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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.

Ajith
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    http://math.stackexchange.com/questions/14970/no-maximumminimum-of-rationals-whose-square-is-lessergreater-than-2 –  Oct 14 '15 at 02:21
  • Also: http://math.stackexchange.com/questions/1448575/intuition-behind-proof-in-the-rudin-book-that-there-is-no-largest-smallest-real and http://math.stackexchange.com/questions/141774/choice-of-q-in-baby-rudins-example-1-1 –  Oct 14 '15 at 02:23
  • What's your question. – fleablood Oct 14 '15 at 02:28
  • @Bungo's first link is really quite a good find for this exact question. – pjs36 Oct 14 '15 at 02:34

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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.

Pablo Herrera
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