Relation Induced by Strict Positivity Property is Transitive
Jump to navigation
Jump to search
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 $<$ is a transitive relation.
Proof
Let $a < b$ and $b < c$.
Thus:
\(\ds \) | \(\) | \(\ds a < b, b < c\) | ||||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map P {-a + b}, \map P {-b + c}\) | Definition of $<$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map P {\paren {-a + b} + \paren {-b + c} }\) | Definition of Strict Positivity Property | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds \map P {-a + c}\) | Properties of $+$ in $D$ | |||||||||||
\(\ds \) | \(\leadsto\) | \(\ds a < c\) | Definition of $<$ |
And so $<$ is seen to be transitive.
$\blacksquare$
Sources
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Theorem $8$