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 1
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:
\(\ds \paren {0 \cdot a} \times b\) | \(=\) | \(\ds 0_F \times b\) | Definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F\) | Definition of Field Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \cdot \paren {a \times b}\) | 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 $\map P 1$:
\(\ds \paren {1 \cdot a} \times b\) | \(=\) | \(\ds a \times b\) | Definition of Integral Multiple: $\forall x \in F: 1 \cdot x = x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 \cdot \paren {a \times b}\) | 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:
\(\ds \paren {\paren {k + 1} \cdot a} \times b\) | \(=\) | \(\ds \paren {a + \paren {k \cdot a} } \times b\) | Definition of Integral Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds a \times b + \paren {k \cdot a} \times b\) | Field Axiom $\text D$: Distributivity of Product over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds a \times b + k \cdot \paren {a \times b}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1} \cdot \paren {a \times b}\) | 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:
\(\ds \paren {m \cdot a} \times \paren {0 \cdot b}\) | \(=\) | \(\ds \paren {m \cdot a} \times 0_F\) | Definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F\) | Definition of Field Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \cdot \paren {a \times b}\) | Definition of Integral Multiple: $\forall x \in F: 0 \cdot x = 0_F$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m 0} \cdot \paren {a \times b}\) |
So $\map P 0$ holds.
Full Result - Basis for the Induction
Next we verify $\map P 1$.
When $n = 1$, we have:
\(\ds \paren {m \cdot a} \times \paren {1 \cdot b}\) | \(=\) | \(\ds \paren {m \cdot a} \times b\) | Definition of Integral Multiple: $\forall x \in F: 1 \cdot x = x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds m \cdot \paren {a \times b}\) | Base Result | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m 1} \cdot \paren {a \times b}\) |
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:
\(\ds \paren {m \cdot a} \times \paren {\paren {k + 1} \cdot b}\) | \(=\) | \(\ds \paren {m \cdot a} \times \paren {k \cdot b + b}\) | Definition of Integral Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m \cdot a} \times \paren {k \cdot b} + \paren {m \cdot a} \times b\) | Field Axiom $\text D$: Distributivity of Product over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m k} \cdot \paren {a \times b} + m \cdot \paren {a \times b}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m k + k} \cdot \paren {a \times b}\) | Integral Multiple Distributes over Ring Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m \paren {k + 1} } \cdot \paren {a \times b}\) |
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.
$\blacksquare$
Proof 2
First let $m = 0$ or $n = 0$.
Without loss of generality, let $m = 0$.
The case where $n = 0$ follows the same lines.
We have:
\(\ds \forall a, b \in R: \forall n \in \Z_{>0}: \, \) | \(\ds \paren {m \cdot a} \times \paren {n \cdot b}\) | \(=\) | \(\ds \paren {0 \cdot a} \times \paren {n \cdot b}\) | Definition of $m$ | ||||||||||
\(\ds \) | \(=\) | \(\ds 0_F \times \paren {n \cdot b}\) | Definition of Integral Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F\) | Definition of Field Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0_F \times \paren {a \times b}\) | Definition of Field Zero | |||||||||||
\(\ds \) | \(=\) | \(\ds 0 \cdot \paren {a \times b}\) | Definition of Integral Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {0 \times n} \cdot \paren {a \times b}\) | $0$ is zero of $\Z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m \times n} \cdot \paren {a \times b}\) | Definition of Integral Multiple |
$\Box$
Next let $m > 0$ and $n > 0$.
\(\ds \forall a, b \in R: \forall m, n \in \Z_{>0}: \, \) | \(\ds \paren {m \cdot a} \times \paren {n \cdot b}\) | \(=\) | \(\ds \paren {\sum_{i \mathop = 1}^m a} \times \paren {\sum_{i \mathop = 1}^n b}\) | Definition of Integral Multiple | ||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{\substack {1 \mathop \le i \mathop \le m \\ 1 \mathop \le j \mathop \le n} } \paren {a \times b}\) | General Distributivity Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^m \paren {\sum_{j \mathop = 1}^n \paren {a \times b} }\) | Definition of Summation | |||||||||||
\(\ds \) | \(=\) | \(\ds m \cdot \paren {n \cdot \paren {a \times b} }\) | Definition of Integral Multiple | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {m n} \cdot \paren {a \times b}\) | Powers of Group Elements: Additive Notation |
$\Box$
The results for $m < 0$ and $n < 0$ follow directly from Powers of Group Elements.
$\blacksquare$
Sources
- 1964: Iain T. Adamson: Introduction to Field Theory ... (previous) ... (next): Chapter $\text {I}$: Elementary Definitions: $\S 2$. Elementary Properties