Question

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

Answer 1

can 1/n = 1/m ?

Answer 2

i assume your text defines natural numbers as positive integers

Answer 3

yes 1/n can equal 1/n and yes N is positive >0 integers.

Answer 4

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

Answer 5

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?

Answer 6

im not familiar with the Archimedean property to be able to comment on its usages

Answer 7

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 \(1-0=1\)

Answer 8

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