The golden ratio $\phi=\frac{1+\sqrt5}2$ is sometimes said to be one of the most difficult numbers to approximate with rational numbers, because its continued fraction development $$\phi = 1 + \cfrac{1}{1 + \cfrac{1}{1 + \cfrac{1}{1 + \cdots}}}$$has all entries equal to $1$, so that its convergents are as far from $\phi$ as possible. This can be quantified and made precise (I guess: I'd love to know the precisely how!) It shares this property with all the numbers $\frac{a+b\phi}{c+d\phi}$ with $\left(\begin{smallmatrix}a&b\\\c&d\end{smallmatrix}\right)\in\mathrm{SL}(2,\mathbb Z)$ (and no others?)

Are there other numbers which are characterized by such extremal properties? Is there a second worst aproximable number?

  • $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ – Qiaochu Yuan Apr 26 '10 at 2:32
  • $\begingroup$ Qiaochu, that's precisely the link I gave! :) $\endgroup$ – Mariano Suárez-Álvarez Apr 26 '10 at 2:42
  • $\begingroup$ Oh. Well, in that case, doesn't the theorem stated count as a quantification of the statement about the golden ratio? (I think there is a strengthening of the statement as follows: if x is an irrational number which is not one of the numbers (a + b phi)/(c + d phi), then the constant in the theorem can be improved from sqrt{5} to some other constant. I remember seeing this on another Wikipedia page, but can't find it.) $\endgroup$ – Qiaochu Yuan Apr 26 '10 at 4:34
  • 3
    $\begingroup$ Qiaochu is right. The next number after sqrt(5) is sqrt(8) and the one after is sqrt(221)/5, and in general you get an infinite sequence, tending to 3. This sequence (or rather its reciprocal) goes by the rather unfortunate name of "the Markoff chain" (see Cassels Ann Math 1949); see for example section 6.3 of Baker's "Concise introduction to the theory of numbers". $\endgroup$ – Kevin Buzzard Apr 26 '10 at 9:46
  • $\begingroup$ Incidentally, in Manfred Schroeder's "Number theory in science and communication" he explains why audio speakers should have dimensions 1 by phi by phi^2, in that they should be as far as possible from resonances. $\endgroup$ – Allen Knutson Apr 26 '10 at 14:27

Certainly there is one type of study of this question, coming under the heading of the Markoff Spectrum, going back to the German tranliteration of Markov. There is a very nice book called "The Markoff and Lagrange Spectra" by Thomas W. Cusick and Mary E. Flahive. It is this topic that lead to the Markoff Numbers, which are any of the triple $x,y,z$ of a positive integer solution to $$ x^2 + y^2 + z^2 = 3 x y z , $$ see


If you are curious, there is my paper with Kaplansky in the Illinois Journal, pdf at: http://zakuski.math.utsa.edu/~jagy/bib.html

I am especially proud of section 8.3, called "Other Families," as I was able to construct a mirror image of our "Markov Ratios" $$ 9 - \frac{4}{m^2} $$ in forms that gave $$ 9 + \frac{4}{m^2} $$

To clarify, any indefinite binary quadratic form with integer coefficients and non-square discriminant has a nonzero "minimum" which is the smallest nonzero absolute value of any value obtained with integral values for the arguments. Our "Markov Ratio" was simply the discriminant divided by the square of this minimum.

Well, your Golden Ratio has the smallest possible Markov Ratio, with 5, the form being equivalent to $x^2 + x y - y^2.$ The second smallest is 8, form $2 x^2 + 4 x y - 2 y^2$ which is visibly imprimitive.

By the way, the "reduced" forms with the rather inflated middle coefficient (see chapter 3, Duncan A. Buell, Binary Quadratic Forms) correspond with purely periodic continued fractions, which I gradually came to realize are the best way to compute things: the cycle of a reduced form eventually arrives back at its precise starting point, so there is no need to compare with a long list of previously computed forms. If you simply compute the continued fraction for a number such as $\sqrt{5}$ you get some non-repeated coefficients or "quotients" $a_0 ; a_1 $ at the start.

If you look at the Cusick and Flahive book, you will see how the Markoff and Lagrange spectra coincide with "Markov Ratio" below 9 but are rather different above, where the description using binary forms is no longer adequate.

For the hasty: http://en.wikipedia.org/wiki/Markov_spectrum

  • $\begingroup$ You're welcome, but in fact I am delighted to have a chance to talk about this. $\endgroup$ – Will Jagy Apr 26 '10 at 3:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.