241-303 Discrete Maths: Induction/1 1 Discrete Maths Objective – –to introduce mathematical induction through examples 241-303, Semester 1 2014-2015 1

241-303 Discrete Maths: Induction/1 1

Discrete Maths

• Objective– to introduce mathematical induction through

examples

241-303, Semester 1 2014-2015

1. Mathematical Induction


Overview

1. Motivation

2. Induction Defined

3. Maths Notation Reminder

4. Four Examples

5. More General Induction Proofs

6. A Fun Tiling Problem

7. Further Information


1. Motivation

• Induction is used in mathematical proofs of many recursive algorithms– e.g. quicksort, binary search

• Induction is used to mathematically define recursive data structures– e.g. lists, trees, graphs

continued


• Induction is often used to derive mathematical estimates of program running time– timings based on the size of input data

• e.g. time increases linearly with the number of data items processed

– timings based on the number of times a loop executes


2. Induction Defined

• Induction is used to solve problems such as:– is S(n) correct/true for all n values?

• usually for all n >= 0 or all n >=1

• Example:– let S(n) be "n2 + 1 > 0"– is S(n) true for all n >= 1?

continued

S(n) can be muchmore complicated,such as a programthat reads in a nvalue.


• How do we prove (disprove) S(n)?

• One approach is to try every value of n:– is S(1) true?– is S(2) true?– ...– is S(10,000) true?– ... forever!!! Not very practical


Induction to the Rescue

• Induction is a technique for quickly proving S(n) true or false for all n– we only have to do two things

• First show that S(1) is true– do that by calculation as before

continued


• Second, assume that S(n) is true, and use it to show that S(n+1) is true– mathematically, we show that

S(n) --> S(n+1)

• Now we know S(n) is true for all n>=1.• Why?

continued

"--> stands for "implies"


• With S(1) and S(n) --> S(n+1)then S(2) is true – S(1) --> S(2) when n == 1



• and so on, for all n


Let’s do it

• Prove S(n): "n2 + 1 > 0" for all n >= 1.

• First task: show S(1) is true– S(1) == 12 + 1 == 2, which is > 0– so S(1) is true

• Second task: show S(n+1) is true by assuming S(n) is true

continued


• Assume S(n) is true, so n2+1 > 0• Prove S(n+1)

– S(n+1) == (n+1)2 + 1== n2 + 2n + 1 +1== (n2 + 1) + 2n + 1

– since n2 + 1 > 0, then (n2 + 1) + 2n + 1 > 0– so S(n+1) is true, by assuming S(n) is true– so S(n) --> S(n+1)

continued


• We have used induction to show two things:– S(1) is true– S(n) --> S(n+1) is true

• From these it follows that S(n) is true for all n >= 1


Induction More Formally

• Three pieces:– 1. A statement S(n) to be proved

• the statement must be about an integer n

– 2. A basis for the proof. This is the statement S(b) for some integer. Often b = 0 or b = 1.

continued


• 3. An inductive step for the proof. We prove the statement “S(n) --> S(n+1)” for all n.

– The statement S(n), used in this proof, is called the inductive hypothesis

– We conclude that S(n) is true for all n >= b• S(n) might not be true for some n < b


3. Maths Notation Reminder

• Summation: means 1+2+3+4+…+n

– e.g. means 4+9+16+…+m2

• Product: means 1*2*3*…*n

n

i

i1

m

j

j2

2

n

i

i1


4. Example 1

• Prove the statement S(n):for all n >= 1– e.g. 1+2+3+4 = (4*5)/2 = 10

• Basis. S(1), n = 1so 1 = (1*2)/2

2

)1(

1

nni

n

i

1

1

1i

i

continued

(1)


• Induction. Assume S(n) to be true.Prove S(n+1), which is:

Notice that:

2

)2)(1(

2

)11(11

1

nnnni

n

i(2)

1

1 1

)1(n

i

n

i

nii (3)

continued


• Substitute the right hand side (rhs) of (1) for the first operand of (3), to give:

= (n2 + n + 2n + 2) /2= (n2+3n+2)/2

)1(2

)1(1

1

nnn

in

i

which is (2)2

)2)(1(1

1

nni

n

i


Example 2

• Prove the statement S(n):for all n >= 0– e.g. 1+2+4+8 = 16-1

• Basis. S(0), n = 0so 20 = 21 -1

122 1

0

n

n

i

i

0

0

022i

i

continued

(1)



Notice that:

122 21

0

nn

i

i(2)

1

0 0

1222n

i

n

i

nii (3)

continued



= 2*(2n+1) - 1

which is (2)

111

0

2122

nnn

i

i

122 2

0

n

n

i

i+1


Example 3

• Prove the statement S(n): n! >= 2n-1

for all n >= 1– e.g. 5! >= 24, which is 120 >= 16

• Basis. S(1), n = 1: 1! >= 20

so 1 >= 1

continued

(1)



(n+1)! >= 2(n+1)-1

>= 2n (2)

Notice that:(n+1)! = n! * (n+1) (3)

continued



(n+1)! >= 2n-1 * (n+1)

>= 2n-1 * 2since (n+1) >= 2

(n+1)! >= 2n which is (2)

why?


Example 4• Prove the statement S(n):

for all n >= 1– This proof can be used to

show that. the limit of the sum:is 1

• Basis. S(1), n = 1so 1/2 = 1/2

1)1(

1

1

n

n

ii

n

i

1

1 11

1

)1(

1

i ii

continued

(1)

...4*3

1

3*2

1

2*1

1



Notice that:

1)1(

)1(

)1(

11

1

n

n

ii

n

i(2)

)2)(1(

1

)1(

1

)1(

11

1 1

nniiii

n

i

n

i

(3)

continued



=1)1(

)1(

n

nwhich is (2)

)2)(1(

1

1)1(

11

1

nnn

n

ii

n

i


5. More General Inductive Proofs

• There can be more than one basis case.

• We can do a complete induction (or “strong” induction) where the proof of S(n+1) may use any of S(b), S(b+1), …, S(n)– b is the lowest basis value


Example

• We claim that every integer >= 24 can be written as 5a+7b for non-negative integers a and b.– note that some integers < 24 cannot be expresse

d this way (e.g. 16, 23).

• Let S(n) be the statement (for n >= 24)“n = 5a + 7b, for a >= 0 and b >= 0”

continued


• Basis. The 5 basis cases are 24 through 28.– 24 = (5*2) + (7*2)– 25 = (5*5) + (7*0)– 26 = (5*1) + (7*3)– 27 = (5*4) + (7*1)– 28 = (5*0) + (7*4)

continued


• Induction: Let n+1 >= 29. Then n-4 >= 24, the lowest basis case.– Thus S(n-4) is true, and we can write

n - 4 = 5a + 7b

– Thus, n+1 = 5(a+1) + 7b, proving S(n+1)


6. A Fun Tiling Problem

• A right tromino is a “corner” shape:

• Use induction to proof that right trominos can be used to tile (cover) any n*n size board, where n is a power of 2 and the board has 1 square missing.

continued


• For example, a 4*4 board (22*22), minus 1 square:


7. Further Information

• DM: section 1.6

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241-303 Discrete Maths: Induction/1 1 Discrete Maths Objective – –to introduce mathematical induction through examples 241-303, Semester 1 2014-2015 1