7
$\begingroup$

I'm trying to describe/define the commutative binary operation on a three-element set which when the operands are the same, gives the same element and when they are different gives the element which is not an operand. So for $\{a,b,c\}$, if we denote the operation $\#$: $$a\#a=a\quad \text{and} \quad a\#b=c \quad\text(etc.)$$

I suppose I could just define it case-by-case, but the notion of "the one that's either both of these or neither of these" is so intuitive and natural that that seems like a clunky solution. It also wouldn't be particularly useful to me. I want to describe it with a notation/definition which will lend itself to arguments regarding member-wise application of the operation on tuples i.e. $$(x_1,y_1,z_1...)\mathbf{\#}(x_2,y_2,z_2,...)=(x_1\#x_2,y_1\#y_2,z_1\#z_2,...)$$ In particular I want to discuss the properties of tuple-chains which, through repeated application of the member-wise operation, transform any given tuple into another. That is, I want to find out, given tuples $\mathbf {s}=(x_s,y_s,z_s,...)$ and $\mathbf{d}=(x_d,y_d,z_d,...)$, and considering $\mathbf{\#}$ to be right-associative, what can I say about the tuples in the sequence $\mathbf{C}=\mathbf{t_0},\mathbf{t_1},..\mathbf{t_n}$ if I know that$$\mathbf{t_0\#t_1\#}...\mathbf{\#t_n\#s}=\mathbf{d}$$

I'm not currently looking for help with that question, as I want to explore it myself for a while, and it's pretty open-ended anyway, but I have some ideas I'm setting out to prove, and I'm having trouble developing a formal definition of $\#$ that is useful in this context.

Any Ideas?

$\endgroup$
  • $\begingroup$ It feels similar to an external direct sum but I'm not sure if that relates to groups only. $\endgroup$ – Karl Aug 24 '15 at 22:00
  • $\begingroup$ @Karl the member-wise application works the same way, yes, but I'm really concerned with how to formally define the $\#$ operation itself in a way that naturally extends to the tuple version and can thus be used to reason about it. $\endgroup$ – Gabriel Burns Aug 24 '15 at 23:07
2
$\begingroup$

It's somewhat unclear what your criterion is for an acceptable definition, but if you want something that "looks" mathematical, you can identify your three-member set with the cube roots of unity, $\{1,e^{2\pi i/3},e^{-2\pi i/3}\}=\{1,{-1+\sqrt{-3}\over2},{-1-\sqrt{-3}\over2}\}$ and then, using complex conjugation, define

$$x\#y= \begin{cases} x\quad\text{if }x\overline y=1\\\\ \overline{xy}\quad\text{if }x\overline y\not=1 \end{cases}$$

If you like, you can let $\phi$ be an arbitrary bijection between your set $\{a,b,c\}$ and the cube roots of unity, in which case the definition is

$$x\#y= \begin{cases} x\quad\text{if }\phi(x)\overline{\phi(y)}=1\\\\ \phi^{-1}(\overline{\phi(x)\phi(y)}\quad\text{if }\phi(x)\overline{\phi(y)}\not=1 \end{cases}$$

(There is a tacit theorem here that the definition truly is a binary operation, and is independent of the choice of $\phi$.)

Added later: Here is a math-y alternative that avoids breaking into cases: Identify your three-member set with the unit vectors $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and then, using the dot and cross product in $\mathbb{R}^3$, define

$$u\#v=(u\cdot v)u+((1,1,1)\cdot(u\times v))u\times v$$

The main trick here is that $u\times v=0$ if $u=v$ and $u\cdot v=0$ is $u\not= v$. The $(1,1,1)\cdot(u\times v)$ is to get the sign right when $u\not=v$.

$\endgroup$
  • $\begingroup$ Thanks for this. You're right in that I didn't define what I was looking for very well, because I didn't really know exactly what I was looking for. I knew that I wanted to prove statements about the tuple-version of this operation (specifically this one) and I was looking for some way to represent the operation which would provide a sort of "way in" to reasoning about it, by analogy to more familiar constructs. Looking "mathy" wasn't the goal, but is a by-product. $\endgroup$ – Gabriel Burns Nov 15 '16 at 19:54
  • $\begingroup$ In any case, both of your suggestions have the potential to provide the sort of foothold I was looking for. $\endgroup$ – Gabriel Burns Nov 15 '16 at 19:56
6
$\begingroup$

I don't really understand the question, but the structure you've discovered is the unique ITSQ on three elements (well-done!).

In detail:

An idempotent totally-symmetric quasigroup (ITSQ) is a set $X$ together with a binary operation $\#$ such that the following hold:

  • $x\#y = y \#x$
  • $(x\#y)\#y=x$
  • $x\#x=x$.

Now let $A = \{a,b,c\}$ and $\#$ be as defined in your question. It's clear that $(A,\#)$ is an ITSQ. I claim that:

Proposition. If $f : A^2 \rightarrow A$ has the property that $(A,f)$ is an ITSQ, then $f=\#$.

Proof. We know that $$f(a,a) = a \qquad f(b,b) = b \qquad f(c,c) = c$$

If we can show that $f(a,b) =c,$ then by symmetry, this completes the proof.

Assume toward a contradiction that $f(a,b) \neq c$. Then we get two cases:

  • Case 0. $f(a,b) = a$.

  • Case 1. $f(a,b) = b$.

Case 1: Suppose $f(a,b)=b$. Then $f(f(a,b),b)=f(b,b)$. So $a=b$, a contradiction.

Case 0: Similar.

$\endgroup$
  • $\begingroup$ It's funny, that was actually my initial attempt to describe the operation, but it was suggested to me that those criteria were not sufficient to describe it. I very much appreciate your proof that they are. $\endgroup$ – Gabriel Burns Nov 21 '16 at 21:50
3
$\begingroup$

I just came across this question from last year. If you're still interested, here's a simple solution:

Use the set $\mathbb{Z}_3$ of integers modulo $3$ as your three-element set, and define $$x\,\#\,y=2(x+y) \pmod{3}.$$


By the way, if you liked @BarryCipra's idea of using the cube roots of unity as your three-element set, you can define $x\,\#\,y$ to be $x^2 y^2$ or, equivalently, $\dfrac1{xy}.$ This is isomorphic to the solution using $\mathbb{Z}_3$ in the first part of this answer.

$\endgroup$
  • $\begingroup$ Thanks. I really like the mod 3 interpretation. That is a helpful way of thinking about it. Much appreciated. $\endgroup$ – Gabriel Burns Nov 28 '16 at 16:43

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.