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:
- $\ds \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \paren {-x} \circ \paren {-x}$
- $\paren {-x} \circ \paren {-x} = x \circ x = \map {\circ^2} x$
Thus:
- $\ds \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \map {\circ^2} x$
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:
\(\ds \map {\circ^n} {-x}\) | \(=\) | \(\ds \paren {-x} \circ \paren {\map {\circ^{n - 1} } {-x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {-x} \circ \paren {\map {\circ^{n - 1} } x}\) | $n-1$ is a strictly positive even integer, so the above case applies | |||||||||||
\(\ds \) | \(=\) | \(\ds -\paren {x \circ \paren {\map {\circ^{n - 1} } x} }\) | Product with Ring Negative | |||||||||||
\(\ds \) | \(=\) | \(\ds -\map {\circ^n} x\) | Definition of $\map {\circ^n} x$ |
$\blacksquare$