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 $m > n$.
By Weak Inequality of Integers iff Strict Inequality with Integer plus One:
- $m \ge n + 1$
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.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): $\S 10$: The well-ordering principle