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Math Help - equivalence relations

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    equivalence relations

    Let R be the relation on the set of ordered pairs of positive integers such that  ((a,b), (c,d))  \in if and only if ad = bc. Show that R is an equivalence relations.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TheRekz View Post
    Let R be the relation on the set of ordered pairs of positive integers such that  ((a,b), (c,d))  \in if and only if ad = bc. Show that R is an equivalence relations.
    this is just a case of applying definitions. Do you know what it means for a relation to be an equivalence relation? reflexive? symmetric? transitive?

    Hint: A relation is said to be an "equivalence relation" if it is reflexive, symmetric and transitive.

    Proving each of those cases are true for this relation should not be very difficult


    Hint 2: A somewhat similar problem was done here. There are also definitions for all the terms mentioned in that thread as well. Use it as a guide
    Last edited by Jhevon; July 7th 2007 at 08:50 AM.
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    I am confused on how to proof an ordered pair is reflexive or not
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by TheRekz View Post
    I am confused on how to proof an ordered pair is reflexive or not
    The transitive part is actually the toughest to prove. but i'll give you the reflexive part just so you see how to work with it.

    Proof (incomplete):

    Let R be a relation on the set \mathbb { N } \times \mathbb { N }, such that (a,b)R(c,d) \Longleftrightarrow ad = bc.

    First, we show that R is reflexive. Since ab = ba we have (a,b)R(a,b). Thus R is reflexive.

    ....now continue
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    MHF Contributor red_dog's Avatar
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    1) Reflexivity:
    (a,b)R(a,b)\Leftrightarrow ab=ab, which is true.
    2) Simmetry:
    (a,b)R(c,d)\Rightarrow ad=bc\Rightarrow cb=da\Rightarrow (c,d)R(a,b)
    3) Tranzitivity:
    (a,b)R(c,d)\Rightarrow ad=bc\Rightarrow \frac{a}{b}=\frac{c}{d} (1)
    (c,d)R(e,f)\Rightarrow cf=de\Rightarrow \frac{c}{d}=\frac{e}{f} (2)
    From (1) and (2) \frac{a}{b}=\frac{e}{f}\Rightarrow af=be\Rightarrow (a,b)R(e,f)
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