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I'm currently self-studying from the book Understanding Analysis by Stephen Abbott. The book offers the proof that $\alpha^2 < 2$ cannot be the case and as an exercise asks to prove that $\alpha^2 > 2$ can also not be the case. The proof that $\alpha^2 < 2$ cannot be the case makes perfect sense to me but I can't seem to understand the logic of the inequalities for $\alpha^2 > 2$ not being the case.

The argument in the instructor's solution manual is as follows: Consider the set $T \subseteq \mathbb{R}$ where $T = \{t \in \mathbb{R}: t^2 < 2\}$ and set $\alpha = sup \space T$. Suppose $\alpha^2 > 2$. To find a number smaller than $\alpha^2$ $$\left(\alpha - \frac{1}{n}\right)^2 = \alpha^2 - \frac{2a}{n} + \frac{1}{n^2} > a^2 - \frac{2a}{n}$$ Choose an $n_1 \in \mathbb{N}$ large enough so that $$\frac{1}{n_1} < \frac{\alpha^2 - 2}{2\alpha}$$ then $$\frac{2\alpha}{n_1} < a^2 - 2$$ $$\left(\alpha - \frac{1}{n_1}\right)^2 > a^2 - \frac{2a}{n_1} = \alpha^2 - (\alpha^2 - 2) = 2$$ Resulting in $$\left(\alpha - \frac{1}{n_1}\right)^2 > 2$$ which means $\alpha$ would not be the least upper bound.

But I can't seem to understand how $$\frac{2\alpha}{n_1} < a^2 - 2$$ implies $$a^2 - \frac{2a}{n_1} = \alpha^2 - (\alpha^2 - 2).$$ I would think $$a^2 - \frac{2a}{n_1} < \alpha^2 - (\alpha^2 - 2).$$ would be correct but with this I don't think it follows that $$\left(\alpha - \frac{1}{n_1}\right)^2 > 2.$$

If we set $$\frac{1}{n_1} > \frac{\alpha^2 - 2}{2\alpha}$$ everything would then make sense to me but my understanding is that the Archimedean property only allows us to pick 1/n < y.

Please, I would be very grateful if anyone could clarify the logic I'm missing.

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Yes, it should be $$\left(\alpha - \frac{1}{n_1}\right)^2 > \alpha^2 - \frac{2\alpha}{n_1} > \alpha^2 - (\alpha^2 - 2) = 2.$$

This comes from $$\frac{2\alpha}{n_1} < \alpha^2-2$$ which implies $$\frac{-2\alpha}{n_1}>-(\alpha^2-2).$$

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  • $\begingroup$ Thank you so much! $\endgroup$ Commented Jun 5, 2023 at 20:58

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