Parity Addition is Commutative
Theorem
Let $R := \struct {\set {\text{even}, \text{odd} }, +, \times}$ be the parity ring.
The operation $+$ is commutative:
- $\forall a, b \in R: a + b = b + a$
Proof 1
From Isomorphism between Ring of Integers Modulo 2 and Parity Ring:
- $\struct {\set {\text{even}, \text{odd} }, +, \times}$ is isomorphic with $\struct {\Z_2, +_2, \times_2}$
the ring of integers modulo $2$.
The result follows from:
and:
$\blacksquare$
Proof 2
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:
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 $r, s \in \Z$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 2 s + \map f b + 2 r + \map f a\) | Integer Addition is Commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds b +_2 a\) | 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 $R$:
- $a + b = b + a$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.2$