Mathematical Induction

Transcription

1 Mathematical Induction Tan Conghui Department of Systems Engineering and Engineering Management, The Chinese University of Hong Kong September 15, 2016 Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

2 Outline 1 Review of Basic Concepts 2 Examples Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

3 Outline Review of Basic Concepts 1 Review of Basic Concepts 2 Examples Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

4 Definition of Predicate Review of Basic Concepts A predicate is a statement whose value depends on one or more variables. Example 1 The following two statements are both predicates depends on n (n is a nonnegative integer): n is a perfect square, n i=1 i = n(n+1) 2, However, the former is only true for some particular choices of n, while the latter is true for all n. Usually, we use P(n) to denote a predicate concerning the variable n. Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

5 Review of Basic Concepts Principle of (Weak) Mathematical Induction Let P(n) be a predicate. If P(0) is true, and P(n) implies P(n + 1) for all nonnegative integers n, then P(n) is true for all nonnegative integers n. Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

6 Review of Basic Concepts A Template for Induction Proofs 1 State that the proof uses induction. 2 Define an appropriate predicate P(n). 3 Prove that P(0) is true (basic step). 4 Prove that P(n) implies P(n + 1) for every nonnegative integer n (inductive step). 5 Invoke induction. Given these facts, the induction principle allows you to conclude that P(n) is true for all nonnegative n. Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

7 Review of Basic Concepts Principle of Strong Mathematical Induction Let P(n) be a predicate. If P(0) is true, and P(0), P(1),..., P(n) together implies P(n + 1) for any nonnegative integer n, then P(n) is true for all nonnegative integers n. Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

8 Outline Examples 1 Review of Basic Concepts 2 Examples Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

9 Problem 1 Examples Question: Prove by induction that if 0 < a < 1, then (1 a) n 1 na for all n = 0, 1,.... Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

10 Examples Problem 1 Question: Prove by induction that if 0 < a < 1, then (1 a) n 1 na for all n = 0, 1,.... Solution: We can prove it by induction. Let P(n) be the predicate (1 a) n 1 na. 1 (Basic Step) When n = 0, LHS = (1 a) 0 = 1 = 1 0 a = RHS, so P(0) is true. 2 (Inductive Step) For any nonnegative integer n, assume P(n) is true, i.e., (1 a) n 1 na. Then we have (1 a) n+1 = (1 a) n (1 a) (1 na)(1 a) (from P(n) and the fact 1 a 0) = 1 (n + 1)a + na 2 > 1 (n + 1)a, which means P(n + 1) is true. Hence, by the principle of induction, P(n) is true for n = 0, 1, 2,.... Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

11 Problem 2 Examples Question: Let T(n) be the number of comparisons needed to sort n cards by merge sort. We already know that T(n) has recursive relationship T(n) = 2T( n 2 ) + n 1 and T(1) = 0. Prove that T(n) n log 2 n for all n = 2 k, where k = 0, 1, 2,.... Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

12 Examples Problem 2 Solution: We prove it by induction. Let P(k) be the predicate T(n) n log 2 n holds where n = 2 k. 1 (Basic step) When k = 0, then n = 2 k = 1 and therefore T(n) = 0 = 1 log 2 1 = n log 2 n, so P(0) is true. 2 (Inductive Step) For any nonnegative integer k, assume P(k) is true, i.e., T(n) n log 2 n where n = 2 k. Note that 2 k+1 = 2n, so we need to prove T(2n) 2n log 2 (2n) for P(n + 1): T(2n) = 2T(n) + 2n 1 2n log 2 n + 2n 1 = 2n(log 2 n + 1) 1 = 2n log 2 (2n) 1 2n log 2 (2n), (from the recursive relationship of T(2n)) (by inductive hypothesis) thus P(k + 1) is also true. Remark: log 2 n + 1 = log 2 n + log 2 2 = log 2 (2n) Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

13 Problem 3 Examples Question: Show that any integer n (n 2) can be written as the product of primes, e.g., 20 = Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

14 Examples Problem 3 Solution: We prove it by strong mathematical induction. Let P(n) be the predicate n can be written as the product of primes. 1 (Basic Step) For n = 2, since 2 itself is a prime number, thus P(2) holds. 2 (Inductive Step) Assume P(k) holds for all 2 k n. Now let s consider n + 1: If n + 1 is not divisible by any positive integer greater than 1 and except itself, then n + 1 itself is a prime number and therefore P(n + 1) is obviously true; Otherwise, then there must exist integers s and t such that n + 1 = s t and 1 < s, t < n + 1. By the inductive hypothesis, we known both s and t can be written as the product of primes, i.e., s = s 1 s 2 s l and t = t 1 t 2 t m, where s i and t j are all primes. Thus, n + 1 = s t = s 1 s 2 s l t 1 t 2 t m and therefore P(n + 1) is true. Hence, by the principle of strong MI, P(n) is true for n = 0, 1, 2,.... Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

15 Problem 4 Examples Question: Recall that the Fibonacci numbers are defined recursively by: 0 if n = 0 F n = 1 if n = 1. F n 1 + F n 2 if n 2 Let φ be the number ( 1 + ) 5 /2 (note it satisfy φ 2 = 1 + φ). Prove F n φ n 1 for all nonnegative integer n by weak mathematical induction. Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

16 Examples Problem 4 Solution: Let P(n) be the predicate F n φ n 1 and F n+1 φ n. (Basic Step): For n = 0, because F 0 = 0 φ 1 and F 1 = 1 1 = φ 0, P(0) is true. (Inductive Step): Assume all P(n) is true, i.e., F n φ n 1 and F n+1 φ n. Now let us look into F n+2 : F n+2 = F n + F n+1 (from def. of F n+2 ) φ n 1 + φ n = φ n 1 (1 + φ) (by IH) = φ n+1 (because φ 2 = 1 + φ). Together with F n+1 φ n (by IH directly), we know P(n + 1) is also true. Obviously, F n φ n 1 for all nonnegative integer n. Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16

17 Examples Thank You! If you have any questions, feel free to contact me. My is Tan Conghui (ENGG 2440B) Tutorial 1 September 15, / 16