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**Unformatted text preview: **Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 Solutions to Homework Assignment #3 Solutions 1. Let F be a field. (a) Fix some a,b,c F . By commutativity of multiplication (ax- iom A1), distributivity of multiplication over addition (axiom AM1), and commutativity of multiplication again (A1), respec- tively, c ( a + b ) = ( a + b ) c = ac + bc = ca + cb . (b) Fix some a,b F such that ab = 0 and assume (without loss of generality) a 6 = 0. Then by the existence of a multiplicative inverse (axiom M4) a has an inverse element a- 1 such that a- 1 a = 1. By a claim proved in class, x 0 = 0 for any x F . Applying this with x = a- 1 and then using associativity (M2) and the defining property (M3) of 1 gives that 0 = a- 1 0 = a- 1 ( ab ) = ( a- 1 a ) b = 1 b = b . Thus, b = 0. (c) Fix some a,b,c F , with ac = bc and c 6 = 0. By the existence of multiplicative inverse, substitution, and assicativity of multipli- cation, ac = bc = ( ac ) c- 1 = ( bc ) c...

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