Talk:Antisymmetric relation

Regarding the terms "equality" and "antisymmetry"
I believe there is a cycle in the definitions: Equality is defined as binary relation which is reflexive, symmetric transitive and antisymmetric. The definition of antisymmetry refers to the notion of equality (a R b and b R a => a = b). I don't know how to fix this.

cheers, chris
 * The '=' above is identity, not equality. Identity is primitive in metaphysics and logic, but is characterised by Leibniz's Law. The reference to equality in the article is more strictly speaking equivalence; two objects can be equivalent without being identical (consider the universal equivalence relation: \forall a, b \in X, a R b). Acanon 20:43, 20 Sep 2004 (UTC)

prakash kumar devta from IIITB


 * I would answer this by saying that the axiom of antisymmetry is an axiom of first-order logic with equality. So equality comes first. One can then prove that equality is the only reflexive, symmetric and antisymmetric relation. Hope that helps. Sam Staton 14:49, 15 January 2007 (UTC)

So, to clarify, the definition of antisymmetry states that if one picks elements a and b, then a R b and b R a implies that a and b are in fact the same element, correct? —Preceding unsigned comment added by Undsoweiter (talk • contribs) 00:05, 20 February 2010 (UTC)

question
suppose R is a subset of A x A where A = {a,b} R={(a,a),(b,b)} is this antisymmetric? if yes then why?
 * The article gives "Greater than or equal to" as an example of antisimmetry. So, the "equal to" "a=b" relation here is just a particular case of it. Futhermore, the article clearly states that the equality relationship is both simmetric and antisimmetric ;) Actually, any xRy=true coincides with xRx, which satisfies both yRx and x=y. These two implications are the only requrements for the antidependency. So, any of your relations are antisimmetric. Javalenok'
 * Am I wrong in saying that.. you're wrong? Greater than or equal to isn't antisymmetric, right? I'm pretty sure its neither symmetric nor antisymmetric. Fresheneesz 04:47, 13 December 2005 (UTC)
 * No, as the article says, the relation "greator than or equal to" is antisymmetric. Paul August &#9742; 05:32, 13 December 2005 (UTC)

Uh, .. where? This article? but greater than or equal to has cases where aRb exists but a does not = b... so .. i'm pretty sure i'm right. 67.161.46.169 02:09, 14 December 2005 (UTC)


 * The fourth line of the article says: Inequalities are antisymmetric, since for different numbers a and b not both a ≤ b and a ≥ b can be true. In other words if both a ≤ b and a ≥ b, then a = b. That is the definition of antisymmetric. Paul August &#9742; 03:03, 14 December 2005 (UTC)

Picture
I had a picture of the equality relation, Arthur Rubin deleted it. Why? First off, we need examples of antisymmetric relations. Secondly, pictures most definately do illustrate the concept. Please tell me why my picture should not be edited back in. Fresheneesz 02:32, 14 December 2005 (UTC)
 * Your picture incorrectly said that for x &ne; y, xRy "must be false". Paul August &#9742; 03:03, 14 December 2005 (UTC)
 * Oh ok yea you're right. Needs a better picture then. Crap.. my picture should be deleted in that case. How do I do that? Fresheneesz 03:10, 14 December 2005 (UTC)
 * I'm scratching my head wondering why you editted your picture as you have. Paul clearly misread the part that says "Must be false if the check mark with the same number (z) is true for it to be an antisymmetric relation" as saying "Must be false if x is the same number as y for it to be an antisymmetric relation" and you seem to have misunderstood which statement he meant and have edited the other statement such that now it is wrong when originally nothing was wrong. I'll revert the picture.86.5.179.220 (talk) 04:44, 27 November 2012 (UTC)

Using a Vin Diagram, an anti-symmetric relation would have self loops around each variable in the relation. —Preceding unsigned comment added by 128.83.230.33 (talk) 00:03, 2 December 2010 (UTC)

Table
The table under examples is IMHO very confusing. The relation "x is even y is odd" is antisymmetric, but I don't understand how the table helps illustrating it. It seems that the table tries to illustrate some conditions for an unspecified relation to be antisymmetric but it is very involved and hardly helps understanding. Can it be removed?

Perhaps it would be helpful to instead have an example of something that's not antisymmetric. E.g., a pre-order like "is cheaper than" for fruits? (if two different fruits have the same price, then it is not anti-symmetric). 130.238.11.101 17:23, 16 February 2006 (UTC)

I don't understand the table either. Does anyone? I guess "the relation x is even, y is odd between a pair (x, y) of integers is antisymmetric" in the vacuous sense, since there is no pair of integers such that the condition, x is odd, y is even and y is odd, x is even, is true. If so, maybe this could be made explicit, as that's quite an unintuitive concept in itself. Fomentalist (talk) 01:14, 12 July 2011 (UTC)

antisymmetric property
please explain me about the antisymmetric relation in greater details with more examples. please include the examples of real time situations if any.

thanks and regards.

Mrs.r.joshi —Preceding unsigned comment added by 119.235.49.2 (talk) 07:31, 30 July 2008 (UTC)

I agree, the first few lines do not tell me what anti-symmetry means. I know what identity, symmetry, asymmetry, (and barely) equality mean, but what is this anti symmetry. The in-my-face formulas are not gettin it through, since I don't know what I'm looking at, or what to look for.- neveos —Preceding unsigned comment added by 70.171.37.195 (talk) 06:48, 13 February 2009 (UTC)


 * Agreed with the other two posters, this topic seriously difficult to understand. 174.113.181.104 (talk) 06:22, 21 July 2009 (UTC)

Proposal to fix ambiguious equivalent mathematical notation
Regarding the sentence beginning "In mathematical notation, this is:", the alternate definition

FOR all a,b elements of X, aRb AND NOT(x = y) then NOT(b)Ra

is ambiguious as it is not clear that NOT(b)Ra equals NOT(bRa). Therefore, I suggest changing the alternate definitioin to

FOR all a,b elements of X, aRb AND NOT(x = y) then NOT(bRa)

DLA (talk) 22:11, 19 December 2008 (UTC)


 * Seems unnecesary to me, as there is no reason to believe that "NOT" is defined on X, making the first interpretation unreasonable.  — Arthur Rubin  (talk) 00:40, 20 December 2008 (UTC)

Divisibility on integers
Divisibility on integers is not antisymmetric. For example, $$-2 | 2 \land 2 | -2$$ does not imply that $$2 = -2$$. I've changed the example to divisibility on natural number, which IS antisymmetric. Catskineater (talk) 19:37, 2 August 2009 (UTC)

Question about vacuous antisymmetric relations
Based on the definition, it would seem that any relation for which $$R(a,b) \wedge R(b,a)$$ never holds would be antisymmetric; an example is the strict ordering < on the real numbers. Such examples aren't considered in the article - are these in fact examples or is the definition missing something? Dcoetzee 23:10, 2 August 2009 (UTC)


 * That's correct. In fact, $$\forall a, b \in X, \lnot(R(a,b) \land R(b,a))$$ is equivalent to "R is asymmetric" (the stronger condition of the two alternatives for "asymmetric"). So every relation that is asymmetric is also antisymmetric, as is already stated in the article for asymmetric relation. Catskineater (talk) 16:44, 5 August 2009 (UTC)

Diagram
Shouldn't the ZV say "to be false for it to be a antisymmetric relation"?
 * Ah, exactly
 * Even and odd antisymmetric relation.png

( Martin | talk • contribs 19:13, 6 May 2016 (UTC))