Positive Rational Numbers under Addition fulfil Naturally Ordered Semigroup Axioms 2 to 4

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

Consider the naturally ordered semigroup axioms:

The algebraic structure:
 * $\struct {\Q_{\ge 0}, +, \le}$

is an ordered semigroup which fulfils the axioms:



but:
 * does not fulfil
 * $\struct {\Q_{\ge 0}, +}$is not isomorphic to $\struct {\N, +}$.

Proof
First we note that from Positive Rational Numbers under Addition form Ordered Semigroup:
 * $\struct {\Q_{\ge 0}, +, \le}$ is an ordered semigroup.

From Rational Numbers under Addition form Infinite Abelian Group, $\struct {\Q, +}$ is a group.

By the Cancellation Laws, all elements of $\struct {\Q, +}$ are cancellable.

From Cancellable Element is Cancellable in Subset, it follows that all elements of $\struct {\Q_{\ge 0}, +}$ are likewise cancellable.

Hence holds.

Let $a, b \in \Q_{\ge 0}$ such that $a \le b$.

Then as $\struct {\Q, +}$ is a group it follows that:
 * $\exists c \in \Q: a + c = b$

and so:
 * $c = b + \paren {-a}$

But as $a \le b$ it follows that $c \ge 0$.

That is:
 * $\exists c \in \Q_{\ge 0}: a + c = b$

Hence holds.

We have that:
 * $0 \in \Q_{\ge 0}$

and:
 * $1 \in \Q_{\ge 0}$

and trivially holds.

We consider the subset $S$ of $\struct {\Q_{\ge 0}, +, \le}$ defined as:


 * $S = \set {x \in \Q_{\ge 0}: x > 0}$

From Smallest Strictly Positive Rational Number does not Exist:
 * there exists no smallest element of $S$.

Hence by definition $\struct {\Q_{\ge 0}, +, \le}$ is not well-ordered by $\le$.

That is, $\struct {\Q_{\ge 0}, +, \le}$ does not fulfil.

From Positive Rational Numbers under Addition not Isomorphic to Natural Numbers:
 * $\struct {\Q_{\ge 0}, +}$is not isomorphic to $\struct {\N, +}$.