hw4_sol

hw4_sol - Problem Set 4 Solutions Problem 1 Base Case: n =...

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Unformatted text preview: Problem Set 4 Solutions Problem 1 Base Case: n = 1 We need to show that P (1) is true; that is, that f (1) = 12 . ￿ Left hand side: f (1) = 1 =1 (2k − 1) = 2(1) − 1 = 1 k Right hand side: 12 = 1 Inductive Step: We assume that P (n) is true. That is, we assume that f (n) = n2 . We want to show that P (n + 1) is true. f (n + 1) = ￿n+1 k=1 (2k − 1) = (2(n + 1) − 1) + = (2n + 2 − 1) + = (2n + 1) + ￿n ￿n k=1 (2k ￿n k=1 (2k k=1 (2k − 1) − 1) − 1) = (2n + 1) + f (n) = (2n + 1) + n2 since, by the inductive hypothesis, f (n) = n2 = n2 + 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 ∈ Z, x + y ∈ O} 3. P = {(x, y ) | x, y ∈ Z, xy ∈ O} 4. {(x, y ) | x, y ∈ O} ⊆ P 5. Note: for grading purposes, there are two possible solutions. The first would be correct 1 if the problem said ”any two integers x and y ” rather than ”any two numbers x and y ”; the second is correct if we interpret ”numbers” as meaning ”real numbers”. (1) S = P (2) {(x, y ) | x, y ∈ R, x + y ∈ O} = {(x, y ) | x, y ∈ R, xy ∈ O} 2 ...
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