Mathematical Induction I.pptx

Mathematical Induction

a means of proving a theorem by showing that if it is true of any particular case it is true of the next case in a series, and then showing

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!!!

• _{In general, mathematical induction can be used to}

prove statements that assert that P(n) is true for all positive integers n, where P(n) is a propositional

function.

• _{When we use mathematical induction to prove a}

theorem, we first show that P(1) is true. Then we know that P(2) is true, because P(1) implies P(2). Further, we know that P(3) is true, because P(2)

• _{Many theorems assert that P(n) is true for all positive integers n,}

where P(n) is a propositional function. Mathematical induction is a technique for proving theorems of this kind. In other words, mathematical induction can be used to prove statements of the form ∀n P(n), where the domain is the set of positive integers. Mathematical induction can be used to prove an extremely wide variety of theorems, each of which is a statement of this form. (Remember, many mathematical assertions include an implicit universal quantifier. The statement “if n is a positive integer, then n3 _{− n is divisible by 3” is an example of this. Making the}

• _{Have you heard of the "Domino Effect"?} • _{Step 1. The first domino falls}

• _{Step 2. When any domino falls, the next}

domino falls

• _{So ... all dominos will fall!}

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

– _{If n = 5, 1+3+5+7+9 = 25 = 5}2

– _{Formally, Show}

• _{Basis step: Show that P(1) is true}

Induction example



n P(n) where P(n)





 

n

i

n i

1

2 1

2

1 1

1 1

) ( 2

1

1

2



 





i

• _{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 that it is true for k+1}

– _{We want to show that:}

Induction example, continued



 

k

i

k i

1

2 1

2





 



1

1

2

) 1 (

1 2

k

i

Induction example, continued

• _{Recall the inductive hypothesis:}

• _{Proof of inductive step:}



2   

  k i k i 1 2 1 2 1 ) 1 ( 2 2 1       



2 k    k 2  k 2  k  1

2 1

2 2

2 _{ k   k  k }

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:





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

Second induction example

– _{Show the sum of the first n positive even integers}

is n2 _{+ n}

– _Rephrased:

• _{The three parts:}

– _{Basis step}

– _{Inductive hypothesis} – _{Inductive step}





 



 n

i

n n

i n

n n

1

2

2 )

P( where )

Second induction example, continued

• _{Basis step: Show P(1):}

• _{Inductive hypothesis: Assume}

• _{Inductive step: Show}

2 2

1 1

) ( 2 )

1

P( 1

1

2

 

 

 



i

i





 

 k i

k k

i k

1

2 2

) P(

)

1

(

)

1

(

2

)

1

Second induction example, continued

• _{Recall our inductive}

hypothesis:







2 _{k   k} 2 _{ k  k } 2 _{ k }

2 3

2

3 2

2 _{ k   k  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



    k i k k i k 1 2 2 ) P( 1 ) 1 ( 2 2 1 1       k k i k i 1 ) 1 ( 2 ) 1 ( 2 2 1      



Third induction example

• _Show

• _{Base case: n = 1}

• _{Inductive hypothesis: assume}

6 ) 1 2 )( 1 ( 1

2 _  





2 _  

Third induction example

• _{Inductive step: show}

) 1 2 )( 1 (

2 _  



6 ) 1 ) 1 ( 2 )( 1 ) 1 )(( 1 ( 1 1

2 _     



2 _     



2 _{ }   





(k  2  k k  k   k  k  k 

) 3 2 )( 2 )( 1 ( ) 1 2 )( 1 ( ) 1 (

6 k  2  k k  k   k  k  k 

6 13 9 2 6 13 9

17

Third induction again: what if your inductive

hypothesis was wrong?

• _Show:

• _{Base case: n = 1:}

• _{But let’s continue anyway…}

• _{Inductive hypothesis: assume}

6 ) 2 2 )( 1 ( 1

2 _  

  n n n i n i 6 7 1 6 7 1 6 ) 2 2 )( 1 1 ( 1 2 1 1 2     



2 _  

Third induction again: what if your inductive

hypothesis was wrong?

• _{Inductive step: show}

) 2 2 )( 1 (

2 _  



6 ) 2 ) 1 ( 2 )( 1 ) 1 )(( 1 ( 1 1

2 _     

  k k k i k i 6 ) 2 ) 1 ( 2 )( 1 ) 1 )(( 1 ( 1 1

2 _     

  k k k i k i 6 ) 4 2 )( 2 )( 1 ( ) 1 ( 1 2

2 _{ }   





(k  2  k k  k   k  k  k 

) 4 2 )( 2 )( 1 ( ) 2 2 )( 1 ( ) 1 (

6 k  2  k k  k   k  k  k 

8 16 10 2 6 14 10

Fourth induction example

• _{Show that n! < n}n _{for all n > 1}

• _{Base case: n = 2}

2! < 22

2 < 4

• _{Inductive hypothesis: assume k! < k}k

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

)! 1

Example: Sum of Odd Integers

 _{Proposition: 1 + 3 + … + (2n-1) = n}2 for all integers n≥1.

 _{Proof (by induction):}

1) Basis step:

The statement is true for n=1: 1=12 _.

2) Inductive step:

Assume the statement is true for some k≥1

Example: Sum of Odd Integers

 _{Proof (cont.):}

The statement is true for k:

1+3+…+(2k-1) = k2 ₍₁₎

We need to show it for k+1:

1+3+…+(2(k+1)-1) = (k+1)2 ₍₂₎

Showing (2):

1+3+…+(2(k+1)-1) = 1+3+…+(2k+1) = 1+3+…+(2k-1)+(2k+1) =

k2_{+(2k+1) = (k+1)}2 _.

We proved the basis and inductive steps,

so we conclude that the given statement true. ■

Proving a divisibility property by

mathematical induction

• _{Proposition: For any integer n≥1,}

7n _{- 2}n _{is divisible by 5. (P(n))}

• _{Proof (by induction):}

1) Basis step:

The statement is true for n=1: (P(1))

71 _{– 2}1 _{= 7 - 2 = 5 is divisible by 5.}

2) Inductive step:

Assume the statement is true for some k≥1 (P(k))

(inductive hypothesis) ;

23

Proving a divisibility property by

mathematical induction

_{Proof (cont.): We are given that}

P(k): 7k _{- 2}k _{is divisible by 5. (1)} Then 7k _{- 2}k _{= 5a for some a}Z . _{(by definition) (2)}

We need to show:

P(k+1): 7k+1 _{- 2}k+1 _{is divisible by 5. (3)}

7k+1 _{- 2}k+1 _{= 7}·₇k _{- 2}·₂k _{= 5}·₇k _{+ 2}·₇k _{- 2}·₂k

= 5·7k _{+ 2}·(₇k _{- 2}k_{) = 5}·₇k _{+ 2}·_{5a (by (2))}

= 5·(7k _{+ 2a) which is divisible by 5. (by def.)}

Strong induction

• _{Weak mathematical induction assumes P(k) is}

true, and uses that (and only that!) to show P(k+1) is true

• _{Strong mathematical induction assumes P(1),}

P(2), …, P(k) are all true, and uses that to show that P(k+1) is true.

Strong induction example 1

• _{Show that any number > 1 can be written as}

the product of primes

• _{Base case: P(2)}

– _{2 is the product of 2 (remember that 1 is not}

prime!)

• _{Inductive hypothesis: P(1), P(2), P(3), …, P(k)}

are all true

Strong induction example 1

• _{Inductive step: Show that P(k+1) is true} • _{There are two cases:}

– _{k+1 is prime}

• _{It can then be written as the product of k+1}

– _{k+1 is composite}

• _{It can be written as the product of two composites, a}

and b, where 2 ≤ a ≤ b < k+1