Littlewood Conjecture

Littlewood Conjecture

Definitions and mathematical statement

For $x\in{\mathbb{R}}$, write $\|x\|=\min_{m\in{\mathbb{Z}}}|x-m|$, its distance to the nearest integer. The assertion is \[ \forall\alpha,\beta\in{\mathbb{R}},\qquad \liminf_{q\to\infty,\ q\in{\mathbb{Z}}_{>0}}q\|q\alpha\|\|q\beta\|=0. \] Equivalently, for every $\varepsilon>0$ and every threshold $Q$ there is $q>Q$ with the displayed product below $\varepsilon$. Both numbers are approximated using the same denominator. No irrationality or independence hypothesis is imposed.

Formulation and concept references: [S1].

Short English statement

Can any two real numbers be simultaneously approximated with a common denominator so that Littlewood's scaled error product becomes arbitrarily small?

Research attempt

Scope and checked context.

This unresolved attempt by GPT-6 Astra Ultra was prepared on 27 September 2026. It proves several classes of pairs satisfy the stated liminf, then tests a proposed strategy using continued-fraction denominators. Finally it compares the ordinary conjecture with the different uniform statement appearing in the recent inherited sources.

Einsiedler--Katok--Lindenstrauss [E1] proved that the set of possible exceptions to the ordinary Littlewood conjecture has Hausdorff dimension zero. That is not an emptiness theorem. The 2026 manuscripts [S2, S3] concern the stronger uniform problem, whose formula is written explicitly below. Their abstracts were inspected for this distinction; their full proofs are not used as premises. The elementary subcases and reductions here are not claimed to be new.

Easy cases must retain unbounded denominators.

If $\alpha=a/b$ is rational, every positive multiple of $b$ makes $\|q\alpha\|=0$. There are such multiples beyond every prescribed threshold, so the liminf is zero for every $\beta$. The same applies with the coordinates exchanged. A single small denominator giving a small product would not be enough; these are arbitrarily large exact witnesses.

For irrational $\alpha$, suppose \[ \inf_{q\geq1}q\|q\alpha\|=0. \tag{95.1} \] All finitely many values with $q\leq Q$ are strictly positive. Thus $95.1$ supplies a sequence of denominators tending to infinity on which $q\|q\alpha\|\to0$. Since $\|q\beta\|\leq1/2$, \[ q\|q\alpha\|\|q\beta\|\leq\tfrac12 q\|q\alpha\| \longrightarrow0. \tag{95.2} \] Consequently any hypothetical counterexample must have both coordinates badly approximable, meaning each has a positive lower bound in $95.1$. This narrows the target without making a claim that all badly approximable pairs are understood.

Continued fractions make the first reduction quantitative.

Write an irrational $\alpha=[a_0;a_1,a_2,\ldots]$, with convergents $p_k/q_k$ and complete quotient $\alpha_{k+1}=[a_{k+1};a_{k+2},\ldots]$. The recurrences and determinant identity give \[ \alpha=\frac{p_k\alpha_{k+1}+p_{k-1}} {q_k\alpha_{k+1}+q_{k-1}},\qquad |q_k\alpha-p_k|=\frac1{q_k\alpha_{k+1}+q_{k-1}}. \tag{95.3} \] Indeed cross-multiplication leaves numerator $p_{k-1}q_k-p_kq_{k-1}$, of absolute value one. In particular \[ q_k\|q_k\alpha\|\leq q_k|q_k\alpha-p_k| <\frac1{a_{k+1}}\leq1. \tag{95.4} \] No assertion that $p_k$ is the nearest integer is needed for this upper bound. Unbounded partial quotients therefore imply the desired Littlewood liminf with every second coordinate. More precisely a subsequence with $a_{k+1}\to\infty$ gives products at most $1/(2a_{k+1})$. Formula $95.3$ supplies the approximation step explicitly rather than using numerical continued fractions as evidence of a limiting property.

All rationally dependent pairs satisfy the conjecture.

Suppose $u\alpha+v\beta+w=0$ with integers $u,v,w$, not all zero. The cases $u=0$ or $v=0$ reduce to a rational coordinate. In the other case assume $\alpha$ irrational and take its convergent denominators $q_k$. Choose the common denominator $q=|v|q_k$. The elementary inequality $\|m t\|\leq|m|\|t\|$ for integers $m$ gives \[ \|q\alpha\|\leq|v|\|q_k\alpha\|, \qquad \|q\beta\|\leq|u|\|q_k\alpha\|. \] For the second inequality use $q\beta=-\operatorname{sgn}(v)u q_k\alpha -\operatorname{sgn}(v)w q_k$; its last term is an integer. By $95.4$, \[ q\|q\alpha\|\|q\beta\| \leq|u|v^2q_k\|q_k\alpha\|^2 <\frac{|u|v^2}{q_k}\longrightarrow0. \tag{95.5} \] Thus the remaining possible obstruction has $1,\alpha,\beta$ rationally independent as well as both coordinates badly approximable. These restrictions follow from proved subcases; they are not assumptions inserted into the original statement.

A denominator-subsequence route and an explicit failure test.

Equation $95.4$ gives another sufficient condition for any irrational $\alpha$: if along some subsequence of its convergent denominators \[ \|q_k\beta\|\to0, \tag{95.6} \] then the Littlewood products tend to zero along that subsequence. It is tempting to argue that the fractional parts $q_k\beta$ must approach zero because $q_k$ grow and $\beta$ is fixed. Growth alone does not justify that assertion.

For an exact test, take $\alpha=1/\sqrt2=[0;1,2,2,\ldots]$ and $\beta=1/2$. Its convergent denominator recurrence starts with $q_0=q_1=1$ and continues as $q_{k+1}=2q_k+q_{k-1}$. Every $q_k$ is odd, so $\|q_k\beta\|=1/2$ for every $k$. Condition $95.6$ fails completely, although the original conjecture is trivial here using even denominators. This example does not address the difficult rationally independent case. It shows that restricting the search to one distinguished denominator sequence can discard all the easiest witnesses.

In particular, proving that $95.6$ holds for every difficult pair would be a stronger intermediate claim requiring an argument of its own. Independent Dirichlet approximations for $\alpha$ and $\beta$ also do not automatically share a denominator. Taking the product of two good denominators can multiply each approximation error by the other denominator; the extra factors cannot be omitted from the scaled product.

The uniform conjecture has a different order of quantifiers.

Define \[ a_n=\|n\alpha\|\|n\beta\|,\qquad U(Q)=Q\min_{1\leq n\leq Q}a_n. \] The uniform Littlewood statement discussed in [S2, S3] asks $U(Q)\to0$ as $Q\to\infty$ for all pairs. The catalogue instead asks $\liminf_{n\to\infty}n a_n=0$.

The uniform assertion implies the ordinary one. Choose a minimizer $n_Q$. If some $a_n=0$, a coordinate is rational and its multiples prove the ordinary statement. Otherwise every finite set of $a_n$ has positive minimum; $U(Q)\to0$ forces the minimizing indices to become unbounded along a subsequence. On that subsequence, \[ n_Qa_{n_Q}\leq Q a_{n_Q}=U(Q)\to0. \] This argument does not reverse: a rare good denominator controls the ordinary liminf but may be much smaller than a later cutoff $Q$, causing the factor $Q/n_Q$ to become large.

An elementary sequence illustrates the logical obstruction without purporting to arise from a pair of real numbers. Let $n_j=2^{2^j}$ and put $a_{n_j}=1/(j n_j)$, taking $a_n=1/4$ at all other indices (start at $j=2$ to avoid initial exceptions). Then $n_j a_{n_j}=1/j\to0$. But at $Q=n_{j+1}-1$, the minimum so far is $1/(j n_j)$ and \[ U(Q)=\frac{n_{j+1}-1}{j n_j}\to\infty. \tag{95.7} \] Even nonnegative bounded values permit this separation of quantifiers. A claimed disproof of the stronger uniform statement therefore cannot by itself disprove the ordinary conjecture.

Why a small exceptional set and finite searches do not finish.

The dimension-zero theorem [E1] is powerful information about any possible exceptional set. However every singleton has Hausdorff dimension zero: covering it by an interval of diameter $\delta$ makes its $s$-dimensional covering cost $\delta^s\to0$ for every $s>0$. Consequently that theorem does not certify a specified pair by itself. No particular point is asserted to be exceptional here.

Likewise a finite search computes only a finite minimum. It can certify that one explicitly bounded interval of denominators contains a witness, but the conjecture asks for witnesses beyond every threshold and below every positive tolerance. A displayed small product, or a long stretch without one, cannot replace either unbounded quantifier.

Verification and outcome.

The companion script checks the determinant and parity recurrences in $95.3$--$95.6$ with exact integers, the coefficient bookkeeping of the rational dependence reduction, and the minima in $95.7$ for finite initial segments. It does not infer an irrational's infinite continued fraction from rounded decimal data. The constants in $95.5$ are algebraic upper bounds proved in the text.

The resulting unconditional subcases include every rational coordinate, every coordinate satisfying $95.1$, and every rational dependence among $1,\alpha,\beta$. The first unresolved step is to find arbitrarily large common denominators with vanishing product for every remaining pair of rationally independent badly approximable coordinates. Neither the convergent-subsequence strategy nor uniform-counterexample literature supplies that missing theorem. The Littlewood conjecture remains UNRESOLVED.

Sources

• S1 Steven Robertson, Combinatorics on Number Walls and the P(t)-adic Littlewood Conjecture, Section 1, 2025 revision. https://arxiv.org/html/2307.00955v3. Formulation source.
• S2 Disproof of the uniform Littlewood conjecture. https://arxiv.org/abs/2603.12611. Further reference; not a proof endorsement.
• S3 The uniform Littlewood conjecture fails on a set of positive Hausdorff dimension. https://arxiv.org/abs/2608.25059. Further reference; not a proof endorsement.

• E1 Manfred Einsiedler, Anatole Katok, and Elon Lindenstrauss, Invariant measures and the set of exceptions to Littlewood's conjecture, Annals of Mathematics 164 $2006$, 513--560, Theorem 1.5. https://annals.math.princeton.edu/2006/164-2/p04. https://annals.math.princeton.edu/wp-content/uploads/annals-v164-n2-p04.pdf. Primary source for the dimension-zero exceptional-set theorem; it does not assert that the exceptional set is empty. Consulted 27 September 2026.
• E2 Johannes Schleischitz, Disproof of the uniform Littlewood conjecture, arXiv:2603.12611v1 (13 March 2026), and Nikita Shulga, The uniform Littlewood conjecture fails on a set of positive Hausdorff dimension, arXiv:2608.25059v1 (25 August 2026). https://arxiv.org/abs/2603.12611v1. https://arxiv.org/abs/2608.25059v1. Version-specific scope checks of [S2, S3]. Abstracts inspected to distinguish the uniform problem from the ordinary liminf conjecture; complete proofs not independently audited or used here. Consulted 27 September 2026.