Number Theory and Counting Method. Divisors -Least common divisor -Greatest common multiple

Transcription

1 Number Theory and Counting Method Divisors -Least common divisor -Greatest common multiple

2 Divisors Definition n and d are integers d 0 d divides n if there exists q satisfying n = dq q the quotient, d the divisor, n the number If d divides n : d n d does not divides n : d n Ex. 21 = , 3 is a divisor (factor) of 21 There exist an integer q such that n = dq, d, n I +, 1 q Therefore, d dq = n.

3 Theorem Let m, n and d be integers (a) if d m and d n then (b) if d m and d n then d (m + n) d (m n) (c) if d m then d mn proof (a) Suppose d m and d n by definition m = dq 1 and n = dq 2 Thus (m + n) = d(q 1 + q 2 ) Therefor There exist q 1 + q 2 is the quotient of it, (a) is proof

4 Prime and Composite number An integer greater than 1 whose only positive divisors are itself and 1 is called prime. An integer greater than 1 that is not prime is called composite Ex. 23 is prime 34 is composite because it is divided by 17 If n > 1 is composite, then d, 1 < d < n To test use 2,3,, n 1 If no integer in this list divides n, then n is prime

5 Composite number theorem (cont.) A positive integer n greater than 1 is composite if and only if n has a divisor d satisfying 2 d n Proof - If n is composite, then n has a divisor d satisfying 2 d n - If n has a divisor d satisfying 2 d n then n is composite

6 Ex1 Determine whether 43 is prime Try 2,3,, 21 = 43 None of these number divides 43 agree with the condition n mod d = 0 43 is prime Ex2. Determine 451 Try 2,3,, 21 = 451 It was founded that 11 is agree with the condition n mod d = 0 451is composite

7 Fundamental Theorem of Arithmetic Any integer greater than 1 can be written as a product of primes. Moreover, if the primes are written in non decreasing order the factorization is unique. The number of primes is infinite Ex. Produce a prime larger than 11 2,3,5,7,11are primes Let m = = is prime

8 The greatest common divisor (GCD) Definition Let m and n be integers with not both m and n zero. The gdc is an integer that divides both m and n. Denote Ex. The positive divisor of 30 are The positive divisor of 105 are The common positive Thus gdc(m, n) 1, 2, 3, 5, 6, 10, 15, 30 1, 3, 5, 7, 15, 21, 35, 105 1, 3, 5, 15 gdc 30,105 = 15 It can be also find by looking carefully at their prime factorization 30 = 2 3 5, 105 = and 5 are the prime factorization of 30 and 105 thus gdc(30,105) = 3 5 = 15

9 Ex. The positive divisors of 30 are 1, 2, 3, 5, 6, 10, 15, 30 The positive divisors of 105 are 1, 3, 5, 7, 15, 21, 35, 105 The common divisors of 30 and 105 are 1,3,5,15 And gcd 30,105 = 15 Ex2. Find the gcd by looking at their prime factorization 30 = = The common prime factorization for 30 and 105 are 3 and 5 Thus gcd(30,105) = 3 5 = 15

10 GDC with prime factorization Theorem Let m and n be integers, m > 1, n > 1 with prime factorization m = p 1 a 1 p 2 a 2 p n a n and n = p 1 b 1 p 2 b 2 p n b n If the prime p i is not a factor of m, we let a i = 0. Similarly, if the prime p i is not a factor of n, we let b i = 0 Then gcd m, n = p 1 min a 1,b 1 p 2 min a 2,b 2 p n min a n,b n Ex. gcd(82320, ) = = gcd 82320, = 2 min 4,2 3 min 1,1 5 min 1,0 7 min 3,4 11 min 0,1 = = 4116

11 The least common multiple Definition Let m and n be positive integer. A common multiple of m and n is an integer that is divisible by both m and n, called the least common multiple. lcm(m, n) Ex. lcm(30,105) = can be divided by both 30 and 105 and it is the least value 30 = 2.3.5, 105 = The lcm must contain 2, 3, 5, 7 thus lcm 30,105 = = 210 Also find lcm() by looking at their prime factorization 30 = 2 3 5, 105 = lcm 30,105 = = 210

12 LCM wit prime factorization Let m and n be integers m > 1, n > 1 with prime factorizations m = p 1 a 1 p 2 a 2 p n a n and n = p 1 b 1 p 2 b 2 p n b n If the prime p i is not a factor of m, we let a i = 0, Similarly, if the prime p i is not a factor of n, we let b i = 0, then lcm m, n = p 1 max a 1,b 1 p 2 max a 2,b 2 p n max a n,b n Ex. lcm(82320, ) = = lcm 82320, = 2 max 4,2 3 max 1,2 5 max 1,0 7 max 3,4 11 = = Note The product of gcd and lcm is equal to the product of the pair of numbers gcd 30,105 lcm 30,105 = = 3150 = max 0,1

13 Representations of Integers and integer algorithm For base 2 bit : a binary digit, value either 0 or 1 For Decimal number 3854 =

14 Representation of integer let n= integer that need to represent by a serie of binary number n = 1 2 k + b k 1 2 k b then and 2 k n n = 1 2 k + b k 1 2 k b k k = 2 k+1 1 < 2 k+1

15 Counting Methods

16 Basic principle Multiplication principle If an activity can be constructed in 1 successive steps and step 1 can be done in n 1 ways, step 2 can then be done in n 2 ways,, and step t can then be done in n t ways, then the number of different possible activities is n 1 n 2 n t Addition principle Suppose that X 1,, X t are sets and that the ith set X i has n i elements. If {X 1,, X t } is a pairwise disjoint family (i.e., if i j, X i X j = ), the number of possible element that can be selected from X 1 or X 2 or or X t is n 1 + n n t

17 How many menu sets can be select from each categories: main course, appetizers, beverages Solution : 24 possible outcome NHT,NHM,NHC,NHR,NCT,NCM,NCC,NCR, NFT,NFM,NFC,NFR,SHT,SHM,SHC,SHR, SCT,SCM,SCC,SCR,SFT,SFM,SFC,SFR

18 Ex1 (a) How many strings of length 4 can be formed using the letters ABCDE if repetitions are not allowed? = 120 (b) How many strings of part (a) begin with the letter B? = 24 (c) How many strings of part (a) do not begin with the letter B? = 96

19 Ex2 In a digital picture, we wish to encode the amount of light at each point as an eight-bit string. How many values are possible at one point? = 2 8 = 256 How many eight-bit strings begin either 101 or 111? Bit 8,7,6 are tied together to form a bit and has 2 option 101 or 111 The 4 successive bit has the possible outcome 2 4 Thus = 64

20 Permutations and Combinations A permutation of n distinct elements x 1,, x n is an ordering of the n elements x 1 x n Theorem : There are n! permutations of n elements n n = n! Ex. How many permutations of the letters ABCDEF contain substring DEF? 4! Ex. How many ways can 6 person be seated around a circular table? (6 1)! = 5! = 120

21 r-permutation An r-permutation of n (distinct) elements x 1, x 2,, x n is an ordering of an r-element subset of x 1, x 2,, x n. The number of r- permutations of a set of n distinct elements is denoted P n, r P n, r = n n 1 n 2 n r + 1, r n n! = n r! Ex. X = {a, b, c}, How many 2-permutations of X? P 3,2 = 3 2 = 6 ab, ac, ba, bc, ca, cb Ex. How many way can we select a chairperson, vice-chairperson, secretary, and treasurer from a group of 10 person? P 10,4 = = 5040

22 r-combination Given a set X = {x 1, x 2,, x n } containing n distinct elements. (a) an r-combination of X is an unordered selection of r-elements of X(i.e., and r-element subset of X) (b) The number of r-combination of a set of n distinct elements is n denotes C(n, r) or r P n, r n! C n, r = = r! n r! r! Ex. A group of five students, Mary, Boris, Rosa, Ahmad, and Nguyen, has decided to talk with the Mathematics Department chairperson about having the Mathematics Department offer more courses in discrete mathematic. The chairperson has said that she will speak with three of the students. How many way can these five students choose three of their group to talk with chairperson? C 5,3 = 10

23 Ex. How many ways can we select a committee of three from a group of 10 distinct person? C 10,3 = 120 Ex. An ordinary deck of 52 cards consist of four suits: clubs, diamonds, hearts, spades, of 13 denominations each ace, 2-10, jack, queen, king. (a)how many (unordered) five card poker hands, selected from an ordinary 52-card deck, are there? C 52,5 = 2,598,960 (b) How many poker hands contain cards all of the same suit? 4 C 13,5 = 5148 (c)how many poker hands contain three cards of one denomination and two cards of a second denomination? C 4,3 C 4,2 = 3744

24 How many routes are there from the lower-left corner of an n n square grid to the upper-right corner if we are restricted to travelling only to the right or upward? C(2n, n)

25 Introduction to discrete probability The probability P(E) of an event E for the finite sample space S is P E = E S Ex. Two fair dice are rolled. What is the probability that the sum of the numbers on the dice is 10? Possible outcome for the sum is 10 : (4,6),(5,5),(6,4) Outcome of the sample space S = 6 6 Thus P E = 3 36 Ex. 5 microprocessors are randomly selected from a lot of 1000 microprocessors among which 20 are defective. Find the probability of obtaining no defective microprocessors. S = C 1000,5, E = C 980,5 P E =

26 Discrete Probability Theory A probability function P assigns to each out come x in a sample space S a number P(x) so that and 0 P x 1, for all x S P(x) = 1 x S Ex. Suppose that a die is loaded so that the numbers 2 through 6 are equally likely to appear but that 1 is three times as likely as any other number to appear. To model this situation. P 2 = P 3 = P 4 = P 5 = P 6, P 1 = 3P 2 And P 2 = 1 8, P 1 = = 3 8

27 Probability of event Let E be an event. The probability of E, P(E) is P E = P x Ex. From the former example, find the probability of an odd number x E P 1 + P 3 + P 5 = = 5 8 Note : For a fair die P odd number = 1 2

28 Probability of complement of event Let E be an event. The probability of തE, the complement of E satisfies P E + P തE = 1

29 Union of event Let E 1 and E 2 be events. Then P E 1 E 2 = P E 1 + P E 2 P(E 1 E 2 ) Corollary: union of mutual event If E 1 and E 2 are mutually exclusive events P E 1 E 2 = P E 1 + P E 2

30 Conditional probability Let E and F be events, and assume that P F > 0. The conditional probability of E given F is Ex. E = die getting a sum of 10 F = at least one die show 5 P E F = P E F P F E F: getting sum of 10 and at least one die show 5 P E F = 1 36 P F = P E F = 1/11

31 Independent conditional event Events E and F are independent if P E F = P E P(F)

32 Bay s Theorem Suppose that the possible classes are C 1,, C n. Suppose further that each pair of classes is mutually exclusive and each item to be classified belongs to one of the classes. For a feature set F, we have P C j F = P F C j P C j n P F C i P C i σ i=1

33 Ex. At the telemarketing firm SellPhone, Dale, Rusty and Lee make calls. The following table shows the percentage of calls each caller makes and the percentage of the persons who are annoyed and hang up on each caller: Let D denote the event Dale made the call, let R denote the event Rusty made the call. let L denote the event Lee made the call and let H denote the event the caller hung up. Find P(D), Caller Dale Rusty Lee Percent of calls Percent of hang-ups P(R), P(L), P(H D), P(H R), P(H L), P(D H), P(R H), P(L H) and P(H) P(D) = 0.4, P(R) = 0.25), P(L) = 0.35 P(H D) = 0.2, P(H R) = 0.55, P(H L) = 0.3 P(D H) use Bay s Theorem P H D P D P D H = P H D P D + P H R P R + P H L P L = Similar for P R H = P L H = 1 P R H P D H = Find P H = P H D P D + P H R P R + P H L P L =

34 Generalized Permutations and Combinations Theorem: Suppose that a sequence S of n items has n 1 identical objects of type 1, n 2 identical objects of type 2,, and n t identical objects type t. Then the number of orderings of S is n! Theorem: n 1! n 2! n t! If X is a set containing t elements, the number of unordered k-elements selections from X, repetitions allowed is C k + t 1, t 1 = C(k + t 1, k)

35 Ex. How many strings can be formed using the following letters? M I S S I S S P P I Choose 2 P: C(11,2) Choose 4 S: C(9,4) Choose 4 I: C(5,4) The outcome: C 11,2 C 9,4 C 5,4 = 34,650

36 Binomial Coefficients and Combinatorial Identities a + b n = (a + b)(a + b) (a + b) n factors Ex. a + b 3 = a + b a + b a + b = aaa + aab + aba + abb + baa + bab + bba + bbb = a 3 + a 2 b + a 2 b + ab 2 + a 2 b + ab 2 + ab 2 + b 3 = a 3 + 3a 2 b + 3ab 2 + b 3 a + b n = n 0 an b 0 + n 1 an 1 b n n 1 a1 b n 1 + n n a0 b n

37 Binomial theorem Theorem : Ex. n = 3 n a + b n = k=0 n k an k b k a + b 3 = 3 0 a3 b a2 b a1 b a0 b 3 Ex. 3x 2y 4 = a 3 + 3a 2 b + 3ab 2 + b 3 = 4 0 3x 4 2y x 3 2y x 2 2y x 1 2y x 0 2y 4 = 3 4 x x 3 2y x y 2 4 3x2 3 y y 4 = 81x 4 216x 3 y + 216x 2 y 2 96xy y 4

38 Ex. Find the coefficient of x 2 y 3 z 4 in the expansion of x + y + z 9 We choose x 2 from 9 term, y 3 from 7 terms and z from 4 terms Thus = 9! 7! 2! 7! 3! 4! = 9! 2! 3! 4! = 1260

39 Theorem n + 1 k = n k 1 + n k, for 1 k n Proof X = n, choose a X. Then C(n + 1, k) is the number of k-element of Y = X {a}. The k-element of Y can be divided into 2 classes 1. Subset of Y not containing a 2. Subset of Y containing a For class 1 are just k-element subset of X = n k For n class 2 consist of (k-1) element subset of X together with a, there are k 1. Therefore n + 1 k = n k 1 + n k

40 Ex. Show that n C(i, k) = C n + 1, k + 1 i=k Use Theorem in the form C i, k = C i + 1, k + 1 C i, k + 1 To obtain C k, k + C k + 1, k + C k + 2, k + + C n, k = 1 + C k + 2, k + 1 C k + 1, k C k + 3, k + 1 C k + 2, k C n + 1, k + 1 C n, k + 1 = C(n + 1, k + 1)

41 Ex. Find the sum n = C 1,1 + C 2,1 + + C(n, 1) = C n + 1,2 n + 1 n = 2

42 The Pigeonhole principle soon