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This feels like a simple question but I am having a sort of writer's block with it.

Show that $L = \{x \in \mathbb{Q}: x^2 < 2 \text{ or } x < 0\}$ has no maximal element.

Any help will be appreciated.

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    Hint: Given a rational number $r$ such that $r^2<2$, there is a rational number $s$ such that $r^2 – 2017-01-17
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    @GeorgeLaw but $s^2$ may be larger than 2. And $\sqrt s$ need not be rational. – 2017-01-17
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    @GeorgeLaw You mean $r^2 – 2017-01-17
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    I'm assuming we are not allowed to assume anything about the existence of irrational reals? Other wise this is easy as for all rational r; $0 < r < \sqrt{2}$ there is a rational s; $0 < r < s < \sqrt{2}$ and $r^2 < s^2 < \sqrt{2}^2$. But usually this problem is an argument for the introduction of the Real numbers. In which case, lhf is the standard (albeit for me arithmetically difficult) answer. – 2017-01-17
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    @AkivaWeinberger ... which would just be a restatement of what has to be proved. – 2017-01-17
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    @AkivaWeinberger Yes, I meant that, thanks. – 2017-01-17

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Hint: It is enough to consider $x>0$ such that $x^2< 2$. Find $h \in (0,1)$ such that $(x+h)^2<2$. You'll see that $h$ is rational when $x$ is rational, and so is $x+h$.

Solution:

It is enough to solve $x^2+2xh+h \le 2$ because then $(x+h)^2=x^2+2xh+h^2 < x^2+2xh+h \le 2$, since $h^2 < h$. Solving $x^2+2xh+h = 2$ gives $h = \dfrac{2-x^2}{1+2x} > 0$, which is rational when $x$ is rational.

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    THank you! I will try to solve this without looking at the solution. – 2017-01-17
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For any $x$ in your set, we need to find another element greater than $x$ also in the set. For nonpositive $x$ we can take $0$; otherwise, we can take:$$f(x)=x+\frac{2-x^2}{100}$$In fact, for any rational $q$ with $q>2\sqrt2$, the formula $x+\frac{2-x^2}q$ works. I'm just playing it safe.
It's clear that $f(x)>x$ if $x^2<2$. In addition, the quadratic is increasing for all $x$ less than $q/2=50$ (since that's where the vertex of the parabola is), so $x<\sqrt2$ implies $f(x)