SampleExam1-Solns

SampleExam1-Solns - Sample Exam 1 Solutions 1(a(Z is a...

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Sample Exam 1 Solutions 1. (a) ( Z , #) is a group. Clearly it is closed under # since the integers are closed under addition. x # - 2 = x and - 2# x = x for all x Z so - 2 is the identity. x #( y # z ) = x #( y + z +2) = x + y + z +4 and ( x # y )# z = ( x + y +2)# z = x + y + z +4 so it is associative. x #( - x - 4) = x - x - 4 + 2 = - 2 and ( - x - 4)# x = - x - 4 + x + 2 = - 2 so x - 1 = - x - 4 for all x , so the group has inverses. (b) ( Z , * ) is not a group. It fails to have associativity. ( x * y ) * z = (2 x +2 y ) * z = 4 x +4 y +2 z but x * ( y * z ) = x * (2 y + 2 z ) = 2 x + 4 y + 4 z . (it also fails to have an identity and thus fails to have inverses). 2. (a) (0 , 0) S so S 6 = Ø. Let ( x 1 ,y 1 ), ( x 2 ,y 2 ) S . ( x 1 ,y 1 )+( - x 2 , - y 2 ) = ( x 1 - x 2 ,y 1 - y 2 ). x 1 - x 2 + y 1 - y 2 = ( x 1 + y 1 ) - ( x 2 + y 2 ) = 0 (Since x 1 + y 1 = 0 and x 2 + y 2 = 0 because ( x 1 ,y 1 ) and ( x 2 ,y 2 ) S ). Thus ( x 1 ,y 1 ) + ( - x 2 , - y 2 ) S which implies
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This note was uploaded on 04/19/2011 for the course MATH 21373 taught by Professor Johnson during the Spring '11 term at Carnegie Mellon.

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SampleExam1-Solns - Sample Exam 1 Solutions 1(a(Z is a...

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