General Distributivity Theorem/Lemma 2
== 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 $b \in R$: :$\displaystyle b * \paren {\sum_{j \mathop = 1}^n a_j} = \sum_{j \mathop = 1}^n \paren {b * a_j}$  where: :$\displaystyle \sum_{j \mathop = 1}^n a_j$ 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: :$\displaystyle b * \paren {\sum_{j \mathop = 1}^n a_j} = \sum_{j \mathop = 1}^n \paren {b * a_j}$ 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: {{begin-eqn}} {{eqn | l = b * \paren {\sum_{j \mathop = 1}^2 a_j}       | r = b * \paren {a_1 \circ a_2}       | c = {{Defof|Composite (Abstract Algebra)|Composite}} }} {{eqn | r = \paren {b * a_1} \circ \paren {b * a_2}       | c = $*$ is distributive over $\circ$ as $\paren {R, \circ, *}$ is a ringoid }} {{eqn | r = \sum_{j \mathop = 1}^2 \paren {b * a_j}       | c = {{Defof|Composite (Abstract Algebra)|Composite}} }} {{end-eqn}} 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: :$\displaystyle b * \paren {\sum_{j \mathop = 1}^k a_j} = \sum_{j \mathop = 1}^k \paren {b * a_j}$ Then we need to show: :$\displaystyle b * \paren {\sum_{j \mathop = 1}^{k + 1} a_j} = \sum_{j \mathop = 1}^{k + 1} \paren {b * a_j}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = b * \paren {\sum_{j \mathop = 1}^{k + 1} a_j}       | r = b * \paren {\paren {\sum_{j \mathop = 1}^k a_j} \circ a_{k + 1} }       | c =  }} {{eqn | r = \paren {b * \paren {\sum_{j \mathop = 1}^k a_j} } \circ \paren {b * a_{k + 1} }       | c = Basis for the Induction }} {{eqn | r = \paren {\sum_{j \mathop = 1}^k \paren {a_j * b} } \circ \paren {b * a_{k + 1} }       | c = Induction Hypothesis }} {{eqn | r = \sum_{j \mathop = 1}^{k + 1} \paren {b * a_j}       | c = Associativity of $\circ$ in $\paren {R, \circ, *}$ }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{> 0}: b * \paren {\sum_{j \mathop = 1}^k a_j} = \sum_{j \mathop = 1}^k \paren {b * a_j}$  == Sources == * {{BookReference|Modern Algebra|1965|Seth Warner|prev = General Distributivity Theorem/Lemma 1|next = Definition:Indexing Set}}: $\S 18$: Theorem $18.8$ * {{BookReference|Algebra Volume 1|1982|P.M. Cohn|edpage = Second Edition|ed = 2nd|prev = Principle of Mathematical Induction/Zero-Based|next = General Distributivity Theorem}}: Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $2$