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Unformatted text preview: 2200 k P(k) 4.5 A proof by mathematical induction that the maximum number of nodes in a binary tree of height P(h) = 2 h+11 is true for all positive integers h. 1. Basis step. Show that proposition P(1) is true. P(1) = 2 1+11 = 3. As the following tree shows, the height is one because the root is not including and there are 3 nodes. 1 N N 2. Inductive step. For every positive k, the implication P(k) → P(k+1) is true. P(k) = 2 k+11 <=> P(k+1) = 2 (k+1)+11 2 k+11 <=> 2 (k+1)+11 (2 k + 2) 1 <=> (2 k + 2 2 ) 1 2 k + 2 <=> 2 k + 4 2<=> 4 2 1 <=> 2 1+1 2 k <=> 2 (k+1) 4.6 A proof by mathematical induction that P(n) filled nodes > P(n+1) null nodes is true for every positive integer n consists of 2 steps. 1. Basis step. Show that proposition P(1) is true. 1 N N 2. Inductive step. For every positive k, the implication P(k) → P(k+1) is true. P(k) for every k is proved by inductive hypothesis . [P(1) ∧ 2200 k (P(k) → P(k+1))] → 2200 k P(k)...
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 Spring '08
 M.McCullen
 Mathematical Induction, Recursion, Natural number, Mathematical logic, Peano axioms, inductive step

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