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Math Help - More Relations on A

  1. #1
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    More Relations on A

    Def:
    R is irreflexive iff. a\nsim a \ \forall a\in A
    Not irreflexive iff. \exists a\in A \ni a\sim a
    R is asymmetric iff. if a\sim b, then  b\nsim a
    Not asymmetric iff. \exists a\sim b and b\sim a
    R is anti-symmetric iff. if a\sim b \ \mbox{and} b\sim a, then a=b.
    Not anti-symmetric iff. \exists a\sim b \ \mbox{and} \ b\sim a \ \mbox{and} \ a\neq b
    \bigtriangleup ={(a,a)|a\in A}

    Proofs:
    R is symmetric implies that R^2 is symmetric.

    Let aR^2b in R. Therefore, (a,x)\in R \ \mbox{and} \ (x,b)\in R. By the symmetric hypothesis, (b,x)\in R. Thus, we have (b,x)\in R \ \mbox{and} \ (x,a)\in R. We now have a path of length from b to x to a in R and R^2 is symmetric.

    If R is asymmetric, then R\cap R^{-1}=\phi

    By contraction: Suppose R is asymmetric and R\cap R^{-1}\neq\phi

    Let (a,b)\in R\cap R^{-1}\rightarrow (a,b)\in R \ \mbox{and} \ (a,b)\in R^{-1}\leftrightarrow (b,a)\in R. Now, we have (a,b)\in R \ \mbox{and} \ (b,a)\in R but R is asymmetric. Therefore, we have reached a contradiction and R\cap R^{-1}=\phi.

    Are this all logically correct?
    Last edited by dwsmith; November 20th 2010 at 11:46 AM.
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    R is symmetric implies that R^2 is symmetric.

    Let aR^2b in R. Therefore, (a,x)\in R \ \mbox{and} \ (x,b)\in R for some x in A. By the symmetric hypothesis, (b,x)\in R and (x,a)\in R. We now have a path of length 2 from b to x to a in R and therefore R^2 is symmetric.

    If R is asymmetric, then R\cap R^{-1}=\phi

    By contraction: Suppose R is asymmetric and R\cap R^{-1}\neq\phi

    Let (a,b)\in R\cap R^{-1}\rightarrow (a,b)\in R \ \mbox{and} \ (a,b)\in R^{-1}\leftrightarrow (b,a)\in R. Now, we have (a,b)\in R \ \mbox{and} \ (b,a)\in R but R is asymmetric. Therefore, we have reached a contradiction and R\cap R^{-1}=\phi.
    (Hint: \emptyset is typed \emptyset.)
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