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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. –  TonyK Dec 7 '10 at 13:19
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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. –  J. J. Dec 7 '10 at 16:50
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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? –  Gunnar Magnusson Dec 7 '10 at 18:01
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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. –  Noah Snyder Dec 8 '10 at 16:35
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@Noah: I did that now, mathoverflow.net/questions/48910/…. –  J. J. Dec 10 '10 at 12:06
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1 Answer

up vote 6 down vote accepted

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

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