# Powers of Ring Elements/General Result

## 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:

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

## Proof

Proof by induction:

For all $n \in \N$, let $\map P n$ be the proposition:

$\paren {m \cdot x} \circ \paren {n \cdot x} = \paren {m n} \cdot \paren {x \circ x}$

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

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

First we verify $\map P 0$.

When $n = 0$, we have:

 $\ds \paren {m \cdot x} \circ \paren {0 \cdot x}$ $=$ $\ds \paren {m \cdot x} \circ 0_R$ $\ds$ $=$ $\ds 0_R$ $\ds$ $=$ $\ds 0 \cdot \paren {x \circ x}$ $\ds$ $=$ $\ds \paren {m 0} \cdot \paren {x \circ x}$

So $\map P 0$ holds.

### Basis for the Induction

Next we verify $\map P 1$.

When $n = 1$, we have:

 $\ds \paren {m \cdot x} \circ \paren {1 \cdot x}$ $=$ $\ds \paren {m \cdot x} \circ x$ $\ds$ $=$ $\ds m \cdot \paren {x \circ x}$ $\ds$ $=$ $\ds \paren {m 1} \cdot \paren {x \circ x}$

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:

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

Then we need to show:

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

### Induction Step

This is our induction step:

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

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

Therefore:

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

$\Box$

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

$\blacksquare$