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Math Help - Upper Bound

  1. #1
    MHF Contributor harish21's Avatar
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    Upper Bound

    S = {x | x in R , x>=0, x^2 < c}

    a. Show that c+1 is an upper bound for S and therefore, by completeness axiom, S has a least upper bound that we denote by b.

    b. Show that if b^2>c, we can show choose a suitably positive number r such that b-r is also an upper bound for S, thus contradicting the choice of b as an upper bound.

    c. If b^2< r, then we can choose a suitable positive number r such that b+r belongs to S, thus contradicting the choice of b as an upper bound of S.

    All I could do here was to show that
    if c<=1, then x<1<1+c. So 1+c is an upper bound for S
    and
    if c>1, x<Sqrt(c)<c<1+c. So 1+c is an upper bound for S.

    How is the rest of part(s) proved?!
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  2. #2
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    b) If b^2 > c, b^2 = c + d for some positive number d.

    Now (b-r)^2 = b^2 - 2rb + r^2 = c + d - 2rb + r^2

    Note that if d - 2rb + r^2 > 0, then we have (b-r)^2 > c.

    Since \lim_{r \to 0} r^2 - 2rb = 0, there exists a positive r such that r^2 - 2rb < d.
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  3. #3
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    Suppose that \left( {\exists y \in S} \right)\left[ {c + 1 < y} \right] then (c+1)^2<y^2<c^2. WHY?
    What is wrong with that picture?
    How does that contradiction show that c+1 is an upper bound on S?
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