integer part of the number 
Solve the equation in real numbers $$[2x]=x-\frac{1}{x}$$ 
  where [x] is the integer part of the number.

I tried to use equality $\{2x\}+[2x]=2x$
And $0<\{2x\}<1$ then $\{2x\}=x+\frac{1}{x}$ 
but inequality $$0<x+\frac{1}{x}<1$$ has no solutions
 A: Drawing the graphical representations of functions $f_1$ (blue curve) and $f_2$ (magenta "staircase') given by equations
$$f_1(x):=x-\dfrac{1}{x} \ \ \text{and} \ \ f_2(x)=[2x]$$

gives 
1) The conviction that there are no solutions to equation $f_1(x)=f_2(x)$.
2) A way to prove it in three steps


*

*as the "staircase pattern" is situated between lines with equations $y=2x-1$ and $y=2x$ (in green), show in a rigorous way that 


$$2x-1 \leq [2x] \leq 2x \tag{1}$$


*

*then show that (1) can be extended, for $x>0$ into 


$$\underbrace{x-\dfrac{1}{x} < 2x-1}_{(I)} \leq [2x] \leq 2x$$ by showing that the quadratic inequation resulting from (I) is verified for all $x>0$. 


*

*do an equivalent reasoning in the case $x<0$ for 


$$ [2x] \leq \underbrace{2x < x-\dfrac{1}{x}}_{(J)}$$
A: We have $$2x-1< [2x]\leq 2x$$ so $$2x-1<x-{1\over x}\leq 2x$$ so $$0\leq x+{1\over x}<1$$ as you already figer  out. Clearly $x$ can not be $0$ or negative. So $x>0$. By AM-GM inequality we have $$x+{1\over x} \geq 2\sqrt {x{1\over x} } =2$$ so there is no solution.
A: For all $x\ne0$ we have
$$\left|2x-\left(x-{1\over x}\right)\right|=\left|x+{1\over x}\right|>1>
\left|2x-\lfloor 2x\rfloor\right|\ .$$
