A=(1/m-1/n: m,neN) show that. infA=1/2 and supA=1

Question

Real analysis, on the topic bounded subset of real numbers

Answer 1

1/m - 1/n=(n-m)/mn, so if m<n, and m,n>0, n-m>1. So the smallest value of n-m=1. The smallest value of mn is 2, and m=1 and n=2, so 1/m - 1/n=1/2.

If m=n, A=0. Also, if m and n are both very large, then A approaches zero.

Neither m nor n can be negative if they both belong to N the set of natural numbers, and neither can be zero.

If n-m=mn (i.e., A=1) then n=m/(1-m), but m cannot be 1 and n cannot be zero. Neither m nor n can be negative.

If n<m, then A is -(m-n)/mn, and the minimum value is -1/2.

It would seem then that the maximum limit (superior) is 1/2 and the minimum (inferior) is -1/2. The range is 1.