Real Number is between Floor Functions

Theorem

 * $\forall x \in \R: \floor x \le x < \floor {x + 1}$

where $\floor x$ is the floor of $x$.

Proof
$\floor x$ is defined as:


 * $\floor x = \sup \set {m \in \Z: m \le x}$

So $\floor x \le x$ by definition.

From Floor plus One:
 * $\floor {x + 1} > \floor x$

Hence by the definition of the supremum:
 * $\floor {x + 1} > x$

The result follows.