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Unformatted text preview: CS 70 Discrete Mathematics for CS Spring 2008 David Wagner Note 4 Induction Induction is an extremely powerful tool in mathematics. It is a way of proving propositions that hold for all natural numbers, such as these: • ∀ k ∈ N . + 1 + 2 + 3 + ··· + k = k ( k + 1 ) / 2. • For every k ∈ N , the sum of the first k odd numbers is a perfect square. 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 for all k ∈ N by proving two facts: Base case: Prove that P ( ) is true. Induction step: Prove that P ( k ) = ⇒ P ( k + 1 ) is true for all k ∈ N . The implication P ( k ) = ⇒ P ( k + 1 ) is in turn normally proven by assuming that P ( k ) is true and showing that P ( k + 1 ) follows as a consequence. So, in practice, induction proofs follow these three steps: Base case: Prove that P ( ) is true. Inductive hypothesis: Assume that P ( k ) is true, where k is arbitrary. Inductive step: Prove that P ( k + 1 ) follows from the inductive hypothesis. The principle of induction formally says that if P ( ) and ∀ n ∈ N . ( P ( n ) = ⇒ P ( n + 1 )) are both true, then ∀ n ∈ N . P ( n ) is true. Intuitively, the base case says that P ( ) holds, while the inductive step says that P ( ) = ⇒ P ( 1 ) and P ( 1 ) = ⇒ P ( 2 ) and P ( 2 ) = ⇒ P ( 3 ) and so on. Intuitively, it makes sense that this is enough to prove that P ( n ) is true for every n : from P ( ) and P ( ) = ⇒ P ( 1 ) , we can deduce P ( 1 ) ; from P ( 1 ) and P ( 1 ) = ⇒ P ( 2 ) , we can deduce P ( 2 ) ; and so on. The principle of induction says that this domino effect eventually shows that P ( n ) is true for every n ∈ N . 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 statement that the k th domino knocks over the k + 1 st domino, for every k , or in other words, that there are no gaps in the line of dominoes. Now, if we knock over the first domino (the one numbered 0), and if there are no gaps in the line of dominoes, then this sets off a chain reaction that knocks down all the dominoes. Let’s see some examples. CS 70, Spring 2008, Note 4 1 Theorem: 2200 k ∈ N . k ∑ i = i = k ( k + 1 ) 2 . Proof : (by induction on k ) Base case: P ( ) states that ∑ i = i = ( + 1 ) 2 . This clearly is true, since the left and right hand sides both equal 0. Therefore, P ( ) is true....
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This note was uploaded on 04/09/2008 for the course CS 70 taught by Professor Papadimitrou during the Spring '08 term at University of California, Berkeley.
 Spring '08
 PAPADIMITROU

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