Course: Discrete structure code :ics252

1. Course: Discrete structurecode :ics252 lecturer:Shamiel Hashim second semester2011-2012 Lecturer: Shamiel Hashim Prepared by: amani Omer

2. Course text book • Discrete Mathematics and its Application-6th Edition - By :Kennth Rosen-McGraw-Hill • Lecturer contact information Email:s.bakheit@uoh.edu.sa Office : BLD 11A -203 lecturer:Shamiel Hashim second semester2011-2012

3. Discrete Structures Definition • Discrete structure deals with discrete objects. Discrete objects are those which are separated from (not connected) each other. Examples: • Integers (whole numbers 5, 10, 15), rational numbers (ones that can be expressed as the share of two integers i.e. 10/5) are discrete object. lecturer:Shamiel Hashim second semester2011-2012

4. Importance of Discrete Structures • •It provides foundation objects for computer science. • •It includes important material from such areas as set theory, logic, graph theory, Relation …etc. • •It enables students to create and understand a proof • isessentialin formal specification, in verification, in databases, and in cryptography. • •The graph theory concepts are used in networks, operating systems, and compilers. • •Set theory concepts are used in software engineering and in databases. • •In engineering, It can be used to control multiproduct batch plants, production of new multifunctional and design of a new class of simulator. lecturer:Shamiel Hashim second semester2011-2012

5. Course Topics : • Ch1: 1.6-1.7 Proof methods and strategy • Ch3 :3.4-3.5-3.7 Number theory • Ch4 :Recursive • Ch8:Relations • Ch12:Modeling computation lecturer:Shamiel Hashim second semester2011-2012

6. ProofAnd Strategies Chapter 1 lecturer:Shamiel Hashim second semester2011-2012

7. Introduction to Proof Sec 1.6 lecturer:Shamiel Hashim second semester2011-2012

8. Introduction to Proofs Outlines • Proof Methods & types of proof • Proof Strategies lecturer:Shamiel Hashim second semester2011-2012

9. Proof Methods • Proof: A proof is a valid argument that establishes the truth of mathematical statement or theorem. There are two types of proofs; • Formal Proof: In this type all steps are supplied (complete) and rules for each step in the arguments are given. Useful theorems can be long and hard to follow. lecturer:Shamiel Hashim second semester2011-2012

10. Proof Methods • Informal Proof: Proof of theorems designed for the human consumption are always informal proofs. More than one rule of inference may be used in each step. Steps may be skipped. Rules of inference are not explicitly stated. • Proof Methods: The following are the proofs methods; 1) Direct Proof: It is a way of showing the truth of a given statement by a straightforward combination of established facts i.e for conditional statement p -> q is true by showing that if p is true then q must also be true lecturer:Shamiel Hashim second semester2011-2012

11. Proof Methods Example: Sum of two even integers is an even number. Proof: Let x and y are two even numbers. Since they are even therefore we can write x= 2a , y = 2b for all integers a and b. x + y = 2a + 2b = 2 (a + b) From this it is clear that x + y has 2 as a factor and therefore is even. Hence, sum of two even integers is an even number. lecturer:Shamiel Hashim second semester2011-2012

12. Proof Methods 2) Indirect Proof: It is also known as proof by contradiction . It is a form of proof that establishes the proof or validity of a proposition by showing that proposition is being false would involve a contradiction. A proposition must be either true or false and its falsity has been shown impossible, then proposition must be true. • Example1: prove that √2 is irrational using proof by contradiction . We start our proof by suppose that the negation of the proposition is true leads to contradiction i.e. it were rational, it could be expressed as a fraction a/b , where a and b are integers, a/b = √2, then a2 = 2b2. Therefore a2 must be even and so is a. let a=2c for some integer c , Now 2b2 = 4c2 Dividing both side by 2 give b2=2c2 by definition of even now b2 is even, and so is b. but we suppose that a/b is rational but 2 divides both a and b as they are even , so our assumption a/b = √2, leads to contradiction that 2 divides both a and b • So a, b is odd –by assumption- and even, a contradiction. Therefore the initial assumption—that √2 can be expressed as a fraction—must be false that mean √2 is irrational. lecturer:Shamiel Hashim second semester2011-2012

13. Proof Methods Example: Give an indirect proof to show that For all integers n, (if 3n + 1 is even, then n is odd). Solution: the first step in a proof by contradition is to assume the both p and not q are true ie 3n+1 is even and n is not odd (even). Or We can write for all integers n, 3n + 1 is even then n is even. If n is even mean n is multiple of 2, therefore n = 2a, for some integer integers a . Then 3n + 1 = 3(2a) +1 = 6a + 1 ………………………. (1) 6a is even because 2(3a). But 6a + 1 is odd. Therefore 3n + 1 is odd from equation (1) and By assuming n is even, we shown that 3n + 1 is odd which is an contradiction (p and ¬pis true) to our assumption . Therefore if n is odd then 3n +1 is even, which is possible. It follows that the original statement (if 3n + 1 is even, then n is odd is true). lecturer:Shamiel Hashim second semester2011-2012

14. Proof Methods 3) Exhaustive Proof : It is also known as proof by cases. It is a special type of proof by cases where each case involves checking a single example. Example 1: Prove that(n + 1)3 > 3n, if n is a positive integer with n ≤ 4. Solution: Proof by exhaustion need only to verify for n=1,2,3,4 n=1, (n + 1)3 = (2)3 = 8 and 3n = 31 = 3; It follows 8 > 3; n = 2, (n + 1)3 = (3)3 = 27 and 3n = 32 = 9; It follows 27 > 9; n = 3, (n + 1)3 = (4)3 = 64 and 3n=33 = 27; It follows 64 > 27; n = 4, (n + 1)3 = (5)3 =125 and 3n = 34 = 81; It follows 125>81 lecturer:Shamiel Hashim second semester2011-2012

15. Proof Methods 4) Proof by cases: A proof by cases must cover all cases that arises in a theorem. Example : Prove that if n is an integer then n2 ≥ n Solution: We can prove for every integer by considering three cases when n=0, n ≥1 and n ≤ -1 Case (i)n=0, because 02=0 therefore n2≥ n holds. Case (ii) n ≥1, multiply both sides of inequality by positive n, we get n. n ≥ n.1, implies n2 ≥ n hold for n ≥1. Case (iii) n ≤ -1, however n2 ≥ 0, implies n2 ≥n. Hence n2≥n hold for all inequalities. lecturer:Shamiel Hashim second semester2011-2012

16. Proof Methods 4) Existence Proof: It is a proof a theorem with a statement involving the existential qualifier It has two types; • Constructive Existence Proof: It proves the existence of a mathematical object with certain properties by creating or providing a method to create this object. Example :Show that there is a positive integer that can be written as the sum of cubes of positive integers in two different ways; Solution: After doing some computation we found that 1729 = 103+ 93 and also 1729= 123+ 13 lecturer:Shamiel Hashim second semester2011-2012

17. Proof Methods • b)Non-constructive Existence Proof: It proves the existence of a mathematical object with certain properties, but does not provide a means of constructing an example. lecturer:Shamiel Hashim second semester2011-2012

18. Proof Methods lecturer:Shamiel Hashim second semester2011-2012 Rather we have shown that either the pair have the desired property and we do not know which of these pairs work.

19. Proof Methods Uniqueness Proofs: It has two fundamental properties; a)Existence: We show that an element x with the desired property exists. b)Uniqueness: We show that if y ≠ x, then y does not have the desired property (of X) . Equivalently, we can show that if x and y both have the desired property, then x = y. lecturer:Shamiel Hashim second semester2011-2012

20. Proof Methods lecturer:Shamiel Hashim second semester2011-2012

21. Proof Strategies Proof Strategies: Generally, if the statement is a conditional statement, you should first try a direct proof; if this fails, you can try an indirect proof if neither of these approaches works , you might try a proof by contradiction. Forward and Backward Reasoning :To begin a direct proof of a conditional statement, you start with the premises (content). Using these premises, together with axioms and known theorems, you can construct a proof using a sequence of steps that leads to the conclusion. This type of reasoning is called forward reasoning (see example in direct proof). But forward reasoning is often difficult to use to prove more complicated results. In such cases it is helpful to use backward reasoning. lecturer:Shamiel Hashim second semester2011-2012

22. Proof Strategies Set to power 2 Multiply by 4 Extract first Add -4xy to each side lecturer:Shamiel Hashim second semester2011-2012 Because (x -y)2 > 0 when x <> y, it follows that the final inequality is true. Since all these inequalities are equivalent, it follows that (x + y)/2 > Sqrt( xy) when x <> y:

23. Proof Strategies Proof Strategies Looking for Counter examples : Counter examples show that certain statements are false. When confronted with a conjecture, you might first try to prove this conjecture, and if your attempts are unsuccessful, you might try to find a counterexample. Example 17:Show that the statement “ Every positive integer is the sum of the square of three integers” is false by finding a counterexample. lecturer:Shamiel Hashim second semester2011-2012

24. Proof Strategies Solution : We first look for a counterexample that “Every positive integer is the sum of three squares of integers” is false, If we find a particular integer that is not the sum of the squares of three integers. To look for a counterexample, we try successive positive integers as a sum of three squares, we find that lecturer:Shamiel Hashim second semester2011-2012

25. Proof Strategies lecturer:Shamiel Hashim second semester2011-2012

26. Proof Strategies lecturer:Shamiel Hashim second semester2011-2012

27. Proof Strategies lecturer:Shamiel Hashim second semester2011-2012

28. Proof Strategies Example 18: Can we tile the standard checkerboard using dominoes. How many ways we have to fill it? Solution: There are the following ways to tile the checkerboard. 1)Tile it by placing 32 dominoes horizontally. 2)Tile it by placing 32 dominoes vertically. 3)Tile it by placing some horizontally and some vertically dominoes. This method is called constructive existence proof lecturer:Shamiel Hashim second semester2011-2012

29. Proof Strategies lecturer:Shamiel Hashim second semester2011-2012

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