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anonymous
 one year ago
m,n are natural numbers, S:={1/n1/m}, find infS, supS.
anonymous
 one year ago
m,n are natural numbers, S:={1/n1/m}, find infS, supS.

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amistre64
 one year ago
Best ResponseYou've already chosen the best response.0i assume your text defines natural numbers as positive integers

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes 1/n can equal 1/n and yes N is positive >0 integers.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0the biggest value we can get is a 1 then; and the smallest approaches 0 so my thoughts are 00 = 0 10 = 1 01 = 1 11 = 0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I want to use the Archimedean property to ensure these extremum, but I am new to the property and an not clear about the corollaries that infer n (sub y) 1<=y<=n (suby)... I believe there will be more to a proof of this conclusion... you know?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.0im not familiar with the Archimedean property to be able to comment on its usages

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider that we can make \(1/n\) arbitrarily close to \(0\) and the maximum value of \(1/m\) is strictly \(1\), so our infimum or greatest lower bound is \(01=1\) for our supremum or least upper bound, consider we can make \(1/n=1\) and then make \(1/m\) arbitrarily close to zero, so we get \(10=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the archimedean property comes in here by guaranteeing that getting arbitrarily close to \(1\) from below or to \(1\) from above precludes there being any other candidate supremum/infimum, since that would suggest that there is some 'infinitesimal' number
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