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MATH 111 m1

# MATH 111 m1 - MODEL ANSWERS TO THE FIRST HOMEWORK 1 Chapter...

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MODEL ANSWERS TO THE FIRST HOMEWORK 1. Chapter 1, § 1: 1. Suppose that a and b are elements of S . By rule (1) a b = a. But by rule (2), a b = b a. Applying rule (1) we get a b = b a = b . Thus a = a b = b . As a and b are arbitrary, S can have at most one element. 1. Chapter 1 § 1: 2. (a) Suppose that a and b are two integers and that a b = b a . Now a b = a - b and b a = b - a so that then a - b = b - a . Applying the standard rules of arithmetic, we get 2 a = 2 b and so a = b . (b) Suppose that a , b and c are integers. Then a ( b c ) = a ( b - c ) = a - ( b - c ) = a + c - b. On the other hand ( a b ) c = ( a - b ) c = ( a - b ) - c = a - ( b + c ) . Thus equality holds i ff a + c - b = a - ( b + c ), that is cancelling c = - c so that c = 0. Thus is not associative. For example, 0 (0 1) = 1 but (0 0) 1 = - 1 . (c) Let a be an integer. Then a 0 = a - 0 = a. (d) Let a be an integer. Then a a = a - a = 0 . 2. Chapter 2, § 1: (a). No, by the question above, this rule of multipli- cation is not associative. 1

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(b) No this is not a group. We consider the three axioms. Suppose that a , b and c are three integers. Then a ( b c ) = a ( b + c + bc ) = a + ( b + c + bc ) + a ( b + c + bc ) = a + b + c + bc + ab + ac + abc.
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MATH 111 m1 - MODEL ANSWERS TO THE FIRST HOMEWORK 1 Chapter...

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