General Distributivity Theorem/Lemma 2

From ProofWiki
Jump to navigation Jump to search

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 $b \in R$:

$\ds b * \paren {\sum_{j \mathop = 1}^n a_j} = \sum_{j \mathop = 1}^n \paren {b * a_j}$

where:

$\ds \sum_{j \mathop = 1}^n a_j$ is the summation $a_1 \circ a_2 \circ \cdots \circ a_n$
$n$ 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 $\circ$:

$\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$


For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:

$\ds b * \paren {\sum_{j \mathop = 1}^n a_j} = \sum_{j \mathop = 1}^n \paren {b * a_j}$


We have that $\struct {R, \circ, *}$ is a ringoid, and so:

$\forall a, b, c \in R: a * \paren {b \circ c} = \paren {a * b} \circ \paren {a * c}$


Basis for the Induction

$\map P 1$ is true, as this just says:

$b * a_1 = b * a_1$


$\map P 2$ is the case:

\(\ds b * \paren {\sum_{j \mathop = 1}^2 a_j}\) \(=\) \(\ds b * \paren {a_1 \circ a_2}\) Definition of Composite
\(\ds \) \(=\) \(\ds \paren {b * a_1} \circ \paren {b * a_2}\) $*$ is distributive over $\circ$ as $\paren {R, \circ, *}$ is a ringoid
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^2 \paren {b * a_j}\) Definition of Composite


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:

$\ds b * \paren {\sum_{j \mathop = 1}^k a_j} = \sum_{j \mathop = 1}^k \paren {b * a_j}$


Then we need to show:

$\ds b * \paren {\sum_{j \mathop = 1}^{k + 1} a_j} = \sum_{j \mathop = 1}^{k + 1} \paren {b * a_j}$


Induction Step

This is our induction step:

\(\ds b * \paren {\sum_{j \mathop = 1}^{k + 1} a_j}\) \(=\) \(\ds b * \paren {\paren {\sum_{j \mathop = 1}^k a_j} \circ a_{k + 1} }\)
\(\ds \) \(=\) \(\ds \paren {b * \paren {\sum_{j \mathop = 1}^k a_j} } \circ \paren {b * a_{k + 1} }\) Basis for the Induction
\(\ds \) \(=\) \(\ds \paren {\sum_{j \mathop = 1}^k \paren {a_j * b} } \circ \paren {b * a_{k + 1} }\) Induction Hypothesis
\(\ds \) \(=\) \(\ds \sum_{j \mathop = 1}^{k + 1} \paren {b * a_j}\) Associativity of $\circ$ in $\paren {R, \circ, *}$


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


Therefore:

$\ds \forall n \in \Z_{> 0}: b * \paren {\sum_{j \mathop = 1}^k a_j} = \sum_{j \mathop = 1}^k \paren {b * a_j}$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=General_Distributivity_Theorem/Lemma_2&oldid=688561"