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A friend of mine introduced me to the following question: Does there exist a smooth function $f: \mathbb{R} \to \mathbb{R}$ ($f \in C^{\infty}$), such that $f$ maps rationals to rationals and irrationals to irrationals and is nonlinear?

He has been able to prove that such a polynomial (with degree at least 2) doesn't exist.

The problem has been asked before at least at http://www.artofproblemsolving.com.

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    That link would be more helpful if it took us straight to the question.2010-12-07
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    @TonyK: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=73&t=107035&p=605167 but there is not much there.2010-12-07
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    Very interesting question: makes you think and it's easy to understand though hard to solve.2010-12-07
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    As a side note: If we required just $f \in C^1$, then looking at the function $f$ defined by $f(x) = \frac{1}{x-1} + 1$ for $x \le 0$ and $f(x) = \frac{1}{x+1} - 1$ for $x \ge 0$ provides an example. There is also an example for the case where we drop the requirement that irrationals are mapped to irrationals.2010-12-07
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    It seems kind of hard to get a grip on these things. That class of functions isn't stable under addition, multiplication (though it is under multiplication by elements of $\mathbb Q$), exponentation or limits, ruling out most of the standard machinery of analysis. It is stable under iteration, though, so maybe dynamics could help?2010-12-07
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    What a nice question! If it turns out no one here can answer it you might eventually want to try mathoverflow too.2010-12-08
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    @Noah: I did that now, http://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational.2010-12-10
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    Are you implicitly excluding trivial solutions like $f(x)=x$?2010-12-10
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    @Raphael: *nonlinear*2010-12-10
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    Did miss that, sorry.2010-12-11

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Sergei Ivanov has given a positive answer for the existence of such functions on MO: https://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational.