이산수학 (Discrete Mathematics)
3.3 수학적 귀납법
(Mathematical Induction)
2006년 봄학기
문양세
강원대학교 컴퓨터과학과
Introduction
3.3 Mathematical Induction
A powerful technique for proving that a predicate P(n) is
true for every natural number n, no matter how large.
(모든 자연수 n에 대해서 P(n)이 true임을 보일 수 있는 매우 유용한 방법임)
Essentially a “domino effect” principle.
(앞에 것이 성립하면(넘어지면), 바로 다음 것도 성립한다(넘어진다.)
Based on a predicate-logic inference rule:
P(0)
n0 (P(n)P(n+1))
n0 P(n)
P(0)가 true이고, P(n)이 true라 가정했을 때
P(n+1)이 true이면, 모든 n에 대해 P(n)이 true이다.
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Discrete Mathematics
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Validity of Induction
3.3 Mathematical Induction
Intuitively, we can prove that induction is correct.
P(1) = T since (P(0) = T)  (P(0)P(1) = T)
P(2) = T since (P(1) = T)  (P(1)P(2) = T)
P(3) = T since (P(2) = T)  (P(2)P(3) = T)
…
P(n) = T since (P(n-1) = T)  (P(n-1)P(n) = T)
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Discrete Mathematics
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Outline of an Inductive Proof
3.3 Mathematical Induction
Want to prove n P(n)…
• E.g. use a direct proof:
• Induction basis (or base case): Prove P(0).
(기본 단계)
• Induction step: Prove n P(n)P(n+1).
(귀납적 단계)
- E.g. use a direct proof:
- Let nN, assume P(n). (induction hypothesis)
- Under this assumption, prove P(n+1).
• Inductive inference rule then gives n P(n).
Can also be used to prove nc P(n) for a given constant
cZ, where maybe c0. (Base로 0이 아닌 상수 c를 사용할 수 있다.)
• In this circumstance, the base case is to prove P(c) rather than P(0),
and the inductive step is to prove nc (P(n)P(n+1)).
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Discrete Mathematics
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Induction Examples (1/4)
3.3 Mathematical Induction
Prove that the sum of the first n odd positive integers is n2.
That is, prove: (처음 n개 홀수의 합은 n2와 동일하다.)
n
n  1 :  ( 2i  1)  n 2
i 1
P(n)
Proof by Induction
• Induction basis: Let n=1. Since 1 = 12, P(1) is true.
• Induction step: Let n1, assume P(n), and prove P(n+1).
 n

( 2i  1)    ( 2i  1)   ( 2( n  1)  1)

i 1
 i 1

By induction
 n 2  2n  1
hypothesis P(n)
2
 ( n  1)
n 1
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Discrete Mathematics
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Induction Examples (2/4)
3.3 Mathematical Induction
Prove that n>0, n<2n. Let P(n)=(n<2n) :
• Induction basis: P(1) = (1 < 21) = (1 < 2) = T.
• Induction step: For n>0, prove P(n)P(n+1).
- Assuming n<2n, prove n+1 < 2n+1.
- Note n + 1 < 2n + 1 (by inductive hypothesis)
< 2n + 2n (because 1 < 2=2∙20  2∙2n-1= 2n)
= 2n+1
- So n + 1 < 2n+1, and we’re done.
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Discrete Mathematics
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Induction Examples (3/4)
3.3 Mathematical Induction
Prove that the sum of the first n positive integers is
n(n+1)/2.
n( n  1) 
 n
i


Let P(n) =  
2
 i 1

1(1  1)  
2
 1
  1    T
• Induction basis: Let P(1) =   i 
2  
2
 i 1
• Induction step: Let n1, assume P(n), and prove P(n+1).
n( n  1)
n( n  1)  2n  2
i   i  ( n  1) 
 ( n  1) 

2
2
i 1
i 1
n 1
n
( n 2  3n  2) ( n  1)( n  2)


2
2
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Discrete Mathematics
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Induction Examples (4/4)
3.3 Mathematical Induction
Aj   Aj when n2. A1  A2  ...  An  A1  A2  ...  An 
Prove 
j 1
n
n
j 1

• Induction basis: n = 2, A1  A2  A1  A2

• Induction step: Assume P(n), and prove P(n+1).
n 1
n
j 1
j 1
 Aj   Aj  An 1
n
  Ak  An 1
j 1
n
  A j  An 1 (by induction hypothesis)
j 1
n 1
  Aj
j 1
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Discrete Mathematics
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Second Principle of Induction (강 귀납법)
3.3 Mathematical Induction
Characterized by another inference rule:
P(0)
n0: (0kn P(k))  P(n+1)
n0: P(n)
P(0)가 true이고, P(0),…,P(n)이 모두 true라 가정했을 때
P(n+1)이 true이면, 모든 n에 대해 P(n)이 true이다.
Difference with 1st principle is that the inductive step
uses the fact that P(k) is true for all smaller k<n+1, not
just for k=n.
(Induction step에서 k=n인 경우 대신에, k<n+1인 모든 k에 대해 P(k)가 true라 가정한다.)
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Discrete Mathematics
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Examples of Second Principle (1/2)
3.3 Mathematical Induction
Show that every n>1 can be written as a product p1p2…ps
of some series of s prime numbers. Let P(n)=“n has that
property”
(모든 양의 정수는 소수의 곱으로 나타낼 수 있다.)
• Induction basis: n=2, let s=1, p1=2. Thus, P(2) = T.
• Induction step: Let n2. Assume
2kn: P(k).
- Consider n+1. If prime, let s=1, p1=n+1. Done.
- Else n+1=ab, where 1<an and 1<bn.
- Then a=p1p2…pt and b=q1q2…qu.
- Then n+1= p1p2…pt q1q2…qu, a product of s=t+u primes.
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Discrete Mathematics
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Examples of Second Principle (2/2)
3.3 Mathematical Induction
Prove that every amount of postage of 12 cents or more
can be formed using just 4-cent and 5-cent stamps.”
(12센트 이상의 우표는 4센트 혹은 5센트 우표만을 사용하여 구성할 수 있다.)
• Induction basis: 12=3(4), 13=2(4)+1(5), 14=1(4)+2(5), 15=3(5),
so 12n15, P(n).
• Induction step: Let n15, assume 12kn P(k).
- Stamps for (n+1) cents = stamps for (n3) cents + 4-cent stamp
By Induction Hypothesis
- Stamps for (n+1) cents = some 4-cent or 5-cent stamps + 4-cent
stamp
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Discrete Mathematics
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