integer part of the number

Question

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

Answer 1

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.

Answer 2

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

$$ [2x] \leq \underbrace{2x < x-\dfrac{1}{x}}_{(J)}$$

Answer 3

For all $x\ne0$ we have $$\left|2x-\left(x-{1\over x}\right)\right|=\left|x+{1\over x}\right|>1> \left|2x-\lfloor 2x\rfloor\right|\ .$$