This question already has an answer here:

Problem Statement:-

Solve: $$\dfrac{1}{\left\lfloor{x}\right\rfloor}+\dfrac{1}{\left\lfloor{2x}\right\rfloor}=\{x\}+\dfrac{1}{3}$$ where $\left\lfloor{x}\right\rfloor$ denotes the integral part of $x$ and $\{x\}=x-\left\lfloor{x}\right\rfloor$

The floor function just eats me up whole whenever I encounter it. I have tried some things to solve it but am just not able to come up with anything useful at all. Here are the things I have tried.

First of all lets define the domain of $x$ which is $x\not\in[0,1)$, because if so happens then $\left\lfloor{x}\right\rfloor=0$ which should not happen for $\dfrac{1}{\left\lfloor{x}\right\rfloor}$ and $\dfrac{1}{\left\lfloor{2x}\right\rfloor}$ to be defined.

After this I was not able to come up with a good line of thought for attempting the problem so, please guide me as to what should be my line of thought for attempting problems related to the integral part of $x$.


merged by quid Mar 11 at 14:37

This question was merged with Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$ because it is an exact duplicate of that question.