Odd Power Function is Strictly Increasing/General Result

Theorem

Let $\struct {R, +, \circ, \le}$ be a totally ordered ring.

Let $n$ be an odd positive integer.

Let $f: R \to R$ be the mapping defined by:

$\map f x = \map {\circ^n} x$

Then $f$ is strictly increasing on $R$.

Proof

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Let $x, y \in R$ such that $0 < x < y$.

By Power Function is Strictly Increasing on Positive Elements:

$\map f x < \map f y$

Suppose that $x < y < 0$.

By Properties of Ordered Ring:

$0 < -y < -x$

By Power Function is Strictly Increasing on Positive Elements (applied to $-y$ and $-x$):

$0 < \map f {-y} < \map f {-x}$

By Power of Ring Negative:

$\map f {-x} = -\map f x$

$\map f {-y} = -\map f y$

Thus:

$0 < -\map f y < -\map f x$

By Properties of Ordered Ring:

$\map f x < \map f y$

By Sign of Odd Power:

$\map f x < 0 = \map f 0$ when $x < 0$

$\map f 0 = 0 < \map f x$ when $0 < x$

Thus we have shown that $f$ is strictly increasing on the positive elements and the negative elements, and across zero.

$\blacksquare$