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:

So $\map P 0$ holds.

Basis for the Induction
Now we verify $\map P 1$:

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:

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

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

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

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:

So $\map P 0$ holds.

Full Result - Basis for the Induction
Next we verify $\map P 1$.

When $n = 1$, we have:

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:

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

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

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