For any additive function, $f: (\mathbb{Q}, +)\to (\mathbb{Q}, +)$, why does $f(nx) = nf(x)$? For the homomorphism, $f: (\mathbb{Q}, +)\to (\mathbb{Q}, +)$, where
$\mathbb{Q}$ is the set of rational numbers.
Show that for all $x\in\mathbb{Q}$ and for all $n\in\mathbb{Z}$, we have $f(nx) = nf(x)$
I can't figure out how to prove this without a definition of $f$.

We have that $f(nx)=f(\underbrace{x+x+\cdots +x}_{n\:\text{times}})=\underbrace{f(x)+f(x)+\cdots+f(x)}_{n\:\text{times}}=nf(x)$.
