Product of Integral Multiples

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$


Sources