Rational homogeneity of an additive function ($f(qx)=qf(x)(q\in \mathbb{Q})$)

For every additive function $f:\mathbb{R}\to \mathbb{R}$, every real number $x$, and every rational number $q$, one has $f(qx)=qf(x)$. This is the scalar multiplication part of being a linear map over the rational numbers.

First $f(0)=f(0)+f(0)$ gives $f(0)=0$, and $f(x)+f(-x)=f(0)$ gives $f(-x)=-f(x)$. Repeated addition gives $f(nx)=nf(x)$ for each positive integer $n$, and the negative case follows from the sign identity. If $q=m/n$ with $m\in \mathbb{Z}$ and $n>0$, thenNo continuity was used. In particular $f(1)=3$ fixes $f(-2/3)=-2$, $f(5/6)=5/2$, and every other rational value, even for a discontinuous additive function.