Is $a \lceil \frac b c \rceil \geq \lceil \frac {ab} c \rceil$? ($a, b \geq 0$ and $c > 0$)

Question

Comparing $a \big\lceil \frac b c \big\rceil$ vs $\big\lceil \frac {ab} c \big\rceil$, where

• $a, b, c$ are integers with $a, b \geq 0$ and $c > 0$
• $a = q_a c + r_a$ where $0 \leq r_a < c$
• $b = q_b c + r_b$ where $0 \leq r_b < c$

Then

$$ \begin{align} a \bigg\lceil \frac b c \bigg\rceil &= a \bigg\lceil \frac {q_b c + r_b} c \bigg\rceil \\ &= a \bigg\lceil q_b + \frac {r_b} c \bigg\rceil \\ &= a \big( q_b + \big[ r_b \neq 0 \big] \big) \\ &= a \big( q_b + \big[ b \not\mid c \big] \big) \\ &= a q_b + a \big[ b \not\mid c \big] \end{align} $$

and

$$ \begin{align} \bigg\lceil \frac {ab} c \bigg\rceil &= \bigg\lceil \frac {a (q_b c + r_b)} c \bigg\rceil \\ &= \bigg\lceil a q_b + \frac {a r_b} c \bigg\rceil \\ &= a q_b + \bigg\lceil \frac {a r_b} c \bigg\rceil \\ &= a q_b + \bigg\lceil \frac {(q_a c + r_a) r_b} c \bigg\rceil \\ &= a q_b + \bigg\lceil q_a r_b + \frac {r_a r_b} c \bigg\rceil \\ &= a q_b + q_a r_b + \bigg\lceil \frac {r_a r_b} c \bigg\rceil \\ &= b q_a + q_b r_a + \bigg\lceil \frac {r_a r_b} c \bigg\rceil \end{align} $$

where $[cond]$ equals $1$ if condition $cond$ is true and $0$ otherwise.

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Case 1. If $r_b = 0$, then

$$ a \bigg\lceil \frac b c \bigg\rceil = a q_b = \bigg\lceil \frac {ab} c \bigg\rceil $$

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Case 2. If $r_b \neq 0$ and $r_a = 0$, then

$$ a \bigg\lceil \frac b c \bigg\rceil = a q_b + a $$

and

$$ \bigg\lceil \frac {ab} c \bigg\rceil = b q_a $$

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Case 3. If $r_b \neq 0$ and $r_a \neq 0$, then

$$ a \bigg\lceil \frac b c \bigg\rceil = a q_b + a $$

and

$$ \bigg\lceil \frac {ab} c \bigg\rceil = a q_b + q_a r_b + \bigg\lceil \frac {r_a r_b} c \bigg\rceil $$

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Can we say anything further about Case 2. and Case 3.?

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Noting that $0 \leq r_a, r_b \leq c - 1$, we can conclude

$$ 0 \leq \bigg\lceil \frac {r_a r_b} c \bigg\rceil \leq c - 1 $$

Answer 1

We have that $\left\lceil\frac{b}{c}\right\rceil \ge \frac{b}{c}$. Multiply by $a$ to get $a\left\lceil\frac{b}{c}\right\rceil \ge \frac{a\ b}{c}$.

Since $a\left\lceil\frac{b}{c}\right\rceil$ is an integer greater than or equal to $\frac{a\ b}{c}$ then it's greater than or equal to the smallest integer greater or equal to $\frac{a\ b}{c}$, that is $a\left\lceil\frac{b}{c}\right\rceil \ge \left\lceil\frac{a\ b}{c}\right\rceil$.

Answer 2

No. If $a <1$, then this may not be true. Consider $a=.5 ,$ $b=c= 1$. Then right side is 1 and left side is .5.

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If we assume a,b, and c are integers, then the statement is true.

Let $b= nc + r$, $ 0 < r < 1$ (if r=0 then we have equality), $n$ a non-negative integer. Then left is $a(n+1)=an + a$ and right is $\lceil a(n+r/c) \rceil=\lceil an+ar/c \rceil$. $a \geq \lceil ar/c \rceil$. The result follows.