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# Week_1_class_notes MATH 470

Texas A&M

GPA 2.69

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This 4 page Class Notes was uploaded by Evelyn Merizalde on Monday September 5, 2016. The Class Notes belongs to MATH 470 at Texas A&M University taught by Eren Kiral in Fall 2016. Since its upload, it has received 4 views. For similar materials see COMM AND CRYPTOGRAPHY in Engineering and Tech at Texas A&M University.

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Date Created: 09/05/16

Definition 1.1. Given two integers a and d with d non-zero, we say that d divides a (written d | a) if there is an integer c with a = cd. If no such integer exists, so d does not divide a, we write d - a. If d divides a, we say that d is a divisor of a. Proposition 1.2.1: Assume that a, b, and c are integers. If a | b and b | c, then a | c. Proposition 1.3. Assume that a, b, d, x, and y are integers. If d | a and d | b, then d | ax + by. Corollary 1.4. Assume that a, b, and d are integers. If d | a and d | b, then d | a + b and d | a − b. Proposition 1.4. Let a, b, c ∈ Z be integers. a) If a | b and b | c, then a | c. b) If a | b and b | a, then a = ±b. c) If a | b and a | c, then a | (b + c) and a | (b − c). Prime: A prime number is an integer p ≥ 2 whose only divisors are 1 and p. A composite number is an integer n ≥ 2 that is not prime. Ex: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. The Division Algorithm: Let a and b be integers with b > 0. Then there exist unique integers q (the quotient) and r (the remainder) so that a = bq + r with 0 ≤ r < b. The greatest common devisor: Assume that a and b are integers and they are not both zero. Then the set of their common divisors has a largest element d, called the greatest common divisor of a and b. We write d = gcd (a, b). 12: 1, 2, 3, 4, 6, and 12. 18: 1, 2, 3, 6, 9, and 18. Then {1, 2, 3, 6} is the set of common divisors of 12 and 18. Gcd(12,18) = 6 Proposition 1.10: If a and b are integers with d = gcd(a, b), then a b gcd , =1 . (d d The Division Algorithm: Let a and b be integers with b > 0. Then there exist unique integers q (the quotient) and r (the remainder) so that b|a a = bq + r with 0 ≤ r < b. Extended Euclidean Algorithm: One of its features is that it allows us to compute gcd’s without factoring. Suppose that we want to compute gcd (123, 456) = 3. The gcd is the devisor when the remainder equals zero. The Consider the following calculation: a a(dividend)|b (devisor) = b Dividend = devisor*quotient +remainder devisor Devisor= quotient*remainder+ remainder (2) = remainder 456 = 3 · 123 + 87 123 = 1 · 87 + 36 87 = 2 · 36 + 15 36 = 2 · 15 + 6 15 = 2 · 6 + 3 6 = 2 · 3 + 0. Theorem 1.11. Let a and b be integers with at least one of a, b non-zero. There exist integers x and y, which can be found by the Extended Euclidean Algorithm, such that gcd(a, b) = ax + by. Prime: Two integers a and b are said to be relatively prime if gcd(a, b) = 1. Remark. If a 6= 0, then gcd(a, 0) = a. However, Other bases: Base 10 3 2 1 0 (1234)5=1∗5 +2∗5 +3∗5 +4∗5 =(194) 10 Rescheduled From base 10 to any other base: We can also convert a number from base 10 to any other base with the use of the Division Algorithm. We proceed by dividing 21963 by 8, then dividing the quotient by 8, and continuing until the quotient is 0. Number = quotient(Number/base) *base + remainder (21963)10−→(52713) 8 21963 = 2745 · 8 + 3 2745 = 343 · 8 + 1 343 = 42 · 8 + 7 42 = 5 · 8 + 2 5 = 0 · 8 + 5. Complexity Adding two integers n & m: n+m≤2log max n,m )bitoperations 2 Writing an integer n in base 10 takes: ?log10 ? digits Writing an integer n in base 2 takes: log2nbits Dividing two integers m|n: ¿of operations=K l og 2n)where K isaconstant Multiplying two integers n*m in binary: n∗m≤K log nlog mbitoperations 2 2 If f is a function of n and g is positive function of n and: |f(n)|≤cg (n)c>0 Then: f n =O(g (n ) If f (n, m) = the number of bit operations a certain algorithm takes to compute n + m, then: f n,m =O(log 2(n)+log 2m )) If h (n, m) is the number of bit operations that takes to compute n*m, then: log (¿¿2(n)log 2m) )number of operations h (n,m )=O¿ Division: dividend|devisor dividend = quotient(devisor) + remainder Devisor is the only thing that would be >> 1 If remainder – devisor >= 0, quotient << 1 If remainder – devisor < 0, do nothing Regardless the sign of remainder – devisor, shift devisor >> 1 Example: 7|2 Iteration Steps Remainder Devisor Quotient

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