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Let $A=\{a_1,\ldots,a_n\}$, $B=\{b_1,\ldots,b_n\}$ and $C=\{\frac{a_1}{b_1},\ldots,\frac{a_n}{b_n}\}$ be three sets of integers. I would like to know when is it true that

$$\min C=\frac{\min A}{\max B}.$$

I tried to prove this equality but I failed.

We have, for positive integers only:

  1. $\min C\leqslant \frac{a_i}{b_i},\,\forall\,i\in\{1,\ldots,n\}$
  2. $\min A\leqslant a_i,\,\forall\,i\in\{1,\ldots,n\}$
  3. $\max B\geqslant b_i,\,\forall\,i\in\{1,\ldots,n\}$

Combining 1., 2. and 3., we get

$$\min C\leqslant \frac{a_i}{b_i},\,\forall\,i\in\{1,\ldots,n\},\\ \frac{\min A}{\max B}\leqslant \frac{a_i}{b_i},\,\forall\,i\in\{1,\ldots,n\}.$$

So I think I cannot conclude anything here, can I? Also, what happens when the integers are negative?

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    Equality occurs if and only if there is an integer $m\in \{1,2,\cdots,n\}$ such that $a_m=\min A$ and $b_m=\max B$ – 2017-02-02
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    As @user160738 said, the minimum of the $a_i$ and the maximum of the $b_i$ have to be attained for the same $i$. Consider for example $$ a_1=2,a_2=4,b_1=1,b_2=2 $$ Then $\min a_i/b_i=2$ and $\min a_i/\max b_i=1$. – 2017-02-02
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    @user160738 If $A=\{0,3\}$ and $B=\{2,3\}$ then $C=\{0,1\}$ and $\min \{c_i\}=0=\frac {\min \{a_i\}}{\max \{b_i\}}$ even though there is no such $m$. Or have I misread? – 2017-02-02
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    Are we to assume that $a_i>0$ for all $i$? That isn't stated, but it would be natural to add the condition. – 2017-02-02

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