1 / 34

Discrete Mathematics

Discrete Mathematics. 4. NUMBER THEORY. Lecture 7. Dr.-Ing. Erwin Sitompul. http://zitompul.wordpress.com. Integers. Integers are whole numbers, without any fractional or decimal components. Example: 8 ; 21 ; 8765 ; –34 ; 0.

roary-diaz
Télécharger la présentation

Discrete Mathematics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Discrete Mathematics 4. NUMBER THEORY Lecture 7 Dr.-Ing. Erwin Sitompul http://zitompul.wordpress.com

  2. Integers • Integers are whole numbers, without any fractional or decimal components. Example: 8 ; 21 ; 8765 ; –34 ; 0. • They are in opposite to real numbers, which posses decimal components. Example: 8.0 ; 34.25 ; 0.02.

  3. Division of Integers • Suppose a and b are integers, a 0. then adivides bwithout remaining if there exists an integer c such that b = ac. • Notation: a | b if b = ac, cZ and a 0. Example: (a) 4 | 12 because 12/4 = 3 (integer) or 12 = 4  3. (b) 4 | 13 because 13/4 = 3.25 (not integer).

  4. Euclidean Theorem Euclidean Theorem 1: Suppose m and n are integers, n > 0. if m is divided by n then there exists a unique integer q (quotient) and r (remainder), such that m = nq + r where 0 r < n. Example: (a)1987/97 = 20, remaining 47 1987 = 9720 + 47 (b) 25/7 = 3, remaining 4 25 = 73 + 4 (c) –25/7 = –4, remaining 3 –25 = 7(–4)+ 3 But not –25 = 7(–3) – 4, because the remainder will be r = –4, while the condition is 0 r < n)

  5. Greatest Common Divisor (GCD) • Suppose a and b are non-zero integers. • The Greatest Common Divisor (GCD) of a and b is the greatest possible integer d such that d | a and d | b. • In this case, it can be written as GCD(a,b) = d. Example: Determine GCD(45,36) ! • Divisors of 45: 1, 3, 5, 9, 15, 45. • Divisors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. • Common divisors of 45 and 36 are 1, 3, 9. For the enumeration above, it can be concluded that GCD(45,36) = 9.

  6. Greatest Common Divisor (GCD) Euclidean Theorem 2: Suppose m and n are integer, n > 0, such that m = nq + r, 0 r < n. Then GCD(m,n) = GCD(n,r). Example: Take the value m = 66, n = 18, 66 = 183 + 12 then GCD(66,18) = GCD(18,12) = 6

  7. Euclidean Algorithm • The objective This algorithm can be used to find the GCD of two integers. • Inventor Euclid (around 300 BC), a Greek mathematician, wrote the algorithm in his book titled, “Element.”

  8. Euclidean Algorithm If m and n are non-negative integers where mn, and suppose r0 = m and r1 = n. Perform the following divisions in sequence to obtain: r0 = r1q1 + r2 0 r2r1, r1 = r2q2 + r3 0 r3r2, ri–2 = ri–1qi–1 + ri 0 riri–1, ri–1 = riqi + 0 … According to Euclidean Theorem 2, GCD(m,n) = GCD(r0,r1) = GCD(r1,r2) = … = GCD(ri–2,ri–1) = GCD(ri–1,ri) = GCD(ri,0) = ri Thus, GCD of m and n is the last non-zero remainder of the above sequence of disions, namely ri.

  9. Euclidean Algorithm Given two non-negative integers m and n (mn), the following Euclidean Algorithm will find the greatest common divisor of m and n. Euclidean Algorithm 1. If n = 0 then m is the GCD(m,n); STOP. If n 0, proceed to Step 2. 2. Divide m with n and obtain r as the remainder. 3. Replace m with n, and n with r, then loop back to Step 1.

  10. Euclidean Algorithm Example: Take m = 80, n = 12, so the condition that mn is fulfilled. 80 = 126 + 8 12 = 81 + 4 8 = 42 + 0 n = 0  m = 4 is the last non-zero remainder GCD(80,12) = 4; STOP.

  11. Linear Combination • GCD(a,b) can be expressed as a linear combination of a and b with the multiplying coefficients that can be freely chosen. Example: GCD(80,12) = 4, then 4 = (–1)80 + 712 Coefficients, can be freely chosen Linear Combination Theorem: Suppose a and b are positive integers, then there exist integers m and n such that GCD(a,b) = ma + nb.

  12. Linear Combination Example: Express GCD(312,70) = 2 as the linear combination of 312 and 70! Solution: Apply Euclidean Algorithm to obtain GCD(312,70) = 2 as follows: 312 = 470 + 32 (1) 70 = 232 + 6 (2) 32 = 56 + 2 (3) 6 = 32 + 0 (4) Rearrange (3) to 2 = 32 – 56 (5) Rearrange (2) to 6 = 70 – 232 (6) Insert (6) to (5) so that 2 = 32 – 5(70 – 232) = 132 – 570 + 1032 = 1132 – 570 (7) Rearrange (1) to 32 = 312 – 470 (8) Insert(8) to (7) so that 2 = 1132 – 570 = 11(312 – 470) – 570 = 11312 – 4970 Thus, GCD(312, 70) = 2 = 11312 – 4970

  13. Modulo Arithmetics Suppose a is an arbitrary integer and m is a positive integer, then a mod m yields the remainder if a is divided by m a mod m = r such that a = mq + r, with0 r < m The result of modulo m lies within the set {0,1,2,…,m–1}

  14. Congruence • Take a long on this 38 mod 5 = 3 and 13 mod 5 = 3. then it can be written that 38  13 (mod 5). Pronounced: 38 is congruent with 13 in modulo 5. • Suppose a and b are integers and m > 0. If m divides a – b without remainder, thenab (mod m). • If a is not congruent with b in modulo m, then it is written as a b (mod m).

  15. Congruence Example: • 17  2 (mod 3)  3 divides 17–2 = 15 without remainder • –7  15 (mod 11)  11 divides –7–15 = –22 without remainder • 12  2 (mod 7)  7 cannot divide 12–2 = 10 • –7  15 (mod 3)  3 cannot divide –7–15 = –22

  16. Congruence ab (mod m) can be written as a = b + km (k integer). Example: • 17  2 (mod 3)  17 = 2 + 53 • –7  15 (mod 11)  –7 = 15 + (–2)11 a mod m = r can also be written as ar (mod m). Example: • 23 mod 5 = 3  23  3 (mod 5) • 6 mod 8 = 6  6  6 (mod 8) • 0 mod 12 = 0  0  0 (mod 12) • –41 mod 9 = 4  –41  4 (mod 9) • –39 mod 13 = 0  –39  0 (mod 13)

  17. Congruence Congruence Theorem: Suppose m is a positive integer. 1. If ab (mod m) and c is an arbitrary integer, then • (a + c)  (b + c) (mod m) • acbc (mod m) • apbp (mod m) , p non-negative 2. If ab (mod m) and cd (mod m), then • (a + c)  (b + d) (mod m) • acbd (mod m)

  18. Congruence Example: Suppose 17  2 (mod 3) and 10  4 (mod 3), then according to the Congruence Theorem, • 17 + 5  2 + 5 (mod 3)  22  7 (mod 3) • 175  25 (mod 3)  85  10 (mod 3) • 17 + 10  2 + 4 (mod 3)  27  6 (mod 3) • 1710  24 (mod 3)  170  8 (mod 3)

  19. Prime Numbers • A positive integer p (p > 1) is called a prime number if its divisors are only 1 and p. Example: 23 is a prime number, because it can only be divided by 1 and 23 to get no remainder. • Numbers which are not prime numbers are called composite numbers. Example: 20 is a composite number, because 20 is divisible by 2, 4, 5, and 10, besides by 1 and 20 itself.

  20. Relative Primes Two integers a and b are said to be relatively prime if they do not have any common factors other than 1, or, GCD(a,b) = 1. Example: • 20 and 3 are relatively prime, since GCD(20,3) = 1. • 7 and 11 are relatively prime, since GCD(7,11) = 1. • 20 and 5 are not relatively prime, since GCD(20,5) = 5 ≠ 1. If a and b are relatively prime, then there exist integers m and n such that ma + nb = 1. Example: • 20 and 3 are relatively prime because GCD(20,3) =1, so that it can be written that 220 + (–13)3 = 1 (m = 2, n = –13). • 20 and 5 are not relatively prime because GCD(20,5) ≠ 1, and thus 20 and 5 cannot be written in the form ofm20 + n5 = 1.

  21. Inverse of Modulo • In real number arithmetics, the inverse of multiplication is division. • As example, the inverse of 4 is 1/4, because 4  1/4 = 1. • In modulo arithmetics, finding the inverse is somehow more difficult. • If a and m are relatively prime and m > 1, then there exists the inverse of “a modulo m”. • The inverse of “a modulo m” is an integer x such that ax 1 (mod m).

  22. Inverse of Modulo Example: Determine the inverse of 4 (mod 9) ! Solution: Because GCD(4,9) = 1, then the inverse of 4 (mod 9) exists. From the Euclidean Algorithm, 9 = 24 + 1. Rearrange the above equation to   –24 + 19 = 1. From the last equation, it can be obtained that –2 is the inverse of 4 (mod 9).  Check that  –24  1 (mod 9)

  23. Inverse of Modulo Remark: Every integer which is congruent with –2 (mod 9) is also the inverse of 4. Example: • 7  –2 (mod 9)  9 divides 7 – (–2) = 9 without remainder • –11  –2 (mod 9)  9 divides –11 – (–2) = –9 without remainder • 16  –2 (mod 9)  9 divides 16 – (–2) = 18 without remainder

  24. Inverse of Modulo Example: Determine the inverse of 17 (mod 7) ! Solution: Since GCD(17,7) = 1, then the inverse of 17 (mod 7) exists. From the Euclidean Algorithm, 17 = 27 + 3 (1) 7 = 23 + 1 (2) 3 = 31 + 0 (3) Rearrange (2) to   1 = 7 – 23 (4) Rearrange (1) to 3 = 17 – 27 (5) Insert (5) to (4) 1 = 7 – 2(17 – 27) 1 = –217 + 57 From the last equation, –2 is the inverse of 17 (mod 7).  Checking, –217  1 (mod 7)

  25. Inverse of Modulo Example: Determine the inverse of 18 (mod 10) ! Solution: Since GCD(18,10) = 2 ≠ 1, then the inverse of 17 (mod 7) does not exist.

  26. Linear Congruence The linear congruence is in the form of : axb (mod m), where m > 0, a and b are arbitrary integers, and x is any integer. The solution can be found in the way: ax = b + km  x = (b + km) / a Try each value of k = 0, 1, 2, … and k = –1, –2, … that delivers integer value for x.

  27. Linear Congruence Example: Determine the solutions for 4x 3 (mod 9) ! Solution: 4x 3 (mod 9)  x = (3 + k9 ) / 4 k = 0 x = (3 + 09) / 4 = 3/4  not a solution k = 1 x = (3 + 19) / 4 = 3  a solution k = 2 x = (3 + 29) / 4 = 21/4  not a solution k = 3, k = 4  no solution k = 5 x = (3 + 59) / 4 = 12  asolution … k = –1 x = (3 – 19) / 4 = –6/4  not a solution k = –2 x = (3 – 29) / 4 = –15/4  not a solution k = –3 x = (3 – 39) / 4 = –6  a solution … k = –7 x = (3 – 79) / 4 = –15  a solution … The set of solutions is: {3, 12, …, –6, –15, …}.

  28. Linear Congruence Example: Determine the solutions for 2x 3 (mod 4) ! Solution: 2x 3 (mod 4)  x = (3 + k4 ) / 2 • Because k4 is always an even number, then 3 + k4 will always be an odd number. • If an odd number is divided by 2, then the result will be a decimal number (never be an integer). • Thus, there is no value of x that can be the solution of 2x 3 (mod 4).

  29. Linear Congruence Example: Find x such that 3x 4 (mod 7) ! Solution: 3x  4 (mod 7) (3)–13x  (3)–14 (mod 7) x  (3)–14 (mod 7) x  –24 (mod 7) x  –8 (mod 7) x  6 (mod 7) x = {..., –8, –1, 6, 13, 19, ...}

  30. Application: ISBN • ISBN (International Standard Book Number) • ISBN consists of 10 characters, commonly separated by space or minus sign, i.e., 0–3015–4561–9. • ISBN is classified into several codes: • Code to identify the language of the book • Code to identify the publisher • Code that uniquely assigned for the book (i.e. titles) • Checksum character or check digit (can be a number of or alphabet X)

  31. Application: ISBN • The check digit is so chosen that Example: ISBN 0–3015–4561–8 0 : Code for English-language country group using, 3015 : Publisher code 4561 : Item number, title of the book 8 : Check digit. The check digit is obtained as follows:   10 + 23 + 30 + 41 + 55 + 64 + 75 + 86 + 91 = 151 Therefore, the check digit is 151 mod 11 = 8.

  32. Application: ISBN Example: ISBN 978-3-8322-4066-0 • Since January 2007, ISBN contains 13 digits. • The way to count the check digit is different. It uses modulo 10. The check digit will be obtained as follows:   91 + 73 + 81 + 33 + 81 + 33 + 21 + 23 + 41 + 03 + 61 + 63 = 100 Thus, the check digit is 100 + x13 0 (mod 10) x13 = 0 How to check the validity of a credit card number?

  33. Homework 7 No.1: Determine GCD(216,88) and express the GCD as a linear combination of 216 and 88. No.2: Given the ISBN-13: 978-0385510455, check whether the code is valid or not. Hint: Verify the check digit of the ISBN numbers.

  34. Homework 7 No.1: Determine the solutions for 5x 7 (mod 11) ! No.2: Given the ISBN-10: 0072880082, check whether the code is valid or not. Hint: Verify the check digit of the ISBN numbers. New No.3: Voluntary for additional 20 points The ISBN-13: 978-007289A054 is valid. What will be the value of A?

More Related