For any additive function, $f: (\mathbb{Q}, +)\to (\mathbb{Q}, +)$, why does $f(nx) = nf(x)$?

Question

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 $f$.

Answer 1

Let $G=(G,*)$ be a group. Remember that an homomorphism $\varphi:G \to G$ is a function that satisfies $\varphi(g*h)=\varphi(g)*\varphi(h)$ for all $g,h \in G$. In this case $G=(\mathbb{Q},+)$, so $$f(nx)=f(\underbrace{x+x+...+x}_{n \: \mathrm{times}})=\underbrace{f(x)+f(x)+...+f(x)}_{n \: \mathrm{times}}=nf(x).$$

Answer 2

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