Page 365 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 351
1 here just to make sure that c is positive, as required in the definition
of O.) Then for all x > a,

| f (x)|= |sf 1 (x) + tf 2 (x)|≤|s|| f 1 (x)|+|t|| f 2 (x)|
≤|s|c 1 |g(x)|+|t|c 2 |g(x)|= (|s|c 1 +|t|c 2 )|g(x)|≤ c|g(x)|.
Therefore f ∈ O(g).
18. (a) Hint: Let h ={(X, y) ∈ A/R × B |∃x ∈ X( f (x) = y)}.
(b) Hint: Use the fact that if xRy then [x] R = [y] R .

Section 5.2
2. (a) f is not a function.
(b) f is not a function. g is a function that is onto, but not one-to-one.
(c) R is one-to-one and onto.
5. (a) Suppose that x 1 ∈ A, x 2 ∈ A, and f (x 1 ) = f (x 2 ). Then we can per-
form the following algebraic steps:
x 1 + 1 x 2 + 1
= ,
x 1 − 1 x 2 − 1
(x 1 + 1)(x 2 − 1) = (x 2 + 1)(x 1 − 1),
x 1 x 2 − x 1 + x 2 − 1 = x 1 x 2 − x 2 + x 1 − 1,
2x 2 − 2x 1 = 0,
x 1 = x 2 .
This shows that f is one-to-one.
To show that f is onto, suppose that y ∈ A. Let
y + 1
x = .
y − 1
Notice that this is defined, since y = 1, and also clearly x = 1, so
x ∈ A. Then
y+1 2y
x + 1 y−1 + 1 y−1
f (x) = = = = y.
x − 1 y+1 − 1 2
y−1 y−1
(b) For any x ∈ A,
x+1 + 1 2x
x−1 x−1
( f ◦ f )(x) = = = x = i A (x).
x+1 − 1 2
x−1 x−1
7. (a) {1, 2, 3, 4}.
(b) f is onto, but not one-to-one.