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 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. 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. 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+… =2·( … )
(1)
by Theorem 2

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 1010 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. 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 = = 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 (P(k+1))

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

9. 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: (P(4))
24 = 16 < 24 = 4! .
2) Inductive step:
Assume the statement is true for some k≥4 ; (P(k))
show that it is true for k (P(k+1))

10. Proving inequalities by mathematical induction
Proof (cont.): We are given that
P(k): 2k < k! (1)
We need to show:
P(k+1): 2k+1 < (k+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 ■

11. Proving inequalities by mathematical induction
Theorem: For all integers n≥5,
n2 < 2n. (P(n))
Proof (by induction):
1) Basis step:
The statement is true for n=5: (P(5))
52 =25 < 32 = 25.
2) Inductive step:
Assume the statement is true for some k≥5 ; (P(k))
show that it is true for k (P(k+1))

12. Proving inequalities by mathematical induction
Proof (cont.): We are given that
P(k): k2 < 2k (1)
We need to show:
P(k+1): (k+1)2 < 2k (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 ■