Parity Multiplication is Commutative/Proof 2

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Theorem

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


The operation $\times$ is commutative:

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


Proof

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.


Then:

\(\ds a \times b\) \(=\) \(\ds \paren {2 r + \map f a} \paren {2 s + \map f b}\) where $r, s \in \Z$
\(\ds \) \(=\) \(\ds \paren {2 s + \map f b} \paren {2 r + \map f a}\) Integer Multiplication is Commutative
\(\ds \) \(=\) \(\ds b \times a\) Definition of Odd Integer and Definition of Even Integer

$\blacksquare$