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IE MS5710 Introduction to Number Theory II

IE MS5710 Introduction to Number Theory II. 5 Feb 20 13 Prof. CHAN Yuen-Yan, Rosanna Department of Information Engineering The Chinese University of Hong Kong. Basic Number Theory Divisor. say a non-zero number b divides a if for some m have a=mb ( a,b,m all integers)

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IE MS5710 Introduction to Number Theory II

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  1. IEMS5710Introduction to Number Theory II 5 Feb 2013 Prof. CHAN Yuen-Yan, Rosanna Department of Information Engineering The Chinese University of Hong Kong

  2. Basic Number TheoryDivisor • say a non-zero number bdividesa if for some m have a=mb (a,b,m all integers) • that is b divides into a with no remainder • denote this b|a • and say that b is a divisor of a • eg. all of 1,2,3,4,6,8,12,24 divide 24 • eg. 13 | 182; –5 | 30; 17 | 289; –3 | 33; 17 | 0 IEMS5710 - Lecture 4

  3. Basic Number TheoryDivisor Properties • If a|1, then a = ±1. • If a|b and b|a, then a = ±b. • Any b ≠ 0 divides 0. • If a | b and b | c, then a | c • e.g. 11 | 66 and 66 | 198  11 | 198 • If b|g and b|h, then b|(mg + nh) for arbitrary integers m and n e.g. b = 7; g = 14; h = 63; m = 3; n = 2 hence 7|14 and 7|63 IEMS5710 - Lecture 4

  4. Basic Number TheoryDivisor Algorithm • if divide a by n get integer quotient q and integer remainder r such that: • a = qn + r where 0 <= r < n; q = floor(a/n) • remainder r often referred to as a residue IEMS5710 - Lecture 4

  5. Basic Number Theory Greatest Common Divisor (GCD) • a common problem in number theory • gcd(a,b) of a and b is the largest integer that divides both a and b • eg gcd(60,24) = 12 • define gcd(0, 0) = 0 • often want no common factors (except 1) define such numbers as relatively prime • eg gcd(8,15) = 1 • hence 8 & 15 are relatively prime IEMS5710 - Lecture 4

  6. Basic Number TheoryModular Arithmetic mod (mod n) • The modfunction • x (mod y) • Inputs a number xand the base y • Outputs xmod y,a number between 0 and y–1 inclusive • i.e. the remainder of ab • In JAVA: use % • The (mod) congruence • Relates two numbers a, a’ to each other relative to some base b • a  a’ (mod b) means that a and a’ are equivalent in “mod b” • a and a’ have the same remainder when dividing by b (mod 7) (mod 3) (mod p) IEMS5710 - Lecture 4

  7. Basic Number TheoryModular Arithmetic Formal Definition: Let a, a’ be integers and b be a positive integer. We say that a is congruent to a’ modulo b (denoted by a  a’ (mod b)) iff b | (a – a’). Equivalently: a mod b = a’ mod b • Which of the following are true? • 3  3 (mod 7) • 3  4 (mod 7) • 3  -2 (mod 5) • 13  15 (mod 5) • -13  13 (mod 26) IEMS5710 - Lecture 4

  8. Basic Number TheoryModular Arithmetic • In the “world of mod”, we can think in this way: • The function “mod 7” will only produce the following outputs {0, 1, 2, 3, 4, 5, 6} • We need to “map” everything to these seven numbers when we are talking about “mod 7” (We will talk about the concept of “field” in subsequent slides) • {0, 1, 2, 3, 4, 5, 6} is a prime field, or F7 • A field has a zero-element: 0 • A field has a unity-element: 1 • Each element in the field has an (multiplicative) inverse • Inverse is an other element in the same field, when an element being multiplied by its inverse, the result equal unity. • From now on, we will have a new concepts on + - * / IEMS5710 - Lecture 4

  9. Basic Number TheoryModular Arithmetic • What are the answers of the following? • 2 + 3 mod 7 • 2 – 3 mod 7 • 2 * 3 mod 7 • How about 2/3 mod 7? • Answer: the steps: • 2/3 mod 7 = 2 * (1/3) mod 7 • 1/3 i.e. 3-1 mod 7 means the inverse of 3 in the world of mod 7 • Ask yourself: which number, when multiplied by 3, results in 1 in the world of mod 7? • Ah~~ the answer is 5, because 5*3 = 15 = 1 mod 7 • Therefore 3-1 = 5 mod 7 • Therefore 2/3 mod 7 = 2*5 mod 7 = 3 mod 7 Remark: we have formal method to obtain modulo inverses, but this is out of the scope of this course. IEMS5710 - Lecture 4

  10. Basic Number TheoryModular Arithmetic • Modular Exponentials • Similar to normal exponentials, but remember to “mod” in the end, for example • 31 = 3 mod 7 • 32 = 9 = 2 mod 7 • 33 = 27 = 6 mod 7 • 34 = 81 = 4 mod 7 • 35 = 243 = 5 mod 7 • 36 = 729 = 1 mod 7 • A primitive element in a group is an element whose powers exhaust the non-zero elements in the group • 3 is a primitive element in 7 • How about 4 (mod 7)? • 41 = 4, 42 = 16 = 2, 43 = 64 = 1, 44 = 256 = 4, 45 = 1024 = 2 … IEMS5710 - Lecture 4

  11. Basic Number TheoryGroups • Almost all cryptography algorithms (RSA, Elliptic Curves, Diffie-Hellman … ) are done in modular arithmetic • Modular arithmetic is operated in finite fields • Group  Ring  Fields • A Group is a set of elements or “numbers” with some operations whose result is also in the set (closure) that obeys: • associative law: (a  b) c = a (b c) • has identity e: e a = a e = a • has inverse a-1: a a-1 = e IEMS5710 - Lecture 4

  12. Basic Number TheoryGroups •  is an abstract operator, and could be any actions including addition or multiplication • E.g. the set of elements can be the fruits, and * is the action of mixing two pieces of fruit together • Abelian group • If the group is commutative i.e. a  b = b a then it is called an Abelian group IEMS5710 - Lecture 4

  13. Basic Number TheoryGroups Example 1 For a set {0, 1, 2, 3} Define an operator  as addition in modulo 4 • For a, b, and c in this set, the set obeys: • associative law: (a +b) +c = a +(b + c) (mod 4) • There exists an identity e=0 where 0 + a = a + 0 = a (mod 4) • There exists inverses a-1 = -a a + a-1 = a + (-a)=0=e (mod 4) • Therefore the set and the operator forms a group • It is commutative: a + b = b + a (mod 4) • Therefore, the above forms an Abelian group IEMS5710 - Lecture 4

  14. Basic Number TheoryGroups Example 2 For a set {0, 1, 2, 3} Define an operator  as multiplication in modulo 4 • a set: {0,1,2,3} with operator(mod 4) • obeys: • associative law: (a b)  c = a (b c) (mod 4) • identity e=1: 1  a = a 1 = a • How about inverses a-1? • First of all, 0 has no inverse • 1 has an inverse (itself) • 3 has an inverse (itself) 3 3=9=1 (mod 4) • 2 has no inverse • Cannot be a group IEMS5710 - Lecture 4

  15. Basic Number TheoryCyclic Groups • define exponentiation as repeated application of operator • example: a3 = aaa • and let identity be: e=a0 • a group is cyclic if every element is a power of some fixed element • i.e.b =ak for some a and every b in group • a is said to be a generator of the group IEMS5710 - Lecture 4

  16. Basic Number TheoryRings • A Ring is a group under addition. Furthermore, a ring satisfies the following properties under multiplication: • Closure: a*b is in the set if both a and b are in the set • Associative: (a*b)*c =a*(b*c) • Distributive over addition: a*(b+c) = a*b + a*c • If multiplication operation is commutative, it forms a commutative ring • The additive identity is denoted by 0 (called the zero-element) • If multiplication operation has an identity and nozero divisors (i.e. if xy = 0 implies x = 0 or y = 0), it forms an integral domain • The multiplicative identity is denote by 1 (called the unity-element) IEMS5710 - Lecture 4

  17. Basic Number TheoryFields A Field F is a nonempty set F together with two binary operations (denoted as addition (+) and multiplication (*) below) such that: • (F,+) is an Abelian group • (F\{0}, *) is an Abelian group • Note: 0 doesn’t have multiplicative inverse • a * (b + c) = a * b + a * c and (a + b) * c = a * c + b * c for all a, b, c in F (left and right distributive laws) • A finite field is a field with a finite field order (i.e., number of elements). The order of a finite field is always a prime or a power of a prime. IEMS5710 - Lecture 4

  18. Basic Number TheoryGroup, Rings, Fields IEMS5710 - Lecture 4

  19. Basic Number TheoryModular Arithmetic Properties IEMS5710 - Lecture 4

  20. Basic Number TheoryFinite (Galois) Fields • finite fields play a key role in cryptography • number of elements in a finite field must be a power of a prime pn • known as Galois fields • denoted GF(pn) • in particular often use the fields: • GF(p) • GF(2n) IEMS5710 - Lecture 4

  21. Basic Number TheoryFinite (Galois) Fields • GF(p) is the set of integers {0,1, … , p-1} with arithmetic operations modulo prime p • For GF(2n), for example: • The elements are expressed in polynomials with coefficients modulo 2 • since coefficients are 0 or 1, can represent any such polynomial as a bit string • E.g. in GF(23) have (x2+1) is 1012 & (x2+x+1) is 1112 • addition becomes XOR of these bit strings • multiplication is shift & XOR • Similar to those in long-hand multiplication IEMS5710 - Lecture 4

  22. E.g. (x2+1) and (x2+x+1) in GF(23) • so addition is • (x2+1) + (x2+x+1) = x • 101 XOR 111 = 0102 • and multiplication is • (x+1)(x2+1) = x (x2+1) + 1.(x2+1) = x3+x+x2+1 = x3+x2+x+1 • 011101 = (101)<<1 XOR (101)<<0 = 1010 XOR 101 = 11112 <<x means shift leftward by x places IEMS5710 - Lecture 4

  23. References • William Stallings, Cryptography and Network Security Principles and Practices, 5/e, Pearson • Chapter 4 IEMS5710 - Lecture 4

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