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Infimim and supremum proof

  1. Feb 2, 2010 #1
    The problem statement, all variables and given/known data

    Find the infimum and supremum of each of the following sets; state whether the infimum and supremum belong to the set E.

    \item 1. ~~~~E={p/q \in \mathbf{Q} | p^2 < 5q^2 \mbox{ and } p,q >0}. \mbox{ Prove your result. }
    \item 2. ~~~~E={2-(-1)^n/n^2|n \in \mathbf{N}. \mbox { Just list your answers}.

    2. Relevant equations

    [tex]sup E - \epsilon < a \leq sup E.[/tex]

    3. The attempt at a solution

    \item 1. ~~~~ \mbox{Here I claim the supremum of E is }\frac{1}{\sqrt{5}}, \mbox{ in E, and the infimum of E is 0, which is not in Q}.
    \item \mbox{Proof: Assume p, q greater than 0. Then } p^2 < 5q^2 \Leftrightarrow p < \sqrt{5} q \Leftrightarrow s = sup E = p/q = \frac{1}{\sqrt{5}} \mbox{how do I prove the infimum part. Hmm...}
    \item 2. ~~~~ \mbox{inf E = 0, sup E = 2 (assume 0 in N). }
  2. jcsd
  3. Feb 2, 2010 #2


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    p/q < √5, not 1/√5.
  4. Feb 2, 2010 #3
    Thanks, it is late. Woo silly mistakes. How do I prove this result, or is that sufficient? I mean, should I just plug in the typical supremum proof for this result?

    Thanks by the way
  5. Feb 2, 2010 #4


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    Yes, you can prove this from the definition of the supremum and by using the fact that between every two distinct real numbers you can find a rational number.
  6. Feb 2, 2010 #5


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    You mean [itex]\sqrt{5}[/itex] not, [itex]1/\sqrt{5}[/itex], but, more importantly, [itex]\sqrt{5}[/itex] is NOT in E since it is not a rational number. 0, on the other hand, is in E: take p= 0, q= 1. And it is the infimum because it is a lower bound (there are no negative numbers in E) while for any n, 1/n, is in E (take p= 1, q= n).

    You had better not "assume 0 in N" because then you are dividing by 0 when n=0.

    When n= 1, 2-(-1)1/12= 3 so 2 is not an upper bound on this set.
    Last edited by a moderator: Feb 2, 2010
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