First, my apologies if this has already been asked/answered. I wasn't able to find this question via search.
My question comes from Rudin's "Principles of Mathematical Analysis," or "Baby Rudin," Ch 1, Example 1.1 on p. 2. In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number $q$ that satisfies equations (3) and (4), plus other conditions needed to show that $q$ is the right number for the proof. As an exercise, I tried to derive his choice of $q$ so that I may learn more about the problem.
If we write equations (3) as$q = p - (p^2 - 2)x$, we can write (4) as
$$q^2 - 2 = (p^2 - 2)[1 - 2px + (p^2 - 2)x^2].$$
Here, we need a rational $x > 0$, chosen such that the expression in $[...]$ is positive. Using the quadratic formula and the sign of $(p^2 - 2)$, it can be shown that we need
$$x \in \left(0, \frac{1}{p + \sqrt{2}}\right) \mbox{ for } p \in A,$$
or, for $p \in B$, $x < 1/\left(p + \sqrt{2}\right)$ or $x > 1/\left(p - \sqrt{2}\right)$.
Notice that there are MANY solutions to these equations! The easiest to see, perhaps, is letting $x = 1/(p + n)$ for $n \geq 2$. Notice that Rudin chooses $n = 2$ for his answer, but it checks out easily for other $n$.
The Question: Why does Rudin choose $x = 1/(p + 2)$ specifically? Is it just to make the expressions work out clearly algebraically? Why doesn't he comment on his particular choice or the nature of the set of solutions that will work for the proof? Is there a simpler derivation for the number $q$ that I am missing?