We have an algebraic number $a$ and a real number $b$. Can the following inequality have infinitely many solutions for $n \in \mathbb{N}$?


$$ \{an\} \in [b - \frac{1}{2^n}, b + \frac{1}{2^n}] $$


Here $\{x\}$ denotes the fractional part of $x$, $\{x\} = x - [x]$.


**Background**. I encountered this problem when I was dealing with some kronecker approximations, I wanted to show that for a given point $(b\_1t \mod 2\pi, b\_2t \mod 2\pi)$ where $b\_1,b\_2$ are algebraic numbers who are linearly independent over rational numbers, one can not get exponentially close to a given fixed point $(\phi\_1,\phi\_2) \in [-1,1]^2$. Some special cases of this problem could be solved by Baker but I was unable to solve it in the general case.



Almost Diophantine approximation
Such $b$ exists for every real $a$.


Define an increasiung sequence of positive integers $n\_1,n\_2,\dots$ as follows. Picj $n\_1$ so that $\{an\_1\}<1/2$. If $n\_i$ has been already defined, choose $n\_{i+1}>n\_i$ such that
$$
 \{an\_{i+1}\}\in\left[\{an\_i\},\{an\_i\}+\frac1{2^{n\_i+1}}\right];
$$
such $n\_{i+1}$ clearly exists (notice here that $\{an\_i\}\leq 1-1/2^{n\_i}$). Then the sequence of segments
$$
 \left[\{an\_i\},\{an\_i\}+\frac1{2^{n\_i}}\right]
$$
is nested, hence its intersection is a desired point $b$.


