<s>[INST] There exists an $\alpha \in R$, with $\alpha^2=2$. Reading Abott's Understanding Analysis. I'm trying to understand each part of this proof that is presented for a theorem. I'll only state the first part of it since it is a division of cases and I'm just confused about the first part.
Theorem 1.4.5. There exists a real number $\alpha \in R$ satisfying $\alpha^2=2$.
$\textbf{Proof}$. Consider the set
$T=\{t\in R : t^2 \lt 2 \}$
and set $\alpha = \sup A$. We are going to prove $\alpha^2 = 2$ by ruling out the possibilities $\alpha^2 \lt 2$ and $\alpha^2 \gt 2$. Lets see what happens if we assume $\alpha^2 \lt 2$. In search of an element of $T$ that is larger than $\alpha$, write
$(\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2}{\alpha} + \frac{1}{n^2}$
So far, I think I get what is going on. We are using the Archimedean Property to find a number that is greater than $\alpha$, but less than 2 (If we have a real number $\alpha$, there exists a natural number such that $\frac{1}{n} \lt \alpha$. We know $\frac{1}{n}$ is smaller than $\alpha$ so we add them in search of $t\in T$, such that t $\gt \alpha$.).
$\textbf{Proof cont.}$
\begin{align*}
(\alpha + \frac{1}{n})^2 &= \alpha^2 + \frac{2}{\alpha} + \frac{1}{n^2} \\
&\lt \alpha^2 + \frac{2}{\alpha} + \frac{1}{n} \\
&= \alpha^2 + \frac{2\alpha + 1}{n} 
\end{align*}
This is where I get a little confused. I understand that $(\alpha + \frac{1}{n})^2 \lt \alpha^2 + \frac{2}{\alpha} + \frac{1}{n}$, but why is this necessary to say? Wouldn't it suffice to say $(\alpha + \frac{1}{n})^2 = \alpha^2 + \frac{2}{\alpha} + \frac{1}{n^2} \lt 2$? This is my question.
$\textbf{Proof cont.}$
But now assuming $\alpha^2 \lt 2$ gives us a little space in which to fit the $\frac{(2\alpha+1)}{n}$ and keep the total less than 2. Specifically, chose $n_0 \in N$ large enough so that
$\frac{1}{n_0} \lt \frac{2- \alpha^2}{2\alpha +1}$.
This implies $\frac{(2 \alpha +1)}{n_0} \lt 2 - \alpha^2$, and consequently that
$(\alpha + \frac{1}{n_0})^2 \lt \alpha^2 + (2- \alpha) = 2$ $\square$.
 [/INST] First of all,
$$ \left( \alpha + \frac1n \right)^2 = \alpha^2 + \frac{2\alpha}{n}+\frac{1}{n^2}.$$
Second, you need to prove that this is $<2$ for some $n$. It doesn't matter how you prove it, but you do need to prove it one way or another.
One way to prove this is to simplify:
$$ \frac{2\alpha}{n}+\frac{1}{n^2} \le \frac{2\alpha}{n}+\frac{1}{n} = \frac{2\alpha + 1}{n}$$
Because we know already that $c/n \to 0$ for any constant $c$.
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