Product of Integral Multiples

Theorem
Let $\left({F, +, \times}\right)$ be a field.

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

Then:
 * $\left({m \cdot a}\right) \times \left({n \cdot b}\right) = \left({m n}\right) \cdot \left({a \times b}\right)$

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:
 * $\left({m \cdot a}\right) \times b = m \cdot \left({a \times b}\right)$

This will be done by induction:

For all $m \in \N$, let $P \left({n}\right)$ be the proposition:
 * $\left({m \cdot a}\right) \times b = m \cdot \left({a \times b}\right)$

First we verify $P \left({0}\right)$.

When $m = 0$, we have:

So $P \left({0}\right)$ holds.

Basis for the Induction
Now we verify $P \left({1}\right)$:

So $P \left({1}\right)$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\left({k \cdot a}\right) \times b = k \cdot \left({a \times b}\right)$

Then we need to show:


 * $\left({\left({k+1}\right) \cdot a}\right) \times b = \left({k+1}\right) \cdot \left({a \times b}\right)$

Induction Step
This is our induction step:

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

Therefore:
 * $\forall m \in \N: \left({m \cdot a}\right) \times b = m \cdot \left({a \times b}\right)$

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

Full Result
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\forall m \in \Z: \left({m \cdot a}\right) \times \left({n \cdot b}\right) = \left({m n}\right) \cdot \left({a \times b}\right)$

First we verify $P \left({0}\right)$.

When $n = 0$, we have:

So $P \left({0}\right)$ holds.

Full Result - Basis for the Induction
Next we verify $P \left({1}\right)$.

When $n = 1$, we have:

So $P \left({1}\right)$ holds. This is our basis for the induction.

Full Result - Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $\left({m \cdot a}\right) \times \left({k \cdot b}\right) = \left({m k}\right) \cdot \left({a \times b}\right)$

Then we need to show:


 * $\left({m \cdot a}\right) \times \left({\left({k+1}\right) \cdot b}\right) = \left({m \left({k+1}\right)}\right) \cdot \left({a \times b}\right)$

Full Result - Induction Step
This is our induction step:

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

Therefore:
 * $\forall m \in \Z: \forall n \in \N: \left({m \cdot a}\right) \times \left({n \cdot b}\right) = \left({m n}\right) \cdot \left({a \times b}\right)$

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