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During class, we were introduced to a proof that used the ceiling function. We assumed (without proof) that:

$$ \left\lceil{\frac{n}{2^i}}\right \rceil= \left\lceil{\frac{\lceil{\frac{n}{2}}\rceil}{2^{i-1}}}\right\rceil,$$

for $i \geq 1$, where $d$ is a positive integer.

I am interested in seeing how this can be proved. I don't know much about the ceiling function, other than its basic properties. I found a general result here , which says for positive integers $m$, $n$ and arbitrary real number $x$.

$$ \left\lceil{\frac{x}{mn}}\right \rceil= \left\lceil{\frac{\lceil{\frac{x}{m}}\rceil}{n}}\right\rceil,$$

Maria
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2 Answers2

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I’ll prove the more general result. Let $m$ and $n$ be positive integers and $x$ a real number. Let $k=\left\lceil\dfrac{x}m\right\rceil$. Then $$k-1<\frac{x}m\le k\;,$$ and hence $$\frac{k-1}n<\frac{x}{mn}\le\frac{k}n\;.$$

We want to show that $$\left\lceil{\frac{x}{mn}}\right \rceil=\left\lceil{\frac{\left\lceil{\frac{x}{m}}\right\rceil}{n}}\right\rceil=\left\lceil\frac{k}n\right\rceil\;.$$

Clearly $\left\lceil{\dfrac{x}{mn}}\right\rceil\le\left\lceil\dfrac{k}n\right\rceil$, so suppose to get a contradiction that $\left\lceil{\dfrac{x}{mn}}\right\rceil<\left\lceil\dfrac{k}n\right\rceil$. Then there must be an integer $\ell$ such that

$$\frac{x}{mn}\le\ell<\frac{k}n$$

and hence $\dfrac{x}m\le n\ell<k$. But then $k=\left\lceil\dfrac{x}m\right\rceil\le n\ell<k$, which is absurd.

Brian M. Scott
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For the case of floor division (ceiling division is analogous), we can get the result directly using the uniqueness of real Euclidean division: $\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}$ $\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}$

For $a$ divided by $b$, the result can be written uniquely as $q,r$ in $a = qb + r$, where quotient $q = \floor{a/b}$ and remainder $r$ satisfies $0 \le r < |b|$.

Let $x$ and $m$ be real numbers ($m$ doesn't have to be integer) and $n$ be a positive integer.

Let $q = \floor{x/m}$, so that $$x = qm + r, \quad 0 \le r < m$$

Let $q' = \floor{\floor{x/m}/n}$, so that $$q = q'n + r', \quad 0 \le r' < n$$

Since $q$ and $n$ are integers, $r'$ is as well, so we have a stronger bound $0 \le r' \le n-1$.

Then $$x = (q'n + r')m + r = q'(nm) + (r'm + r)$$

From our remainder bounds, $0 \le r' m \le (n-1)m $ and $0 \le r < m$, so together $0 \le r'm + r < mn$, that is $r'm + r$ is the remainder of $x / mn$ and $q'$ is the quotient. Conclude $q' = \floor{x/mn}$.

Using floor and ceiling bounds is more flexible, as shown in Concrete Mathematics.

Let $f$ be any continuous, monotonously increasing function with the property that $f(x)$ integer implies $x$ integer. Then $$\floor{f(x)} = \floor{f(\floor x)} \quad \text{and} \quad \ceil{f(x)} = \ceil{f(\ceil x)}$$

Not immediately obvious is that the nested floor and ceiling identities are a special case of function $f(x) = x/n$ evaluated at $x/m$.

qwr
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  • Why was this question bumped and what does this answer cover that the previous answer does not? – Mike Jul 29 '24 at 20:07
  • The bumping happens automatically. This answer covers Euclidean division, which can be extended into Euclidean domains with suitable choice of valuation function. It also cites a useful generalization. – qwr Jul 29 '24 at 20:12
  • There's also a small difference in that it is a direct proof from bounds, rather than a proof by contradiction. – qwr Jul 30 '24 at 19:18