Prove a floor function is onto/surjective I have function $u(x) = \lfloor x \rfloor$ mapped from $\mathbb{R}$ to $\mathbb{Z}$ which I need to prove is onto.
I know that standard way of proving a function is onto requires that for every $Y$ in the co-domain there should exist an $x$ in the domain such that $u(x) = y$
I usually go about this by finding the inverse of the function and then plugging the inverse into the function itself to show that the function $u(x) = y$
Intuitively, I know that $u$ is onto because for every integer $y$, there exists a real number $x$ that can plugged into $u$ that returns $y$,
I just have no clue how to prove this since I don't understand how one would take the inverse of a floor function. How should I approach this problem in order to prove it?

Note that for an integer $z$,  $$ \lfloor z \rfloor=z$$ 
Thus the function $$u(z) = \lfloor z \rfloor$$ is onto.
