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Math Help - Abstract Algebra - Equivalnce Relation

  1. #1
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    Abstract Algebra - Equivalnce Relation

    Let R be an integral domain.

    Let Q = { (a, b) | a, b ∈ R and b 0 }


    Define a relation ∼ on Q by (a, b) ∼ (c, d) if and only if ad = bc.


    Prove that ∼ is an equivalence relation (that is, ∼ is reflexive, symmetric, and transitive).

    So for reflexivity, for instance, you must prove that (a, b) ∼ (a, b).

    Would this be correct:

    1.) For all (a, b) ∈ Q, we have (a, b) ~ (a, b) since ab = ba. So ~ is reflexive.

    2.) If (a, b), (c, d) ∈ Q and (a, b) ~ (c, d), then

    ad = bc
    da = cb
    cb = da
    (c, d) ~ (a, b)
    So, ~ is symmetric

    3.) If (a, b), (c, d), (e, f) ∈ Q and (a, b) ~ (c, d), (c, d) ~ (e, f), then

    ad = bc and cf = de

    adf = bcf and bcf = bde

    daf = dbe

    d(af - be) = 0

    From (c, d) ∈ Q, we know that d is not equal to 0, since R has no zero divisors, we obtain af - be = 0. Then, af = be, so (a, b) ~ (e, f) and ~ is transitive

    So ~ is an equivalnce relation on Q
    Last edited by AwesomeHedgehog; April 17th 2014 at 05:58 AM.
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  2. #2
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    Re: Abstract Algebra - Equivalnce Relation

    probably a good start would be looking up what reflexive, symmetric, and transitive relations are
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  3. #3
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    Re: Abstract Algebra - Equivalnce Relation

    Quote Originally Posted by romsek View Post
    probably a good start would be looking up what reflexive, symmetric, and transitive relations are
    I posted my answer above.
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  4. #4
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    Re: Abstract Algebra - Equivalnce Relation

    Quote Originally Posted by AwesomeHedgehog View Post
    Let R be an integral domain.

    Let Q = { (a, b) | a, b ∈ R and b 0 }


    Define a relation ∼ on Q by (a, b) ∼ (c, d) if and only if ad = bc.
    Prove that ∼ is an equivalence relation (that is, ∼ is reflexive, symmetric, and transitive).
    So for reflexivity, for instance, you must prove that (a, b) ∼ (a, b).

    1.) For all (a, b) ∈ Q, we have (a, b) ~ (a, b) since ab = ba. So ~ is reflexive.
    correct

    2.) If (a, b), (c, d) ∈ Q and (a, b) ~ (c, d), then

    ad = bc
    da = cb
    cb = da
    (c, d) ~ (a, b)
    So, ~ is symmetric
    not quite.

    $(a,b)\sim (c,d) \Rightarrow ac=bd$

    show $(c,d)\sim (a,b) \Rightarrow ac=db$ is True

    3.) If (a, b), (c, d), (e, f) ∈ Q and (a, b) ~ (c, d), (c, d) ~ (e, f), then

    ad = bc and cf = de
    no. look at your relation again
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