# Power Function Preserves Ordering in Ordered Semigroup

## Theorem

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

Let $x, y \in S$ such that $x \preceq y$.

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

Then:

- $x^n \preceq y^n$

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

## Proof 1

By definition of ordered semigroup:

- $\preceq$ is compatible with $\circ$.

By definition of ordering:

- $\preceq$ is transitive.

Thus by Transitive Relation Compatible with Semigroup Operation Relates Powers of Related Elements:

- $x^n \preceq y^n$

$\blacksquare$

## Proof 2

The proof proceeds by induction.

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

- $x \preceq y \implies x^n \preceq y^n$

$\map P 1$ is the case:

- $x \preceq y \implies x \preceq y$

which is trivially true.

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

### Basis for the Induction

We have:

\(\ds x\) | \(\preceq\) | \(\ds y\) | ||||||||||||

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

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 as $\preceq$ is an ordering and hence transitive:

- $x \preceq y \implies x \circ x \preceq y \circ y$

That is:

- $x \preceq y \implies x^2 \preceq y^2$

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:

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

from which it is to be shown that:

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

### Induction Step

This is the induction step:

\(\ds x\) | \(\preceq\) | \(\ds y\) | ||||||||||||

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

and:

\(\ds x\) | \(\preceq\) | \(\ds y\) | ||||||||||||

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

Hence as $\preceq$ is an ordering and hence transitive:

- $x \preceq y \implies x \circ x^k \preceq y \circ y^k$

That is:

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

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

Therefore:

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

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.4$