Integral Multiple of Ring Element

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}$

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

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

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

Proof
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:

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

Basis for the Induction
Now we verify $P \left({1}\right)$:

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:

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.

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

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