Fskik 1 nota

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MATHEMATICAL INDUCTION

MTK3013DISCRETE STRUCTURES


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


• 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

Induction example

11

11)(21

1

2

i

i)1P(


4

• 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:

k

i

ki1

212

1

1

2)1(12k

i

ki

Induction example, continued


5

12121)1(2 2

1

kkikk

i

121)1(2 22 kkkk

Induction example, continued

• Recall the inductive hypothesis:

• Proof of inductive step:

k

i

ki1

212

1212 22 kkkk

21

1

)1(12

kik

i


6

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: )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


8

Second induction example

– Show the sum of the first n positive even integers is n2 + n

– Rephrased:

• The three parts:– Base case– Inductive hypothesis– Inductive step

n

i

nninnn1

22)P( where)P(


9

Second induction example, continued

• Base case: Show P(1):

• Inductive hypothesis: Assume

• Inductive step: Show

22

11)(2)1P(1

1

2

i

i

k

i

kkik1

22)P(

)1()1(2)1P( 21

1

kkikk

i


2323 22 kkkk

1)1()1(2 22 kkkkk

1)1(2)1(2 2

1

kkikk

i

1)1(2 21

1

kkik

i

Second induction example, continued

• Recall our inductive hypothesis:

k

i

kkik1

22)P(


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

k

i

kkik1

22)P(

2323 22 kkkk

1)1()1(2 22 kkkkk

1)1(2)1(2 2

1

kkikk

i

1)1(2 21

1

kkik

i


12

Third induction example

• Rosen, question 7: Show

• Base case: n = 1

• Inductive hypothesis: assume

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


13

Third induction example

• Inductive step: show 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:

• Base case: n = 1:

• But let’s continue anyway…• Inductive hypothesis: assume

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


15

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

• Inductive step: show 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

• Rosen, question 14: 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

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

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