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Least Upper Bounds

  1. Mar 9, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the least upper bound and greatest lower bound (if they exist) of the following sets and state whether they belong to the set:

    a. {1/n:n[itex]\in[/itex]"Natural Number"}
    b. {x[itex]\in[/itex]"Rational Number":0≤x≤√5
    c. {x irrational:√2≤x2}
    d. {(1/n)+(-1)n:n[itex]\in[/itex]"Natural Number"}

    2. Relevant equations

    Not applicable.

    3. The attempt at a solution

    a. least upper bound does not exist; greatest lower bound is 1 and does belong to the set
    b. least upper bound is √5 and does not belong to the set; greatest lower bound is 0 and does belong to the set.
    c. least upper bound is 2 and does not belong to the set; greatest lower bound is √2 and does belong to the set.
    d. I am not sure about this one, I don't know what the graph would look like.

    Am I getting the right idea here? Any ideas for d.?

    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 9, 2012 #2

    radou

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    For a. Is 1 <= x, for all x in your set? Revise the definitions of "greatest lower bound", i.e. "infimum".
     
  4. Mar 9, 2012 #3
    Well, I thought it was, because x has to be a natural number.
     
  5. Mar 9, 2012 #4

    radou

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    Unless I'm missing something, n has to be a natural number. Numbers of the form 1/n, where n is natural, are rational numbers.
     
  6. Mar 9, 2012 #5

    HallsofIvy

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    ??? There is NO mention of an "x" in problem (a). There is mention of a natural number, n, and the problem talks about the number 1/n for each n:
    if n= 1, 1/n= 1
    if n= 2, 1/n= 1/2
    if n= 3, 1/n= 1/3
    if n= 4, 1/n= 1/4
    ...

    I strongly recommend that you write out at least a few of the numbers in each problem.
     
  7. Mar 9, 2012 #6
    This is why I thought that the greatest lower bound was one.

    Because n=1, 1/n=1.
    as far as I know n cannot be smaller than one as a natural number. Doesn't this make 1 the greatest lower bound??
     
  8. Mar 9, 2012 #7

    LCKurtz

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    As Halls mentioned, ##S = \{1,1/2,1/3,\, ...\}##. Plot a few of those points on the ##x## axis. Then remember that a greatest lower bound of a set is at least a lower bound. Then ask yourself if 1 is a lower bound for ##S##.
     
  9. Mar 9, 2012 #8
    Okay! Thanks!
     
  10. Mar 9, 2012 #9

    LCKurtz

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    To whom are you replying? And what have you decided about this problem?
     
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