1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Equivalence Relation Problem

  1. Jun 4, 2007 #1
    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. Jun 4, 2007 #2
    Can you show that "=" is an equivalence relation on the integers?
  4. Jun 4, 2007 #3
    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. Jun 4, 2007 #4
    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. Jun 4, 2007 #5

    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. Jun 4, 2007 #6
    Thanks fopc! but what is the use of condition x1+y1 = x2+y2 ?
  8. Jun 4, 2007 #7

    matt grime

    User Avatar
    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. Jun 4, 2007 #8

    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. Jun 5, 2007 #9
    Thanks a lot... :-)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook