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Thread: Equivalence relation?

  1. #1
    Junior Member Fnus's Avatar
    Nov 2006

    Equivalence relation?

    I've added a picture with the information about the relation.

    Reflexive: I'd say it's reflexive since x*x would stiil belong to the rational numbers, right?
    But is it enough to just write that, or do I need to do something more?

    Symmetric: Again I'd say it is symmetric because multiplication is commutative.
    Is that enough? xD

    Transitive: No idea, I'm so bad at transitive, so any help would be appreciated, thanks!
    Attached Thumbnails Attached Thumbnails Equivalence relation?-math.bmp  
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  2. #2
    Member Haven's Avatar
    Jul 2009
    Your idea for the reflexivity of the relation is good

    $\displaystyle xTx = \{x\in\\Q $ ^ $\displaystyle {x^2}\in\\Q\} $

    which is obvious if we let $\displaystyle x = a/b\rightarrow\\{x^2} = {a^2}/{b^2} $. which will be in Q


    $\displaystyle xTy = \{x\in\\Q $ ^ $\displaystyle {x*y}\in\\Q\}\rightarrow\{y\in\\Q $ ^ $\displaystyle {y*x}\in\\Q\} = yTx $

    since multiplication in Q is communative, i.e., $\displaystyle xy = yx $.
    If $\displaystyle x = a/b $ and $\displaystyle xy = c/d $, since both x and xy are rational then it follows there must be a y in Q such that $\displaystyle xy = {a/b}*{y}= c/d $. Apply the same argument but for x and you show $\displaystyle xTy = yTx $


    if $\displaystyle xTy $ and $\displaystyle yTz $ then $\displaystyle xTz $

    by $\displaystyle xTy $, we know x and xy are in Q, and by $\displaystyle yTz $ we know y and yz are in Q. So we want to know if xz is in Q. It follows if y and yz are in Q then z must be in Q, by the same argument for the Symmetric. So if x is in Q and z is in Q, then xz must be in Q.
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