Parity Addition is Commutative/Proof 2

Theorem

Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.

$\forall a, b \in R: a + b = b + a$

Let $a, b \in R$.

That is, $a$ and $b$ are both either $\text{even}$ or $\text{odd}$.

By definition of odd:

$\text{odd} = 2 m + 1$

for some $m \in \Z$.

By definition of even:

$\text{even} = 2 n + 0$

for some $n \in \Z$.

Thus we can define the mapping $f: R \to \Z$ as:

$\forall x \in R: \map f x := \begin{cases}

0 & : x \text { is even} \\ 1 & : x \text { is odd} \end{cases}$

Thus an element of $R$ can be expressed as an arbitrary integer of the form:

$x = 2 k + \map f x$

where:

$k \in \Z$ is an integer

$\map f x$ is either $0$ or $1$ according to whether $x$ is even or odd.

Let $+_2$ be used to denote the operation of $+$ in $R$, that is, the addition of two parities.

Then:

$\ds a +_2 b$

$=$

$\ds 2 r + \map f a + 2 s + \map f b$

where $r, s \in \Z$

$\ds $

$=$

$\ds 2 s + \map f b + 2 r + \map f a$

$\ds $

$=$

$\ds b +_2 a$

Definition of Odd Integer and Definition of Even Integer

The result follows by the identification of $+_2$ to be used to denote the operation of $+$ in $R$:

$a + b = b + a$

$\blacksquare$