E = {p/q ∈ Q : $p^2$ < $5q^2$ and p, q < 0}
I know that inf(E) = 0 and sup(E) = sqrt(5), but I'm struggling to prove this in any real way. I can, of course, articulate why it's the case, but simply explaining how a very large q yields the infimum or how a bit of algebra yields the supremum isn't quite rigorous enough.
Let's try to prove $\sup(E)=\sqrt{5}$. It is clear that $\sqrt{5}$ is a upper bound for $E$. We now want to show that it is the least upper bound. In other words, we need to show that for all $\epsilon>0$, there exists $p/q\in E$ with $p/q>\sqrt{5}-\epsilon$, i.e. showing that $$E \cap(\sqrt{5}-\epsilon,\sqrt{5})\neq \varnothing.$$ Now, notice that $\sqrt{5}-(\sqrt{5}-\epsilon)=\epsilon>0$. Thus, by Archimedean property of real numbers, there exists an integer $n$ large enough so that $$n(\sqrt{5}-(\sqrt{5}-\epsilon))>1\iff n\sqrt{5}>n(\sqrt{5}-\epsilon)+1.$$ But then this implies $n\sqrt{5}$ and $n(\sqrt{5}-\epsilon)$ are two numbers that are more than distance $1$ apart, so there exists an integer $m\in (n(\sqrt{5}-\epsilon),n\sqrt{5})$, giving us $$n\sqrt{5}>m>n(\sqrt{5}-\epsilon).$$ Dividing both sides by $n$ we get $$\sqrt{5}>\frac{m}{n}>\sqrt{5}-\epsilon.$$ So we conclude that indeed $\sup(E)=\sqrt{5}$.
$\inf(E)=0$ can be proved with a similar strategy.