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The problem is,

Find all continuous functions $f: \mathbb{R} \to \mathbb{R}$ such that if $x-y$ is rational, $f(x)-f(y)$ is rational.

I was thinking that only constant functions would work. How would I prove this??

neth
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zxcvber
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    Certainly more than constant functions work; for instance, any affine transformation works. – neth Apr 14 '16 at 06:23
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    @neth: Isn't $x \mapsto \sqrt{2}x$ an affine transformation that doesn't work? –  Apr 14 '16 at 06:29
  • @Bungo: sorry, you're right. Any affine transformation with rational coefficients. – neth Apr 14 '16 at 06:40
  • A good starting point is http://math.stackexchange.com/questions/167620/functions-that-take-rationals-to-rationals and maybe some other questions listed under "Related". – Gerry Myerson Apr 14 '16 at 07:24
  • Certainly all linear functions with rational coefficients would work. But I was looking for group of functions that satisfy the given condition, without any extra conditions for the coefficients. – zxcvber Apr 14 '16 at 07:51
  • What is a "group"? They do form a group in the sense of group theory; apparently, you mean it in some other sense. What is it? – Ivan Neretin Apr 14 '16 at 08:04
  • Yes they would but I didn't mean the mathematical 'group'. I meant something like 'set of linear functions', 'set of exponential function' etc. – zxcvber Apr 14 '16 at 08:10
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    I found the answer here. [link] (http://math.stackexchange.com/questions/1004280/continuos-function-preserving-rational-difference?rq=1) – zxcvber Apr 14 '16 at 08:11
  • @zxcvber Yes. This should be closed as duplicate. – almagest Apr 14 '16 at 11:00

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