Math 310 Class Notes 5: The Recursion Theorem for N In this set of notes we will tighten a loose end that has been left dangling. Namely, in Class Notes 3 we used the word "assign" when we defined the notion of + for N (see Definition 2 of the notes). The word "assign" actually means we have a function that does the assignment. To be precise, we have the following theorem. Theorem 1. (The recursion theorem) Let c be a fixed natural number and let f : N ! N be a function. Then there is a unique function g : N ! N such that g(1) = c and g(x0) = f(g(x)) for all x 2 N. Before proving this theorem, let us see a few applications. Application 1. Consider the function f : N ! N given by f : x 7! x0: Fix a number m 2 N. Then by the recursion theorem there is a function g such that g(1) = m0 and g(x0) = f(g(x)) for all x 2 N. Let us now define m + n := g(n) for all n 2 N. Theorem 2. Fix m 2 N. We have m + n0 = (m + n)0 for all n 2 N. Proof. m + n0 = g(n0) = f(g(n)) = f(m + n) = (m + n)0. Slightly harder to verify is the following. Theorem 3. Fix m 2 N. We have m0 + n = (m + n)0 for all n 2 N. Proof. m + n = g(n) for all n 2 N, where g(1) = m0 and g(n0) = f(g(n)). Likewise, m0 + n = h(n) for all n 2 N, where h(1) = (m0)0 and h(n0) = f(h(n)). Let S be the set of all n 2 N such that m0 + n = (m + n)0. We first show 1 2 S. Now m0 + 1 = h(1) = (m0)0 while m + 1 = g(1) = m0, so that m0 + 1 = (m0)0 = (m + 1)0. Thus 1 2 S. Next suppose n 2 S. We establish that n0 2 S. That n 2 S means m0 + n = (m + n)0; which is the induction hypothesis. Therefore, by the preceding theorem and the induction hypothesis we have m0 + n0 = (m0 + n)0 = ((m + n)0)0 = (m + n0)0: Therefore, n0 2 N and so S = N. That is m0 + n = (m + n)0 for all n 2 N. 1 2 This theorem is what we have seen in class. The advantage of this approach via the recursion formula is that we can take care of m + n for all m and n at one stroke. Application 2. Fix m 2 N. Consider the function f : N ! N given by f : x 7! x + m: Then by the recursion theorem there is a function g : N ! N such that g(1) = m and g(x0) = f(g(x)). We define m · n := g(n) for all n 2 N. Exercise. Prove that m · n0 = m · n + m and m0 · n = m · n + n for all n 2 N. Application 3. Fix a 2 N. Consider the function f : N ! N given by f : x 7! ax: Then by the recursion theorem there is a function g : N ! N such that g(1) = a and g(x0) = f(g(x)). We define an := g(n): Exerxise. By the mathematical induction, prove that am+n = am · an and (am)n = amn. Now let us look at the proof of the recursion theorem. Proof. Existence of g. By definition, a function g : N ! N is a special sort of subset of N × N. Recall that g(1) = c means (1; c) 2 g and g(x0) = f(g(x)) means that if we let y = g(x), i.e., let (x; y) 2 g, then (x0; f(y)) 2 g as this just says g(x0) = f(g(x)) = f(y): Motivated by this, let us begin to construct the function g by consid- ering the set S, where S := fA ⊂ N × N : (1; c) 2 A and (x; y) 2 A implies (x0; f(y)) 2 Ag: S is not empty since N × N 2 S. Consider the set g := \A2S A; which consists of elements common to all A 2 S. Now (1; c) 2 g since it is in all A 2 S. Moreover, if (x; y) 2 g then (x; y) 2 A for all A 2 S. But then (x0; f(y)) 2 A for all A 2 S so that 3 (x0; f(y)) 2 g. We conclude therefore that g is in fact an element of S; it is the smallest element of S. We now prove that g is a function from N to N. First we show that the domain of g is N. Let T be the subset of n 2 N such that there is a z 2 N for which (n; z) 2 g. Now 1 2 T since (1; c) 2 g. Suppose n 2 T . Then (n; z) 2 g for some z 2 N. This gives (n0; f(z)) 2 g. Hence n0 2 T . It follows by mathematical induction that T = N. That is, the domain of g is N. Secondly, we show that each domain element of g is mapped to a unique target element. To this end, consider the set U := fx 2 N : (x; y); (x; z) 2 g imply y = zg: We show 1 2 U. Suppose this is not the case. That is, suppose we have (1; z), in addition (1; c) 2 g, where z =6 c. Then the set g nf(1; z)g would be a member of S (WHY?). But this contradicts the fact that g is already the smallest element of S. Thus, 1 2 U. We show that if x 2 U then x0 2 U. Suppose this is not the case. Then there is an m 2 U so that m0 2= U. Since m 2 U, there is a unique n 2 N with (m; n) 2 g. As a consequence (m0; f(n)) 2 g. Since m0 2= U, there is an (m0; z) 2 g, in addition to (m0; f(n)) 2 g, where z =6 f(n). But then g n f(m0; z)g is an element of S that is even smaller than g (WHY?), which is absurd. The mathematical induction then says that U = N. In conclusion, g is a function. Uniqueness of g. Suppose we have two functions g1 and g2 satisfying g1(1) = c = g2(1) 0 0 and g1(x ) = f(g1(x)) and g2(x ) = f(g2(x)) for all x 2 N. Let us consider the set V := fn 2 N : g1(n) = g2(n)g: 1 2 V because g1(1) = g2(1) = c. If n 2 V , then g1(n) = g2(n), so that 0 0 g1(n ) = f(g1(n)) = f(g2(n)) = g2(n ); which implies n0 2 V . Hence V = N by the mathematical induction. That is, g1 = g2. .