In the proof for existence of $$\alpha^2 = 2$$
following steps are used:
Choose $\alpha$ as upper bound for the set T
$\{\alpha \in T : \alpha^2 < 2\}$
$$(\alpha + 1/n)^2 = \alpha^2 + 2\alpha/n + 1/n^2$$
$$< \alpha^2 + 2\alpha/n + 1/n$$
$$= \alpha^2 + (2\alpha + 1)/n$$
Choose $n_0$ such that
$$1/n_0<(2-\alpha^2)/(2\alpha + 1)$$
…
which goes on to show that $\alpha$ wasn't the upper bound.
My question is how did we end up choosing that specific $n_0$, because anyone can see that, that particular choice of $n_0$ goes on to only prove that $\alpha$ wasn't the upper bound.
What was the inspiration to choose that particular $n_0$
This is from Stephen Abbott's Understanding Analysis Chapter 1
Best Answer
This is really a proof by contradiction. We will assume that $\alpha^2 <2$ and derive a contradiction. That contradiction will be to prove that for small enough $n$, $$ (\alpha+1/n)^2 <2 $$ This will contradict $\alpha$ being the sup of this set. Choosing the value for $n$ now is done in reverse. We know we want to end up with a $2$ because that's what the structure of the proof suggests. Looking at the last line, we have $$ \alpha^2 + (2α+1)/n $$ so 1/n had better cancel the $(2α+1)$ on top. Then we want it to cancel the $\alpha^2$ and leave us with a $2$. Putting all of these together, we end up with exactly the inequality given, that is $$ 1/n_0<\frac{2-\alpha^2}{2(\alpha+1)} $$