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