Simple analysis problem that is giving me some grief.

I am given a function $f:\mathbb{R}\rightarrow\mathbb{R}$ that has the property that $f(u+v)=f(u)+f(v)$ for all $u,v\in\mathbb{R}$. Then we define $m=f(1)$ and I am asked to prove that $f(x)=mx$ for all rational numbers $x$.

This should be pretty straightforward but I can't quite seem to nug this one out.

What I wrote out so far is as follows:
$$f(x)=f(\frac{p}{q})=f(\frac{1_1}{q}+...+\frac{1_p}{q})=f(\frac{1_1}{q})+...+f(\frac{1_p}{q})=pf(\frac{1}{q}).$$
This is where I am stuck.

Answer

Continue the hint in the comment: $qf(1/q) = \underbrace{f(1/q) + \cdots + f(1/q)}_{\mbox{$q$ times}} = f(q/q) = f(1) = m$

$f(1/q) = m/q \implies f(p/q) = \underbrace{f(1/q) + \cdots + f(1/q)}_{\mbox{$p$ times}} = pf(1/q) = pm/q = m \cdot p/q$