# 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.

$\blacksquare$