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Show Set is Bounded

  1. May 5, 2007 #1
    Hi everyone. I'm a math student still learning to do
    proofs. Here is a problem I encountered that seems easy but
    has me stuck.

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

    Let a be a positive rational number. Let A = {x e Q (that
    is, e is an element of the rationals) | x^2 < a}. Show that
    A is bounded in Q. Find the least upper bound in R of this
    set.

    2. Relevant equations
    None.

    3. The attempt at a solution
    So I want to show that there exists an M such that x < or =
    to M for all x in A.
    So for all x in A, x^2<a.
    => x < (a/x) if x>0 or x > (a/x) if x < 0
    So it seems like I find an upper bound for x if x is
    positive and a lower bound for x if x is negative but what
    havn't acounted for the other cases.

    Thanks for your help.
     
    Last edited: May 5, 2007
  2. jcsd
  3. May 5, 2007 #2

    quasar987

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    Gold Member

    It does not suffice to find an individual bound for all the elements of A. You must find ONE bound that fits ALL the elements of A.

    I suggest you treat the two cases a< or =1 and a>1 and find a bound for A is both cases.
     
  4. May 5, 2007 #3
    So the set seems to be bounded by -1,1 if a<1 and sqrt(a) and -sqrt(a) if a>1. But I got this through taking square roots, which arn't there when x is an element of the rationals.

    Hm, I guess I have to think about this some more.
     
  5. May 6, 2007 #4

    quasar987

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    why not just use a if a>1 ?
     
  6. May 6, 2007 #5

    HallsofIvy

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    Staff Emeritus
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    Notice that you are also asked to find the least upper bound in R. That should be obvious.
     
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