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Least Upper Bounds/Supremum Proof

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Give a formal proof based on the definition of least upper bounds that the least upper bound of the set {1/2,2/3,3/4,4/5,...,n/(n+1),...} is 1.


    2. Relevant equations
    None.


    3. The attempt at a solution
    Basically, the best attempt at a solution that I have worked out is as follows:
    assume 1 is not the least upper bound of the set. then, let p be a real number such that p<1 and p>k/(k+1) for some k. (i.e. let p be the "least upper bound"). now, i need to find a number that would somehow bring up a contradiction proving p>1 or that there is a number of the form n/(n+1)>p thus showing p cannot be the "least upper bound". this is where i get stuck. i can't seem to determine a number greater than p which is in the set. i have also tried to start with the trivial fact that n+1>n and try to make something appear, but i get nowhere.

    if anyone could give some pointers as to how to continue along with the approach i have already used or lead me towards any approach, it would be greatly appreciated.

    thanks.
     
  2. jcsd
  3. Sep 21, 2009 #2

    Office_Shredder

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    You want to find n such that
    [tex]\frac{n}{n+1} > p[/tex]

    Take the reciprocal of both sides. This is equivalent to finding n such that

    [tex] \frac{n+1}{n}< \frac{1}{p}[/tex]

    which is the same as finding n such that

    [tex]1+\frac{1}{n} < \frac{1}{p}[/tex]

    What can you do with that?
     
  4. Sep 21, 2009 #3
    I dont really see how this how this could help the problem. Could you please elaborate?

    Thanks a lot.
     
  5. Sep 21, 2009 #4

    Office_Shredder

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    We can change this last line to be

    [tex] \frac{1}{n} < \frac{1}{p} - 1[/tex]

    But we know [itex] \frac{1}{p}-1>0[/itex] so we can look at its reciprocal. So finding an n equivalent to what you want is finding an n such that (taking the reciprocals of the first inequality in this post

    [tex] n > \frac{1}{ \frac{1}{p}-1}[/tex]

    And we know the right hand side is a positive real number
     
  6. Sep 21, 2009 #5
    so basically taking any n greater than 1/(1/p - 1) should do the trick right? from what I understand of your suggestion is that taking any n greater than 1/(1/p - 1) would make p>1. am i right?
     
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