mathematical induction Sept. 29, 2017 1 For all 1 , 1 + 2 + 3 + - PowerPoint PPT Presentation

COMP 250 Lecture 10 mathematical induction Sept. 29, 2017 1

For all 𝑜 ≥ 1 , 1 + 2 + 3 + … . + 𝑜 − 1 + 𝑜 = 𝑜(𝑜 + 1) 2 How to prove such a statement ? By “proof”, we mean a formal logical argument that convincely shows the statement is true. Note that “ convincely ” is itself not well defined. 2

1 + 2 + … + 𝑜 − 1 + 𝑜 Rewrite by considering 𝑜/2 pairs : 1 + 2 + ⋯ + 𝑜 2 + ( 𝑜 2 + 1) … + 𝑜 − 1 + 𝑜 If 𝑜 is even, then adding up the 𝑜/2 pairs gives 𝑜/2 *( 𝑜 + 1). What if 𝑜 is odd? 3

What if 𝑜 is odd? Then, 𝑜 -1 is even. So, 1 + 2 + … + 𝑜 − 1 + 𝑜 𝑜−1 ( 2 ∗ 𝑜 ) + 𝑜 = 𝑜−1 = ( 2 + 1) ∗ 𝑜 = 𝑜+1 2 ∗ 𝑜 which is the same formula as before. 4

Mathematical Induction Consider a statement of the form: “For all 𝑜 ≥ 𝑜 0 , 𝑄 𝑜 is true” where 𝑜 0 is some constant and proposition 𝑄(𝑜) has value true or false for each 𝑜 . Mathematical induction is a general technique for proving such a statement. 5

“ For all 𝑜 ≥ 𝑜 0 , 𝑄 𝑜 is true” For all 𝑜 ≥ 1 , 𝑜(𝑜+1) 1 + 2 + … + 𝑜 − 1 + 𝑜 = 2 6

Mathematical induction requires proving two things: Base case: “P( 𝑜 0 ) is true.” Induction step: “For any 𝑙 ≥ 𝑜 0 , if P(𝑙) is true, then P(𝑙 + 1) is also true.” 7

1 2 𝑜 0 k k+1 Base case: “P( 𝑜 0 ) is true .” Induction step: “For any k ≥ 𝑜 0 , if P(k) is true , then P(k+1) is also true.” The statement “P(k) is true” is called the “induction hypothesis”. 8

Base case: Induction step: For any k >= 𝑜 0 , if P(k) is true P( 𝑜 0 ) is true. then P(k+1) is true. 1 2 n0 k k+1 Thus we have proved: For any n >= 𝑜 0 , P(n) is true. 1 2 n0 k k+1

For all 𝑜 ≥ 1 , Statement: 1 + 2 + 3 + … . + 𝑜 − 1 + 𝑜 = 𝑜(𝑜 + 1) 2 Proof (base case, n=1): 10

For all 𝑜 ≥ 1 , Statement: 1 + 2 + 3 + … . + 𝑜 − 1 + 𝑜 = 𝑜(𝑜 + 1) 2 Proof (base case, n=1): 1(1+1) 1 = (true) 2 11

induction hypothesis is that P(𝑙) is true: 1 + 2 + … + 𝑙 = 𝑙(𝑙 + 1) 2 Base case: Induction step: For any k >= 𝑜 0 , if P(k) is true P( 1 ) is true. then P(k+1) is true. 1 2 k k+1

Proof of Induction Step: ( 1 + 2 + 3 + … . + 𝑙) + 𝑙 + 1 by induction hypothesis 𝑙(𝑙+1) + 𝑙 + 1 = 2 = 13

Proof of Induction Step: ( 1 + 2 + 3 + … . + 𝑙) + 𝑙 + 1 by induction hypothesis 𝑙(𝑙+1) + 𝑙 + 1 = 2 𝑙 2 + 1) (𝑙 + 1) = ( 1 2 ( 𝑙 + 2 ) (𝑙 + 1) = Thus, 𝑄 𝑙 𝑗𝑡 𝑢𝑠𝑣𝑓 𝑗𝑛𝑞𝑚𝑗𝑓𝑡 𝑄 𝑙 + 1 𝑗𝑡 𝑢𝑠𝑣𝑓. 14

Possible confusion 𝑄 𝑙 has value true or false (Boolean). So, 𝑄 𝑙 𝑗𝑡 𝑢𝑠𝑣𝑓 means what? 15

Examples “ 3 = 2 + 1 ” is true. “ 3 = 2 + 2 ” is false. 16

Examples “ 3 = 2 + 1 ” is true. “ 3 = 2 + 2 ” is false. “If 3 = 2 + 2 then 5 > 7” is true. If this is a mystery to you, then I strongly advise you to take MATH 240 or MATH 318 (logic). 17

Mathematical Induction: Example 2 Prove the following statement: For all 𝑜 ≥ 3 , 2𝑜 + 1 < 2 𝑜 . 2 𝑜 2𝑜 + 1 𝑜 0 1 2 3 4 18

For all 𝑜 ≥ 3 , 2𝑜 + 1 < 2 𝑜 . Statement: Note: P( 𝑜 ) is false for n= 1, 2. But that has nothing to do with what we need to prove. Proof (base case, 𝑜 =3): 19

For all 𝑜 ≥ 3 , 2𝑜 + 1 < 2 𝑜 . Statement: Note: P( 𝑜 ) is false for n= 1, 2. But that has nothing to do with what we need to prove. Proof (base case, 𝑜 =3): 2*3 + 1 < 8 (true) 20

For all 𝑜 ≥ 3 , 2𝑜 + 1 < 2 𝑜 . Statement: Proof of Induction Step: We want to show that P(k) implies P(k+1). 21

For all 𝑜 ≥ 3 , 2𝑜 + 1 < 2 𝑜 . Statement: Proof of Induction Step: We want to show that P(k) implies P(k+1). 2 𝑙 + 1 + 1 = ? 2𝑙 + 2 + 1 (This is 2𝑜 + 1 where 𝑜 = 𝑙 + 1 .) 22

For all 𝑜 ≥ 3 , 2𝑜 + 1 < 2 𝑜 . Statement: Proof of Induction Step: We want to show that P(k) implies P(k+1). 2 𝑙 + 1 + 1 = 2𝑙 + 2 + 1 by induction hypothesis < 2 𝑙 + 2 23

For all 𝑜 ≥ 3 , 2𝑜 + 1 < 2 𝑜 . Statement: Proof of Induction Step: We want to show that P(k) implies P(k+1). 2 𝑙 + 1 + 1 = 2𝑙 + 2 + 1 by induction hypothesis < 2 𝑙 + 2 < 2 𝑙 + 2 𝑙 , for k ≥ 3 This inequality is also true for k >=2 but we don’t care because we are trying to = 2 𝑙+1 prove for k>=3. 24

Base case: Induction step: For any k >= 𝑜 0 , if P(k) then P(k+1). P( 𝑜 0 ) is true. 1 2 n0 k k+1 Thus, For any n >= 𝑜 0 , P(n) is true. 1 2 n0 k k+1

Example 3 Statement: For all 𝑜 ≥ 5 , 𝑜 2 < 2 𝑜 . 2 𝑜 𝑜 2 𝑜 0 1 2 3 4 5 26

Statement: For all 𝑜 ≥ 5 , 𝑜 2 < 2 𝑜 . Base case (n = 5): Induction step: 27

Statement: For all 𝑜 ≥ 5 , 𝑜 2 < 2 𝑜 . Base case (n = 5): 25 < 32 Induction step: What do we assume ? What do we want to prove ? 28

Statement: For all 𝑜 ≥ 5 , 𝑜 2 < 2 𝑜 . Base case (n = 5): 25 < 32 Induction step: What do we assume ? 𝑙 2 < 2 𝑙 , 𝑙 ≥ 5 What do we want to show ? (𝑙 + 1) 2 < 2 𝑙+1 29

Statement: For all 𝑜 ≥ 5 , 𝑜 2 < 2 𝑜 . Base case (n = 5): 25 < 32 Induction step: (𝑙 + 1) 2 = 𝑙 2 + 2𝑙 + 1 30

Statement: For all 𝑜 ≥ 5 , 𝑜 2 < 2 𝑜 . Base case (n = 5): 25 < 32 Induction step: (𝑙 + 1) 2 = 𝑙 2 + 2𝑙 + 1 by induction hypothesis < 2 𝑙 + 2𝑙 + 1 31

Statement: For all 𝑜 ≥ 5 , 𝑜 2 < 2 𝑜 . Base case (n = 5): 25 < 32 Induction step: (𝑙 + 1) 2 = 𝑙 2 + 2𝑙 + 1 by induction hypothesis < 2 𝑙 + 2𝑙 + 1 by Example 2 < 2 𝑙 + 2 𝑙 = 2 𝑙+1 32

Example 4 : Fibonacci Sequence 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, …. 𝐺 0 = 0 𝐺 1 = 1 𝐺 𝑜 + 2 = 𝐺 𝑜 + 1 + 𝐺(𝑜) , for 𝑜 ≥ 0. Statement: For all 𝑜 ≥ 0, 𝐺 𝑜 < 2 𝑜 33

𝐺 0 = 0 𝐺 1 = 1 𝐺 𝑜 + 2 = 𝐺 𝑜 + 1 + 𝐺(𝑜) , for 𝑜 ≥ 0. Base case(s): 𝑜 = 0: 0 < 2 0 is true. 𝑜 = 1: 1 < 2 1 is true. 34

𝐺 0 = 0 𝐺 1 = 1 𝐺 𝑜 + 2 = 𝐺 𝑜 + 1 + 𝐺(𝑜) , for 𝑜 ≥ 0. Induction step: 𝐺 𝑙 + 1 = 𝐺 𝑙 + 𝐺 𝑙 − 1 by induction hypothesis < 2 𝑙 + 2 𝑙−1 35

𝐺 0 = 0 𝐺 1 = 1 𝐺 𝑜 + 2 = 𝐺 𝑜 + 1 + 𝐺(𝑜) , for 𝑜 ≥ 0. Induction step: 𝐺 𝑙 + 1 = 𝐺 𝑙 + 𝐺 𝑙 − 1 by induction hypothesis < 2 𝑙 + 2 𝑙−1 < 2 𝑙 + 2 𝑙 = 2 𝑙+1 36