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Unformatted text preview: UCLA Mathematics 110A: selected solutions from homework #1 David Wihr Taylor July 2, 2010 Introduction When reading these solutions always keep in mind the common techniques being used. The point of home work, and subsequently these solutions, is to give you experience and competence in solving abstract algebra problems on your own ! By this point you all have the capacity to memorize algorithms and the statements of theorema. However developing both an understanding of what a subject really means and how one can use it requires a more specialized type of effort. You must practice problem solving and also try to abstract , that is find the general themes inherent in, your solutions. I’ll try to make my own ideas and strategies clear in the solutions that follow. 1 Problems from Section 1.1 The main result of this section is theorem 1 . 1. This is the familiar division algorithm that we all learn in grade school. You probably just memorized this many years ago and never gave much thought as to why it works. The main technical tool in its proof is to use the fact that every nonempty subset of the integers, which is bounded below has a smallest element. 1.1 Problem 1.1.7: Claim: Let n be a positive integer. Prove that a and c leave the same remainder when divided by n if and only if a c = nk for some integer k . Proof: We’re proving an if and only if statement here, so we’re going to have to write two arguments: ⇒ If a and c leave the same remainder when divided by n , then a c = nk for some integer k . ⇐ If a c = nk for some integer k , then a and c leave the same remainder when divided by n . Remember, we only have theorem 1 . 1 and its corollary to work with, so we’ll have to pay close attention to what it really says at each step of our proof. ⇒ : Since we assume that a and c leave the same remainder when divided by n we can use the division algorithm to write a = pn + r and b = qn + r (1) for integers p,q,r ∈ Z . The we can compute the difference a c by using equations (1): a c = pn + r ( qn + r ) = ( p q ) n = kn 1 where we’ve let n := p q . This is what we wanted to show, so we’re finished with this part of the problem....
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This note was uploaded on 11/03/2010 for the course MATH 26233820 taught by Professor Gieseker,d. during the Fall '10 term at UCLA.
 Fall '10
 GIESEKER,D.
 Math, Algebra

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