# Product of Integral Multiples

## 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 1

Let the zero of $F$ be $0_F$.

### Base Result

First we need to show that:

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

This will be done by induction:

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

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

First we verify $\map P 0$.

When $m = 0$, we have:

 $\ds \paren {0 \cdot a} \times b$ $=$ $\ds 0_F \times b$ Definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ $\ds$ $=$ $\ds 0_F$ Definition of Field Zero $\ds$ $=$ $\ds 0 \cdot \paren {a \times b}$ Definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$

So $\map P 0$ holds.

#### Basis for the Induction

Now we verify $\map P 1$:

 $\ds \paren {1 \cdot a} \times b$ $=$ $\ds a \times b$ Definition of Integral Multiple: $\forall x \in F: 1 \cdot x = x$ $\ds$ $=$ $\ds 1 \cdot \paren {a \times b}$ Definition of Integral Multiple: $\forall x \in F: 1 \cdot x = x$

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:

$\paren {k \cdot a} \times b = k \cdot \paren {a \times b}$

Then we need to show:

$\paren {\paren {k + 1} \cdot a} \times b = \paren {k + 1} \cdot \paren {a \times b}$

#### Induction Step

This is our induction step:

 $\ds \paren {\paren {k + 1} \cdot a} \times b$ $=$ $\ds \paren {a + \paren {k \cdot a} } \times b$ Definition of Integral Multiple $\ds$ $=$ $\ds a \times b + \paren {k \cdot a} \times b$ Field Axiom $\text D$: Distributivity of Product over Addition $\ds$ $=$ $\ds a \times b + k \cdot \paren {a \times b}$ Induction Hypothesis $\ds$ $=$ $\ds \paren {k + 1} \cdot \paren {a \times b}$ Definition of Integral Multiple

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

Therefore:

$\forall m \in \N: \paren {m \cdot a} \times b = m \cdot \paren {a \times b}$

$\Box$

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

$\blacksquare$

### Full Result

Proof by induction:

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

$\forall m \in \Z: \paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$

First we verify $\map P 0$.

When $n = 0$, we have:

 $\ds \paren {m \cdot a} \times \paren {0 \cdot b}$ $=$ $\ds \paren {m \cdot a} \times 0_F$ Definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ $\ds$ $=$ $\ds 0_F$ Definition of Field Zero $\ds$ $=$ $\ds 0 \cdot \paren {a \times b}$ Definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ $\ds$ $=$ $\ds \paren {m 0} \cdot \paren {a \times b}$

So $\map P 0$ holds.

#### Full Result - Basis for the Induction

Next we verify $\map P 1$.

When $n = 1$, we have:

 $\ds \paren {m \cdot a} \times \paren {1 \cdot b}$ $=$ $\ds \paren {m \cdot a} \times b$ Definition of Integral Multiple: $\forall x \in F: 1 \cdot x = x$ $\ds$ $=$ $\ds m \cdot \paren {a \times b}$ Base Result $\ds$ $=$ $\ds \paren {m 1} \cdot \paren {a \times b}$

So $\map P 1$ holds.

This is our basis for the induction.

#### Full Result - 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:

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

Then we need to show:

$\paren {m \cdot a} \times \paren {\paren {k + 1} \cdot b} = \paren {m \paren {k + 1} } \cdot \paren {a \times b}$

#### Full Result - Induction Step

This is our induction step:

 $\ds \paren {m \cdot a} \times \paren {\paren {k + 1} \cdot b}$ $=$ $\ds \paren {m \cdot a} \times \paren {k \cdot b + b}$ Definition of Integral Multiple $\ds$ $=$ $\ds \paren {m \cdot a} \times \paren {k \cdot b} + \paren {m \cdot a} \times b$ Field Axiom $\text D$: Distributivity of Product over Addition $\ds$ $=$ $\ds \paren {m k} \cdot \paren {a \times b} + m \cdot \paren {a \times b}$ Induction Hypothesis $\ds$ $=$ $\ds \paren {m k + k} \cdot \paren {a \times b}$ Integral Multiple Distributes over Ring Addition $\ds$ $=$ $\ds \paren {m \paren {k + 1} } \cdot \paren {a \times b}$

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

Therefore:

$\forall m \in \Z: \forall n \in \N: \paren {m \cdot a} \times \paren {n \cdot b} = \paren {m n} \cdot \paren {a \times b}$

$\Box$

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

$\blacksquare$

## Proof 2

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.

$\blacksquare$