Let A be a set of positive rationals $p$ such that $p^2<2$. Now this set contains no upper bound. To prove this, for every rational $p$, a number $p- \frac{p^2-2}{p+2}$ is associated. This number (let's call it $q$) is greater than $p$. Also, it can be proved that $q^2<2$. Now the proof is complete.
I don't know if this is the only proof of this theorem but anyway how could one come up with that number to be associated with $p$. I am interested in the thought process behind.
First prove $\sqrt{2} \notin \mathbb{Q} $ then let $T:=\{p \in \mathbb{Q} : p^2<2\}$
Then prove that $\sup T = \sqrt{2} $ .Finally $\sup T \notin \mathbb{Q}$
$\mathbb{Q}$ is not complete.