My goal in this post is to describe a surprising and beautiful method (the Stern-Brocot tree) for generating all positive reduced fractions. I’ll then discuss how properties of the tree yield a simple, direct proof of a famous result in Diophantine approximation due to Hurwitz. Finally, I’ll discuss how improvements to Hurwitz’s theorem led Markoff to define another tree with some mysterious (and partly conjectural) similarities to the Stern-Brocot tree.
The Stern-Brocot tree
We first describe what I’ll christen the “extended Stern-Brocot tree” . It’s an ordered binary tree whose root note is labeled by the triple . Each node in the tree is labeled with an ordered triple of rational numbers (plus ), written in lowest terms, and each node has two children. If a node is labeled by the triple , then its left child is labeled and its right child is labeled , where denotes the mediant of and . For example, the two children of the root are and . And the two children of are and . Here is a depiction of all descendents of in the extended Stern-Brocot tree:
To get the Stern-Brocot tree , we keep just the mediants (the middle entry) and throw away the first and last entry in each triple:We can construct the Stern-Brocot tree T without reference to the extended tree by the following recursive rule: each fraction is the mediant of its closest ancestor to the left and its closest ancestor to the right, where a “left ancestor” of is an ancestor which is less than and a “right ancestor” is one which is greater than .
The Stern-Brocot tree has a number of remarkable properties, the most notable of which is that every positive reduced fraction appears exactly once in the tree. Let’s see why this is the case…
Properties of the Stern-Brocot tree
We record the following properties of the Stern-Brocot and extended Stern-Brocot trees:
(a) Every triple occurring in has .
Proof: A simple calculation shows that the mediant of two distinct fractions and is always strictly between and .
(b) For every triple occurring in , we have , where .
Proof: This holds for the root node, and a simple calculation shows that if it holds for a given node then it also holds for both children; the result follows by induction.
(c) Every fraction appearing in is reduced.
Proof: If are reduced and , then and thus the GCD of and is 1. The result now follows from (b) by induction.
(d) Given , let be the least common ancestor of and . Then iff is a descendent of (or equal to) the left child of and is a descendent of (or equal to) the right child of .
Proof: Again, this is a straightforward induction.
(e) No fraction appears more than once in .
Proof: This follows immediately from (d).
(f) For every triple occurring in , has a smaller denominator than any other fraction between and .
Proof: Let be any fraction between and . Then
It follows that , and the right-hand side is the denominator of . If equality holds, then , which implies that and .
(g) Every reduced positive fraction appears in .
Proof: (Sketch) Assume for simplicity that (the general case is similar). We claim that appears at the latest in row of . Indeed, if has not already appeared by row , then there is a triple in row of with . By induction and the uniqueness part of (f), it follows that .
Combining (c), (e), and (g), we find:
Theorem: Every reduced fraction appears exactly once in
Stern-Brocot expansions of real numbers
Every positive real number has a “Stern-Brocot expansion” consisting of a (possibly empty, possibly infinite) string of L’s and R’s determined by starting at the root and moving downwards through . At each stage, we move to the node labeled by unique triple with , stopping if , and we record whether we went left or right in order to reach .
For example, the Stern-Brocot expansions of and are and , respectively. In this way, every positive rational number corresponds to a unique finite string (with 1 corresponding to the empty string), and every positive irrational number corresponds to a unique infinite string. The inverse-golden ratio corresponds to the infinite string and a result of Euler implies that the irrational number corresponds to the infinite string
(If you’re familiar with continued fractions, there is a simple dictionary between Stern-Brocot and continued fraction expansions: a finite Stern-Brocot expansion of the form or with all corresponds to the finite continued fraction , and an infinite Stern-Brocot expansion of the form corresponds to the infinite continued fraction )
Given a real number , the Stern-Brocot triples corresponding to are the triples appearing as vertices in the rooted path in which we just described. The mediants of these triples are called the semiconvergents of . For example, when the corresponding Stern-Brocot triples are , and , and the semiconvergents are .
For the inverse-golden ratio the semiconvergents are (ratios of consecutive Fibonacci numbers). The last semiconvergent of a rational number is equal to , and if is irrational then the semiconvergents converge to .
(The name semiconvergent comes from the fact that every convergent of the corresponding continued fraction is a semiconvergent, but not necessarily vice-versa.)
The most important fact about semiconvergents is that they give the closest approximations to with a given denominator, i.e., a reduced fraction is a semiconvergent of iff is closer to than any other fraction of the form with .
This leads us into the theory of Diophantine approximation, which is precisely concerned with such approximations by rational numbers. The cornerstone of this theory is Dirichlet’s Approximation Theorem, which is the following assertion:
Theorem (Dirichlet): If is a positive irrational number, there are infinitely many reduced fractions with
- The “trivial” bound in Dirichlet’s theorem would be , since for every positive integer there is a rational number with (and this holds whether or not is irrational).
- The converse of Dirichlet’s theorem is true as well: if is rational then there are only finitely many reduced fractions with So Dirichlet’s theorem allows us to distinguish rational numbers from irrational ones in a rather interesting and non-trivial way.
- The exponent of 2 in Dirichlet’s theorem cannot be improved, since by a famous theorem of Klaus Roth (for which he won the Fields Medal), if is an algebraic number then for each and every constant , there are only finitely many reduced fractions with
- The constant term in Dirichlet’s theorem can, however, be improved. This is the subject of Hurwitz’s theorem, to which we turn in the next section.
- Dirichlet’s Approximation Theorem has a number of applications, e.g. it can be used to prove that Pell’s equation has infinitely many integer solutions for every non-square positive integer . And the Stern-Brocot tree (or, equivalently, the theory of continued fractions) can be used to explicitly find all solutions.
As a motivation for Hurwitz’s theorem, we note that if is the inverse-golden ratio, a simple calculation shows that for any there are only finitely many reduced fractions with
Indeed, let denote the other root of the quadratic polynomial satisfied by . Then since has no rational roots, for any rational number we have . Writing , we see that if then After a bit of algebra, we see that this implies , which (since ) cannot hold when is large.
Adolf Hurwitz proved that , and the corresponding constant of , is the “worst-case scenario” in Dirichlet’s theorem:
Theorem (Hurwitz): If is a positive irrational number, there are infinitely many reduced fractions with
Proof of Hurwitz’s theorem
As a warm-up, we use the Stern-Brocot tree to give a quick proof of Dirichlet’s theorem. Indeed, we claim that in any Stern-Brocot triple corresponding to , either or must satisfy the conclusion of Dirichlet’s Approximation Theorem. The proof of property (g) above shows that each rational number can appear in only finitely many Stern-Brocot triples, so we’re done once we establish the claim. And the claim is easy to prove: if and , then since and the denominator of is at least as large as the denominator of . Similarly, if then satisfies the conclusion of Dirichlet’s theorem.
To prove Hurwitz’s theorem, we refine the above observation by proving that in any Stern-Brocot triple corresponding to , either , , or must satisfy the conclusion of Hurwitz’s Theorem. It suffices to prove this in the case , since if we may replace by , by , and by .
Assuming, therefore, that , if the conclusion of Hurwitz’s theorem fails for each of then (setting ):
Adding (1) and (3) and simplifying using property (b) gives Similarly, adding (2) and (3) gives
Adding these two inequalities gives Using the fact that and simplifying, this is equivalent to where is the golden mean. But then , which is impossible since the is irrational.
Hurwitz in fact proved more than just the above. He also showed that one could improve on the above theorem for irrational numbers which are not -equivalent to (e.g. ). More precisely, if are real numbers, we write if there exist integers with such that If we replace with any real number , it is still true that for any there are only finitely many reduced fractions with However, Hurwitz showed that if , then one can do better, with an optimal constant of instead of . This prompts the following definition:
If is a positive irrational number, its Lagrange number is the supremum of all such that there are infinitely many reduced fractions with
It’s not difficult to show that -equivalent irrationals have the same Lagrange number.
The Lagrange spectrum is the set of all numbers of the form for a positive irrational number. By what we’ve already said, we know that and belong to the Lagrange spectrum.
Markoff and his equation
Andrei Markoff (also spelled Markov – he’s the same mathematician for whom Markov chains are named) discovered a remarkable connection between the Lagrange spectrum and integer solutions to the Diophantine equation
now known as the Markoff equation.
In order to state Markoff’s result, we define a Markoff number to be a positive integer such that for some integers , i.e., such that is a solution to the Markoff equation. The sequence of Markoff numbers begins
1. The Lagrange spectrum is discrete in the interval . On the other hand, 3 is a limit point of the Lagrange spectrum.
2. More precisely, the Lagrange spectrum below 3 is precisely the set of real numbers where is a Markoff number.
As one of the steps toward proving this theorem, Markoff established the remarkable fact that all integer solutions to his equation can be described by an algorithmic procedure similar to the one which we used to define the (extended) Stern-Brocot tree!
The extended Markoff tree
The observation which gets everything going is that if is a solution to the Markoff equation with , then and are also solutions (in the same numerical order). We’ll call and the degenerate Markoff triples, and all other solutions with non-degenerate Markoff triples. The smallest non-degenerate Markoff triple is .
The extended Markoff tree is a rooted tree in which each node is labeled with a Markoff triple, with the root node labeled . The unique child of is , and the unique child of is . The subtree consisting of the node labeled with the non-degenerate Markoff triple and all its descendants is an ordered binary tree: if a node is labeled by the non-degenerate Markoff triple , its left child is labeled and its right child is labeled :The following result is surprising, but not very difficult to prove:
Theorem (Markoff): Every integer solution to the Markoff equation with appears precisely once in the extended Markoff tree.
The Markoff tree and the uniqueness conjecture
We obtain the Markoff tree M from the extended Markoff tree in the same way that we obtained the Stern-Brocot tree from its extended version: by forgetting and and just keeping the middle element of each Markoff triple. The Markoff numbers are (by the symmetry of the Markoff equation) precisely those numbers which appear somewhere in the Markoff tree.
The major unsolved problem in the subject is:
Conjecture (Markoff’s Uniqueness Conjecture, Version 1): Every Markoff number appears exactly once in the Markoff tree.
The interest in this conjecture stems partly from its apparent degree of difficulty and partly from its interesting implications. For example, the uniqueness conjecture is equivalent to the following statement which does not mention Markoff’s equation at all but relates directly to the Lagrange spectrum:
Conjecture (Markoff’s Uniqueness Conjecture, Version 2): If are irrational numbers with , then .
For an extensive history and mathematical discussion of the Markoff uniqueness conjecture (in these and other forms), see the lovely book .
- The basic properties of the Stern-Brocot tree yield a constructive proof (which does not make explicit use of the extended Euclidean algorithm) that if are relatively prime positive integers then there are positive integers such that . (Take to be the triple corresponding to the rational number and let .)
- Achille Brocot was a French clockmaker who used the Stern–Brocot tree to design systems of gears with a gear ratio close to a particular desired value.
- Another characterization of equivalence is that are -equivalent if and only if their continued fraction expansions coincide from some point onward (i.e., they have the same “tail”); see  for additional information.
- There is a real number , called Freiman’s constant, such that every real number greater than belongs to the Lagrange spectrum but for every there is a real number in which does not belong to the Lagrange spectrum. Explicitly (as you would no doubt guess), In the interval the Lagrange spectrum is complicated (a fractal-like blend of discrete and continuous parts).
- The Stern-Brocot tree is closely related to Farey sequences, which are in turn closely related to Ford circles. Ford circles allow for an appealing geometric visualization of the Stern-Brocot tree:
 M. Aigner, Markov’s Theorem and 100 Years of the Uniqueness Conjecture, Springer-Verlag, 2013.
 R. Graham, D.E. Knuth, and O. Patashnik, Concrete Mathematics (2nd edition), Addison-Wesley, 1994.
 J.D. Sally and P.J. Sally, Jr., Roots to Research, American Mathematical Society, 2007.