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?


Claim: If $q \in \mathbb Q$ then $q \in A$ if and only if $-\sqrt 5 < q < \sqrt 5$.

Proof:
If $q \in A$ then $q = \frac nm$ for some integers $n,m$ with $n^2 < 5m^2$.  That means $q^2 = (\frac nm) = \frac {n^2}{m^2} < 5$ which means $|q| < \sqrt 5$ which means $-\sqrt 5 < q < \sqrt 5$.
If on the other hand if $-\sqrt 5 < q < \sqrt 5$, the if we let $q = \frac nm; n,m\in \mathbb Z$ then $-\sqrt 5 < q < \sqrt 5$ so $q^2 = \frac {n^2}{m^2} < 5$.  So $n^2 < 5m^2$.  So $q \in A$.
.....
Now.... can you use the definitions of $\inf$ and $\sup$ to determine and prove what $\inf A, \sup A$ are?

 If $q \in A$ then $q < \sqrt 5$.  So $\sqrt 5$ is an upper bound of $A$.
 If $w < \sqrt 5$ then by the archimedian principal there exists a rational number $r$ so that $w < r < \sqrt 5$.  So as $r \in A$. ... well, it is if $r > -\sqrt 5$.  .... Let's do this again.

 If $w< \sqrt 5$ then either $w < 2$ or $2 \le w < \sqrt 5$.  If $w < 2$ then $w$ is not an upper bound of $A$ because $2 \in A$ and $w < 2$.  If $2 \le w < \sqrt 5$ then there exists a rational $r$ so that $w < r < \sqrt 5$.  So $-\sqrt 5 < r < \sqrt 5$ so $r \in A$. But $r > w$ so $w$ is not an upper bound of $r$.  So either way... if $w < \sqrt 5$ then $w$ is not an upper bound $r$.

 So we have $\sqrt 5$ is an upper bound of $A$; and if $w < \sqrt 5$ then $w$ is not an upper bound.  And that is the definition that $\sup A = \sqrt 5$.

