Product of Integral Multiples

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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.

$\blacksquare$


Sources