I am reading a proof for the existence of $\sqrt 2$. The first half of the proof goes as follows:
Consider the set $T = \{t \in \mathbb{R} : t^2 \lt 2\}$. Let $\alpha = \sup T$.
Case 1: Show $\alpha^2 \lt 2$ is impossible by implying $\alpha$ is not an upper bound for $T$.
We want to find an element that is larger than $\alpha$. So
$(\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2 \alpha}{n} + \frac{1}{n^2}$
$\lt \alpha^2 + \frac{2 \alpha}{n} + \frac{1}{n} = \alpha^2 + \frac{2 \alpha + 1}{n}$
The author states:
But now assuming $\alpha^2 < 2$ gives us a little space in which to fit the $\frac{(2 \alpha + 1)}{n}$ term and keep the total less than 2. Specifically, choose $n_0 \in \mathbb{N}$ large enough so that
$ \frac{1}{n_0} \lt \frac{(2 – \alpha^2)}{2 \alpha + 1}$
This is where I get lost in the proof. I understand that his goal with this is that in order to preserve $a^2 \lt 2$ it must be the case the difference $2 – \alpha^2$ needs to be some fraction that doesn't push $a^2$ outside of $2$. That is, the "extra stuff" being added to $\alpha^2$ here can't push it over our assumed bound.
The confusion for me is understanding how he constructed $ \frac{1}{n_0} \lt \frac{(2 – \alpha^2)}{2 \alpha + 1}$ from the assumption $\alpha^2 < 2$. I had initially thought he rearranged the inequality so that $0 < 2 -\alpha^2$, but this isn't correct.
What exactly did he use to choose this? I took a look in the solution manual and he didn't really describe the methodology there either. It seems arbitrary – beneficial to the problem. I'm not exactly sure how it's constructed from the assumed inequality.
Best Answer
This is just goal-oriented work: The author wants $$\tag1\left(\alpha+\frac1n\right)^2<2.$$ Equivalently, $$\alpha^2+\frac{2\alpha}n+\frac1{n^2}<2. $$ Rearrange to $$\frac{2\alpha}n+\frac1{n^2}<2-\alpha^2. $$ This looks promising because, by assumption, the right hand side is positive and we only have to find $n$ that makes the left (positive and) small enough. As $\frac1n\le\frac1{n^2}$, we can boldly strengthen our task to find $n$ that makes (even) $$\frac{2\alpha}n+\frac1{n}<2-\alpha^2. $$ (The advantage of this is that the dependence on $n$ is now simpler than with the square part).
Now multiply with the positive(!) $n$ $$ 2\alpha+1<(2-\alpha^2)n$$ and divide by the positive(!) $2-\alpha^2$ to arrive at $$ \frac{2\alpha+1}{2-\alpha^2}<n$$ as a sufficient (and readily fulfilled) condition for $n$ to make $(1)$ true.