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Is my proof valid?

  1. Oct 5, 2013 #1
    Can someone check if my proof is correct.Please exscuse the bad notation, i've no idea how to type the symbols.
    The question was prove that between any 2 rational number , there is a third rational.

    x,y ,z are elements of Q
    (for all x ) (for all y) (there exist z)[x>z>y] <->
    (for all x ) (for all y) (there exist z)[(x>z) ^ (z>y)]

    Proof by contradiction:
    Suppose its false that for any x and y , there exists a z between x and y

    ~((for all x ) (for all y) (there exist z)[x>z>y])
    (there exists x) (there exists y)( for all z)[ (x< or = z) V (z < or = y)]
    There is no x that is smaller than or equals to any z.
    There is no y that is larger than or equals to any z.
    Both are false, the disjunction is false.
    Therefore the statement (there exists x) (there exists y)( for all z)[ (x< or = z) V (z < or = y)]
    is false and the statement (for all x ) (for all y) (there exist z)[x>z>y] is true.
     
  2. jcsd
  3. Oct 5, 2013 #2

    haruspex

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    Since you have not used any facts about rational numbers, it seems vanishingly unlikely that your proof is valid.
    How about doing something really simple and obvious: given two rationals p/q and r/s construct a rational that lies between them.
     
  4. Oct 5, 2013 #3
    If i construct a rational in between p/q and r/s , i doesnt apply to any other rationals, so it doesnt really prove anything. Am i misinterpreting your statement ( im really bad at math so please excuse my lack of ability)?
     
  5. Oct 5, 2013 #4

    haruspex

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    P, q, r and s can be any integers (q, s nonzero). If you construct a rational between p/q and r/s then you will have provided a general construction for any given pair of rationals.
     
  6. Oct 5, 2013 #5

    Ibix

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    p/q and r/s are arbitrary rational numbers. Haruspex is suggesting that you construct an expression in terms of p, q, r and s that is rational and guaranteed to lie between the the two.
     
  7. Oct 6, 2013 #6
    Ah ok i see , thanks guys.
     
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