Real Analysis Beginning Proof... Alright, I've been assigned to work through a proof in my RA course and it just has me bogged down at this point. We're trying to show that If $b^2 > c$ then there exists a positive real number $r$ such that $b-r>c$. This is an intermediate step in showing that the set $S=\\{x|x \in \mathbb{R}, x \geq 0, x^2<c\\}$ - essentially that square roots exist. Thus I can't use anything concerning square roots in my proof. Obviously I can't choose $r<b-c$ because I can't show that $b-c$ is positive. I think what I need to choose is some $r$ such that $(b-r)^2 > c$ however I'm just not seeing the how to find this $r$. Any help on moving forward/direction with this problem is very much appreciated.

After showing it is bounded above, you get a supremum, call it $b$, exists. What you want is to show neither $b^2c$ can hold, so it must be the case $b^2=c$, that is $\sqrt b=c$. One can always use Archimedianity. The claim is we may take $r=n^{-1}$, $n$ a natural number we'll specify in what follows. You want $\left(b-\dfrac 1 n \right)^2>c $. This is $b^2-\dfrac{2b}n+\dfrac 1 {n^2}>c$ or $b^2-c>\dfrac{2b}n-\dfrac 1{n^2}$. Now $b^2-c>0$, so there certainly exists $n$ such that $$n(b^2-c)>2b>2b-\frac 1n$$

Which gives what you want. You can do something completely analogous when assuming $b^2