# Power of Ring Negative

## Theorem

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

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

Let $x \in R$.

Then:

If $n$ is even:
$\map {\circ^n} {-x} = \map {\circ^n} x$
If $n$ is odd:
$\map {\circ^n} {-x} = -\map {\circ^n} x$

## Proof

First, suppose that $n$ is even.

Then for some $m \in \N_{>0}$:

$n = 2 m = m + m$

Thus since $\circ$ is associative:

$\displaystyle \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \paren {-x} \circ \paren {-x}$
$\paren {-x} \circ \paren {-x} = x \circ x = \map {\circ^2} x$

Thus:

$\displaystyle \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \map {\circ^2} x$
$\map {\circ^n} {-x} = \map {\circ^{2 m} } x = \map {\circ^n} x$

$\Box$

Now suppose instead that $n$ is odd.

If $n = 1$, then:

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

Otherwise:

 $\displaystyle \map {\circ^n} {-x}$ $=$ $\displaystyle \paren {-x} \circ \paren {\map {\circ^{n - 1} } {-x} }$ $\displaystyle$ $=$ $\displaystyle \paren {-x} \circ \paren {\map {\circ^{n - 1} } x}$ $n-1$ is a strictly positive even integer, so the above case applies $\displaystyle$ $=$ $\displaystyle -\paren {x \circ \paren {\map {\circ^{n - 1} } x} }$ Product with Ring Negative $\displaystyle$ $=$ $\displaystyle -\map {\circ^n} x$ Definition of $\map {\circ^n} x$

$\blacksquare$