Lesson 80 - Mathematical Induction

HL2 - Santowski

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

2

Opening Exercise – An Example 310

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

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

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

Induction example #1, continued

• Recall the inductive hypothesis:

• Proof of inductive step:

10

2(k 1) 1 2i 1i1

k

k2 2k 1

2(k 1) 1k2 k2 2k 1

2i 1 k2

i1

k

k2 2k 1 k2 2k 1

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)• Because of P(k) → P(k+1), if it’s true for P(2), then it’s true for P(3)• Because of P(k) → P(k+1), if it’s true for P(3), then it’s true for P(4)• Because of P(k) → P(k+1), if it’s true for P(4), then it’s true for P(5)• 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

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Why speling is not so important…

I cdnuolt blveieetaht I cluod aulaclty uesdnatnrd waht I was rdanieg. The phaonmneal pweor of thehmuan mind. Aoccdrnig to a rscheearch at Cmabrigde Uinervtisy, it deosn't mttaer in waht oredr the ltteers in a wrod are, the olny iprmoatnt tihng is taht thefrist and lsat ltteer be in the rghit pclae. The rset can be a taotl mses andyou can sitll raed it wouthit a porbelm. Tihs is bcuseae the huamn mnid deosnot raed ervey lteter by istlef, but the wrod as a wlohe. Amzanig huh? yaeh and I awlyas thought slpeling was ipmorantt.

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Second induction example

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

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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

Second induction example, continued

• Base case: Show P(1):

• Inductive hypothesis: Assume

• Inductive step: Show

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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 3k 2 k2 3k 2

2(k 1)k2 k (k 1)2 k 1

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

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P(k) 2i k2 ki1

k

k2 3k 2 k2 3k 2

2(k 1)k2 k (k 1)2 k 1

2(k 1) 2ii1

k

(k 1)2 k 1

2ii1

k1

(k 1)2 k 1

Third induction example

• Show

19

6

)12)(1(

1

2

nnni

n

i

Third induction example

• Show

• Base case: n = 1

• Inductive hypothesis: assume

20

6

)12)(1(

1

2

nnni

n

i

116

61

6

)12)(11(1

2

1

1

2

i

i

6

)12)(1(

1

2

kkki

k

i

Third induction example

• 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

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

• Show:

22

6

)22)(1(

1

2

nnni

n

i

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

• 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

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

• 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

Fourth induction example

• Show that n! < nn for all n > 1

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Fourth induction example

• Show that n! < nn for all n > 1

• Base case: n = 22! < 22

2 < 4

• Inductive hypothesis: assume k! < kk

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

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Fourth induction example

• Show that n! < nn for all n > 1

• Base case: n = 22! < 22

2 < 4

• Inductive hypothesis: assume k! < kk

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

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

Proving De Moivre’s Theorem by Induction

Proving De Moivre’s Theorem by Induction

Further Examples - #1 & #2

Further Examples - #3

IB Examples

IB Examples

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?

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Mathematical Induction: Example

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

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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.

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5¢ 3¢ 3¢

Replace 5¢ coin by two 3¢ coinsk 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

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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

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43

kkkk 22

kkiikikkiiiiki 22222 2222

22

2

2

22222 kkiikikkiiiiki

)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)