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Math Help - Least upper bound and positivity axioms

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    Least upper bound and positivity axioms

    I have a rather simple question here which however I feel needs a lot of background info to ask properly. So I apologize in advance!

    I'm self-studying with Fitzpatrick's Advanced Calculus and am working on a problem from the very first section, where no calculus has been introduced yet:

    Define S\equiv \{x\mid x\in\mathbb{R},\,x\geq0,\,x^2<c\}.
    a. 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.
    b. Show that if b^2>c, then we can choose a suitably small positive number r such that b-r is also an upper bound for S, thus contradicting the choice of b as the least upper bound of S.
    c. Show that if b^2<c, then we can choose a suitably small positive number r such that b+r belongs to S, thus contradicting the choice of b as an upper bound of S.
    d. Use parts (b) and (c) and the positivity axioms for \mathbb{R} to conclude that b^2=c.

    The problem was asked previously in this thread: Upper Bound.

    From the hints on that thread I think I have answered parts a-c, however, I have a question on part d. Parts b and c showed that b^2\leq c$ and c\leq b^2, respectively. So it seems obvious that b^2 must equal c, but I am confused by the reference to the positivity axioms. As stated in the books, they are:

    1. If a and b are positive, then ab and a+b are also positive.
    2. For a real number a, exactly one of the following three alternatives is true: a is positive, -a is positive, a=0.

    How do these allow us to conclude that $b^2=c$? Sorry for the involved question, it's probably really simple but I have no one to ask.
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    Re: Least upper bound and positivity axioms

    Quote Originally Posted by Ragnarok View Post
    . As stated in the books, they are:

    1. If a and b are positive, then ab and a+b are also positive.
    2. For a real number a, exactly one of the following three alternatives is true: a is positive, -a is positive, a=0.

    How do these allow us to conclude that b^2=c? Sorry for the involved question, it's probably really simple but I have no one to ask.
    b^2\le c means c-b^2 is positive or zero.

    c\le b^2 means b^2-c=-(c-b^2) is positive or zero.

    Therefore, the only possible alternative is that b^2-c=0 or b^2=c.
    Thanks from Ragnarok
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    Re: Least upper bound and positivity axioms

    Ah! Got it! Thanks so much again, Plato.
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    Re: Least upper bound and positivity axioms

    Since people here are so helpful, I thought maybe I'd ask to make sure I have parts b and c right. This is my proof for part b; c is similar. I guess where I get confused is what allows us to assume that we can always choose an r such that 2rb-r^2<d, since the concept of limit has not been developed yet.

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

    Suppose b^2>c. Then b^2 can be written as c+d for some positive number d. Then  (b-r)^2 = b^2-2rb+r^2 = c+d-2rb+r^2. So (b-r)^2 will be larger than c when d-2rb+r^2>0, i.e. when 2rb-r^2<d. So if we choose r such that 2rb-r^2<d, then we get (b-r)^2>c. Then by the definition of S, x<b-r for all x\in S. So b-r is an upper bound for S which is a smaller upper bound than b. Since supposing that b^2>c contradicts our assumption that b=\text{sup }S, we must have b^2\leq c.
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