Rational Numbers

**tukeywilliams** · July 14th 2007, 03:25 PM

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."