Integers under Addition form Totally Ordered Group

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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.

$\blacksquare$