Power Function is Strictly Increasing on Positive Elements

Theorem
Let $\struct {R, +, \circ, \le}$ be an ordered ring.

Let $x, y \in R$.

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

Let $0 < x < y$.

Then:
 * $0 < \map {\circ^n} x < \map {\circ^n} y$

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $0 < \map {\circ^n} x < \map {\circ^n} y$

Basis for the Induction
$\map P 1$ is the case:
 * $0 < \map {\circ^1} x < \map {\circ^1} y$

which is just:
 * $0 < x < y$

Thus $\map P 1$ 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:
 * $0 < \map {\circ^k} x < \map {\circ^k} y$

from which it is to be shown that:
 * $0 < \map {\circ^{k + 1}} x < \map {\circ^{k + 1}} y$

Induction Step
This is the induction step:

We have:
 * $0 < x < y$
 * $0 < \map {\circ^k} x < \map {\circ^k} y$

By Ring Product preserves Inequalities on Positive Elements:
 * $0 < \map {\circ^k} x \circ x < \map {\circ^k} y \circ y$

Hence:
 * $0 < \map {\circ^{k + 1}} x < \map {\circ^{k + 1}} y$

So $\map P k \implies \map P {k + 1}$ and thus it follows by the Principle of Mathematical Induction that:


 * $\forall n \in \N_{> 0}: 0 < \map {\circ^n} x < \map {\circ^n} y$

Also see

 * Odd Power Function is Strictly Increasing