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Does projective transformation from $\mathbb{R}$ to $\mathbb{Q}$ exist or not? $f:\mathbb{R}\rightarrow \mathbb{Q},\forall x,y\in \mathbb{R},f(x)+f(y)=f(x+y),f(1)=1.$ Does $f$ exist?

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closed as off-topic by M. Winter, Aqua, kingW3, Davide Giraudo, Xam Nov 14 '17 at 19:51

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    $\begingroup$ "Projective transformation from $\mathbb{R}$ to $\mathbb{Q}$ exists or not." This is a true statement. $\endgroup$ – 5xum Nov 14 '17 at 11:50
  • $\begingroup$ The issue of the question is not the "exist or not" and the logician joke. One issue is that the wording "projective transformation" is confusing. Transformation is usually used for bijections, and "projective" refers to something else too. $\endgroup$ – YCor Nov 14 '17 at 17:57
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And the other side: such an example cannot be proved to exist using only ZF set theory. Some form of choice is required.

There is a model for ZF due to Solovay where the only solutions of Euler's functional equation $f(x+y)=f(x)+f(y)$ from $\mathbb R$ to $\mathbb R$ are those of the form $f(x) = cx$. The function requested by the OP, then, cannot exist in that model.

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Using the Axiom of Choice, the answer is "yes". All you need is to extend $\{1\}$ to a basis of the $\Bbb Q$-vector space $\Bbb R$ and declare that $f$ is zero on all other basis vectors.

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This problem is somehow interesting. To solve the problem you need to look $\mathbb R$ as a infinite dimension linear space based on $\mathbb Q$. At once you do this, by Zorn's lemma you could choose a basis , let us call it $B_I=\{e_i|i\in I\}$. Now consider the value of $f(e_i)$, some of them are 0, and the others are not 0. Just take them by randomness but remember $f(1)=1$ and we are done!

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