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Sets and Algebraic Structures, help with equivalence relations

  1. Aug 13, 2011 #1
    Let Q be the group of rational numbers with respect to addition. We define a
    relation R on Q via aRb if and only if a − b is an even integer. Prove that this is an
    equivalence relation.

    I am very stumped with this and would welcome any help
    Thank you
     
  2. jcsd
  3. Aug 13, 2011 #2
    What are the defining characteristics of an equivalence relation?
     
  4. Aug 13, 2011 #3
    An equivalence relation is a relation that is reflexive, symmetric and transitive.
     
  5. Aug 13, 2011 #4
    Can you prove that R is reflexive, symmetric and transitive?
     
  6. Aug 14, 2011 #5
    I think it is reflexive and symmetric but not transitive but im not really sure im looking at i the correct way around.
     
  7. Aug 14, 2011 #6
    I would guess it IS transitive because the question asks you to prove it is an equivalence relation, it doesn't ask if it is or isn't.

    Let a, b, and c be any rational number. Let j and k be any integer.
    To show transitivity:
    A-b=2k
    B-c=2j
    So C=b-2j
    So a-c can be written a-b-2j=(a-b)-2j=2k-2j=2(k-j)

    Which is 2 times some integer, thus even.

    Can you do the other two (reflexive and symmetric)?

    Just a tip, you would get more help and sooner if you show what work you have already, even if it's wrong. The more you show you know, the less we ask you silly questions that you do already know, like define an equivalence relation. This will help you get to the answer faster. We love to help here, we just like to make it as much your own discovery as possible, kind of like what you teacher or professor might do.
     
  8. Aug 14, 2011 #7
    would reflexive be just as simple as, let a be any rational number. a=a
    and symmetric, let a,b be any rational numbers. and again simply a*b=b*a

    Thank you for the help and advice =]
     
  9. Aug 14, 2011 #8
    Not quite that simple.
    aRa iff a-a= 2k. Since a=a, a-a=0 which is even. That is why it is reflexive, not because a=a, but because a relates a. Imagine if the relation was aRb iff a<b. Then, saying a=a implies the relation is reflexive would be wrong, since a does not relate a. So, even if you understood a=a to mean a-a=0 so a relates a, you still have to write it all the way out. That's why a proof is different than just asking if it's true or false.

    I'm not sure why you talk about a*b for symmetric. I take that to mean a times b, is that what you mean? Because you are not showing a relates b by talking about multiplying. a relates b iff a-b= even integer. You must show that if a-b = even integer, then b-a = even integer.
     
  10. Aug 14, 2011 #9
    ok thank you,
    so for the symmetric could you say (let k,j be any integer) a relates b iff a-b=even integer

    a=2k and b=2j and so a-b=2k-2j=even integer
    and b-a=2j-2k=even integer
     
  11. Aug 14, 2011 #10
    You can't say a=2k where k is any integer because a is not necessarily an even integer, it is only necessarily a rational number. For example, 5.2-1.2=4, so 5.2 relates 1.2.
    So, if a-b=2k, what can you say about b-a=? Just move things around algebraically.
     
  12. Aug 14, 2011 #11
    b-a=-2k
    yes??
     
  13. Aug 14, 2011 #12
    Yes :) and since -2k is also an even integer, if aRb, then bRa.
     
  14. Aug 14, 2011 #13
    great thank you very much, it may of taken some time but i got there =]
     
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