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How does $A$ relate to $B$ if $A - lfloor A/B rfloor - lceil A/B rceil leq lfloor A/B rfloor times (B+1)$?

Mathematics Asked by Nicholas on December 14, 2020

For $A geq B$, both are strictly positive integers, what is the relationship between $A$ and $B$ such that the following is true?
$$A – lfloor A/B rfloor – lceil A/B rceil leq lfloor A/B rfloor times (B+1)$$

Previously I asked this question can be here, and a counterexample has been shown to disprove it. Now I’d like to ask whether we can find the conditions (an expression in terms of $A$ and $B$) such that the above is true.

One thing I noticed (a generalisation from @Clement Yung’s answer in my original post – thanks!) is that if $B = lceil A/k rceil$ (for any constant $k$), then the above is false. I wonder if there are any other cases such that it’s false, or if better if there’s condition(s) for when it is always true.

2 Answers

Consider firstly the case in which $A=B$ and then $A/B=1$. In this case, $lfloor A/Brfloor=lceil A/Brceil=1$, so that the inequality of the OP would reduce to

$$A-3lfloor A/B rfloor leq B lfloor A/B rfloor$$ $$A-3leq A $$

which is trivially true.

If $A/B>1$, then $lfloor A/Brfloor+1=lceil A/Brceil$, so that the inequality becomes

$$A-3lfloor A/B rfloor -1leq B lfloor A/B rfloor$$ $$A-(B+3)lfloor A/B rfloor -1leq 0$$ $$lfloor A/B rfloorgeq frac{A-1}{B+3}$$

This is the condition needed to satisfy the initial inequality of the OP.


For example, if $A=5$ and $B=2$, then the condition is satisfied since $$lfloor 5/2 rfloor=2 > frac{5-1}{2+3}=frac 45$$

Accordingly, for these values the initial inequality holds, as it gives

$$5-2-3leq 2cdot 3$$ $$0leq 6$$

As another example, if $A=12$ and $B=7$, then the condition is not satisfied since $$lfloor 12/7 rfloor=1 < frac{12-1}{7+3}=frac {11}{10}$$

Accordingly, for these values the initial inequality does not hold, since it would give

$$12-1-2leq 1cdot 7$$ $$9leq 7$$

Correct answer by Anatoly on December 14, 2020

$ newcommand{f}[1]{leftlfloor #1 rightrfloor} newcommand{c}[1]{leftlceil #1 rightrceil} $ Consider writing $A = NB + k$ for some $N in Bbb{Z}^+$ and $0 leq k < B$. We consider two cases.

If $k = 0$ (i.e. $A$ is a multiple of $B$), then we can rewrite the inequality as: begin{align*} A - f{A/B} - c{A/B} leq f{A/B}(B + 1) &iff NB - 2N leq N(B + 1) \ &iff -2N leq N \ &iff N geq 0 end{align*} which always holds. If $k > 0$, then: begin{align*} A - f{A/B} - c{A/B} leq f{A/B}(B + 1) &iff (NB + k) - N - (N + 1) leq N(B + 1) \ &iff k - 2N - 1 leq N \ &iff 3N + 1 geq k end{align*} For a fixed $B in Bbb{Z}^+$, we can now classify all integers $A$ such that the inequality is satisfied by considering the value of $k$ (i.e. the remainder of $A$ when divided by $B$, which takes finitely many possible values). In particular, if $3N + 1 geq B - 1$, then the inequality is immediately satisfied.

Answered by Clement Yung on December 14, 2020

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