# Least upper bound and positivity axioms

• May 7th 2013, 10:53 AM
Ragnarok
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. • May 7th 2013, 11:37 AM Plato 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$\displaystyle b^2=c$. • May 7th 2013, 07:20 PM Ragnarok Re: Least upper bound and positivity axioms Ah! Got it! Thanks so much again, Plato. • May 7th 2013, 07:40 PM Ragnarok 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\$.