I did my best to search for this question on the site- but I did not find it. Here it is:
If a function $f:\mathbb{R}^2\to\mathbb{R}$ satisfies $f(u+v)=f(u)+f(v)$ for all $u,v\in\mathbb{R}^2$, then $f(\frac{p}{q}v)=\frac{p}{q}f(v)$ for all $p\in \mathbb{Z}$, $q\in \mathbb{Z}\setminus \{0\}$.
So far, I have tried to use argument substitutions and "abuse" of additivity to achieve a homogeneity condition. For example: $$ f\bigg(\frac{p}{q}v\bigg)=f\bigg(\frac{p-q}{q}v+v\bigg)=f(v)+f\bigg(\frac{p-q}{q}v\bigg)$$ and other similar things- but I have been unable to show this.
Note: I only want a gentle nudge in the right direction - I explicitly request that no one give me the answer.
Hint: $f$ satisfies $$ f(nv)=nf(v) $$ and $$ nf(\frac{1}{n}v)=f(v) $$ for every positive integer $n$ and for every $v\in\mathbb{R}^2$. (Why?)