Integral Multiple of Ring Element

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

Let $$n x$$ be defined as in Power of an Element:

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

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

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

Base Result
Then $$\forall n \in \mathbb{Z}: \forall x \in R: \left({n x}\right) \circ x = n \left({x \circ x}\right) = x \circ \left({n x}\right)$$.

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

Base Proof
Proof by induction:

For all $$n \in \mathbb{N}$$, let $$P \left({n}\right)$$ be the proposition $$\left({n x}\right) \circ x = n \left({x \circ x}\right) = x \circ \left({n x}\right)$$.


 * First we verify $$P(0)$$. When $$n = 0$$, we have:

So $$P(0)$$ holds.

Basis for the Induction

 * Now we verify $$P(1)$$:

So $$P(1)$$ 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:

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

Then we need to show:

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

Induction Step
This is our induction step:

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

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

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


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

Generalized Proof
Proof by induction:

For all $$n \in \mathbb{N}^*$$, let $$P \left({n}\right)$$ be the proposition $$\left({m x}\right) \circ \left({n x}\right) = \left({m n}\right) \left({x \circ x}\right)$$.

In what follows, we make extensive use of the base result (proved above):

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


 * First we verify $$P(0)$$. When $$n = 0$$, we have:

So $$P(0)$$ holds.

Basis for the Induction

 * Next we verify $$P(1)$$. When $$n = 1$$, we have:

So $$P(1)$$ 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:

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

Then we need to show:

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

Induction Step
This is our induction step:

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

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


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