I am not sure whether this is a suitable question for MO. We know the classical version of Dirichlet's approximation theorem that if $x$ is a real number and $Q>0$ there exist $p,q\in \mathbb{Z}$ with $0<q\le Q$ such that $|x-p/q|<1/qQ$.
I am looking for a version of this theorem where $q$ is only a prime power, say $2^n$. Then by binary approximation of real number one can immediately say that $|x-p/2^n|<1/2^n$. But can we do better along the line of Dirichlet?
Thanks for any reference.
You cannot do much better than what you observed, because the fractional parts $\{2^n x\}$ are essentially the tails in the binary expansion of $x$, so they can be bounded away from zero (even for rational numbers $x$).
However, Furstenberg (1967) proved that the fractional parts $\{2^m3^n x\}$ are dense in $(0,1)$ for any irrational $x$, and in fact the same holds for any nonlacunary semigroup. A simple proof was given by Boshernitzan (Proceedings of the AMS, 1994).
Minor observations:
If $x$ is rational with non-prime power denominator, you certainly can't do much better. If $x = 1/6$, then $|x-p/q| \geq 1/(6q)$ for any prime power $q$.
If $x$ is irrational, then Vinogradov showed that $\{ x q \}$ is equidistributed as $q$ runs over the primes. So, for any $\epsilon>0$, we can find a prime $q$ and an integer $p$ such that $|xq-p| < \epsilon$ or, in other words, $|x-p/q|<\epsilon/q$.
I can't decide whether or not I think, for all irrational $x$, that there should be a prime $q$ such that $|x-p/q|<c q^{-2}$. (For some constant $c$ independent of $x$.) I definitely think it should be true except for $x$ of measure zero -- heuristically, the odds that a particular $q$ works are $2c/q$, and the sum $\sum 1/q$ diverges.
