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I cannot think of a non-constant smooth function which maps all real numbers into rational numbers.

Can anyone give a simple example ? The simpler, the better !

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No. There isn't even a non-constant continuous function $\mathbb R \rightarrow \mathbb Q$. Recall a continuous function maps connected sets to connected sets. We know that $\mathbb R $ is connected, but $\mathbb Q$ is totally disconnected: each point is its own connected component. So any continuous map $\mathbb R \rightarrow \mathbb Q$ must be constant.

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    $\begingroup$ Classic example where searching for the duplicates is far less rewarding than writing a short answer. $\endgroup$ – Asaf Karagila Jun 6 '13 at 5:52
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However, you didn't mention these two creatures as they are in Group Theory; I want to note some points about them from Group Theory point of views. We know that $G_1=(\mathbb Q,+)$ and $G_2=(\mathbb R, +)$ are both groups and of course abelian, i.e; $x+y=y+x$. Moreover in $G_1,G_2$ and for every $n\in\mathbb N$ and $a\in G_1~~\text{or}~~a\in G_2$ we can find some $x\in G_1~~\text{or}~~\in G_2$ such that the equation $a=nx$ has at least one solution (Divisibility). Also both groups are torsion free and so they are both vector spaces over $\mathbb Q$. If $G_1\cong\ G_2$ so both should be have the same dimensions over $\mathbb Q$. But this not true. These points just show that Rational and Real numbers have another differences as well.

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    $\begingroup$ Nice, Babak! +1 $\endgroup$ – Namaste Jun 7 '13 at 0:01

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