Real Number is Integer iff equals Floor

Theorem
Let $x \in \R$.

Then:
 * $x = \left \lfloor {x} \right \rfloor \iff x \in \Z$
 * $x = \left \lceil {x} \right \rceil \iff x \in \Z$

where $\left \lfloor {x} \right \rfloor$ is the floor of $x$ and $\left \lceil {x} \right \rceil$ is the ceiling of $x$.

Proof
Let $x = \left \lfloor {x} \right \rfloor$.

As $\left \lfloor {x} \right \rfloor \in \Z$, then so must $x$ be.

Now let $x \in \Z$.

We have:
 * $\left \lfloor {x} \right \rfloor = \sup \left({\left\{ {m \in \Z: m \le x}\right\} }\right)$

As $x \in \sup \left({\left\{{m \in \Z: m \le x}\right\}}\right)$, and there can be no greater $n \in \Z$ such that $n \in \sup \left({\left\{{m \in \Z: m \le x}\right\}}\right)$, it follows that:
 * $x = \left \lfloor {x} \right \rfloor$

The result for $\left \lceil {x} \right \rceil$ follows similar lines.