MATH
HomeworkSolutions2

# HomeworkSolutions2 - HOMEWORK 2 SOLUTIONS MATT RATHBUN MATH...

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HOMEWORK 2 SOLUTIONS MATT RATHBUN MATH 310, SECTION 3 ABSTRACT ALGEBRA 1.3 #2, 6, 13, 23; 2.1 #2, 8, 11, 27 Section 1.3. 2. Suppose that p is prime, and let a be an integer. Then ( a, p ) is a positive divisor of p , so it is either 1 or | p | . If it is 1, then we are done. Otherwise, | p | is a divisor of a , so p | a as well. Conversely, suppose that for any a Z , either ( a, p ) = 1 or p | a . Let us suppose, by way of contradiction, that p is not prime. Then p has a non-trivial factorization into primes, p = p 1 · p 2 · · · p k for some k 2. Let 1 l < k . Consider a = p 1 · p 2 · · · p l . Since l < k , p - a . But ( a, p ) = (( p 1 · · · p l ) , ( p 1 · · · p k )) = p 1 · · · p l 6 = 1. This contradicts our hypothesis, so p must be prime. 6. Suppose p is prime and p | a n . Then by Corollary 1.9, p | a , so p n | a n . 13. Consider the factorizations of a and b into primes: a = p n 1 1 · · · p n r r , b = q m 1 1 · · · q m s s , with n i , m j 1. Then c 2 = ab = p n 1 1 · · · p n r r · q m 1 1 · · · q m s s . Taking the square root of both sides, we arrive at c = p n 1 2 1 · · · p

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• Summer '08
• KUMAR
• Algebra, Prime number, Divisor, 6k, MATT RATHBUN

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