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Real Number Proof

  1. Apr 11, 2006 #1

    "Suppose X, Y in R (Real Numbers), X < Y prove there exists Z in R such that X < Z < R."

    I'm currently trying to prove that X + Y / 2 satisfies this but I'm getting stuck. I first show that X + Y / 2 cannot be = to either X or Y. I then try to show that X + Y / 2 is > X since X < Y but I cannot seem to tie this in. Any help would be appreciated.
  2. jcsd
  3. Apr 11, 2006 #2
    Nvm I think I got it.

    Since X < Y
    Y - X must be in Positive Reals
    2 is in Positive Reals thus 1/2 is in Positive Reals
    Y - X / 2 is thus in Positive Reals
    You can then rearrange such that you can the equation

    X + Y / 2 - X in positive reals
    Thus X < X + Y / 2.

  4. Apr 11, 2006 #3

    matt grime

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    Dunno about that (you ought to bracket things up), but if x<y why don't you just add something to both sides?
  5. Apr 11, 2006 #4


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    In case you missed it, matt's point is: x+ y/2 is NOT between x and y:
    (x+ y)/2 IS!
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