Relation Induced by Strict Positivity Property is Trichotomy

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Theorem

Let $\struct {D, +, \times}$ be an ordered integral domain where $P$ is the (strict) positivity property.

Let the relation $<$ be defined on $D$ as:

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


Then $\forall a, b \in D:$ exactly one of the following conditions applies:

$a < b$
$a = b$
$a > b$

That is, $<$ is a trichotomy.


Proof

Take any $a, b \in D$ and consider $-a + b$.

From the trichotomy law of ordered integral domains, exactly one of the following applies:

$(1): \quad \map P {-a + b}$
$(2): \quad \map P {-\paren {-a + b} }$
$(3): \quad -a + b = 0$

Taking each of these in turn and taking into account that $\struct {D, +}$ is the additive group of $\struct {D, +, \times}$:


\(\text {(1)}: \quad\) \(\ds \) \(\) \(\ds \map P {-a + b}\)
\(\ds \) \(\leadsto\) \(\ds a < b\) Definition of $<$

$\Box$


\(\text {(2)}: \quad\) \(\ds \) \(\) \(\ds \map P {-\paren {-a + b} }\)
\(\ds \) \(\leadsto\) \(\ds \map P {-b + a}\)
\(\ds \) \(\leadsto\) \(\ds b < a\) Definition of $<$

$\Box$


\(\text {(3)}: \quad\) \(\ds \) \(\) \(\ds -a + b = 0\)
\(\ds \) \(\leadsto\) \(\ds a = b\) adding $a$ to both sides

$\Box$

The result has been proved.

$\blacksquare$


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