H: and n is a positive integer

C:

Proof: We will show that . Let .

Therefore where k = 0, 1, 2,..., n - 1.

Therefore where k = 0, 1, 2,..., n - 1 and j is any integer.Therefore

. But = ln r and thus

where k = 0, 1, 2,..., n - 1 and j is any integer. But

where l is any integer. However nj + k where k = 0, 1, 2,..., n - 1 and j

is any integer is the same as the set of all integers. To see this let l be any integer

and divide l by n. Thus l = nj + k where j is the quotient and k is the remainder

which must be in the list . Therefore which says that .