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Find-inf-m-n-m-n-N-m-lt-n-2-




Question Number 208103 by depressiveshrek last updated on 05/Jun/24
Find inf{(m/n) ∣ m, n ∈ N, m<n−2}
$$\mathrm{Find}\:\mathrm{inf}\left\{\frac{{m}}{{n}}\:\mid\:{m},\:{n}\:\in\:\mathbb{N},\:{m}<{n}−\mathrm{2}\right\} \\ $$
Answered by A5T last updated on 05/Jun/24
Since m,n∈N, (m/n)>0,so 0 is a lower bound.  (m/n)≥(1/n) and (1/n) is in the set.  For any ε>0, one can always find n such that  (1/n)−ε<0. Take n>(1/ε) which is possible by the  Archimedean property, hence 0 is the infimum  of the set.
$${Since}\:{m},{n}\in\mathbb{N},\:\frac{{m}}{{n}}>\mathrm{0},{so}\:\mathrm{0}\:{is}\:{a}\:{lower}\:{bound}. \\ $$$$\frac{{m}}{{n}}\geqslant\frac{\mathrm{1}}{{n}}\:{and}\:\frac{\mathrm{1}}{{n}}\:{is}\:{in}\:{the}\:{set}. \\ $$$${For}\:{any}\:\epsilon>\mathrm{0},\:{one}\:{can}\:{always}\:{find}\:{n}\:{such}\:{that} \\ $$$$\frac{\mathrm{1}}{{n}}−\epsilon<\mathrm{0}.\:{Take}\:{n}>\frac{\mathrm{1}}{\epsilon}\:{which}\:{is}\:{possible}\:{by}\:{the} \\ $$$${Archimedean}\:{property},\:{hence}\:\mathrm{0}\:{is}\:{the}\:{infimum} \\ $$$${of}\:{the}\:{set}. \\ $$

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