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Unformatted text preview: MATH 30A FALL 04 Homework This is homework from Math 30a, Fall 2004. I was teaching only one course. So I had a lot of time to prepare the course. • The book was John Fraleigh ”A first course in Abstract Alge bra,” 7th edition. We covered the first four chapters. • About 3/4 of the course was planned to be on group theory (Chapters 1,2,3). The last quarter (Chapter 4) covers basic concepts of rings and fields. • There were several practice quizzes, a midterm and a takehome final. Date : August 30, 2006. 1 1. Homework 1 State the questions and answer them in complete sentences or as you would speak. (Symbols are nouns and equations are also sentences. E.g, ” a = b but c 6 = d ” is a sentence.) 12) Decide whether these relations are (1) functions, (2) 11, (3) onto. a) is a function. It is not 11 since f ( a ) = f (2). It is not onto since 2 is not in the range. b) is a function but it is not 11 and not onto. c) is not a function since 1 maps to more than one element and 2,3 don’t map to anything. d) is a bijection (a function which is 11 and onto). e) is a function which is neither 11 nor onto. f) is not a function since 2 ∈ A maps to more than one element of B . 15 Show that the open interval S = (0 , 1) has the same cardinality as R . Proof. We will show that the function f : (0 , 1) → R is given by f ( x ) = tan(( x 1 / 2) π ) is a bijection. ( function ) tan y ∈ R is defined for all real y so f is a function. ( 11 ) The derivative of f ( x ) is , f ( x ) = πsec 2 ( x 1 / 2) π = π cos 2 ( x 1 / 2) π . This is finite and positive as long as the denominator is nonzero. But cos( x 1 / 2) π = 0 only if an odd multiple of π/ 2, i.e., when x is an integer. So f is 11. (The rule is that a differentiable function on an interval is 11 if (but not only if) its derivative is always positive on the interval.) ( onto ) Take any y ∈ R . Then π/ 2 < tan 1 y < π/ 2. So, < tan 1 y + π/ 2 π < 1 If we let x = tan 1 y + π/ 2 π then π ( x 1 / 2) = πx π/ 2 = tan 1 y and y = f ( x ). So, f is onto. 16 (a) The empty set has one subset (itself). In general a set with n elements has 2 n subsets. This was explained in class. 1 2. Homework 2 9) Calculate (1 i ) 5 using the binomial expansion. (1 i ) 5 = 1 5 i + 10 i 2 10 i 3 + 5 i 4 i 5 = 1 5 i 10 + 10 i + 5 i = 4 + 4 i 21 Find all solutions of the equation z 6 = 64. One solution is z = 2 e πi/ 6 = √ 3+ i . There are three other conjugates of this number given by √ 3 i, √ 3 + i, √ 3 i And there are two more solutions: ± 2 i . There are no other solutions since the ratio of any two solutions is a 6th root of unity: z z 6 = z 6 z 6 = 64 64 = 1 ....
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This note was uploaded on 03/24/2012 for the course MATH 203 taught by Professor Ankit during the Spring '12 term at Evergreen.
 Spring '12
 ankit
 Algebra

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