Prove that if $m$ and $n$ are positive integers, and $x$ is a real number, then: $ceiling(\frac{ceiling(x)+n}{m}) = ceiling(\frac{x+n}{m})$

Question

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,

$= ceiling(\frac{x+n}{m})$

$= ceiling(\frac{a+b+n}{m})$

$= ceiling(\frac{x}{m}+\frac{n}{m})$

$= ceiling(\frac{b}{m}+\frac{a+n}{m}) \dots (2)$

Answer 1

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 $1+a+n$.

Answer 2

This is straightforward using the universal property of the ceiling function, viz. $$\rm k\ge \lceil r \rceil \color{#c00}\iff k\ge r,\ \ \ for\ \ \ k\in \mathbb Z,\ r\in \mathbb R$$ Therefore for $\rm\:0 < m,n\in \mathbb Z,\ x\in \mathbb R,\ $
$${\rm\begin{eqnarray} &\rm \qquad\qquad k &\ge&\:\rm\ \lceil (\lceil x \rceil + n)/ m\rceil \\ \color{#c00}\iff& \qquad\qquad\rm k &\ge&\ \ \ \ \rm (\lceil x \rceil + n)/m \\ \iff&\qquad \rm mk\!-\!n\ &\ge&\ \ \rm\ \ \ \lceil x \rceil \\ \color{#c00}\iff&\qquad \rm mk\!-\!n &\ge&\ \ \rm\ \ \ \ \ x \\ \iff&\qquad\qquad \rm k &\ge&\ \ \rm \ \ \ \ (x+n)/m \\ \color{#c00}\iff&\qquad\qquad \rm k &\ge&\ \ \rm \ \ \lceil(x+n)/m \rceil \\ \\ \Rightarrow\ \ &\rm \lceil (\lceil x \rceil + n)/ m\rceil\ &=&\rm\ \ \lceil(x+n)/m \rceil \end{eqnarray}}$$

If you know category theory you can view this universal property of ceiling as a left adjoint to the inclusion $\,\Bbb Z\to \Bbb R,\,$ e.g. see Arturo's answer here, or see most any textbook on category theory. But, of course, one need not know any category theory to understand the above proof. Indeed, I've had success explaining this (and similar universal-inspired proofs) to high-school students.