
#19
Mar1204, 01:49 PM

P: 150

you just need to show that if you have some Natural number x , that you infact have a partition.
so x can be in some set A and in some set B iff A = B. how it is partitioned is not as important as showing that it is an equivilence relation because once you establish the EQ Relation on the set, you have no need to proove that a partition exists. 



#20
Mar1204, 07:41 PM

Sci Advisor
P: 1,190

"Is y allowed to be x? For then every reflexive relation is total."  Matt
Yes, I didn't see any restriction saying y had to be distinct from x. So you make a good point: every reflexive relation has to be a total relation. 



#21
Mar1204, 07:52 PM

Sci Advisor
P: 1,190

I am not sure if mathematicians use the term "antiparallel," but I think physicists do. I believe vectors A and B are said to be antiparallel by physicists if they point in opposite directions. So 'antiparallel' is a binary relation on a vector space which is not reflexive, is symmetric, is not transitive.




#22
Mar1204, 07:54 PM

P: 150

a reflexive relation is an equivilence relation because it is reflexive, and vaccuosly transative and vaccuosly symetric.
reflexivity is the only aspect that MUST have a component for all x in A. 



#23
Mar1204, 08:11 PM

Sci Advisor
P: 1,190

Modman, it almost sounds like you are claiming that all reflexive relations are equivalence relations, but we have given examples in this thread where that is not the case. I must be misunderstanding your point.




#24
Mar1204, 08:48 PM

Sci Advisor
HW Helper
P: 9,398

a note for either ed or janitor, modman's replies seem eminently ignorable to be honest.
to show that the numbers 1,..n1 form a complete set of equivalence classes mod n, one must only note that [p]=[q] iff n divides pq, and if p an q are both less than n (and greater than zero) that that implies p=q 



#25
Mar1304, 12:48 AM

P: 150

when I made the post, I had this type of reflexive set in mind:
{ (1,1) , (2,2) } so I most certainly did not have all cases that are reflexive in mind. I most certainly know that the class { (1,1), (2,2), (2,3) } is not an equivilence....but then if you read carfuly, you would see I used the term "vaccuos" in relation to symetric and transitive, meaning that no ordered pair (like the(2,3) I used above) exists in the set, so all elements are examples of reflexivity, nothing more. reflexive is a bidirectional proposition, so you MUST have at least one ordered pair in the set, and it must represent for all x : (x,x) but Symetric is an implication, so it the antecedent is false, it is symetric. and Transitive is an implication so in the same way as symetric, it will be true. 



#26
Mar1304, 03:50 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

The "vacuously" part did not clarify, it's just wrong: not all reflexive relations are transitive or symmetric. 



#27
Mar1504, 12:46 AM

Sci Advisor
P: 1,190

"Identity relation" sounds like a better term for what modman was talking about.




#28
Mar1704, 06:24 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

And, of course, the information that every "identity" relation (x is related to y if and only if x= y) is an equivalence relation is not news. Identity is the prototype of all equivalence relations.



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