8
$\begingroup$

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?

$\endgroup$
  • 5
    $\begingroup$ I think this satisfies all conditions (1)-(4) if I am not too sleepy: $$x*y:=|x|\,y$$ for all $x,y\in\mathbb{R}$. $\endgroup$ – Batominovski Aug 25 '18 at 0:37
  • $\begingroup$ @Batominovski: You should type that up as an answer by checking the requirements. $\endgroup$ – Adrian Keister Aug 25 '18 at 0:39
  • 2
    $\begingroup$ @AdrianKeister Please do it. I don't mind. I'm going to bed. $\endgroup$ – Batominovski Aug 25 '18 at 0:39
  • $\begingroup$ @Batominovski: You got it, thanks! $\endgroup$ – Adrian Keister Aug 25 '18 at 0:46
7
$\begingroup$

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.

$\endgroup$
6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.