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R is symmetric iff R = R^-1

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    In Velleman's "How to Prove it", he gives a proof that "R is symmetric iff R = R-1, which I find to be confusing when he is proving that [tex]R^{-1}\subseteq{R}[/tex]:

    Now suppose [tex](x,y)\in R^{-1}[/tex]. Then [tex](y,x)\in R[/tex], so since R is symmetric, [tex](x,y)\in R[/tex]. Thus, [tex]R^{-1}\subseteq R[/tex] so R=R-1

    It seems to me that he is saying that since [tex]xRy\rightarrow yRx[/tex] and yRx, xRy, which makes no sense.

    Basically my question is this: how this part of his proof could be correct?
     
    Last edited: Mar 8, 2010
  2. jcsd
  3. Mar 8, 2010 #2

    HallsofIvy

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    It would help if you would tell us what "R" is! A relation?

    If R is a relation, then it is a set of ordered pairs. [itex]R^{-1}[/itex] is defined as the set of pairs [itex]\{(x, y)| (y, x)\in R\}[/itex].

    What he is saying is that if (x,y) is in R-1, then (y, x) is in R. Since R is symmetric, (x, y) is in R and so [itex]R^{-1}\subset R[/itex].
     
  4. Mar 8, 2010 #3
    Ah, okay. I guess I was stuck thinking that (x,y) was in R and didn't consider that R being symmetric could mean that if yRx then xRy.

    And yes, R was a relation haha.

    Thanks!
     
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