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Math Help - [SOLVED] Equivalence Relations

  1. #1
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    [SOLVED] Equivalence Relations

    Prove that if R is a symmetric, transitive relation on A and the domain of R is A, then R is reflexive on A.

    I'm not really sure where to start.
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  2. #2
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    Quote Originally Posted by Ryaη View Post
    Prove that if R is a symmetric, transitive relation on A and the domain of R is A, then R is reflexive on A.
    If x \in A then by the fact that \mbox{dom}R=A \Rightarrow \quad \left( {\exists y \in A} \right)\left[ {\left( {x,y} \right) \in R} \right].
    By symmetry we know that {\left( {y,x} \right) \in R}.
    Now you finish.
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  3. #3
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    Quote Originally Posted by Plato View Post
    If x \in A then by the fact that \mbox{dom}R=A \Rightarrow \quad \left( {\exists y \in A} \right)\left[ {\left( {x,y} \right) \in R} \right].
    By symmetry we know that {\left( {y,x} \right) \in R}.
    Now you finish.
    Would you then go on to say:

    Since {\left( {x,y} \right) \in R} and {\left( {y,x} \right) \in R} and since transitivity holds on R, then xRy and yRx \Rightarrow xRx (which is the definition of being reflexive)?

    Therefore R is reflexive on A.
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  4. #4
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    Quote Originally Posted by Ryaη View Post
    Would you then go on to say:

    Since {\left( {x,y} \right) \in R} and {\left( {y,x} \right) \in R} and since transitivity holds on R, then xRy and yRx \Rightarrow xRx (which is the definition of being reflexive)?

    Therefore R is reflexive on A.
    yup

    i'd word it to fit the definition more, you know, with quantifiers and all that, but yes, that's the basic idea
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  5. #5
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    How could I word this to fit the definition better?

    Let x \in A. \mbox{dom}R=A \Rightarrow \left( {\exists y \in A} \right)\left[ {\left( {x,y} \right) \in R} \right].
    By symmetry we know {\left( {y,x} \right) \in R}.
    Since {\left( {x,y} \right) \in R} and {\left( {y,x} \right) \in R}, then by transitivity we know (xRy and yRx) \Rightarrow xRx. (definition of being reflexive)

    Therefore R is reflexive on A.

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Ryaη View Post
    How could I word this to fit the definition better?

    Let x \in A. \mbox{dom}R=A \Rightarrow \left( {\exists y \in A} \right)\left[ {\left( {x,y} \right) \in R} \right].
    By symmetry we know {\left( {y,x} \right) \in R}.
    Since {\left( {x,y} \right) \in R} and {\left( {y,x} \right) \in R}, then by transitivity we know (xRy and yRx) \Rightarrow xRx. (definition of being reflexive)

    Therefore R is reflexive on A.

    look at how the definitions are in your text, for instance, they usually end with phrases like "for all x,y \in R"
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  7. #7
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    It's getting kind of late here. Now that I have the main idea down, I'll work on refining it tomorrow. Thank you for all the help!
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