Ordering in terms of Addition

Theorem

Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.

Then $\forall m, n \in S$:

$m \preceq n \iff \exists p \in S: m \circ p = n$

Proof

Necessary Condition

From Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product, we have:

$\forall m, n \in S: m \preceq n \implies \exists p \in S: m \circ p = n$

$\Box$

Sufficient Condition

Suppose that $m \circ p = n$.

\(\ds 0\)

\(\preceq\)

\(\ds p\)

Definition of Zero of Naturally Ordered Semigroup

\(\ds \leadsto \ \ \)

\(\ds m \circ 0\)

\(\preceq\)

\(\ds m \circ p\)

$\preceq$ is compatible with $\circ$

\(\ds \leadsto \ \ \)

\(\ds m\)

\(\preceq\)

\(\ds n\)

Zero is Identity in Naturally Ordered Semigroup

$\Box$

So $\forall m, n \in S$:

$m \preceq n \iff \exists p \in S: m \circ p = n$

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.2$