Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 '14 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 '14 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 '14 at 12:16
up vote 5 down vote accepted

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.

share|cite|improve this answer

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.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.