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$f(f(x))=x \ \forall x \in \mathbb{R}$. I am trying to prove there exists an irrational $t$ such that $f(t)$ is also irrational.

I have been trying things like assume $t$ irrational implies $f(t)$ is rational and then $f(f(t)+t)$ is rational but $f(f(f(t)+t))=f(t)+t$ but I can't come up with a contradiction. Is there a less mind-boggling approach?

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Hint: Observe that $f$ is a bijection. – Zhen Lin Aug 22 '11 at 17:41
Such a function is called an "involution", by the way. It's one of my favorite fancy-math-words-describing-a-simple-concept. – MartianInvader Aug 22 '11 at 19:07
up vote 26 down vote accepted

Suppose that $f$ is its own inverse. Then $f$ is injective: $f(a)=f(b)$ implies $a=f(f(a))=f(f(b))=b$.

Ok, now suppose that $f(t)$ is rational for every irrational $t$. Then we have an injection from the set of irrationals into the set of rationals, given by $t\mapsto f(t)$. But this is impossible, because the set of rational numbers is countable and the set of irrationals is not.

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typo: $f(f(a))=f(f(b))$ (too small for me to edit) – robjohn Aug 22 '11 at 18:03
@robjohn: Danke! – Andrés E. Caicedo Aug 22 '11 at 18:19

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