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**Discrete Mathematics I Lectures Chapter 6**

Some material adapted from lecture notes provided by Dr. Chungsim Han and Dr. Sam Lomonaco Dr. Adam Anthony Spring 2011

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**… and now for something completely different…**

Set Theory Actually, you will see that logic and set theory are very closely related.

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**Set Theory Set: Collection of objects (“elements”)**

aA “a is an element of A” “a is a member of A” aA “a is not an element of A” A = {a1, a2, …, an} “A contains…” Order of elements is meaningless It does not matter how often the same element is listed – repeats are OK!.

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Set Equality Sets A and B are equal if and only if they contain exactly the same elements. Examples: A = {9, 2, 7, -3}, B = {7, 9, -3, 2} : A = B A = {dog, cat, horse}, B = {cat, horse, squirrel, dog} : A B A = {dog, cat, horse}, B = {cat, horse, dog, dog} : A = B

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**Examples for Sets “Standard” Sets: Natural numbers N = {0, 1, 2, 3, …}**

Integers Z = {…, -2, -1, 0, 1, 2, …} Positive Integers Z+ = {1, 2, 3, 4, …} Real Numbers R = {47.3, -12, , …} Rational Numbers Q = {1.5, 2.6, -3.8, 15, …} (correct definition will follow)

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**Examples for Sets A = “empty set/null set”**

A = {z} Note: zA, but z {z} A = {{b, c}, {c, x, d}} A = {{x, y}} Note: {x, y} A, but {x, y} {{x, y}} A = {x | P(x)} “set of all x such that P(x)” A = {x | xN x > 7} = {8, 9, 10, …} “set builder notation”

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Examples for Sets We are now able to define the set of rational numbers Q: Q = {a/b | aZ bZ+} or Q = {a/b | aZ bZ b0} And how about the set of real numbers R? R = {r | r is a real number} That is the best we can do.

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**Exercise 1 What are the members of the following sets?**

A = {x| x R x2 = 5} B = {n | n Z n 2} Use set-builder notation to describe the following sets: The set of all positive integers that are divisible by 3 The set of all animals with black and white stripes

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**Why Sets? Sets add another layer of notational convenience**

Call it Logic 3.0! We can use sets to simplify logic expressions Let BW = {x | x is an animal and x has black and white stripes} All animals with black and white stripes are easy to see a BW, easy-to-see(a) There exists an animal without black and white stripes that is easy to see aBW, easy-to-see(a)

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**Subsets A B “A is a subset of B”**

A B if and only if every element of A is also an element of B. We can completely formalize this: A B x (xA xB) Examples: A = {3, 9}, B = {5, 9, 1, 3}, A B ? true A = {3, 3, 3, 9}, B = {5, 9, 1, 3}, A B ? true A = {1, 2, 3}, B = {2, 3, 4}, A B ? false

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**Subsets Useful rules: A = B (A B) (B A)**

(A B) (B C) A C (see Venn Diagram) U C B A

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**Subsets Useful rules: A for any set A A A for any set A**

Proper subsets: A B “A is a proper subset of B” A B x (xA xB) x (xB xA) or A B x (xA xB) x (xB xA)

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**Exercise 2 Is A B? Is A D? Is C A? Is D A? Is A = C? Is A = D?**

Let A = {1,3,4}, B = {1,2,4}, C = {4,3,1,1}, D = {1,2,3,4} Is 3A? Is {3} A? Is {3} A? Is 3 A? Is 3 B? Is 3 B? Is A B? Is A D? Is C A? Is D A? Is A = C? Is A = D?

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Exercise 3 Let A = {a | a Z a -3}, B = {b | b Z bk = 5 for some k Z+}, C = {c | c Z c is prime 7 < c < 10} Describe A, B and C by listing their elements Is B A? Is A B? Is C A?

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Cardinality of Sets If a set S contains n distinct elements, nN, we call S a finite set with cardinality n. Examples: A = {Mercedes, BMW, Porsche}, |A| = 3 B = {1, {2, 3}, {4, 5}, 6} |B| = 4 C = |C| = 0 D = { xN | x 7000 } |D| = 7001 E = { xN | x 7000 } E is infinite!

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**The Power Set (A) “power set of A”**

(A) = {B | B A} (contains all subsets of A) Examples: A = {x, y, z} (A) = {, {x}, {y}, {z}, {x, y}, {x, z}, {y, z}, {x, y, z}} A = (A) = {} Note: |A| = 0, | (A)| = 1

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The Power Set Cardinality of power sets: | (A) | = 2|A| Imagine each element in A has an “on/off” switch Each possible switch configuration in A corresponds to one element in 2A z y x 8 7 6 5 4 3 2 1 A For 3 elements in A, there are 222 = 8 elements in P(A)

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Exercise 4 Find the power sets ({1,2}) and ({1,2,3})

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Ordered Pairs We are familiar with ordered pairs in algebra when we plot on a Cartesian coordinate system: The pair (x,y) is ordered so that we know how to plot it on the axes Important: (a,b) = (c,d) a = c and b = d The order matters! (x,y)

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Cartesian Product The ordered n-tuple (a1, a2, a3, …, an) is an ordered collection of objects. Just a general version of the ordered pair: Two ordered n-tuples (a1, a2, a3, …, an) and (b1, b2, b3, …, bn) are equal if and only if they contain exactly the same elements in the same order, i.e. ai = bi for 1 i n. The Cartesian product of two sets is defined as: AB = {(a, b) | aA bB} Example: A = {x, y}, B = {a, b, c} AB = {(x, a), (x, b), (x, c), (y, a), (y, b), (y, c)}

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Cartesian Product The Cartesian product of two sets is defined as: AB = {(a, b) | aA bB} Example: A = {good, bad}, B = {student, prof} AB = { (good, student), (good, prof), (bad, student), (bad, prof)} BA = { (student, good), (prof, good), (student, bad), (prof, bad)}

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**Cartesian Product Note that: A = A = **

For non-empty sets A and B: AB AB BA |AB| = |A||B| The Cartesian product of two or more sets is defined as: A1A2…An = {(a1, a2, …, an) | aiA for 1 i n}

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**Set Operations Union: AB = {x | xA xB}**

Example: A = {a, b}, B = {b, c, d} AB = {a, b, c, d} Intersection: AB = {x | xA xB} AB = {b}

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Set Operations Two sets are called disjoint if their intersection is empty, that is, they share no elements: AB = If a set C = A B and A and B are disjoint, then the set P = {A,B} is referred to as a partition of C The difference between two sets A and B contains exactly those elements of A that are not in B: A-B = {x | xA xB} Example: A = {a, b}, B = {b, c, d}, A-B = {a}

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Set Operations Given a Universe identified by the set U, the complement of a set A contains exactly those elements under consideration that are not in A: Ac = U-A Example: U = N, B = {250, 251, 252, …} Bc = {0, 1, 2, …, 248, 249}

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Exercise 5 Let the Universe U = {0,1,2,3,…,9} and: A = {0,2,4,6,8} B = {1, 2, 4, 7} C = {1,5,7} Find the following: A B A B Bc A - B (A B)c A (B C) (A B) C Ac Bc

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Exercise 6 Let A = {1,2,3,4,5,6,7} Which collections form a partition of A? P1 = {1,3}, {2,5}, {4,6,7} P2 ={1}, {2,4,6}, {3,5} P3 ={1,3,5,7}, {2,4,6}, {3,6} Let B = {x | x R x -2 }, C = {x | x R -2 < x 4}, D = {x | x R x > 4} Is P4 = {B,C,D} a partition of R?

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**Venn Diagrams of Operations**

NOTE: The amount that A and B overlap is not always guaranteed A B

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**Venn Diagrams of Operations**

B

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**Venn Diagrams of Operations**

B A B

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**Venn Diagrams of Operations**

A B A B

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**Venn Diagrams of Operations**

A B

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**Venn Diagrams of Operations**

A - B

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**Venn Diagrams of Operations**

Ac

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Exercise 7 Draw a Venn Diagram to represent the following expressions over the sets A,B and C: A B = A B = A and C B C A B Ac – C A B and A C and B C BUT A B C

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**Section 6.2,6.3: Properties of Sets**

Unifying Logic and Sets Proofs with sets Set identities

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**Logic and Sets We can interpret our set operations logically:**

A B x A x B (Set-based implication) A B x A x B (Set-based OR) A B x A x B (Set-based AND) x Ac x A (Set-based Negation) A – B x A x B (NEW expressive power!) (x,y) A x B x A y B (NEW expressive power!)

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**Element Arguments Prove that for all sets A and B, A B A**

Normal starting Point: “Suppose A and B are arbitrarily chosen sets” Where does this get us? We make more progress using an ‘element argument’: Note that A B A translates to “If x A B then x A” From this, we get the statement: “Suppose A and B are arbitrarily chosen sets and x A B” Idea: focusing on a single arbitrary element in a set lets us prove something about every single element in the set!

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**Transitivity Revisited**

Prove, using logical interpretations, the transitive property for subsets: For all sets A, B and C, if A B and B C, then A C.

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**Proving Equality How can we Prove A(BC) = (AB)(AC)?**

Recall: A = B A B B A To prove A = B, you must show both subset claims are true So an equality proof really requires 2 element proofs!

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**Prove A(BC) = (AB)(AC).**

Prove that A(BC) (AB)(AC): Suppose that x is an arbitrarily chosen element of A(BC). There are 2 cases for x: x A or x (BC) Case 1: x A Since x A, by definition of union, x (A B) and x (A C). Therefore, by definition, x (AB)(AC). Case 2: x (BC). By definition of intersection, x B and x C. By definition of union, x (AB) and x (AC). Therefore, by definition of intersection, x (AB)(AC)

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**Prove A(BC) = (AB)(AC).**

Prove that (AB)(AC) A(BC) Drawing a Venn diagram might help us think this through… Suppose that x is an arbitrarily chosen element of (AB)(AC). We’ll analyze this one element in two cases: x A and x A. Case 1: x A It immediately follows that x A(BC) by definition of union. Case 2: x A Since x (AB)(AC), it must be true that x (AB) and x (AC). Since x A in this case, it must follow that x B and x C or else x would not be a member of (AB)(AC). Therefore, by definition of intersection, x (BC) and by definition of Union, x A(BC)

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**OR…Use logic instead! How can we prove A(BC) = (AB)(AC)?**

Suppose x is an arbitrary element of A(BC). Then by definition, the following statements are true: x(A(BC)) (xA) (x(BC)) (xA) (xB xC) (xA xB) (xA xC) (distributive law for logical expressions) (x(AB)) (x(AC)) x((AB)(AC)) Since each statement is a biconditional with the one before it, x(A(BC)) x((AB)(AC)) and x((AB)(AC)) x(A(BC)) . So by definition, (A(BC)) ((AB)(AC)) and ((AB)(AC)) (A(BC)). Therefore, A(BC) = (AB)(AC)

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Which method to use? You may use an element argument or logic conversions to prove set identities on assignments and exams Which one you choose depends on your strengths Neither approach is ‘easier’ than the other Element arguments provide more direction about where you’re headed Logic arguments have less direction, but tend to be more concise

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**Set Identities—Look Familiar?**

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Exercise 1 Use Venn Diagrams to verify the following set-theoretic DeMorgan’s law: (A B)c = Ac Bc

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Exercise 2 Recall that two sets are disjoint if their intersection is the empty set (A B = ). Draw venn diagrams for the sets A – B and B – A Use Theorem to construct a proof that A – B and B – A are disjoint

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**Exercise 3 Use Theorem 6.2.2 to prove the following:**

(A (B C) ((A B) – C) = A B

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**A Technology Tie-In Sailors**

Don’t forget your set theory—You’ll use it in Databases! Each ‘table’ in a database is called a relation, filled with tuples Database theory says that a relation is actually a subset of the universe of all possible tuples Sailors

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A Technology Tie-In SQL (structured query language): Select SID, SNAME, RATING, AGE FROM SAILORS WHERE AGE > 40 S = {X SAILORS | AGE > 40} See the similarity? Remembering set-theory will really help you work with databases Sailors

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EE1J2 – Discrete Maths Lecture 7

EE1J2 – Discrete Maths Lecture 7

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