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


*

*The operation $\circ$ defined by $x \circ y=x$.

*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 
  
  
*
  
*associative,
  
*non-commutative,
  
*nontrivial (operators of the form $x\circ y=f(x)$ or $x\circ y=g(y)$ are trivial), and
  
*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?

 A: Ok, Batominovski's got an answer in the comments. I will type up the checking:
Our candidate is $x\circ y=|x|y$. Then:


*

*Associative? We have $(x\circ y)\circ z=(|x|y)\circ z=||x|y|z=|xy|z.$ On the other hand, $x\circ(y\circ z)=x\circ(|y|z)=|x||y|z=|xy|z.$

*Non-commutative? $x\circ y=|x|y\not=|y|x=y\circ x$. 

*Nontrivial? Well, it is not a function of $x$ or $y$ only. 

*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.
A: If you want (1)-(3), and (4'), but not (4),  then you can take $$x*y:=\lfloor x\rfloor+y$$ for all $x,y\in\mathbb{R}$.  Then, $*$ is continuous almost everywhere, except on the set $\mathbb{Z}\times\mathbb{R}$ which is a subset of measure $0$ of $\mathbb{R}\times\mathbb{R}$.  Also, for any $y\in\mathbb{R}$, the function $\_*y$ is continuous on $\mathbb{R}\setminus\mathbb{Z}$, whereas $x*\_$ is continuous on the whole $\mathbb{R}$ for any $x\in\mathbb{R}$.
It would be interesting to add the condition (4'') which demands that the binary operation be an everywhere differentiable (or even smooth) map from $\mathbb{R}\times\mathbb{R}$ to $\mathbb{R}$.  The only examples we have so far do not satisfy (1)-(3) and (4''), although they satisfy a weaker condition, which demands that the binary operation be an almost everywhere smooth map.
