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** $\displaystyle S\equiv \{x\mid x\in\mathbb{R},\,x\geq0,\,x^2<c\}$**.**

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

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

**c.** Show that if $\displaystyle b^2<c$, then we can choose a suitably small positive number $\displaystyle r$ such that $\displaystyle b+r$ belongs to $\displaystyle S$, thus contradicting the choice of $\displaystyle b$ as an upper bound of $\displaystyle S$.

**d.** Use parts (b) and (c) and the positivity axioms for $\displaystyle \mathbb{R}$ to conclude that $\displaystyle b^2=c$.

The problem was asked previously in this thread: http://mathhelpforum.com/discrete-ma...per-bound.html.

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 $\displaystyle b^2\leq c$$ and $\displaystyle c\leq b^2$, respectively. So it seems obvious that $\displaystyle b^2$ must equal $\displaystyle c$, but I am confused by the reference to the positivity axioms. As stated in the books, they are:

1. If $\displaystyle a$ and $\displaystyle b$ are positive, then $\displaystyle ab$ and $\displaystyle a+b$ are also positive.

2. For a real number $\displaystyle a$, exactly one of the following three alternatives is true: $\displaystyle a$ is positive, $\displaystyle -a$ is positive, $\displaystyle a=0$.

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.

Re: Least upper bound and positivity axioms

Quote:

Originally Posted by

**Ragnarok** . As stated in the books, they are:

1. If $\displaystyle a$ and $\displaystyle b$ are positive, then $\displaystyle ab$ and $\displaystyle a+b$ are also positive.

2. For a real number $\displaystyle a$, exactly one of the following three alternatives is true: $\displaystyle a$ is positive, $\displaystyle -a$ is positive, $\displaystyle a=0$.

How do these allow us to conclude that $\displaystyle b^2=c$? Sorry for the involved question, it's probably really simple but I have no one to ask.

$\displaystyle b^2\le c$ means $\displaystyle c-b^2$ is positive or zero.

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

Therefore, the only possible alternative is that $\displaystyle b^2-c=0$ or $\displaystyle b^2=c$.

Re: Least upper bound and positivity axioms

Ah! Got it! Thanks so much again, Plato.

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 $\displaystyle r$ such that $\displaystyle 2rb-r^2<d$, since the concept of limit has not been developed yet.

Quote:

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

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