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CS 70 Lecture 03

# CS 70 Lecture 03 - CS 70 Spring 2010 Induction Discrete...

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CS 70 Discrete Mathematics and Probability Theory Spring 2010 Alistair Sinclair 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 ( 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. Let’s see some examples. Theorem: k N , k i = 0 i = k ( k + 1 ) 2 . Proof (by induction on k ): • Base Case: P ( 0 ) asserts: 0 i = 0 i = 0 ( 0 + 1 ) 2 . This clearly holds, since the left and right hand sides both equal 0. CS 70, Spring 2010, Note 3 1

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• Inductive Hypothesis: Assume P ( k ) is true. That is, k i = 0 i = k ( k + 1 ) 2 . • Inductive Step: We must show P ( k + 1 ) . That is, k + 1 i = 0 i = ( k + 1 )( k + 2 ) 2 : k + 1 i = 0 i = ( k 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 . Hence, by the principle of induction, the theorem holds. Note the structure of the inductive step. You try to show P ( k + 1 ) under 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 ) . We will now look at another proof by induction, but first we will introduce some notation and a definition for divisibility. We say that integer a divides b (or b is divisible by a ), written as a | b , if and only if for some integer q , b = aq .
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