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I am not sure how this number is constructed.
I know that $2>q^2>p^2$, I then also know that q>p, hence I could construct a q such that q>p. I therefore choose q=p+x, where x could be an element of of rationals and then given I know that $2>q^2$ then $2>(p+x)^2$, so I should be able to solve for x. But since I cant use square root( have not created the real number line yet) I need to figure out how to solve $2>p^2+2 p x+x^2$
How would I proceed?
Or am I completely off track?
Thanks
This is what the OP says in a comment: Yes, I would like to see the construction of $\,q$ . If that's all, then we are lucky; I've recently posted a derivation of the formula as an answer to this question:
Contrary to a statement in an answer by Paramanand Singh, there is nothing mysterious / magical about the formula by Rudin. Seems that another author has been re-inventing the wheel, though :-(Summary of the know how. To understand why the above link should be followed.
Essential ingredient is the mediant
of two fractions. And how the Stern-Brocot tree
is formed with help of these mediants. We initialize $p < \sqrt{2}$ and $q > \sqrt{2}$ as:
$$
p = \frac{m}{n} \quad ; \quad q = \frac{2n}{m}
$$
Where $m$ and $n$ are positive integers. Now form the mediant of $p$ and $q$ , two times:
$$
q := \frac{m+2n}{n+m} \quad ; \quad q := \frac{m+(m+2n)}{n+(n+m)} = \frac{2m+2n}{m+2n} = \frac{2p+2}{p+2}
$$
The rightmost formula is already Rudin's.
This is a serious problem with the way Rudin complicates a simple problem. In order to prove that the set $A$ has no largest element it is not really necessary to find an explicit formula for a member $q\in A$ with $q > p$ where $p$ is a given member of $A$. We only need to prove that if $p \in A$ then there is a $q \in A$ with $q > p$ and for that that we don't need any mysterious / magical formula for $q$ in terms of $p$.
Hardy in his textbook A Course of Pure Mathematics does it so much better and also teaches the way things work in real-analysis. The crux of the argument by Hardy is that the set of positive rationals is partitioned into $A, B$ such that $A \cup B = \mathbb{Q}^{+}, A\cap B = \emptyset$ and further that we can find a member of $A$ and a member of $B$ which are as close to each other as we please.
Clearly $1 \in A, 2 \in B$ and given any positive integer $n$ we have the following chain of increasing rational numbers $$1, 1 + \frac{1}{n}, 1 + \frac{2}{n}, \ldots, 1 + \frac{n}{n} = 2$$ such that successive numbers in the above chain differ by $1/n$. Since the first member of the chain lies in $A$ and the last member of the chain lies in $B$ it follows that there is a last number in the chain which lies in $A$ and the next one belongs to $B$. Thus we have found two rationals $q, r$ with $q \in A, r \in B$ such that $r - q = 1/n$. Thus we can find a member of $A$ and a member of $B$ which are as close to each other as we please. Moreover we can choose $q, r$ both less than 2.
Now suppose we have a $p \in A$ then by definition of $A$ we have $p^{2} < 2$ and hence $\epsilon = 2 - p^{2}$ is a positive rational number. We can now find a positive integer $n$ such that $1/n < \epsilon/4$. Then by argument in previous paragraph we can find $q, r$ with $q \in A, r \in B$ with $r - q = 1/n < \epsilon/4$. Also both $q, r$ can be chosen to be less than $2$ so $r + q < 4$ and hence $r^{2} - q^{2} = (r - q)(r + q) < \epsilon$. And this means that $$(r^{2} - 2) + (2 - q^{2}) < \epsilon$$ Since $q\in A, r \in B$ each expression in parentheses in above equation is positive and their sum is less than $\epsilon$ so that each expression itself is less than $\epsilon$. Therefore we have $$r^{2} - 2 < \epsilon, 2 - q^{2} < \epsilon$$ By definition of $\epsilon$ it now follows that $$2 - q^{2} < 2 - p^{2}$$ or $q > p$ and thus we have found $q > p, q \in A$.
Note that the most of the deep significant theorems in real-analysis are existential in nature where it is of utmost importance to focus on the existence of a quantity with certain specific desired properties rather than explicitly finding such a quantity and the above proof is a typical example of proofs seen in real-analysis where it shows the existence of an element $q \in A, q > p$ given an element $p \in A$ without explicitly giving a formula for $q$ in terms of $p$.
Rudin on the other hand uses some sort of numerical technique to have explicit formula for $q$ in terms of $p$. Hardy also gives this approach by asking his readers to prove the following simple theorem:
If $x$ is a positive approximation to $\sqrt{2}$ then $(x + 2)/(x + 1)$ is a better approximation to $\sqrt{2}$ but in a different direction so that if $x < \sqrt{2}$ then $(x + 2)/(x + 1) > \sqrt{2}$.
Applying this rule twice we get $$x \to \frac{x + 2}{x + 1}\to\dfrac{\dfrac{x + 2}{x + 1} + 2}{\dfrac{x + 2}{x + 1} + 1} = \frac{3x + 4}{2x + 3}$$ so we can choose $q = (3p + 4)/(2p + 3)$ also.
I'd do it by trying to get $(p+1/n)^2 < 2$ for some $n\in \mathbb N.$ Note that for any $n\in \mathbb N,$
$$(p+1/n)^2 = p^2 +2p/n + 1/n^2 < p^2 +4/n + 1/n = p^2 + 5/n,$$
where we've used $p<2$ and $1/n^2 \le 1/n.$ So we'll be done if we can make $p^2 + 5/n < 2.$ Can we? Sure, it's the same as saying $n > 5/(2-p^2).$
