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Math Help - Discrete set maths

  1. #1
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    Question Discrete set maths

    On the set
    <br />
\mathbb{P}\times{}\mathbb{P}<br />
    define a relation by (x,y)~(u,v) if and only if xv = yu.
    (a) Show that ~ is an equivalence relation.
    (b) Find the equivalence class of (2,3)


    thanks
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  2. #2
    Super Member Gamma's Avatar
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    not sure what your sets P are, but...

    Im assuming we are in a commutative setting here.
    Reflexive
    (a,b)\sim(a,b)?
    ab=ba
    Symmetry
    (a,b)\sim (c,d)\Rightarrow ad=bc \Rightarrow cb=da \Rightarrow (c,d) \sim (a,b)

    Think you can do transitive?

    (2,3), so the things equivalent are of the form (a,b) and must satisfy 2b=3a\Rightarrow b=\frac{3}{2}a, so the equivalence class is
    \{(a, \frac{3}{2}a)|a\in \mathbb{P})\}
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  3. #3
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    I'm sorry but could you explain that again? I'm confused, ab = ba, how does that show that its reflexive? and I thought that (x,y) =/= (y,x)

    Thank you =]
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  4. #4
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    Quote Originally Posted by Gamma View Post
    Im assuming we are in a commutative setting here.
    Reflexive
    (a,b)\sim(a,b)?
    ab=ba
    Symmetry
    (a,b)\sim (c,d)\Rightarrow ad=bc \Rightarrow cb=da \Rightarrow (c,d) \sim (a,b)

    Think you can do transitive?

    (2,3), so the things equivalent are of the form (a,b) and must satisfy 2b=3a\Rightarrow b=\frac{3}{2}a, so the equivalence class is
    \{(a, \frac{3}{2}a)|a\in \mathbb{P})\}
    I think p is positive integer.
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  5. #5
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    Quote Originally Posted by yoonsi View Post
    I'm sorry but could you explain that again? I'm confused, ab = ba, how does that show that its reflexive? and I thought that (x,y) =/= (y,x)

    Thank you =]
    (x,y)\sim(u,v) \Leftrightarrow xv = yu
    ~ is reflexive iff \forall {a,b} \in {\mathbb{P}} :  (a,b)\sim(a,b) \
     (a,b) \sim (a,b) \Leftrightarrow ab = ba
    Since \mathbb{P} is commutative, ab=ba for any {a,b} \in \mathbb{P}
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  6. #6
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    oh well I understand the concept of reflexivity buuuut, how come P is commutative? I thought in ordered pairs, (2,3) does not equal (3,2)?

    Sorry I must seem really silly...

    Could someone also step me through showing how that is transitive?
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  7. #7
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    Quote Originally Posted by Gamma View Post
    Im assuming we are in a commutative setting here.
    Reflexive
    (a,b)\sim(a,b)?
    ab=ba
    Symmetry
    (a,b)\sim (c,d)\Rightarrow ad=bc \Rightarrow cb=da \Rightarrow (c,d) \sim (a,b)

    Think you can do transitive?

    (2,3), so the things equivalent are of the form (a,b) and must satisfy 2b=3a\Rightarrow b=\frac{3}{2}a, so the equivalence class is
    \{(a, \frac{3}{2}a)|a\in \mathbb{P})\}
    In other words (a, b) is in the same equivalence class as (2, 3) if and only if \frac{b}{a}= \frac{3}{2}. You could think of that class as all fractions that are equivalent to the fraction \frac{3}{2} and so representing the rational number \frac{3}{2}.
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