Talk:Equivalence relation

message: Hmmmm. Let's see:
*by 2, a~b, then b~a
*take a=c
*So, by 3, a~a
QED
Yes, I know this is wrong, but where's the error?
--User:Mdob
Given a fixed a, there need not exist a b such that a~b. For example, the empty relation is not an equivalence relation. It is symmetric and transitive but not reflexive. -- Fropuff 01:08, 12 October 2005 (UTC)
replies: text: I would think that an empty relation on an empty set would vacuuously be an equivalence relation, but that an empty relation on a non-empty set would not be an equivalence relation because it does not satisfy x ~ x for any x in that set. 129.110.241.254 21:18, 22 August 2006 (UTC)

text: I misread the relation "a ~ b" iff a and b are both even as "a ~ b" iff a = b (mod 2), which isn't quite the same.  Sorry about the misedit there.  JamesBWatson restored it correctly, although I'm still not sure that it should be in the definition section. Antares5245 (talk) 09:00, 12 June 2010 (UTC)
thread_title:  Doesnt 2 and 3 imply 1? 

message: I hereby invite someone to add a section on groups and equivalence relations. Lattice theory is the algebraic structure of partial orderings. What is the algebraic structure of equivalence relations? Bas Van Frassen (1989) convinced me that it is (non-Aberlian) group theory, but algebra texts are typically totally silent about this fact, with the possible exception of Birkhoff and MacLane (1979: 72). The mathematicians whom I've queried about this tell me they have no idea what I am talking about. A curious situation indeed... On equivalence relations and group theory, see  and the references therein.202.36.179.65 10:22, 18 March 2006 (UTC)
thread_title:  Partial Order : Lattice theory :: Equivalence relations : ? 

message: If it were, then it would lead to people being their own sibling.  A sibling of B implies B sibling of A.  Then by transitivity, A would be a sibling of A, which is false.  It may sound like I am committing the error "2 and 3 imply 1" above, but I'm not.  If the relation were transitive, then anyone that had a sibling would be considered a sibling of themself as well, which is clearly undesirable. —The preceding unsigned comment was added by 67.124.203.53 (talk • contribs) 17:07, May 25, 2006  (UTC)
:Correct. Paul August ☎ 17:38, 25 May 2006 (UTC)
replies: text: This further illustrates the dangers in extending the idea of equivalence relation (a mathematical concept) to collections of objects and relationships between those objects which are not mathematical objects. I don't like the idea of the "set" of human beings because I'm not really sure that there is such a thing. Having examples which are not mathematical means that the concepts involved must be really well-defined, or they fall apart (as in the case of sibling). I have never seen the term "sibling" mean "having the same parents as;" in general, as far as I am aware, it means, "is a sister or brother to," and in this sense "sibling" would not be an equivalence relation, as has already been pointed out. Xantharius 03:37, 14 September 2007 (UTC)

text: Why is this article to blame for half your class getting it wrong? It states quite clearly that “sibling” is not an equivalence relation, and it explains why. So the reason must lie elsewhere, is my guess. Hanche (talk) 10:46, 12 February 2012 (UTC)
thread_title:  "Sibling of" is NOT Transitive 

message: The following doesn't make any sense:--345Kai 23:30, 7 September 2006 (UTC)
replies: text: Relations that are transitive, symmetric, but irreflexive are very rare (e.g., the empty relation). The previous example illustrates why. More generally, if a relation R is symmetric and transitive, then for any x, y in the field of R, if xRy, then yRx (by symmetry), and xRx (by transitivity). Hence if there exist an x and y in the field of R, such that y≠x and xRy both come out true, the reflexivity of R follows and R must be an equivalence relation. R can be transitive, symmetric but irreflexive iff x is an "island," namely for an x that is not related to anything at all.
What you are objecting to is legacy material I found when I began my major edit of this entry. Feel free to edit or delete.202.36.179.65 14:59, 5 November 2006 (UTC)
thread_title:  transtive, symmetric and irreflexive 

message: I think this paragraph is incomprehensible, misleading and even wrong
* what could be a "reflexive strict order" ?? strict means non-reflexive !
* what means "antisymmetry enhanced to full symmetry" ??
I think it's better to remove it... — MFH:Talk 21:41, 16 October 2006 (UTC)
replies: text: Here's what I meant. Begin with a strict order. Then replace non-reflexivity with reflexivity and the result is an equivalence relation. Likewise, start with a partial order, then replace antisymmetry with symmetry to obtain, again, an equivalence relation. It is also valid to go the other way. A variety of of types of relations can be derived from an equivalence relation by suitably relaxing parts of the definition of an equivalence relation. I see relation types as forming a hierarchy in a manner similar to the way normal modal logics form a hierarchy. Equivalence relations are at the top of the relations hierarchy is a manner analogous to the way S5 is at the top of the normal modal logic hierarchy.202.36.179.65 15:11, 5 November 2006 (UTC)
replies: text: you say "Begin with a strict order. Then replace non-reflexivity with reflexivity", but is that well-defined? In a 2-element set {a, b} if I define the strict partial order to be (a— Preceding unsigned comment added by 79.183.12.73 (talk) 17:15, 1 May 2012 (UTC)
thread_title: From order to equivalence via symmetry

message: I think that the example (quote begins)
* The empty relation R on a non-empty set X (i.e. aRb is never true) is vacuously symmetric and transitive, but not reflexive (except when X is also empty).
(quote ends)
could be supplied by the concrete case of the relation a R b = "a and b are both even numbers" on the set X of natural numbers (all natural numbers, not just the even numbers). This relation is clearly symmetric (if a and b are both even, b and a are both even), it is transitive (if a and b are both even and if in addition b and c are both even, a and c are both even), but it is not reflexive because 3 and 3 are not both even numbers. 129.132.146.184 09:13, 27 February 2007 (UTC)
replies: text: Yup, that works. Paul August ☎ 02:37, 8 March 2007 (UTC)
replies: text: not really.  Not considering the headline is "irreflexive." The only relation with those 3 qualities is the empty relation.24.58.63.18 (talk) 22:30, 15 March 2009 (UTC)
replies: text: Your example is a good one for 'transitive, symmetric, but not reflexive'. Unfortunately, 'irreflexive' doesn't mean 'not reflexive', it means 'never reflexive'. Adammanifold (talk) 21:27, 14 May 2009 (UTC)

text: The problem is that such a relation isn't transitive either. 2A01:CB10:21:8500:F110:46D6:DB03:EAAB (talk) 21:56, 23 July 2022 (UTC)
replies: text: That is right. Also, 24.58.63.18's post of 15 Mar 2009 above is right: if R is transitive, symmetric, and non-empty, then aRb holds for some a,b. By symmetry, bRa also holds, Hence, aRa by transitivity from aRb and bRa. That is, R cannot be irreflexive. - Jochen Burghardt (talk) 14:44, 24 July 2022 (UTC)
thread_title:  More concrete example of transitive, symmetric, but irreflexive 

message: Just a quick note: The relation (a,b) equivalent to (c,d) if ad=bc on N x N does not give all rational numbers, but only the positive rational numbers. To get all rationals you have to look at this relation as defined on Z x Z^*, but I'll leave it up to others to make this precise in the article.  —Preceding unsigned comment added by 136.152.181.253 (talk) 17:40, 19 October 2007 (UTC)
thread_title:  Rational numbers 

message: What about complex, topological or conformal conjugacy ? Is it a Equivalence relation ? --Adam majewski (talk) 08:11, 31 January 2009 (UTC)
thread_title:  conjugacy 

message: if this isn't a shorthand, please wikilink it. Thanks!  —Preceding unsigned comment added by 128.32.35.76 (talk) 00:46, 13 August 2009 (UTC)
replies: text: Just random vandalism. Algebraist 00:51, 13 August 2009 (UTC)
thread_title:  subs --> subsets? 

message: This heading is clearly a complete fabrication and should be deleted. Even the editor who added it admitted to giving Euclid a very generous interpretation. Unless I hear strong objections to the deletion of the section in question I will delete it promptly. ReneGMata (talk) 14:46, 13 January 2010 (UTC)
replies: text: I agree that in its present form the section is a stretch. However, "complete fabrication" seems a little strong, and it may be worth a mention that Euclid did make use of (essentially) the idea of equivalence in his deductive system. Perhaps a drastic rewrite and abbreviation, rather than total deletion, but I don't have time to do it now. JamesBWatson (talk) 17:03, 13 January 2010 (UTC)

text: I was not suggesting rewriting speculation: I was suggesting rewriting the element of truth in there. Euclid certainly did use axioms which are now part of the definition of an equivalence relation, and I see no reason not to mention the fact. JamesBWatson (talk) 14:57, 18 January 2010 (UTC)
thread_title:  Euclid anticipated equivalence 

message: Euclid aside, where is the first recorded explicit defnition of equivalence relation? I think it would be worth adding to the article. Simplifix (talk) 12:15, 10 February 2010 (UTC)
replies: text: The relation of equipollence used by Giusto Bellavitis is an early example with a suggestive label.Rgdboer (talk) 23:01, 3 November 2014 (UTC)
thread_title: History

message: Jacobson mention that monoid isomorphisms define an equivalence relation (in Basic Algebra I, p. 37). So, is he considering "the set of all monoids"? Is this really a set? (is Mon a small category?)
If not, there should be a note about equivalence relation over more general classes? Helder (talk) 21:32, 13 May 2010 (UTC)
thread_title:  Relations seems to be defined for other classes than "sets" 

message: It is currently written:
:Theorem on projections: Let the function f: X → B be such that a ~ b → f(a) = f(b). Then there is a unique function g : X/~ → B, such that f = gπ. If f is a surjection and a ~ b ↔ f(a) = f(b), then g is a bijection.
I think the last sentence is wrong, and here is a counter-example. Let the set X have two elements. Let the equivalence relation be the equality, i.e. a ~ b ↔ a = b. Let the set B be a singleton. The function f: X → B is a constant function and is a surjection. But g : X/~ → B can not be a bijection since X/~ has two elements but B only one.
Frozsyn (talk) 12:45, 16 November 2010 (UTC)
replies: text: In your example, if f is a constant function from a two element set {a,b} to a one element set, since a is not equivalent to b, f does not have the property that a ~ b ↔ f(a) = f(b). That extra property in the last sentence is important. — Carl (CBM · talk) 12:55, 16 November 2010 (UTC)
replies: text: Thanks for the reply! Somehow, I re-read the statement many times without noticing the property a ~ b ↔ f(a) = f(b). I assumed only a ~ b → f(a) = f(b) as written in the first part. My bad. Frozsyn (talk) 14:16, 16 November 2010 (UTC)
thread_title:  Theorem on projections partially wrong 

message: Hmmmmmmm?
Alt 247  —Preceding unsigned comment added by 173.218.85.222 (talk) 20:06, 16 February 2011 (UTC)
replies: text: No. Algebraist 20:07, 16 February 2011 (UTC)
thread_title:  Isn't it ≈  & NOT ~ 

message: A partition of X is a set P of nonempty subsets of X such that every element of X is an element of a single element of P.
: Correct, fixed. McKay (talk) 01:47, 21 February 2011 (UTC)
thread_title:  Imprecise definition of partition 

message: The link to the Projection (relational algebra) (main article) in the "Projection" chapter, is it relevant at all? Should not the "projection" be called the "quotient map"? I do not see how projections and quotient maps are related. --Beroal (talk) 22:59, 1 March 2011 (UTC)
thread_title:  Projection (relational algebra) 

message: The technical term 'cell' appears to have no entry in the disambiguation page for 'cell'. Jim Bowery (talk) 22:17, 1 January 2016 (UTC)
replies: text: I've never seen "cell" used in this context in the mathematical literature.  A quick Google search finds only a few Wikipedia articles and some quotes from Wikipedia.  Objects in "cell"s or "bin"s are sometimes used as examples of equivalence relations.  — Arthur Rubin  (talk) 20:13, 2 January 2016 (UTC)

text: The Encyclopedia of Mathematics has no such entry.  I think it may be original to Wikipedia.   — Arthur Rubin  (talk) 20:25, 2 January 2016 (UTC)

text: "Cell" and "equivalence relation" are an unusual combination, but "cell" and "partition" (of a set) are often used together. If you search at Scholar for "cell of the partition" and "cells of the partition" (with quotes) you'll find many examples. Given that partitions and equivalence relations are more or less the same concept, that's enough to justify the mention of cells here, imo. McKay (talk) 01:24, 17 March 2016 (UTC)
thread_title:  Cell 

message: Hello fellow Wikipedians,
I have just modified  on Equivalence relation. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:
*Added archive https://web.archive.org/web/20130509233055/http://planetmath.org/equivalencerelation to http://planetmath.org/equivalencerelation
When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.
Cheers.—InternetArchiveBot (Report bug) 07:10, 25 December 2016 (UTC)
thread_title:  External links modified 

message: Hello fellow Wikipedians,
I have just modified one external link on Equivalence relation. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:
*Added archive http://arquivo.pt/wayback/20160514115224/http://www.bangor.ac.uk/r.brown/topgpds.html to http://www.bangor.ac.uk/r.brown/topgpds.html
When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.
Cheers.—InternetArchiveBot (Report bug) 05:56, 22 September 2017 (UTC)
thread_title:  External links modified 

message: Skimming this article at 9:01 am, US Central Time, on Friday, July 23rd, 2021, I found the following mistake. I'd correct it immediately if I knew how to do that, but regrettably, I don't know how to edit or write formulas on Wikipedia. In any case, here is the error:
Symmetric and transitive: The relation R on N, defined as aRb ↔ ab ≠ 0. Or any partial equivalence relation;
Transitive means that if aRb and bRc, then aRc. Consider a=2, b=3, and c=2. This counter-example proves that ≠ is not transitive, because 2≠3 is true, and 3≠2 is true, yet 2≠2 is false.
We therefore require a replacement relation that is symmetric and transitive but not reflexive. I would recommend a binary relation that I call the "department head" relation. What ever request you make, the answer is always "no." In other words, for all x and for all y, xRy is false. This binary relation is not reflexive, but it is symmetric, and it is transitive. (Any one who has ever worked in education will find this example intuitive and familiar.)
Respectfully submitted,
Gregorybard (talk) 14:09, 23 July 2021 (UTC)
replies: text: You are right that ≠ itself isn't transitive. However, R defined by aRb ↔ ab ≠ 0 <U>is</U> transitive. (Note that R is different from ≠, e.g. 2R2, but not 2≠2, and e.g. 0≠1, but not 0R1.) The definition of R can be re-phrased as aRb ↔ a≠0 ∧ b≠0. Then aRb and bRc implies a≠0 and b≠0 and c≠0 by the alternate definition, hence a≠0 ∧ c≠0, hence aRc. BTW: your "department head" relation is known as the empty relation. I feel it is too particular to be the only example for a partial equivalence. - Jochen Burghardt (talk) 14:42, 23 July 2021 (UTC)
replies: text: I find this example and the one given in the article interesting. However they both look like ternary relations in disguise (between a, b and 0, in this case). Which seems inevitable for a non-empty non-reflexive relation to be both symmetric and transitive. I don't know if this is true, have no idea how it could be proved and realize that it doesn't make the example any less correct or relevant. I just thought it would be cool if someone more knowledgeable could explain why. 2A01:CB10:21:8500:CD8:3FB4:5BD5:E3AE (talk) 11:28, 25 July 2022 (UTC)
replies: text: I'd like to spend some effort on trying to prove your conjecture - if I'd understand what exactly it consists of. Maybe "" But then, how to formalize ""? - Jochen Burghardt (talk) 17:52, 25 July 2022 (UTC)
thread_title: Major Mistake

message: If (a, b), (b, c) (a, c) belongs R, and (b, b) doesn't belongs to R then will it be transitive or not. If not then why? 2409:4063:420B:CD60:75BF:7362:1CFF:DB7E (talk) 05:12, 5 December 2021 (UTC)
thread_title:  Transitive Relation 

message: Writing the composed relations R and S in the backward fashion of function composition is not recommended. A section using calculus of relations institutes this notion. It ignors the left to right reading of relations. Seeking consensus here to repress this perverse initiative. The greater than (>) and less than (<) relations show importance of direction of text. Habits from calculus of real and  complex variables need reformation when addressing relations. Rgdboer (talk) 04:05, 8 June 2022 (UTC)
thread_title: Composition notation

message: :What are the square brackets in the "simple example" and why are these equivalent classes?
:Quote (more or less):
::On set X={a,b,c}, R={(a,a),(b,b),(c,c),(b,c),(c,b)} is an equivalence relation. 
::The following sets are equivalence classes of this relation: 
::[a]={a}, [b]=[c]={b,c}. [a]={a}, [b]=[c]={b,c}. 
:פשוט pashute ♫ (talk) 18:55, 16 November 2022 (UTC)
thread_title:  "simple" example - what are the square brackets?