Page 367 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Appendix 1: Solutions to Selected Exercises 353
Therefore f ◦ g = i R and g ◦ f = i R , and by Theorems 5.3.4 and 5.3.5 it
follows that f is one-to-one and onto and f −1 = g.
5. f −1 (x) = 2 − log x.
9. Suppose that f : A → B, g : B → A, and f ◦ g = i B . Let b be an
arbitrary element of B. Let a = g(b) ∈ A. Then f (a) = f (g(b)) =
( f ◦ g)(b) = i B (b) = b. Since b was arbitrary, this shows that f is onto.
11. (a) Suppose that f is one-to-one and f ◦ g = i B . By part 2 of Theo-
rem 5.3.3, f is also onto, so f −1 : B → A and f −1 ◦ f = i A . This
gives us enough information to imitate the reasoning in the proof of
Theorem 5.3.5:

g = i A ◦ g = ( f −1 ◦ f ) ◦ g = f −1 ◦ ( f ◦ g) = f −1 ◦ i B = f −1 .
(b) Hint: Imitate the solution to part (a).
(c) Hint: Use parts (a) and (b), together with Theorem 5.3.3.

14. (a) Suppose x ∈ A = Ran(g). Then we can choose some b ∈ B such
that g(b) = x. Therefore (g ◦ f )(x) = g( f (g(b))) = g(( f ◦ g)(b)) =
g(i B (b)) = g(b) = x.

(b) By the given information, ( f A ) ◦ g = i B , and by part (a),
g ◦ ( f A ) = i A . Therefore by Theorem 5.3.4, f A is a one-to-one,

−1

onto function from A to B, and by Theorem 5.3.5, g = ( f A ) .
16. Hint: Suppose x ∈ R. To determine whether or not x ∈ Ran( f ), you must
see if you can find a real number y such that f (y) = x. In other words, you
2
must try to solve the equation 4y − y = x for y in terms of x. Notice that
this is similar to the method we used in part 1 of Example 5.3.6. However,
in this case you will find that for some values of x there is no solution for
y, and for some values of x there is more than one solution for y.
Chapter 6
Section 6.1
1. Base case: When n = 0, both sides of the equation are 0.
Induction step: Suppose that n ∈ N and 0 + 1 + 2 +· · · + n =
n(n + 1)/2. Then

0 + 1 + 2 +· · · + (n + 1) = (0 + 1 + 2 +· · · + n) + (n + 1)
n(n + 1)
= + (n + 1)
2
n (n + 1)(n + 2)

= (n + 1) + 1 = ,
2 2
as required.