Q:

Help explaining the existence of a square root - Proof

From stephen abbott's book, there is a section in the text where he aims to show
that
There exists a real number $α ∈ \mathbb{R}$ satisfying $α^2 =2$
Consider the set $T=\{t∈R:t^2<2\}$
Set $\alpha = \sup T$ 
He approaches the proof by first showing that $\alpha < 2$ violates the condition that $\alpha$ is an upper bound. I need help in understanding the following:
"
Assume $\alpha < 2$. In search of an element of $T$ that is larger than $\alpha$, write:
$(\alpha + \frac{1}{n})^2 = \alpha ^2 + \frac{2 \alpha}{n} + \frac{1}{n^2} < \alpha^2 + \frac{2 \alpha + 1}{n}$
"
What justifies the choice of writing $(\alpha + \frac{1}{n})^2$? If I had to guess I'd say it's somehow related to the Archimedean property , but really I am unsure.
To continue on with my query, abbott says
"But now assuming $\alpha^2< 2$ gives us a little space in which to fit the $(2\alpha +1)/n$ term  and keep the total less than 2. Specifically, choose $n_0 \in \mathbb{N}$ large enough so that $ \frac{1}{n_0} < \frac{2- \alpha^2}{2\alpha + 1}$
Why this final inequality? How is it even formulated. Again it looks like it may have to do with the Archimedean property, but I am unsure. If someone could shed light onto these two issues, I'd appreciate it.
Thanks.

A:

In the authors proof of $\alpha^2 = 2$, he defines $\alpha$ to be the supremum of the set of numbers which have a square smaller than 2.
He now writes "In search of an element of $T$ that is larger than $\alpha$", since any number of the form $\alpha + 1/n$ is larger than $\alpha$, this is an appropiate choice.

The inequality
$$\frac{1}{n_0} < \frac{2- \alpha^2}{2\alpha + 1}$$
was formed when the author got the idea of keeping the total less than 2, here the total refers to $(\alpha + 1/n)^2$
$$\alpha^2+\frac{2\alpha+1}n<2$$
then solving for $1/n$ and renaming $n$ to $n_0$
$$\frac{2\alpha+1}{n_0}<2-\alpha^2$$
$$\frac{1}{n_0}<\frac{2-\alpha^2}{2\alpha+1}$$

