Real Number is between Floor Functions

Theorem
$$\forall x \in \mathbb{R}: \left \lfloor {x} \right \rfloor \le x < \left \lfloor {x + 1} \right \rfloor$$

Proof
$$\left \lfloor {x} \right \rfloor$$ is defined as:

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

So $$\left \lfloor {x} \right \rfloor \le x$$ by definition.

Now $$\left \lfloor {x + 1} \right \rfloor > \left \lfloor {x} \right \rfloor$$, so by the definition of the supremum, $$\left \lfloor {x + 1} \right \rfloor > x$$.

The result follows.