# 'Fixed Point' Irrationals

I found this interesting problem which turns out to be more difficult than it first appears:

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function such that $f(f(x))=x$ for all $x \in \mathbb{R}$. Prove that there exists an irrational number $t$ such that $f(t)$ is irrational.

The $f(f(x))=x$ condition reminds me of fixed point problems but as nothing else about $f$ is known I'm not sure how to apply this. Instead, I thought about the standard 'irrational to irrational power being rational' problem. So I thought about trying something along the lines of taking $x \in \mathbb{R}$ irrational then looking at $f(x)=y$. If $y$ is irrational we are done. If not, then I feel like trying something like $\sqrt{2}y$ as an input would work but nothing really panned out.

Then I observed if $g(x)=(f\circ f)(x)$, we have $$g(xy)=xy=g(x)g(y)$$ and $$g(x+y)=x+y=g(x)+g(y)$$ but am unsure what this gets me. Any clues as to how I might proceed or perhaps an alternative route?

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So far three essentially identical answers have appeared. Shall we make bets on what sorts of other answers will appear and how many of each? –  Michael Hardy Jun 16 at 4:28
@MichaelHardy: I predict an uncountable number of answers, but only a countable quantity of patience to read them. –  Eric Towers Jun 16 at 4:34
@EricTowers: I think the site policy allows only for a finite number of answers, even though maybe no concrete bound is imposed. –  Marc van Leeuwen Jun 16 at 12:16

If $f(f(x))=x$ for every $x\in\mathbb R$, then $f$ is one-to-one.

If $f$ is one-to-one then the set $\{f(x) : x\text{ is irrational}\}$ is uncountably infinite. Therefore that set cannot be a subset of the set of all rational numbers.

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The function $f$ is invertible, in fact it is its own inverse, and is therefore a bijection. Therefore the images of the (uncountably many) irrationals cannot be only the (countably many) rationals.

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The condition is $f\circ f$ is the identity, and thus $f^{-1}=f$, and in particular $f$ is injective. If $f(x)$ was rational for every irratioanl $x$, then restricting $f$ to the set of irrationals, one would obtain an injection into the set of ratioanls. That is impossible though due to cardinalities.

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