hw3_solutions

hw3_solutions - Math 110A homework 3 solutions October 21,...

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Math 110A homework 3 solutions October 21, 2010 § 2.2 1 (c) + 0 1 2 3 4 5 6 0 0 1 2 3 4 5 6 1 1 2 3 4 5 6 0 2 2 3 4 5 6 0 1 3 3 4 5 6 0 1 2 4 4 5 6 0 1 2 3 5 5 6 0 1 2 3 4 6 6 0 1 2 3 4 5 · 0 1 2 3 4 5 6 0 0 0 0 0 0 0 0 1 0 1 2 3 4 5 6 2 0 2 4 6 1 3 5 3 0 3 6 2 5 1 4 4 0 4 1 5 2 6 3 5 0 5 3 1 6 4 2 6 0 6 5 4 3 2 1 2 (a) In Z 8 , 0 2 = 0 1 2 = 1 2 2 = 4 3 2 = 9 = 1 4 2 = 16 = 0 5 2 = 25 = 1 6 2 = 36 = 4 7 2 = 49 = 1 So the solutions in Z 8 to x 2 = 1 are 1, 3, 5, and 7. 1
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(b) In Z 5 , 0 4 = 0 1 4 = 1 2 4 = 16 = 1 3 4 = ( - 2) 4 = 16 = 1 4 4 = ( - 1) 4 = 1 So the solutions in Z 5 to x 4 = 1 are 1, 2, 3, and 4. (c) Note that in Z 6 , x 2 + 3 x + 2 = ( x + 1)( x + 2). In Z 6 , (0 + 1)(0 + 2) = 1 · 2 = 2 (1 + 1)(1 + 2) = 2 · 3 = 6 = 0 (2 + 1)(2 + 2) = 3 · 4 = 12 = 0 (3 + 1)(3 + 2) = 4 · 5 = 4 · ( - 1) = - 4 = 2 (4 + 1)(4 + 2) = 5 · 6 = 5 · 0 = 0 (5 + 1)(5 + 2) = 6 · 7 = 0 · 1 = 0 So the solutions in Z 6 to ( x + 1)( x + 2) = 0 are 1, 2, 4, and 5. (d)
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This note was uploaded on 11/03/2010 for the course MATH 262-338-20 taught by Professor Gieseker,d. during the Fall '10 term at UCLA.

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hw3_solutions - Math 110A homework 3 solutions October 21,...

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