Let $\displaystyle A$ be the set of all positive rationals $\displaystyle p$ such that $\displaystyle p^2 < 2$ and let $\displaystyle B$ consist of all positive rationals $\displaystyle p$ such that $\displaystyle p^2 > 2$. We want to show that there is no maximum element in $\displaystyle A$ and no minimum element in $\displaystyle B$. So:
$\displaystyle q = p - \frac{p^{2}-2}{p+2} = \frac{2p+2}{p+2}$. Then $\displaystyle q^{2} -2 = \frac{2(p^{2}-2)}{(p+2)^{2}}$.
First of all, how do we even know to do this? Just by intuition? Why did we set $\displaystyle q = p - \frac{p^{2}-2}{p+2} = \frac{2p+2}{p+2}$? Could we have set it to something else? So we basically have shown that $\displaystyle p < q$ for every $\displaystyle p$ in $\displaystyle A$, and $\displaystyle q < p$ for every $\displaystyle p$ in $\displaystyle B$. By the way, this is from Rudin's "Principles of Mathematical Analysis."
2. Suppose that $\displaystyle a \in A$ then let $\displaystyle b = \frac{{4a}}{{a^2 + 2}}$. Now show that $\displaystyle b \in A\quad \& \quad a < b$.