Condition for Constant Operation to be Distributive over Another Operation
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Theorem
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\sqbrk c$ denote the constant operation for some $c \in S$.
Then:
- $\sqbrk c$ is distributive over $\circ$
- $c \circ c = c$
Proof
Sufficient Condition
Let $\sqbrk c$ be distributive over $\circ$
\(\ds \forall x, y, z \in S: \, \) | \(\ds \paren {x \sqbrk c y} \circ \paren {x \sqbrk c z}\) | \(=\) | \(\ds c circ c\) | Definition of Constant Operation | ||||||||||
\(\ds \forall x, y, z \in S: \, \) | \(\ds x \sqbrk c \paren {y \circ z}\) | \(=\) | \(\ds c\) | Definition of Constant Operation |
- $c \circ c = c$
$\Box$
Necessary Condition
Let $\forall c \circ c = c$.
Then:
\(\ds \forall x, y, z \in S: \, \) | \(\ds \paren {x \sqbrk c y} \circ \paren {x \sqbrk c z}\) | \(=\) | \(\ds c \circ c\) | Definition of Constant Operation | ||||||||||
\(\ds \) | \(=\) | \(\ds c\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds x \sqbrk c \paren {y \circ z}\) | Definition of Constant Operation |
and:
\(\ds \forall x, y, z \in S: \, \) | \(\ds \paren {x \sqbrk c z} \circ \paren {y \sqbrk c z}\) | \(=\) | \(\ds c \circ c\) | Definition of Constant Operation | ||||||||||
\(\ds \) | \(=\) | \(\ds c\) | by hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ y} \sqbrk c z\) | Definition of Constant Operation |
That is, $\sqbrk c$ is distributive over $\circ$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.24$