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construction some qus

# construction some qus - MATH 30A FALL 04 Homework This is...

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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 take-home final. Date : August 30, 2006. 1

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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 = d ” is a sentence.) 12) Decide whether these relations are (1) functions, (2) 1-1, (3) onto. a) is a function. It is not 1-1 since f ( a ) = f (2). It is not onto since 2 is not in the range. b) is a function but it is not 1-1 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 1-1 and onto). e) is a function which is neither 1-1 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. ( 1-1 ) 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 1-1. (The rule is that a differentiable function on an interval is 1-1 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, 0 < 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 0 = 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 0 6 = z 6 z 6 0 = - 64 - 64 = 1 . 35 if φ : U 8 Z 8 is an isomorphism with φ ( ζ ) = 5 then find φ ( ζ m ) for m = 0 , 3 , 4 , 5 , 6 , 7 For all m , ζ m = ζ m - 1 ζ . So, φ ( ζ m ) = φ ( ζ m - 1 + 8 φ ( ζ ) = φ ( ζ m - 1 + 8 5 We know that φ ( ζ 2 ) = 2. So, φ ( ζ 3 ) = φ ( ζ 2 ) + 8 5 = 2 + 8 5 = 7 φ ( ζ 4 ) = 7 + 8 5 = 12 - 8 = 4 φ ( ζ 5 ) = 4 + 8 5 = 9 - 8 = 1 φ ( ζ 6 ) = 1 + 8 5 = 6 φ ( ζ 7 ) = 6 + 8 5 = 11 - 8 = 3 φ ( ζ 0 ) = φ ( ζ 8 ) = 3 + 8 5 = 0 37 If φ : U 6 Z 6 is an isomorphism then explain why φ ( ζ ) = 4.

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