Simultaneous approximation by rationals with relatively prime numerators The following seems hard to me (or perhaps just not true), but perhaps I am mistaken.  It is known that given irrational numbers $x_1$ and $x_2$, there are infinitely many simultaneous rational approximations:
\begin{align*}
|x_1 - p_1/q| &< q^{-3/2},\\
|x_2 - p_2/q| &< q^{-3/2}.
\end{align*}
If $x_1$ and $x_2$ are $\mathbb{Q}$-linearly independent, is there any chance I can take $p_1$ and $p_2$ relatively prime in these approximations?
Thanks. 
 A: It is false. Consider the numbers of the form
$$
   \sum_{n=0}^{\infty} c_n 10^{-n!}
$$
where $c_n\in\{3,9\}$. I claim that any two numbers from the set yield a counterexample. (As long as one avoids $\mathbb{Q}$-dependent pairs, which is easy since there are uncountably many numbers of this form.) 
Let $\alpha$ be such a number, and consider rational approximations to $\alpha$. There are obvious rational approximations with denominator $10^{n!}$; denote them by $p_{n,\alpha}/q_n$. Here $q_n=10^{n!}$. We have
$$
  \left\lvert\alpha-\frac{p_{n,\alpha}}{q_n}\right\rvert\leq \frac{10}{q_{n+1}}.
$$
Recall that $$\lvert p_{n,\alpha}/q_n-p/q\rvert\geq 1/q_nq$$ if the two rational numbers are unequal. Hence if $$\lvert \alpha-p'/q'\rvert<(q')^{-3/2},$$ then either $q'<Cq_n^{1/2}$ or $q'>Cq_{n+1}/q_n$. In view of spacing in the sequence $\{q_n\}$, that implies that the only good rational approximations to $\alpha$ are $\{p_{n,\alpha}/q_n\}$. Note however, the numerator in such an approximation is always divisible by $3$. 
