## Equivalence Relations

I have a question. For something like x=ymodn, how would you prove something which is intuitively obvious like for this relation, n distinct equivalence classes exist ([0],[1],...[n-1]).
 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.
 Recognitions: Science Advisor "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.
 Recognitions: Science Advisor 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.
 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.
 Recognitions: Science Advisor 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.
 Recognitions: Homework Help Science Advisor a note for either ed or janitor, modman's replies seem eminently ignorable to be honest. to show that the numbers 1,..n-1 form a complete set of equivalence classes mod n, one must only note that [p]=[q] iff n divides p-q, and if p an q are both less than n (and greater than zero) that that implies p=q
 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.

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