Dirichlet approximation theorem and BA numbers I think some version of the following definition given in "Diophantine Approximations" by W.M. Schmidt:

Let $\sigma \in (0,1)$ and $x \in \mathbb{R}$. Then $x$ is 
  $\sigma$-improvement for Dirichlet theorem  if there is $Q_0$ so
  that for all $Q > Q_0$ there are $p,q \in \mathbb{Z}$, so that $ 0 < q
 < \sigma Q$ and $\| x - \frac{p}{q} \| \leq \frac{\sigma}{qQ}$.

I'm having a bit of trouble understanding the following question, making use of the above mentioned definition:

If $x$ is an irrational $\sigma$-improvement for Dirichlet theorem for
  some $\sigma \in (0,1)$, then $x \in BA$ (there is $c > 0$ so that for
  all $p,q \in \mathbb{Z}$, $\|x - \frac{p}{q}\| \geq \frac{c}{q^2}$).

I think it's something that has to do with continued fractions, but I'm not sure how to use it here. And even before that, I'm not sure where is the contradiction if I assume the opposite.
And the other direction (kind of):

If $x$ is $\sigma$-improvement for Dirichlet theorem for all $\sigma \in (0,1)$, then $x$ is rational.

Intuitively, this one seems true, as we approximate $x$ as tightly as we want. I'm not sure what theorem makes it true, though.
I'll appreciate any clue.  
 A: You are saying that this is homework, so I won't develop a complete solution yet, but I think I can set you in the right direction.
First of all, note that the second question is, as you almost wrote, really easy given the first, as being a $\sigma$-improvement of Dirichlet's theorem implies the existence of infinitely many high quality approximations:
$$\left|x-\frac{p}{q}\right| \le \frac{\sigma}{q^2}$$
If those exist for all $\sigma\in(0,1)$, then $x$ certainly can't be badly-approximable, but the first part proved that if it's irrational it must be BA, so $x$ can't be irrational altogether.
As for the first part, I think your continued fraction intuition is correct. Recall that a number $x$ is BA if and only if its continued fraction approximation coefficients are bounded. Thus, if you assume $x\notin BA$, there must be infinitely many arbitrarily large coefficients.
If you observe that $q_n=a_nq_{n-1} + q_{n-2}$, where $q_k$ is the denominator of the $k$th continued fraction approximation of $x$ and $a_n$ is the $n$th coefficient, you can see that if $a_n > \frac{1}{\sigma}$, then $q_{n-1}<\sigma q_n$, and then the $\sigma$-improvement property will apply for the previous continued fraction approximation, as it is the best approximation with denominator lower than $q_n$.
I believe that you can then a lower bound on $\left|x-\frac{p_n}{q_n}\right|$ using the condition imposed by the $\sigma$-improvement and the equality
$$\left|x-\frac{p_k}{q_k}\right|=\frac{1}{x_{k+1}q_k^2+q_kq_{k-1}}$$
where $x_{n+1}$ denotes the tail $[a_{n+1},a_{n+2},\ldots]$ of the approximation of $x$, for a few values of $k$ around $n$.
