A binary operation, closed over the reals, that is associative, but not commutative
I am aware that matrix multiplication as well as function composition is associative, but not commutative, but are there any other binary operations, specifically that are closed over the reals, that holds this property? And can you give a specific example?
Solution 1:
We can define $x \oplus y=y$. Then $(x \oplus y) \oplus z =z= x \oplus (y \oplus z)$ but $y=x \oplus y \neq y \oplus x=x$
Solution 2:
If you already know that matrix multiplication is associative, but not commutative, then you can just choose your favorite bijection $f:\mathbb R\rightarrow M_2(\mathbb R)$, since the two sets are equinumerous. Then, define $a\oplus b = f^{-1}(f(a)f(b))$ to get an operation on $\mathbb R$ which is associative, but not commutative. If you want to have inverses as well, then you can replace $f$ with your favorite map from $\mathbb R$ to the invertible $2\times 2$ matrices.
In general, without more structure, you are equivalently asking, "Is there any associative, but not commutative operation defined on a domain of cardinality $|\mathbb R|$?" since the role of $\mathbb R$ in the question is nothing more than a set.
Solution 3:
Batominovski has another answer in the comments to this question. 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? $x\circ y=|x|y\not=|y|x=y\circ x$.
So this solution of Batominovski's fits the bill.