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Let $G$ be the set of rational numbers of the form $m/n$ , where $m,n$ are positive integers and $n \leq g $ for some possitive integer $g$. Suppose it is bounded by $1/k$ , k is a positive integer less than g.

Prove or disprove that, cardinality of set $G$ is finite.

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Hint: Is G bounded? –  peoplepower Aug 12 '12 at 15:55
    
yes it should be bounded I did some error –  mehdi Aug 12 '12 at 15:56
    
"Bounded by $1/k$"? Is that an upper bound, or a lower bound, or what? The set $G$ as defined in the first sentence is not bounded. If there's a boundedness condition, it should be part of the stated definition of $G$. And it should be stated clearly. As now written, it's hard to be sure what is meant. –  Michael Hardy Aug 12 '12 at 16:03
    
lower bound is 1/g (from definition no other smaller rational can be written), upper bound is 1/k since k is les than g –  mehdi Aug 12 '12 at 16:06
    
$\cfrac1 g$ is a strictly smaller rational than $\cfrac1 k$, and (as far as I can tell from the definition given) is an element of $G$. $2$ is a strictly larger rational than $\cfrac1 k$, and again seems to be an element of $G$. –  Cameron Buie Aug 12 '12 at 16:31
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1 Answer

up vote 2 down vote accepted

More generally, let $g$ be a positive integer and $M$ a positive real number, and let $G$ be the set of rational numbers $m/n$ such that $m$ and $n$ are positive integers, $m/n\le M$, and $n\le g$; then $G$ is finite.

To see this, note first that there are only $g$ possible denominators for members of $G$, namely, the integers $1,2,\dots,g$. Let $n$ be one of these denominators. Then $m/n\le M$ if and only if $m\le Mn$. In other words, $m/n\in G$ if and only if $m$ is one of the $Mn$ integers $1,2,\dots,Mn$. In particular, since $n\le g$, we must have $Mn\le Mg$. Thus, there are only $g$ possible denominators, and for each of them there are at most $Mg$ possible numerators, so altogether there are at most $g(Mg)=Mg^2$ possible fractions in $G$.

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Thanks this was better –  mehdi Aug 12 '12 at 16:10
    
Why are we assuming that $G$ is bounded above? –  Cameron Buie Aug 12 '12 at 16:34
    
@Cameron: Because mehdi added that hypothesis to the problem. –  Brian M. Scott Aug 12 '12 at 16:37
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