construct two rational numbers

Question

Can one explain me a bit or more about how to construct the two rational numbers?

From 《Principles of Mathematical Analysis》page 2 in proving $\sqrt{2}$ is not a rational number.

$p>0$

$$q=p-\frac{p^2-2}{p+2}=\frac{2p+2}{p+2}.~~~(1)$$

Then

$$q^2-2=\frac{2\left(p^2-2\right)}{(p+2)^2}~~~(2)$$

If $p$ is in $A$ then $p^2-2<0$, (1) shows that $q>p$, and (2) shows that $q^2<2$. Thus $q$ is in $A$.

If $p$ is in $B$ then $p^2-2>0$,(2) shows that $0<q<p$, and (2) shows that $q^2>2$. Thus $q$ is in $B$.

Let $A$ be the set of all positive rationals $p$ such that $p^2<2$ and let $B$ consist of all positive rationals suth that $p^2>2$.

Answer 1

Consider the case $p^2>2$ (so that $p>\sqrt{2}$).

We want to find a distance shorter than $p-\sqrt{2}$. We have: $$p-\sqrt{2}=\frac{p^2-2}{p+\sqrt{2}}>\frac{p^2-2}{p+2}>0$$ so that $$p>p-\frac{p^2-2}{p+2}>\sqrt{2}$$ and the intermediate term is called $q$. A similar reasoning could be done for $p<\sqrt{2}$.