The discussion centers on manipulating inequalities involving the expression 1/(n-2) and the variable epsilon. The key conclusion is that for the inequality 1/(n-2) < epsilon to hold true, it is necessary that n - 2 > 1/epsilon, assuming n > 2 and epsilon > 0. The participants clarify that epsilon is typically a small positive number, which means that 1/epsilon will generally exceed epsilon itself, thus impacting the inequality's validity.
PREREQUISITESMathematicians, students studying calculus, and anyone interested in understanding inequalities and their applications in mathematical proofs.
Epsilon is typically a small positive number that is much closer to 0 than it is to 1. Unless epsilon is negative, epsilon + 2 will be larger than 1.mjjoga said:so I have 1/(n-2). I have that n>max(epsilon+2,1). I need to get 1/(n-2) < epsilon. I know that 1/(n-2)<1/(epsilon+2-2)=1/epsilon. but 1/epsilon is not always less than epsilon. can you see any errors?