JakobS

was just thinking the same (for latex...)

Your answer has the same problem that is stated in the question: The corner case where $x\in (\ceil{x}-\varepsilon,\ceil{x})$

I know that. But I'm still not convinced. Suppose $x=1-\frac{\varepsilon}{2}$ then $y$ has to be $1$ (as $1-\frac{\varepsilon}{2}>1-\varepsilon$), but it should be $0$.

sorry, my mistake... Had ceil and not floor function in mind... Does this also work if $x<0$ (especially the corner cases with $x\pm\varepsilon<0$)?

With the "strictly less than" constraint you'll run into problems when implementing this in a real model to be solved with a solver. You'll then have to do the epsilon trick that was already mentioned in the question.

@smileycreations15 See here en.wikipedia.org/wiki/Operations_research for an overview of the topics discussed in the site. The questions posted should also give you some indication what to ask. Many questions will probably be about en.wikipedia.org/wiki/Mathematical_optimization