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366 Appendix 1: Solutions to Selected Exercises
(b) Suppose A is finite. Then by definition of “finite,” we know that there
is at least one n ∈ N such that I n ∼ A. To see that it is unique, sup-
pose that n and m are natural numbers, I n ∼ A, and I m ∼ A. Then by
Theorem 7.1.3, I n ∼ I m , so by part (a), n = m.
8. (a) We use induction on n.
Base case: n = 0. Suppose A ⊆ I n = ∅. Then A = ∅,so |A|= 0.
Induction step: Suppose that n ∈ N, and for all A ⊆ I n , A is finite,
|A|≤ n, and if A = I n then |A| < n. Now suppose that A ⊆ I n+1 .If
A = I n+1 then clearly A ∼ I n+1 ,so A is finite and |A|= n + 1. Now
suppose that A = I n+1 .If n + 1 /∈ A, then A ⊆ I n , so by inductive
hypothesis, A is finite and |A|≤ n.If n + 1 ∈ A, then there must be
some k ∈ I n such that k /∈ A. Let A = (A ∪{k}) \{n + 1}. Then by

matching up k with n + 1 it is not hard to show that A ∼ A. Also,

A ⊆ I n ,sobyinductivehypothesis, A isfiniteand|A |≤ n.Therefore

by exercise 7, A is finite and |A|≤ n.
(b) Suppose A is finite and B ⊆ A. Let n =|A|, and let f : A → I n be
one-to-one and onto. Then f (B) ⊆ I n , so by part (a), f (B)isfi-
nite, | f (B)|≤ n, and if B = A then f (B) = I n ,so | f (B)| < n. Since
B ∼ f (B), the desired conclusion follows.
12. It will be helpful first to verify two facts about the function f. Both of the
facts below can be checked by straightforward algebra:
+
(1) For all j ∈ Z , f (1, j + 1) − f (1, j) = j.
+
(2) For all i, j ∈ Z , f (1, i + j − 1) ≤ f (i, j) < f (1, i + j). It follows
that i + j is the smallest k ∈ Z such that f (i, j) < f (1, k).
+
To see that f is one-to-one, suppose that f (i 1 , j 1 ) = f (i 2 , j 2 ). Then by fact
(2) above,
i 1 + j 1 = the smallest k ∈ Z such that f (i 1 , j 1 ) < f (1, k)
+
+
= the smallest k ∈ Z such that f (i 2 , j 2 ) < f (1, k)
= i 2 + j 2 .

Using the definition of f, it follows that
(i 1 + j 1 − 2) (i 1 + j 1 − 1)
i 1 = f (i 1 , j 1 ) −
2
(i 2 + j 2 − 2) (i 2 + j 2 − 1)
= f (i 2 , j 2 ) −
2
= i 2 .

But then since i 1 = i 2 and i 1 + j 1 = i 2 + j 2 , we must also have j 1 = j 2 ,
so (i 1 , j 1 ) = (i 2 , j 2 ). This shows that f is one-to-one.