Real Number lies between Unique Pair of Consecutive Integers

Theorem
Let $x$ be a real number.

Right-open Interval
There exists a unique integer $n\in\Z$ such that:
 * $n \leq x < n + 1$

Left-open Interval
There exists a unique integer $n\in\Z$ such that:
 * $n - 1 < x \leq n$

Existence
By Set of Integers Bounded Above by Real Number has Greatest Element, the set:
 * $S = \left\{{m \in \Z: m \le x}\right\}$

has a greatest element, say $n$.

Because $n+1>n$, $n+1\notin S$.

Thus $n+1> x$.

Thus $n\leq x < n+1$.

Uniqueness
Let $n\in\Z$ be such that:
 * $n \leq x < n + 1$

We show that $n$ is a greatest element of the set:
 * $S = \left\{{m \in \Z: m \le x}\right\}$

so that the uniqueness follows from Greatest Element is Unique.

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

Let $m \in S$.

Because $m \leq x < n+1$, $n+1 > m$.

By Weak Inequality of Integers iff Strict Inequality with Integer plus One:
 * $n \geq m$.

Because $m$ was arbitrary, $n$ is a greatest element of $S$.

Also see

 * Definition:Floor Function
 * Definition:Ceiling Function