Real Number is between Floor Functions

Theorem

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

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

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


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

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.