Binomial Theorem/Ring Theory

Theorem
Let $\struct {R, +, \odot}$ be a ringoid such that $\struct {R, \odot}$ is a commutative semigroup.

Let $n \in \Z: n \ge 2$.

Then:


 * $\ds \forall x, y \in R: \odot^n \paren {x + y} = \odot^n x + \sum_{k \mathop = 1}^{n - 1} \binom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y} + \odot^n y$

where $\dbinom n k = \dfrac {n!} {k! \ \paren {n - k}!}$ (see Binomial Coefficient).

If $\struct {R, \odot}$ has an identity element $e$, then:


 * $\ds \forall x, y \in R: \odot^n \paren {x + y} = \sum_{k \mathop = 0}^n \binom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y}$

Proof
First we establish the result for when $\struct {R, \odot}$ has an identity element $e$.

For $n = 0$ we have:


 * $\ds \odot^0 \paren {x + y} = e = {0 \choose 0} \paren {\odot^{0 - 0} x} \odot \paren {\odot^0 y} = \sum_{k \mathop = 0}^0 {0 \choose k} x^{0 - k} \odot y^k$

For $n = 1$ we have:


 * $\ds \odot^1 \paren {x + y} = \paren {x + y} = {0 \choose 1} \paren {\odot^{1 - 0} x} \odot \paren {\odot^0 y} + {1 \choose 1} \paren {\odot^{1 - 1} x} \odot \paren {\odot^1 y} = \sum_{k \mathop = 0}^1 {1 \choose k} x^{1 - k} \odot y^k$

Basis for the Induction
For $n = 2$ we have:

This is the basis for the induction.

Induction Hypothesis
This is our inductive hypothesis:


 * $\ds \forall n \ge 2: \odot^n \paren {x + y} = \odot^n x + \sum_{k \mathop = 1}^{n - 1} \dbinom n k \paren {\odot^{n - k} x} \odot \paren {\odot^k y} + \odot^n y$

Induction Step
This is the induction step:

The result follows by the Principle of Mathematical Induction.