
#1
Jun407, 01:51 AM

P: 4

Hi,
Here is my question. I need to prove the following an equivalence relation. Let A = {1,2,3,4,5} X {1.2,3,4,5} and define a relation R on A by (x1,y1)R{ x2,y2) if x1+y1=x2+y2. I am bit confused how to use the condition x1+y1=x2+y2 to prove for transitive, symmetric and reflexive properties. Please help. Thanks 



#2
Jun407, 02:36 AM

P: 74

Can you show that "=" is an equivalence relation on the integers?




#3
Jun407, 02:48 AM

P: 4

Thanks for your reply John.
Ya sure. If I have got your question correctly. I have done this. Let S be a nonempty set of integers and let equality = be our relation. Then = is an equivalence relation on S since (i) a = a for all a S, (ii) if a = b, then b = a (for all a,b), (iii) if a = b and a = b then a = c (for all a,b,c ). I am eager to know whether I am correct on this or not. 



#4
Jun407, 03:04 AM

P: 74

Equivalence Relation Problem
I think on you probably meant: (iii) If a = b and b = c then a = c. The point is, at any rate, that it's easy to see that x_{1}+y_{1} and x_{2}+y_{2} from your original post are both integers.




#5
Jun407, 03:15 AM

P: 90

The relation R that you defined is an equivalence relation on A; it is not, however, the equality relation on A. I suggest you state the reflexive, symmetric and transitive properties for binary relations in general. Then verify that R satisfies all three. 



#6
Jun407, 04:00 AM

P: 4

Thanks fopc! but what is the use of condition x1+y1 = x2+y2 ?




#7
Jun407, 05:34 AM

Sci Advisor
HW Helper
P: 9,398

Use? Who knows. It's just an example to see if you can prove something is an equivalence relation.
You've not written a proof yet. So, does (x,y)R(x,y)? Does (x,y)R(u,v) imply (u,v)R(x,y)? Does (x,y)R(u,v) and (u,v)R(s,t) imply (x,y)R(s,t)? 



#8
Jun407, 06:55 AM

P: 90

Here's some supplementary information to consider. R is a subset of AxA. x1+y1 = x1+y2 is the property or predicate (call it P) that defines R, i.e., R = {((x1,y1),(x2,y2))  x1+y1 = x2+y2} (loosely stated). What use is P? P must be used to establish the three properties in question. There is no escape here. It must be used. What I think you'll see is that the properties in question will be inherited from P itself. Think about '='. Invariably, for me at least, generating specific examples gives confidence that something can be established in general. Here's an example for each property. Reflexivity: Ask yourself, is every member of A in relation with itself (all 25 of them)? Of course, yes. For example, is ((1,2),(1,2)) in R? Symmetry: Example: Clearly, ((3,4),(5,2)) is in R. Then we must have that ((5,2),(3,4)) in R and it is. Transitivity: Example: Clearly, ((5,1),(4,2)) and ((4,2),(3,3)) are in R. Then we must have ((5,1),(3,3)) in R and it is. 



#9
Jun507, 02:10 AM

P: 4

Thanks a lot... :)



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