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Littlewood's well-known conjecture about simultaneous rational approximation is that for all $x, y \in \mathbb{R}$, $\liminf_{n \to \infty} n \Vert nx \Vert \Vert ny \Vert = 0$ (where $\Vert x \Vert$ denotes the distance from $x$ to the nearest integer).

A heuristic argument for this (mentioned in this survey article by Akshay Venkatesh) is as follows. Write $q_n(x)$ for the denominator of the $n$th convergent of the continued-fraction expansion of $x$. We know that $q_n(x) \Vert q_n(x) x \Vert$ is bounded, and it is reasonable to expect that $\Vert q_n(x) y \Vert$ will be small sometimes. Venkatesh points out that this argument doesn't work: in fact, given any sequence $q_n$ such that $\liminf q_{n+1} / q_n > 1$, we can find a $y \in \mathbb{R}$ such that $\Vert q_n y \Vert$ is bounded away from zero.

However, this doesn't quite rule out using the continued-fraction expansions of $x$ and $y$ to prove Littlewood's conjecture. The result would follow if it could be shown that for all $x, y \in \mathbb{R}$, either $\liminf_{n \to \infty} \Vert q_n(x) y \Vert = 0$ or $\liminf_{n \to \infty} \Vert q_n(y) x \Vert = 0$.

My question is whether this can be shown to be false. That is: do there exist $x$ and $y$ such that both $\Vert q_n(x) y \Vert$ and $\Vert q_n(y) x \Vert$ are bounded away from zero?

In light of gowers's answer below, let me add the condition that $x$ and $y$ are badly-approximable, i.e. that they have bounded partial quotients in their continued fractions. (It is still the case that a negative answer would imply Littlewood's conjecture.)

Here is an example from the realm of badly-approximable numbers. Let $x = \frac12 (1 + \sqrt{5})$ and $y = \frac12 (1 - 1/\sqrt{5})$. Then $q_n(x) = F_n$ (Fibonacci number), and $q_n(y) = 2F_n + F_{n+1}$ for $n \geq 1$. In this case, $\liminf \Vert q_n(x) y \Vert = \frac15 > 0$, but $\Vert q_n(y) x \Vert \to 0$.

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There's a recent paper of Badziahin, Pollington and Velani that answers a question that's similar in spirit to yours, but I'm not sure that it actually implies it. –  gowers Aug 14 '10 at 20:43
Oops, I meant to give a link: arxiv.org/pdf/1001.2694v2 –  gowers Aug 14 '10 at 20:43

2 Answers 2

up vote 4 down vote accepted

Yes, there are badly approximable numbers $x$ and $y$, in fact quadratic irrationals, such that $||q_n(x)y||$ and $||q_n(y)x||$ are bounded away from zero. Specifically, we can take $x = \sqrt{2}/2$ and $y = \sqrt{2} + 1/2$. It's straightforward to check that $q_n(\sqrt{2}/2)$ is always odd, from which it follows that $||q_n(x)y||$ tends to $1/2$ as $n \to \infty$. Unfortunately, I don't know of a similarly easy way to analyze $||q_n(y)x||$, but one can do it as follows. If we start numbering with $n=1$, then $q_n(y)$ is the nearest integer to $(1+\sqrt{2})^n$ times $(-2+3\sqrt{2})/8$, $\sqrt{2}/8$, $(2+3\sqrt{2})/8$, or $\sqrt{2}/4$, according to whether $n$ is $1$, $2$, $3$, or $4$ modulo $4$. (Specifically, $q_n(y)$ equals this number plus its conjugate over $\mathbb{Q}(\sqrt{2})$.) If we multiply by $\sqrt{2}/2$, then we get an integer divided by $8$, $8$, $8$, or $4$. One can check using the recurrence that this integer is always $2$ mod $4$, from which it follows that $||q_n(y)x||$ tends to $1/2$ when $n$ is $0$ mod $4$ and $1/4$ otherwise.

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Nice. Thank you! –  Alec Edgington Jul 5 '11 at 7:10

I think your question may need to be reformulated, to add the condition that x and y are badly approximable. Otherwise you could take x to be a number like 1.0001000000000000000000000100...0001000.... The convergents to this will be a rapidly growing sequence of powers of 10. Now we need to choose y. To do that, we can (I think) choose a number to satisfy various constraints. We need $10^ky$ to be far from an integer when $10^k$ is a convergent of x. We also need to make $m_ry$ very close to an integer for some rapidly growing sequence $(m_r)$ of numbers that we choose such that $m_rx$ is far from an integer. This gives us so much flexibility that I think we can achieve it just by making small adjustments to y to force it to satisfy each condition in turn. (For instance, if $m_r$ is much bigger than the last multiple we worried about, then we just move $m_ry$ so that it is extremely close to an integer, and the effect on earlier multiples of y is not enough to mess up what we've done so far.)

As ever, I'm not 100% certain that the above is correct, but I get a reasonably good feeling about it.

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Ah yes, that's quite persuasive! Thanks -- I have edited the question. –  Alec Edgington Aug 15 '10 at 7:27
This is suggestive of finite (or no) injury priority arguments. I wonder if an exact parallel can be made so that a number of "almost-computable" results in number theory can arise. Gerhard "From Diophantine To Zero Jump" Paseman, 2011.07.04 –  Gerhard Paseman Jul 4 '11 at 18:20

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