# Positive Rational Numbers under Addition form Ordered Semigroup

## Theorem

Let $\Q_{\ge 0}$ denote the set of positive rational numbers.

$\struct {\Q_{\ge 0}, +, \le}$

forms an ordered semigroup.

## Proof 1

It is necessary to ascertain that $\struct {\Q_{\ge 0}, +, \le}$ fulfils the ordered semigroup axioms:

An ordered semigroup is an algebraic system $\struct {S, \circ, \preceq}$ which satisfies the following properties:

 $(\text {OS} 0)$ $:$ Closure $\ds \forall a, b \in S:$ $\ds a \circ b \in S$ $(\text {OS} 1)$ $:$ Associativity $\ds \forall a, b, c \in S:$ $\ds a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$ $(\text {OS} 2)$ $:$ Compatibility of $\preceq$ with $\circ$ $\ds \forall a, b, c \in S:$ $\ds a \preceq b \implies \paren {a \circ c} \preceq \paren {b \circ c}$ where $\preceq$ is an ordering $\ds a \preceq b \implies \paren {c \circ a} \preceq \paren {c \circ b}$

From Rational Numbers form Ordered Field, $\struct {\Q, +, \times, \le}$ is an ordered field.

Hence $\struct {\Q, +, \le}$ is an ordered group, and so an ordered semigroup.

From Positive Rational Numbers are Closed under Addition we have that $\text {OS} 0$ holds.

From Restriction of Associative Operation is Associative we have that $\text {OS} 1$ holds.

From Restriction of Ordering is Ordering we have that $\text {OS} 2$ holds.

The result follows.

$\blacksquare$

## Proof 2

From Rational Numbers form Ordered Field, $\struct {\Q, +, \times, \le}$ is an ordered field.

Hence $\struct {\Q, +, \le}$ is an ordered group, and so an ordered semigroup.

From Positive Rational Numbers are Closed under Addition we have that $\struct {\Q_{\ge 0}, +}$ is closed.

Hence from Subsemigroup Closure Test, $\struct {\Q_{\ge 0}, +}$ is a subsemigroup of $\struct {\Q, +}$.

From Subsemigroup of Ordered Semigroup is Ordered, $\struct {\Q_{\ge 0}, +, \le}$ is an ordered semigroup.

$\blacksquare$