I'm trying to solve the following question:
Given function $ \displaystyle f\colon \mathbb Z \rightarrow \mathbb Z,$ where $f(n) = \lfloor {\frac {n}{3}} \rfloor$, determine if it's $1:1$, and onto; and prove why.
I could say that it is onto, because it will always have an element in the codomain that maps to the domain. It is also not one to one, because the floor would result in multiple numbers with the same floor (like $ \frac {1}{3}$ and $ \frac {2}{3}$ both having the floor $0$). But how can I actually prove these?