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How can one prove that $$\left\lceil\frac xy\right\rceil = \left\lceil\frac {\lceil x\rceil}y\right\rceil$$ with $x\in \Bbb R$ and $y\in \Bbb Z$.

Liam Martens
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    This looks like a special case of http://math.stackexchange.com/questions/652947/prove-that-if-m-and-n-are-positive-integers-and-x-is-a-real-number-then-ceili for $y > 0$. For $y < 0$ it seems to be wrong (e.g. $x = 3.5$, $y = -1$). – Martin R Nov 03 '15 at 20:23

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You can't prove it because it is not true. For a counterexample, choose $x = -4.8 \in \Bbb{R}$ and $y = -2 \in \Bbb{Z}$. Then: $$ \left \lceil \frac{x}{y} \right \rceil = \left \lceil \frac{-4.8}{-2} \right \rceil = \left \lceil 2.4 \right \rceil = 3 $$ whilst $$ \left \lceil \frac{ \left \lceil x \right \rceil}{y} \right \rceil = \left \lceil \frac{ -4}{-2} \right \rceil = \left \lceil 2 \right \rceil = 2 $$

If you require, instead, that $y \in \Bbb{Z}^+$ (that is, that $y$ be a positive integer, then the proof of that statement goes as follows: $$ \left \lceil \frac{x}{y} \right \rceil = k \in \Bbb{Z} \\ \frac{x}{y} = k - \phi : 0 \leq phi < 1 \\ x = yk - y\phi \\ y\phi = \rho + m \in \Bbb{Z} : 0 \leq rho < 1, \text{ and } 0 \leq m < y \\ x = yk - m - \rho \rightarrow \left \lceil x \right \rceil = yk -m \\ \frac{\left \lceil x \right \rceil }{y} = k - \frac{m}{y} $$ but since $0 \leq m < y$, $0 \leq \frac{m}{y} < 1$ thus $$ \left \lceil \frac{\left \lceil x \right \rceil }{y} \right \rceil =k = \left \lceil \frac{x}{y} \right \rceil $$

Mark Fischler
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