Quote:
Originally Posted by CaptainBlack  Yes, because the inductive step does not hold as you have the product
of the first k r's equal to 1, the k+1 st r must be 1. So this proof only works if
all the roots are -1, but this is in general not true.
RonL |
I wasn't arguing that the r_{x} roots must be 1 (that is that every root must be 1), but that if the roots for the n = k case are r_1, r_2, ..., r_k, where r_1, r_2, ..., r_k are real numbers, and that the first "k" roots of the n = k + 1 case are r_1, r_2, ... , r_k, where r_1, r_2, ..., r_k are the same values from the n = k case, then the r_{k + 1} root must be 1.
[Edit] There is a problem with this logic that just occurred to me: The roots of n = k + 1 don't have to be the same as those of n = k.