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.
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