Associative, non-commutative, nontrivial operation on the real numbers

associativitybinary operationsexamples-counterexamples

This MSE question asks about binary operations on the real numbers which are associative, but not commutative. Two answers are given:

  1. The operation $\circ$ defined by $x \circ y=x$.
  2. Letting $f:\mathbb R\to M_n(\mathbb R)$ be a bijection, then $x*y=f^{-1}(f(x)\cdot f(y))$, where $\cdot$ is matrix multiplication.

Operation 1 is good, but is what I would call a trivial binary operation as it only depends on one of its inputs. Operation 2 is far from satisfying since it does not respect the structure of the reals at all. So my question is,

Does there exist a binary operation $\star$ on the real numbers which is

  1. associative,
  2. non-commutative,
  3. nontrivial (operators of the form $x\circ y=f(x)$ or $x\circ y=g(y)$ are trivial), and
  4. continuous (with respect to the usual topologies on $\mathbb R^2$ and $\mathbb R$)?

I would also be satisfied with an operator where condition 4 was relaxed to

 4'. continuous almost everywhere?

Best Answer

Ok, Batominovski's got an answer in the comments. I will type up the checking:

Our candidate is $x\circ y=|x|y$. Then:

  1. Associative? We have $(x\circ y)\circ z=(|x|y)\circ z=||x|y|z=|xy|z.$ On the other hand, $x\circ(y\circ z)=x\circ(|y|z)=|x||y|z=|xy|z.$
  2. Non-commutative? $x\circ y=|x|y\not=|y|x=y\circ x$.
  3. Nontrivial? Well, it is not a function of $x$ or $y$ only.
  4. Continuous or continuous almost everywhere? $f(x)=x$ and $g(x)=|x|$ are both continuous everywhere, hence their product is.

So this solution of Batominovski's fits the bill.

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