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1 Mathematical Induction (cont.). 2 Example (of sum of the first n integers).  In a round-robin tournament each of the n teams plays every other team.

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Presentation on theme: "1 Mathematical Induction (cont.). 2 Example (of sum of the first n integers).  In a round-robin tournament each of the n teams plays every other team."— Presentation transcript:

1 1 Mathematical Induction (cont.)

2 2 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.

3 3 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.

4 4 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·( … ) (1) by Theorem 2

5 5 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 people.(3) What is the conclusion based on (1), (2), (3)?

6 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)

7 7 Proving a divisibility property by mathematical induction Proposition: For any integer n≥1, 7 n - 2 n is divisible by 5. (P(n)) Proof (by induction): 1) Basis step: The statement is true for n=1: (P(1)) 7 1 – 2 1 = = 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))

8 8 Proving a divisibility property by mathematical induction  Proof (cont.): We are given that P(k): 7 k - 2 k is divisible by 5. (1) Then 7 k - 2 k = 5a for some a  Z. (by definition) (2) We need to show: P(k+1):7 k k+1 is divisible by 5. (3) 7 k k+1 = 7 · 7 k - 2 · 2 k = 5 · 7 k + 2 · 7 k - 2 · 2 k = 5 · 7 k + 2 ·( 7 k - 2 k ) = 5 · 7 k + 2 · 5a (by (2)) = 5 ·( 7 k + 2a) which is divisible by 5. (by def.) Thus, P(n) is true by induction. ■

9 Proving inequalities by mathematical induction Theorem: For all integers n≥4, 2 n < n!. (P(n)) Proof (by induction): 1) Basis step: The statement is true for n=4: (P(4)) 2 4 = 16 < 24 = 4!. 2) Inductive step: Assume the statement is true for some k≥4 ; (P(k)) show that it is true for k+1. (P(k+1))

10 10 Proving inequalities by mathematical induction  Proof (cont.): We are given that P(k): 2 k < k! (1) We need to show: P(k+1): 2 k+1 < (k+1)! (2) 2 k+1 = 2·2 k < 2·k! (based on (1)) < (k+1)·k! (since k≥4) = (k+1)! Thus, P(n) is true by induction. ■

11 Proving inequalities by mathematical induction Theorem: For all integers n≥5, n 2 < 2 n. (P(n)) Proof (by induction): 1) Basis step: The statement is true for n=5: (P(5)) 5 2 =25 < 32 = ) Inductive step: Assume the statement is true for some k≥5 ; (P(k)) show that it is true for k+1. (P(k+1))

12 12 Proving inequalities by mathematical induction  Proof (cont.): We are given that P(k): k 2 < 2 k. (1) We need to show: P(k+1): (k+1) 2 < 2 k+1. (2) (k+1) 2 = k 2 +2k+1 < k 2 +2 k (since k≥5) < 2 k + 2 k (based on (1)) = 2·2 k = 2 k+1. Thus, P(n) is true by induction. ■


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