December 6th, 2011, 07:59 AM  #1 
Newbie Joined: Dec 2011 Posts: 15 Thanks: 0  Equivalence Relation
Set S := {1, 2, 3, . . .} and A := S x S. Now define a relation R on A by (x, y)R(a, b) :, x^y = a^b. (a) Show that R is an equivalence relation (b) Determine the equivalence class [(9, 2)] of (9, 2). (c) Determine an equivalence class with exactly 4 elements. 
December 6th, 2011, 05:18 PM  #2 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Equivalence Relation
What have you done, besides copy the problem? (a) Show symmetric, reflexive, and transitive properties hold. (b) ... (c) If I understand the given information properly, then we have [(4, 2)] = [(2, 4)] = [(16, 1)] since 4^2 = 2^4 = 16^1 ( = 16). This class has no other elements in S x S. So I have found a class with three elements. 
December 7th, 2011, 10:19 AM  #3 
Newbie Joined: Dec 2011 Posts: 15 Thanks: 0  Re: Equivalence Relation
Thanks for responding and yes I havent dont anything on this problem because I'm confused on to how to do it. How do you show all those three properties hold? And I'm not sure what an equivalence class is exactly. That is why when people on here do answer my question I can learn off of them and know how to do future alike questions.

December 7th, 2011, 10:42 AM  #4 
Global Moderator Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4  Re: Equivalence Relation
Not to be trite, but you really should research (websearch) what an equivalence class is. In this case, we write [1] to denote the equivalence class of 1 = 1^1, [2] for the equivalence class of 2, as so forth. We could (and maybe should) write this as [(1, 1)], [(2, 1)], etc. The equivalence class of a number "n" (IN THIS EXAMPLE ONLY!) is all the positive integer pairs of numbers (x, y) such that x^y = n. So for any "n", we can pick a representative n^1 > (n, 1). This will have equivalence class [(n, 10]. Here I literally just put brackets around the pair and called it an equivalence class. (a)Now we wish to show that (x, y) is equivalent to itself (reflexive) By definition of this equivalence, that means we have to show that (x^y) = (x^y). There is nothing to show. We wish to show that if (x, y) is equivalent to (a, b), then (a, b) is equivalent to (x, y). (symmetric) In other words, if x^y = a^b, then a^b = x^y. Again, nothing to show, since equality is symmetric. We wish to show the transitive property. If (x, y) is related to (a, b) and (a, b) is related to (c, d), then (x, y) is related to (c, d). In other words, If x^y = a^b and a^b = c^d, then x^y = c^d. There is nothing to show, no algebra or calculus or anything else. Since equality is transitive, this statement is true. QED. RIP part (a). 

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