Real Number lies between Unique Pair of Consecutive Integers

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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:

$n - 1 < x \leq n$

Proof

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 $n\leq x < n+1$.

$\Box$

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 $S$.

$\blacksquare$

Also see

• Definition:Floor Function
• Definition:Ceiling Function