Proving the sup and the inf $A=\{\frac{n}{m}: n,m \in \mathbb{Z}, n^2<5m^2\}$. $A=\{\frac{n}{m}: n,m \in \mathbb{Z}, n^2<5m^2\}$.
My goal is to find and prove the $\inf$ and $\sup$ of $A$. I can't really use anything beyond the definition of inf and sup. I have been trying to find a way to generate integers $n,m$ for all $\epsilon>0$ such that the following is satisficed $\sqrt{5}-\epsilon<\frac{n}{m}<\sqrt{5}$.
How would you go about proving this?

Notice that for $\frac nm\in A$, we have $\frac{n^2}{m^2}<5$, so that $-\sqrt 5<\frac nm<\sqrt 5$. Thus, $-\sqrt 5$ and $\sqrt 5$ are lower and upper bounds of $A$, respectively.
Let's prove that $\sqrt 5$ is the lowest upper bound of $A$. Choose $\epsilon>0$ for which we wish to find $\frac nm\in A$ satisfying $\frac nm>\sqrt 5-\epsilon$. Since $\mathbb Q$ is dense in $\mathbb R$, there is some $\frac n m\in\mathbb Q_{\geq 0}$ for which $\sqrt 5-\epsilon<\frac nm<\sqrt 5$. One easily sees that $\frac nm\in A$. This shows that $\sqrt 5$ is the supremum of $A$.
Can you do the same for the infimum?
