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Unformatted text preview: CS 70 Discrete Mathematics for CS Spring 2004 Papadimitriou/Vazirani Lecture 2 This lecture covers induction, which is the most common technique for proving universally quantified sen tences over the natural numbers and other discrete sets of objects (lists, trees, tilings, graphs, programs, etc.). We need induction, or something like it, because any attempt at proof by enumeration of possible worlds is doomed to failurewith infinitely many objects, there are infinitely many possible worlds that can be defined. The principle of induction The principle of induction is an axiom of the natural numbers. Recall from Lecture 1 that the natural numbers satisfy Peanos axioms (where we have written n + 1 instead of s ( n ) for clarity): 0 is a natural number If n is a natural number, n + 1 is a natural number These axioms dont quite define the natural numbers, because they include no statement of the kind, ... and by the way, there are no other natural numbers. This makes it impossible to use these axioms alone to prove that any property holds for all natural numbers n . The principle of induction essentially fills this gap. Informally, it says the following: If you can prove that some property holds for 0, and you can prove that the property holds for n + 1 if it holds for n , then you have proved that the property holds for all the natural numbers. To state this principle formally, let P ( n ) be an arbitrary proposition about the natural number n . (For exam ple, P ( n ) might be 2 n > n .) The principle of induction is as follows: Axiom 2.1 (Induction) : For any property P , if P ( ) and 2200 n N ( P ( n ) = P ( n + 1 )) , then 2200 n N P ( n ) . This says that if P ( ) holds, and P ( ) = P ( 1 ) , and P ( 1 ) = P ( 2 ) , and so on, then P ( n ) must be true for all n . This seems pretty reasonable. (Later we will see that the axiom has an alternative form that seems even more reasonable.) Inductive proofs The standard first example is to verify the wellknown formula for the sum of the first n positive integers: Theorem 2.1 : 2200 n N n i = 1 i = n ( n + 1 ) / 2 [Note: n i = 1 i means 1 + 2 + + n but is more precise. When n = 0 the summation has no terms so is 0; when n = 1 it has just the term i = 1 so the sum is 1.] In many, but not all, cases of proof by induction, the property P used in the induction is exactly the one we want to prove. That is the case here: P ( n ) is the proposition that n i = 1 i = n ( n + 1 ) / 2. CS 70, Spring 2004, Lecture 2 1 To construct an inductive proof, first we need to establish the base case P ( ) , then we need to prove that BASE CASE P ( n + 1 ) follows from P ( n ) . The latter proof is called the inductive step , and usually involves all the work....
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This note was uploaded on 08/27/2008 for the course CS 1050 taught by Professor Huang during the Spring '05 term at Georgia Institute of Technology.
 Spring '05
 HUANG

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