Is this linear? Are you sure???

Let $f:𝕉\to 𝕉$ be a continuous function such that whenever $a$ , $b\in 𝕉$ are such that $a-b\in 𝕈$ , then also $f\left(a\right)-f\left(b\right)\in 𝕈$ . Prove that $f$ is of the form $f\left(x\right)=qx+r$ , and that $q$ is rational.

Update (16/04/2008): Solution posted! (show solution)
Consider the function $g_1\left(x\right)=f(x+1)-f\left(x\right)$ , and note that it is continuous, and it only allows rational values. Thus, it has to be a constant $g_1\left(x\right)=K_1$ (because of the mean-value theorem, and that you can always find an irrational number between two rationals). We can also consider $g_{1/2}=f(x+1/2)-f(x)$ , this must also be a constant $g_{1/2}\left(x\right)=K_{1/2}$ , and note that

${\displaystyle K_1=f(x+1)-f\left(x\right)=}$

${\displaystyle \hspace{0.33em}=f(x+1)-f(x+1/2)+f(x+1/2)-f\left(x\right)=}$

${\displaystyle \hspace{0.33em}=K_{1/2}+K_{1/2}=}$

${\displaystyle \hspace{0.33em}=2K_{1/2}}$

And we can repeat this with $K_{1/4},K_{1/8},...,K_{2^{-n}}$ , and get that $K_1=2^n{K}_{2^{-n}}$ . So, for every diadic rational number $p{2}^{-n}$ , we have that $f\left(p{2}^{-n}\right)-f\left(0\right)=p{K}_{2^{-n}}=p{2}^{-n}{K}_1$ , ie. that $f\left(x\right)-f\left(0\right)$ is linear on diadic rationals, and by continuity it is linear everywhere. Thus $f\left(x\right)=qx+r$ , and of course $f\left(1\right)-f\left(0\right)=q\in 𝕈$ .