# Prove there exists a real number $\alpha \in \mathbb{R}$ satisfying $\alpha^2 = 2$ by finding $\frac{1}{n_0} < \frac{\alpha^2 - 2}{2\alpha}$.

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.

## 1 Answer

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 $$\frac{-2\alpha}{n_1}>-(\alpha^2-2).$$

• Thank you so much! Commented Jun 5, 2023 at 20:58