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anonymous

  • one year ago

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

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  1. amistre64
    • one year ago
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    can 1/n = 1/m ?

  2. amistre64
    • one year ago
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    i assume your text defines natural numbers as positive integers

  3. anonymous
    • one year ago
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    yes 1/n can equal 1/n and yes N is positive >0 integers.

  4. amistre64
    • one year ago
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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

  5. anonymous
    • one year ago
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    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?

  6. amistre64
    • one year ago
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    im not familiar with the Archimedean property to be able to comment on its usages

  7. anonymous
    • one year ago
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    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 \(1-0=1\)

  8. anonymous
    • one year ago
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    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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