Proving $\sup\left\{ r\in\mathbb{Q}:r^{2}<3\right\}=\sqrt{3}$ 
Let $E=\left\{ r\in\mathbb{Q}:r^{2}<3\right\}$.  Prove that $\sup E=\sqrt{3}$.

Since $E$  is bounded from above by $\sqrt{3}$  and is nonempty, $\alpha:=\sup E$  must exist by the Least Upper Bound Principle. Now I am stuck. Suppose $\alpha<\sqrt{3}$. Then what?
Edit: It just hit me.  If $\alpha < \sqrt{3}$ then since the rationals are dense in $\mathbb{R}$, there exists $q\in\mathbb{Q}$ such that $\alpha < q <\sqrt{3}$ and hence $\alpha^2 <q^2 <3$ which contradicts the fact that $\alpha$ is the sup of $E$
 A: Let $r\in \bf E$ , $r = \dfrac m n $.
Then we need to prove that there always exists a $\dfrac{p}{q}$ such that
$$ \frac{m}{n}<\frac p q < \sqrt 3$$
By exploting the special properties of $\sqrt 3 $,we can do it:
We start with 
$$\eqalign{
  & r < \sqrt 3   \cr 
  & r + 1 < \sqrt 3  + 1  \cr 
  & \frac{1}{{r + 1}} > \frac{1}{{\sqrt 3  + 1}}  \cr 
  & \frac{1}{{r + 1}} > \frac{{\sqrt 3  - 1}}{2}  \cr 
  & \frac{{r + 3}}{{r + 1}} > \sqrt 3  \cr} $$
But now we've gotten a number greater than $\sqrt 3 $. So what we'll do is apply the process again, and the relation will be reversed:
$$\frac{{Q + 3}}{{Q + 1}} < \sqrt 3 $$
Letting ${Q = \frac{{r + 3}}{{r + 1}}}$ will give
$$\frac{{2r + 3}}{{r + 2}} < \sqrt 3 $$
All we need to do now is prove that
$$r < \frac{{2r + 3}}{{r + 2}}$$
But since $r+2>0$ we get that
$$\eqalign{
  & {r^2} + 2r < 2r + 3  \cr 
  & {r^2} < 3 \cr} $$
which is true by hypothesis. Thus, given any rational $r$, there exists another rational $q$ such that
$$r<q<\sqrt 3$$
where $$q = \frac{{2r + 3}}{{r + 2}}$$
Just to make things clear, I'll add some extra information. 
The recursion defined as $r_0=1$ $$r_{n+1}=\frac{2 r_n+3}{r_n +2} $$ converges monotonically (increasing) to $\sqrt 3 $. This in particular means that given any $\epsilon>0$ there is an $r \in \rm E$ such that $$\sqrt 3 -\epsilon < r$$ which means $\sqrt 3 $ is the supremum of the set. 
This also plays the role of showing that $3$ is irrational.
