Prove the function cannot be injective 
Let $f: \{x \in \mathbb{Q}, x \gt 0 \} \rightarrow \mathbb{Q}$ such
  that $f(xy)=f(x) + f(y), \forall x,y$.Prove:
  
  
*
  
*$f$ cannot be injective
  
*Can $f$ be surjective?
  


For $x=y=1$ I get $f(1)=0$. Also it's easy to prove 
$$f(x^n)=n\,f(x),\quad \forall n \in \mathbb{Z}$$
I could not get further, any help is appreciated
UPDATE
I've found the proof for 1).
Let $p, q$ be distinct primes, and $a,b \in \mathbb{Z}$ such that $\frac {f(p)}{f(q)}=\frac{b}{a}$. Then $f(p^a)=af(p)=bf(q)=f(q^b)$.
 A: Answer to 1.
Suppose $f:(\mathbb{Q}^{>0},\cdot)\to(\mathbb{Q},+)$ is an injective function satisfying $f(x\cdot y)=f(x)+f(y)$. Then, $f(1)=0$ and for every $p\neq 1$, $f(p)\neq f(1)=0$. For two different primes $p,q$ we would have:
\begin{align*}
f(p)=\frac{m}{n}, f(q)=\frac{r}{s} &\Rightarrow n\cdot f(p)=m \text{ and } s\cdot f(q)=r \\
&\Rightarrow m\cdot r=r\cdot n\cdot f(p)=m\cdot s\cdot f(p)\\
&\Rightarrow f(p^{n\cdot r})=f(q^{m\cdot s})
\end{align*}
and since $f$ is injective this means $p^{nr}=q^{ms}$, which is only possible if both $r=s=0$, implying $f(p)=0$, absurd.
Answer to 2
Notice that such a function $f$ is completely characterized by its value on prime numbers, as for primes $p_1,\ldots,p_k,q_1,\ldots,q_\ell$ and natural numbers $m_1,\ldots,m_k,n_1,\ldots,n_\ell$ we will have
$$f\left(\dfrac{p_1^{m_1}\cdots p_{k}^{m_k}}{q_1^{n_1}\cdots q_\ell^{n_\ell}}\right)=\sum_{i=1}^k m_i\cdot f(p_i)-\sum_{j=1}^{\ell}n_j\cdot f(q_j).$$
Since $(\mathbb{Q},+)$ is an abelian group, it can be seen as a $\mathbb{Z}$-module, and we can choose a countable base $\{e_p:p\in\mathbb{P}\}$, indexed by the set of prime numbers. We can then define a homomorphism $f:(\mathbb{Q}^{>0},\cdot)\to \langle e_n:n\in\mathbb{N}\rangle_\mathbb{Z}\cong (\mathbb{Q},+)$ be putting $$f\left(\dfrac{p_1^{m_1}\cdots p_{k}^{m_k}}{q_1^{n_1}\cdots q_\ell^{n_\ell}}\right)=\sum_{i=1}^k m_i\cdot e_{p_i}-\sum_{j=1}^{\ell}n_j\cdot e_{q_j}$$
It is easy to show that $f$ defines a homomorphism, and if $r\in\mathbb{Q}$ then $r$ can be written as an integer combination of finitely many elements $e_{p}$. By arranging positive and negative scalars separately we obtain
$$r=m_{p_1}\cdot e_{p_1}+\cdots + m_{p_k}\cdot e_{p_k}-n_{q_1}\cdot e_{q_1}-\cdots-n_{q_\ell}\cdot e_{q_\ell}$$ and we would have 
$$f\left(\dfrac{p_1^{m_1}\cdots p_{k}^{m_k}}{q_1^{n_1}\cdots q_\ell^{n_\ell}}\right)=\sum_{i=1}^k m_i\cdot e_{p_i}-\sum_{j=1}^{\ell}n_j\cdot e_{q_j}=r,$$
which shows it is surjective.
A: It can be surjective, at least assuming the axiom of choice.
Such a function is uniquely determined by its values on the prime numbers, which may be arbitrary rational numbers. Choose a bijection $f$ from the set of prime numbers to the set of numbers of the form $1/p^n$ for $p$ prime and $n$ a positive integer (both sets are countably infinite). Since every rational number may be expressed as an integer linear combination of numbers of the form $1/p^n$, this defines a surjective function $f$. It might be interesting to ask if you can find constructible (in some sense) function $f$ of this type.
