analysis - If $f(x + y) = f(x) + f(y)$ showing that $f(cx) = cf(x)$ holds for rational $c$

For $f:\mathbb{R}^n \to \mathbb{R}^m$, if $f(x + y) = f(x) + f(y)$ for then for rational $c$, how would you show that $f(cx) = cf(x)$ holds?

I tried that for $c = \frac{a}{b}$, $a,b \in \mathbb{Z}$ clearly

$$f\left(\frac{a}{b}x\right) = f\left(\frac{x}{b}+\dots+\frac{x}{b}\right) = af\left(\frac{x}{b}\right)$$
but I can't seem to finish it, any help?

Answer

Try computing $bf(x/b){}{}{}$.