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(This is a follow-up to Need help with proof for Dedekind cuts on $\mathbb{Q}^+$ posted December 23.)

I am still working on the same proof about Dedekind cuts on the positive rational numbers. Now I am stuck on the final step of the proof on the following point and would appreciate any help.

Given $x\in \mathbb{Q}^+$ such that $x<2$, how can I prove the existence of $y\in \mathbb{Q}^+$ such that $y^2 < \frac{x^2}{2}$ and $y^2 < 2$. I suspect the last requirement may be redundant.

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    $y=\frac{x}2$ works, but are you sure that this is what you actually want? – 2017-01-04
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    That should do it! Thanks. Post it as an answer and I will formally accept it. – 2017-01-04
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    You’re welcome; done! – 2017-01-04

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Probably the simplest way is to let $y=\frac{x}2$: then $y^2=\frac{x^2}4<\frac{x^2}2$.

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    Downvoting a correct answer that was posted at the explicit request of the OP? Tsk. – 2017-01-04
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    It seems your suspicion that I was asking the wrong question was correct. The inequality should have been $\frac{x^2}{2} < y^2 < 2$ where $x<2$. Do you have any suggestions on how I might prove the existence of such a $y\in \mathbb{Q}^+$? – 2017-01-05
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    @Dan: (Sorry to take so long, but it’s been a busy few days.) I’m going to have to give it quite a bit more thought; I don’t immediately see how to do it. – 2017-01-05
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    I've probably taken enough of your time, Brian. Someone at Quora posted an informal proof to related problem that looks promising. I haven't had a chance to look at the details yet. – 2017-01-07