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C OURSE : D ISCRETE STRUCTURE CODE : ICS 252 Lecturer: Shamiel Hashim 1 lecturer:Shamiel Hashim second semester2011- 2012 Prepared by: amani Omer.

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Presentation on theme: "C OURSE : D ISCRETE STRUCTURE CODE : ICS 252 Lecturer: Shamiel Hashim 1 lecturer:Shamiel Hashim second semester2011- 2012 Prepared by: amani Omer."— Presentation transcript:

1 C OURSE : D ISCRETE STRUCTURE CODE : ICS 252 Lecturer: Shamiel Hashim 1 lecturer:Shamiel Hashim second semester2011- 2012 Prepared by: amani Omer

2 C OURSE TEXT BOOK Discrete Mathematics and its Application-6 th Edition - By :Kennth Rosen-McGraw-Hill Lecturer contact information Office : BLD 11A -203 2 lecturer:Shamiel Hashim second semester2011-2012

3 D ISCRETE S TRUCTURES D EFINITION 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. 3 lecturer:Shamiel Hashim second semester2011-2012

4 I MPORTANCE OF D ISCRETE S TRUCTURES 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 is essential in 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. 4 lecturer:Shamiel Hashim second semester2011-2012

5 C OURSE T OPICS : 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 5 lecturer:Shamiel Hashim second semester2011-2012

6 Proof And Strategies Chapter 1 6 lecturer:Shamiel Hashim second semester2011- 2012

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

8 I NTRODUCTION TO P ROOFS Outlines Proof Methods & types of proof Proof Strategies 8 lecturer:Shamiel Hashim second semester2011-2012

9 P ROOF M ETHODS 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. 9 lecturer:Shamiel Hashim second semester2011-2012

10 P ROOF M ETHODS 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 10 lecturer:Shamiel Hashim second semester2011-2012

11 P ROOF M ETHODS 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. 11 lecturer:Shamiel Hashim second semester2011-2012

12 P ROOF M ETHODS 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 a 2 = 2 b 2. Therefore a 2 must be even and so is a. let a=2c for some integer c, Now 2b 2 = 4c 2 Dividing both side by 2 give b 2 =2c 2 by definition of even now b 2 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. 12 lecturer:Shamiel Hashim second semester2011-2012

13 P ROOF M ETHODS 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). 13 lecturer:Shamiel Hashim second semester2011-2012

14 P ROOF M ETHODS 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 > 3 n, 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 3 n = 3 1 = 3; It follows 8 > 3; n = 2, (n + 1) 3 = (3) 3 = 27 and 3 n = 3 2 = 9; It follows 27 > 9; n = 3, (n + 1) 3 = (4) 3 = 64 and 3 n =3 3 = 27; It follows 64 > 27; n = 4, (n + 1) 3 = (5) 3 =125 and 3 n = 3 4 = 81; It follows 125>81 14 lecturer:Shamiel Hashim second semester2011-2012

15 P ROOF M ETHODS 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 n 2 ≥ 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 0 2 =0 therefore n 2 ≥ n holds. Case (ii) n ≥1, multiply both sides of inequality by positive n, we get n. n ≥ n.1, implies n 2 ≥ n hold for n ≥1. Case (iii) n ≤ -1, however n 2 ≥ 0, implies n 2 ≥n. Hence n 2 ≥n hold for all inequalities. 15 lecturer:Shamiel Hashim second semester2011-2012

16 P ROOF M ETHODS 4) Existence Proof: It is a proof a theorem with a statement involving the existential qualifier It has two types; a) 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 = 10 3 + 9 3 and also 1729 = 12 3 + 1 3 16 lecturer:Shamiel Hashim second semester2011-2012

17 P ROOF M ETHODS 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. 17 lecturer:Shamiel Hashim second semester2011-2012

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

19 P ROOF M ETHODS 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. 19 lecturer:Shamiel Hashim second semester2011-2012

20 P ROOF M ETHODS 20 lecturer:Shamiel Hashim second semester2011-2012

21 P ROOF S TRATEGIES 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. 21 lecturer:Shamiel Hashim second semester2011-2012

22 P ROOF S TRATEGIES Set to power 2 Multiply by 4 Extract first Add -4xy to each side 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: 22 lecturer:Shamiel Hashim second semester2011-2012

23 P ROOF S TRATEGIES 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. 23 lecturer:Shamiel Hashim second semester2011-2012

24 P ROOF S TRATEGIES 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 24 lecturer:Shamiel Hashim second semester2011-2012

25 P ROOF S TRATEGIES 25 lecturer:Shamiel Hashim second semester2011-2012

26 P ROOF S TRATEGIES 26 lecturer:Shamiel Hashim second semester2011-2012

27 P ROOF S TRATEGIES 27 lecturer:Shamiel Hashim second semester2011-2012

28 P ROOF S TRATEGIES 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 28 lecturer:Shamiel Hashim second semester2011-2012

29 P ROOF S TRATEGIES 29 lecturer:Shamiel Hashim second semester2011-2012

30 END 30 lecturer:Shamiel Hashim second semester2011-2012

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