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Homework Help: Proving an equivalence relation

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data
    On the set of integers, define the relation R by: aRb if ab>=0.

    Is R an equivalence relation?

    2. Relevant equations

    3. The attempt at a solution

    R is an equivalence relation if it satisfies:

    1) R is reflexive
    Show that for all a∈Z, aRa.

    Let a∈Z. Then if a is a negative integer, aa>=0. If a is a positive integer, aa>=0. And if a = 0, aa>=0.
    Hence aRa

    I feel like it is too simple.. lacking something??

    2) R is symmetric
    Show that for all a∈Z, aRb --> bRa

    Let a∈Z, b∈Z such that aRb. By the definition of R, ab>=0.
    This is not symmetric. Take a = -1, b = 2.
    Then we have ab = -2 which is not >= 0.

    3) R is transitive
    if aRb and bRc implies aRc for all a,b,c ∈ Z

    Let a, b, c ∈ Z s/t aRb, bRc --> aRc

    Now I think this one is true.. but I'm not sure. But since aRb, and bRc, then you would always have ab or bc >=0 yea? so that means aRc must be true..
    How would I prove it properly if it is correct?

    Any help is appreciated! :) Thanks.
  2. jcsd
  3. Jan 24, 2010 #2


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    1) is too simplistic because you didn't take account of the case a<0. 2) That (-1)R(2) is false doesn't prove anything. You want to prove if ab>=0 THEN ba>=0. For 3) suppose one of a,b and c is 0. Can you show with a simple example that it's not true?
  4. Jan 24, 2010 #3
    1) I had that if it was a negative integer, which is the case of a<0 no?

    The problem was I was doing them backwards, assuming the relation first rather than the property.

    Let a, b ∈ Z s/t ab>=0 --> a>=0, b>=0, so ab>=0 then ba>=0

    So I think this is actually symmetric. Im not sure though, because I used the inequality to solve for a and b, but am I allowed to do that??

    3) Let a, b, c ∈ Z s/t ab>=0 and bc>=0
    So I should prove ac>=0
    Let a = 1, b = 0, c = -1
    Then we have ab>=0, bc>=0, but ac<0
    so its false.

    Thanks for the help!
    Last edited: Jan 24, 2010
  5. Jan 24, 2010 #4


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    Oh yeah, I see you did cover the negative case for 1). Sorry, somehow I didn't see that. But then for 2) ab>=0 doesn't imply a>=0 and b>=0. But it certainly does imply ba>=0. And yes, 3) is false. Just as you say.
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