Integral Multiple of Ring Element/General Result

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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 Integral Multiple of Ring Element:

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


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.

$\blacksquare$

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