Induction3

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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
i0
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)?
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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))
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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 aZ . (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.
■
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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.
■
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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):
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