Inversion Mapping Reverses Ordering in Ordered Group/Corollary/Proof 1
Jump to navigation
Jump to search
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 Inversion Mapping Reverses Ordering in Ordered Group:
\(\ds \forall x \in G: \, \) | \(\ds x \preccurlyeq e\) | \(\iff\) | \(\ds e^{-1} \preccurlyeq x^{-1}\) | |||||||||||
\(\ds e \preccurlyeq x\) | \(\iff\) | \(\ds x^{-1} \preccurlyeq e^{-1}\) | ||||||||||||
\(\ds x \prec e\) | \(\iff\) | \(\ds e^{-1} \prec x^{-1}\) | ||||||||||||
\(\ds e \prec x\) | \(\iff\) | \(\ds x^{-1} \prec e^{-1}\) |
Since $e^{-1} = e$, the theorem holds.
$\blacksquare$