How can we prove that $f(x) = \lfloor 1.5x \rfloor$ is not onto? How can we prove that $f:\mathbb Z \to \mathbb Z, f(x) = \lfloor 1.5x \rfloor$ is not onto? While it's very obvious to see, how can I actually prove that this is in fact the case? Traditional methods that I would use don't seem applicable because of the floor part of the equation...
 A: For all $x \in \mathbb{Z}$, $f(x) = \lfloor 1.5 x \rfloor$ is either equal to $1.5x$ or equal to $1.5x - 0.5$. So $2f(x) = 3x \text{ or } 3x - 1$. In particular:
\begin{align*}
2f(x) \equiv 0 \text{ or } 2 \pmod{3} \\
\implies f(x) \equiv 0 \text{ or } 1 \pmod{3}.
\end{align*}
Thus $f$ does not attain any value $\equiv 2 \pmod{3}$, so it cannot be onto.
A: Notice that $\lfloor 1.5x \rfloor $ is non-decreasing.
$f(1)=1$, $f(2)=3$.
We do not have a preimage for $2$.
A: You actually haven't defined the function completely. A proper definition needs to show the domain and codomain of the function i.e. $f:\mathbb R\rightarrow \mathbb R$ or $f:\mathbb R\rightarrow \mathbb N$ etc. In general $f:A\rightarrow B$. 
I'm assuming that you mean $f:\mathbb Z\rightarrow \mathbb Z$. 
To prove the function isn't onto, just find one element $y \in B$ s.t. there is no element in $A$ that maps to $y$. Trying a few values, $f(0) = 0,\ f(1) = 1,\ f(2) = 3$. Seems we missed $2$. Can you prove that no future value will ever map to 2?
A: Hint: Find an element in the codomain that is not in the image (use your intuition here). Suppose it is $c\in\Bbb Z$. Then find the possible solutions for $\lfloor 1.5x\rfloor=c$. If the possible values for $x$ contain no integers, then there is no $x\in\Bbb Z$ such that $\lfloor 1.5x\rfloor=c$.
A: I'm assuming your function is $f:\mathbb{R} \to \mathbb{R}$.
$f$ isn't onto. Just choose any value $x \notin \mathbb{Z}$ and there won't be any correspondence in the domain of $f$.
Suppose by contradiction that given $y \notin \mathbb{Z}$ there is $x \in \mathbb{R} \text{ }st. f(x)=y$. Then $\lfloor{1.5x} \rfloor \notin \mathbb{Z}$, which is absurd by definition of floor function.
edit: 
OP updated that it is $f:\mathbb{Z} \to \mathbb{Z}$.
Then, firstly note that $f$ is non-decreasing. Then take a look at the image set $Im (f) = \{ ..., -3, -2, 0, 1, 3, ... \}$ and you can see that there are integers in between those elements that aren't in the image set. Just take one of them, say $2$. Suppose there is $x$ in the domain of $f$ such that $f(x)=2$. Then $\lfloor 1.5x \rfloor = 2$, which means $ 2 \leq 1.5x < 3$ which is absurd since $x$ is an integer.
