Would someone please give me some help with my case 2 below?
The problem: Let x 0, x such that for any > 0, x . Prove that x = 0
Solution:
Case 1: Let x > 0 and =
then
= > 0 because x > 0
since > 0 by given, x
Case 2: Let x = 0 and =
then
= 0/2 = 0
since = 0 = x, I cannot conclude x
unless I let = x + y (where y ). But can I do that? Some thing does not look right in my opinion, but I am stuck and unsure.
Thanks,
SlipEternal,
Thank you so much for your help. I am understood now. It was a logical contradiction for the case of x > 0; therefore, that case was false. That left x = 0 is true for x <= espilon.