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.