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'sAdvanced Calculusand 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: 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 $\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.