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Equivalence Relations Simple Question

  1. Jan 30, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    For each of the relations defined on ℚ, either prove that it is an equivalence relation or show which properties it fails.

    x ~ y whenever xy ∈ Z

    2. Relevant equations


    3. The attempt at a solution

    Here's my problem: I am starting off the proof with the first condition of reflexivity.
    Now, do I let x ∈ ℚ ? I would think so, if that is the case, then x can be 2/3.
    So xx is thus 4/9 which does not exist in the integers.
    Thus, the proof would fail.

    But I think I am missing something here.


    Did I do this right or am I making a fatal error by suggesting that x can be any ℚ?
     
  2. jcsd
  3. Jan 30, 2016 #2

    PeroK

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    I'm not sure why you doubt what you've done.
     
  4. Jan 30, 2016 #3

    RJLiberator

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    I search for many proofs of equivalence relations online and just feel like something might be off here.

    Let me give you an example.

    The first part of this question, part a is
    a) x~y whenever x-y∈ℤ
    I proved the reflective part by showing x-x = 0 always.
    II symetric part by stating x-y must be an integer so y-x = -(x-y) thus that must be an integer.
    III transitive, by showing that (x-y) + (y-z) = x-z and since the first two are integers, this must be an integer also.

    I guess, in this part I did not need to suggest a rational number anywhere. Hm...

    Okay, back to the original question.

    We say x~y whenever x*y ∈ℤ
    An example would be x = 1/9 and y = 9/1 since x*y = 9 and that is an integer.

    So if we do 1. reflexive, does it have to be an x such that x*x is an integer by the rule? Or can I pick any x value I want out of the rational numbers. This is my problem.
    We say x~y whenever xy ∈ℤ.
    Must we also say that x~x whenever xx ∈ℤ ?
     
  5. Jan 30, 2016 #4

    PeroK

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    The reflexive rule must apply for all ##x## (not just for some ##x##). So, you only need to find one counterexample, as you did, to show that the rule does not, in general, hold.
     
  6. Jan 30, 2016 #5

    RJLiberator

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    Excellent. Thank you for your help.
     
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