I'm having trouble understanding this proof regarding Supremum. I am trying to understand this proof, I understand the general form of the argument, and I understand what they are trying to show, Im just having trouble following the steps they've used. if anyone could explain why they've done what they've done I'd be really grateful. Also, this is only half the proof, but hopefully if I can get this part ill be able to figure the rest out myself.
Define the set $A = \left \{ x\in \mathbb{R}:x> 0\:  and\: x^{2}< 2 \right \}$, let $a= SupA$.
Show that $a^{2}= 2$.
First rule out that $a^{2}< 2\; or\; a^{2}> 2$.
Suppose that $a^{2}< 2$
then we can show that $\left ( a+\frac{1}{n} \right )^{2}< 2$ for large enough $n$. We have: $$\left ( a+\frac{1}{n} \right )^{2}-2=\left ( a^{2} -2\right )+\frac{2a}{n}+\frac{1}{n^{2}}\leq \left ( a^{2} -2\right )+\frac{2a+1}{n}$$
( (1) here I'm assuming that in the final term the reason we can drop one of the $n$'s is because of the less than sign?)
By the archimedian property there exists $n\in \mathbb{N}$ such that $$\frac{1}{n}< \frac{2-a^{2}}{2a+1}$$
( (2) ok, where on earth did they get that last fraction from? why is 2a+1 suddenly on the bottom, and how did $\left ( a^{2} -2\right )$ get flipped around and put on top?)   
so that $a^{2}-2+\frac{2a+1}{n}< 0$. Therefore $a+\frac{1}{n}\in A$ so a is not an upper bound for A.
 A: The reason of dropping $n^2$ is
$$ \frac{2a}{n} + \frac{1}{n^2} = \frac{2a + n^{-1}}{n} \leq \frac{2a}{n}
$$
$n$ is defined to be "large enough". That is, a upper bound. If $n < 1$, then $n = 1$ provides another upper bound, so it is safe to assume that $n \geq 1$. The idea is to make $n$ even larger to meet some other criteria.
In the second part, Archimedean property is used here since $\frac{2-a^2}{2a+1}$ is nonzero and positive. To see this, notice that $2 - a^2 > 0$ by hypothesis, and $2a+1 > 0$ since $a > 0$. Its derivation is not related to the algebra above at this point.
A: Question 1
Yeah, I believe you have this one right. Let me write it out in detail for completeness, where I have inserted an extra step:
$\left ( a+\frac{1}{n} \right )^{2}-2=\left ( a^{2} -2\right )+\frac{2a}{n}+\frac{1}{n^{2}}\leq (a^2-2) + \frac{2a}{n} + \frac{1}{n}=\left ( a^{2} -2\right )+\frac{2a+1}{n}.$
The extra step can be seen as follows: Since $n^2\geq n$, we have $\frac{1}{n^2}\leq\frac{1}{n}$. Replacing $\frac{1}{n^2}$ by $\frac{1}{n}$ thus gives us something larger.
Question 2
The expression $\frac{2-a^2}{2a+1}$ has been carefully chosen, such that when we use $\frac{1}{n}<\frac{2-a^2}{2a+1}$ in the last expression $(a^2-2)+\frac{2a+1}{n}$ from your first question, we get exactly that the expression is less than $0$. In detail, whenever $\frac{1}{n}<\frac{2-a^2}{2a+1}$, we have
\begin{align*}
(a^2-2)+\frac{2a+1}{n}&=(a^2-2)+(2a+1)\frac{1}{n}\\
&<(a^2-2)+(2a+1)\frac{2-a^2}{2a+1}\\
&= (a^2-2) + (2-a^2)\\
&= 0.
\end{align*}
This shows that whenever $\frac{1}{n}<\frac{2-a^2}{2a+1}$ (which happens for large enough $n$), we have $(a+\frac{1}{n})^2-2<0$, and thus $a+\frac{1}{n}\in A$.
A: A different method of showing that $\neg (\sup A<2)$: Apply the following to your Q with $K=2$:
As noted by Heron (a.k.a. Hero) of Alexandria (circa 60 AD), for any $K>0,$ if $x_1>0$ and  $(x_1)^2> K,$ then let $$x_{n+1}=\frac {1}{2}(x_n + K/x_n)\; \text {for each } n\in N.$$ Then for all $n\in N$ we have $x_n>x_{n+1}$ and $(x_n)^2>K.$ (In fact, as Heron noted, the sequence $((x_n)^2-K)_{n\in N}$ converges  to $0.$) 
So if $a>0$  and $a^2<K$ let $x_1=K/a.$ We obtain  the strictly decreasing sequence $(x_1,x_2,x_3,...)$ with  $(x_n)^2>K$ for all $n\in N.$ Then $(K/x_1,K/x_2,K/x_3,...)$ is a strictly increasing sequence whose first term $K/x_1$ is equal to $a$, and we  also have $(K/x_n)^2< K$ for all $n\in N.$
In particular, $a=K/x_1<K/x_2$ and $a^2<(K/x_2)^2<K.$
Remark: We have $x_{n+1}^2-K=(x_n-K/x_n)^2/4(x_n)^2>0.$
