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Mathematical Induction (cont.) 1 Example (of sum of the first n integers). In a round-robin tournament each of the n teams plays every other team exactly once. What is the total number of games played? 2 ways to solve: 1) Each of the n teams plays n-1 games; this gives a total of n(n-1) games. But each game was counted exactly twice; Thus, the total number of games is n(n-1)/2. 2 Example (of sum of the first n integers). 2) Team 1 plays n-1 games; Team 2 plays n-2 games (not counting the game with team 1); Team 3 plays n-3 games (not counting the games with teams 1 and 2); …. Team n-1 plays 1 game with team n (not including the games counted before). Thus, the total number of games is 1+2+…+(n-2)+(n-1) = n(n-1)/2 by Theorem 1. 3 Example (of sum of a geometric sequence). If all of your ancestors were distinct, what would be the total number of your ancestors for the past 40 generations? Solution: The total number is 2+4+8+…+239+240=2·(20+21+22+…+238+239) 40 39 2 1 (1) i 41 2 2 2 2 2 2 1 i0 by Theorem 2 4 Example (of sum of a geometric sequence). Assuming that each generation represents 30 years, how long is 40 generations? Answer: 30·40 = 1200 years (2) The total number of people ever lived is approximately 10 billion, which equals 1010 people. (3) What is the conclusion based on (1), (2), (3)? 5 Connection of Mathematical Induction to traditional principles of Induction and Deduction Steps of logical reasoning: Conjecture a general principle after observing it in a large number of specific instances. (traditional induction) Prove the conjecture by mathematical induction. Use the (proved) general principle to infer a conclusion for any specific instance. (traditional deduction) Proving a divisibility property by mathematical induction • Proposition: For any integer n≥1, 7n - 2n is divisible by 5. (P(n)) • Proof (by induction): 1) Basis step: The statement is true for n=1: (P(1)) 71 – 21 = 7 - 2 = 5 is divisible by 5. 2) Inductive step: Assume the statement is true for some k≥1 (P(k)) (inductive hypothesis) ; show that it is true for k+1 . (P(k+1)) 7 Proving a divisibility property by mathematical induction Proof (cont.): We are given that P(k): 7k - 2k is divisible by 5. (1) Then 7k - 2k = 5a for some aZ . (by definition) (2) We need to show: P(k+1): 7k+1 - 2k+1 is divisible by 5. (3) 7k+1 - 2k+1 = 7·7k - 2·2k = 5·7k + 2·7k - 2·2k = 5·7k + 2·(7k - 2k) = 5·7k + 2·5a (by (2)) = 5·(7k + 2a) which is divisible by 5. (by def.) Thus, P(n) is true by induction. ■ 8 Proving inequalities by mathematical induction • Theorem: For all integers n≥4, 2n < n! . (P(n)) • Proof (by induction): 1) Basis step: The statement is true for n=4: 24 = 16 < 24 = 4! . 2) Inductive step: (P(4)) Assume the statement is true for some k≥4 ; (P(k)) show that it is true for k+1 . (P(k+1)) Proving inequalities by mathematical induction Proof (cont.): We are given that P(k): 2k < k! We need to show: P(k+1): 2k+1 < (k+1)! (1) (2) 2k+1 = 2·2k < 2·k! (based on (1)) < (k+1)·k! (since k≥4) = (k+1)! Thus, P(n) is true by induction. ■ 10 Proving inequalities by mathematical induction • Theorem: For all integers n≥5, n 2 < 2 n. • Proof (by induction): 1) Basis step: The statement is true for n=5: 52 =25 < 32 = 25. 2) Inductive step: (P(n)) (P(5)) Assume the statement is true for some k≥5 ; (P(k)) show that it is true for k+1 . (P(k+1)) Proving inequalities by mathematical induction Proof (cont.): We are given that P(k): k2 < 2k. (1) We need to show: (k+1)2 < 2k+1. (2) (k+1)2 = k2+2k+1 < k2 +2k (since k≥5) < 2k + 2k (based on (1)) = 2·2k = 2k+1. Thus, P(n) is true by induction. ■ P(k+1): 12