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In Rudin's Principles of Mathematical Analysis (3rd ed), he proves (at the very beginning: example 1.1) that the set $A$ of all positive rationals $p$ such that $p^2<2$ contains no largest number and the set $B$ of all positive rationals $p$ such that $p^2>2$ contains no smallest number.

To show this, he starts by associating with each rational $p>0$, the number $q=p-(p^2-2)/(p+2) = (2p+2)/(p+2)$. And thus $q^2-2=2(p^2-2)/(p+2)^2$.

My question is actually simple: from where does he get the expression for $q$ and how does one get the intuition to come up with the form of the number $q$?

Antonio Vargas
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ahron
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    This is something you have to used to in Rudin. His writing is terse and to the point, and his proofs, while elegant, are often difficult to digest. I would not recommend this book if you are learning analysis for the first time. – Math1000 Sep 23 '15 at 20:06
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    I am used to it, just wondering how the magic comes about.. – ahron Sep 23 '15 at 20:11
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    According to a (now retired) mathematician I was talking to after a seminar a couple years ago: "I knew Rudin, and his wife told me that when he wrote a book, he would pace around the house for hours at a time without writing anything, and then suddenly sit down and write an entire chapter, with the proofs exactly as they appear in the published book." So, I guess...the intuition is...be Walter Rudin? – Ben Sheller Sep 23 '15 at 20:14
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    The "magic" comes about in working with the algebra until you find something that fits. What you see in a printed book are the end result of these computations, not the hours (or even years!) of work that went into them. – Math1000 Sep 23 '15 at 20:16
  • Ok, so if I understand correctly: no real logic to how it came about, just try a bunch of stuff (leveraging on one's knowledge of course) till something works.. ? – ahron Sep 23 '15 at 20:20
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    The function relating $p$ to $q$, namely $f(p) = {2p+2\over p+2}$, is a rational function (with integer coeffcients, ensuring that $f(p)$ is rational if $p$ is rational) with certain properties required for the argument: (i) $f$ is strictly increasing on $[0,\infty)$; (ii) $f(0)=1$, ensuring (with (i)) that $f(p)>0$ for all $p\ge 0$, and (iii) $f(\sqrt{2})=\sqrt{2}$. Rudin's choice for $f$ is the simplest (indeed the only) such function that I can find. – John Dawkins Sep 23 '15 at 20:36
  • @JohnDawkins thank you! This is just what I was looking for. Probably an odd question, but is there a text that includes this kinda explanation (while maintaining the relative terseness of Rudin), or is it best to stick with Rudin (I like the style) and ask on SE? Thanks again. I can accept this as an answer if you write it as one. – ahron Sep 23 '15 at 21:11
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    I know of no such text, but it can be fun to try to get into Rudin's head sometimes. ASCII Advocate has now "answered" your question in detail, so there's no need for me to do so. – John Dawkins Sep 23 '15 at 21:24

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Rudin needs a rational function $R(x)$ with $R(x)=x$ at $x=\pm \sqrt{2}$, and with rational coefficients, and within that class of functions, one that moves points closer to $+\sqrt{2}$.

The basic idea for finding nice examples, is that $R(x)$ can be taken to be a linear fractional transformation $\frac{ax+b}{cx+d}$. There are theoretical reasons, that Rudin was surely well aware of, why reasonable solutions exist in that class of functions. Given that idea, it's a few lines of simple algebra to identify all possible solutions in that class, and pick out the simplest one.

Why should a linear-fractional solution exist? Ignoring the requirement about the parameters $a,b,c,d$ being rational, the fixed-point conditions are a system of 2 homogeneous linear equations in 4 variables. Solutions therefore exist, as is also clear from projective geometry where any 3 points of the line can be mapped to any 3 other points by a linear-fractional transformation. And the theory of continued fractions, in which $\sqrt{2}$ has a periodic fraction, shows that the rational coefficients requirement can be satisfied in that class of functions; no need to go to degree $2$ or higher. Or take any linear fractional transformation that sends the (unordered) set of two roots to itself, and any rational number $a$ to another rational number $b$. By Galois theoretic symmetry considerations the set of what-maps-to-what is invariant under the conjugation in $\mathbb{Q}(\sqrt{2})$ and hence the function has rational coefficients.

So if you (which is to say, Rudin) knew of those arguments, the first thing to try is linear fractional transformations. Here is the computation:

$a + bX = X(c+dX)$ has solutions $X = \pm \sqrt{2}$.

$dx^2 + (c-b)X - a$ divisible by $(X^2 - 2)$

$a = 2d \quad ; \quad b=c$.

Taking $d=1$ with no lost generality (rescaling of $(a,b,c,d)$ doesn't change the fraction), all such linear fractional function are of the form $\frac{2 + bx}{b + x}$ which is Rudin's function when $b=2$.

To get attraction toward the positive root we need $|f'(\sqrt{2})| \leq 1$. The difference quotient is

$$|\frac{\frac{2+bx}{b+x} - x}{x - \sqrt{2}}| = |\frac{\sqrt{2}+x}{b+x}|$$ so that $|b+x|$ should exceed $|\sqrt{2}+x|$ at $x = \sqrt{2}$. That means $b \geq \sqrt{2}$ which excludes the smallest integer value $b=1$, but $b=2$ works and this is the parameter in the exercise.

The conclusion is that Rudin selected the smallest working example with integer parameters. Because his linear fractional function is attracting at the positive square root, and is continuous on the interval between the roots, it is increasing between them, so it also fits the requirement of sending positive numbers to positive numbers. Hence the ability to simplify some aspects of the problem (the complication that the distance to $\sqrt{2}$ will not be decreased by $R(x)$ for all real numbers) by restricting to positive real numbers.

That is one explanation of how Rudin engineered the problem.

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