Example:Antiassociative Structure

From ProofWiki
Jump to navigation Jump to search

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.

$\blacksquare$