I am working out of Advanced Calculus Second Edition by Patrick M. Fitzpatrick (page 9) :
Proposition 1.3 Let $c$ be a positive number. Then there is a (unique) positive number whose square is $c$; that is, the equation
$$x^2 = c, \ \ x > 0$$
has a unique solution.
The question defines $$ S\equiv\{x|x\in\mathbb{R}, x\geq 0\text{ and }x^2<c\}. $$ It then breaks the proof into four parts. Part a was establishing a least upper bound by showing that $c+1$ was an upper bound. This part wasn't too bad.
The second part says: 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 $S$. I am struggling to argue this precisely. Any help or direction is greatly appreciated.
Suppose that $b^2>c$. You wnat to prove that there's a $r>0$ such that $(b-r)^2>c$. Note that$$(b-r)^2>c\iff b^2-2br+r^2>c\iff b^2-c>2br-r^2.$$So, pick $r>0$ such that $2br<b^2-c\left(\iff r<\frac{b^2-c}{2b}\right)$. Then, since $-r^2<0$, it will be automatically true that $2br-r^2<b^2-c$.