TITLE: Proving a set in $\mathbb{Q}$ that is bounded above to have no largest number (greatest element)
QUESTION [15 upvotes]: 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.

REPLY [2 votes]: 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 $p^2 < q^2 < M$.