Most people are only familiar with a handful of numbers that can’t be written as fractions, like $latex sqrt{2}$ or $latex pi$. But such numbers, called irrational numbers, are far more plentiful than fractions or rational numbers.
How easy are they to approximate with fractions? If you use a fraction with an arbitrarily large denominator, you can get arbitrarily close. (As is well known, 22/7 gives a decent approximation of $latex pi$; 355/113 is even better.) But some irrational numbers are harder to approximate than others, meaning that you need to use a very big denominator to get a close approximation. The very toughest turns out to be the golden ratio, $latex phi$, or $latex (1+ sqrt{5})/2$. It is, in a specific mathematical sense, the number that is “farthest” from being rational.
What’s the next-farthest? And the next? The sequence of tough-to-approximate irrational numbers turns out to be given by the integer solutions to a deceptively simple equation that has no obvious connection to approximating irrational numbers. This connection was proved by Andrey Markov, a prescient Russian mathematician, in 1879.
Markov is famous for coming up with a concept in probability theory called Markov chains, which are used in everything from Google’s PageRank algorithm to models of DNA evolution. But though the solutions to his equation, called Markov numbers, are not nearly as well known, they arise in a vast range of mathematical disciplines, including combinatorics, number theory, geometry and graph theory.
“It’s not just an equation, it’s a kind of method,” said Oleg Karpenkov, a mathematician at the University of Liverpool. “These numbers are central, deep inside mathematics … structures like this one are the kinds of ideas that are rare.”
His equation, $latex x^2+y^2+z^2=3xyz$, has an obvious integer solution when x, y and z are all 1 (since 1 + 1 + 1=3 × 1). It turns out that all the integer solutions to the equation are connected by a simple rule. Start with a solution (a, b, c). Then the related triplet (a, b, 3ab − c) is also a solution. The first two numbers stay the same, while c, the third, is replaced by 3ab − c. Apply this rule to (1, 1, 1) and you get (1, 1, 2). (It’s easy to check that inputting these values makes both sides of the equation equal to 6.) Apply the rule again, and you’ll be back where you started, since 3 − 2=1. But if you flip the order of the numbers in the triple before applying the rule, it creates a whole universe of solutions. Input (1, 2, 1) and you’ll get (1, 2, 5).
Up until now, because of the identical 1s, the tree (shown in the illustration at the beginning of this story) doesn’t branch — the first few steps grow the trunk of the tree, so to speak. But if you start with a solution with three different numbers, like (1, 2, 5), the branches start to proliferate. Input (5, 1, 2) and you get (2, 5, 29). But (2, 5, 1) results in (1, 5, 13). (If you input (1,2,5) then the rule takes you back down to a lower branch of the tree.) From this point on, every solution has three different numbers, so every branch of the tree leads to two new branches.
The leftmost branch of the tree might look familiar — it contains every other number in the Fibonacci sequence, one of the best known in mathematics (each number in this sequence is the sum of the two previous terms: 1, 1, 2, 3, 5, 8, 13, 21, 34, …). The rightmost branch similarly contains every other term in the Pell sequence, a related, if slightly less famous, sequence. The way these sequences appear in the tree of solutions is “one of the most beautiful things in mathematics I know,” said Alexander Gamburd, a professor at the City University of New York.
Markov’s 1879 theorem, relating each triplet to a difficult-to-approximate irrational number, was the first hint that this equation might resonate deeply throughout math. In a 2013 book on the subject, Martin Aigner, an Austrian mathematician who died in October, called the theorem “without doubt one of the all-time classics in number theory.”
In 1913, Georg Frobenius, a German mathematician who made wide-ranging contributions to algebra, number theory, and the study of differential equations, noticed something curious about the Markov triples. Each largest number seemed to uniquely determine the two smaller ones. A number — take 5 for example — might appear in many triplets, such as (1, 2, 5), (1, 5, 13), (2, 5, 29) and so forth. But, he observed, if you look only at the largest number in each triplet, it will be affiliated with only one pair of smaller numbers.
Because the numbers grow so quickly, it is far from obvious that this should be true. For example, take the triplet (5, 433, 6,466). It’s not readily apparent that if you set z to 6,466, the only possible x and y that will solve the equation are 5 and 433. But as far as Frobenius could tell, the largest number always uniquely determined the two smaller ones. In the 110 years since, despite an enormous body of research connecting the Markov numbers to other problems, nobody has been able to prove what has come to be known as the uniqueness conjecture.
The relative simplicity of the conjecture illustrates a common mathematical paradox. Tools like the Markov equation can be used to prove subtle and complicated results even while basic questions about their properties remain unresolved.
However, in the last few years, there has been some notable progress toward proving the uniqueness conjecture. It has long been known that it’s possible to create a correspondence between every Markov triple and all the fractions between zero and 1. For each fraction p/q, which is called an index, you can assign a Markov number mp/q by following a particular mathematical procedure. For example, m2/3 is 29, and m3/5 is 433.
In 2013, Aigner made three conjectures about how the triples can be ordered using this correspondence. These conjectures are stepping stones on the way to proving the uniqueness conjecture. He hypothesized that if you keep the numerator of the index constant and increase the denominator (as in 1/2, 1/3, 1/4, 1/5, …), the respective Markov numbers will keep getting bigger. Likewise, he thought that if you increase the numerator but keep the same denominator (as in 1/17, 2/17, 3/17, 4/17, …), you should also get a string of ever-larger Markov numbers. The same pattern of increasing numbers, he thought, should hold if the sum of the numerator and denominator is kept constant (as in 1/100, 2/99, 3/98, …).
The constant-numerator conjecture was proved in a 2020 paper in Advances in Mathematics by Michelle Rabideau of the University of Hartford and Ralf Schiffler of the University of Connecticut. In February 2023, together with two other collaborators, Rabideau and Schiffler published a proof of the other two conjectures as well.
Because of these and other advances, Karpenkov is optimistic that a proof of Frobenius’ uniqueness conjecture might finally be in the works. “I know people who say they are near proving it,” he said. “I think we are pretty close — maybe within the next five years.”
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