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Rationals in the reals.

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data
    If we have the set [itex] P= \{ \frac{X_m}{X_n} \} [/itex]
    where [itex] X_m=1+2+3+4......+m [/itex]
    and [itex] X_n=1+2+3+...+n [/itex] and m<n determine all the limit points for this set.
    m and n are positive integers
    3. The attempt at a solution
    It seems to me that we might be able to construct all the rationals between 0 and 1
    with this set. So i would think this set will be dense on (0,1)
    And since we cant generate every integer with just [itex] X_m [/itex] alone
    [itex] X_m [/itex] and [itex]X_n [/itex] need to share common factors for this to happen. Or maybe we don't need to show that we can construct every rational on that interval but just show that we can get arbitrarily close to every rational on that interval.
     
  2. jcsd
  3. Mar 11, 2012 #2

    tiny-tim

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    hi cragar! :smile:

    hmm … is it all the rationals, or all the reals, or … ?

    anyway, it'll be lot easier to solve if you write each one as m(m+1)/n(n+1) :wink:
     
  4. Mar 11, 2012 #3
    I don't think it would be all the reals, well maybe it could if we let m and n go to infinity.
    so if we write it as that formula that you said. should I try to algebraically manipulate
    it and get to where I could see what kind of numbers it would generate. Or do some limit argument to say that we could construct all the reals between 0 and 1.
     
  5. Mar 11, 2012 #4

    Dick

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    Well, try it. Here's a hint to make it easier. First show numbers of the form m^2/n^2 are limit points of P. Now can you show, for example, that 1/2 is a limit point of numbers of the form m^2/n^2?
     
  6. Mar 11, 2012 #5

    tiny-tim

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    first, make sure that it really does do all the rationals

    then see whether that automatically means it does all the reals (because the rationals are dense)
     
  7. Mar 11, 2012 #6
    I guess i couldn't get 1/2 as a limit point with [itex] \frac{m^2}{n^2} [/itex]
    becuase no integer squared is 2 .
     
  8. Mar 11, 2012 #7

    Dick

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    It says "limit point" it doesn't say "equals". You can approximate sqrt(2) with rationals, can't you?
     
  9. Mar 12, 2012 #8
    ok yes, so I need to show that between any 2 reals that there exists
    [itex] a< \frac{m^2}{n^2} < b[/itex]
     
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