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Here's a question i would be curious to know the answer

The question is: what is the set of all entire functions $f: \mathbb{C} \to \mathbb{C}$ such that $f(\mathbb{Q})\subseteq \mathbb{Q}$.

Water
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  • I'd imagine all polynomials with rational coefficients qualify? – orion Mar 29 '14 at 12:50
  • @orion No, even considering all polynomials with rational coefficients only, they can have $f(\mathbb{Q})\subsetneq \mathbb{Q}$. – Daniel Fischer Mar 29 '14 at 12:51
  • We evidently(?) have $f(z) = az+b$ with $a\in \mathbb{Q}\setminus{0}$ and $b\in\mathbb{Q}$. I strongly suspect these are all, but proving it isn't obvious. – Daniel Fischer Mar 29 '14 at 12:55
  • Oh right, we have to avoid any functions with irrational inverses... In that case, I belive @DanielFischer is right. I suppose that if we ignore "entire function" requirement, we get all the rational Mobius transformations? – orion Mar 29 '14 at 12:58
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    There's a lot of them, see Functions that take rationals to rationals. The question is about $\mathbb R\to\mathbb R$, but the intermediate function $g$ I construct in my answer is analytic $\mathbb C\to\mathbb C$ and maps rationals to rationals, but is not a polynomial. It should be reasonably clear that there's continuum many similar functions. – hmakholm left over Monica Mar 29 '14 at 12:59
  • @orion,Daniel Fisher: sorry i was thinking of $\subseteq$ and i wrote =. I've edited. – Water Mar 29 '14 at 13:03
  • In that case, all rational polynomials and quotients of rational polynomials are alright. – orion Mar 29 '14 at 13:10
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    @orion: Quotients of polynomials are not entire. – hmakholm left over Monica Mar 29 '14 at 14:04
  • Sure, I just expanded on my previous answer (Mobius transformations) but of course among entire functions, only polynomials apply. – orion Mar 29 '14 at 14:08

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Somewhat surprisingly, even much more can be achieved:

MR0301195 Barth, K. F.; Schneider, W. J. Entire functions mapping arbitrary countable dense sets and their complements onto each other. J. London Math. Soc. (2) 4 (1971/72), 482–488.

It is proved that if A, B are two countable dense sets in the complex plane, then there exists an entire function w=f(z) such that f(z)∈B if and only if z∈A. This rather surprising result answers a question first raised by P. Erdős.

This applies to "complex rational" numbers, but in another paper they treat the real case:

Barth, K. F.; Schneider, W. J. Entire functions mapping countable dense subsets of the reals onto each other monotonically. J. London Math. Soc. (2) 2 1970 620–626.

Alexandre Eremenko
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