# 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

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:

 $\displaystyle \paren {0 \cdot a} \times b$ $=$ $\displaystyle 0_F \times b$ Definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ $\displaystyle$ $=$ $\displaystyle 0_F$ Definition of Field Zero $\displaystyle$ $=$ $\displaystyle 0 \cdot \paren {a \times b}$ 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 $\map P 1$:

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

 $\displaystyle \paren {\paren {k + 1} \cdot a} \times b$ $=$ $\displaystyle \paren {a + \paren {k \cdot a} } \times b$ Definition of Integral Multiple $\displaystyle$ $=$ $\displaystyle a \times b + \paren {k \cdot a} \times b$ Distributivity of $\times$ over $+$ $\displaystyle$ $=$ $\displaystyle a \times b + k \cdot \paren {a \times b}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \paren {k + 1} \cdot \paren {a \times b}$ 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:

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

So $\map P 0$ holds.

#### Full Result - Basis for the Induction

Next we verify $\map P 1$.

When $n = 1$, we have:

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

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:

 $\displaystyle \paren {m \cdot a} \times \paren {\paren {k + 1} \cdot b}$ $=$ $\displaystyle \paren {m \cdot a} \times \paren {k \cdot b + b}$ Definition of Integral Multiple $\displaystyle$ $=$ $\displaystyle \paren {m \cdot a} \times \paren {k \cdot b} + \paren {m \cdot a} \times b$ Distributivity of $\times$ over $+$ $\displaystyle$ $=$ $\displaystyle \paren {m k} \cdot \paren {a \times b} + m \cdot \paren {a \times b}$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \paren {m k + k} \cdot \paren {a \times b}$ Integral Multiple Distributes over Ring Addition $\displaystyle$ $=$ $\displaystyle \paren {m \paren {k + 1} } \cdot \paren {a \times b}$

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.

$\blacksquare$