m,n are natural numbers, S:={1/n-1/m}, find infS, supS.

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m,n are natural numbers, S:={1/n-1/m}, find infS, supS.

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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can 1/n = 1/m ?
i assume your text defines natural numbers as positive integers
yes 1/n can equal 1/n and yes N is positive >0 integers.

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the biggest value we can get is a 1 then; and the smallest approaches 0 so my thoughts are 0-0 = 0 1-0 = 1 0-1 = -1 1-1 = 0
I 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?
im not familiar with the Archimedean property to be able to comment on its usages
consider 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 \(0-1=-1\) 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 \(1-0=1\)
the 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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