
#1
Mar3012, 05:11 PM

P: 33

I'm wondering if someone could briefly explain how I can determine the equivalence class of relation?
I understand that first you must test the relation to see if is true for the properties, reflexive, symmetric, and transitive. But my main problem is once that is done how can I get the equivlant class of that. My book that I'm using does a terrible job in explaining the theorm, and hence it is very difficult for me to solve the practice problems. Thanks! 



#2
Mar3012, 07:35 PM

P: 1,623

Do you understand what the definition of equivalence class is but are confused about how to picture it / describe it? If this is the case, then it would help if you posted an example of the problem that you're working on, since it is difficult in general to get a good picture of what an equivalence class looks like.




#3
Mar3012, 08:08 PM

P: 33

Yeah, I'm a very visual learner (hence why discrete math isn't my strongest subject), I try to picture a lot of problems and I find that helps me to 1) understand/memorize concepts easily and 2) apply the knowledge in tests. Anyways,
I was working on this problem and it drove me crazy until I had to look at the solution manual: 1) Define P on the set R × R of ordered pairs of real numbers as follows: For all (w, x), (y, z) ∈ R × R, (w, x) P (y, z) ⇔ w = y. 2) Let A be the set of all statement forms in three variables p, q, and r . R is the relation defined on A as follows: For all P and Q in A, P R Q ⇔ P and Q have the same truth table. 



#4
Mar3012, 08:19 PM

P: 799

Equivalence ClassesHere's a perfect one. Now let's define an equivalence class based on this pic. Let's say that our set is the set of cow particles, where a particle is just some tiny part of the cow  like a molecule, say. Two cow particles x and y are equivalent if they are in the same section of the cow. Is this reflexive? Symmetric? Transitive? Yes, yes, and yes, but you should walk through the logic for yourself. Write down the proof that the relation of two cow molecules being in the same numbered section is an equivalence relation. That's all an equivalence relation is. It's a partition of a set into a collection of mutually disjoint subsets. Every element of the set goes to exactly one subset; and each of the subsets is an equivalence class. In other words if you have any equivalence relation, and for some element x you defined the equivalence class of x, denoted [x], as the set of all elements that are equivalent to x; then the set of all the equivalence classes are a partition of the original set. Another way to say this is that if you have two equivalence classes [a] and [b], then either [a] = [b] or [a] and [b] are disjoint. You should prove that. An equivalence relation gives you a partition; and every partition gives you an equivalence relation. Equivalence classes and partitions are just two ways of looking at the same thing. Once you get this, equivalence relations are easy. And it's totally visual. [Note: I think that's a bull, not a cow. Fortunately we're not on the Biology forum] 


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