lesson 81- mathematical induction...2017-11-24 1 lesson 75 - mathematical induction hl2 - santowski...

2017-11-24

1

Lesson 75 - Mathematical Induction

HL2 - Santowski

Opening Exercise - Challenge

• Take a pile of n chips (where 10 < n < 20 for now)

• Split the pile into two smaller piles of size r and s

• Repeat until you have n piles of 1 chip each

• Take the product of all the splits

• So all the r’s and s’s from each split

• Sum up each of these products

2

Opening Exercise – An Example 3

10

3

14

7

1 12 2 2 1

23

1 1 1 1 1 1

21

12 2

4 2 1

1 1 1

10

2

9*104511112421221

2017-11-24

2

Opening Exercise - Challenge

• Take a pile of n stones• Split the pile into two smaller piles of size r and s

• Repeat until you have n piles of 1 stone each

• Take the product of all the splits• So all the r’s and s’s from each split

• Sum up each of these products

• Prove that this product equals

4

2

)1( nn

What is induction?• A method of proof

• It does not generate answers: it only can prove them

• Three parts:• Base case(s): show it is true

for one element

• Inductive hypothesis: assume it is true for any given element• Must be clearly labeled!!!

• Show that if it true for the next highest element

6

2017-11-24

3

Induction example #1

• Show that the sum of the first n odd integers is n2

7

Induction example #1

• Show that the sum of the first n odd integers is n2

• Example: If n = 5, 1+3+5+7+9 = 25 = 52

• Formally, Show

• Base case: Show that P(1) is true

8

2i 1 n2

i1

n

11

11)(21

1

2

i

i

)P( where)P( nnn

)1P(

Induction example #1, continued

• Inductive hypothesis: assume true for k• Thus, we assume that P(k) is true, or that

• Note: we don’t yet know if this is true or not!

• Inductive step: show true for k+1• We want to show that:

9

2i 1 k2

i1

k

2i 1 (k 1)2

i1

k1

2017-11-24

4

Induction example #1, continued

• Recall the inductive hypothesis:

• Proof of inductive step:

10

2(k 1)1 2i 1i1

k

k2 2k 1

2(k1)1k2 k2 2k1

2i 1 k2

i1

k

k2 2k1 k2 2k1

2i 1i1

k1

(k 1)2

What did we show• Base case: P(1)

• If P(k) was true, then P(k+1) is true• i.e., P(k) → P(k+1)

• We know it’s true for P(1)

• Because of P(k) → P(k+1), if it’s true for P(1), then it’s true for P(2)




• And onwards to infinity

• Thus, it is true for all possible values of n

• In other words, we showed that:

11

)P()1P()P()1P( nnkkk

The idea behind inductive proofs

• Show the base case

• Show the inductive hypothesis

• Manipulate the inductive step so that you can substitute in part of the inductive hypothesis

• Show the inductive step

12

2017-11-24

5

Why speling is not so important…

I cdnuolt blveieetaht I cluod aulaclty uesdnatnrd waht Iwas rdanieg. The phaonmneal pweor of thehmuanmind. Aoccdrnig to a rscheearch at CmabrigdeUinervtisy, it deosn't mttaer in waht oredr the ltteers ina wrod are, the olny iprmoatnt tihng is taht thefrist andlsat ltteer be in the rghit pclae. The rset can be a taotlmses andyou can sitll raed it wouthit a porbelm. Tihs isbcuseae the huamn mnid deosnot raed ervey lteter byistlef, but the wrod as a wlohe. Amzanig huh? yaeh andI awlyas thought slpeling was ipmorantt.

13

Second induction example

• Example #2:• Show the sum of the first n positive even integers is n2 + n

14

Second induction example

• Example #2:• Show the sum of the first n positive even integers is n2 + n

• Rephrased:

• The three parts:• Base case

• Inductive hypothesis

• Inductive step

15

n P(n) where P(n) 2i n2 ni1

n

2017-11-24

6

Second induction example, continued

• Base case: Show P(1):

• Inductive hypothesis: Assume

• Inductive step: Show

16

P(1) 2(i) 12 1i1

1

2 2

P(k) 2i k2 ki1

k

P(k 1) 2ii1

k1

(k 1)2 (k 1)

Second induction example, continued

• Recall our inductive hypothesis:

17

k2 3k2 k2 3k2

2(k1)k2 k (k1)2 k1

2(k 1) 2ii1

k

(k 1)2 k 1

2ii1

k1

(k 1)2 k 1

P(k) 2i k2 ki1

k

Notes on proofs by induction

• We manipulate the k+1 case to make part of it look like the k case

• We then replace that part with the other side of the k case

18

P(k) 2i k2 ki1

k

k2 3k2 k2 3k2

2(k1)k2 k (k1)2 k1

2(k 1) 2ii1

k

(k 1)2 k 1

2ii1

k1

(k 1)2 k 1

2017-11-24

7

Third induction example

• Show

19

6

)12)(1(

1

2

nnni

n

i


• Show

• Base case: n = 1

• Inductive hypothesis: assume

20

6

)12)(1(

1

2

nnni

n

i

11

6

61

6

)12)(11(1

2

1

1

2

i

i

6

)12)(1(

1

2

kkki

k

i


• Inductive step: show

21

6

)1)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)32)(2)(1()1(

1

22

kkkik

k

i

6139261392 2323 kkkkkk

)32)(2)(1()12)(1()1(6 2 kkkkkkk

6

)32)(2)(1(

6

)12)(1()1( 2

kkkkkk

k

6

)1)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)12)(1(

1

2

kkki

k

i

2017-11-24

8

Third induction again: what if your inductive hypothesis was wrong?

• Show:

22

6

)22)(1(

1

2

nnni

n

i


• Show:

• Base case: n = 1:

• But let’s continue anyway…

• Inductive hypothesis: assume

23

6

)22)(1(

1

2

nnni

n

i

6

71

6

71

6

)22)(11(1

2

1

1

2

i

i

6

)22)(1(

1

2

kkki

k

i


• Inductive step: show

24

6

)2)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)42)(2)(1()1(

1

22

kkkik

k

i

816102614102 2323 kkkkkk

)42)(2)(1()22)(1()1(6 2 kkkkkkk

6

)42)(2)(1(

6

)22)(1()1( 2

kkkkkk

k

6

)2)1(2)(1)1)((1(1

1

2

kkki

k

i

6

)22)(1(

1

2

kkki

k

i

2017-11-24

9

Fourth induction example

• Show that n! < nn for all n > 1

25



• Base case: n = 22! < 22

2 < 4

• Inductive hypothesis: assume k! < kk

• Inductive step: show that (k+1)! < (k+1)k+1

26



• Base case: n = 22! < 22

2 < 4

• Inductive hypothesis: assume k! < kk

• Inductive step: show that (k+1)! < (k+1)k+1

27

)!1( k !)1( kk kkk )1( kkk )1)(1( 1)1( kk

2017-11-24

10

Proving De Moivre’s Theorem by Induction

Proving De Moivre’s Theorem by Induction

Further Examples - #1 & #2

2017-11-24

11

Further Examples - #3

2017-11-24

12

IB Examples

IB Examples

2017-11-24

13

IB Examples

Mathematical Induction: Example

Show that any postage of ≥ 8¢ can be

obtained using 3¢ and 5¢ stamps.

First check for a few particular values:

8¢ = 3¢ + 5¢

9¢ = 3¢ + 3¢ + 3¢

10¢ = 5¢ + 5¢

11¢ = 5¢ + 3¢ + 3¢

12¢ = 3¢ + 3¢ + 3¢ + 3¢

How to generalize this?

38

Mathematical Induction: Example

• Let P(n) be the sentence “n cents postage can be

obtained using 3¢ and 5¢ stamps”.

• Want to show that

“P(k) is true” implies “P(k+1) is true”

for any k ≥ 8¢.

• 2 cases: 1) P(k) is true and

the k cents contain at least one 5¢.

2) P(k) is true and

the k cents do not contain any 5¢.

39

2017-11-24

14

Mathematical Induction: Example• Case 1: k cents contain at least one 5¢ coin.

• Case 2: k cents do not contain any 5¢ coin.

Then there are at least three 3¢ coins.

40

5¢ 3¢ 3¢

Replace 5¢ coin by two 3¢ coins

k cents k+1 cents

3¢3¢

3¢ 5¢ 5¢

Replace three 3¢ coins by two 5¢ coins

k cents k+1 cents

Opening Exercise - Solution

• We will show it is true for a pile of k stones, and show it is true for k+1 stones• So P(k) means that it is true for k stones

• Base case: n = 1• No splits necessary, so the sum of the products = 0

• 1*(1-1)/2 = 0

• Base case proven

41

Opening Exercise - Solution

• Inductive hypothesis: assume that P(1), P(2), …, P(k) are all true• This is strong induction!

• Inductive step: Show that P(k+1) is true• We assume that we split the k+1 pile into a pile of i stones and a pile

of k+1-i stones

• Thus, we want to show that (i)*(k+1-i) + P(i) + P(k+1-i) = P(k+1)

• Since 0 < i < k+1, both i and k+1-i are between 1 and k, inclusive

42

2017-11-24

15

43

kkkk 22

kkiikikkiiiiki 22222 2222

22

2

2

22222 kkiikikkii

iiki

)1P()1P()P()1(*)( kikiiki

22

)11)(1()1P(

2 kkkkk

2

2

2

)11)(1()1P(

22 iikikkikikik

2)P(

2 iii

Thus, we want to show that (i)*(k+1-i) + P(i) + P(k+1-i) = P(k+1)

lesson 81- mathematical induction...2017-11-24 1 lesson 75 - mathematical induction hl2 - santowski...

Documents