Floor Functions and Disjoint Intervals

Start from $\lfloor x \rfloor (2^{x-\lfloor x \rfloor}-1)≤1$ since $\lfloor x \rfloor$ is a positive integer $(2^{x-\lfloor x \rfloor}-1)≤\frac{1}{\lfloor x \rfloor}$ $2^{x-\lfloor x \rfloor}≤\frac{1}{\lfloor x \rfloor}+1$ $2^{x-\lfloor x \rfloor}≤\frac{1+\lfloor x \rfloor}{\lfloor x \rfloor}$ $x-\lfloor x \rfloor≤\log_2{(\frac{1+\lfloor x \rfloor}{\lfloor x \rfloor})}$ $x≤\lfloor x \rfloor+\log_2{(\frac{1+\lfloor x \rfloor}{\lfloor x \rfloor})}$

From this, it can be seen that the length of solution for $i≤x<i+1$ for any positive integer $i$ is $\log_2{(\frac{1+\lfloor i \rfloor}{\lfloor i \rfloor})}$ .

Using the properties of the floor function, this is equivalent to $\log_2{(\frac{1+i}{i})}$ .

Since $a≤x<b$ , we sum this from $a$ to $b-1$ to get the sum of the lengths of all the disjointed unions: $c=\displaystyle\sum_{i=a}^{b-1}\log_2{(\frac{1+i}{i})}$

Use the rule of the sum of logarithms $\log{a}+\log{b}=\log{ab}$ : $c=\log_2{(\displaystyle\prod_{i=a}^{b}\frac{1+i}{i})}$ $c=\log_2{\frac{1+a}{a} \frac{2+a}{1+a} \frac{3+a}{2+a}\dots\frac{b-1}{b-2}\frac{b}{b-1}}$

Most of the terms cancel out, leaving $c=\log_2{\frac{b}{a}}$ .

Then, the answer $2^{c}=\frac{b}{a}$ .