# lecture22 - MA1100 Lecture 22 Number Theory Fundamental...

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MA1100 Lecture 22 Number Theory Fundamental Theorem of Arithmetic Number and sum of divisors Perfect numbers Infinitude of primes Different types of primes Chartrand: 11.6, 11.7

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Lecture 22 2 Announcement ± HW4 scores in Gradebook ± HW5 due next Tuesday ± Return clickers THIS Friday
Lecture 22 3 Fundamental Theorem of Arithmetic Part 1: Every integer greater than 1 is either a prime number or a product of prime numbers . Part 2: For any integer greater than 1 , the prime factorization is unique . Existence of prime factorization Uniqueness of prime factorization

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Lecture 22 4 Fundamental Theorem of Arithmetic Part 2: For any integer greater than 1, the prime factorization is unique . Rephrase: n = p 1 ä p 2 ä ä p r Then with p 1 § p 2 § § p r Let n > 1 be an integer. n = q 1 ä q 2 ä ä q s Suppose n has two prime factorizations: with q 1 § q 2 § § q s r = s and p j = q j for all j = 1, 2, …, r Prove by Strong Mathematical Induction P(n): n has a unique prime factorization
Lecture 22 5 Fundamental Theorem of Arithmetic Proof If n = p 1 ä p 2 ä ä p r and n = q 1 ä q 2 ä ä q s Then r = s and p j = q j for all j = 1, 2, …, r Base case : Prove P(2) is true Since 2 is a prime number, its prime factorization is just 2 itself. So P(2) is true. Inductive step : [P(2) P(3) ... P(k)] P(k+1) for all k ¥ 2 hypothesis: want to get: Each of 2, 3, …, k has unique prime factorization. k + 1 has unique prime factorization P(n): n has a unique prime factorization

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Lecture 22 6 Fundamental Theorem of Arithmetic If k + 1 = p 1 ä p 2 ä ä p r and k + 1 = q 1 ä q 2 ä ä q s Then r = s and p j = q j for all j = 1, 2, …, r hypothesis want to get P(k+1): Each of 2, 3, …, k has unique prime factorization. k + 1 has unique prime factorization p 1 | k + 1 p 1 | q 1 ä q 2 ä ä q s By Euclid’s lemma, p 1 | q j for some q j Since both are prime numbers q 1 § p 1 p 1 = q j We first prove: p 1 = q 1 p 1 § q 1 and p 1 ¥ q 1 q 1 | k + 1 q 1 | p 1 ä p 2 ä ä p r By Euclid’s lemma, q 1 | p j for some p j Since both are prime numbers p 1 § q 1 q 1 = p j We may assume k + 1 is composite. (If k + 1 is prime, there is nothing to prove.)
Lecture 22 7 Fundamental Theorem of Arithmetic Then r = s and p j = q j for all j = 1, 2, …, r hypothesis want to get P(k+1): Each of 2, 3, …, k has unique prime factorization.

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## This note was uploaded on 04/15/2010 for the course MATHS MA1101R taught by Professor Vt during the Spring '10 term at National University of Singapore.

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lecture22 - MA1100 Lecture 22 Number Theory Fundamental...

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