Claim: There exists a number $\displaystyle \alpha$ such that $\displaystyle \alpha ^2 = 2$

Proof:

So, we look at the set $\displaystyle S = \{x \in R | x^2 < 2\}$. We want to consider $\displaystyle \alpha = sup(S)$. Claim that $\displaystyle \alpha ^2 = 2$, and prove by contradiction.

There are two possibilites: $\displaystyle \alpha ^2 < 2$ or $\displaystyle \alpha ^2 > 2$.

For $\displaystyle \alpha ^2 < 2$: Let $\displaystyle n \in R$ be large enough so that $\displaystyle n > \frac{2\alpha + 1}{2 - \alpha ^2}$ (<---Do not understand choice of n here). Then $\displaystyle \frac{1}{n} < \frac{2 - \alpha^2}{2\alpha + 1}$. Then

$\displaystyle (\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n^2} < \alpha^2 + \frac{2\alpha}{n} + \frac{1}{n} = \alpha^2 + \frac{2\alpha + 1}{n}$

So $\displaystyle \alpha^2 + \frac{2\alpha}{n} < \alpha^2 + (2\alpha + 1) \frac{2 - \alpha^2}{2\alpha + 1} = 2$

But then $\displaystyle \alpha + \frac{1}{n} \in S$, and $\displaystyle \alpha + \frac{1}{n} > \alpha$, contradicting the fact that $\displaystyle \alpha$ is an upper bound for S.

This isn't the complete proof. You then have to use P.B.C. for the other case, $\displaystyle \alpha > 2$. But, my problem is that I don't understand the choice of n here. Of course it makes sense looking at the proof, but how does one know how to choose n for any other problem, say, for $\displaystyle \alpha^3 = 4$?