# Power of Ring Negative

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## 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:

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

By associativity:

- $\map {\circ^n} {-x} = \map {\circ^{2 m} } x = \map {\circ^n} x$

$\Box$

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 $\map {\circ^n} x$ |

$\blacksquare$