Find the real numbers $x$ such that $\{x\} =\frac {x-3}{2\lfloor x\rfloor-5}.$

Question

Determine the real numbers that verify the relation

$$\{x\} =\frac {x-3}{2\lfloor x\rfloor-5}$$

where $\{x\}$ represents the fractional part, respectively $\lfloor x\rfloor$ represents the whole part of the real number $x.$
I tried writing $\{x\}=x-\lfloor x\rfloor$ or $\lfloor x \rfloor=x-\{x\} $ but I don't know what to do with them. I need a idea to start.

Thank you and I hope one of you can help me!

Answer 1

Let $n=\lfloor x\rfloor$ and $f=x-n.$ Then, $$\{x\} =\frac {x-3}{2\lfloor x\rfloor-5}\iff f=\frac{n+f-3}{2n-5}.$$

• If $f\ne\frac12,$ $$\begin{align}f(2n-5)=n+f-3&\iff2nf-5f=n+f-3\\&\iff2nf-n=5f+f-3\\&\iff n(2f-1)=6f-3\\&\iff n=\frac{6f-3}{2f-1}\\&\iff n=3.\end{align}$$
• $f=\frac12,$ $f(2n-5)=n+f-3.$

Hence the set of solutions is $[3,4)\cup(\Bbb Z+\frac12).$