Powers of Ring Elements

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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:

$n \cdot x = \underbrace {x + x + \cdots + x}_{\text {$n$ times} }$

For $n < 0$ we use:

$-n \cdot x = n \cdot \paren {-x}$


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

General Result

$\forall m, n \in \Z: \forall x \in R: \paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$.


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:

\(\ds \paren {0 \cdot x} \circ x\) \(=\) \(\ds 0_R \circ x\)
\(\ds \) \(=\) \(\ds 0_R\)
\(\ds \) \(=\) \(\ds 0 \cdot \paren {x \circ x}\)
\(\ds \) \(=\) \(\ds x \circ 0_R\)
\(\ds \) \(=\) \(\ds x \circ \paren {0 \cdot x}\)

So $\map P 0$ holds.

Basis for the Induction

Now we verify $\map P 1$:

\(\ds \paren {1 \cdot x} \circ x\) \(=\) \(\ds x \circ x\)
\(\ds \) \(=\) \(\ds 1 \cdot \paren {x \circ x}\)
\(\ds \) \(=\) \(\ds x \circ \paren {1 \cdot x}\)

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:

\(\ds \paren {\paren {k + 1} \cdot x} \circ x\) \(=\) \(\ds \paren {x + \paren {k \cdot x} } \circ x\)
\(\ds \) \(=\) \(\ds x \circ x + \paren {k \cdot x} \circ x\) Ring Axiom $\text D$: Distributivity of Product over Addition
\(\ds \) \(=\) \(\ds x \circ x + k \cdot \paren {x \circ x}\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \paren {k + 1} \cdot \paren {x \circ x}\)

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.


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


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