Product of Integral Multiples/Proof 2

Theorem

Let $\struct {F, +, \times}$ be a field.

Let $a, b \in F$ and $m, n \in \Z$.

Then:

$\paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$

where $m \cdot a$ is as defined in Integral Multiple.

Proof

First let $m = 0$ or $n = 0$.

Without loss of generality, let $m = 0$.

The case where $n = 0$ follows the same lines.

We have:

\(\ds \forall a, b \in R: \forall n \in \Z_{>0}: \, \)

\(\ds \paren {m \cdot a} \times \paren {n \cdot b}\)

\(=\)

\(\ds \paren {0 \cdot a} \times \paren {n \cdot b}\)

Definition of $m$

\(\ds \)

\(=\)

\(\ds 0_F \times \paren {n \cdot b}\)

Definition of Integral Multiple

\(\ds \)

\(=\)

\(\ds 0_F\)

Definition of Field Zero

\(\ds \)

\(=\)

\(\ds 0_F \times \paren {a \times b}\)

Definition of Field Zero

\(\ds \)

\(=\)

\(\ds 0 \cdot \paren {a \times b}\)

Definition of Integral Multiple

\(\ds \)

\(=\)

\(\ds \paren {0 \times n} \cdot \paren {a \times b}\)

$0$ is zero of $\Z$

\(\ds \)

\(=\)

\(\ds \paren {m \times n} \cdot \paren {a \times b}\)

Definition of Integral Multiple

$\Box$

Next let $m > 0$ and $n > 0$.

\(\ds \forall a, b \in R: \forall m, n \in \Z_{>0}: \, \)

\(\ds \paren {m \cdot a} \times \paren {n \cdot b}\)

\(=\)

\(\ds \paren {\sum_{i \mathop = 1}^m a} \times \paren {\sum_{i \mathop = 1}^n b}\)

Definition of Integral Multiple

\(\ds \)

\(=\)

\(\ds \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a \times b}\)

General Distributivity Theorem

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 1}^m \paren {\sum_{j \mathop = 1}^n \paren {a \times b} }\)

Definition of Summation

\(\ds \)

\(=\)

\(\ds m \cdot \paren {n \cdot \paren {a \times b} }\)

Definition of Integral Multiple

\(\ds \)

\(=\)

\(\ds \paren {m n} \cdot \paren {a \times b}\)

Powers of Group Elements: Additive Notation

$\Box$

The results for $m < 0$ and $n < 0$ follow directly from Powers of Group Elements.

$\blacksquare$