# Real Number lies between Unique Pair of Consecutive Integers

Jump to navigation
Jump to search

This article is incomplete.In particular: the proof is for the right-open case onlyYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by expanding it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Stub}}` from the code.If you would welcome a second opinion as to whether your work is correct, add a call to `{{Proofread}}` the page. |

## 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 $S$.

$\blacksquare$