Why my double can contain a value below the machine epsilon?

Question

I was solving an equation using double precision and I got -7.07649e-17 as a solution instead of 0.

I agree it\'s close enough that I can s

Answer 1

A common solution for the floating point precision problem is to define an epsilon value yourself and compare to that instead of zero.

e.g.

double epsilon = 0.00001;
if (abs(value) < epsilon) // treat value as 0 in your code