Powers of Ring Elements

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Let $\left({R, +, \circ}\right)$ be a ring whose zero is $0_R$.

Let $n \cdot x$ be an integral multiple of $x$:

$n \cdot x = \begin{cases} 0_R & : n = 0 \\ x & : n = 1 \\ \left({n - 1}\right) \cdot x + x & : n > 1 \end{cases}$

... i.e. $n \cdot x = x + x + \cdots \left({n}\right) \cdots x$.

For $n < 0$ we use $-n \cdot x = n \cdot \left({-x}\right)$.


$\forall n \in \Z: \forall x \in R: \left({n \cdot x}\right) \circ x = n \cdot \left({x \circ x}\right) = x \circ \left({n \cdot x}\right)$

General Result

$\forall m, n \in \Z: \forall x \in R: \left({m \cdot x}\right) \circ \left({n \cdot x}\right) = \left({m n}\right) \cdot \left({x \circ x}\right)$.


Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ be the proposition:

$\left({n \cdot x}\right) \circ x = n \cdot \left({x \circ x}\right) = x \circ \left({n \cdot x}\right)$

First we verify $P \left({0}\right)$.

When $n = 0$, we have:

\(\displaystyle \left({0 \cdot x}\right) \circ x\) \(=\) \(\displaystyle 0_R \circ x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0_R\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 0 \cdot \left({x \circ x}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ 0_R\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ \left({0 \cdot x}\right)\) $\quad$ $\quad$

So $P \left({0}\right)$ holds.

Basis for the Induction

Now we verify $P \left({1}\right)$:

\(\displaystyle \left({1 \cdot x}\right) \circ x\) \(=\) \(\displaystyle x \circ x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle 1 \cdot \left({x \circ x}\right)\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ \left({1 \cdot x}\right)\) $\quad$ $\quad$

So $P \left({1}\right)$ holds.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\left({k \cdot x}\right) \circ x = k \cdot \left({x \circ x}\right) = x \circ \left({k \cdot x}\right)$

Then we need to show:

$\left({\left({k+1}\right) \cdot x}\right) \circ x = \left({k+1}\right) \cdot \left({x \circ x}\right) = x \circ \left({\left({k+1}\right) \cdot x}\right)$

Induction Step

This is our induction step:

\(\displaystyle \left({\left({k+1}\right) \cdot x}\right) \circ x\) \(=\) \(\displaystyle \left({x + \left({k \cdot x}\right)}\right) \circ x\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ x + \left({k \cdot x}\right) \circ x\) $\quad$ Distributivity of $\circ$ over $+$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x \circ x + k \cdot \left({x \circ x}\right)\) $\quad$ Induction Hypothesis $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({k+1}\right) \cdot \left({x \circ x}\right)\) $\quad$ $\quad$

A similar construction shows that $\left({k+1}\right) \cdot \left({x \circ x}\right) = x \circ \left({\left({k+1}\right) \cdot x}\right)$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.


$\forall n \in \N: \left({n \cdot x}\right) \circ x = n \cdot \left({x \circ x}\right) = x \circ \left({n \cdot x}\right)$


The result for $n < 0$ follows directly from Powers of Group Elements.