathbb R,f(x)=x^2$ and $g:\mathbb R_{\ge0}\to\mathbb R,g(x)=x^2$ are identical, and they both have inverses. See this thread for more details. – Joe Dec 22 '21 at 21:29
  • I guess you make a fair point. You can always assume a restriction to the image in order to invert it, anyway. Thank you for the reference, by the way! – Gauss Dec 22 '21 at 22:15
  • @Gauss: No problem! By the way, my above comment had a typo: I meant to write that the functions $f:\mathbb R_{\ge0}\to\mathbb R,f(x)=x^2$ and $g:\mathbb R_{\ge0}\to\color{\red}{\mathbb R_{\ge0}},g(x)=x^2$ are identical in this definition of function. This means it no longer makes sense to talk about whether a function is "surjective" or not, and so it is understable that in some areas of mathematics the set-theoretic definition of "function" is not preferred. – Joe Dec 22 '21 at 22:18