$\mathbb{Q}$ has the topology induced from $\mathbb{R}$, therefore it is in principle possible to talk about power series and define analytic functions on $\mathbb{Q}$ to be power series (with coefficients in $\mathbb{Q}$) that converge on some open subset of $\mathbb{Q}$ and only take values in $\mathbb{Q}$. I have a strong feeling that the only possible examples should be polynomials, but I cannot think of an easy proof. Any hints or counterexamples?
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2Even there exists a non-polynomial entire function $f$ such that $f(\mathbb{Q}) \subseteq \mathbb{Q}$, e.g., see this posting. – Sangchul Lee Apr 03 '19 at 13:06
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Thanks, very interesting! – 57Jimmy Apr 03 '19 at 13:43
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I know of one counterexampe: $f(x) = 1 + x + x^2 + \ldots$ which on the appropriate open subset of $\mathbb{Q}$ equals $f(x) = \frac{1}{1-x}$, and hence sends rationals to rationals.
Still the modified question "Can the set of power series that are convergent on some open subset of $\mathbb{Q}$ and converge to rational numbers only on that set be characterized in a different, appealing way?" is very interesting I think.
Vincent
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$f(x)=\dfrac{1}{1+x^2}$ has a power series with rational coefficients which is convergent in $(-1,1)$, and maps rationals to rationals.
TonyK
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