Shortest decimal expansion within binary interval
Decimal numbers with k digits to the right of the decimal point are $10^{-k}$ apart. So if you choose k large enough that $10^{-k} &lt; 2·2^n$, then there is a decimal number with k decimals inside the interval. You can for example calculate it as $round (x·10^k) / 10^k$. 

You may be lucky, and a number with fewer decimals may be inside the interval. Rounding the result to k-1 or fewer decimals may give a result in that interval.