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I'm trying to convince myself how one can hypothesize that

$$ \left \lceil \frac{a}{b} \right \rceil \leq \frac{a+b-1}{b}$$ or alternatively that $$ \left \lceil \frac{a}{b} \right \rceil -\frac{a}{b}\leq 1-\frac{1}{b}. $$

My best answer is to roll into the proof, but I would not try to prove it because I'm not sure I would hypothesize the claim. (I probably will see that it is not tight, and then try to generate a better claim). Last but not least, it seems to me that equality occurs whenever $a+b+1$ is a multiply of $b$. Can you articulate what this condition means?

Proof. The proof splits into two cases:

  • $\left \lceil \frac{a}{b} \right \rceil =\frac{a}{b}$, then trivially the difference is zero and the claim follows.
  • $\left \lceil \frac{a}{b} \right \rceil >\frac{a}{b}$. Then, $\frac{a}{b} > \left \lceil \frac{a}{b} \right \rceil - 1$ multplying by $b$ yields: $ a>\left \lceil \frac{a}{b} \right \rceil b-b$, substract $a$ to obtain $\left \lceil \frac{a}{b} \right \rceil b-b-a < 0$, and using the fact that the rhs must be an integer to deduce $\left \lceil \frac{a}{b} \right \rceil b-b-a \leq 1$. Finally devide everything by $b$ and get $$\left \lceil \frac{a}{b} \right \rceil-\frac{a}{b} \leq 1 - \frac{1}{b}.$$ Q.E.D.
Jyrki Lahtonen
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Proper Illumination
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  • Hint: \lfloor \frac ab \rfloor = n = \frac{nb}b$ for some special integer $n$. What would happen if the inequality were violated? – Greg Martin Jul 02 '22 at 22:36
  • @Martin R. Kinda, if we write $a=qb+r$ then $\left \lceil \frac{a}{b} \right \rceil -\frac{a}{b}=\left \lceil \frac{qb+r}{b} \right \rceil-\frac{qb+r}{b}=q+1-q-r/b=1-r/b$, as long as $0 < r \leq b-1$. But, still I don't know why $r\leq 1$? – Proper Illumination Jul 02 '22 at 22:40
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    Are $a$ and $b$ supposed to be integers? – jjagmath Jul 02 '22 at 23:23
  • I see nothing wrong with your proof, assuming $a$ and $b$ are positive integers. (I think it even works for negative integers but I didn't scrutinize edge cases. – Eric Snyder Jul 02 '22 at 23:52

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