Operation over which Every Commutative Associative Operation is Distributive is either Left or Right Operation/Lemma
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Theorem
Let $S$ be a set.
Let $A \subsetneqq S$ be a proper subset of $S$.
Let $a \in A$ and $b \in \relcomp S A$.
Let $\odot$ be the operation on $S$ defined as:
- $\forall x, y \in S: x \odot y = \begin {cases} a & : \set {x, y} \subseteq A \\ b & : \set {x, y} \not \subseteq A \end {cases}$
Then $\odot$ is commutative and associative.
Proof
Commutativity
Trivially:
\(\ds \forall x, y \in S: \, \) | \(\ds y \odot x\) | \(=\) | \(\ds \begin {cases} a & : \set {y, x} \subseteq A \\ b & : \set {y, x} \not \subseteq A \end {cases}\) | Definition of $\odot$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \begin {cases} a & : \set {x, y} \subseteq A \\ b & : \set {x, y} \not \subseteq A \end {cases}\) | Definition of Set | |||||||||||
\(\ds \) | \(=\) | \(\ds x \odot y\) | Definition of Set |
$\Box$
Associativity
There are $2$ cases to attend to:
- Case $1$
- $\set {x, y, z} \subseteq A$
\(\ds x \odot \paren {y \odot z}\) | \(=\) | \(\ds x \odot \paren {\begin {cases} a & : \set {y, z} \subseteq A \\ b & : \set {y, z} \not \subseteq A \end {cases} }\) | Definition of $\odot$ | |||||||||||
\(\ds \) | \(=\) | \(\ds x \odot a\) | as $\set {y, z} \subseteq A$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin {cases} a & : \set {x, a} \subseteq A \\ b & : \set {x, a} \not \subseteq A \end {cases}\) | Definition of Set | |||||||||||
\(\ds \) | \(=\) | \(\ds a\) | as $\set {x, a} \subseteq A$ |
\(\ds \paren {x \odot y} \odot z\) | \(=\) | \(\ds z \odot \paren {x \odot y}\) | as $\odot$ is a priori commutative | |||||||||||
\(\ds \) | \(=\) | \(\ds a\) | from above | |||||||||||
\(\ds \) | \(=\) | \(\ds x \odot \paren {y \odot z}\) | from above |
That is:
- $x \odot \paren {y \odot z} = \paren {x \odot y} \odot z = a$
- Case $2$
- $\set {x, y, z} \not \subseteq A$
Then at least one of $\set {x, y, z}$ is in $\relcomp S A$.
Without loss of generality, suppose $x \in \relcomp S A$.
Then:
- $\forall y \in S: \set {x, y} \not \subseteq A$
and so:
- $\forall y \in S: x \odot y = b$
Then we note that similarly:
- $\forall y \in S: \set {b, y} \not \subseteq A$
and so:
- $\forall y \in S: y \odot b = b$
Hence:
- $x \odot \paren {y \odot z} = \paren {x \odot y} \odot z = b$
Both cases have been examined, and we see that:
- $\forall x, y, z \in S: x \odot \paren {y \odot z} = \paren {x \odot y} \odot z$
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.26 \ \text {(b)}$