Math Induction & Recursive Definitions in Discrete Math (CS 2233 Lecture 16), Assignments of Discrete Mathematics

Download Math Induction & Recursive Definitions in Discrete Math (CS 2233 Lecture 16) and more Assignments Discrete Mathematics in PDF only on Docsity! 1 Discrete Mathematical Structures CS 2233 Lecture Sixteen Prof. William Winsborough March 19, 2009 19 March 2009 Winsborough CS 2233 Lecture 16 2 Business • Homework 7 – 4.1: 4, 6 – 4.2: 4 • Read sections 4.1, 4.2 and 4.3 • Midterm II is in two weeks on 4/2 19 March 2009 Winsborough CS 2233 Lecture 16 3 Mathematical Induction • How can we show that a proposition P(n) holds for all natural numbers n ∈ N? • Proof technique called mathematical induction: – Basis: show P(0) – Inductive Step: show that for all k ∈ N, P(k) → P(k+1) • The proposition P(k) is called the induction hypothesis • (P(0) ∧ ∀k ∈ N.(P(k) → P(k+1))) → ∀n ∈ N.P(n) 19 March 2009 Winsborough CS 2233 Lecture 16 4 Example of Induction • Theorem: P(n) ≡ Σ 2i = 2n+1 – 1 • Proof by induction – Basis: P(0) ≡ Σ 2i = 20+1 – 1 ≡ 20 = 2-1, which clearly holds – Step: We assume P(k) ≡ Σ 2i = 2k+1 – 1 and show P(k+1) ≡ Σ 2i = 2k+2 – 1 as follows: Σ 2i = Σ 2i + 2k+1 = (2k+1 – 1) + 2k+1 by the induction hypothesis = 2· 2k+1 – 1 = 2k+2 – 1 0 ≤ i ≤ n 0 ≤ i ≤ 0 0 ≤ i ≤ k 0 ≤ i ≤ k+1 0 ≤ i ≤ k+1 0 ≤ i ≤ k 19 March 2009 Winsborough CS 2233 Lecture 16 5 Validity of Induction • Induction is valid because the natural numbers are well founded – Definition: A set is well ordered if each of its subsets has a least element • Once basis and step are shown, the assumption that the property fails for some values yields a contradiction – Assume for contradiction that P(0) ∧ ∀k(P(k) → P(k+1)) ∧ ¬∀nP(n) – Consider the least m ∈ N such that ¬P(m) • Case 1: if m = 0, the contradiction is immediate • Case 2: if m = k+1 for some k ∈ N, then by minimality of m P(k) holds. But this, together with the step, implies that P(k+1) ≡ P(m) holds, giving us the desired contradiction 19 March 2009 Winsborough CS 2233 Lecture 16 6 Strong Induction • To show ∀n∈N (P(n)), the following is sufficient: – Basis: show P(0) – Inductive Step: show that for all k ∈ N, [P(0) ∧…∧ P(k)] → P(k+1) • This gives us a stronger induction hypothesis to use in the step • It is valid for similar reasons to those shown on the previous slide