Q:

A discontinuous linear function over the rationals

Possible Duplicate:
On sort-of-linear functions 

It is an interesting exercise to show that $\operatorname{Gal}(\mathbb R ~\colon \mathbb Q)$ is trivial. The only solution I know hinges on the fact that the automorphism is order-preserving, which in turn depends on the fact that $\theta(xy)=\theta(x)\theta(y)$ for $\theta \in \operatorname{Aut} \mathbb R$. 
Now, a function $L:\mathbb R \to \mathbb R$ with just the property that $L(x+y)=L(x)+L(y)$ can be shown to preserve multiplication on the rationals. And, I have been unsuccessful in trying to extend this fact to the reals. This could be because such a function could be discontinuous, an example of which I also have failed to construct (it's a bad day).
My question:

Can you give me an example of a discontinuous function $L:\mathbb R \to \mathbb R$ with the property that $L(x+y)=L(x)+L(y)$ and $L|_{\mathbb Q}= \text{identity}$? 

An idea: Perhaps we could consider a function that preserves order on rationals but reverses it on irrationals.

A:

This is impossible to do in ZF alone, but possible with a Hamel basis for $\mathbb{R}$ as a $\mathbb{Q}$ vector space.  Hence no "explicit" example can be expected.  Complete $\{1\}$ to a Hamel basis $B$, let $x\in B\setminus\{1\}$, and let $f:\mathbb{R}\to\mathbb{R}$ be the unique linear function such that $f(x)=0$ and $f(b)=b$ for all $b\in B\setminus\{x\}$; in particular, $f(1)=1$ implies that $f$ is the identity on the rationals.  Because $x$ can be approximated by rationals (not to mention with rational multiples of other elements of $B$), $f$ is not continuous at $x$.  In fact, $f$ is not continuous anywhere.

