TITLE: Prove the function cannot be injective
QUESTION [8 upvotes]: 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 $f(p^a)=af(p)=bf(q)=f(q^b)$.

REPLY [2 votes]: 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.