Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 2
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Theorem
Let $\struct {G, \circ, \preccurlyeq}$ be an ordered group with identity $e$.
Let $x \in G$.
Then the following equivalences hold:
| \(\ds \forall x \in G: \, \) | \(\ds x \preccurlyeq e\) | \(\iff\) | \(\ds e \preccurlyeq x^{-1}\) | |||||||||||
| \(\ds e \preccurlyeq x\) | \(\iff\) | \(\ds x^{-1} \preccurlyeq e\) | ||||||||||||
| \(\ds x \prec e\) | \(\iff\) | \(\ds e \prec x^{-1}\) | ||||||||||||
| \(\ds e \prec x\) | \(\iff\) | \(\ds x^{-1} \prec e\) |
Proof
By the definition of an ordered group, $\preccurlyeq$ is a relation compatible with $\circ$.
Thus by Inverses of Elements Related by Compatible Relation: Corollary:
| \(\ds \forall x \in G: \, \) | \(\ds x \preccurlyeq e\) | \(\iff\) | \(\ds e \preccurlyeq x^{-1}\) | |||||||||||
| \(\ds e \preccurlyeq x\) | \(\iff\) | \(\ds x^{-1} \preccurlyeq e\) |
By Reflexive Reduction of Relation Compatible with Group Operation is Compatible, $\prec$ is also compatible with $\circ$.
Thus again by Inverses of Elements Related by Compatible Relation: Corollary:
| \(\ds \forall x \in G: \, \) | \(\ds x \prec e\) | \(\iff\) | \(\ds e \prec x^{-1}\) | |||||||||||
| \(\ds e \prec x\) | \(\iff\) | \(\ds x^{-1} \prec e\) |
$\blacksquare$