Page 378 - HOW TO PROVE IT: A Structured Approach, Second Edition

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364 Appendix 1: Solutions to Selected Exercises
(b) Let C =∪ n∈Z B n . Clearly B = B 1 ⊆ C ⊆ A. To see that C is closed
+
+
under f, suppose x ∈ C. Then for some n ∈ Z , x ∈ B n ,so f (x) ∈
B n+1 and therefore f (x) ∈ C. Finally, to see that C is smallest, suppose
that B ⊆ D ⊆ A and D is closed under f. We prove by induction that
∀n ∈ Z , B n ⊆ D, from which it follows that C ⊆ D.
+
Base case: n = 1. Then B n = B ⊆ D by assumption.
Induction step: Suppose n ∈ Z and B n ⊆ D. Now let y ∈ B n+1 be
+
arbitrary. Then by the definition of B n+1 , we can choose x ∈ B n such
that y = f (x). Since B n ⊆ D, x ∈ D, and since D is closed under
f, y = f (x) ∈ D. Thus B n+1 ⊆ D.
5. {n ∈ Z | n ≥ 2}.
8. (a) R ∩ S ⊆ R and R ∩ S ⊆ S. Therefore by exercise 7, for every pos-
n n n n n
itive integer n, (R ∩ S) ⊆ R and (R ∩ S) ⊆ S ,so(R ∩ S) ⊆
n
n
R ∩ S . However, the two need not be equal. For example,
if A ={1, 2, 3, 4}, R ={(1, 2), (2, 4)}, and S ={(1, 3), (3, 4)}, then
2
2
2
(R ∩ S) = ∅ but R ∩ S ={(1, 4)}.
n
n
n
(b) R ∪ S ⊆ (R ∪ S) , but they need not be equal. (You should be able
to prove the first statement, and find a counterexample to justify the
second.)
10. (a) We use induction.
1
Base case: n = 1. Suppose (a, b) ∈ R = R. Let f ={(0, a),
(1, b)}. Then f is an R-path from a to b of length 1. For the other direc-
tion, suppose f is an R-path from a to b of length 1. By the definition
of R-path, this means that f (0) = a, f (1) = b, and ( f (0), f (1)) ∈ R.
1
Therefore (a, b) ∈ R = R .
n
Induction step: Suppose n is a positive integer and R ={(a, b) ∈
A × A | there is an R-path from a to b of length n}. Now suppose
n
1
(a, b) ∈ R n+1 = R ◦ R . Then there is some c such that (a, c) ∈ R n
and (c, b) ∈ R. By inductive hypothesis, there is an R-path f from a to
c of length n. Then f ∪{(n + 1, b)} is an R-path from a to b of length
n + 1. For the other direction, suppose f is an R-path from a to b of
length n + 1. Let c = f (n). Then f \{(n + 1, b)} is an R-path from
n
a to c of length n, so by inductive hypothesis (a, c) ∈ R . But also
n
1
(c, b) = ( f (n), f (n + 1)) ∈ R,so(a, b) ∈ R ◦ R = R n+1 .
(b) This follows from part (a) and Theorem 6.5.2.
14. We use induction on n.
Base case: n = 1. Then x = 2! + 2 = 4. The only value of i we have
to worry about is i = 0, and for this value of i we have i + 2 = 2 and
x + i = 4. Since 2 | 4, we have (i + 2) | (x + i), as required.
Induction step: Suppose that n is a positive integer, and for ev-
ery integer i,if0 ≤ i ≤ n − 1 then (i + 2) | ((n + 1)! + 2 + i). Now let