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Appendix 1: Solutions to Selected Exercises 365
x = (n + 2)! + 2, and suppose that 0 ≤ i ≤ n.If i = n then we have

x + i = (n + 2)! + 2 + i = (i + 2)! + (i + 2) = (i + 2) ((i + 1)! + 1),
so (i + 2) | (x + i). Now suppose 0 ≤ i ≤ n − 1. By inductive hypoth-
esis, we know that (i + 2) | ((n + 1)! + 2 + i), so we can choose some
integer k such that (n + 1)! + 2 + i = k(i + 2), and therefore (n + 1)! =
(k − 1)(i + 2). Therefore

x + i = (n + 2)! + 2 + i = (n + 2)(n + 1)! + (i + 2)
= (n + 2)(k − 1)(i + 2) + (i + 2) = (i + 2) ((n + 2) (k − 1) + 1),
so (i + 2) | (x + i).

Chapter 7

Section 7.1

1. (a) Define f : Z → N by the formula f (n) = n − 1. It is easy to check
+
that f is one-to-one and onto.
(b) Let E ={n ∈ Z | n is even}, and define f : Z → E by the formula
f (n) = 2n. It is easy to check that f is one-to-one and onto, so Z ∼ E.
+
+
But we already know that Z ∼ Z, so by Theorem 7.1.3, Z ∼ E, and
therefore E is denumerable.
4. (a) No. Counterexample: Let A = B = C = Z and D ={1}.
+
(b) No. Counterexample: Let A = B = N, C = Z , and D = ∅.
−
6. (a) We proceed by induction on n.
Base case: n = 0. Suppose that m ∈ N and there is a one-to-one,
onto function f : I n → I m . Since n = 0, I n = ∅. But then since f is
onto, we must also have I m = ∅,so m = 0 = n.

Induction step: Suppose that n ∈ N, and for all m ∈ N,if I n ∼ I m
then n = m. Now suppose that m ∈ N and I n+1 ∼ I m . Let f : I n+1 →
I m be a one-to-one, onto function. Let k = f (n + 1), and notice that
1 ≤ k ≤ m,so m is positive. Using the fact that f is onto, choose some
j ≤ n + 1 such that f ( j) = m.
We now define g : I n → I m−1 as follows:

f (i) if i = j,

g(i) =
k if i = j.
We leave it to the reader to verify that g is one-to-one and onto. By
inductive hypothesis, it follows that n = m − 1, so n + 1 = m.