Let $N > 10$ be an integer.
Let $N_{b}$ be its base-$b$ representation.
Consider the decimal expansions of $\frac{1}{N}$ and $\left( \frac{1}{N} \right)_{b}$.
(1) Will there be $\lfloor \log n \rfloor$ and $\lfloor \log_{b} n \rfloor$ consecutive zeros immediately following the decimal point of $\frac{1}{N}$ and $\left( \frac{1}{N} \right)_{b}$, respectively?
(2) If so, how would one prove this?
Thank you.
There would be $m$ consecutive zeros following the decimal point of a number $x \in [0, 1)$ in base $b$ iff $$0.1_b=\frac{1}{b} \le x \cdot b^m < 1$$
Or in other words, if the number created by shifting the decimal point right $m$ times would be exactly one more right shift from being above $1$, then $m$ is the number of consecutive zeros following the decimal point.
This means that $\frac{1}{bx} \le b^m \to m\ge \log_b\left( \frac{1}{bx}\right)$. Since we are trying to find the minimal integer value of $m$ satisfying this condition, we can take the ceiling to get $$m = \left\lceil\log_b\left( \frac{1}{bx}\right)\right\rceil = \left\lceil -1+\log_{b}\left(\frac{1}{x}\right) \right\rceil$$
Since $x = \frac{1}{N}$, this is $$\left\lceil -1+\log_{b}\left(N\right) \right\rceil$$ which would only disagree with your results when $N$ is an integer power of $b$.