# Powers of Ring Elements/General Result

## Theorem

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

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

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

that is:

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

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

Then:

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

## Proof

Proof by induction:

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

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

In what follows, we make extensive use of Powers of Ring Elements:

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

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

When $n = 0$, we have:

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

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

### Basis for the Induction

Next we verify $P \left({1}\right)$.

When $n = 1$, we have:

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

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:

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

Then we need to show:

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

### Induction Step

This is our induction step:

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

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

Therefore:

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

$\Box$

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

$\blacksquare$