A question involving fractional part of a number

Question

Let $n \in \mathbb{N}$ and consider $2^n = r \cdot 10^l +$ terms involving lower powers of $10$. Then we have the following relations: $$r 10^l \le 2^n < (r+1)10^l$$ $$\Leftrightarrow \log_{10} r + l \le n \log_{10} 2 < \log_{10} (r+1) + l$$ $$\Leftrightarrow \log_{10} r \le \{ n \log_{10} 2 \} < \log_{10} (r+1),$$ where $\{\cdot\}$ denotes the fractional part. I don't understand the last equivalence, namely why $\log_{10} r \le \{ n \log_{10} 2 \} < \log_{10} (r+1)$?

Answer 1

If $r$ is a non-zero single digit number, then $10^{l+1}> 2 ^n \geq 10^l$. Taking logarithms, $l+1 > n \log_{10}2 \geq n$. This means that the greatest integer less than or equal to $n \log_{10} 2 $ is $l$. Therefore, $\{n \log_{10} 2\} = n \log_{10}2 - l$. So the last equivalence follows by simply subtracting $l$ from each of the terms in the inequality above that.

If $r$ is not a single digit number, such a statement would clearly be false, since $\log_{10} r \geq 1$ but $\{n \log_{10} 2\} < 1$. I am pointing this out merely because it was implicitly assumed in the question that this was the case, but it was key to the above step.

Answer 2

Note that $\log_{10} r < 1$ ,as $r < 10$. Then as $l \in \mathbb{N}$ all the numbers are between $l$ and $l+1$. Therefore as they differ only in their fractional parts, the fractional parts have to keep the order as original numbers.