Suppose $\phi: \mathbb{Q} \to \mathbb{Z}$ is a homomorphism, where both $\mathbb{Q}$ and $\mathbb{Z}$ are under addition. Prove $\phi$ is the zero map 
Suppose $\phi: \mathbb{Q} \to \mathbb{Z}$ is a homomorphism, where
both $\mathbb{Q}$ and $\mathbb{Z}$ are under addition. Prove $\phi$ is
the zero map.

Suppose $\phi$ is a non zero map. Meaning there is at least one $x \in \mathbb{Q}$ such that $\phi(x) = a$ for some non zero integer a. Since the definition of homomorphism involves two elements from the domain, I tried taking another element $y \in \mathbb{Q}$ such that $\phi(y) = b$ for some integer b. Since $\phi$ is homomorphism, $\phi(x+y) =\phi(x) + \phi(y)= a + b$.
My intention is to arrive at a contradiction that $\phi$ is not a homomorphism. But I am stuck.

Let $k$ be the smallest/largest integer $>0$/$<0$ such that for some $x\in\mathbb{Q}$, $\phi(x)=k$. Then $\phi(\frac{x}{2})+\phi(\frac{x}{2})=k$, so $\phi(\frac{x}{2}) = \frac{k}{2}$ which is smaller/greater than $k$ and greater/smaller than $0$.
