This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Induction II Margaret M. Fleck 1 March 2010 In this lecture, we see more examples of mathematical induction (section 4.1 of Rosen). 1 Announcements Homework 5 is hard. Even if you plan to do most of your work at the last minute, browse through it earlier to see if you understand how to get started. Were in the middle of grading the midterms. Watch the newsgroup for exactly when theyll be returned. 2 Recap If we want to prove a claim P ( n ), for all integers n , a proof by induction has the following outline: Proof: We will show P ( n ) is true for all n , using induction on n . Base : We need to show that P (1) is true. Induction : Suppose that P ( k ) is true, for some integer k . We need to show that P ( k + 1) is true. 1 3 Why is this legit? There are several ways to think about mathematical induction, and under stand why its a legitimate proof technique. Different people prefer different motivations at this point, so Ill offer several. A proof by induction of that P ( k ) is true for all positive integers k involves showing that P (1) is true (base case) and that P ( k ) P ( k + 1) (inductive step). Domino Theory: Imagine an infinite line of dominoes. The base step pushes the first one over. The inductive step claims that one domino falling down will push over the next domino in the line. So dominos will start to fall from the beginning all the way down the line. This process continues forever, because the line is infinitely long. However, if you focus on any specific domino, it falls after some specific finite delay. Recursion fairy: The recursion fairy is the mathematicians version of a programming assistant. Suppose you tell her how to do the proof for P (1) and also why P ( k ) implies P ( k + 1). Then suppose you pick any integer (e.g. 1034) then she can take this recipe and use it to fill in all the details of a normal direct proof that P holds for this particular integer. That is, she takes P (1), then uses the inductive step to get from P (1) to P (2), and so on up to P (1034). Smallest counterexample: Lets assume weve established that P (1) is true and also that P ( k ) implies P ( k +1). Lets prove that P ( j ) is true for all positive integers j , by contradiction. That is, we suppose that P (1) is true, also that P ( k ) implies P ( k + 1), but there is a counterexample to our claim that P ( j ) is true for all j . That is, suppose that P ( m ) was not true for some integer m . Now, lets look at the set of all counterexamples. We know that all the counterexamples are larger than 1, because our induction 2 proof established explicitly that P (1) was true. Suppose that the smallest counterexample is s . So P ( s ) is not true. We know that s > 1, since P (1) was true. Since s was supposed to be the smallest counterexample, s...
View
Full
Document
This note was uploaded on 09/21/2011 for the course CS 173 taught by Professor Erickson during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Erickson

Click to edit the document details