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Thread: Reflexive and symmetric but not an equivalence relation

  1. #1
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    Reflexive and symmetric but not an equivalence relation

    Define a binary relation R on {n Z: ≥ 2} such that aRb if and only if a and b havea common factor not equal to 1.
    Show R is reflexive and symmetric but not an equivalencerelation.

    hmm so n Z: ≥ 2
    and a and b does not have a common factor of 1.

    so if we make a all even squares ≥ 2 and b, all squares powers..from 1 upwards
    {1,2,3}

    we use a subset we could have would be all Z ≥ 2 a ={2,4,8} b ={1,2,3}



    Is this on the right track
    any help would be appreciated thanks

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    Re: Reflexive and symmetric but not an equivalence relation

    Quote Originally Posted by bee77 View Post
    so if we make a all even squares ≥ 2 and b, all squares powers..from 1 upwards
    {1,2,3}

    we use a subset we could have would be all Z ≥ 2 a ={2,4,8} b ={1,2,3}

    Is this on the right track
    any help would be appreciated thanks

    no.

    your relation is

    $aRb \Rightarrow a = u v,~b = u w,~u \neq 1$

    First show it's reflexive and symmetric.

    Reflexive means $aRa$

    does $aRa$ ?

    well $a$ is a common factor and by assumption $a\geq 2$

    so yes $aRa$ and $R$ is reflexive

    Can you show that $R$ is symmetric, i.e. that $a R b \Rightarrow b R a$ ?

    Then you have to show that despite the above $R$ is not an equivalence relation.

    What further property must an equivalence relation have?

    Show that $R$ does not possess this further property and thus is not an equivalence relation.
    Thanks from bee77
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