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:

 $\displaystyle \left({0 \cdot a}\right) \times b$ $=$ $\displaystyle 0_F \times b$ by definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ $\displaystyle$ $=$ $\displaystyle 0_F$ by definition of zero $\displaystyle$ $=$ $\displaystyle 0 \cdot \left({a \times b}\right)$ by definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$

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

Basis for the Induction

Now we verify $P \left({1}\right)$:

 $\displaystyle \left({1 \cdot a}\right) \times b$ $=$ $\displaystyle a \times b$ by definition of Integral Multiple: $\forall x \in F: 1 \cdot x = x$ $\displaystyle$ $=$ $\displaystyle 1 \cdot \left({a \times b}\right)$ by definition of Integral Multiple: $\forall x \in F: 1 \cdot x = x$

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:

 $\displaystyle \left({\left({k+1}\right) \cdot a}\right) \times b$ $=$ $\displaystyle \left({a + \left({k \cdot a}\right)}\right) \times b$ by definition of Integral Multiple $\displaystyle$ $=$ $\displaystyle a \times b + \left({k \cdot a}\right) \times b$ Distributivity of $\times$ over $+$ $\displaystyle$ $=$ $\displaystyle a \times b + k \cdot \left({a \times b}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({k+1}\right) \cdot \left({a \times b}\right)$ by definition of Integral Multiple

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

Therefore:

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

$\Box$

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

$\blacksquare$

Full Result

Proof by induction:

For all $n \in \N$, let $P \left({n}\right)$ 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:

 $\displaystyle \left({m \cdot a}\right) \times \left({0 \cdot b}\right)$ $=$ $\displaystyle \left({m \cdot a}\right) \times 0_F$ by definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ $\displaystyle$ $=$ $\displaystyle 0_F$ by definition of zero $\displaystyle$ $=$ $\displaystyle 0 \cdot \left({a \times b}\right)$ by definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ $\displaystyle$ $=$ $\displaystyle \left({m 0}\right) \cdot \left({a \times b}\right)$

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

Full Result - Basis for the Induction

Next we verify $P \left({1}\right)$.

When $n = 1$, we have:

 $\displaystyle \left({m \cdot a}\right) \times \left({1 \cdot b}\right)$ $=$ $\displaystyle \left({m \cdot a}\right) \times b$ by definition of Integral Multiple: $\forall x \in F: 1 \cdot x = x$ $\displaystyle$ $=$ $\displaystyle m \cdot \left({a \times b}\right)$ from the base result $\displaystyle$ $=$ $\displaystyle \left({m 1}\right) \cdot \left({a \times b}\right)$

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:

 $\displaystyle \left({m \cdot a}\right) \times \left({\left({k+1}\right) \cdot b}\right)$ $=$ $\displaystyle \left({m \cdot a}\right) \times \left({k \cdot b + b}\right)$ by definition of Integral Multiple $\displaystyle$ $=$ $\displaystyle \left({m \cdot a}\right) \times \left({k \cdot b}\right) + \left({m \cdot a}\right) \times b$ Distributivity of $\times$ over $+$ $\displaystyle$ $=$ $\displaystyle \left({m k}\right) \cdot \left({a \times b}\right) + m \cdot \left({a \times b}\right)$ Induction Hypothesis $\displaystyle$ $=$ $\displaystyle \left({m k + k}\right) \cdot \left({a \times b}\right)$ Integral Multiple Distributes over Ring Addition $\displaystyle$ $=$ $\displaystyle \left({m \left({k+1}\right)}\right) \cdot \left({a \times b}\right)$

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

Therefore:

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

$\Box$

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

$\blacksquare$