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Let $E=\{q\in\mathbb{Q}|q^2<2\}\subseteq\mathbb{Q}$. If $\alpha$ is an upper bound for $E$ in $\mathbb{Q}$, then why does it need to satisfy $\alpha^2>2$?

The context of the question: I am reading a proof that $\mathbb{Q}$ has no least upper bound and this is the argument used in the proof. Can you have a look at http://www.math.ubc.ca/~feldman/m320/qsup.pdf? (page 2, step 2) I do not understand why we need to make $\widetilde{q}^2>2$?

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    Because it cannot satisfy $\alpha^2 < 2$ (you can do this by explicit construction of a counterexample) or $\alpha^2 = 2$ (infinite descent proof) – 2017-02-14
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    Example 1.1 in Rudin's Principles of Mathematical Analysis goes over this. – 2017-02-14
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    What is the reason for a downvote? this is not a homework question; it is a part of the argument of a proof that I need to understand. Please see http://www.math.ubc.ca/~feldman/m320/qsup.pdf (step 2, page 2) – 2017-02-14

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If $\alpha^2< 2$, it lies in $E$ and thus you can find a rational number greater than $\alpha$ which also lies in $E$ by density of rationals thus disproving $\alpha $ to be an upper bound.

If $\alpha^2=2$, $\alpha=\sqrt{2}$ which you can prove to be irrational. (Assume it be rational and thus $\alpha=p/q$ and proceed)

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    Please see my edited question. – 2017-02-14
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    $\tilde{q}^2>2$ implies that it is an upper bound and as we chose it in such way that it was less than $\alpha$, proves that $\alpha$ is not l.u.b. – 2017-02-14
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    My question is why we need $\tilde{q}^2>2$? – 2017-02-14
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    First step was to choose $\tilde{q}=\alpha-\epsilon$ so that we have a number smaller than $\alpha$. Then if we could show that $\tilde{q}$ is positive and $\tilde{q}^2>2$, it would prove that $\tilde{q}$ is also an upper bound which is smaller than $\alpha$. In short, $\tilde{q}^2>2$ together with $\tilde{q}>0$ was needed to show that $\tilde{q}$ is an upper bound. – 2017-02-14
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you have 3 possibilities: less than, equal or greater than

Equal is out since root 2 is irrational.

Let $x$ be rational with $x^2<2.$ Let $y= 2-x^2.$ S $y>0.$ Now consider $$ x_n = x+ n^{-1}. $$ $$ x_{n}^{2} = x^2 + 2 x/n + {1/n^2} = 2 - y + 2 x/n + n^{-2} $$ So $$ 2-x_{n}^{2} = y - 2x/n - n^{-2} > y - 4/n - 1/n = y - 5/n. $$ Let $n$ be bigger than $5y.$ We have $$ 2-x_{n}^{2} > 0. $$ So $x_{n}^{2} < 2$ and bigger than $x$. So $x$ is not an upper bound.

The only case left is $x^2 >2.$

(see my book "proof patterns" for more discussion)

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    please see my edited question. – 2017-02-14
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    I think my answer answers the edited question. – 2017-02-14