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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.

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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 $y$.

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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)

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  • $\begingroup$ How is this answer derived? If I saw this on an exam for the first time I would never come up with that. $\endgroup$ – JohanLiebert Aug 27 '13 at 5:29
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    $\begingroup$ @SeniAdeyemi: I recommend following the links in my comment for more information on this method from Rudin's book. $\endgroup$ – Jonas Meyer Aug 27 '13 at 5:48

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