How did Rudin come up with this exact expression?

Question

To show that for every rational $p>0$ such that $p^2<2$ one can get a rational $q$ with $p^2<q^2<2$ and $p<q$ Rudin writes,

We now examine this situation a little more closely. 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 $$q=p-\frac{p^2-2}{p+2}=\frac{2p+2}{p+2}.$$ Then $$q^2-2=\frac{2(p^2-2)}{(p+2)^2}.$$

This appears many times in analysis, so if the answer can be generalized/a general method can be given then that would be the best thing. My question is how does one come up with the quantity

$$p-\frac{p^2-2}{p+2}$$

My intuition is that can come from any of the rational approximation possibles, by for example using Newton's method, though I'm lacking when it comes to the details.

This is a good question. I wonder if it's possible to use analogous methods to prove $a^{1/b}$ is irrational whenever $a$ isn't a $b$th-power. — MathematicsStudent1122, Oct 15 '19 at 21:14
Newton's method offshoots the side to the left of the root of $x^2-2$ to the right side, so you won't be able to show that $A$ has no largest element with this directly. (Note, however, that this works for $B$.) As the answer of Bill Dubuque in the linked question by J.W. Tanner mentions, you can use the secant method instead, and this will yield what Rudin is doing. — Aloizio Macedo, Oct 15 '19 at 21:48
My answer to the question I linked was essentially that $(2-\sqrt2)(\sqrt2-p)=(p+2)\sqrt2-(2p+2)<\sqrt2-p\implies \sqrt2-q=\sqrt2-\dfrac{2p+2}{p+2}<\sqrt2-p$ — J. W. Tanner, Oct 15 '19 at 22:04

score 1 · Answer 1 · answered Oct 15 '19 at 21:28

It's always seemed easier to me, as a matter of intuition, to find $n\in \mathbb N$ such that $(p+1/n)^2<2.$ This is the same as making $(2p)/n+1/n^2<2-p^2.$ And because $1/n^2\le 1/n,$ it suffices to find $n$ such that

$$(2p)/n+1/n= (2p+1)/n<2-p^2.$$ Now you're just a manipulation away from applying the Archimedian principle.

How did Rudin come up with this exact expression?

1 Answers1