CSE115/ENGR160 Discrete Mathematics 03/13/124.1 Divisibility and modular arithmeticExampleTheorem and corollaryThe division algorithmModular arithmeticSlide 7TheoremSlide 9Slide 10Corollary4.2 Integer representations and algorithmsBase conversionSlide 14Modular exponentiationSlide 16AlgorithmSlide 184.3 Primes and greatest common divisionsSlide 20Slide 21Slide 22Procedure for prime factorizationSlide 24CSE115/ENGR160 Discrete Mathematics03/13/12Ming-Hsuan YangUC Merced14.1 Divisibility and modular arithmetic•Number theory: the branch of mathematics involves integers and their properties•If a and b are integers with a≠0, we say that a divides b if there is an integer c s.t. b=ac•When a divides b we say that a is a factor of b and that b is a multiple of a •The notation a | b denotes a divides b. We write a ∤ b when does not divide b2Example•Let n and d be positive integers. How many positive integers not exceeding n are divisible by d?•The positive integers divisible by d are all integers of them form dk, where k is a positive integer•Thus, there are positive integers not exceeding n that are divisible by d3 dn /Theorem and corollary•Theorem: Let a, b, and c be integers, then–If a | b and a| c, then a | (b+c)–If a | b, and a | bc for all integers c–If a | b and b | c, then a | c•Corollary: If a, b, and c are integers s.t. a | b and a | c, then a | mb+nc whenever m and n are integers4The division algorithm•Let a be integer and d be a positive integer. Then there are unique integers q and r with 0 ≤ r < d, s.t. a=dq+r•In the equality, q is the quotient, r is the remainder q = a div d, r = a mod d•-11 divided by 3•-11=3(-4)+1, -4=-11 div 3, 1=-11 mod 3•-11=3(-3)-2, but remainder cannot be negative5Modular arithmetic•If a and b are integers and m is a positive integer, then a is congruent to b modulo m if m divides a-b•We use the notation a≡b (mod m) to indicate that a is congruent to b modulo m •If a and b are not congruent modulo m, we write a ≢b (mod m)•Let a and b be integers, m be a positive integer. Then a≡b (mod m) if and only if a mod m = b mod m6Example•Determine whether 17 is congruent to 5 modulo 6, and whether 24 and 14 are not congruent modulo 6–17-5=12, we see 17≡5 (mod 6)–24-14=10, and thus 24≢14 (mod 6)7Theorem•Karl Friedrich Gauss developed the concept of congruences at the end of 18th century•Let m be a positive integer. The integer a and b are congruent modulo m if and only if there is an integer k such that a=b+km–() If a=b+km, then km=a-b, and thus m divides a-b and so a≡b (mod m)–() if a≡b (mod m), then m | a-b. Thus, a-b=km, and so a=b+km8Theorem•Let m be a positive integer. If a ≡ b (mod m) and c ≡ d (mod m), then a+c=b+d (mod m) and ac ≡ bd (mod m)–Since a ≡ b (mod m) and c ≡ d (mod m), there are integers s.t. b=a+sm and d=c+tm–Hence, b+d=(a+c)+m(s+t), bd=(a+sm)(c+tm)=ac+m(at+cs+stm)–Hence a+c ≡ b+d (mod m), and ac ≡ bd (mod m)9Example•7 ≡ 2 (mod 5) and 11 ≡ 1 (mod 5), so–18=7+11 ≡ 2+1=3 (mod 5)–77=7 11 ≡2 1=2(mod 5)∙ ∙10Corollary•Let m be a positive integer and let a and b be integers, then (a+b) mod m=((a mod m) +(b mod m)) mod m ab mod m = ((a mod m)(b mod m)) mod m•Proof: By definitions mod m and congruence modulo m, we know that a≡(a mod m)(mod m) and b≡(b mod m)(mod m). Hence–(a+b) ≡((a mod m)+(b mod m)) (mod m)–ab ≡ (a mod m)(b mod m)(mod m)114.2 Integer representations and algorithms•Base b expansion of n•For instance, (245)8=2*82+4*8+5=165•Hexadecimal expansion of (2AE0B)16 (2AE0B)16=2*164+10*163+14*162+0*16+11=175627•Constructing base b expansion12Base conversion•Constructing the base b expansion n=bq0+a0, 0 ≤a0<b•The remainder a0, is the rightmost digit in the base b expansion of n•Next, divide q0 by b to obtain q0=bq1+a1, 0≤a1<b•We see a1 is the second digit from the right in the base b expansion of n•Continue this process, successively dividing the quotients by b, until the quotient is zero13Example•Find the octal base of (12345)10•First, 12345=8*1543+1•Successively dividing quotients by 8 gives 1543=8*192+7 192=8*24+0 24=8*3+0 3=8*0+3•(12345)10=(30071)814Modular exponentiation•Need to find bn mod m efficiently in cryptography•Impractical to compute bn and then mod m•Instead, find binary expansion of n first, e.g., n=(ak-1 … a1 a0)•To compute bn , first find the values of b, b2, …, (b4)2=b8, …•Next multiple the where aj=115012211011122222 aaaaaaanbbbbbbkkkkkkjb2Example•To compute 311 •11=(1011)2 ,So 311=38 32 31 . First compute 32=9, and then 34=92=81, and 38=(34)2=(81)2=6561, So 311=6561*9*3=177147•The algorithm successively finds b mod m, b2 mod m, b4 mod m, …, mod m, and multiply together those terms1612kbAlgorithm•procedure modular exponentiation (b:integer, n=(ak-1ak-2a1a0, …, an)2, m:positive integer) x := 1 power:=b mod m for i:=0 to k-1 if ai =1 then x:=(x⋅ power) mod m power:=(power⋅power) mod m end {x equals bn mod m}•It uses O((log m)2 long n) bit operations17Example•Compute 3644 mod 645–First note that 644=(1010000100)2–At the beginning, x=1, power=3 mod 645 = 3–i=0, a0=0, x=1, power=32 mod 645=9–i=1, a1=0, x=1, power=92 mod 645=81–i=2, a2=1, x=1*81 mod 645=81, power=812 mod 645=6561 mod 645=111–i=3, a3=0, x=81, power=1112 mod 645=12321 mod 645=66–i=4, a4=0, x=81, power=662 mod 645=4356 mod 645=486–i=5, a5=0, x=81, power=4862 mod 645=236196 mod 645=126–i=6, a6=0, x=81, power=1262 mod 645=15876 mod 645=396–i=7, a7=1, x=(81*396) mod 645=471, power=3962 mod 645=156816 mod 645=81–i=8, a8=0, x=471, power=812 mod 645=6561mod 645=111–i=9, a9=1, x=(471*111) mod 645=36•3644 mod 645=36184.3 Primes and greatest common divisions•Prime: a positive integer p greater than 1 if the only positive factors of p are 1 and p•A positive integer greater than 1 that is not prime is called composite •Fundamental theorem of arithmetic: Every positive integer greater than 1 can be written uniquely as a prime or as the product of two or more primes when the prime factors are written in order of non-decreasing size19Example•Prime factorizations of integers–100=2 2 5 5=2∙ ∙ ∙25∙2–641=641–999=3 3 3 37=3∙ ∙ ∙337∙–1024=2 2 2 2 2 2 2 2 2 2=2∙ ∙ ∙ ∙ ∙ ∙ ∙ ∙