# Equivalence of Definitions of Floor Function

## Theorem

Let $x$ be a real number.

The following definitions of the concept of Floor Function are equivalent:

### Definition 1

The floor function of $x$ is defined as the supremum of the set of integers no greater than $x$:

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

where $\le$ is the usual ordering on the real numbers.

### Definition 2

The floor function of $x$, denoted $\floor x$, is defined as the greatest element of the set of integers:

$\set {m \in \Z: m \le x}$

where $\le$ is the usual ordering on the real numbers.

### Definition 3

The floor function of $x$ is the unique integer $\floor x$ such that:

$\floor x \le x < \floor x + 1$

## Proof

### Definition 1 equals Definition 2

Follows from Supremum of Set of Integers equals Greatest Element.

$\Box$

### Definition 1 equals Definition 3

Let $S$ be the set:

$S = \set {m \in \Z: m \le x}$

Let $n = \sup S$.

By Supremum of Set of Integers is Integer, $n \in \Z$.

By Supremum of Set of Integers equals Greatest Element, $n\in S$.

Because $n \in S$, we have $n \le x$.

Because $n + 1 > n$, we have by definition of supremum:

$n + 1 \notin S$

Thus $n + 1 > x$.

Thus $n$ is an integer such that:

$n \le x < n + 1$

So $n$ is the floor function by definition 3.

$\Box$

### Definition 3 equals Definition 2

Let $n$ be an integer such that:

$n \le x < n + 1$

We show that $n$ is the greatest element of the set:

$S = \set {m \in \Z: m \le x}$

Let $m \in \Z$ such that $m \le x$.

We show that $n \ge m$.

Aiming for a contradiction, suppose $m > n$.

$m \ge n + 1$

and so from the definition of $g$ it follows that $m > x$.

By Proof by Contradiction it follows that $m \le n$.

Because $m \in S$ was arbitrary, $n$ is the greatest element of $S$.

Thus $n$ is the floor function by definition 2.

$\blacksquare$