# Integers under Addition form Totally Ordered Group

## Theorem

Let $\left({\Z, +}\right)$ denote the additive group of integers.

Let $\le$ be the usual ordering on $\Z$.

Then the ordered structure $\left({\Z, +, \le}\right)$ is a totally ordered group.

## Proof

### $\left({\Z, +, \le}\right)$ is an Ordered Structure

#### $(1)$

By Integer Addition is Closed, $\left({\Z, +}\right)$ is an algebraic structure.

#### $(2)$

$\le$ is an ordering on $\Z$.

Thus, $\left({\Z, \le}\right)$ is an ordered set.

#### $(3)$

By Ordering is Preserved on Integers by Addition and Integer Addition is Commutative, $\le$ is compatible with $+$.

Thus, $\left({\Z, +, \le}\right)$ is an ordered structure.

$\Box$

### $\left({\Z, +, \le}\right)$ is a Totally Ordered Structure

By definition, the ordered structure $\left({\Z, +, \le}\right)$ is a totally ordered structure if and only if $\le$ is a total ordering.

This follows from Ordering on Integers is Total Ordering.

$\Box$

### $\left({\Z, +, \le}\right)$ is a Totally Ordered Group

By definition, the totally ordered structure $\left({\Z, +, \le}\right)$ is a totally ordered group if and only if $\left({\Z, +}\right)$ is a group.

This follows from Integers under Addition form Abelian Group.

$\blacksquare$