hw4_solutions

hw4_solutions - f ( n ) = n 2 = n 2 + 2 n + 1 = ( n + 1) 2...

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Problem Set 4 Solutions Problem 1 Base Case: n =1 We need to show that P (1) is true; that is, that f (1) = 1 2 . Left hand side: f (1) = ° 1 k =1 (2 k 1) = 2(1) 1=1 Right hand side: 1 2 =1 Inductive Step: We assume that P ( n ) is true. That is, we assume that f ( n )= n 2 .W ewan t to show that P ( n + 1) is true. f ( n +1)= ° n +1 k =1 (2 k 1) =(2( n +1) 1) + ° n k =1 (2 k 1) =(2 n +2 1) + ° n k =1 (2 k 1) =(2 n +1)+ ° n k =1 (2 k 1) =(2 n +1)+ f ( n ) =(2 n +1)+ n
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Unformatted text preview: f ( n ) = n 2 = n 2 + 2 n + 1 = ( n + 1) 2 Problem 2 1. H ⊆ B 2. B ∩ H ° ° = ∅ 3. ( B ∩ H ) ⊆ E 4. ( E ° ∩ H ) ⊆ B 5. B ⊆ ( H ∪ E ° ) 6. ( H ° ∪ B ) ∩ E ° ° = ∅ Problem 3 1. T = { x + y | x,y ∈ O } 2. S = { ( x,y ) | x + y ∈ O } 3. P = { ( x,y ) | xy ∈ O } 4. { ( x,y ) | x,y ∈ O } ⊆ P 5. S = P 1...
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This note was uploaded on 11/21/2011 for the course CSE 20 taught by Professor Foster during the Spring '08 term at UCSD.

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