Littlewood conjecture

A fraction can approximate a number without equalling it exactly. Littlewood's conjecture concerns approximating two numbers with fractions that share a denominator. Multiply both numbers by a positive whole number, measure how far each result is from the nearest whole number, and multiply those two distances by that same multiplier. Can this score get as close to zero as we like, using arbitrarily large multipliers, for every pair of numbers?[1] [2]

Littlewood's conjecture asks whether a shared denominator always permits arbitrarily small multiplicative approximation scores for two real numbers. Write $\lVert x\rVert$ for the distance to the nearest integer. The classical statement is

$$\liminf_{n\to\infty} n\lVert n\alpha\rVert\lVert n\beta\rVert=0$$

for every $\alpha,\beta\in\mathbb{R}$, with $n$ a positive integer. If $p$ and $q$ are nearest integers to $n\alpha$ and $n\beta$, the score equals $n^3|\alpha-p/n|\,|\beta-q/n|$. Thus the two approximating fractions use the same denominator. For every positive tolerance and every starting threshold, the conjecture requires some larger $n$ with a smaller score. It does not require the score to stay small for all subsequent $n$.[1] [2]

The assertion for every pair is unproved. Possible exceptions form an extremely small set, but that set could still contain points.[1] [2]

The dimension-zero bound does not exclude every exceptional pair. The August 2026 preprint’s counterexamples to a stronger uniform formulation do not disprove this classical statement.[1] [2]

Approximating each number separately lets you choose two different denominators. Sharing one denominator couples the choices. This problem asks whether there is always enough freedom left to make the combined score very small, even for pairs that resist especially good approximation.[1] [2]

This is a universal statement about a coupled approximation problem, rather than two separate one-dimensional estimates. Measure-based methods have reduced the possible exceptional set to Hausdorff dimension zero, but that leaves the all-pairs assertion unsettled. Closing that gap would determine whether any pair can defeat the multiplicative tradeoff indefinitely.[1] [2]