<s>[INST] Prove that if $m$ and $n$ are positive integers, and $x$ is a real number, then: $ceiling(\frac{ceiling(x)+n}{m}) = ceiling(\frac{x+n}{m})$ I can't get eq (1) and eq (2)
Question: Prove that if $m$ and $n$ are positive integers, and $x$ is a real number, then: $ceiling(\frac{ceiling(x)+n}{m}) = ceiling(\frac{x+n}{m})$
Answer: 
Let us define the real number $x$ as the sum of an integer ‘$a$’ and a positive real number ‘$b$’ which is lesser than $1$. Therefore, $x = a + b$, where $a$ is an integer and $b$ is a real number lesser than $1$. 
Therefore, 
$=ceiling(\frac{ceiling(x)+n}{m})$
$=ceiling(\frac{ceiling(a+b)+n}{m})$
$=ceiling(\frac{ceiling(b)+a+n}{m})$
$=ceiling(\frac{1+a+n}{m})$
$=ceiling(\frac{1}{m}+\frac{a+n}{m}) \dots (1)$
Evaluating the other expression,
$= ceiling(\frac{x+n}{m})$
$= ceiling(\frac{a+b+n}{m})$
$= ceiling(\frac{x}{m}+\frac{n}{m})$
$= ceiling(\frac{b}{m}+\frac{a+n}{m}) \dots (2)$
 [/INST] There can be no integer between $\frac{b+a+n}{m}$ and $\frac{1+a+n}m$ because there is no integer (and even less so a multiple of $m$) between $b+a+n$ and $1+a+n$.
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