# Strict Positivity Property induces Total Ordering

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## Theorem

Let $\struct {D, +, \times}$ be an integral domain whose zero is $0_D$.

Let $D$ be endowed with a (strict) positivity property $P: D \to \set {\mathrm T, \mathrm F}$.

Then there exists a total ordering $\le$ on $\struct {D, +, \times}$ induced by $P$ which is compatible with the ring structure of $\struct {D, +, \times}$.

## Proof

By definition of the strict positivity property:

 $(P \, 1)$ $:$ Closure under Ring Addition: $\displaystyle \forall a, b \in D:$ $\displaystyle \map P a \land \map P b \implies \map P {a + b}$ $(P \, 2)$ $:$ Closure under Ring Product: $\displaystyle \forall a, b \in D:$ $\displaystyle \map P a \land \map P b \implies \map P {a \times b}$ $(P \, 3)$ $:$ Trichotomy Law: $\displaystyle \forall a \in D:$ $\displaystyle \map P a \lor \map P {-a} \lor a = 0_D$ For $P \, 3$, exactly one condition applies for all $a \in D$.

Let us define a relation $<$ on $D$ as:

$\forall a, b \in D: a < b \iff \map P {-a + b}$

Setting $a = 0$:

$\forall b \in D: 0 < b \iff \map P b$

demonstrating that (strictly) positive elements of $D$ are those which are greater than zero.

From Relation Induced by Strict Positivity Property is Compatible with Addition we have that $<$ is compatible with $+$.

From Relation Induced by Strict Positivity Property is Transitive we have that $<$ is transitive.

From Relation Induced by Strict Positivity Property is Asymmetric and Antireflexive we have that $<$ is asymmetric and antireflexive.

Thus by definition, $<$ is a strict ordering.

Let the relation $\le$ be defined as the reflexive closure of $<$.

From Reflexive Closure of Strict Ordering is Ordering we have that $\le$ is an ordering on $D$.

From Relation Induced by Strict Positivity Property is Trichotomy, and from the Trichotomy Law (Ordering), we have that $\le$ is a total ordering.

Hence the result.

$\blacksquare$