If $(\mathbb{Q},+)$ is homomorphic to $( \mathbb{Q}^+, \times)$, then $f(x)=1$ for all $x\in \mathbb{Q}$ 
If $(\mathbb{Q},+)$ is homomorphic to $( \mathbb{Q^+}, \times)$, then $f(x)=1$ for all $x\in \mathbb{Q}$.

This is one of questions from my assignment. I've been working on it for 2 days, but still don't have a clue. Can somebody give me a hint?
 A: HINT. Given $x\in \mathbb{Q}$ and $n\gt 0$, there always exist $y\in\mathbb{Q}$ such that $ny = x$. What will happen to this equation when you apply $f$? What does that tell you?

Note. The original problem had $(\mathbb{Q},+)$ and $(\mathbb{Q},\times)$. Hence the following:
As stated, the claim is incorrect. Since $(\mathbb{Q},\times)$ is not a group, the only reasonable interpretation is that we are considering these objects as either semigroups, or monoids. But if we consider them as semigroups,  then $f(x)=0$ for all $x$ is a perfectly fine semigroup homomorphism from the additive semigroup of rationals to the multiplicative semigroup of rationals, so the claim is false.
If we consider them as monoids (so that $f(0)=1$ necessarily is true), then we do have that any $f\colon\mathbb{Q}\to\mathbb{Q}$ such that $f(a+b) = f(a)f(b)$ is necessarily of the given form. 
(If these are meant to be groups, then the second one should be either $(\mathbb{Q}-\{0\},\times)$, or $(\mathbb{Q}_{\gt 0},\times)$. )
