How to show that $\left \lceil x\right \rceil-\left \lceil y \right \rceil=\left \lceil x-y \right \rceil-1$ with $x,y \in R^+$

Question

Recently, I have found this inequality: $\left \lceil x \right \rceil + \left \lceil y \right \rceil\geq \left \lceil x+y \right \rceil$. Im' trying to understand also if this similar equality holds: $\left \lceil x\right \rceil-\left \lceil y \right \rceil=\left \lceil x-y \right \rceil-1$ where $x,y \in R^+$. I've already tried to substitute into $x$ and $y$ some values and it seems correct, but I don't know how to prove it. Any idea?

Answer 1

You cannot show that, because that is not true. It fails for integers, the ceiling and floor of integers are themselves.

Answer 2

Note that $\lceil p\rceil-\lceil q\rceil-(\lceil p-q\rceil-1)=\lceil x\rceil-\lceil y\rceil-(\lceil x-y\rceil-1)$ when $p=x-\lceil x\rceil$ and $q=y-\lceil y\rceil$.

Thus, we just need to consider when $-1<x,y\le 0$.

Since $\lceil x\rceil,\lceil y\rceil=0$ here, we need to show $0\ge\lceil x-y\rceil-1$ and this is trivial since $x-y<1$.

(I think you probably asked $\lceil x\rceil-\lceil y\rceil\ge\lceil x-y\rceil-1$ not $\lceil x\rceil-\lceil y\rceil=\lceil x-y\rceil-1$ as that's whay you wrote first, but for the current question we have simple counterexample of $-1<x<y<0$, which can be motivated by above proof)