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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?

Paul
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  • Matrix multiplication is function composition (of linear functions), so you really only have one example. – rschwieb Jun 19 '16 at 18:44

4 Answers4

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We can define $x \oplus y=y$. Then $(x \oplus y) \oplus z =z= x \oplus (y \oplus z)$ but $y=x \oplus y \neq y \oplus x=x$

Ross Millikan
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    A rare example of the good one-line answer. I was going for the left-handed version, but then, I'm left-handed. – hardmath Jun 20 '16 at 13:31
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    @hardmath: I am left handed, too. I started with $\max$, but the symmetry made it commutative. Maybe because large numbers are on the right I was prompted by the right hand version. – Ross Millikan Jun 20 '16 at 15:49
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    For an underlying set ${0, 1}$ of two elements, this operation and its left-handed version (to which it is anti-isomorphic) are the only examples. The other associative operations, $0, 1, \wedge, \vee, \veebar, \not\veebar$, are all commutative. – Travis Willse Nov 03 '18 at 00:15
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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.

Milo Brandt
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  • FTR, that operation would not only be noncommutative, but also extremely discontinuous and strongly dependent on the choice of bijection. This makes it no less of a valid example, but it's certainly a bit on the pathological side. – leftaroundabout Jun 19 '16 at 21:39
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    If a question has a valid pathological answer, and if you find that disturbing, then the question was wrong. – Ryan Reich Jun 19 '16 at 22:23
  • @RyanReich One question I would like to see answered here is if there is a nonabelian topological group operation on the reals. – Mario Carneiro Jun 20 '16 at 08:00
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    To partially answer my own question: according to @PseudoNeo, any $C_1$ topological group operation on $\Bbb R$ is of the form $\phi^{-1}(\phi(x)+\phi(y))$ for a $C_1$ diffeomorphism $\phi$, and hence is abelian. – Mario Carneiro Jun 20 '16 at 08:15
  • @MarioCarneiro I suggest that you ask the question on this site, it's what it's for after all. I'd be interested in an answer too. – Najib Idrissi Jun 20 '16 at 15:06
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    @NajibIdrissi I was hoping some "overachiever" would answer the extended question here, since it subsumes the OP's assuming the answer is yes, but barring that I agree it is worth separate focus. Asked. – Mario Carneiro Jun 20 '16 at 15:25
  • @RyanReich: I don't find it disturbing. Only, if there's a valid simple (even trivial) answer, like Ross Millikan's, and a valid pathological answer, then I rather prefer the simple one. – leftaroundabout Jun 20 '16 at 16:45
  • @MarioCarneiro Yes, I would also liked to have had an answer that could be a group operation. I am aware that my accepted answer is not a group operation, but I failed to specify that in my question, so I accepted it anyway, since it answered the given question quite elegantly. – Paul Jun 20 '16 at 18:23
  • @MarioCarneiro I am fairly new to group theory, are you saying in your second comment that it is impossible to have such a group operation that I have described? – Paul Jun 20 '16 at 18:25
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    @Paul No, it is possible to have a nonabelian group operation on $\Bbb R$ - the easiest way is to map a known nonabelian group with the same cardinality as the reals across a bijection (which is exactly Milo's example). It becomes impossible if you also require that the group operation be continuous (i.e. "simple / non-pathological"), that is, a topological group. The question I asked reveals that $(\Bbb R,+)$ is the only topological group structure on $\Bbb R$ up to homeomorphism. – Mario Carneiro Jun 20 '16 at 18:34
  • The only thing that it not clear to me from Milo's answer, is that how do you know that a bijection exists from the set of reals to the set of 2x2 matrices? – Paul Jun 20 '16 at 18:44
  • Could he or someone else give an example of such a bijection along with its inverse? That is essentially the question that I was asking at the beginning. – Paul Jun 20 '16 at 18:45
  • @Paul The idea is that the set of $2\times 2$ matrices is described by four numbers. If you know that there is a bijection $f:\mathbb R\rightarrow \mathbb R^2$, then you can easily make a bijection $g:\mathbb R\rightarrow \mathbb R^4$ by applying $f$ to $x$ to get some $(y,y')$. Then, apply $f$ to $y$ and $y'$ to get pairs $(z,z')$ and $(z'',z''')$ respectively. Then $(z,z',z'',z''')\in\mathbb R^4$ and this process forms a bijection. – Milo Brandt Jun 23 '16 at 00:34
  • For invertible matrices, one notes that there are at least $|\mathbb R|$ many such matrices and no more than $|\mathbb R^4|$, but since these cardinalities are equal, there are equally many invertible matrices. (One can also note that the inverse of the mapping I described before is an injection from the invertible 2x2 matrices to $\mathbb R$ to the invertible matrices, and then note that $x\mapsto \begin{bmatrix}1 & x \ 0 & 1\end{bmatrix}$ is an injection the other way, and then use the Cantor-Schoder-Bernstein theorem) – Milo Brandt Jun 23 '16 at 00:37
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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:

  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? $x\circ y=|x|y\not=|y|x=y\circ x$.

So this solution of Batominovski's fits the bill.

Adrian Keister
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Yes, there are many other bilinear products on a given vector space which are associative but not commutative. For example, real associative algebras (not necessarily commutative), see also this MO-question.

Furthermore the quaternion algebra is a real division algebra which is associative but not commutative.

Edit: You added that the algebra must be part of the real numbers. Then also your example with the matrix algebra does no longer work.

Community
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Dietrich Burde
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  • I edited my question. I don't think I stated it correctly. – Paul Jun 19 '16 at 16:05
  • Are quaternions part of the Real numbers? Or am I completely missing what you're trying to say? I only just started learning abstract algebra. – Paul Jun 19 '16 at 16:18
  • Isn't that what "closed under the reals" mean? – Paul Jun 19 '16 at 16:21
  • I am not sure, what "closed under the reals" means. Before your edit the answer was OK. Now you may have changed the meaning, of course. – Dietrich Burde Jun 19 '16 at 16:21
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    @DietrichBurde "closed under the reals" means that the operation has the form $\circ:\Bbb R\times\Bbb R\to\Bbb R$, that is, it is a binary operation on the real numbers. The quaternions are not real numbers, unless you mean to map them to reals by some bijection as in Milo Brandt's answer. – Mario Carneiro Jun 20 '16 at 07:57
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    @DietrichBurde According to the edit history it used to say "binary operation over the reals" which if anything is more clear. It is true that OP's examples do not match his own criteria, but I assume that's why he's asking the question. – Mario Carneiro Jun 20 '16 at 08:18