Why is $f(1)=0$ on $f:(\mathbb{Q},+) \rightarrow (\mathbb{Z},+)$, where $f$ is a homomorphism?

Question

Why is $f(1)=0$ on $f:(\mathbb{Q},+) \rightarrow (\mathbb{Z},+)$, where $f$ is a homomorphism?

My professor stated that $f(1)=0$ in the such case because:

$$f(1)=f\left(\frac{1}{n}+...+\frac{1}{n}\right)=nf(1/n)$$

so we have $f(1)/n=f(1/n)$ as $f(\cdot) \in \mathbb{Z}$ he told that $f(1)=0$ but I see no reason why $f(1)=0$, why can't I just take $f(1)=kn$ with $k \in \mathbb{Z}$? If the neutral element from the first group was $1$ and $0$ for the second it would be easier to get.

Answer 1

Since $f(1/n)\in \Bbb Z$, we have that $n\mid f(1)$. But $n$ was arbitrary. Hence $f(1)=0$.