1. ## Rational Numbers

Let $A$ be the set of all positive rationals $p$ such that $p^2 < 2$ and let $B$ consist of all positive rationals $p$ such that $p^2 > 2$. We want to show that there is no maximum element in $A$ and no minimum element in $B$. So:

$q = p - \frac{p^{2}-2}{p+2} = \frac{2p+2}{p+2}$. Then $q^{2} -2 = \frac{2(p^{2}-2)}{(p+2)^{2}}$.

First of all, how do we even know to do this? Just by intuition? Why did we set $q = p - \frac{p^{2}-2}{p+2} = \frac{2p+2}{p+2}$? Could we have set it to something else? So we basically have shown that $p < q$ for every $p$ in $A$, and $q < p$ for every $p$ in $B$. By the way, this is from Rudin's "Principles of Mathematical Analysis."

2. Suppose that $a \in A$ then let $b = \frac{{4a}}{{a^2 + 2}}
$
. Now show that $b \in A\quad \& \quad a < b$.