台湾交大离散数学8

The Inclusion-Exclusion Principle
Theorem If A1 , A2 ,
n [
, An are …nite sets and n
1, we have
Ai
=
i =1
i =1
∑ jAi j
.
1 i <j n
∑
jAi \ Aj j +
1 i <j <k n
∑
jAi \ Aj \ Ak j
Discrete Mathematics Chapter 4 Induction and Recursion §4.2 Strong Induction and Well-Ordering
Example of Second Principle
Example Show that every n > 1 can be written as a product p1 p2 . . . ps of some series of s prime numbers. Solution Let P (n) =“n has that property”.
3
Discrete Mathematics Chapter 4 Induction and Recursion §4.2 Strong Induction and Well-Ordering
Another 2nd Principle Example
Example Prove that every amount of postage of 12 cents or more can be formed using just 4-cent and 5-cent stamps. Solution Let P (n) =“n can be formed using 4-cent and 5-cent stamps."
Harmonic Numbers
Theorem The harmnoic numbers Hj for j = 1, 2, 3, Hj = 1 + Then, H2n Proof. BASIS STEP INDUCTIVE STEP CONCLUSION 1+ n 2. 1 1 + + 2 3 are 1 + . j
Proof. BASIS STEP INDUCTIVE STEP CONCLUSION
Discrete Mathematics Chapter 4 Induction and Recursion §4.2 Strong Induction and Well-Ordering
§4.2 Strong Induction and Well-Ordering
Discrete Mathematics
Discrete Mathematics (2009 Spring) Induction and Recursion (Chapter 4, 3 hours)
Chih-Wei Yi
Dept. of Computer Science National Chiao Tung University
Generalizing Induction
Can also be used to prove 8n c 2 Z, where maybe c 6= 0.
c P (n) for a given constant
In this circumstance, the base case is to prove P (c ) rather than P (0), and the inductive step is to prove 8n c (P (n) ! P (n + 1)).
n +1 i =1
∑ (2i
1) =
= n + 2n + 1 = (n + 1)2
So, according to the inductive inference rule, the property is proved.
i =1 2
∑ (2i
n
1)
+ (2 (n + 1)
1)
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
Let n 2 N, assume P (n ). (inductive hypothesis) Under this assumption, prove P (n + 1).
3
Inductive inference rule then gives 8n P (n).
Example Prove that the sum of the …rst n odd positive integers is n2 . That is, prove:
8n
Leabharlann Baidu
1:
i =1
∑ (2i
n
1) = n 2
P (n )
|
{z
}
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
Proof. Base Case: Let n = 1. The sum of the …rst 1 odd positive integer is 1 which equals 12 . Inductive Step: Prove 8n 1 : P (n) ! P (n + 1). Let n 1, assume P (n), and prove P (n + 1). !
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
Outline of an Inductive Proof
Want to prove 8n P (n). . .
1 2
Base case (or basis step): Prove P (0). Inductive step: Prove 8n 0 P (n) ! P (n + 1). E.g. use a direct proof:
8n
z }| { 0 : (80 k n P (k )) ! P (n + 1) ) 8n 0 : P (n )
P is true in all previous cases
Di¤erence with the 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.
n \
2, then
Aj =
j =1
j =1
n [
Aj and
j =1
n [
Aj =
j =1
n \
Aj .
Proof. BASIS STEP INDUCTIVE STEP CONCLUSION
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
< 2n + 2n (because 1 < 2 = 2 20 = 2n +1
3
2 2n
1
= 2n )
So n + 1 < 2n +1 , and we’ re done.
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
April 17, 2009
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
§4.1 Mathematical Induction
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
1
2
Base case: 12 = 3 (4), 13 = 2 (4) + 1 (5), 14 = 1 (4) + 2 (5), 15 = 3 (5), so 812 n 15, P (n).
Inductive step: Let n 15. Assume 812 k n P (k ). Note n + 1 = (n 3) + 4 and 12 n 3 n. Since P (n 3), add a 4-cent stamp to get postage for n + 1. So, P (n) is true for all n 12.
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
De Morgan’ s Law
Theorem If A1 , A2 , we have , An are subsets of a universal set U and n
Discrete Mathematics Chapter 4 Induction and Recursion §4.2 Strong Induction and Well-Ordering
§4.2: Strong Induction Second Principle of Induction
Characterized by another inference rule: P (0)
§4.1 Mathematical Induction
A powerful, rigorous technique for proving that a predicate P (n) is true for every natural number n, no matter how large. Essentially a “domino e¤ect” principle. Based on a predicate-logic inference rule: BasisStep : P (0) InductiveStep : 8n 0, (P (n) ! P (n + 1)) Conclusion : ) 8n 0, P (n) The First Principle of Mathematical Induction.
Induction can also be used to prove 8n arbitrary series fan g.
c P (an ) for an
Can reduce these to the form already shown.
Discrete Mathematics Chapter 4 Induction and Recursion §4.1 Mathematical Induction
1 2
Base case: For n = 2, let s = 1, p1 = 2. Inductive step: Let n 2. Assume 82 k n: P (k ). Consider n + 1. If prime, let s = 1, p1 = n + 1. Else n + 1 = ab, where 1 < a n and 1 < b n. Then a = p1 p2 . . . pt and b = q1 q2 . . . qu . Then n + 1 = p1 p2 . . . pt q1 q2 . . . qu , a product of s = t + u primes. So, P (n) is true for all n > 1.
Example Prove that 8n > 0, n < 2n . Solution Let P (n) = (n < 2n ).
1 2
Base case: P (1) = 1 < 21 = (1 < 2) = T. Inductive 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)