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Appendix 1: Solutions to Selected Exercises 367
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To see that f is onto, suppose n ∈ Z . Let k be the smallest positive
integer such that f (1, k) > n, and notice that f (1, 1) = 1 ≤ n,so k ≥ 2.
Since k is smallest, f (1, k − 1) ≤ n, and therefore by fact (1),

0 ≤ n − f (1, k − 1) < f (1, k) − f (1, k − 1) = k − 1.
Adding 1 to all terms, we get

1 ≤ n − f (1, k − 1) + 1 < k.

Thus, if we let i = n − f (1, k − 1) + 1 then 1 ≤ i < k. Let j = k − i, and
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notice that i ∈ Z and j ∈ Z . With this choice for i and j we have
(i + j − 2)(i + j − 1)
f (i, j) = + i
2
(k − 2)(k − 1)
= + n − f (1, k − 1) + 1
2
(k − 2)(k − 1)
(k − 2)(k − 1)
= + n − + 1 + 1 = n.
2 2
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14. (a) If B \{ f (m) | m ∈ Z , m < n}= ∅ then B ={ f (m) | m ∈ Z ,
m < n}, so by exercise 10, B is finite. But we assumed that B was
infinite, so this is impossible.
(b) We use strong induction. Suppose that ∀m < n, f (m) ≥ m.Now
suppose that f (n) < n. Let m = f (n). Then by inductive hypoth-
esis, f (m) ≥ m. Also, by the definition of f (n), m = f (n) ∈ B \
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{ f (k) | k ∈ Z , k < n}⊆ B \{ f (k) | k ∈ Z , k < m}.Butsince f (m)
is the smallest element of this last set, it follows that f (m) ≤ m. Since
we have f (m) ≥ m and f (m) ≤ m, we can conclude that f (m) = m.
But then m /∈ B \{ f (k) | k ∈ Z , k < n}, so we have a contradiction.
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(c) Suppose that i ∈ Z , j ∈ Z , and i = j. Then either i < j or
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j < i. Suppose first that i < j. Then according to the defini-
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tion of f ( j), f ( j) ∈ B \{ f (m) | m ∈ Z , m < j}, and clearly f (i) ∈
{ f (m) | m ∈ Z , m < j}. It follows that f (i) = f ( j). A similar ar-
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gument shows that if j < i then f (i) = f ( j). This shows that f is
one-to-one.
To see that f is onto, suppose that n ∈ B. By part (b), f (n + 1) ≥
n + 1 > n. But according to the definition of f, f (n + 1) is the
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smallest element of B \{ f (m) | m ∈ Z , m < n + 1}. It follows that
n /∈ B \{ f (m) | m ∈ Z , m < n + 1}. But n ∈ B, so it must be the
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case that also n ∈{ f (m) | m ∈ Z , m < n + 1}. In other words, for
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some positive integer m < n + 1, f (m) = n.