Was learning a bit about discrete math and was trying to get an expression for when the above equality is true/false.

I have this so far: For any real numbers $x$, $y$: (I know I should avoid division by zero here, but otherwise I am unsure if I need to narrow the domain to just positive numbers, et cetera)

Separating $x$ and $y$ into an integer and decimal part means that $\displaystyle \left\lfloor \frac{\lfloor \operatorname{int}_x + \operatorname{decimal}_x \rfloor}{\lfloor \operatorname{int}_y + \operatorname{decimal}_y \rfloor} \right\rfloor$ will simplify to $\displaystyle \left\lfloor \frac{\operatorname{int}_x}{\operatorname{int}_y} \right\rfloor$, since the floor function would remove the decimal parts.

So now I am trying to simplify $\displaystyle \left\lfloor \frac{\lfloor \operatorname{int}_x + \operatorname{decimal}_x \rfloor}{\lfloor \operatorname{int}_y + \operatorname{decimal}_y \rfloor} \right\rfloor$ to $\displaystyle \left\lfloor \frac{\operatorname{int}_x}{\operatorname{int}_y} \right\rfloor$, but got stuck there.

Thanks for the help!



Allow me to limit the discussion to the first quadrant.

Geometrically, the LHS equals a given integer $n$ in a sector such that

$$n\le\frac xy<n+1$$ or

$$ny-x\le 0\le (n+1)y-x.$$

As regards the RHS, these sectors contain a number of integer points, and every point in a unit square with the bottom-left corner at these integer points maps to the integer point, i.e. for integers $p,q$,

$$p\le x<p+1\land q\le y<q+1\land nq-p\le0<(n+1)q-p.$$

Hence my guess is that the solution set is the whole quadrant, minus the parts of the tiles that extend out of the sector, forming triangles.


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