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Unformatted text preview: Math 3124 Thursday, September 1 First Homework Solutions 1. We have f : R R defined by f ( x ) = ( x 1 ) 2 . (a) f is onto. Indeed given a R , then a is well defined, 1 + a R , and f ( 1 + a ) = a . (b) f is not onetoone because, for example, f ( ) = f ( 2 ) = 1. (c) We claim that f ( R 2 ) = R 1 . If x 2, then f ( x ) 1. Conversely if a 1, then a is well defined, 1 + a 2, and f ( 1 + a ) = a . 2. Exercise 2.23 on page 19. Prove that if : S T , : T U , : T U , is onto, and = , then = . We are given = and that is onto. We need to prove that = , equiva lently ( t ) = ( t ) for all t T . But if t T , we may write t = ( s ) for some s S because is onto (of course, s depends on t ). We now have ( t ) = ( s ) = ( s ) = ( t ) which completes the proof. 3. Exercise 3.4 on page 23. Does the following define an operation on the set of integers? If yes, is it associative, is it commutative, and does it have an identity element? We have * defined on Z according to the formula m * n = mn 2 . (a) * is an operation because if m , n Z , then so is mn 2 . (b) * is not associative. We have 1 * ( 1 * 2 ) = 1 * 4 = 16, but ( 1 * 1 ) * 2 = 1 * 2 = 4. (c) * is not commutative. We have 1 * 2 = 4, but 2 * 1 = 2. (d) * does not have an identity. Suppose x is an identity for * . Then x * 2 = 2 (by definition of identity), yet x * 2 = 4 x (using the formula for * ), and we have a contradiction. Math 3124 Thursday, September 8 Second Homework Solutions 1. Exercise 4.2 on page 28. Let S = { a , b , c } and let A = { , , , } , where , , , and are the elements of M ( S ) defined as follows. ( a ) = a ( b ) = b ( c ) = c ( a ) = b ( b ) = a ( c ) = c ( a ) = a ( b ) = a ( c ) = a ( a ) = b ( b ) = b ( c ) = b (a) (b) Composition of mappings is an associative operation (Theorem 4.1(a) in the book). (c) is not commutative; for example = , yet = . (d) The identity element for is . 2. Problem 5.14 on page 34 Let H = { f : R R  f ( x ) = 0 for all x R } . For f , g H , define f g by ( f g )( x ) = f ( x ) g ( x ) for all x R . Then f g H . Verify that with this operation H is a group. How does this group differ from the group of invertible mappings in M ( R ) ? Also, is this group H abelian? First note that we do indeed have a binary operation, because if f , g H , then f g : R R , and f g ( x ) = f ( x ) g ( x ) = 0 for all x R . We now have the three axioms for a group to check....
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 Fall '08
 PARRY
 Math, Algebra

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