Power Function with Cancellable Element Preserves Strict Ordering in Ordered Semigroup

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Theorem

Let $\struct {S, \circ, \preceq}$ be an ordered semigroup.

Let $x, y \in S$ be such that:

$(1): \quad x \prec y$
$(2): \quad$ either $X$ or $y$ (or both) is cancellable for $\circ$.

Let $n \in \N_{>0}$ be a strictly positive integer.


Then:

$x^n \prec y^n$

where $x^n$ is the $n$th power of $x$.


Proof

Without loss of generality, suppose $x$ is cancellable for $\circ$.


The proof proceeds by induction.

For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:

$x \prec y \implies x^n \prec y^n$


$\map P 1$ is the case:

$x \prec y \implies x \prec y$

which is trivially true.


Thus $\map P 1$ is seen to hold.


Basis for the Induction

We have:

\(\ds x\) \(\prec\) \(\ds y\)
\(\ds \implies \ \ \) \(\ds x \circ x\) \(\prec\) \(\ds x \circ y\) Strict Ordering Preserved under Product with Cancellable Element

and:

\(\ds x\) \(\preceq\) \(\ds y\)
\(\ds \implies \ \ \) \(\ds x \circ y\) \(\preceq\) \(\ds y \circ y\) Ordered Semigroup Axiom $\text {OS} 2$: Compatibility of $\preceq$ with $\circ$


Hence:

$x \prec y \implies x \circ x \prec y \circ y$

That is:

$x \prec y \implies x^2 \prec y^2$


Thus $\map P 2$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

$x \preceq y \implies x^k \prec y^k$

from which it is to be shown that:

$x \preceq y \implies x^{k + 1} \prec y^{k + 1}$


Induction Step

This is the induction step:

\(\ds x\) \(\prec\) \(\ds y\)
\(\ds \implies \ \ \) \(\ds x \circ x^k\) \(\prec\) \(\ds x \circ y^k\) Strict Ordering Preserved under Product with Cancellable Element and Induction Hypothesis

and:

\(\ds x\) \(\prec\) \(\ds y\)
\(\ds \implies \ \ \) \(\ds x \circ y^k\) \(\preceq\) \(\ds y \circ y^k\) Ordered Semigroup Axiom $\text {OS} 2$: Compatibility of $\preceq$ with $\circ$


Hence:

$x \prec y \implies x \circ x^k \prec y \circ y^k$

That is:

$x \prec y \implies x^{k + 1} \prec y^{k + 1}$


So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 0}: x \prec y \implies x^n \prec y^n$

$\blacksquare$


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