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Unformatted text preview: 3.1. Let P ( n ) be “ n < 100”. 3.2. Let Q ( n ) be ¬ P ( n ). By induction, Q ( n ) is true for all n ∈ N , so P ( n ) is false for all n ∈ N . 3.6. This is false. Let P ( n ) be “ n > 1.” 3.11. We use induction on n . If n = 1, then a set with n elements has precisely 2 = 2 1 subsets, namely itself and ∅ , which establishes the basis step. Now suppose the result holds for sets containing n elements and let S be a set containing n +1 elements. Then S 6 = ∅ , so there is some x ∈ S . Set T = P ( S{ x } ). By our induction hypothesis, T has 2 n elements since the elements of T are precisely the subsets of S{ x } , which is a set containing n elements. Define T x = { A ∪ { x } : A ∈ T } . We claim two things: ( i ) T x contains 2 n elements. ( ii ) P ( S ) is the disjoint union of T and T x . First we show ( i ). Clearly T x contains no more than 2 n elements since every element of T x is obtained by adding x to an element of T . On other hand, T x contains no fewer than 2...
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 Spring '10
 Math, Mathematical Induction, Recursion, Inductive Reasoning, Natural number, Mathematical logic, Structural induction

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