ics141-lecture06-Sets

# ics141-lecture06-Sets - University of Hawaii ICS141:...

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6-1 ICS 141: Discrete Mathematics I (Spr 2008) University of Hawaii ICS141: Discrete Mathematics for Computer Science I Department of Information and Computer Sciences University of Hawaii Stephen Y. Itoga

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6-2 ICS 141: Discrete Mathematics I (Spr 2008) University of Hawaii Lecture 6 Chapter 1. The Foundations 1.6 Introduction to Proofs Chapter 2. Basic Structures 2.1 Sets
6-3 ICS 141: Discrete Mathematics I (Spr 2008) University of Hawaii Some material in these slides were taken/adapted from the slides made by Prof. Michael P. Frank and Prof. Jonathan L. Gross which are provided through the publisher of “Discrete Mathematics and Its Applications” written by Kenneth H. Rosen. Some material was done by Prof. Baek

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6-4 ICS 141: Discrete Mathematics I (Spr 2008) University of Hawaii Announcements Homework assignment #2 Due on Tuesday, Jan. 29 by start of class Quiz I Thursday, Jan. 31 Topic #1.0 – Propositional Logic: Operators
6-5 ICS 141: Discrete Mathematics I (Spr 2008) University of Hawaii Proof Methods for Implications For proving implications p q , we have: Direct proof: Assume p is true, and prove q . Indirect proof: Proof by Contraposition : Assume ¬ q , and prove ¬ p . Proof by Contradiction Vacuous proof: Prove ¬ p by itself. Trivial proof: Prove q by itself.

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6-6 ICS 141: Discrete Mathematics I (Spr 2008) University of Hawaii Direct Proof Example Definition: An integer n is called odd iff n =2 k +1 for some integer k ; n is even iff n =2 k for some k . Theorem: Every integer is either odd or even. This can be proven from even simpler axioms. Theorem: (For all numbers n ) If n is an odd integer, then n 2 is an odd integer. Proof: If n is odd, then n = 2 k +1 for some integer k . Thus, n 2 = (2 k +1) 2 = 4 k 2 + 4 k + 1 = 2(2 k 2 +2 k ) +
6-7 ICS 141: Discrete Mathematics I (Spr 2008) University of Hawaii Indirect Proof Example: Proof by Contraposition Theorem: (For all integers n ) If 3 n +2 is odd, then n is odd. Proof: Suppose that the conclusion is false, i.e. , that n is even. Then n =2 k for some integer k . Then 3 n +2 = 3(2 k )+2 = 6 k +2 = 2(3 k +1) . Thus 3 n +2 is even, because it equals 2 j for integer j = 3 k +1 . So 3 n +2 is not odd.

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## This note was uploaded on 09/12/2008 for the course ICS 141 taught by Professor Idk during the Fall '08 term at Hawaii.

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ics141-lecture06-Sets - University of Hawaii ICS141:...

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