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Math Help - Equivalence Relations - I am stuck, help!

  1. #1
    Junior Member
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    Apr 2009
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    Equivalence Relations - I am stuck, help!

    Hey all,

    I am getting stuck proving the following relation:

    x, y are elements of Q^+. x~y iff x = y(2^n) for some integer n.

    Here's what I have so far:

    Reflexive:
    Let x be an element in Q^+.
    x = x(2^n).
    x/x = 2^n
    1 = 2^n
    1 = 2^0
    1 = 1, Ok so does this prove xRx or do I need further work?

    Symmetric:
    Let x,y be elements in Q+.
    Suppose xRy.
    x = y(2^n)
    Now suppose yRx.
    y = x(2^m)
    Plugging in for y and solving for x, I get
    x = (x(2^n))(2^m)
    x = x(2^n+m)
    x/x = 2^(n+m), This is what I don't know what to do. What exactly does this result tell me? Do I need further work, or is this completely the wrong approach? Please somebody.

    Transitive:
    Let x,y,z be elements in Q+
    Suppose xRy.
    x = y(2^n)
    Suppose yRz.
    y = z(2^m).
    Pluggin in and solving.
    x = (z(2^m))(2^n)
    x = z(2^m+n), Once again, now what?

    Please explain my faults and show how to correct. Explanations would be very helpful. Thanks
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  2. #2
    Junior Member
    Joined
    Apr 2009
    Posts
    44

    Thumbs up

    Quote Originally Posted by spearfish View Post
    Hey all,

    I am getting stuck proving the following relation:

    x, y are elements of Q^+. x~y iff x = y(2^n) for some integer n.

    Here's what I have so far:

    Reflexive:
    Let x be an element in Q^+.
    x = x(2^n).
    x/x = 2^n
    1 = 2^n
    1 = 2^0
    1 = 1, Ok so does this prove xRx or do I need further work?

    Symmetric:
    Let x,y be elements in Q+.
    Suppose xRy.
    x = y(2^n)
    Now suppose yRx.
    y = x(2^m)
    Plugging in for y and solving for x, I get
    x = (x(2^n))(2^m)
    x = x(2^n+m)
    x/x = 2^(n+m),
    This is what I don't know what to do. What exactly does this result tell me? Do I need further work, or is this completely the wrong approach? Please somebody.

    Transitive:
    Let x,y,z be elements in Q+
    Suppose xRy.
    x = y(2^n)
    Suppose yRz.
    y = z(2^m).
    Pluggin in and solving.
    x = (z(2^m))(2^n)
    x = z(2^m+n), Once again, now what?

    Please explain my faults and show how to correct. Explanations would be very helpful. Thanks
    Actually what you have done to prove reflexive and transient property of R is correct ,
    But you have made a mistake on the proof of symmetric property of R .
    You have taken that if (x,y) belongs to R x = y (2^n) , y = x (2^m)
    Mistake is there
    if (x,y) belongs to R x = y (2^n) ,
    y = x (2^-n) we know that -n is also an Integer
    therefore R is symmetric .
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  3. #3
    Junior Member
    Joined
    Apr 2009
    Posts
    70
    Thanks a bunch for the help.
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