Power Function with Cancellable Element Preserves Strict Ordering in Ordered Semigroup
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:
- $\forall n \in \Z_{\ge 0}: x \prec y \implies x^n \prec y^n$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.4$