General Distributivity Theorem

Theorem

Let $\left({R, \circ, *}\right)$ be a ringoid.

Then for every sequence $\left \langle {a_k} \right \rangle_{1 \le k \le n}$ of terms of $R$, and for every $b \in R$:

• $\left({a_1 \circ \cdots \circ a_n}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_n * b}\right)$
• $b * \left({a_1 \circ \cdots \circ a_n}\right) = \left({b * a_1}\right) \circ \cdots \circ \left({b * a_n}\right)$

Consequently, in the context of a ring, this can be translated into:

Let $x, y \in \left({R, +, \circ}\right)$. Then:

$\forall n \in \Z^*: \left({n \cdot x} \right) \circ y = n \cdot \left({x \circ y}\right) = x \circ \left({n \cdot y}\right)$

Proof

We will prove that:

$\forall n \in \N^*: \left({a_1 \circ \cdots \circ a_n}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_n * b}\right)$

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

$\left({a_1 \circ \cdots \circ a_n}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_n * b}\right)$

$P(1)$ is true, as this just says $a_1 * b = a_1 * b$.

Basis for the Induction

• $P(2)$ is the case:

$\left({a_1 \circ a_2}\right) * b = \left({a_1 * b}\right) \circ \left({a_2 * b}\right)$

which is true by dint of $\left({R, \circ, *}\right)$ being a ringoid.

This is our basis for the induction.

Induction Hypothesis

Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:

$\left({a_1 \circ \cdots \circ a_k}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_k * b}\right)$

Then we need to show:

$\left({a_1 \circ \cdots \circ a_k \circ a_{k+1}}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_k * b}\right) \circ \left({a_{k+1} * b}\right)$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({a_1 \circ \cdots \circ a_k \circ a_{k+1} }\right) * b\)	\(=\)	\(\displaystyle \left({\left({a_1 \circ \cdots \circ a_k}\right) \circ a_{k+1} }\right) * b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\left({a_1 \circ \cdots \circ a_k}\right) * b}\right) \circ \left({a_{k+1} } * b \right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Basis for the Induction
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({\left({a_1 * b}\right) \circ \cdots \circ \left({a_k * b}\right)}\right) \circ \left({a_{k+1} } * b \right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Induction Hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({a_1 * b}\right) \circ \cdots \circ \left({a_k * b}\right) \circ \left({a_{k+1} } * b \right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Associativity of $\circ$ in $\left({R, \circ, *}\right)$

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

Therefore:

$\forall n \in \N^*: \left({a_1 \circ \cdots \circ a_n}\right) * b = \left({a_1 * b}\right) \circ \cdots \circ \left({a_n * b}\right)$

$\blacksquare$

The result:

$b * \left({a_1 \circ \cdots \circ a_n}\right) = \left({b * a_1}\right) \circ \cdots \circ \left({b * a_n}\right)$

is proved in the same way.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965)... (previous)... (next): $\S 18$: Theorem $18.8$