Rudin's proof on the Analytic Incompleteness of Rationals

I've always thought Rudin is kind of lame here. He has a fondness for pulling rabbits out of a hat, and this is not the last time you will see it.

It seems to me easier to consider

$$(p+1/n)^2 <2,\, n=1,2,\dots $$

The intuition is then clear: Surely this will be true for large enough $n,$ let's go find one, call it $n_0,$ and then $p+1/n_0$ does the job.

The idea is that as $p \to \sqrt{2}^-$, you want to add something scaling like $2-p^2$ to $p$, so that what you add goes to zero as $p \to \sqrt{2}^-$. But you can't just add $2-p^2$ to $p$. Consider for instance $p=0$, then $p+2-p^2=2$ which is too big.

How much is it too big by? Well, $p^2-2=(p+\sqrt{2})(p-\sqrt{2})$, so it is too big by a factor of $p+\sqrt{2}$. So it would be enough to divide it by any rational number greater than $p+\sqrt{2}$. $p+2$ is just what you get when you use the trivial estimate $\sqrt{2}<2$. Numerous other options would have worked, though, for example $q=p+\frac{2-p^2}{4}$.

This cries out Newton's method looking for the root of $f(x)=x^2-2$. But Newton's method will give a rational number bigger than $\sqrt{2}$, i.e., outside of $A$. All the iterations of Newton will give a rational point greater than $\sqrt{2}$ (due to the positive second derivative at the root) (thanks to Ian for correcting this). This shows incompleteness of the set of rationals greater than $\sqrt{2}$.

We simply take a rational number bigger than the square root of 2, say 2. We form the chord between the point $(p,f(p))$ and $(2,f(2))=(2,2)$. We take the intersection of the chord with the $x$-axis as our new point. I drew the picture with geogebra: https://ggbm.at/nkfcPUB4 You simply need to verify that the Rudin's formula gives you the intersection of the chord with the $x$-axis.

Convexity of the graph of $y=x^2-2$ ensures the new point is in $A$.