Question
I need the solution for problem#1.
For a positive integer n let sigma (n) denote the sum of the set of all positive integral divisors of n (including n itself). Recall that sigma has the following properties: (i) A positive integer N is perfect if and only if sigma (N) = 2N; (ii) sigma (2k - 1) = 2k - 1 for any positive integer k: ( A positive integer p is prime if and only if sigma(p) = p + 1 and (iv) sigma (mn) = sigma (m) sigma (n) if m and n are relatively prime. Use these properties of sigma to give an alternative argument that if 2n - 1 is prime then N = 2n - 1(2n - 1) is perfect. (Hint: Your argument should be about one line long and should use each of the above properties of sigma precisely once.) Fix a positive integer n and let D denote the set of positive integral divisors of n. That is, D = {k: k N and K divides n}. Define the set epsilon by epsilon = {m : m = n/k for some k D). (a)Prove that epsilon = D. (Hint: Recall that this means you must show that every element of epsilon belongs to D and that every element of D belongs to epsilon.) (b)Use your answer to part (a) to conclude that the sum of the reciprocals of all the positive integral divisors of n is sigma (n)/n. where is the function described m Problem 1. Use your answer to part (b) to conclude that if n is perfect then the sum of the reciprocals of all the positive integral divisors of n is 2. Carefully explain the error in our "proof" that 64 = 65. Be specific and explicit about where the "extra" unit of area comes from. (See the attached file for how the 8 times 8 square is divided up.) Problem 2.5.(e) on page 59 of our textbook. (Hint: Use the fact that the center of a circle lies on the perpendicular bisector of any chord of the circle and note that the sides of the triangle of lengths 50 and 60 are two chords of the circumscribing circle. Similar triangles should come in handy.)
Explanation / Answer
N=(2^n - 1)(2^(n-1)) sigma((2^n - 1)(2^(n-1))=sigma(2^n - 1)sigma(2^(n-1)) from 4 sigma(2^(n-1)) = 2^n -1 from 2 sigma(2^n - 1) = 2^n from 3 sigma((2^n - 1)(2^(n-1)) = (2^n) (2^n -1) =2*N
