Floor Function is Integer

Theorem
Let $f: \R \to \R$ be the mapping defined as:
 * $\forall x \in \R: f \left({x}\right) = \left\lfloor{x}\right\rfloor$

where $\left\lfloor{x}\right\rfloor$ denotes the floor of $x$.

Then $f \left({x}\right)$ is an integer.

Proof
By definition of floor function:


 * $z = \left \lfloor {x} \right \rfloor \iff z \in \Z \land x \in \left\{ {y \in \R: z \le y < z + 1}\right\}$

That is, if $f \left({x}\right)$ is the floor of $x$, then:
 * $f \left({x}\right) \in \Z$

where $\Z$ denotes the set of integers.

Also see

 * Ceiling Function is Integer