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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 "Princicples 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?

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3 Answers

up vote 8 down vote accepted

In the interest of making this question and answer more self-contained, here is the example in question. My answer is below.

enter image description here

I think you hit the nail on the head. He was looking for a rational $y$ such that $q = p+y$ will have the desired properties in both cases.

First, if $p \in A$ we want $p<q \Leftrightarrow y > 0$ and if $p \in B$ we want $p>q \Leftrightarrow y < 0$. We might as well take advantage of the sign of $p^2-2$ in each case to achieve this by searching for a positive quantity $x$ such that $q = p - (p^2-2)x$.

As you showed, any choice $0 < x < 1/(p+\sqrt{2})$ will satisfy the requirements $p \in A \Rightarrow q \in A$ and $p \in B \Rightarrow q \in B$. He wanted to ensure $x$ was rational, and the easiest way to do this is to take $x = 1/(p+k)$ where $k$ is an integer larger than or equal to $2$. There's no need to complicate matters further than that, so he simply chooses the smallest $k$ which works, namely $2$.

My derivation was the same as yours, and I doubt you could get it more simple than that.

As for why he didn't comment on his choice: well, that's kind of just how Rudin is. He will rarely (if ever?) comment on the motivation for his proofs! It is endearing to some and perhaps a little infuriating to others.

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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.

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+1, Interesting! –  Antonio Vargas May 6 '12 at 21:56
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Given a positive rational $p$ with $p^{2} < 2$, Rudin finds a rational $q$ with $q > p$ and $q^{2} < 2$. He finds an expression for $q$ in terms of $p$ which is definitely based on numerical techniques for finding square root of a number. Giving a formula for $q$ without any explanation makes it all the more mysterious and therefore this question came into being.

A much simpler approach is to show that such a $q$ exists without giving a direct formula for it. This is what Hardy does in his first chapter of "A Course of Pure Mathematics". Since $1^{2} < 2 < 2^{2}$, we can find find $n + 1$ rationals between $1$ and $2$ namely $1, 1 + 1/n$, $1 + 2/n, \cdots, 1 + n/n = 2$. In this sequence of rationals there will be a last whose square is less than $2$ and the next one will have its square greater than $2$.

Thus we have two rationals positive rationals $x, y$ such that $x^{2} < 2 < y^{2}$ and $y - x = 1/n$. By taking $n$ large enough it is easy to see that given any positive rational $\epsilon$ we can find positive rationals $x, y$ with $x^{2} < 2 < y^{2}, x < 2, y < 2$ and $y - x < \epsilon$. It thus follows that $y^{2} - x^{2} = (y + x)(y - x) < 4\epsilon$. This means that $(y^{2} - 2) + (2 - x^{2}) < 4\epsilon$ and hence $(y^{2} - 2) < 4\epsilon, (2 - x^{2}) < 4\epsilon$ as both the expressions $(y^{2} - 2), (2 - x^{2})$ are positive.

Now we choose $4\epsilon = 2 - p^{2}$ and then we can find positive rational $x$ such that $2 - x^{2} < 4\epsilon = 2 - p^{2}$ so that $x > p$ and we already have $x^{2} < 2$.

See the smartness of the above technique. Ideally what we need is an approximation (on the lower side) for $\sqrt{2}$ which is better than existing approximation $p$. So we just need to choose a rational between $p$ and $\sqrt{2}$. This is possible without even defining the symbol $\sqrt{2}$ because we have access to numbers $1$ (lower approx to $\sqrt{2}$) and $2$ (higher approx) and then we can divide the gap between $1$ and $2$ as finely as possible to obtain approximations to $\sqrt{2}$ which are as good as we need.

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