Choice of qq in Baby Rudin’s Example 1.1

First, my apologies if this has already been asked/answered. I wasn’t able to find this question via search.

My question comes from Rudin’s “Principles of Mathematical Analysis,” or “Baby Rudin,” Ch 1, Example 1.1 on p. 2. In the second version of the proof, showing that sets A and B do not have greatest or lowest elements respectively, he presents a seemingly arbitrary assignment of a number q that satisfies equations (3) and (4), plus other conditions needed to show that q is the right number for the proof. As an exercise, I tried to derive his choice of q so that I may learn more about the problem.

If we write equations (3) as
q = p – (p^2 – 2)x, we can write (4) as

q^2 – 2 = (p^2 – 2)[1 – 2px + (p^2 – 2)x^2].

Here, we need a rational x > 0, chosen such that the expression in […] is positive. Using the quadratic formula and the sign of (p^2 – 2), it can be shown that we need

x \in \left(0, \frac{1}{p + \sqrt{2}}\right) \mbox{ for } p \in A,

or, for p \in B, x < 1/\left(p + \sqrt{2}\right) or x > 1/\left(p - \sqrt{2}\right).

Notice that there are MANY solutions to these equations! The easiest to see, perhaps, is letting x = 1/(p + n) for n \geq 2. Notice that Rudin chooses n = 2 for his answer, but it checks out easily for other n.

The Question: Why does Rudin choose x = 1/(p + 2) specifically? Is it just to make the expressions work out clearly algebraically? Why doesn't he comment on his particular choice or the nature of the set of solutions that will work for the proof? Is there a simpler derivation for the number q that I am missing?


Rudin's approximation to \sqrt{2} arises simply by applying by the secant method - a difference analog of Newton's method for finding successively better approximations to roots.enter image description here

As the linked Wikipedia article shows, the recurrence relation for the secant method is as below.

\rm S_{n+1}= \dfrac{S_{n-1}\ f\:(S_n) - S_n\ f\:(S_{n-1})}{f\:(S_n)-f\:(S_{n-1})}\qquad\qquad\qquad\qquad

For \rm\ (S_{n-1},S_n,S_{n+1}) = (q,p,p')\ and \rm\ f\:(x) = x^2-d\:,\: we obtain

\rm p'\ =\ \dfrac{q\:(p^2-d) - p\:(q^2-d)}{p^2-d-(q^2-d)}\ =\ \dfrac{(p-q)\:(p\:q+d)}{p^2-q^2}\ =\ \dfrac{p\:q+d}{p+q}

Finally specializing \rm\: q = 2 = d\: yields Rudin's approximation \rm\displaystyle\ p'\ =\ \frac{2\:p+2}{\ \:p+2}

The secant method has stunningly beautiful connections with the group law on conics.
To learn about this folklore, I highly recommend Sam Northshield's Associativity of the Secant Method. The reader already familiar with the group law on elliptic curves, but unfamiliar with the degenerate case of conics, might also find helpful some of Franz Lemmermeyer's expositions, e.g. Conics - a poor man's elliptic curves.

Source : Link , Question Author : Rachel , Answer Author : user153330

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