Real Number is Integer iff equals Floor

Theorem
Let $$x \in \R$$.

Then: where $$\left \lfloor {x} \right \rfloor$$ is the floor of $$x$$ and $$\left \lceil {x} \right \rceil$$ is the ceiling of $$x$$.
 * $$x = \left \lfloor {x} \right \rfloor \iff x \in \Z$$
 * $$x = \left \lceil {x} \right \rceil \iff x \in \Z$$

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 $$n \in \sup \left({\left\{{m \in \Z: m \le x}\right\}}\right)$$, $$x = \left \lfloor {x} \right \rfloor$$.


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