Solution set for $\lfloor x\rfloor\{x\}=1$

Question

What is the solution set for $\lfloor x\rfloor\{x\}=1$ , where $\{x\}$ and $\lfloor x\rfloor$ are respectively fractional part and greatest integer function of $x$.

P.S.: the answer is $\{m+1/m:m\in\Bbb N\setminus\{1\}\}$. Please explain me the procedure and logic behind these kind of problems .

Answer 1

Suppose $[x]\{x\}=1 $. $x$ can not be an integer, or else $\{x\}=0$.

Also, $x>0$ or else $[x] < 0$ and $\{x\} > 0$.

Therefore, let $[x] = a$ and $\{x\} = b$, where $0 < b < 1$.

Then $ab = 1$, so $b = \dfrac1{a}$, so $x = a+b =a+\dfrac1{a} $.

Answer 2

Notice that no matter what number is $x$ we have that $\lfloor x\rfloor\in\Bbb Z$. Equating then $\{x\}\in\{1/z:z\in\Bbb Z\setminus\{0\}\}$, and notice that we can write

$$\{x\}=x-\lfloor x\rfloor$$

Then we can write the equation system

$$x-\lfloor x\rfloor=\frac1z\quad\text{ and }\quad\lfloor x\rfloor=z$$

then we get that

$$z=x-\frac1z\quad\to\quad x=z+\frac1z,\quad\text{ with }z\in\Bbb Z\setminus\{0\}$$

P.S.: the definition of the function $\{x\}$ for negative numbers is conflictive, see here, I was using the simpler one.

Answer 3

$$\lfloor x\rfloor\{x\}=1$$

$\lfloor x\rfloor\not =0$. Let $\lfloor x\rfloor=k, k\in Z$. Then $$\{x\}=\frac1k$$ $0\le\{x\}<1 \Rightarrow k>0$ T Then $$x=\lfloor x\rfloor+\{x\}=k+\frac1k, k\in \mathbb N$$