Proving a set in $\mathbb{Q}$ that is bounded above to have no largest number (greatest element) I'm studying the set $Q$. At this point I don't know the real numbers. This is the theorem: 

Let $A$ be the set such that $A=\{p\in \mathbb{Q}^{+}:p^{2}<2\}$. Then $A$ contains no largest number.

The prove given in Rudin's book of analysis goes something like this: 

Given $p\in A$, let's consider the number
  $q=p-\dfrac{p^2-2}{p+2}=\dfrac{2p+2}{p+2}$. Then
  $q^2-2=\dfrac{2(p^2-2)}{(p+2)^2}$.  This prove that if $p\in A$ then
  $p^2-2<0$, $q>p$, and $q^2<2$. Thus $q\in A$ and $q>p$. Therefore, for
  every $p\in A$ we can find a rational $q\in A$ such that $p<q$.

Now here is my problem. First of all I understand the proof perfectly, but I don't see the motivation for finding $q$. Let's suppose I wanted to find $q$ in the set $B=\{p\in Q^{+}: p^{2}<3\}$. I wouldn't now how to find it. 
Also I would like to know some 'standard', if there is some, process to apply in this cases. For example in my try, before looking at the solution, I was thinking in this way: 

I need to find a number $q$ such that  $p^{2}<q^{2}<2$ in such a way that $p<q$". If I set this numbers on the real line I can observe essentially two ways to follow. 
In the first one, if try to find the number $q^2$ such that $p^{2}<q^{2}<2$ I can think of some of the form $q^{2}=p^{2}+(\dfrac{2-p^2}{n})$ where $n$ is a natural number sufficiently large as necessary (here I'm using Archimedian property) to find a perfect square. 
In my second approach I'd like to find the number $q$ such that $p<q$. In this case I would think almost in the same way as before. I can think of $q$ as something sufficiently close to $p$ in the form $q=p+\dfrac{1}{n}$ with $n$ sufficiently large so that $p^2<q^{2}<2$. 

So far I haven't been able to solve the problem with any of my tries and I don't know if I'm in the right direction or it's just an obsession to use the Archimedian property. 
 A: Note that $$\frac{2p+2}{p+2}=p$$ gives $$2p+2=p^2+2p$$ $$p^2=2$$
Thus, we're playing around the fixed point of $f(x)=\dfrac{2x+2}{x+2}$ which is $\sqrt 2$. If we choose $x^2<2$, then iterating with $f$ will get use closer to $\sqrt 2$, and because the fixed point is attractive, and the function is monotone increasing, we get closer to $\sqrt 2$ from below, with increasing values.

This is another example. Suppose $r$ is rational, positive, and $$r<\sqrt 2$$
Then $$r+1<\sqrt 2+1$$
 $$\frac{1}{r+1}>\sqrt 2-1$$
$$\frac{r+2}{r+1}>\sqrt 2 $$
We used $(\sqrt 2-1)(\sqrt 2+1)=1$.  Thus, we have obtained a new rational $q=\frac{r+2}{r+1}$ with $q>\sqrt 2$. Repeating the process we arrive at $$\frac{q+2}{q+1}<\sqrt 2$$ and replacing what $q$ was we obtain $$\frac{{3r + 4}}{{2r + 3}} < \sqrt 2 $$
It turns out this new rational, call it $r$, is such that $r<r'$ and $r'^2<2$. The motivation: we played with the fixed point of $\dfrac{3x+4}{2x+3}$ which is $\sqrt 2$ again.
A: I’ll do something almost as good as the two answers so far, namely I’ll describe an explicit way of finding a sequence of negative rationals whose squares are all greater than $2$, that is rationals less than $-\sqrt2$, upper bounded (by $-1$), with no least upper bound. Here’s my method, using Newton-Raphson, which I’m very fond of, and which is (or should be) standard first-year Calculus material.
Apply Newton-Raphson to $f(x)=x^2-2$ to find a negative root. The procedure is
$$
z_i\mapsto z_{i+1}= z_i-f(z_i)/f'(z_i) =\frac{z_i^2+2}{2z_i}\,.
$$
When you look at the picture of what’s going on, you see that if you start with a negative $z$ so that $(z,0)$ is under (to the left of) the graph, all the resulting approximations will also be to the left of the graph, so that the successive $z$’s will all be less than $-\sqrt2$, i.e. their squares will be bigger than $2$. If you start with $z_0=-2$, for instance, you get the sequence
$$
-2, -3/2,-17/12,-577/408\ldots
$$
A: Suppose you found $~p=\max\limits_{x^2<M,x\in\mathbb Q}x~$ for some $M$. $M=2$ in your example, but it doesn't matter. Let $\Delta = \sqrt{M} - p > 0$. There is always a number $n$ such that $\frac 1 n<\Delta$, because $\frac 1 n \to 0$ (for example, $n={1+\left\lceil\frac{1}{\Delta}\right\rceil}$). But then $p < p + \frac 1 n = q < p + \Delta = \sqrt{M}$, so $p^2 < q^2 < M$.
A: Eventually we’re going to have the reals, and we know how we want them to behave, so pretend for a moment that we do have them; then we want to find a rational number $q$ such that $p<q<\sqrt2$. Setting $r=q-p$, we want to find a positive rational $r<\sqrt2-p$. This inequality is equivalent to the inequality $r(\sqrt2+p)<2-p^2$, so what we want is a rational $r$ such that
$$0<r<\frac{2-p^2}{\sqrt2+p}\;.\tag{1}$$
$(1)$ still has that pesky $\sqrt2$ that we don’t really have yet. However, we know that it will be true that $\sqrt2<2$, so it will also be true (when we actually do have $\sqrt2$) that
$$0<\frac{2-p^2}{2+p}<\frac{2-p^2}{\sqrt2+p}\;.$$
Thus, if we set
$$r=\frac{2-p^2}{2+p}\;,$$
we ought to be able to prove that $p<p+r$ and $(p+r)^2<2$. And indeed the proof that you included in the question shows that we can.
If you follow the same heuristic reasoning for your set $B$, you find that setting 
$$r=\frac{3-p^2}{3+p}$$
and hence
$$q=p+r=p+\frac{3-p^2}{3+p}=p-\frac{p^2-3}{p+3}$$
ought to work, and indeed it does.
