The inoffensive floor function

Let us represent $x=\left\lfloor x\right\rfloor +r$ where $0\leq r<1$ . Multiplying both sides by $2^{2015}$ we obtain $2^{2015}x=2^{2015}\left\lfloor x\right\rfloor +2^{2015}r.$ Therefore $\left\lfloor2^{2015}x\right\rfloor=2^{2015}\left\lfloor x\right\rfloor +\left\lfloor 2^{2015}r\right\rfloor\quad\quad (*)$ Obviously $0\leq \left\lfloor 2^{2015}r\right\rfloor < 2^{2015}.$ Since $2^{2015}$ is a factor of $\left\lfloor 2^{2015}x\right\rfloor-\left\lfloor x\right\rfloor$ , using $( * )$ we obtain that $2^{2015}$ is a factor of the difference $\left\lfloor 2^{2015} r \right\rfloor -\left\lfloor x\right\rfloor$ where both terms are not negative integer numbers less than $2^{2015}$ . Thus that difference must be $0$ and then $\left\lfloor 2^{2015}r \right\rfloor = \left\lfloor x\right\rfloor$ . Substituting in $(*)$ we get $\left\lfloor 2^{2015}x\right\rfloor =2^{2015}\left\lfloor x\right\rfloor+\left\lfloor x\right\rfloor$ Since $\left\lfloor x\right\rfloor$ is not equal to zero, $\frac{\left\lfloor 2^{2015}x\right\rfloor - \left\lfloor x \right\rfloor}{\left\lfloor x\right\rfloor}=2^{2015}$ . Hence $\frac{a}{b}=2$ , that implies that $a=2$ and $b=1.$ Then the answer is 3.