Example:Antiassociative Structure
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Theorem
Let $\left({\R_{>0}, \circ}\right)$ be an algebraic structure where $\R_{>0}$ denotes the strictly positive real numbers.
Define:
- $\forall x, y \in \R_{>0}: x \circ y = x y + y$
Then $\circ$ is antiassociative on $\R_{>0}$:
- $\forall x, y, z \in \R_{>0}: \left({x \circ y}\right) \circ z \ne x \circ \left({y \circ z}\right)$
Proof
Let $a, b, c \in \R_{>0}$:
Then:
\(\ds a \circ \left({b \circ c}\right)\) | \(=\) | \(\ds a \circ \left({b c + c}\right)\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a \left({b c + c}\right) + b c + c\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a b c + a c + b c + c\) |
and:
\(\ds \left({a \circ b}\right) \circ c\) | \(=\) | \(\ds \left({a b + b}\right)\circ c\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \left({a b + b}\right)c + c\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a b c + b c + c\) |
As $ac \ne 0$:
- $\forall x, y, z \in \R_{>0}: \left({x \circ y}\right) \circ z \ne x \circ \left({y \circ z}\right)$
Hence the result.
$\blacksquare$