There was a question whether associative, but non-commutative binary operation over the real numbers exist. A trivial answer is the binary operation $x\circ y = x$ or $x\circ y = y$.

As a followup, there was another question, which specifically excluded trivial binary operations of the type $x\circ y = f(x)$ and $x\circ y = g(y)$ and also requiring it to be continuous (almost) everywhere. One answer was $x\circ y = \vert x \vert y$. None of the answers (so far) provided an analytic function.

So, the question is whether there is a binary operation $\circ:S \to \mathbb R$ (where $S \subseteq \mathbb R\times \mathbb R$) which is:

  1. associative: $(x \circ y) \circ z = x \circ (y \circ z)$
  2. non-commutative: $\exists (x,y)$ such that $x \circ y \ne y \circ x$
  3. non-trivial: $\nexists f(x)$ such that $x\circ y = f(x)$ or $x\circ y= f(y)$
  4. analytic: $$x \circ y = \sum_{m,n=0}^\infty a_{mn}(x-x_0)^m (y-y_0)^n$$
  • $\begingroup$ Restricting $S$ to $\{ -1, 1\}^2$ suffices, but isn't satisfying. This way $x^2 = |x|,$ and so the analytic function $x \circ y = x^2 y$ works. I presume you want $S$ an open set or something. $\endgroup$ Sep 6 '18 at 6:18
  • $\begingroup$ @ArtimisFowl $x \circ y = x^2 y$ is not associative: $(x\circ y)\circ z = x^4 y^2 z$, while $x\circ(y\circ z)=x^2 y^2 z$. $\endgroup$
    – Danijel
    Sep 6 '18 at 11:36
  • $\begingroup$ it is on the given domain. $x^4 = x^2$ when $x$ is $\pm 1.$ there are only 4 inputs, try them. $\endgroup$ Sep 6 '18 at 13:39
  • $\begingroup$ @ArtimisFowl You are right, I thought you wrote $[-1,1]^2$, my mistake. However, the output of this binary operation depends only on one argument because $x^2 = |x| = 1$ when restricted to $\{-1, 1\}$ and therefore $x\circ y = y$. $\endgroup$
    – Danijel
    Sep 6 '18 at 13:50
  • $\begingroup$ You are right, so pick $S ={-1,0,1}.$ now if $x$ or $y$ is 0 matters, so it is not trivial. It's not commutative, analytic, and associative by previous work. But still unsatisfying I think? $\endgroup$ Sep 6 '18 at 13:54

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