1. Lecture 3.1: Mathematical Induction
CS 250, Discrete Structures, Fall 2015
Nitesh Saxena
Adopted from previous lectures by Cinda Heeren, Zeph Grunschlag

2. Lecture 3.1 -- Mathematical Induction
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Lecture Mathematical Induction

3. Lecture 3.1 -- Mathematical Induction
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Lecture Mathematical Induction

4. Lecture 3.1 -- Mathematical Induction
Outline
Mathematical Induction Principle
Examples
Why it all works
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Lecture Mathematical Induction

5. Mathematical Induction
Suppose we have a sequence of propositions which we would like to prove:
P (0), P (1), P (2), P (3), P (4), … P (n), …
EG: P (n) =
“The sum of the first n positive odd numbers is equal to n2”
We can picture each proposition as a domino:
Book typically starts with n=1 instead of n=0. Starting point is arbitrary.
P (n)
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Lecture Mathematical Induction

6. Mathematical Induction
So sequence of propositions is a sequence of dominos.
P (0)
P (1)
P (2)
P (n)
P (n+1) … …
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Lecture Mathematical Induction

7. Mathematical Induction
When the domino falls, the corresponding proposition is considered true:
P (n)
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Lecture Mathematical Induction

8. Mathematical Induction
When the domino falls (to right), the corresponding proposition is considered true:
P (n)
true
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Lecture Mathematical Induction

9. Mathematical Induction
Suppose that the dominos satisfy two constraints.
Well-positioned: If any domino falls (to right), next domino (to right) must fall also.
First domino has fallen to right
P (n)
P (n+1)
P (0)
true
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Lecture Mathematical Induction

10. Mathematical Induction
Suppose that the dominos satisfy two constraints.
Well-positioned: If any domino falls to right, the next domino to right must fall also.
First domino has fallen to right
P (n)
P (n+1)
P (0)
true
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Lecture Mathematical Induction

11. Mathematical Induction
Suppose that the dominos satisfy two constraints.
Well-positioned: If any domino falls to right, the next domino to right must fall also.
First domino has fallen to right
P (n)
true
P (n+1)
true
P (0)
true
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Lecture Mathematical Induction

12. Mathematical Induction
Then can conclude that all the dominos fall!
P (0)
P (1)
P (2)
P (n)
P (n+1) …
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Lecture Mathematical Induction

13. Mathematical Induction
Then can conclude that all the dominos fall!
P (0)
P (1)
P (2)
P (n)
P (n+1) …
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Lecture Mathematical Induction

14. Mathematical Induction
Then can conclude that all the dominos fall!
P (1)
P (2)
P (n)
P (n+1)
P (0)
true …
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Lecture Mathematical Induction

15. Mathematical Induction
Then can conclude that all the dominos fall!
P (2)
P (n)
P (n+1)
P (0)
true
P (1)
true …
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Lecture Mathematical Induction

16. Mathematical Induction
Then can conclude that all the dominos fall!
P (n)
P (n+1)
P (0)
true
P (1)
true
P (2)
true …
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Lecture Mathematical Induction

17. Mathematical Induction
Then can conclude that all the dominos fall!
P (n)
P (n+1)
P (0)
true
P (1)
true
P (2)
true …
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Lecture Mathematical Induction

18. Mathematical Induction
Then can conclude that all the dominos fall!
P (n+1)
P (0)
true
P (1)
true
P (2)
true …
P (n)
true
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Lecture Mathematical Induction

19. Mathematical Induction
Then can conclude that all the dominos fall!
P (0)
true
P (1)
true
P (2)
true …
P (n)
true
P (n+1)
true
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Lecture Mathematical Induction

20. Mathematical Induction
Principle of Mathematical Induction:
If:
[basis] P (0) is true
[induction] k P(k)P(k+1) is true
Then: n P(n) is true
This formalizes what occurred to dominos.
P (0)
true
P (1)
true
P (2)
true …
P (n)
true
P (n+1)
true
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Lecture Mathematical Induction

21. Lecture 3.1 -- Mathematical Induction
Exercise 1
Use induction to prove that the sum of the first n odd integers is n2.
Prove a base case (n=1)
Prove P(k)P(k+1)
Base case (n=1): the sum of the first 1 odd integer is 12. Yup, 1 = 12.
Assume P(k): the sum of the first k odd ints is k … + (2k - 1) = k2
Prove that … + (2k - 1) + (2k + 1) = (k+1)2
… + (2k-1) + (2k+1) =
k2 + (2k + 1)
= (k+1)2
By arithmetic
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Lecture Mathematical Induction

22. Lecture 3.1 -- Mathematical Induction
Exercise 2
Prove that 11! + 22! + … + nn! = (n+1)! - 1, n
Base case (n=1): 11! = (1+1)! - 1?
Yup, 11! = 1, 2! - 1 = 1
Assume P(k): 11! + 22! + … + kk! = (k+1)! - 1
Prove that 11! + … + kk! + (k+1)(k+1)! = (k+2)! - 1
11! + … + kk! + (k+1)(k+1)! =
(k+1)! (k+1)(k+1)!
= (1 + (k+1))(k+1)! - 1
= (k+2)(k+1)! - 1
= (k+2)! - 1
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Lecture Mathematical Induction

23. Exercises 3 and 4 (have seen before?)
Recall sum of arithmetic sequence:
Recall sum of geometric sequence:
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Lecture Mathematical Induction

24. Mathematical Induction - why does it work?
Proof of Mathematical Induction:
We prove that
(P(0)  (k P(k)  P(k+1)))  (n P(n))
Assume
P(0)
k P(k)  P(k+1)
n P(n)
Proof by contradiction.
n P(n)
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Lecture Mathematical Induction

25. Mathematical Induction - why does it work?
Assume
P(0)
k P(k)  P(k+1)
n P(n)
n P(n)
Since N is well ordered, S has a least element. Call it k.
Let S = { n : P(n) }
What do we know?
P(k) is false because it’s in S.
k  0 because P(0) is true.
P(k-1) is true because P(k) is the least element in S.
But by (2), P(k-1)  P(k). Contradicts P(k-1) true, P(k) false.
Done.
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Lecture Mathematical Induction

26. Lecture 3.1 -- Mathematical Induction
Today’s Reading
Rosen 5.1
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Lecture Mathematical Induction