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Mathematical Induction.

• Induction is the most important proof method in computer science.

• Suppose you want to prove that the proposition P (n) is true for all n N, where N = {0, 1, 2, …} (an infinite set). ( i. e. every natural number n has some property P)

• You might be able to prove it for 0, 1, 2, … But you can’t checkone-by-one that all natural numbers have property P.

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• The key idea of mathematical induction is start with 0 and repeatedly add 1.

• Suppose you can show that i) 0 has property P and ii) whenever you add 1 to a number that has property P the resulting number also has property P.

This guarantees that as you go through the list of all natural numbers, every number you encounter must have property P.

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• The proof method of mathematical induction says that you just need to do two things: i) prove it for n = 0 (basis case) ii) prove “if it’s true for n=k, then it’s true for n =k+1”

Induction Hypothesis:• fix some k 0 • assume P (k)

Induction Step• Using assumption P (k) prove that P (k+1)

In this way we prove that for any n [P(n) P(n+1)]

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Why does this prove it for all n N ?

Actually it relies upon the property of integers N={0, 1, 2, …}

The pattern is:

To prove n 0 (P (n)) it’s enough to show:1) (Base) P (0) 2) n 0 (P(n) P (n +1))

So, in the second step we say:“take some k 0 and assume that property P holds for n = k“(Induction Hypothesis).Then, (based upon this assumption) we need to show that the property P can be implied for n = k +1.

How can we prove n 0 (P(n) P (n +1)) ?

Pick arbitrary k 0 and prove it for n = k.

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Suppose we want to prove the formula for the sum of ‘geometric progression’

q

qqqqq

nn

1

1...1

132

where 10 q , and n is any integer, n0

How can we prove that the formula is correct for any n0 ?

• Show it for arbitrary fixed n0• Use induction on n0

Notation:

n

i

inn qqqqS

0

2 ...1

q

qS

n

n

1

1 1

6

• Show it for some n01212 1)...1(... nnn

n qqqqqqqSq

nS

11 nnn qSSq

11)1( nn qqS

q

qS

n

n

1

1 1

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• Proof by induction on n0

1) Base: n=0

;100

00

qqSi

i 11

1

1

1 10

q

q

q

q

2) Induction Hypothesis: Assume the formula is correct for n=k, where k is some integer, k0

q

qS

k

k

1

1 1

Induction Step: prove the formula for n=k+1, i. e. q

qS

k

k

1

1 1)1(

1

1121 ...1

kk

kkk qSqqqqS

q

q

q

qqq

qq

q

k

kkk

kk

1

1

1

1

1

1

1)1(

1)1(11

11

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There is a variant if you want to prove it for n n0 :

To prove n n0 (P (n)) it’s enough to show:1) (Base) P ( n0) 2) n n0 (P(n) P (n +1))

1) In the base case we explicitly check n = n0

2) Pick arbitrary k n0 and assume that the property P holds for n = k . (Induction Hypothesis) Based upon this assumption prove that the property P can be implied for n = k +1. (Induction Step)

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Example. Prove that for all n1 n! 2n1.

Proof by induction on n1. 1). Basis: n =1. 1!=1=211=20. So, for the case n=1, we checked that n!2n1.2). Assume that for n=k, k is some integer k1, inequality holds, i. e. k! 2k1 (IH).We want to prove that n! 2n1 for n = k +1, i. e. we want to show that

(k+1)! 2(k+1)1 Using the definition of factorial we have (k+1)!= (k+1)k! (k+1)2k1 ……by IH

(1+1)2k1 ……because k1 = 2 2k1 = 2k……QED, because 2(k+1)1= 2k

By Induction Principle we conclude that n! 2n1 for all n1

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Induction proof for the summation formula

Let’s prove the formula for the sum of integers from 1 to n:

n 0

2

)1(...3210

nnn

P (n)

Alternative notation for the LHS: ni

n

i

ii00

or

Thus we want to prove: n 0 2

)1(

0

nni

n

i

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Let’s stop for a moment and check that the formula we are to prove makes sense:

n 0 2

)1(

0

nni

n

i

Consider a few particular cases:• n =1

• n = 0

2

)11(110...210

0

nin

i

2

)10(00...210

0

nin

i

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In accordance with induction pattern: 1) (Base) P (0) 2) n 0 (P(n) P (n +1))

P (n) here is the predicate:

n

i

nni

0 2

)1(

Prove: n 0

• (Base)

• (Inductive step) n 0

n

i

nni

0 2

)1(

0

0 2

)10(0

i

i

n

i

n

i

nni

nni

0

1

0 2

)2)(1(

2

)1(

P (n) P (n+1)

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1) Basis: check that the formula is correct for the case of no terms

2) fix some k 0 and assume that proposition holds for n = k

k

i

kki

0 2

)1( P(n=k) (induction hypothesis, IH)

1

0 2

)2)(1(k

i

kkiShow that the P(n=k+1): can be implied (IS)

2

)2)(1(1

2)1(

)1(2

)1(

)1(

)1(...210

0

1

0

kkkk

kkk

ki

kki

k

i

k

i

……..by IH

….algebra

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Why induction proofs work? Intuitively, because we can reach any number n if we start withthe smallest one and add 1 step-by-step. So, induction principle express this obvious fact.

Induction Principle . If A is a subset of N that satisfies twoproperties:

1) 0A 2) n 0 [n A(n+1)A],

then A =N

More formally, induction principle is formulated in terms of a set A = {n N | P (n)}

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Although it sounds very obvious, the induction principle hinges on the property of integers, so called Well-Ordering principle (accept it as an axiom).

Well-Ordering Principle. Any nonempty subset of N contains a smallest element.

Neither rational, no real numbers has this property!

Induction principle is equivalent to the Well-ordering principle.

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Theorem. Well-ordering principle(WOP) implies Induction Principle(IP)

Proof. Suppose WOP is true, i.e. any nonempty subset of N contains a smallest element. We want to prove that IP is true, i. e. if any set A satisfies properties

i) 0Aii) n 0 [nA(n+1)A],

then A =N.We are going to prove it by contradiction. 1) Assume that A satisfies both properties, but A N . 2) Then N A = B , i. e. B is nonempty subset of N .3) By WOP B contains the smallest element. Let denote it s, s B, so sA. 4) Since s is the smallest element of B, s1B. 5) Since 0A by i), 0B, so s >0 and s1 0, i. e. s1N.6) s1N and s1B imply s1A .7) By ii) s1A sA , contradiction with sA.

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n

012

.

.

.

i) 0A

A = {n N | P(n)}

ii) n 0 [nA(n+1)A]

B = N A

Proof by contradiction:Assume that A N.

s 1 A, since s is the smallest

B

s A by ii)

sB, since B = N A

sB (by WOP)

in contradiction with sB

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Example. Prove that the number of different binary strings of length n is 2n.

Proof by induction on n1.We want to prove: n1 [P (n)], where P (n): the number of binary strings of length n is 2n.

1) Basis. n=1. We have two strings of the length one: 0 and 1.

2) Assume P (n) for n=k, where k is some integer, k 1 (IH).

So, we assume that for some k 1 there are 2k binary strings.

In the IS we need to prove P (n) for n=k+1

I. e. we need to show that there are 2(k+1) binary stringsof the length (k +1).

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Here is the proof that there are 2(k+1) binary strings of length (k+1).

Any binary string of length (k+1) can be represented as a binarystring of length k with one extra bit (let it be the last one). There are two choices for the last bit: 0 or 1.

By IH there are 2k strings of length k. Then, we have 2k strings of length (k+1) that end with 0 and 2k strings of length (k+1) that end with 1, so the total number is 2k + 2k =2 2k =2(k+1)

QED.

By IP we conclude, that for any n 1 the number of binary strings is 2n.

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Let’s use the induction to prove the following known fact.

Example. Prove that for any set A cardinality of a power set is |Power(A)|=2|A|.To apply induction we need to chose the integer variable.Here we can prove by induction on n = |A|. So, we want to prove that the propositionP= |Power(A)|=2|A| is true for any n=|A| 0.

1) Basis: n=|A|=0, A= , There is only one subset of the empty set, so |Power()|=1=2

2) IH: Assume that for n=k, k is any integer k 0, we have that any set A with |A|=k has 2k subsets. IS: we need to imply that any set B with cardinality |B|=k+1 has the property P, i. e. has 2(k+1) subsets.

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Here we need to do the job…Take any set B of k+1 elements. Pick any element x. Then all subsets of B can be divided into two parts:

i) subsets that do not contain x. How many such subsetsexist? We can argue, that subsets of B that do not contain x constitute subsets of the set B {x}, that has cardinality k. |B{x}|=k and by IH there exist 2k subsets.

ii) subsets that include x. How many such subsets exist?Again, the number of such subsets equals the number of Subsets of the set B {x}, this number is 2k. Now, the total number of subsets is the sum of two groups, 2k + 2k =2 2k= 2k+1. So, from assumption that the proposition is true for n=k ( k is any integer k 0) we can imply that the proposition is true for n =k+1.By Induction principle we conclude that the proposition is true for all integer n 0, i. e. for sets of any size |A|=n.