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Unformatted text preview: CS 70 Discrete Mathematics for CS Fall 2006 Papadimitriou & Vazirani Lecture 3 Induction Induction is an extremeley powerful tool in mathematics. It is a way of proving propositions that hold for all natural numbers (i.e., universally quantified ∀ k ∈ N ): 1) ∀ k ∈ N , 0 + 1 + 2 + 3 + ··· + k = k ( k +1) 2 2) ∀ k ∈ N , the sum of the first k odd numbers is k 2 . 3) Any graph with k vertices and k edges contains a cycle. Each of these propositions is of the form ∀ k ∈ N P ( k ). For example, in the first proposition, P ( k ) is the statement 0 + 1 + ··· + k = k ( k +1) 2 , P (0) says 0 = 0(0+1) 2 , P (1) says 0 + 1 = 1(1+1) 2 , etc. The principle of induction asserts that you can prove P ( k ) is true ∀ k ∈ N , by following these three steps: Base Case: Prove that P (0) is true. Inductive Hypothesis: Assume that P ( k ) is true. Inductive Step: Prove that P ( k + 1) is true. The principle of induction formally says that if P (0) and ∀ n ∈ N ( P ( n ) = ⇒ P ( n + 1)), then ∀ n ∈ N P ( n ). Intuitively, the base case says that P (0) holds, while the inductive step says that P (0) = ⇒ P (1), and P (1) = ⇒ P (2), and so on. The fact that this domino effect eventually shows that ∀ n ∈ N P ( n ) is what the principle of induction (or the induction axiom) states. In fact, dominoes are a wonderful analogy: we have a domino for each proposition P ( k ). The dominoes are lined up so that if the k th domino is knocked over, then it in turn knocks over the k +1 st . Knocking over the k th domino corresponds to proving P ( k ) is true. So the induction step corresponds to the fact that the k th domino knocks over the k + 1 st domino. Now, if we knock over the first domino (the one numbered 0), then this sets off a chain reaction that knocks down all the dominoes. Theorem: ∀ k ∈ N , k summationdisplay i =0 i = k ( k + 1) 2 . Proof (by induction on k ): Base Case: P (0) asserts: summationdisplay i =0 i = 0(0 + 1) 2 . This clearly holds, since the left and right hand sides both equal 0. 1 Inductive Hypothesis: Assume P ( k ) is true. That is, k summationdisplay i =0 i = k ( k + 1) 2 . Inductive Step: We must show P ( k + 1). That is, k +1 summationdisplay i =0 i = ( k + 1)( k + 2) 2 : k +1 summationdisplay i =0 i = ( k summationdisplay i =0 i ) + ( k + 1) = k ( k + 1) 2 + ( k + 1) (by the inductive hypothesis) = ( k + 1)( k 2 + 1) = ( k + 1)( k + 2) 2 Note the structure of the inductive step. You try to show P ( k +1) with the assumption that P ( k ) is true. The idea is that P ( k +1) by itself is a difficult proposition to prove. Many difficult problems in computer science are solved by breaking the problem into smaller, easier ones. This is precisely what we did in the inductive step  P ( k + 1) is difficult to prove, but we were able to recursively define it in terms of P ( k )....
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 Fall '08
 HAULLGREN
 Mathematical Induction, Recursion, Inductive Reasoning, Natural number, Mathematical proof, inductive hypothesis

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