Let E = $\{x\in \mathbb{Q}: x>0$ and $x^2 <2\}$. Prove E does not have a largest element. I'm pretty sure the way to go is with a contradiction, letting $x = \sup E$. Then I think I need to either show that $x < x^2\in E$ (contradicting $x = \sup E$ in that an element of E is greater than x) or show that $({x^2})^2 > 2$ (contradicting the terms of E). As I wrote this I realized showing $({x^2})^2 > 2$ would be redundant since that would only mean something if $x^2\in E$ in the first place... anyway, I'm not sure how to go about this proof.

For every $p \in E$, consider
$$q = p - \frac{p^2 - 2}{p + 2} = \frac{2p + 2}{p + 2}$$
Note that, $q > p$. Now,
$$q^2 - 2 = \frac{2(p^2 - 2)}{(p+2)^2}$$
Therefore, $q^2 < 2$. This implies $q \in E$.
Therefore, for every $p \in E$, we can find a $q > p$ such that $q \in E$. This implies that E has no largest element.
(This is a proof basically from Rudin)
