General Distributivity Theorem/Lemma 1
Lemma
Let $\struct {R, \circ, *}$ be a ringoid.
Then for every sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$, and for every $b \in R$:
- $\ds \paren {\sum_{j \mathop = 1}^n a_j} * b = \sum_{j \mathop = 1}^n \paren {a_j * b}$
where:
- $\ds \sum_{j \mathop = 1}^n a_j$ is the composite $a_1 \circ a_2 \circ \cdots \circ a_n$
- $n$ is a strictly positive integer: $n \in \Z_{> 0}$.
Proof
The proof proceeds by the Principle of Mathematical Induction.
Recall that as $\struct {R, \circ, *}$ is a ringoid, $*$ is distributive over $\circ$:
- $\forall a, b, c \in S: \paren {a \circ b} * c = \paren {a * c} \circ \paren {b * c}$
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
- $\ds \paren {\sum_{j \mathop = 1}^n a_j} * b = \sum_{j \mathop = 1}^n \paren {a_j * b}$
Basis for the Induction
$P(1)$ is true, as this just says:
- $a_1 * b = a_1 * b$
$P(2)$ is the case:
\(\ds \paren {\sum_{j \mathop = 1}^2 a_j} * b\) | \(=\) | \(\ds \paren {a_1 \circ a_2} * b\) | Definition of Composite (Abstract Algebra) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a_1 * b} \circ \paren {a_2 * b}\) | $*$ is distributive over $\circ$ as $\struct {R, \circ, *}$ is a ringoid | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^2 \paren {a_j * b}\) | Definition of Composite (Abstract Algebra) |
This is our basis for the induction.
Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
- $\ds \paren {\sum_{j \mathop = 1}^k a_j} * b = \sum_{j \mathop = 1}^k \paren {a_j * b}$
Then we need to show:
- $\ds \paren {\sum_{j \mathop = 1}^{k + 1} a_j} * b = \sum_{j \mathop = 1}^{k + 1} \paren {a_j * b}$
Induction Step
This is our induction step:
\(\ds \paren {\sum_{j \mathop = 1}^{k + 1} a_j} * b\) | \(=\) | \(\ds \paren {\paren {\sum_{j \mathop = 1}^k a_j} \circ a_{k + 1} } * b\) | Definition of Composite (Abstract Algebra) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {\sum_{j \mathop = 1}^k a_j} * b} \circ \paren {a_{k + 1} * b}\) | Basis for the Induction | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\sum_{j \mathop = 1}^k \paren {a_j * b} } \circ \paren {a_{k + 1} * b}\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^{k + 1} \paren {a_j * b}\) | Associativity of $\circ$ in $\struct {R, \circ, *}$ |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\ds \forall n \in \Z_{> 0}: \paren {\sum_{j \mathop = 1}^k a_j} * b = \sum_{j \mathop = 1}^k \paren {a_j * b}$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.8$