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.

Suppose that $x$ and $y$ are rational numbers such that $x^2<y$. For each rational $h$ with $0<h<1$, $$(x+h)^2=x^2+2xh+h^2<x^2+h(2|x|+1).$$
If $h$ is chosen such that $h\leq\dfrac{y-x^2}{2|x|+1}$ (e.g., $h$ could be taken to be equal to that quantity or $\frac12$, whichever is smaller), then $(x+h)^2<y$, and therefore $x+h$ is a rational number larger than $x$ whose square is still less than $y$.
