1 Examples of Weak Induction

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1 More About Mathematical Induction Mathematical induction is designed for proving that a statement holds for all nonnegative integers (or integers beyond an initial one). Here are some extra examples of proofs by induction. 1 Examples of Weak Induction [ ] n(n + 1) Theorem 1.1. For every integer n > 0, n 3 =. [ ] 1(1 + 1) Base case: When n = 1, we have 1 3 = 1 and = 1. Inductive step: [ Let ] k 1 be an arbitrary integer, and suppose that k(k + 1) k 3 =. Then [ ] k(k + 1) k 3 + (k + 1) 3 = + (k + 1) 3 (by induction hypothesis) = k (k + 1) (k + 1)(k + 1) + = (k + k + )(k + 1) = (k + [ ] ) (k + 1) (k + 1)(k + ) =. Theorem 1.. For every integer n, n n. Base case: When n =, we have n = 16 = n. 1

2 Inductive step: Let k be an arbitrary integer, and suppose that k k. Then k+1 = k k (by induction hypothesis) = k + k k + k (since k ) = k + k + k k + k + 1 = (k + 1). Theorem 1.3. For every positive integer n, n 1 = 10 n 1. Base case: n = 1. Then 9 10 n 1 = 9 whereas 10 n 1 = 9. Inductive step: Suppose k 1 = 10 k 1 for some positive integer k. Then k k = 10 k k = k 1 1 = 10 k 1. Theorem 1.. For every positive integer n, ( 1 1 ) ( 1 1 ) ( 1 1 ) (1 1 ) 1 8 n + 1. n+1 Base case: n = 1. Then 1 1 n = 1 while n+1 = = 1 1.

3 ( Inductive step: Suppose 1 1 ) ( 1 1 ) (1 1 ) k some positive integer k. Then ( 1 1 ) ( 1 1 ) (1 1 k ) ( 1 1 k for k+1 ) ( ) ( 1 1 ) k+1 k+1 = k+1 1 k+1 k+1 = k+1 1 k+1 k+1 = 1 ( ) 1 k+1 k ( 3 1 ) 1 k+1, where we have used the fact that k 1 implies that k + 1, which in turn implies that k+1 1, which then implies that 1, which then finally k+1 implies that 1 1. So k+1 ( 1 1 as desired. ) ( 1 1 ) (1 1 k ) ( 1 1 k+1 ) 1 ( ) 1 = k+1 1 = 1 k+1 + 1, k+ 1 k+1 Examples of Strong Induction If ordinary induction is not good enough to prove that a statement is true for every positive integer, then, hopefully, strong induction or the well-ordering principle will work instead. Definition.1. The Fibonacci numbers are defined recursively by F 0 = 0; F 1 = 1; 3

4 for every integer n 1, F n+1 = F n + F n 1. Theorem.. For every n 1, F n ϕ n, where ϕ is the larger zero of the polynomial x x 1. (The constant ϕ is called the golden ratio.) Proof. We use strong induction. Base case 1: n = 1. Then F n = F 1 = 1 and ϕ n = ϕ 1 = 1/ϕ. Given the above characterization of ϕ, observe that ϕ ϕ 1 = 0 so that ϕ = ϕ + 1. Dividing both sides of this equation by 1/ϕ yields 1 = 1/ϕ + 1/ϕ 1/ϕ. So F 1 = 1 ϕ 1. Base case : n =. Then F n = F = F 1 + F 0 = = 1 and ϕ n = ϕ 0 = 1. The theorem holds in this case. Inductive step: Let k be an arbitrary integer larger than 1. Assume that for every positive integer j k, we have that F j ϕ j. Then, since k + 1 >, we may use the recursive definition of the Fibonacci numbers to conclude that F k+1 = F k + F k 1 ϕ k + ϕ k 3 = ϕ k 3 (ϕ + 1) = ϕ k 3 ϕ = ϕ k 1. (by induction hypothesis) Note that in the proof of Theorem., we needed two base cases. That is because since the recursive definition of the Fibonacci numbers rely on the previous two numbers of the sequence, we must independently verify the first two cases of the theorem in order to apply the recurrence validly.

5 This is also the reason why we need strong induction. When working out the inductive step, we need to assume that the theorem holds for more than just the previous integer. Theorem.3. (Existence and uniqueness of m-ary representations.) Let m be an integer, m. Every natural number n can be uniquely written in the form n = a 0 + a 1 m + a m + + a j m j, where j is a nonnegative integer, 1 a j m 1, and 0 a i m 1 for i = 0, 1,..., j 1. Proof. We proceed by strong induction on n. Base case: n = 0. Then take a 0 = 0 and k = 0. This case trivially holds. Inductive step: Let k be an arbitrary positive integer, k, and suppose that for every positive integer r < k, r can be uniquely written in the form r = a 0 + a 1 m + + a j m j, where j is a nonnegative integer, 1 a j m 1, and 0 a i m 1 for i = 0, 1,..., j 1. Let j be the power of m such that m j k < m j +1. Then 0 k m j < m j +1 m j = m j (m 1). Furthermore, because 0 k m j < k, we can apply the induction hypothesis on k m j to get a unique m-ary representation k m j = a 0 + a 1 m + a m + + a j m j, with j j. If j < j, then k = a 0 + a 1 m + a m + + a j m j + m j is an m-ary representation of k. If j = j, then the fact that k m j < (m 1)m j tells us that a j m so that a j + 1 m 1 and, thus, k = a 0 + a 1 m + a m + + (a j + 1)m j is an m-ary representation of k. In any event, k has its own m-ary representation. We shall now show that this representation is unique. Let k = b 0 + b 1 m + b m + + b l m l be another m-ary representation of k, where 0 b j m 1 5

6 for all j = 1,,..., l and b l 1. If l j + 1, then k < m j +1 b l m l n, which is impossible. If instead l j 1, then k = b 0 + b 1 m + b m + + b l m l (m 1) + (m 1)m + (m 1)m + + (m 1)m l = m l+1 1 (geometric sum formula from calculus) < m j k, which is again impossible. Therefore j = l. If a j < b j, then k = a 0 + a 1 m + a m + + a j 1m j 1 + a j m j (m 1) + (m 1)m + (m ), + + (m 1)m j 1 + a j m j = (m j 1) + a j m j < (a j + 1)m j b j m j n, which is also impossible. Hence b j a j. By symmetry, we must also have a j b j, and so a j = b j. Then k a j m j = a 0 +a 1 m+a m ++a j 1m j 1 = b 0 +b 1 m+b m ++b j 1m j 1, which is incidentally a nonnegative integer smaller than k. By the inductive hypothesis, a i = b i for i = 0, 1,..., j 1. It follows that the m-ary representation of k is unique. 6