p.8 - ex. 12 Show containment both ways: If x is in E\(Int)A_j, x is in E , and x is not in some A_j (call it A_k). Therefore x is in E\A_k, hence in the indicated union. if x is in the union, it is in some E\A_j, call it E\A_v hence x is in E, and not in A_v, hence not in (intersection)A_j hence x is in E\(intersection)A_j A similar argument will show the other equality. p.16 - ex. 5 If x is in f^(-1)(G union H), f(x) is in G or H, hence x is in f^(-1)G or f^(-1)H If x is in f^(-1)G union f^(-1)H, f(x) is in G or f(x) is in H, hence f(x) is in G union H, hence x is in f^(-1)(G union H) The other equality is shown similarly. p.16 - ex. 8 If x is in A, f^(-1)(f((x))) = x, since f is injective, hence the same is true for any x in E since E is a subset of A. Hence E is a subset of f^(-1)(f(E)). If x is in f^(-1)(f(E)), because f is a function, it is associated with a unique elment of f(E), which by injectiveness is associated with a unique element of E. Hence f^(-1)(f(E)) is a subset of E. There fore the sets are equal. Let A=Z, B=Z+ union {0}, and E=Z+. p.21 - ex. 4 Let P(n) be: 6 is a factor of n^3 + 5n 6 divides 0 (n=0) and 6 (n=1, if that is where you start) Assume 6 divides k^3 + 5k note (k+1)^3 + 5(k+1) = k^3 + 3k^2 + 3k + 1 + 5k + 5 = [k^3 + 5k] + 3k(k+1) + 6, which is divisible by 6 because each summand is. (the first summand by assumption, the second since k or k+1 is even, the third because 6 divideds 6. Therefore P(k) -> P(k+1), and since P(1) is true P(n) is true for all n. p.21 - ex. 9 1/3, 1/3 + 1/15 = 6/15 = 2/5, 2/5 + 1/35 = 15/35 = 3/7, 3/7 + 1/63 = 28/63 = 4/9 P(n): (sum)_{k=1}^n 1/(2k-1)(2k+1) = k/(2k+1) P(1) through P(4) have been verified. assume P(k) is true: (sum)[k] = k/(2k+1) sum[k+1] = k/(2k+1) + 1/(2(k+1)-1)(2(k+1)+1) = (k(2k+3) + 1)/(2k+1)(2k+3) = (2k^2 + 3k + 1)/(2k+1)(2k+3) = (2k+1)(k+1)/(2k+1)(2(k+1)+1) which is of the desired form. since P(1) is true and P(k) -> P(k+1) for all k, P(n) is true for all n. Russell Campbell
Mon Sep 1 1997