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Yesterday I was out running in the park, and like many others I always run in a counterclockwise direction around the central lake. There are also strange people who always run in a clockwise direction. I realized that all the runners I can recognize by face run clockwise, because they are the ones I actually meet regularly. I never meet the people who run in the same direction as I do.

So we have a set of runners that can be divided into two distinct subsets, and runner A and runner B can recognize each other iff they are not in the same group.

Is there a name for such a relation?

RemcoGerlich
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    @GitGud : Such a relationship would not be reflexive. – Prahlad Vaidyanathan Jul 02 '14 at 14:24
  • I thought it was sort of the opposite of an equivalence relation, the equivalence relation would be the case where A and B were related if they were in the same group? It's also not reflexive or transitive. Does a "nonequivalence relation" exist? – RemcoGerlich Jul 02 '14 at 14:24
  • You're all right. I misread the question (rather I didn't read it completely). – Git Gud Jul 02 '14 at 14:24
  • @GitGud: renamed the groups to distinct subsets – RemcoGerlich Jul 02 '14 at 14:24
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    @RemcoGerlich the term "anti-equivalence relation" seems to be giving some results when I search it. – Ben Grossmann Jul 02 '14 at 14:26
  • I don't think this has a name, but mathematically, the relationship looks to be $$A\times A^c \bigcup A^c \times A $$ where $A$ is the set of runners who run clockwise. – Prahlad Vaidyanathan Jul 02 '14 at 14:29
  • @GitGud: sure! there should be contact details on my profile. – RemcoGerlich Jul 02 '14 at 14:37
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    discrepancy relations is the term you're looking for :)

    An example, would be xor, ${(p,q) ,|, \text{ exactly one of } p,q \text{ is true }}$.

     – Musa Al-hassy Jul 03 '14 at 00:32
  • @Omnomnomnom: the results I get in Google for anti-equivalence relation talk about relations that don't have any of the properties of equivalence relations, but this one is symmetric. – RemcoGerlich Jul 03 '14 at 20:14
  • @Moses: I like the name a lot, but Googling doesn't show anything. If it's been used anywhere it should be an answer. – RemcoGerlich Jul 03 '14 at 20:15
  • I recall seeing it mentioned by Backhouse et al as the name for xor, but I don't recall more general usage. I think it may be fine to use the term as it brings about connotations of distinctness ---just as the term equivalence brings connotations of similarity or identity.

    I don't know; hope that helps!

     – Musa Al-hassy Jul 05 '14 at 00:16

3 Answers3

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We call such a relation usually 'an' orthogonality relation, e.g.: $$A\perp B:\iff A\cap B=\varnothing$$

C-star-W-star
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  • In this particular case, we probably want $$p \bot q ;:\equiv; See(p) \cap See(q) = \emptyset$$ where $See(x)$ is the set of people that $x$ sees :) – Musa Al-hassy Jul 03 '14 at 00:26
  • Yes, for example that one ;) but that might be not enough as you'd maybe like to ask for more people simultaneously as do Eric, Gina and Ellison run all opposite to Luis, Carl and Anna... – C-star-W-star Jul 03 '14 at 08:02
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Let $J$ be the set of joggers, runners, and walkers etc. that run around the lake at the park. Define the relation $S$ on $J$ as

$$jSk\iff\text{ }j\text{ sees }k \text{ while jogging, running, or walking around the lake}.$$

Now a reflexive relation $R$ on a set $A$ is one such that $\forall a\in A,aRa$. Clearly $S$ is not reflexive since a person does not see themselves as they run.

A symmetric relation $R$ is one such that if $jRk$, then $kRj$. Note that $S$ meets this criteria because for each person you see, they see you. Thus $S$ is symmetric.

A transitive relation $R$ is one such that $jRk$ and $kRl$ implies that $jRl$. Note that the relation $S$ is not transitive. $jSk$ implies $kSj$ by the symmetry of $S$. That together with $kSl$ show that $k$ sees both $j$ and $l$. Thus $j$ and $l$ run in the same direction and do not see each other.

Finally, an antisymmetric relation $R$ is one such that $jSk$ and $kSj$ imply that $j=k$. Note that $S$ is not antisymmetric, If $jSk$ and $kSj$, the runners $j$ and $k$ see each other, and thus are not the same person.

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As other people have said, it's exactly the opposite of an equivalence relation, so it should be called "not equivalent".

It can be justified as follow:

-The relationship where object have the relationship when they are in the same partition is an equivalent class.

-The relationship where object have the relationship when they are not in the same partition is independence of the object to represent the class: that is if you pick $A,B$ from a class and $C,D$ from a class, the relationship of $A$ to $C$ is the same as that of $B$ to $D$.

-Hence you can factor out the equivalent class and made them into object.

-Once factored out, the relationship where object have the relationship when they are not in the same partition turn into the relationship where object have relationship with everything except itself: ie. "not identical" relationship.

Gina
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