jasperE3 wrote: Determine all real numbers $x$ such that $$\frac{2002\lfloor x\rfloor}{\lfloor-x\rfloor+x}>\frac{\lfloor2x\rfloor}{x-\lfloor1+x\rfloor}.$$ Obviously $x\notin\mathbb Z$ (else LHS would be undefined). So let $x=n+y$ with $n\in\mathbb Z$ and $y\in(0,1)$ Inequation becomes $\frac{2002n}{y-1}>\frac{2n+\lfloor 2y\rfloor}{y-1}$ $\iff$ $2002n<2n+\lfloor 2y\rfloor$ $\iff$ $2000n<\lfloor 2y\rfloor$ And so $\boxed{x\in\left((-\infty,0)\cup\left[\frac 12,1\right)\right)\setminus\mathbb Z}$

Nice .....