Baby Rudin: relationship between Diophantine approximation of $\sqrt{2}$ and algebraic geometry?

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From near the beginning of baby Rudin there is the following passage.

Let $A$ be the set of all positive rationals $p$ such that $p^2 < 2$ and let $B$ consist of all positive rationals $p$ such that $p^2 > 2$. We shall show that $A$ contains no largest number and $B$ contains no smallest.

More explicitly, for every $p$ in $A$ we can find a rational $q$ in $A$ such that $p < q$, and for every $p$ in $B$ we can find a rational $q$ in $B$ such that $q < p$.

To do this, we associate with each rational $p > 0$ the number$$q = p - {{p^2 - 2}\over{p + 2}} = {{2p + 2}\over{p + 2}}.\tag*{(1)}$$Then$$q^2 - 2 = {{2(p^2 - 2)}\over{(p + 2)^2}}.\tag*{(2)}$$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 $p^2 - 2 > 0$, $(1)$ shows that $0 < q < p$, and $(4)$ shows that $q^2 > 2$. Thus $q$ is in $B$.

Given that there is something which looks like polynomial division that shows up here in the proof, I can't help up but wonder two related things.

  1. Translated into the language of algebraic geometry (https://en.wikipedia.org/wiki/Algebraic_geometry), what would this statement or salient features of the proof offered by Rudin look like?
  2. What is the significance of this particular simple phenomena in algebraic geometry? As in, when does something like this show up, and why is it important?

Thank you, I know this is a vague question, but in the past plenty of vague questions like this on MSE have gotten very excellent responses, so here's to hoping that a seasoned algebraic geometry student or practitioner might be able to provide their perspective.