Equivalence Relation Problem

  1. 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.
    Last edited: Jun 4, 2007
  2. jcsd
  3. Can you show that "=" is an equivalence relation on the integers?
  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.
  5. 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 x1+y1 and x2+y2 from your original post are both integers.

  6. 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.
  7. Thanks fopc! but what is the use of condition x1+y1 = x2+y2 ?
  8. matt grime

    matt grime 9,395
    Science Advisor
    Homework Helper

    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)?

  9. 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.

    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?

    Example: Clearly, ((3,4),(5,2)) is in R. Then we must have that ((5,2),(3,4)) in R and it is.

    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.
    Last edited: Jun 4, 2007
  10. Thanks a lot... :-)
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