1. 1 Discrete and Combinatorial Mathematics R. P. Grimaldi, 5 th edition, 2004 Chapter 4 Properties of the Integers

2. 2 Well-Ordering Principle ( 良序原理 ) Every nonempty subset A of  + = {1,2,…} contains a smallest element. Note: A can be finite or infinite. Some sets are not well-ordered:  : A=  ,  : A=(0,1) It is the basis of a prove technique -- Mathematical Induction.

3. 3 Principal of Mathematical Induction Let S(n) denote an open mathematical statement that involves one or more occurrences of the variable n   +. (a) If S(1) is true; and (b) If S(k) is true for some particular k   +, then S(k+1) is true; then S(n) is true for all n   +.

4. 4 Proof of Theorem 4.1 Proof by contradiction: Define the set F = { m | S(m) does not hold}   +. If F is non-empty, then F must have a smallest element m (well-ordering of  + ), with  S(m). Because we know that S(1), it must hold that m>1. Because m is the smallest value, it must hold that S(m–1), which contradicts our proof for all k   + : S(k)  S(k+1). Thus F has to be empty. Therefore, S holds for all  +. #

5. 5 Principal of Strong Form Mathematical Induction Let S(n) denote an open mathematical statement that involves one or more occurrences of the variable n   +. Let n 0, n 1   + with n 0  n 1. (a) If S(n 0 ), S(n 0 +1), S(n 0 +2), …, S(n 1 -1), and S(n 1 ) are true; and (b) If whenever S(n 0 ), S(n 0 +1), …, S(k-1),and S(k) are true for some particular k   +, where k  n 1, then the statement S(k+1) is also true; Then S(n) is true for all n  n 0.

6. 6 Mathematical Induction If we want to prove a property P(n) for all n   +, we can do that in the following inductive way: Basis step: Prove it for n=1: does P(1) hold? Inductive step: Assuming that P(1), P(2), …, P(k) holds for some k   +, prove P(k+1). By the structure of  +, this proves it for all n: P(1) holds, hence P(2), hence P(3), et cetera.

7. 7 Example of Induction Show that n!  2 n-1, for n   +. Proof. Induction Basis. n = 1  1!  2 1-1 = 1. Inductive step. Assume that i!  2 i-1, for i = 1, 2, …, k. (k+1)! = (k+1)  k!. Then (k+1)! = (k+1)  k!  (k+1)  2 k-1  2  2 k-1 = 2 k. Therefore, the assertion is true. #

8. 8 More Example of Induction For every n   + where n  14, prove S(n) : n can be written as a sum of 3’s and/or 8’s. Proof. Induction Basis. n = 14  n = 3 + 3 + 8. n = 15  n = 3 + 3 + 3 + 3 + 3. n = 16  n = 8 + 8. Inductive step. Assume that S(14), S(15), …, S(k) are true for some k   + where k  16. k+1 = (k-2) + 3.  S(k+1) is true. Thus, S(n) is true for all n  14. #

9. 9 Recursive Definition Fibonacci sequence: 0,1,1,2,3,5,8,13,21,… It is sequence number A000045 in The On-Line Encyclopedia of Integer Sequences at http://www.research.att.com/~njas/sequences/index.html Recursive base. F 0 = 0, F 1 = 1. Recursive process. F n = F n-1 + F n-2, n  2. Explicit formula: For large n, it grows like F n ≈ 0.447214  1.61803 n. This (1+√5)/2 ≈ 1.61803 is the Golden Ratio.

10. 10 Fibonacci in Nature Shape of shells: Golden ratio: Perfect shape : total height / navel height = 1.618

11. 11 Property of Fibonacci Observation. Conjecture.

12. 12 Property of Fibonacci Proof. Induction Basis. Inductive step. Assume that for some k   + where k  1. #

13. 13 Other Sequences Harmonic sequence:  Lucas sequence: 

14. 14 More Sequences Binomial sequence: Eulerian sequence:

15. 15 Number System Binary digits: 0 and 1, called bits. Review of decimal system: Example: 45,238 is equal to 8ones8 x 1 =8 3tens3 x 10 = 30 2 hundreds2 x 100 = 200 5thousands5 x 1000 = 5000 4ten thousands4 x 10000 = 40000

16. 16 From Binary to Decimal The number 1101011 is equivalent to 1 one1 x2 0 =1 1 two1x2 1 = 2 0 four0x2 2 =0 1 eight1x2 3 =8 0 sixteen0x2 4 = 0 1 thirty-two1x2 5 = 32 1 sixty-four1x2 6 = 64 107 in decimal base

17. 17 From Decimal to Binary The number 73 10 is equivalent to 73 = 2 x 36 + remainder 1 36 = 2 x 18 + remainder 0 18 = 2 x 9 + remainder 0 9 = 2 x 4 + remainder 1 4 = 2 x 2 + remainder 0 2 = 2 x 1 + remainder 0  73 10 = 1001001 2 (write the remainders in reverse order preceded by the quotient 1)

18. 18 Adding Binary Numbers Example: add 100101 2 + 110011 2 1 1 1  carry ones 100101 2 +110011 2 1011000 2

19. 19 Hexadecimal Number System Decimal system 0123456789101112131415 0123456789ABCDEF Hexadecimal system

20. 20 Hexadecimal to Decimal The hexadecimal number 3A0B 16 is 11 x 16 0 =11 0 x 16 1 = 0 10 x 16 2 = 2560 3 x 16 3 = 12288 14859 10

21. 21 Two’s Complement One’s complement: Replace each 0(1) in the binary representation by 1(0). 100101  011010 Two’s complement: Add 1 to one’s complement. 011010 + 1 011011

22. 22 Subtraction 33 - 15 33 = 00100001 2, 15 = 00001111 2 -15 = 11110001 2 00100001 2 +11110001 2 (1)00010010 2 = 18

23. 23 More Subtraction 15 - 33 33 = 00100001 2, 15 = 00001111 2 -33 = 11011111 2 00001111 2 +11011111 2 11101110 2  Two’s complement 00010010 2 = 18  15 – 33 = -18.

24. 24 Divisibility of Integers Let x,y  Z and y  0, x is a multiple of y if and only if there exist and integer m  Z such that x = y  m. We also say y divides x or y is a divisor of x. Notation: y|x.

25. 25 Division Properties For all a,b,c,x,y,z  Z: a) 1|a and a|0 b) [(a|b) and (b|a)]  a = ±b c) [(a|b) and (b|c)]  a|c d) a|b  a|bx for all x e) x = y + z, a|x and a|y  a|z, a|y and a|z  a|x f) [(a|b) and (a|c)]  a|(bx+cy) bx+cy is a linear combination of b and c. g) a|c i, c j  Z, for 1 ≤ i ≤ n  a|(c 1 x 1 +c 2 x 2 + … + c n x n ) for all x j  Z.

26. 26 Proof of (f) f) [(a|b) and (a|c)]  a|(bx+cy). Proof. a|b  Let b = ma for some m  Z. a|c  Let c = na for some n  Z.  bx+cy = xma+yna = a(xm+yn)  a| (bx+cy). #

27. 27 Division Algorithm Let a,b  Z and b>0, there exist unique q,r  Z with 0≤r<b, such that a = qb + r. We call a the dividend, b the divisor, q the quotient, and r the remainder.

28. 28 Primes An integer p>1 is prime if and only if its two positive divisors are 1 and p. An integer n>1 that is not prime is composite. The first primes are: 2,3,5,7,11,… Euclid: there are infinitely many primes. Prime factorization is unique for each n  {2,3,4,…}.

29. 29 Property of Composite Integers Lemma 4.1 If n  Z + is composite, then there is a prime p such that p|n. Proof. Let S be the set of all composite integers that have no prime divisors. Assume S is not empty. Let m be the least member of S. m is composite   m 1,m 2 [m = m 1  m 2 ].   p [p|m 1 ], where p is prime.  p| m 1  m 2  p|m.  #

30. 30 Infinitely Many Primes Theorem 4.4 There is an infinite number of primes. Proof. Assume there are finite primes p 1, p 2, …,p k. Consider N = p 1  p 2  …  p k + 1.  N must be a composite. By Lemma 4.1,  p i [p i | N]. However p i  N for 1  i  k.  #

31. 31 Property of Composite Integers If n  Z + is composite, then there is a prime p such that p|n and. Proof. n is composite   n 1,n 2  Z + [n = n 1  n 2 ]. Assume and.  n 1  n 2 > n.  Without loss of generality, we assume By Lemma 4.1,  a prime p [p | n 1 ].  p|n and. #

32. 32 Common Divisor For a,b  , a positive integer c is said to be a common divisor of a and b if c|a and c|b. The greatest common divisor c, denoted by gcd(a,b), is the common divisor of a and b such that for any common divisor d of a and b satisfies d|c. Examples: gcd(121,33) = 11; gcd(6,35)=1, gcd(8,16)=8.

33. 33 Uniqueness of GCD For all a,b  Z +, there exists a unique c  Z + that is the greatest common divisor of a and b. Proof. Given a,b  Z +, let S = {as+bt | s,t  , as+bt>0}. We claim that the smallest element c  S is gcd(a,b). c  S   x,y  Z [c = ax + by]. Assume c  a.  a = qc + r, q  N,r  Z +, 1  r < c.  r = a – qc = a – q(ax+by) = (1-qx)a – (qy)b.  r  S contradicts to that c is the smallest in S. Thus, c|a. Similarly, c|b. If d  Z and d|a and d|b, then d|c. (Rule (f)) #

34. 34 Properties of GCD For all a,b   +, gcd(a,b) is the smallest positive integer we can write as a linear combination of a and b. Integers a and b are called relatively prime when gcd(a,b) = 1. That is,  x,y   [ax + by = 1]. Examples: gcd(42,70) = 14   x,y   [42x + 70y = 14]   x,y   [3x + 5y = 1]  x = 2-5k, y = -1+3k. (Infinite solutions.) Theorem 4.8 If a,b,c   +, the Diophantine equation ax+by=c has an integer solution iff gcd(a,b)|c.

35. 35 Euclidean Algorithm Let a,b   +. gcd(a,b) is calculated by setting r 0 = a, r 1 = b, and applying the division algorithm n times as follows: r 0 = q 1 r 1 + r 2,0<r 2 <r 1 r 1 = q 2 r 2 + r 3,0<r 3 <r 2 … r i = q i+1 r i+1 + r i+2,0<r i+2 <r i+1 … r n-2 = q n-1 r n-1 + r n,0<r n <r n-1 r n–1 = q n r n.gcd(a,b)=r n

36. 36 Correctness of Euclidean Alg. Goal: gcd(a,b) = r n. step 1.  c [c|a and c|b]  c|r n. step 2. r n |a and r n |b. Proof. Let c be a positive integer such that c|r 0 and c|r 1. r 0 = q 1 r 1 + r 2  c|r 2 c|r 1 and c|r 2 and r 1 = q 2 r 2 + r 3  c|r 3 … c|r n-2 and c|r n-1 and r n-2 = q n-1 r n-1 + r n  c|r n. r n–1 = q n r n  r n |r n-1 r n |r n-1 and r n-2 = q n-1 r n-1 + r n  r n |r n-2  …  r n |r 1 and r n |r 0. #

37. 37 Least Common Multiple For a,b,c   +, c is a common multiple of a and b if a|c and b|c. Furthermore, c is the least common multiple of a,b, denotd by lcm(a,b), is the smallest of all common multiples of a and b. The lcm(a,b) always exists and it is unique. For any common multiple d of a and b, lcm(a,b)|d. For all a,b   +, a  b=gcd(a,b)  lcm(a,b).

38. 38 Prime Divisor If a,b   + and p is a prime, then p|ab  p|a or p|b. Proof. p|a  finished. p  a  gcd(p,a) = 1 (p is a prime)   x,y   [px + ay = 1]  (p)bx + (ab)y = b p|p and p|ab  p|b. # If a i   + for all 1  i  n. If p is a prime and p|a 1 a 2 …a n, then p|a i for some 1  i  n.

39. 39 Irrrational Number is irrational. Proof. (Aristotle,384-322 B.C.) Assume = a / b, for some a,b   + and gcd(a,b)=1.  2 = a 2 / b 2  2b 2 = a 2  2|a 2  2|a Let a = 2c.  2b 2 = a 2 = 4c 2  b 2 = 2c 2  2|b  gcd(a,b)  2.  Thus, is irrational. #

40. 40 Prime Factoring Prime factoring of 980220 is 2 2  3  5  17  31 2. The number of positive divisors of 980220 is (2+1)(1+1)(1+1)(1+1)(2+1) = 72. Let m = 2 2  3  5  17  31 2, n = 2 3  7  5 2  31. gcd(m,n) = 2 2  5  31 lcm(m,n) = 2 3  3  5 2  7  17  31 2.

41. 41 Uniqueness of Prime Factoring Every integer n > 1 can be written as a product of primes uniquely. (The Fundamental Theorem of Arithmetic) Proof. Existence. Assume m is the smallest integer not expressible as a product of primes.  m is not a prime   m 1,m 2   + [m = m 1  m 2 ] m 1,m 2 < m  m 1 and m 2 can be written as products of primes  m can be written as a product of primes. 

42. 42 Uniqueness of Prime Factoring Uniquness. Basis step: n = 2 can be uniquely written as a product of primes. Inductive step: Assuming that 2, 3, …, n-1 can be uniquely written as products of primes. Suppose, where p 1 | n  p 1 |  p 1 | q j for some 1  j  r  p 1 = q 1 by contradiction. (p 1 < p e = q 1 < q j = p 1 ) By induction, n can be uniquely written as a product of primes. #

43. 43 Brainstorm 題目源自 1981 年柏林「德國邏輯思考學院」的考題改編， 98% 的測試者無法解題，國內某家半導體設計公司曾以此題目招 考員工， 題目如下 : 有五位小姐排成一列 ; 所有的小姐穿的衣 服顏色都不一樣 ; 所有的小姐姓也不同 ; 所有的小姐都養不同 的寵物, 喝不同的飲料, 吃不同的水果. 錢小姐穿紅色的衣服 ; 翁小姐養了一隻狗 ; 陳小姐喝茶 ; 穿綠衣服的站在穿白衣服的左邊 ; 穿綠衣服的小姐喝咖啡 ; 吃西瓜的小姐養鳥 ; 穿黃衣服的小姐吃柳丁 ; 站在中間的小姐喝 牛奶 ; 趙小姐站在最左邊 ; 吃橘子的小姐站在養貓的隔壁 ; 養魚的 小姐隔壁吃柳丁 ; 吃蘋果的小姐喝香檳 ; 江小姐吃香蕉 ; 趙小姐站 在穿藍衣服的隔壁 ; 只喝開水的小姐站在吃橘子的隔壁。 問題：請問那位小姐養蛇 ?