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I don't get it

  1. Dec 15, 2005 #1

    Päällikkö

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    Let
    [tex]S = \{x | x \in \mathbb{R}, x \ge 0, x^2 < c\}[/tex]

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

    Pretty much the only tools I've got are the Field Axioms.
    I think I'm supposed to do something like:
    x2 [itex]\ge[/itex] 0. Thus c > 0.
    x2 < c < c + 1

    Thus c + 1 is an upper bound.

    By the Completeness axiom, S has a least upper bound that we denote by b.

    QED


    It can't be just this, can it? I'm totally lost in maths, these things were dealt with ages ago and I still can't quite grasp the logic.

    The part "Thus c + 1 is an upper bound" is where I think my logic fails. If this was the way (which I think is not the case) to prove c + 1 was an upper bound, couldn't we just have concluded that c is an upper bound, and thus b exists?
     
  2. jcsd
  3. Dec 15, 2005 #2

    arildno

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    Remember that c is NOT an upper bound of S if c<1!!

    But in the case of c<1, you'll get that x<1<1+c, i.e, 1+c is an upper bound for S

    If c>1, then [itex]x<\sqrt{c}<c<c+1[/itex] and thus, c+1 is an upper bound for S.
     
  4. Dec 15, 2005 #3

    HallsofIvy

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    No, you just proved that c+ 1 is larger than x2, not x.
    Since, for some x, x2< x, it does not follow that an upper bound on x2 is an upper bound on x.

    Try a proof by contradiction. Suppose c+ 1 is NOT an upper bound on S. That is, suppose there is x in S such that x> c+ 1. Now compare
    x2 and c.
     
  5. Dec 15, 2005 #4

    Päällikkö

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    I don't think I can do this, as I have yet to prove that there exists a c, so that x2 = c.

    How's this:
    x2 < c < (c+1)2
    Thus x < c+1

    Or by contradiction, maybe:
    x > c+1
    { x2 > (c+1)2
    { x2 < c
    (c+1)2 > c, so x > c+1 cannot hold.

    In an unassisted problem, how do I come up with the upper bound (like the c+1 here)? Just make one up?
     
  6. Dec 15, 2005 #5

    arildno

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    c is either less than one, equal to one, or greater than 1.
    You can break up your proof in special cases.
     
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