'Fixed Point' Irrationals

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.