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Uniqueness proof

  1. Oct 31, 2012 #1
    To get the following proof I followed another similar example, but I'm not sure if it's correct. Does this proof properly show existence and uniqueness?

    Show that if x is a nonzero rational number, then there is a unique rational number y such that xy = 2

    Solution:
    Existence: The nonzero rational number y = 2/x is a solution of xy = 2 because x(2/x) = 2 = x(2/x) - 2 = x - x = 0.

    Uniqueness: Suppose s is a nonzero rational number such that xs = 2. Then, xy =2 = xy - 2 = 0 and xs = 2 = xs - 2 = 0. Then:

    xy - 2 = xs - 2
    xy = xs
    y = s

    This would be a complete proof wouldn't it?
     
  2. jcsd
  3. Oct 31, 2012 #2
    It could be cleaner.

    First of all, use -> instead of = (for implication) since
    xy = 2 = xy - 2=0 is a contradiction. Whereas xy = 2 -> xy-2=0 works.

    Second, for the first part. It is sufficient to say xy=2. That is, let y=2/x, then xy=(x)(2/x)=2. You must make the restriction x =/= 0.

    Third, Try to prove the uniqueness part of the proof more cleanly. Think about what you've done versus xs=2 and what xy is.
     
  4. Nov 1, 2012 #3

    Bacle2

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    Science Advisor

    If you're allowed to use the field properties of the rationals, you can then also show that

    multiplicative inverses are unique , so that :

    xy=2 -> x-1xy=x-12 , so that y=2x-1 (by

    commutativity).
     
    Last edited: Nov 1, 2012
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