hw_3_sol_s16(1) - MAT 243 HW 3 SOLUTIONS(1(4 pts Fill in...

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MAT 243 HW 3 SOLUTIONS (1) (4 pts) Fill in the blank in the statements below: (a) P ( S ) is the set containing all subsets of S i.e., P ( S ) = { X | X S } (b) A - B = { x | x A x / B } (c) A function f : A B is one-to-one if and only if a A b A ( f ( a ) = f ( b ) a = b ) (d) A sequence is a function whose domain is the set N 0 , i.e., whose domain is the set of positive integers and zero. (2) (10 pts) Let A = { 1 , 2 } and B = { 2 , 3 } . Then A B = { 1 , 2 , 3 } and P ( A ) = {∅ , { 1 } , { 2 } , { 1 , 2 }} and P ( B ) = {∅ , { 2 } , { 3 } , { 2 , 3 }} . Then P ( A ) P ( B ) = {∅ , { 1 } , { 2 } , { 3 } , { 1 , 2 } , { 2 , 3 }} and P ( A B ) = {∅ , { 1 } , { 2 } , { 3 } , { 1 , 2 } , { 1 , 3 } , { 2 , 3 } , { 1 , 2 , 3 }} Thus the relation is: P ( A ) P ( B ) ⊆ P ( A B ) (3) (10 pts) Prove that f : Z Z ; f ( n ) = 3 m - 5 is one-to-one but not onto. (a) Suppose that n and m are arbitrary integers such that f ( n ) = f ( m ). Then f ( n ) = 3 n - 5 = 3 m - 5 = f ( m ), so 3 n = 3 m , thus n = m . Therefore f ( n ) is one-to-one. (b) Let m = 0. Suppose n is an arbitrary integer. Case 1: n 1. Then 3 m - 5 3(1) - 5 = - 2. Case 2: n 2. Then 3 m - 5 3(2) - 5 = 1. Thus 3 m - 5 6 = 0 for all integer n . Therefore f ( n ) is not onto. (4) (10 pts) Prove that g : R R ∪ { 0 } ; g (
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  • Spring '06
  • Callahan
  • Math, Sets, pts, Natural number, arbitrary real number

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