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?



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.
