Self-Distributive Structure/Examples/Arithmetic Mean

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Example of Self-Distributive Structure

Let $\Q$ denote the set of rational numbers.

Let $\circ$ be the operation defined on $\Q$ as:

$\forall x, y \in \Q: x \circ y := \dfrac {x + y} 2$

That is, $x \circ y$ is the arithmetic mean of $x$ and $y$ in $\Q$.

Then the algebraic structure $\struct {\Q, \circ}$ so formed is a self-distributive quasigroup.


Proof

\(\ds \forall a, b, c \in \Q: \, \) \(\ds a \circ \paren {b \circ c}\) \(=\) \(\ds \dfrac {a + \frac {b + c} 2} 2\) Definition of $\circ$
\(\ds \) \(=\) \(\ds \dfrac a 2 + \dfrac b 4 + \dfrac c 4\) simplification
\(\ds \) \(=\) \(\ds \paren {\dfrac a 4 + \dfrac b 4} + \paren {\dfrac a 4 + \dfrac c 4}\) rearrangement
\(\ds \) \(=\) \(\ds \dfrac {\paren {a \circ b} + \paren {a \circ c} } 2\) Definition of $\circ$
\(\ds \) \(=\) \(\ds \paren {a \circ b} \circ \paren {a \circ c}\) Definition of $\circ$


As Rational Addition is Commutative, it follows immediately from Left Distributive and Commutative implies Distributive that:

$\forall a, b, c \in \Q: \paren {a \circ b} \circ c = \paren {a \circ c} \circ \paren {b \circ c}$


To demonstrate that $\struct {\Q, \circ}$ is a quasigroup, it remains to be shown that:

$\forall a, b \in \Q: \exists ! x \in \Q: x \circ a = b$
$\forall a, b \in \Q: \exists ! y \in \Q: a \circ y = b$


We have:

\(\ds x \circ a\) \(=\) \(\ds b\)
\(\ds \leadstoandfrom \ \ \) \(\ds \dfrac {x + a} 2\) \(=\) \(\ds b\)
\(\ds \leadstoandfrom \ \ \) \(\ds x\) \(=\) \(\ds 2 b - a\)

and similarly:

\(\ds a \circ y\) \(=\) \(\ds b\)
\(\ds \leadstoandfrom \ \ \) \(\ds \dfrac {a + y} 2\) \(=\) \(\ds b\)
\(\ds \leadstoandfrom \ \ \) \(\ds y\) \(=\) \(\ds 2 b - a\)

Hence both $x$ and $y$ are determined uniquely by $a$ and $b$.

Hence by definition $\struct {\Q, \circ}$ is a quasigroup.

$\blacksquare$


Sources