TITLE: Spivak, Prologue, Chapter 2, problem 16c
QUESTION [2 upvotes]: This problem asks you to prove:
if $$m, n \in \mathbb{N}$$
and $$\frac{m^2}{n^2} < 2$$
then $$\frac{(m+2n)^2}{(m+n)^2} > 2$$
and also $$\frac{(m+2n)^2}{(m+n)^2} - 2 < 2 - \frac{m^2}{n^2}$$
Both of these inequalities can be shown by simply manipulating/simplifying the inequalities and at some point having to make the assumption that $\frac{m^2}{n^2} < 2$.
The question I have is about item c in this question which asks us to prove:
if $$\frac{m}{n} < \sqrt{2}$$
then there is another rational number $$\frac{m'}{n'} \: with \: \frac{m}{n} < \frac{m'}{n'} < \sqrt{2}$$
I provide my solution below, and the solutions manual solution; I'd like to know if mine is correct, and where the solutions manual one came from as it is not clear at all.
My own solution is:
Given
$$\frac{m^2}{n^2} < 2 \tag{1}$$
We've proven that
$$\frac{(m+2n)^2}{(m+n)^2} - 2 < 2 - \frac{m^2}{n^2}$$
Therefore
$$\frac{(m+2n)^2}{(m+n)^2} < 4 - \frac{m^2}{n^2}$$
And given $(1)$, we can see that the right side is strictly less than 4:
$$\frac{(m+2n)^2}{(m+n)^2} < 4 - \frac{m^2}{n^2} < 4$$
$$\frac{(m+2n)}{(m+n)} < 2$$
Also, we know that
$$\frac{(m+2n)^2}{(m+n)^2} > 2$$
$$\frac{(m+2n)}{(m+n)} > \sqrt{2} > \frac{m}{n}$$
Therefore, for $m' = m +2n$ and $n' = m + n$
$$\frac{m}{n} < \frac{m'}{n'} < 2$$
The solutions manual gives a solution without really explaining how it reached that solution. That solution is:
Let $$m_1 = m + 2n$$
And $$n_1 = m + n$$
Then choose $$m' = m_1 + 2n_1 = 3m + 4n$$
And
$$n' = m_1 + n_1 = 2m + 3n$$
Is my solution correct and where does the solution manual solution come from; is there a trick or strategy employed in coming up with that solution?

REPLY [3 votes]: Note: We aren't trying to find a rational number $m'/n'$ with $m/n < m'/n' < 2$, but rather $m/n < m'/n' < \sqrt 2$.
The problem begins asking us to prove that for natural numbers $m$ and $n$ if $m^2/n^2 < 2$, then $(m+2n)^2 / (m+n)^2 > 2$ and moreover that
$$\frac{(m+2n)^2}{(m+n)^2} - 2 < 2 - \frac{m^2}{n^2}$$
Let's summarize what this means. Given a positive rational number $m/n$ that's less than $\sqrt 2$, we can generate a different rational with a square that's even closer to $2$ than the square of our first number, and this new number "sits on the other side" of $\sqrt 2$, i.e. it's greater than $\sqrt 2$.
The next step shows that the same results hold with the inequality signs reversed, that is, starting with a positive rational $m/n$ greater than $\sqrt 2$ we can again find a different rational who's square is even closer to $2$ than $m^2/n^2$, but that again sits on the other side of $\sqrt 2$ (this time the new number is less than $\sqrt 2$.)
The book solution is to do the process twice.
We begin with $m/n < \sqrt 2$.
We generate a new number $m_1/n_1$ with its square $m_1^2/n_1^2$ closer to $2$ than is $m^2/n^2$. This new number is greater than $\sqrt 2$.
We then use this new number to generate a third number $m'/n'$. Like our original number, this new number is back over on the left side of $\sqrt 2$. The square of this new number is closer to $2$ than the square of our original number. As a result, this new number is closer to $\sqrt 2$ than the original number:
$$\ 2 - \frac{m^2}{n^2} > \frac{m_1^2}{n_1^2} - 2 > 2 -  \frac{m'^2}{n'^2} > 0$$
$$ \frac{m^2}{n^2} - 2 < \frac{m'^2}{n'^2} - 2 < 0$$
$$ \frac{m^2}{n^2} < \frac{m'^2}{n'^2}  < 2$$
$$ \frac{m}{n} < \frac{m'}{n'}  < \sqrt 2$$
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