HomeworkSolutions2 - HOMEWORK 2 SOLUTIONS MATT RATHBUN MATH...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern