Prove that the greatest integer function: $\mathbb{R} \rightarrow \mathbb{Z}$ is onto but not $1-1$ Statement: the greatest integer function int: $\mathbb{R} \rightarrow \mathbb{Z}$ is onto but not $1-1$
Proof: let $x \in \mathbb{R}$, then $int(x) \leq x$ and is an element of $\mathbb{Z}$. Since $\mathbb{Z}$ is an element of $\mathbb{R}$, the greatest integer function maps onto $\mathbb{Z}$. However, it is not one-to-one, because $int(0.2)=int(0.3)=0$
Is my proof valid? 
 A: You'd want to more clearly show how given $x \in \mathbb{Z}$ you could find a real number $r \in \mathbb{R}$ such that $\text{int}(r)=x$. Then you have onto. As for $1:1$, you have shown that it cannot be $1:1$. Note that a function that can be plotted in $\mathbb{R}^2$ can only be $1:1$ if every horizontal line drawn in the plane intersects the graph at most one point--clearly not the case here.
A: Yes, your reasoning is correct, but you should say $\Bbb Z$ is a subset of $\Bbb R$. The element relation does not hold between $\Bbb Z$ and $\Bbb R$. (Some students confuse containment and membership early on.)
It looks like you are using the fact that integers map to themselves to demonstrate the map is onto, and that's good. However, it might strengthen what you wrote if you stated that a little more explicitly.
(Incidentally, the question should also be more like "determine if this function is 1-1 and/or onto," but I can tell that's really what you were interested in because of your solution.) (Obsolete due to intervening edits.)
