By Herbert Amann

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13), we have fn+1 (k + 1) = fn (k + 1) = Vk+1 fn (0), . . , fn (k) = Vk+1 fn+1 (0), . . , fn+1 (k) for 0 < k + 1 ≤ n, and hence fn+1 (n + 1) = Vn+1 fn (0), . . , fn (n) = Vn+1 fn+1 (0), . . , fn+1 (n) . This completes the induction step and proves the existence of the functions fn for all n ∈ N. 5 The Natural Numbers 41 (c) After these preliminary steps we ﬁnally deﬁne f : N → X by f: N→X , a, fn (n) , f (n) := n=0, n ∈ N× . Because of the properties of the functions fn proved in (b), we have f (n + 1) = fn+1 (n + 1) = Vn+1 fn+1 (0), .

Operations A function : X × X → X is often called an operation on X. In this case we write x y instead of (x, y). For nonempty subsets A and B of X we write A B for the image of A × B under , that is, A B = {a b ; a ∈ A, b ∈ B } . 1) If A = {a}, we write a B instead of A B. Similarly A b = {a b ; a ∈ A }. A nonempty subset A of X is closed under the operation , if A A ⊆ A, that is, if the image of A × A under the function is contained in A. 8 Examples (a) Let X be a set. Then composition ◦ of functions is an operation on Funct(X, X).

These sets contain n men each. Let c be in the intersection of Ma and Mb . Since a, c ∈ Mb , the induction hypothesis implies that a and c are equal. Similarly, since b, c ∈ Ma , we have that b and c are equal. The claim then follows from the transitivity of equality. What is wrong with this proposition? n ) + 2(2 n+1 ) for all n ∈ N. 10 Show that 7 divides 1 + 2(2 11 Fix some g ∈ N with g ≥ 2. 16) j=0 where yk ∈ {0, . . , g − 1 } for k ∈ {0, . . , } and y > 0. 16) is unique, that is, if n = m j=0 zj g with zk ∈ {0, .