Page 364 - HOW TO PROVE IT: A Structured Approach, Second Edition

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350 Appendix 1: Solutions to Selected Exercises
10. (a) Suppose b ∈ B. Since Dom(S) = B, we know that there is some c ∈ C
such that (b, c) ∈ S. To see that it is unique, suppose that c ∈ C and

(b, c ) ∈ S. Since Ran(R) = B, we can choose some a ∈ A such that
(a, b) ∈ R.Butthen(a, c) ∈ S ◦ R and(a, c ) ∈ S ◦ R,andsince S ◦ R

is a function, it follows that c = c .

(b) A ={1}, B ={2, 3}, C ={4}, R ={(1, 2), (1, 3)}, S ={(2, 4), (3, 4)}.
12. (a) No. Example: A ={1}, B ={2, 3}, f ={(1, 2)}, R ={(1, 1)}.
(b) Yes. Suppose R is symmetric. Suppose (x, y) ∈ S. Then we can choose
some u and v in A such that f (u) = x, f (v) = y, and (u, v) ∈ R. Since
R is symmetric, (v, u) ∈ R, and therefore (y, x) ∈ S.
(c) No. Example: A ={1, 2, 3, 4}, B ={5, 6, 7}, f ={(1, 5), (2, 6),
(3, 6), (4, 7)}, R ={(1, 2), (3, 4)}.
16. (a) Let a = 3 and c = 8. Then for any x > a = 3,
2
| f (x)|=|7x + 3|= 7x + 3 < 7x + x = 8x < 8x = c|g(x)|.
This shows that f ∈ O(g).
Now suppose that g ∈ O( f ). Then we can choose a ∈ Z + and
+
c ∈ R such that ∀x > a(|g(x)|≤ c| f (x)|), or in other words, ∀x > a
2
(x ≤ c(7x + 3)). Let x be any positive integer larger than both a
and 10c. Multiplying both sides of the inequality x > 10c by x,
2
we can conclude that x > 10cx. But since x > a, we also have
2
x ≤ c(7x + 3) ≤ c(7x + 3x) = 10cx, so we have reached a contra-
diction. Therefore g /∈ O( f ).
(b) Clearly for any function f ∈ F we have ∀x ∈ Z (| f (x)|≤ 1 ·| f (x)|),
+
so f ∈ O( f ), and therefore ( f, f ) ∈ S. Thus, S is reflexive. To see that
it is also transitive, suppose ( f, g) ∈ S and (g, h) ∈ S. Then there are
positive integers a 1 and a 2 and positive real numbers c 1 and c 2 such
that ∀x > a 1 (| f (x)|≤ c 1 |g(x)|) and ∀x > a 2 (|g(x)|≤ c 2 |h(x)|). Let
a be the maximum of a 1 and a 2 , and let c = c 1 c 2 . Then for all x > a,

| f (x)|≤ c 1 |g(x)|≤ c 1 c 2 |h(x)|= c|h(x)|.
Thus, ( f, h) ∈ S,so S is transitive. Finally, to see that S is not a partial
order, we show that it is not antisymmetric. Let f and g be the functions
from Z to R defined by the formulas f (x) = x and g(x) = 2x. Then
+
for all x ∈ Z , | f (x)|≤|g(x)| and |g(x)|≤ 2| f (x)|,so f ∈ O(g) and
+
also g ∈ O( f ). Therefore ( f, g) ∈ S and (g, f ) ∈ S,but f = g.
(c) Since f 1 ∈ O(g), we can choose a 1 ∈ Z and c 1 ∈ R such that ∀x >
+
+
a 1 (| f 1 (x)|≤ c 1 |g(x)|). Similarly, since f 2 ∈ O(g) we can choose a 2 ∈
+
Z and c 2 ∈ R such that ∀x > a 2 (| f 2 (x)|≤ c 2 |g(x)|). Let a be the
+
maximum of a 1 and a 2 , and let c =|s|c 1 +|t|c 2 + 1. (We have added