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:

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

By Product of Ring Negatives:

$\paren {-x} \circ \paren {-x} = x \circ x = \map {\circ^2} x$

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 $\map {\circ^n} x$

$\blacksquare$