Question about the proof for "The rational line Q does not have the completeness property" Referring to the attached png file, I have questions about the inequality:
$$(n^2)/(2n+1) > (p^2)/(2q^2 - p^2)$$
How is this inequality derived?  How did the n get into the inequality?  Where did $n^2/(2n+1)$, $p^2/(2q^2 - p^2)$, and later in the proof, $2q^2/(p^2 - 2q^2)$ come from?   There were no derivations for these 3 expressions.  Could someone please provide the derivations?


 A: The development of a mathematical proof often has an "exploratory" phase in which various formulas (or other facts) are discovered, and then those formulas are used in the formal proof.
What often happens when the proof gets written up (in a textbook, or in a research journal) is that the "exploratory" phase is ignored and only the formal proof is written up, and the formulas appear to pop out of nowhere. The justification for this is that it doesn't matter where those formulas came from (see the comment of @fleablood for a guess as to their origin), only that they actually work.
This can be unsatisfying to a student who is trying to understand a proof. "Where did that formula come from?" is a common question. And if the proof writer does not say where it came from, we are left to guess; it's impossible to know for sure.
In this case, though, there's maybe just enough information to guess where the formula came from. Let me try to explain what could have happened when writing the paragraph of the proof which starts with the sentence "Suppose first that $x^2 < 2$."
In that paragraph, one is assuming that $x = \frac{p}{q}$ is a positive rational number (so $p,q \in \mathcal N$), and that $x^2 < 2$, and using those assumptions one wishes to prove that $x$ is not an upper bound for the set $A$.
This will follow if one can prove the existence of $n \in \mathcal N$ such that
$$y = \left( \frac{n+1}{n} \right) \frac{p}{q} \in A
$$
because that quantity $y$ is clearly larger than $x=\frac{p}{q}$.
To say that $y \in A$ is equivalent to the inequality
$$(*) \qquad \left( \frac{n+1}{n} \right)^2  \, \frac{p^2}{q^2} < 2
$$
At this point, the author pretty clearly manipulated the inequality $(*)$ with the purpose of isolating all the $n$'s on one side and all the $p$'s and $q$'s on the other side, rewriting the inequality in the form
$$\frac{n^2}{2n+1} > \frac{p^2}{2q^2-p^2}
$$
And then the author, who is very good with sequences, probably thought:

The left hand side of this inequality diverges to $+\infty$ as $n \to +\infty$. But $p$ and $q$ are fixed. The definition of divergence to $+\infty$ therefore guarantees the existence of $n \in \mathcal N$ which makes this inequality true. So that's how I'll start my proof, and then I'll use that to derive the inequality $(*)$, by working backwards through my exploratory manipulations.

or something to that effect.
And later, when cleaning up their desktop, the author tossed all that lovely exploratory derivation in the garbage pan, leaving us to make our wild guesses as to the origin of that final formula.

Something quite similar is going on in the later paragraph which starts "It follows that we must have $x^2 \ge 2$." I'll leave that for you to think about, and if you like to try to reconstruct the possible exploratory phases which led to the proof in that paragraph.
A: You have $\frac{p^2}{2q^2-p^2}$ fixed. Since $\lim_{n \rightarrow \infty} \frac{n^2}{2n+1}=\infty$, using definition of limit at $\infty$ the inequality involving $n$ is obtained.
A: I will provide some explanation from the point when the author makes an assumption that $x^2<2$.
The key here is to observe that any upper bound $x$ of $A$ must satisfy $x^2\geq 2$ (a good guess because all the $x$ with $x^2<2$ are in $A$). To show that one assumes $x^2<2$ and derives a contradiction.
The way to get a contradiction here is to find some $y\in A$ such that $y>x$ (a contradiction is obtained because $x$ is an upper bound for $A$ and no member of $A$ can exceed $A$).
Now the author takes a suitable form for $y$ as $y=(n+1)x/n$ for some positive integer $n$. This automatically ensures that $y>x$, but more importantly $y$ is near $x$ (in terms of limits $\lim_{n\to\infty} y=x$). Another form could have been $y=x+(1/n)$ and a proof may be given based on this.
To obtain the contradiction we not only need $y>x$ but also $y\in A$ ie $y^2<2$. This means that we need $$\frac{(n+1)^2}{n^2}<\frac{2}{x^2}$$ or $$\frac{(n+1)^2} {n^2} -1<\frac {2} {x^2} -1$$ or $$\frac{2n+1}{n^2}<\frac{2q^2-p^2}{p^2}$$ or $$\frac{n^2}{2n+1}>\frac {p^2}{2q^2-p^2}$$
Also note that this is not an inequality which is logical consequence of our assumptions, but this is a target/goal which needs to be achieved if we want our contradiction. If we can't find an $n$ which can satisfy this inequality then we have no hope.
Since we have $$\frac {n^2}{2n+1}>\frac{4n^2-1}{4(2n+1)}=\frac{2n-1}{4}>\frac{n}{2}-1$$ we can see that the desired goal is achieved if we take $$n>2\left(\frac{p^2}{2q^2-p^2}+1\right)=\frac{4q^2}{2q^2-p^2}$$
The inequality in question arose from the special form of $y$ as $y=(n+1)x/n$  which ensures $y>x$ and an additional requirement that $y^2<2$.
Also the entire argument above does not need $p, q$. It could have worked by just using the symbol $x$. This is typical of many textbook authors who uses too much symbols instead of natural language.
Using this desired value of $n$ we obtain the contradiction and hence we must have $x^2\geq 2$.
The next part of the proof mentions that since $x$ is rational we can't have $x^2=2$ and therefore we must have $x^2>2$. Here we again obtain a contradiction by finding a $y<x$ such that $y\notin A$. Can you figure out why finding such a $y$ would lead to a contradiction? Also you should be able to guess that the chosen form of $y$ should now be $y=nx/(n+1)$ because it automatically ensures $y<x$ (another option would be have $y=x-1/n$).

I personally do not like the way the proof has been presented by the author. It is best if the proof and its ideas are developed by author taking the reader also into confidence. No point trying to play smart with your readers.
A: Yeah....
So we have $(\frac pq)^2 < 2$ and we want to show $\frac pq$ is not an upper bound so we need to show there is $S > \frac pq$ so that $(\frac pq)^2 < S^2 < 2$.
$S$ is rational, so there is a rational $r$ so that $S =r\frac pq$ and $r > 1$.
We need to show that $r^2\frac {p^2}{q^2} < 2$ or that $r^2 p^2 < 2q^2$ for some rational $r$.
Now we know $\frac {p^2}{q^2} < 2$ so $2q^2 > p^2$ and so we need to show that
$(r^2 -1) p^2 < 2q^2-p^2$ or that $r^2 -1 < \frac {2q^2-p^2}{p^2}$.
Now the RHS of that is a constant fixed positive (albeit it possibly quite small) so we need to show that we can have an $r^2-1$ that is smaller than that.
And it is sufficient to show we can have $r$ so that $r^2$ is arbitrarily close to $1$.
Now we usually can show rationals are arbitrarily close but showing that if $n$ is arbitrarily large then $\frac 1n$ is arbitrarily small.
So if we let $r = 1+ \frac 1n$ the want to show, we can choose and $n$ so that
$(1 + \frac 1n)^2 -1 = \frac 2n + \frac 1{n^2} < \frac {2p^2-p^2}{p^2}$ we will be done.
And that's a matter of making $\frac 2n + \frac 1{n^2} = \frac {2n+1}{n^2}$ arbitrarily small
or
making $\frac {n^2 }{2n+1}$ arbitrarily large.  Or at least, $\frac {n^2}{n+1} > \frac {p^2}{2q^2 -p^2}$.
We can do that as $\frac {n^2}{n+1} \to \infty$
And that's it:  If $\frac {n^2}{n+1} > \frac {p^2}{2q^2 -p^2}$ it's just algebraic manipluation that
For $r = 1+\frac 1n = \frac {n+1}n$ then $r\frac pq = \frac {p(n+1)}{qn}$ when squared will be less than $2$.

$(r\frac pq)^2 = (1+\frac 1n)^2\frac {p^2}{q^2}=$
$(1 + \frac {2n+1}{n^2})\frac {p^2}{q^2}=$
$\frac {p^2}{q^2} + \frac {2n+1}{n^2}\frac {p^2}{q^2} < $
$\frac {p^2}{q^2} + \frac {2q^2-p^2}{p^2}\frac {p^2}{q^2} =$
$\frac {p^2}{q^2} + \frac {2q^2 -p^2}{q^2}=$
$\frac {p^2 + 2q^2 -p^2}{q^2}=\frac {2q^2}{q^2}=2$.

We are done.
.....
But this was a bass-ackward proof in that we started with what we needed and worked our way back to show we can have it.  That's not an invalid way of doing a proof because we never assumed what we wanted to prove, it's okay to say "If we need this, we can do that by this, and, hey, we can do that" but it's not an ideal argument.  We like proofs to be direct and forward.
So we start with:  "We can always find an $n$ so that $\frac {p^2}{2q^2 - q^2} < \frac {n^2}{2n+1}$".
To which the reader reacts with ".... well, sure, but where the #@%! did that come from"
And the proofer can say "It doesn't matter.  We can do it, right?"
And the reader says "well, sure, but..."
And the proofer goes "ninh, ninh, ni..."
And the reader goes "...but..."
And the proofer goes "NINH! .... so if we do that then....."  And lo and behold, it works.
