Archimedean Property - The use of the property in basic real anaysis proofs I've been looking for something like this in the previous answers on the topic, but I didn't enounter anything similar, so here there is my problem.
First of all, here there is the definition of Archimedean Property (AP) that I found in the book I am self-studying:
$\forall (a,b)\in \mathbb{R}_{++} \times \mathbb{R}, \exists m \in \mathbb{N} (b<ma)$
My problem is with the way in which it is presented a proof of the fact that $\sup\{q\in \mathbb{Q}:q^2<2\}=\sqrt{2}$.
Indeed, the author first of all points out that the Completeness Axiom ensures that $s:=\sup S$ is a real number (where $S:=\{q\in \mathbb{Q}:q^2<2\}$). Then, he supposes that $s^2>2$, from which we have that $s^2 - 2>0$. At this point, he states that by AP there exists an $m \in \mathbb{N}$ s.t. $m(s^2 - 2) >2s$.
Problem 1
Here, the author is using the AP in the following way: $m$ is the $m$ of the original formulation, $(s^2 - 2)$ is the $a$ of the original formulation, and $2s$ is the $b$. Am I right?
Problem 2
Where does the $2s$ come from?
Anyway, the author moves on and states that, from $m(s^2 - 2) >2s$, we have that 
$$ \left( s- \frac{1}{m} \right)^2=s^2 - \frac{2s}{m} + \frac{1}{m^2} > s^2 - (s^2 - 2) = 2$$
which means that $(s-\frac{1}{m})^2>q^2$.
Problem 3
Where does  $(s-\frac{1}{m})^2>q^2$ come from?
To me it really looks a bit a rabbit-out-of-the-hat proof.
What am I missing exactly? Is my way of looking at the AP the correct one (see Problem 1)? How can I figure out in this kind of cases (that looks to me quite common in Real Analysis) where the author takes the elements of his proof?
Any help, hint or feedback is more than welcome. Thanks a lot.
 A: First of all, you are right: AP states that for any $a,b>0$ there exists $m$ such that $am>b$. In your case you would like to pick up $m$ in such a way, that $m(s^2-2)>2s$, so your choice of $a$ and $b$ is correct. Second, we want to show that if we assume that $s = \sup S$ is such that $s^2>2$, than something meaningless will follow. For example, in your case the author shows that $s^2>2$ would imply that $(s-\frac1m)^2>2$ and hence $(s-\frac1m)^2>q^2$ for all $q\in S$. As a result, it means that $s-\frac1m$ is an upper bound for $S$ as well, which can't be true since $s$ was the least upper bound and $s-\frac1m<s$ is even less. However, to show all latter facts, as vadim123 mentioned, the author had to use some tricks which at the first glance are indeed magical. On the other hand, after reading 3-5 proofs on the similar topic you'll get the intuition which tricks to try if you shall prove something yourself just by looking for analogies. 
Let's take a backward look on how this magic works, and why is it intuitive. We assume that $s^2>2$ where the strict inequality holds, so intuitively there is a gap between $2$ and $s^2$ and hence we shall be able to insert there another positive number $x^2$ such that 
$$
  s^2>x^2>2. \tag{1}
$$ 
That would mean exactly that $s\neq\sup S$ since $x\geq \sup S$ from $(1)$, and $s>x$. Now, instead of looking for $x$ that satisfies $(1)$ it is notationally more convenient to denote $x = s-v$ for some $v>0$ and look for some $v$ small enough such that $(1)$ holds. We have
$$
  (s-v)^2 = s^2 - 2sv + v^2\geq s^2 - 2sv.
$$
Now, to choose $v>0$ in such a way that $(s-v)^2>2$, we have to have that
$$
  s^2 - 2sv>2 \iff \frac1v(s^2 - 2)>2s.
$$
The question now is how to pick up $v>0$ so that LHS is greater than the RHS. Here AP comes to help us: we just denote $m = \frac1v$. Now, after we followed this backward logic, we know how to prove the theorem. However, it is more neat (though less understandable) to do proof in forward direction (keeping all the results of the backward one that we already found).
A: *

*Your interpretation is correct.

*$2s$ was chosen to make the subsequent algebra work out right.  It is common in mathematics to work out a messy calculation, then rewrite it to avoid any messy parts and instead have "magic" choices like this one.

*$q^2<2$ for all $q$ in the set.  $(s-1/m)^2=2>q^2$
The purpose of this calculation is to find a positive separation, namely $\frac{1}{m}$, between $s$ and all the elements of the set.  Consequently, $s$ can't be a limit point.
