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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.

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Discrete Mathematics

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  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?

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