This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CS 702 Discrete Mathematics and Probability Theory Spring 2009 Alistair Sinclair, David Tse Note 3 Induction Induction is an extremely powerful tool in mathematics. It is a way of proving propositions that hold for all natural numbers: 1) ∀ k ∈ N , 0 + 1 + 2 + 3 + ··· + k = k ( k + 1 ) 2 2) ∀ k ∈ N , the sum of the first k odd numbers is a perfect square. 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 ( ) says 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 ( ) 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 ( ) and ∀ n ∈ N ( P ( n ) = ⇒ P ( n + 1 )) , then ∀ n ∈ N P ( n ) . Intuitively, the base case says that P ( ) holds, while the inductive step says that P ( ) = ⇒ 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. Let’s see some examples. Theorem: ∀ k ∈ N , k ∑ i = i = k ( k + 1 ) 2 . Proof (by induction on k ): • Base Case: P ( ) asserts: ∑ i = i = ( + 1 ) 2 . This clearly holds, since the left and right hand sides both equal 0. CS 702, Spring 2009, Note 3 1 • Inductive Hypothesis: Assume P ( k ) is true. That is, k ∑ i = i = k ( k + 1 ) 2 . • Inductive Step: We must show P ( k + 1 ) . That is, k + 1 ∑ i = i = ( k + 1 )( k + 2 ) 2 : k + 1 ∑ i = i = ( k ∑ i = i )+( k + 1 ) = k ( k + 1 ) 2 +( k + 1 ) (by the inductive hypothesis) = ( k + 1 )( k 2 + 1 ) = ( k + 1 )( k + 2 ) 2 . Hence, by the principle of induction, the theorem holds. ♠ 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 ) ....
View
Full Document
 Spring '08
 PAPADIMITROU
 Mathematical Induction, Inductive Reasoning, Natural number, Mathematical proof, inductive hypothesis

Click to edit the document details