Parity Multiplication is Commutative
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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 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:
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 |
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.2$