Solution of $x-1=(x-\lfloor x \rfloor)(x-\{x\})$

Question

Find all solutions for $$x-1=(x-\lfloor x \rfloor)(x-\{x\})$$

$$$$My approach: $$x-1=\lfloor x \rfloor\{x\}$$

$$\dfrac{x-1}{\lfloor x \rfloor}=\{x\}$$ $$\Rightarrow 0\le \dfrac{x-1}{\lfloor x \rfloor}<1$$

$$Case 1: 0\le \dfrac{x-1}{\lfloor x \rfloor}$$ This gives me the solution set $(-\infty,0)\cup[1,\infty)$$$$$ $$Case 2:\dfrac{x-1}{\lfloor x \rfloor}<1$$ $$\Rightarrow\dfrac{x-1-\lfloor x \rfloor}{\lfloor x \rfloor}<0$$ $$\dfrac{\{x\}-1}{\lfloor x \rfloor}<0$$ Since $0\le\{x\}<1\Rightarrow \{x\}-1<0 $. Thus to make the above condition true, $\lfloor x \rfloor$ must be positive. Thus, $x\ge 1$

Taking the intersection of the solution sets for both cases, $x\in[1,\infty)$$$$$ However this answer is clearly wrong; the given answer is $x\in[1,2)$

Could somebody please tell me where $I've$ made a mistake? Many thanks!

Answer 1

This simplifies to $\lfloor x \rfloor \{x\} =\lfloor x \rfloor + \{x\} - 1$, or $(\lfloor x \rfloor - 1)(\{x\}-1) = 0$.

Since we cannot have $\{x\}=1$, we have $\lfloor x \rfloor = 1$, which gives all solutions.

Answer 2

Let $\lfloor x \rfloor=n,\{x\}=\epsilon$, then $x=n+\epsilon$. The equation (1) $$x-1=(x-\lfloor x \rfloor)(x-\{x\})\tag{1}$$

becomes

$$n+\epsilon-1=\epsilon n\tag{2}$$ or

$$n(1-\epsilon)=(1-\epsilon)\tag{3}$$

Since $0\le\epsilon<1$, we have $1-\epsilon>0$, thus from (3) we can solve for $n$ and result is $n=1$.

Consequently $x=1+\epsilon$ where $0\le\epsilon<1$.