1 Intro to Induction Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong.

1

Intro to Induction

Supplementary Notes

Prepared by Raymond WongPresented by Raymond Wong

2

e.g.1 (Page 4)

Illustration of “Proof by Contradiction”We are going to prove that a claim C is correct

Proof by Contradiction:

Suppose “NOT C”

….

Derive some results, which may contradict to 1. “NOT C”, OR

2. some facts

e.g., we derived that C is true finally

e.g., we derived that “1 = 4”

3

e.g.1

Illustration of “Proof by smallest counter example”

We are going to prove the following claim C: statement P(m) is true for each non-negative integer m, namely 0, 1, 2, …

P(0) true

P(1) true

P(2) true

P(3) true

P(4) true

If we can prove that statement P(m) is true for each non-negative integer separately, then we can prove the above claim C is correct.

Suppose that I want to prove that the above claimis correct by “Proof by Contradiction”.

… true

4

e.g.1



P(0) true

P(1) true

P(2) true

P(3) true

P(4) true

We can assume that there exists a non-negative integer k’ such that P(k’) is false


… true

false

Suppose “NOT C”.

There may exist another non-negative integer k such that P(k) is false

false

5

e.g.1



P(0) true

P(1) true

P(2) true

P(3) true

P(4) true


… true

false


false

We can assume that there exists a smallest non-negative integer k such that P(k) is false

Why?

This is called by “Proof by smallest counter example”.

6

e.g.2 (Page 5)Steps for “Proof by smallest counter example”


We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, …

P(0) true

P(1) true

P(2) true

P(3) true

P(4) true

… true

false

false

Step 1: Suppose that claim C is not true.

Step 2: there exists a smallest non-negative integer m such that P(m) is false.

Step 3: We want to show that this value m must be greater than the smallest value (i.e., 0)

Step 4: We derive that P(i) is true for 0 i < m

Step 5: We consider a special case that P(m-1) is true.

Step 6: Consider the LHS (or some components) of P(m) Prove that P(m) is true (by using P(m-1))

Step 7: We have a contradiction that P(m) is false

Step 8: Thus, by the principle of proof by contradiction,claim C is correct.

e.g., P(n) is“0+1+2+…+n = ”

n(n+1)

2

7

e.g.3 (Page 14)

Illustration of “Proof by mathematical induction”


P(0) true

P(1) true

P(2) true

P(3) true

P(4) true

If we can prove that statement P(n) is true for each non-negative integer separately, then we can prove the above claim C is correct.

… true

8

e.g.3



P(0) true

P(1)

P(2)

P(3)

P(4)

…

Step 1: Prove that P(0) (i.e., the base case) is true.

Verify that P(0) is true

9

e.g.3



P(0) true

P(1)

P(2)

P(3)

P(4)

…


Step 2: Prove that “P(n-1) P(n)” is true for all n > 0.

true

true

true

true

true

Step 2(a): Assume that P(n-1) is true for n > 0.


Step 2(b): According to P(n-1), we deduce that P(n) is true.

Inductive Hypothesis

Inductive Step

10

e.g.4 (Page 15)Prove that n 0, 2n+1 n2+2


Let P(n) be “2n+1 n2+2“

Consider 20+1 = 2

= 0+2

We want to show that 20+1 02+2

02+2

Thus, P(0) is true.

11

e.g.4Prove that n 0, 2n+1 n2+2

Let P(n) be “2n+1 n2+2“

12

e.g.4

Prove that n 0, 2n+1 n2+2

Let P(n) be “2n+1 n2+2“Step 2: Prove that “P(n-1) P(n)” is true for all n > 0.



That is, 2(n-1)+1 (n-1)2+2 for n > 0.

2n (n-1)2+2 for n > 0.

Consider 2n+1

We want to show that 2n+1 n2+2

= 2n . 2

[(n-1)2+2] . 2

= 2(n-1)2+2. 2

= 2(n2-2n+1)+4

= 2n2-4n+2+4

= 2n2-4n+6

= (n2+n2) – 4n + (2+4)

What should I do next?

= n2+2 + n2 – 4n + 4

= (n2+2) + (n2 – 4n + 4)

= (n2+2) + (n – 2)2

n2+2 (Since (n-2)2 0)

That is, P(n) is true.

We prove that “P(n-1) P(n)” is true for all n > 0

By Mathematical Induction, n 0, 2n+1 n2+2

Thus, 2n+1 n2+2

13

e.g.5 (Page 16)Prove that n 2, 2n+1 > n2+3


Let P(n) be “2n+1 > n2+3“

Consider 22+1 = 23

= 8

We want to show that 22+1 > 22+3

> 7

Thus, P(2) is true.

= 22+3

14

e.g.5Prove that n 2, 2n+1 > n2+3

Let P(n) be “2n+1 > n2+3“

15

e.g.5

Prove that n 2, 2n+1 > n2+3

Let P(n) be “2n+1 > n2+3“Step 2: Prove that “P(n-1) P(n)” is true for all n > 2.



That is, 2(n-1)+1 > (n-1)2+3 for n > 2.

2n > (n-1)2+3 for n > 2.

Consider 2n+1

We want to show that 2n+1 > n2+3

= 2n . 2

>[(n-1)2+3] . 2

= 2(n-1)2+3. 2

= 2(n2-2n+1)+6

= 2n2-4n+2+6

= 2n2-4n+8

= (n2+n2) – 4n + (3+5)


= n2+3 + n2 – 4n + 5

= (n2+3) + (n2 – 4n + 5)

= (n2+3) + (n2 – 4n + 4 + 1)

> n2+3 (Since (n-2)2 + 1 0)

Thus, P(n) is true.

We prove that “P(n-1) P(n)” is true for all n > 2

By Mathematical Induction, n 2, 2n+1 > n2+3

= (n2+3) + (n – 2)2 + 1

16

e.g.6 (Page 18)Prove that k Z+, 1+3+5+…+(2k-1) = k2


Let P(k) be “1+3+5+…+(2k-1) = k2”

Consider 1 = 12

We want to show that 1 = 12

Thus, P(1) is true.

17

e.g.6Prove that k Z+, 1+3+5+…+(2k-1) = k2

Let P(k) be “1+3+5+…+(2k-1) = k2”

18

e.g.6

Prove that k Z+, 1+3+5+…+(2k-1) = k2

Let P(k) be “1+3+5+…+(2k-1) = k2”Step 2: Prove that “P(n-1) P(n)” is true for all n > 1.



That is, 1+3+5+…+(2(n-1)-1) = (n-1)2 for n > 1.

1+3+5+…+(2n-3) = (n-1)2 for n > 1.

Consider 1+3+5+…+(2n-1)

We want to show that 1+3+5+…+(2n-1) = n2

= 1+3+5+…+(2n-3)+(2n-1)

= (n-1)2 + (2n-1)

= (n2 – 2n + 1) + (2n – 1)

= n2

Thus, P(n) is true.

We prove that “P(n-1) P(n)” is true for all n > 1 By Mathematical Induction, k Z+, 1+3+5+…+(2k-1) = k2

19

e.g.7 (Page 19)For what positive integer values of n is “2n > n2”?

Step 1: Prove that P( ? ) (i.e., the base case) is true.

Let P(n) be “2n > n2”

Consider P(1) We want to see whether P(1) is true.

(i.e., whether “21 > 12” is true.)Consider 21

We don’t know the base case. Thus, we need to test the “smallest” value of n for the base case.

= 2

> 12

Thus, P(1) is true.


(i.e., whether “22 > 22” is true.)Consider 22 = 4

Thus, P(2) is false.

> 22

P(1) is true.

P(2) is false.

20

e.g.7For what positive integer values of n is “2n > n2”?






= 8





> 42

> 32

P(1) is true.

P(2) is false.

P(3) is false.

P(4) is false.

21







= 32

Thus, P(5) is true.



Thus, P(6) is true.

> 62

> 52

Thus, we guess that P(7), P(8), …are also true.

Thus, we think that the base case is P(5).

P(5)

P(1) is true.

P(2) is false.

P(3) is false.

P(4) is false.

P(5) is true.

P(6) is true.

22




P(5)

Prove that n 5, 2n > n2

Thus, we think that the base case is P(5).

23

e.g.7



P(5)


24

e.g.7



P(5)


We want to show that “25 > 52”

Consider 25 = 32

Thus, P(5) is true.

> 52

25

e.g.7



26

e.g.7


Let P(n) be “2n > n2”Step 2: Prove that “P(n-1) P(n)” is true for all n > 5.



That is, 2n-1 > (n-1)2 for n > 5.

Consider 2n

We want to show that 2n > n2

= 2n-1 . 2

> (n-1)2 . 2

= (n2 – 2n + 1) . 2

= 2n2 – 4n + 2

Thus, P(n) is true.

We prove that “P(n-1) P(n)” is true for all n > 5 By Mathematical Induction, n 5, 2n > n2

= n2 + n2 – 4n + 2


> n2 + n2 – 4n

> n2 + n2 – 5.n

> n2 + n2 – n.n (Since n > 5)= n2

Thus, 2n > n2

27

e.g.7


Let P(n) be “2n > n2”Step 2: Prove that “P(n-1) P(n)” is true for all n > 5.



That is, 2n-1 > (n-1)2 for n > 5.

Consider 2n

We want to show that 2n > n2

= 2n-1 . 2

> (n-1)2 . 2= (n2 – 2n + 1) . 2

= 2n2 – 4n + 2

Thus, P(n) is true.We prove that “P(n-1) P(n)” is true for all n > 5 By Mathematical Induction, n 5, 2n > n2

= n2 + n2 – 4n + 2

= n2 + n2 – 4n + 4 – 4 +2

= n2 + (n2 – 4n + 4) – 2

= n2 + (n – 2)2 – 2

Thus, 2n > n2

Alternative Derivation

Since (n-2)2 is increasing when n> 5,

we have (n-2)2 > (5-2)2

= 32

= 9> n2 + 9 – 2

= n2 + 7> n2

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e.g.8 (Page 21)

Illustration of “Proof by mathematical induction” (Weak Induction)


P(0) true

P(1)

P(2)

P(3)

P(4)

…



29

e.g.8

Illustration of “Proof by mathematical induction” (Weak Induction)


P(0) true

P(1)

P(2)

P(3)

P(4)

…


Step 2: Prove that “P(n-1) P(n)” is true for all n > 0.

true

true

true

true

true





Inductive Step

30

e.g.8

Illustration of “Proof by mathematical induction” (Strong Induction)

We are going to prove the following claim C: statement P(n) is true for each non-negative integer n, namely 0, 1, 2, …P(0) true

P(1)

P(2)

P(3)

P(4)

…



31

e.g.8



P(1)

P(2)

P(3)

P(4)

…


Step 2: Prove that “P(0)P(1)…P(n-1) P(n)” is true for all n > 0.

true

Step 2(a): Assume that P(0)P(1)…P(n-1) is true for n > 0.


Step 2(b): According to P(0)P(1)…P(n-1), we deduce that P(n) is true.


Inductive Step

32

e.g.8



P(1)

P(2)

P(3)

P(4)

…



true

true





Inductive Step

33

e.g.8



P(1)

P(2)

P(3)

P(4)

…



true

true





Inductive Step

true

34

e.g.8



P(1)

P(2)

P(3)

P(4)

…



true

true





Inductive Step

true

true

35

e.g.9 (Page 26)Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers


Consider 1 = 20

which is a power of a prime number (i.e., 2).

Thus, P(1) is true.

Let P(n) be “n is a power of a prime number OR a product of powers of prime numbers.”We want to show that 1 is a power of a prime number OR a product of powers of prime numbers

e.g., 4 = 22

12 = 22 . 3

36

e.g.9Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers

Let P(n) be “n is a power of a prime number OR a product of powers of prime numbers.”

37

e.g.9

Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers

Let P(n) be “n is a power of a prime number OR a product of powers of prime numbers.”Step 2: Prove that “P(1)P(2)…P(n-1) P(n)” is true for all n > 1.



That is, 1 is a power of a prime number OR a product of powers of prime numbers. 2 is a power of a prime number OR a product of powers of prime numbers. … n-1 is a power of a prime number OR a product of powers of prime numbers.

We want to show that n is a power of a prime number OR a product of powers of prime numbers.Consider two cases.

Case 1: n is a prime number.

Case 2: n is not a prime number.

n is a power of a prime numbern is a product of two smaller numbers, namely a and b (where a < n and b < n)Thus, n = a . bNote that a is a power of a prime number or a product of powers of prime

numbers.Note that b is a power of a prime number or a product of powers of prime numbersThus, n is a power of a prime number or a product of powers of prime numbers

38

e.g.9

Prove that every positive integer n is a power of a prime number OR a product of powers of prime numbers

Let P(n) be “n is a power of a prime number OR a product of powers of prime numbers.”Step 2: Prove that “P(1)P(2)…P(n-1) P(n)” is true for all n > 1.



That is, 1 is a power of a prime number OR a product of powers of prime numbers. 2 is a power of a prime number OR a product of powers of prime numbers. … n-1 is a power of a prime number OR a product of powers of prime numbers.

Thus, P(n) is true.

We prove that “P(1)P(2)…P(n-1) P(n)” is true for all n > 1

By Strong Mathematical Induction, every positive integer n is a power of a prime number OR a product of powers of prime numbers

1 Intro to Induction Supplementary Notes Prepared by Raymond Wong Presented by Raymond Wong.

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