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.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $9: \ \text{O} 3$