n = p 1 p 2 p r = q 1 q 2 q m, then r = m (i.e. the number of primes in any prime decomposition

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1 Department of Mathematical Sciences Instructor: Daiva Pucinskaite Discrete Mathematics Factoring Recall. Fundamental Theorem of Arithmetic. Let n be a positive integer n > 1. Then n can be represented in (up to order) exactly one way as a product of prime numbers. Remark: If p 1, p 2,..., p r, q 1, q 2,..., q m are primes, and n = p 1 p 2 p r = q 1 q 2 q m, then r = m (i.e. the number of primes in any prime decomposition is the same) and the only difference of (p 1, p 2,..., p r ) and (q 1, q 2,..., q r ) is the order. This product is the prime factorization of n Factor the following positive integers into primes. a. 25 = 5 5 b = c = (25) 10 = (2 5) (2 5) (2 5) = } 2 2 {{ 2} 5 } 5 {{ 5}. 10-times 10-times 10-times d. 19 = equals to the empty product Let x be an integer. Prove that 2 x and 3 x if and only is 6 x. We have to prove two statements: I. If 2 x and 3 x, then 6 x II. If 6 x, then 2 x and 3 x

2 Proof of I. By the assumption 2 x and 3 x: 2 x and 3 x = x = 2 c and x = 3 z for certain c, z Z = 2 appears in the prime factorization of x and 3 appears in the prime factorization of x = the prime factorization of x is of the form x = 2 3 further prime factors = x = 6 further prime factors = 6 is a factor of x = 6 x (this is what we had to show) Proof of II. By the assumption 6 x 6 x = x = 6 c for certain c Z, = x = 2 3 c ( because 6 = 2 3), = x = 2 3 c and x = 3 2 c, = Z Z 2 is a factor of x and 3 is a factor of x, = 2 x and 3 x (this is what we had to show). Generalisation: Let p and q be unequal primes. Then p x and q x if and only if (pq) x Let a be an positive integer, and p a prime. Prove that p a if and only if the prime factorization of a contains p (or in other words: p a if and only if p appears in the prime factorization of a). We have to prove two statements: I. If p x, then p appears in the prime factorization of a II. If p appears in the prime factorization of a, then p x. Proof of I. By the assumption p x: 2

3 p x = x = p n for some n N, ( n is positive, because a is positive by the assumption) n N = n = p 1 p 2 p r is the prime factorization of n. (this follows form the Fundamental Theorem of Arithmetic ) = x = p p 1 p 2 p r n is the prime factorization of x (all numbers in this multiplication are prime, thus Fundamental Theorem of Arithmetic provides, x = p p 1 p 2 p r is the (one and only) prime factorization of x) = p appears in the prime factorization of a ( this is what we had to show). Proof of II. If p appears in the prime factorization a = p 1 p i 1 p i p i+1 p r of a, then p = p i for some 1 i r. Therefore implies p x. x = p i (p 1 p i 1 p i+1 p r ) p (Example. The prime p = 4 appears in the prime factorization of a = 2 p 1 p 2 p 3 4 p 4 61, p 5 because p = p 4. Therefore a = 4 ( 2 61 ) implies that 4 is a divisor p p 1 p 2 p 3 p 5 of a) Let a, b N, and p is a prime. If p (a b), then p a or p b. Proof. Since a, b N, by the Fundamental Theorem of Arithmetic there is (up to order) unique list of primes (p 1, p 2,..., p r ) such that a = p 1 p 2 p r, and unique list of primes (t 1, t 2,..., t m ) such that b = t 1 t 2 t m. Therefore a b = p 1 p 2 p r t 1 t 2 t m is the prime factorization of ab (because all numbers in this product are primes). The assumption p (a b) implies that p appears in the prime factorization of a b. We have 3

4 a b = p 1 p 2 p r t 1 t 2 t m = p further prime factors. Since the the prime factorization of a b is (up to order) unique, we have p = p i for some i {1,..., r} or p = t j for some j {1,..., m}. If p = p i, i.e. p appears in the prime factorization of a, then p a (statement 39.4). If p p i for all i {1,..., r}, then p = t j for some j. In this case p b, because p appears in the prime factorization of b. 1. Let n N, and let p be a prime. Show that the following statements are true (1) p appears an even number (or zero) of times in the prime factorization of n 2. Proof. Let n = p 1 p 1 p 1 p 2 p 2 p 2 n 1 -times n 2 -times p m p m p m n m-times the prime factorization of n, where p 1, p 2,..., p r are pairwise different (i.e. p i p j if i j). Example: In the prime factorization of n = = we have m = 4, because p 1 = 2, p 2 =, p 3 = 23, p 4 = 41 are pairwise different primes, and n 1 = 3, n 2 = 4, n 3 = 2, n 4 = 5, because p 1 = 2 appears n 1 = 3 times in the prime factorization of n, p 2 = appears n 2 = 4 times in the prime factorization of n, p 3 = 23 appears n 3 = 2 times in the prime factorization of n, p 4 = 41 appears n 4 = 5 times in the prime factorization of n. Then n 2 = p 1 p 1 p 2 p 2 p m p m p 1 p 1 p 2 p 2 p m p }{{ m } n 1 -times n 2 -times n m-times n 1 -times n 2 -times n m-times = p 1 p 1 p 2 p 2 p m p }{{ m } 2n 1 -times 2n 2 -times 2n m-times is the prime factorization of n 2. If p p i for all i {1, 2,..., r}, then p appears 0 times in the prime factorization of n 2. If p = p i, then p appears 2 n i times be 4

5 in the prime factorization of n. Since 2 n i is even, we showed what we had to show. (2) p appears an odd number of times in the prime factorization of m = p n 2. Proof. From (1) we know that p appears even number (or zero) of times in the prime factorization of n 2. If p appears 0 times in the prime factorization of n 2 (i.e. p don t appears in the prime factorization of n 2 ), then p appears 1 time in the prime factorization of m = p n 2. If p appears 2 k times in the prime factorization of n 2, then p appears 2 k + 1 times in the prime factorization of m = p n 2. Since 1 as well as 2 k + 1 are odd, we showed what we had to show. Let n = , then n 2 = the prime factorization of n 2, because all numbers in this product are prime. Thus 1 appears 3 times in the prime factorization of 1 n 2, 19 appears 5 times in the prime factorization of 19 n 2, 4 appears times in the prime factorization of 4 n 2, p appears 1 time in the prime factorization of p n 2 if p 1, 19, Prove the following statements. (1) There are no numbers a, b such that a b =. This is: For all numbers a, b, we have a b. This statement is logically equivalent to the statement If a b =, then a N or b N. We will show that if a b =, then a 2 has two prime factorizations. In the first prime factorization of a 2 the prime appears an even number of times, and in the second prime factorization 5

6 of a 2 the prime appears an odd number of times. By the Fundamental Theorem of Arithmetic this is not possible if a 2 is a number. This implies that a 2 N and therefore a N (because if a N, then a 2 N). Proof. Assume a b =, we have a b = = a2 b 2 = = a 2 = b 2 = appears an even number of times in the prime factorization of a 2 (Statement 1(1)) and appears an odd number of times in the prime factorization of b 2 = a 2 (Statement 1(2)) = a 2 can t be a number by the Fundamental Theorem of Arithmetic. = a can t be a number (2) There are no numbers a, b such that a b = 23. This is: For all numbers a, b, we have a b 23. Proof. Assume a b = 23, we have a b = 23 = a2 b 2 = 23 = a 2 = 23 b 2 = the prime 23 appears an even number of times in the prime factorization of a 2 (Statement 1(1)) and the prime 23 appears an odd number of times in the prime factorization of 23 b 2 = a 2 (Statement 1(2)) = a 2 can t be a number by the Fundamental Theorem of Arithmetic. = a can t be a number (3) Let p be a prime. There are no numbers a, b such that a b = p. 6

Midterm Exam. There are 6 problems. Your 5 best answers count. Please pay attention to the presentation of your work! Best 5

Midterm Exam. There are 6 problems. Your 5 best answers count. Please pay attention to the presentation of your work! Best 5 Department of Mathematical Sciences Instructor: Daiva Pucinskaite Modern Algebra June 22, 2017 Midterm Exam There are 6 problems. Your 5 best answers count. Please pay attention to the presentation of