Let $\alpha:\mathbb{R}^n\to\mathbb{R}^n$, $n\geq 2$, be a $\mathbb{Q}$-linear bijection with the following properties:

1) $\alpha$ sends straight affine $\mathbb{R}$-lines to straight affine $\mathbb{R}$-lines

2) If $r$ and $s$ are parallel straight affine $\mathbb{R}$-lines, then so are $\alpha(r)$ and $\alpha(s)$.

**Question:** does $\alpha$ have to be $\mathbb{R}$-linear?

This (perhaps not exactly research level...) question came to me when I tried to answer the following (also elementary - Edit: as the comment by Misha points out, it was not elementary indeed! Just classical from nineteenth Century) question: does a set-theoretical bijection $\alpha:\mathbb{R}^n\to\mathbb{R}^n$ that sends lines to lines and preserves parallellism have to be $\mathbb{R}$-affine? When $n=1$ the answer is vacuously no, as there is only one line and every set-theoretical bijection would do. But when $n\geq 2$, assuming $\alpha(0)=0$, one can prove${}^*$ that $\alpha$ is actually $\mathbb{Q}$-linear, and then -if one assumes continuity- also that $\alpha$ is $\mathbb{R}$-linear. But what if we don't assume continuity?

${}^*$: First let's prove $\alpha$ is additive, that is $\alpha(u+v)=\alpha(u)+\alpha(v)$. If $u$ and $v$ are not multiples of each other, then the nondegenerate parallelogram corresponding to the points $0,u,v,u+v$ is sent to the nondegenerate parallelogram $0,\alpha(u),\alpha(v),\alpha(u+v)$, hence $\alpha(u+v)$ is the vector sum of $\alpha(u)$ and $\alpha(v)$. If $u,v$ are multiples of each other, write $u=u_1+u_2$ and $v=v_1+v_2$ generically enough that none of the following pairs $(w,w')$ of points is such that $w,w'$ are multiples of each other: $(u_1+v_1, u_2+v_2)$, $(u_1,v_1)$, $(u_2,v_2)$, $(u_1,u_2)$, $(v_1,v_2)$. Now,

$$\alpha(u+v)=\alpha(u_1+u_2+v_1+v_2)=\alpha(u_1+v_1)+\alpha(u_2+v_2)=$$

$$\alpha(u_1)+\alpha(v_1)+\alpha(u_2)+\alpha(v_2)=\alpha(u_1+u_2)+\alpha(v_1+v_2)=\alpha(u)+\alpha(v).$$

So $\alpha$ is additive. For every $u$, by additivity $\alpha(u)=n \alpha (\frac{1}{n} u)$, so for every $u$, $\alpha(\frac{1}{n} u)=\frac{1}{n} \alpha(u)$. The above implies $\mathbb{Q}$-linearity.