The irrationality exponent of π is 2

The explanation

Write π as a fraction and you will always be a little wrong. 22/7 misses by about 0.0013. 355/113 misses by about 0.00000027. Bigger bottom numbers buy better accuracy. The real question is how fast that accuracy can improve.

There is a basic yardstick. Take any number that is not a fraction. A simple counting argument guarantees endless fractions p/q that miss it by less than 1/q². Count the misses for q = 113 in millionths. The 1/q² yardstick allows 78 millionths. The much stricter 1/q³ allows 0.69 millionths. The actual miss of 355/113 is 0.27 millionths. That is about 290 times better than the basic yardstick, and it beats the strict one too. So π has at least one lucky fraction.

Mathematicians measure this luck with one number, the irrationality exponent. If a number keeps producing fractions that beat 1/q³ forever, its exponent is at least 3. An exponent of 2 means the opposite. Pick any power a little above 2, such as 2.01. Then only finitely many fractions beat 1/q^2.01, and past some size of bottom number the lucky streak is over. No number that is not a fraction can go below 2. Experts expected π to sit exactly there.

Knowing that π is not a fraction puts no limit on its luck at all. The earliest limit this paper cites is Kurt Mahler's, from 1953: the exponent is at most 42. Later work cut it down. Salikhov reached 7.606308… in 2008. Zeilberger and Zudilin reached 7.103205334137… in 2020. In September 2026, a preprint by Yufei Bai claimed µ(π) < 7.101862832357, where µ(π) is π's exponent. Each step came from a cleverer special integral built for the job. The gains had shrunk to the third decimal place, while the expected answer, 2, sat about 5 lower.

This paper proves the full answer: π's exponent is exactly 2. It does not shave the old bound. The earlier bounds are context, not inputs to its proofSource 1.

The argument runs by contradiction. Suppose π had endless fractions that beat 1/q^ν for some power ν above 2. Pick several of them, one after another. Use them to fill a large square grid of numbers. Then compute one number from the whole grid, called its determinant. The determinant works like a trap. Once fractions are cleared, it is built from whole numbers. So if it is not zero, it cannot be tinier than a size the proof can write down. Yet fractions that sit very close to π force it to be tiny. For large enough choices, both cannot be true. So the too-good fractions cannot exist.

The hard part is making sure the determinant is not zero. A zero would spring no trap, and the argument would collapse. The longest part of the paper is a new theorem that rules this out. It shows that certain polynomials can match any requested values and slopes at several chosen points at once. The proof must also avoid a circle. The grid's design is fixed before anyone knows where the lucky fractions areSource 2. Only then are the fractions chosen to fit itSource 3.

So the theorem forbids endless luck of the 355/113 kind, even luck as mild as beating 1/q^2.01. It still allows finitely many lucky fractions for each power above 2. The proof shows that a last lucky fraction exists. It gives no way to compute how large its bottom number isSource 4. It says the streak ends. It does not say when.

A side effect settles an old puzzle, the Flint–Hills series. It adds up 1/(n³·sin²n) for n = 1, 2, 3 and onward. Whole numbers that land very close to a multiple of π make sin n tiny and the terms largeSource 5. In 2022, Alex Meiburg showed that the sum converges if π's exponent is below 2.5. Converging means it settles to a finite total. An exponent of 2 clears that line, so the sum converges.

How sure is this? A computer proof checker called Lean has checked a formal version of the main theorem: the exponent is exactly 2. The Flint–Hills step is outside that check. No specialists have published a review yet.

What it doesn't show. This does not prove that π is irrational or transcendental. Those facts were proved in the 1760s and in 1882.