Real Number is Integer iff equals Floor

Theorem
Let $$x \in \mathbb{R}$$.

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

Proof

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

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


 * Now let $$x \in \mathbb{Z}$$.

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

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


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