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This question already has an answer here:

Let A be a set of positive rationals $p$ such that $p^2<2$. Now this set contains no upper bound. To prove this, for every rational $p$, a number $p- \frac{p^2-2}{p+2}$ is associated. This number (let's call it $q$) is greater than $p$. Also, it can be proved that $q^2<2$. Now the proof is complete.

I don't know if this is the only proof of this theorem but anyway how could one come up with that number to be associated with $p$. I am interested in the thought process behind.

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marked as duplicate by Cameron Buie, pjs36, Claude Leibovici, user147263, Alex M. Oct 14 '15 at 6:42

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First prove $\sqrt{2} \notin \mathbb{Q} $ then let $T:=\{p \in \mathbb{Q} : p^2<2\}$

Then prove that $\sup T = \sqrt{2} $ .Finally $\sup T \notin \mathbb{Q}$

$\mathbb{Q}$ is not complete.

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    $\begingroup$ I think their main question is about the number being called $q$. $\endgroup$ – pjs36 Oct 14 '15 at 2:23

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