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](http://math.stanford.edu/~akshay/research/eklexp.pdf) 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 $\Vert q\_n(y) x \Vert \to 0$.



Simultaneous rational approximation of two reals using their continued fractions
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.


