# Bài giảng Discrete structures for computer science - Chapter 2: Proving methods

Introduction Definition A proof is a sequence of logical deductions from - axioms, and - previously proved theorems that concludes with a new theorem.

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Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.1 Chapter 2 Proving Methods Discrete Structure for Computing (CO1007) (Materials drawn from Chapter 2 in: “Michael Huth and Mark Ryan. Logic in Computer Science: Modelling and Reasoning about Systems, 2nd Ed., Cambridge University Press, 2006.”) Nguyen An Khuong, Huynh Tuong Nguyen Faculty of Computer Science and Engineering University of Technology, VNU-HCM Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.2 Contents 1 Proof Methods 2 Homeworks and Exercises Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.3 Introduction Definition A proof is a sequence of logical deductions from - axioms, and - previously proved theorems that concludes with a new theorem. Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.4 Terminology • Theorem (định lý ) = a statement that can be shown to be true • Axiom (tiên đề ) = a statement we assume to be true • Hypothesis (giả thiết) = the premises of the theorem Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.5 • Lemma (bổ đề ) = less important theorem that is helpful in the proofs of other results • Corollary (hệ quả ) = a theorem that can be established directly from a proved theorem • Conjecture (phỏng đoán) = statement being proposed to be true, when it is proved, it becomes theorem Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.6 Proving a Theorem Many theorem has the form ∀xP (x)→ Q(x) Goal: • Show that P (c)→ Q(c) is true with arbitrary c of the domain • Apply universal generalization ⇒ How to show that conditional statement p→ q is true. Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.7 Methods of Proof • Direct proofs (chứng minh trực tiếp) • Proof by contraposition (chứng minh phản đảo) • Proof by contradiction (chứng minh phản chứng) • Mathematical induction (quy nạp toán học) Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.8 Direct Proofs Definition A direct proof shows that p→ q is true by showing that if p is true, then q must also be true. Example Ex.: If n is an odd integer, then n2 is odd. Pr.: Assume that n is odd. By the definition, n = 2k + 1, k ∈ Z. n2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 is an odd number. Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.9 Proof by Contraposition Definition p→ q can be proved by showing (directly) that its contrapositive, ¬q → ¬p, is true. Example Ex.: If n is an integer and 3n+ 2 is odd, then n is odd. Pr.: Assume that “If 3n+ 2 is odd, then n is odd” is false; or n is even, so n = 2k, k ∈ Z. Substituting 3n+ 2 = 3(2k) + 2 = 6k + 2 = 2(3k + 1) is even. Because the negation of the conclusion of the conditional statement implies that the hypothesis is false, Q.E.D. Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.10 Proofs by Contradiction Definition p is true if if can show that ¬p→ (r ∧ ¬r) is true for some proposition r. Example Ex.: Prove that √ 2 is irrational. Pr.: Let p is the proposition “ √ 2 is irrational”. Suppose ¬p is true, which means √ 2 is rational. If so, ∃a, b ∈ Z,√2 = a/b, a, b have no common factors. Squared, 2 = a2/b2, 2b2 = a2, so a2 is even, and a is even, too. Because of that a = 2c, c ∈ Z. Thus, 2b2 = 4c2, or b2 = 2c2, which means b2 is even and so is b. That means 2 divides both a and b, contradict with the assumption. Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.11 Problem Assume that we have an infinite domino string, we want to know whether every dominoes will fall, if we only know two things: 1 We can push the first domino to fall 2 If a domino falls, the next one will be fall We can! Mathematical induction. Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.12 Mathematical Induction Definition (Induction) To prove that P (n) is true for all positive integers n, where P (n) is a propositional function, we complete two steps: • Basis Step: Verify that P (1) is true. • Inductive Step: Show that the conditional statement P (k)→ P (k + 1) is true for all positive integers k Logic form: [P (1) ∧ ∀kP (k)→ P (k + 1))]→ ∀nP (n) What is P (n) in domino string case? Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.13 Example on Induction Example Show that if n is a positive integer, then 1 + 2 + . . .+ n = n(n+ 1) 2 . Solution Let P (n) be the proposition that sum of first n is n(n+ 1)/2 • Basis Step: P (1) is true, because 1 = 1(1+1) 2 • Inductive Step: Assume that 1 + 2 + . . .+ k = k(k+1) 2 . Then: 1 + 2 + . . .+ k + (k + 1) = k(k + 1) 2 + (k + 1) = k(k + 1) + 2(k + 1) 2 = (k + 1)(k + 2) 2 shows that P (k + 1) is true under the assumption that P (k) is true. Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.14 Example on Induction Example Prove that n < 2n for all positive integers n. Solution Let P (n) be the proposition that n > 2n. • Basis Step: P (1) is true, because 1 > 21 = 2 • Inductive Step: Assume that P (k) is true for the positive k, that is, k < 2k. Add 1 to both side of k < 2k, note that 1 ≤ 2k. k + 1 < 2k + 1 ≤ 2k + 2k = 2 · 2k = 2k+1. shows that P (k + 1) is true, namely, that k + 1 < 2k+1, based on the assumption that P (k) is true. Proving Methods Nguyen An Khuong, Huynh Tuong Nguyen Contents Proof Methods Homeworks and Exercises 2.15 Homeworks and Exercises 1 Cauchy inequality on means 2 Fibonacci in Pascal’s Triangle: Prove that Fn = C(n, 0) +C(n− 1, 1) +C(n− 2, 2) + . . .+C(dn/2e, bn/2c), where Fn is the nth Fibonacci number, F0 = F1 = 1. Notice that if C(a, b) = 0 for b > a, we can rewrite the desired result as Fn = ( n 0 ) + ( n− 1 1 ) + ( n− 2 2 ) + . . .+ ( 1 n− 1 ) + ( 0 n ) in order to have a simpler version to work with, and avoid considerations of whether n is even or odd. 3 Solve Exercises 7-11 in Huth and Ryan’s book 4 Solve exercises in the attachment 5 Try to solve as much as possible related exercises in Rosen’s book.