Integral Multiple of Ring Element

Theorem
Let $\struct {R, +, \circ}$ 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 \\ \paren {n - 1} \cdot x + x & : n > 1 \end {cases}$

that is $n \cdot x = x + x + \cdots \paren n \cdots x$.

For $n < 0$ we use:
 * $-n \cdot x = n \cdot \paren {-x}$

Then:
 * $\forall n \in \Z: \forall x \in R: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$

Proof
Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition:
 * $\paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$

First we verify $\map P 0$.

When $n = 0$, we have:

So $\map P 0$ holds.

Basis for the Induction
Now we verify $\map P 1$:

So $\map P 1$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:


 * $\paren {k \cdot x} \circ x = k \cdot \paren {x \circ x} = x \circ \paren {k \cdot x}$

Then we need to show:


 * $\paren {\paren {k + 1} \cdot x} \circ x = \paren {k + 1} \cdot \paren {x \circ x} = x \circ \paren {\paren {k + 1} \cdot x}$

Induction Step
This is our induction step:

A similar construction shows that $\paren {k + 1} \cdot \paren {x \circ x} = x \circ \paren {\paren {k + 1} \cdot x}$.

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

Therefore:
 * $\forall n \in \N: \paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$

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