2. Empty Set, Partitions, Power Set

2. Empty Set, Partitions, Power Set
5.1.1 Chapter 5 - Set Theory 1. Basic Definitions 2. Empty Set, Partitions, Power Set 3. Properties of Sets

Section 1. Basic Definitions
5.1.2 Section 1. Basic Definitions A Set is a collection of items, called elements. {1, 2, 3} {x Î R | x2 > 5} S = {Tom, Sue, Jim}

We use ellipses to simplify things: {1, 2, 3, …, 10} {1, 2, 3, …}
5.1.3 We use ellipses to simplify things: {1, 2, 3, …, 10} {1, 2, 3, …} {…, -2, -1, 0, 1, 2, …} Be careful! ({1, 2, …}???)

5.1.4 We relate an item in the set with the set using the “Î (element of)” relation. x Î {x, y, z} {1, 2} Î {{1, 2}, {1, 2, 3}}.

5.1.5 Special Sets We refer to specific sets of numbers so often that we give them special names. These sets, and their corresponding symbols, will be referenced throughout this course.

We define the Natural Numbers to be: N = {0, 1, 2, 3, …}
5.1.6 Natural Numbers We define the Natural Numbers to be: N = {0, 1, 2, 3, …} Note that the Naturals are “closed” under addition and multiplication.

We define the Integers to be: Z = {…, -2, -1, 0, 1, 2, 3, …}
5.1.7 The Integers We define the Integers to be: Z = {…, -2, -1, 0, 1, 2, 3, …} Note that Z is “closed” under addition, subtraction, and multiplication.

We define the Rationals to be: Q = {p/q | p,q Î Z and q ¹ 0}
5.1.8 The Rational Numbers We define the Rationals to be: Q = {p/q | p,q Î Z and q ¹ 0} Note that Q is “closed” under addition, subtraction, multiplication, and non-zero division.

5.1.9 The Rational Numbers Alternatively, we can view Q as the set of all infinite, repeating decimal expansions. 7.35 = … Î Q … Î Q However, p Ï Q

The Irrational Numbers
5.1.10 The Irrational Numbers I = {all infinite, nonrepeating decimals} Obviously, irrational numbers are impossible to write down exactly. We use symbols to represent special values such as p, e, and Ö2. The Irrationals are not closed under + or ´.

The Real Numbers R = {all decimal expansions}
5.1.11 The Real Numbers R = {all decimal expansions} The Real Numbers are created by adjoining the Rationals with the Irrationals. The Reals are closed under all operations and satisfy the Field Axioms (see Appendix A, p. 695). The Reals form a continuum: we use the Real Number Line to represent this.

5.1.12 The Complex Numbers The Reals fall short when solving simple polynomial equations like x2 + 1 = 0. The Complex Numbers patch this hole. C = {a + bi | a,b Î R and i = Ö (-1)} Use the Complex Plane to represent these numbers. The Complex Numbers are also a field.

5.1.13 Subsets If A and B are sets, A is called a subset of B, denoted A Í B, provided every element of A is an element of B. So, A Í B means "x, if x Î A, then x Î B. We also say, “A is contained in B” or “B contains A” to show this relationship. Equivalently, we denote A Ë B provided $x ' x Î A and x Ï B.

Examples of Subsets {{1}, {2}} Í {{1}, {2}, {1,2}}.
5.1.14 Examples of Subsets If A = {1, 2, 3} and B = {0, 1, 2, 3, 4}, then clearly A Í B. {{1}, {2}} Í {{1}, {2}, {1,2}}. Q Í R and Z Í Q and N Í Z. {a, b, c} is a proper subset of {a, b, c, d}. {a, b, c} is an improper subset of {a, b, c}. We denote interval subsets of R as [a, b) = {x Î R | a £ x < b}. So [2, 5) Í [0,5].

5.1.15 Set Equality We say sets A and B are equal (A = B) if every element of A is in B and every element of B is in A. Thus, A = B means A Í B and B Í A. For example {1, 2, 3} = {1, 2, 3}, but A = {1, 2, 3} ¹ {1, 2, 3, 4} = B, since 4 Î B but 4 Ï A. Also, [a, b) ¹ [a, b] since b is only in [a, b].

5.1.16 Operations on Sets Given sets A and B, which are subsets of a universal set, U, we define the following: (Union) A È B = {x Î U | x Î A or x Î B}. (Intersection) A Ç B = {x Î U | x Î A and x Î B}. (Difference or Relative Complement) A - B = {x Î U | x Î A and x Ï B}. (Complement) Ac = {x Î U | x Ï A}. Note that Ac = U - A.

Examples of Set Operations
5.1.17 Examples of Set Operations Let U = R, A = [1, 3] and B = (2, 4). A È B = [1, 4) A Ç B = (2, 3] A - B = [1, 2] B - A = (3, 4) Ac = (-¥, 1) È (3, ¥) Bc = (-¥, 2] È [4, ¥)

5.1.18 Cartesian Products Given two sets, A and B, we define the Cartesian Product, A ´ B = {(a, b) | a Î A and b Î B}. The element (a, b) is called an ordered pair, since (a, b) and (b, a) are distinct if a ¹ b. If A = {1, 2, 3} and B = {8, 9}, then: A ´ B = {(1, 8), (1, 9), (2, 8), (2, 9), (3, 8), (3, 9)} B ´ A = {(8, 1), (8, 2), (8, 3), (9, 1), (9, 2), (9, 3)}

Generalized Cartesian Products
5.1.19 Generalized Cartesian Products Given three sets, A, B, and C, we define their Cartesian Product by A ´ B ´ C = {(a, b, c) | a Î A, b Î B and c Î C}. Although similar, we note that A ´ B ´ C and (A ´ B) ´ C are not, technically, the same since one contains (a, b, c) and the other ((a, b), c). In general, we define A1Á2Á3´...Án to be {(a1,a2,a3,…,an) | a1ÎA1,a2ÎA2,a3ÎA3,…, anÎAn}

5.1.20 Formal Languages Let S be a finite set, which we will, henceforth, call an alphabet. A string of characters of the alphabet S (or a string over S ) is either: (1) an ordered n-tuple of elements of S written without parentheses or commas, or (2) the null string e, which has no characters.

Formal Languages (cont’d.)
5.1.21 Formal Languages (cont’d.) If S = {1, 2, 3}, then is a string of length 6 over S. ® (1, 3, 1, 2, 2, 1) Î S´S´S´S´S´S. Clearly, the length of a string, s, over an alphabet S, is the number of characters of S that are in s. We denote the function L(s) to be this length. Hence, if S = {0, 1}, then L( ) = 9.

Formal Languages (cont’d.)
5.1.22 Formal Languages (cont’d.) Any set of strings over an alphabet is called a formal language over the alphabet. Let S be an alphabet and n Î N : Sn = {strings over S with L(s) = n}; Gn = {strings over S with L(s) £ n}; S* = {strings over S of finite length}.

Examples Let S = {0, 1}: S3 = {000, 001, 010, 011, 100, 101, 110, 111}
5.1.23 Examples Let S = {0, 1}: S3 = {000, 001, 010, 011, 100, 101, 110, 111} G3 = {e, 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111} S* = {e, 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111, …, , …, , ...}.

Section 5.3 The Empty Set Partitions Power Sets Boolean Algebras
5.3.24 Section 5.3 The Empty Set Partitions Power Sets Boolean Algebras

5.3.25 The Empty Set The unique set containing no elements is called the empty set, denoted Æ or {}. Theorem: If A is a set, then Æ Í A. Corollary: Æ is unique. Why? Since Æ1 Í Æ2 and Æ2 Í Æ1 we conclude Æ1 = Æ2 .

Set Operations With Æ For any set A from a universal set U: A Ç Æ = Æ
5.3.26 Set Operations With Æ For any set A from a universal set U: A Ç Æ = Æ A È Æ = A A Ç Ac = Æ A È Ac = U Uc = Æ and Æc = U.

5.3.27 Partitions of a Set Two sets are called disjoint if they have no elements in common. That is, A and B are disjoint provided A Ç B = Æ. Theorem: If A and B are any sets, then (A - B) and B are disjoint. A collection of sets {A1,A2,…,An} is called mutually or pairwise disjoint if Ai Ç Aj = Æ whenever i ¹ j.

Partitions of a Set (cont’d.)
5.3.28 Partitions of a Set (cont’d.) A collection of sets {A1,A2,…,An} is called a partition of a set A provided: 1. {A1,A2,…,An} is mutually disjoint; 2. A1 È A2 È ... È An = A. {{1, 2, 3},{4, 5},{6, 7, 8}} partitions? {{1,2,3,...},{-1,-2,-3,...},{0}} partitions? {Q, I} partitions?

5.3.29 Power Sets If A is a set, the Power Set of A, denoted P (A), is the set of all subsets of A. Since Æ Í A, we conclude Æ Î P (A), and A Í A implies A Î P (A). If A = {0,1}, P (A) = {Æ,{0},{1},{0,1}} Theorem: If A and B are sets with A Í B, then P (A) Í P (B). Theorem: If |A| = n, then | P (A) | = 2n.

5.3.30 Boolean Algebras If A is a set, the collection {A, +, ×} is called a Boolean Algebra if: 1. " a,bÎA, a + b = b + a and a × b = b × a 2. " a,b,cÎA, (a + b) + c = a + (b + c) and (a × b) × c = a × (b × c) 3. " a,b,cÎA, a + (b × c) = (a + b) × (a + c) and a × (b + c) = (a × b) + (a × c) 4. $! 0,1ÎA ' " aÎA, a + 0 = a and a × 1 = a 5. " aÎA, $ bÎA ' a + b = 1 and a × b = 0

Chapter 1. Symbolic Logic
1.1.31 Chapter 1. Symbolic Logic Logical Form and Equivalence Conditional Statements Valid and Invalid Arguments Digital Logic Circuits (Boolean Polynomials)

Logic of Compound Statements
1.1.32 Logic of Compound Statements A statement (or proposition) is a sentence that is true (T) or false (F), but not both or neither. Examples: Today is Monday. x is even and x > 7. If x2 = 4, then x = 2 or x = -2.

1.1.33 Counterexamples If a sentence cannot be judged to be T or F or is not even a sentence, it cannot be a statement. Examples: Open the door! (imperative) Did you open the door? (interrogative) If x2 = 4. (fragment)

Compound Statements Denote statements using the symbols p, q, r, ...
1.1.34 Compound Statements Denote statements using the symbols p, q, r, ... Denote the operations Ù, Ú, ~, ® (to be defined shortly), where: p Ù q - conjunction of p and q (p and q); p Ú q - disjunction of p and q (p or q); ~ p - negation of p (not p); p ® q - implication of p and q (p implies q);

Compound Statements (cont’d.)
1.1.35 Compound Statements (cont’d.) A Compound statement (or statement form) is a statement which includes at least one operation and one other “atomic” statement. For example, “x = 7 and y = 2” is a compound statement based on the “atomic” statements p = “x = 7” and q = “y = 2”. In this instance, we can symbolize the compound statement as r = p Ù q.

Compound Statements (cont’d.)
1.1.36 Compound Statements (cont’d.) The Truth Table of a compound statement is the collection of all the output truth values corresponding to all possible combinations of input truth values of the atomic statements. Since each atomic statement can take on 1 of 2 values, 2 inputs have 4 combinations, 3 inputs have 8, 4 inputs have 16, 5 inputs have 32, etc.

Logical Operations Negation: p ~p T F F T
1.1.37 Logical Operations Negation: p ~p T F F T Conjunction: • Disjunction: p q (p Ù q) p q (p Ú q) T T T T T T T F F T F T F T F F T T F F F F F F

1.1.38 Example: (p Ú q) Ù ~r Proceed from left to right: p q r (p Ú q) ~r (p Ú q) Ù ~r T T T T F F T T F T T T T F T T F F T F F T T T F T T T F F F T F T T T F F T F F F F F F F T F

1.1.39 Logical Equivalence Two compound statements are logically equivalent if they have the same truth table. We denote this as p º q. p ~p ~(~p) T F T F T F hence p º ~(~p). ~(p Ù q) º ~p Ù ~q ? No, since ~(T Ù F) º T, but (~T Ù ~F) º F.

Tautology & Contradiction
1.1.40 Tautology & Contradiction A statement whose truth table is all “T” is called a tautology, denoted as p º t. A statement whose truth table is all “F” is called a contradiction, denoted as p º c. Clearly, ~t º c and ~c º t. Are all logical statements either tautology or contradiction?

Algebra of Symbolic Logic
1.1.41 Algebra of Symbolic Logic Commutative Laws: p Ù q º q Ù p p Ú q º q Ú p Associative Laws: (p Ù q) Ù r º p Ù (q Ù r) (p Ú q) Ú r º p Ú (q Ú r) Distributive Laws: p Ù (q Ú r) º (p Ù q) Ú (p Ù r) p Ú (q Ù r) º (p Ú q) Ù (p Ú r)

1.1.42 Algebra of Symbolic Logic Identity Laws: p Ù t º p p Ú c º p Negation Laws: p Ù ~p º c p Ú ~p º t Double Negative Laws: ~(~p) º p Negations of t and c: ~t º c ~c º t

1.1.43 Algebra of Symbolic Logic Idempotent Laws: p Ù p º p p Ú p º p DeMorgan’s Laws: ~(p Ù q) º ~p Ú ~q ~(p Ú q) º ~p Ù ~q Universal Bound Laws: p Ù c º c p Ú t º t Absorption Laws: p Ù (p Ú q) º p p Ú (p Ù q) º p

Section 1.2 Conditional Statements
1.2.44 Section 1.2 Conditional Statements Logical Equivalences Involving Conditionals Converses, Inverses, and Contrapositives Biconditional Statements

Conditional Statements
1.2.45 Conditional Statements If p and q are statement variables, the conditional or implication of q by p is “If p then q” or “p implies q” and is denoted by p ® q. The truth table of the implication operator is: p q p ® q T T T T F F F T T F F T Example: If you mow my lawn, I’ll pay $20.

Hypotheses & Conclusions
1.2.46 Hypotheses & Conclusions In the form “If p then q” the statement p is called the hypothesis and the statement q is the conclusion. Conditionals form the basis of “deductive” reasoning. (Aristotilean Logic) In looking at the truth table, we consider the cases where the hypothesis is false to yield vacuous results. The interesting cases are when the hypothesis is true.

1.2.47 Logical Equivalences In the framework of symbolic logic, the implication operator would seem to be a new and distinct process. However, this is not the case! Theorem: p ® q º ~p Ú q. Thus, we can always rewrite an implication as a disjunction. Corollary: ~(p ® q) º p Ù ~q.

Negation of a Conditional
1.2.48 Negation of a Conditional From the previous corollary, the negation of p ® q is p Ù ~q. For example, the negation of If today is Sunday, then I wash my car. is: Today is Sunday and I do not wash my car.

Converse of a Conditional
1.2.49 Converse of a Conditional Given the statement p ® q, we define its converse to be the statement q ® p. For example, the converse of If today is Sunday, then I wash my car. is: If I wash my car, then today is Sunday.

Contrapositive of a Conditional
1.2.50 Contrapositive of a Conditional Given the statement p ® q, we define its contrapositive to be the statement ~q ® ~p. For example, the contrapositive of If today is Sunday, then I wash my car. is: If I do not wash my car, then today is not Sunday.

Inverse of a Conditional
1.2.51 Inverse of a Conditional Given the statement p ® q, we define its inverse to be the statement ~p ® ~q. For example, the inverse of If today is Sunday, then I wash my car. is: If today is not Sunday, then I do not wash my car.

1.2.52 Equivalent Forms Theorem: Given the statement p ® q, we have that p ® q º ~q ® ~p. Corollary: Given the statement p ® q, we have that q ® p º ~p ® ~q. Therefore from the above, we see that a conditional and its contrapositive are logically equivalent. Moreover, the statement’s converse and inverse forms are logically equivalent to each other.

Biconditional Statements
1.2.53 Biconditional Statements Definition: Given the statement variables p and q, the biconditional of p and q is read, “p if and only if q,” denoted p « q and means that both p ® q and q ® p . By direct calculation: p q p « q T T T T F F F T F F F T

Using the Biconditional
1.2.54 Using the Biconditional Looking closely at the truth table, we see that p « q is T whenever p and q have the same truth value. Theorem: p « q is a tautology implies p º q and conversely, p º q implies p « q is a tautology. This gives us a systematic way to calculate logical equivalence, rather then just scan the matches of truth values by eye.

Section 1.3 Valid and invalid argument forms.
1.3.55 Section 1.3 Valid and invalid argument forms. Special valid argument forms. Dilemmas Fallacies. Contradictions and valid arguments.

Valid and Invalid Arguments
1.3.56 Valid and Invalid Arguments An argument (or argument form) is a sequence of statements. All statements but the final one are called premises, assumptions, or hypotheses. The final statement is called the conclusion. An argument form is valid provided its conclusion is always true whenever all of its premises are true.

Valid and Invalid Arguments
1.3.57 Valid and Invalid Arguments An argument (or argument form) is a sequence of statements. All statements but the final one are called premises, assumptions, or hypotheses. The final statement is called the conclusion. An argument form is valid provided its conclusion is always true whenever all of its premises are true. The truth of the conclusion follows inescapably from the truth of the hypotheses.

Testing for Validity Identify the premises and conclusion.
1.3.58 Testing for Validity Identify the premises and conclusion. Construct a truth table for the premises and conclusion. Find the critical rows, where all premises are T. For each critical row, if the conclusion is also T, then the argument is valid. If at least one critical row leads to a conclusion being F, the argument is invalid. If there are no critical rows, the argument is vacuously valid.

A Valid Argument p Ú (q Ú r) ~r \ (p Ú q)
1.3.59 A Valid Argument p Ú (q Ú r) ~r \ (p Ú q) Truth Table: p q r [p Ú (q Ú r)] ~r (p Ú q) T T T T F T T T F T T T T F T T F T T F F T T T F T T T F T F T F T T T F F T T F F F F F F T F

An Invalid Argument p Ú (q Ú r) ~r \ (p Ú r)
1.3.60 An Invalid Argument p Ú (q Ú r) ~r \ (p Ú r) Truth Table: p q r [p Ú (q Ú r)] ~r (p Ú r) T T T T F T T T F T T T T F T T F T T F F T T T F T T T F T F T F T T F F F T T F T F F F F T F

Special Argument Forms
1.3.61 Special Argument Forms Modus Ponens: p ® q p \ q Truth Table: p q p ® q p q T T T T T T F F T F F T T F T F F T F F Premises: If today is Sunday, then I was my car. Today is Sunday. Conclusion: I wash my car.

Modus Tollens Modus Tollens: p ® q ~q \ ~p
1.3.62 Modus Tollens Modus Tollens: p ® q ~q \ ~p Truth Table: p q p ® q ~q ~p T T T F F T F F T F F T T F T F F T T T Premises: If today is Sunday, then I was my car. I do not wash my car. Conclusion: Today is not Sunday.

Disjunctive Addition Disjunctive Addition: p \ p Ú q
1.3.63 Disjunctive Addition Disjunctive Addition: p \ p Ú q Truth Table: p q p Ú q T T T T F T F T T F F F Premise: Today is Sunday. Conclusion: Today is Sunday or I wash my car.

Conjunctive Simplification
1.3.64 Conjunctive Simplification Conjunctive Simplification: p Ù q \ p also \ q Truth Table: p q p Ù q T T T T F F F T F F F F Premise: Today is Sunday and I wash my car. Conclusion 1: Today is Sunday. Conclusion 2: I wash my car.

Disjunctive Syllogism
1.3.65 Disjunctive Syllogism Disjunctive Syllogism: p Ú q p Ú q ~p ~q \ q \ p Truth Table: p q p Ú q ~p T T T F T F T F F T T T F F F T Premises: Today is Sunday or Saturday. Today is not Sunday. Conclusion: Today is Saturday.

Hypothetical Syllogism
1.3.66 Hypothetical Syllogism Hypothetical Syllogism: p ® q q ® r \ p ® r Premises: If x is an integer, then x is a rational. If x is a rational, then x is a real. Conclusion: If x is an integer, then x is real.

Dilemma: Division Into Cases
1.3.67 Dilemma: Division Into Cases Dilemma: p Ú q p ® r q ® r \ r Premises: x is positive or x is negative. If x is positive , then x2 is positive. If x is negative, then x2 is positive. Conclusion: x2 is positive.

Application: Find My Glasses
1.3.68 Application: Find My Glasses 1. If my glasses are on the kitchen table, then I saw them at breakfast. 2. I was reading in the kitchen or I was reading in the living room. 3. If I was reading in the living room, then my glasses are on the coffee table. 4. I did not see my glasses at breakfast. 5. If I was reading in bed, then my glasses are on the bed table. 6. If I was reading in the kitchen, then my glasses are on the kitchen table.

Find My Glasses (cont’d.)
1.3.69 Find My Glasses (cont’d.) Let: p = My glasses are on the kitchen table. q = I saw my glasses at breakfast. r = I was reading in the living room. s = I was reading in the kitchen. t = My glasses are on the coffee table. u = I was reading in bed. v = My glasses are on the bed table.

Find My Glasses (cont’d.)
1.3.70 Find My Glasses (cont’d.) Then the original statements become: 1. p ® q 2. r Ú s 3. r ® t 4. ~q 5. u ® v 6. s ® p and we can deduce (why?): 1. p ® q 2. s ® p 3. r Ú s 4. r ® t ~q ~p ~s r \ ~p \ ~s \ r \ t Hence the glasses are on the coffee table!

1.3.71 Fallacies A fallacy is an error in reasoning that results in an invalid argument. Three common fallacies: Using vague or ambiguous premises; Begging the question; Jumping to a conclusion. Two dangerous fallacies: Converse error; Inverse error.

Converse Error If Zeke cheats, then he sits in the back row.
1.3.72 Converse Error If Zeke cheats, then he sits in the back row. Zeke sits in the back row. \ Zeke cheats. The fallacy here is caused by replacing the impication (Zeke cheats ® sits in back) with its biconditional form (Zeke cheats « sits in back), implying the converse (sits in back ® Zeke cheats).

Inverse Error If Zeke cheats, then he sits in the back row.
1.3.73 Inverse Error If Zeke cheats, then he sits in the back row. Zeke does not cheat. \ Zeke does not sit in the back row. The fallacy here is caused by replacing the impication (Zeke cheats ® sits in back) with its inverse form (Zeke does not cheat ® does not sit in back), instead of the contrapositive (does not sit in back ® Zeke does not cheat).

1.3.74 Contradiction Rule If you can show that assuming statement p is false leads logically to a contradiction, then you can conclude that p is true. In argument form: ~p ® c \ p This is the logical heart of the proof method called Proof by Contradiction.

Section 1.4 Digital Logic Circuits Boolean Polynomials
1.4.75 Section 1.4 Digital Logic Circuits Boolean Polynomials Normal Forms (Disjunctive/Conjunctive) Designing Circuits with Specified Conditions Showing Two Circuits Are Equivalent

Digital Logic Circuits
1.4.76 Digital Logic Circuits Developed by Claude Shannon in 1938 to model telephone switching circuits: x AND y x y Series Switch x x OR y y Parallel Switch

1.4.77 Logical Gates Instead of working with switches, we model digital circuits using gates: AND-gates, OR-gates, and NOT-gates. We draw these as: x y x + y OR x x’ NOT x y xy AND

1.4.78 Notation Modeling digital circuits leads to the equivalent analysis of symbolic logic. Symbolic Logic Digital Circuits T, t 1, 1 F, c 0, 0 p, q, r, x, y, z, ... ~p x’ p Ù q xy p Ú q x + y

1.4.79 Boolean Polynomials When modeling, we use Boolean polynomials to describe algebraically the function of a combinatorial circuit. A combinatorial circuit is one in which the output at any time depends on the inputs at the previous time. (i.e. no feedback loops) A Boolean polynomial is a function which takes 0,1 inputs and outputs a 0 or 1 using the operations AND, OR, and NOT.

Examples of Boolean Polynomials
1.4.80 Examples of Boolean Polynomials When working with Boolean polynomials, we must first know the specific input variables. Examples: f(x,y,z) = x + y + z f(x,y) = x’ + xy f(x,y,z) = x(y + z’)

Evaluating Boolean Polynomials
1.4.81 Evaluating Boolean Polynomials Using: x y x’ y’ (x + y) xy Examples: Find f(x,y) = x’ + xy x y x’ xy (x’ + xy)

1.4.82 Normal Forms Expressing a Boolean Polynomial in its normal form provides an easy method to calculate its truth table. We can create two different normal forms for Boolean Polynomials: the disjunctive and the conjunctive normal form. These forms are made up of special terms called minterms or maxterms.

Disjunctive Normal Form
1.4.83 Disjunctive Normal Form A minterm is a Boolean polynomial that is only the product of each variable or its negation (but not both). Examples: f(x,y) = xy’ f(x,y,z) = x’yz’ f(w,x,y,z) = wx’y’z The disjunctive normal form (DNF) is a Boolean polynomial that is the sum of minterms (sum of products).

Disjunctive Normal Form (cont’d.)
1.4.84 Disjunctive Normal Form (cont’d.) Express f(x,y,z) = x + x’z in its DNF. f(x,y,z) = x + x’z = x(y + y’)(z + z’) + x’(y + y’)z = (xy + xy’)(z + z’) + (x’y + x’y’)z = xyz + xyz’ + xy’z + xy’z’ + x’yz + x’y’z The thing to note here is that each minterm has an output of 1 at only a single, particular line of the truth table. i.e. xy’z = 1 at 101 and = 0 elsewhere.

Disjunctive Normal Form (cont’d.)
1.4.85 Disjunctive Normal Form (cont’d.) We can now think of the inputs, in fact, as their associated minterms to get outputs: x y z x y z f(x,y,z) 1 1 1 x y z 1 1 0 x y z’ 1 0 1 x y’z 1 0 0 x y’z’ 0 1 1 x’y z 0 1 0 x’y z’ 0 0 1 x’y’z 0 0 0 x’y’z’

Designing Circuits with Specified Conditions
1.4.86 Designing Circuits with Specified Conditions In the other direction: x y z f(x,y,z) ® f(x,y,z) = xy’z + x’yz + x’yz’ + x’y’z

Conjunctive Normal Form
1.4.87 Conjunctive Normal Form In a similar fashion, we can analyze functions using the conjunctive normal form - the product of sums. In this case, we look for the 0’s in the function’s output and associate each with a maxterm, whose output is 0 at that row.

1.4.88 Equivalent Circuits Two logical circuits are equivalent if and only if they have the same truth table. This can be thought similarly as holding when the two circuits have the same disjunctive (conjunctive) normal form.

Section 5.2 Properties of sets
5.2.89 Section 5.2 Properties of sets Methods to show one set is a subset of another Set identities Methods to show two sets are equal

Some Subset Relations For all sets A, B, and C:
5.2.90 Some Subset Relations For all sets A, B, and C: 1. A Ç B Í A and A Ç B Í B 2. A Í A È B and B Í A È B 3. If A Í B and B Í C, then A Í C.

Procedural Versions of the Set Operations
5.2.91 Procedural Versions of the Set Operations x Î A È B means x Î A or x Î B. x Î A Ç B means x Î A and x Î B. x Î A - B means x Î A and x Ï B. x Î Ac means x Ï A. (x, y) Î A ´ B means x Î A and y Î B.

Example: Show A Ç B Í A Let x Î A Ç B. Show x Î A.
5.2.92 Example: Show A Ç B Í A Let x Î A Ç B. Show x Î A. x Î A Ç B means x Î A and x Î B. In particular, this means x Î A. Hence, given x Î A Ç B, we deduce that x Î A. Therefore A Ç B Í A.

Set Identities Commutative Laws A Ç B = B Ç A and A È B = B È A
5.2.93 Set Identities Commutative Laws A Ç B = B Ç A and A È B = B È A Associative Laws (A Ç B) Ç C = A Ç (B Ç C) (A È B) È C = A È (B È C) Distributive Laws A È (B Ç C) = (A È B) Ç (A È C) A Ç (B È C) = (A Ç B) È (A Ç C)

Set Identities (cont’d.)
5.2.94 Set Identities (cont’d.) Intersection with U A Ç U = A Universal Bound A È U = U Double Complement Law (Ac)c = A Idempotent Laws A È A = A and A Ç A = A

Set Identities (cont’d.)
5.2.95 Set Identities (cont’d.) DeMorgan’s Laws (A Ç B)c = Ac È Bc and (A È B)c = Ac Ç Bc Set Difference Law A - B = A Ç Bc Absorption Laws A È (A Ç B) = A A Ç (A È B) = A

Basic Method to Show Set Equality
5.2.96 Basic Method to Show Set Equality Let sets A and B be given. Show A = B. First, show A Í B. Second, show B Í A. If the “Í” holds in both directions, then we can conclude that A = B.

Example 1: A È (B Ç C) = (A È B) Ç (A È C)
5.2.97 Example 1: A È (B Ç C) = (A È B) Ç (A È C) First, show A È (B Ç C) Í (A È B) Ç (A È C). Then, show (A È B) Ç (A È C) Í A È (B Ç C).

Example 2: If A Í B, then A È B = B and A Ç B = A
5.2.98 Example 2: If A Í B, then A È B = B and A Ç B = A First, show A Ç B Í A. Then, show A Í A Ç B.

5.2.99 (A È B) - C = (A - C) È (B - C) To show these sets are equal, we will simply apply the Properties of Sets. (A È B) - C = (A È B) Ç Cc = (A Ç Cc) È (B Ç Cc ) = (A - C) È (B - C )

Chapter 2. The Logic of Quantified Statements
Chapter 2. The Logic of Quantified Statements Predicates Quantified Statements Valid Arguments and Quantified Statements

Section 1. Predicates and Quantified Statements I
Section 1. Predicates and Quantified Statements I In Chapter 1, we studied the logic of compound statements, but the argument reasoning in there cannot show the validity of the following simple argument: All men are mortal. Socrates is a man. Therefore, Socrates is mortal.

Predicates To study these types of logical arguments, we turn to predicate calculus. A predicate is a sentence that contains a finite number of variables and becomes a statement when specific values are substituted for the variables. The domain of a predicate variable is the set of all values that may be substituted in place of the variable.

Predicate Notation If P(x) is a predicate and x has a domain D, the truth set of P(x) is the set of all elements of D that make P(x) true when substituted for x. The truth set is denoted {x Î D | P(x)}. If P(x) and Q(x) are predicates and the common domain of x is D, then the notation P(x) Þ Q(x) denotes that the truth set of P(x) is a subset of the truth set of Q(x). If P(x) and Q(x) have the same truth set, we denote this as P(x) Û Q(x).

The Universal Quantifier
The Universal Quantifier We often find predicates involved when we are making claims about properties that some or all the elements of a set obey. This leads us to look at statements using one of two quantifiers. The Universal Quantifier: If P(x) is a predicate over a domain D, we say a universal statement is one of the form “"x Î D, P(x).” This universal statement is true provided P(x) is true for every x in D. Any x Î D with P(x) false, is a counterexample.

Examples Example 1: Let D = {1,2,3,4,5} and let P(x) be the predicate x2 ³ x. Using the Method of Exhaustion, we find that 12 ³ 1, 22 ³ 2, 32 ³ 3, 42 ³ 4, and 52 ³ 5 are all true, hence the universal statement "x Î {1,2,3,4,5}, x2 ³ x is true. Example 2: If we change this universal statement to: "x Î R, x2 ³ x, it is no longer true since x = 1/2 is a counterexample.

The Existential Quantifier
The Existential Quantifier The Existential Quantifier: If P(x) is a predicate over a domain D, we say an existential statement is one of the form “$x Î D ' P(x).” This existential statement is true provided P(x) is true for at least one x in D, and is false if P(x) is false for every x in D. From this, we see that the negation of an existential statement is a universal statement, and, likewise, the negation of a universal statement is an existential one.

More Examples Consider: $x Î D ' x2 < 0.
More Examples Consider: $x Î D ' x2 < 0. Example 1: If D = C (the Complex numbers), then x = i yields i2 = (-1) < 0, hence the existential statement is true. Example 2: If D = R, then by the properties of R, we know that x2 ³ 0 for all x in R, hence the existential statement is false. This second example show us the negation of $ x Î R ' x2 < 0 is the universal statement "x Î R, x2 ³ 0.

Negations of Quantifiers
Negations of Quantifiers As seen in the previous example, the negation of an existential statement is a universal statement. Formally, we denote: ~[$x Î D ' P(x)] º "x Î D, ~P(x). By the same process, we have that: ~["x Î D, P(x)] º $x Î D ' ~P(x). Intuitively, the first says the opposite of at least one thing satisfying a property is that none do, and the opposite of all things satisfying the property is that at least one does not.

Examples of Negations The negation of: Some people are sad. is All people are not sad. The negation of: All integers are rational. is At least one integer is irrational. Which of each pair is true?

Universal Conditional
Universal Conditional The statement: "x, if P(x), then Q(x) is called the universal conditional. Many mathematical statements are universal conditionals. Example: "x Î R, if x > 2 then x2 > 4 (formal) is equivalent to: (informally) Every real number greater than 2 has a square greater than 4. The square of any real number greater than 2 is greater than 4.

Negation of Quantified Conditionals
Negation of Quantified Conditionals Since we see the properties of symbolic logic carry over when dealing with quantified logic, we deduce that: ~["x Î D, if P(x), then Q(x)] is $x Î D ' P(x) and ~Q(x). Similarly, ~[$x Î D ' if P(x), then Q(x)] is "x Î d, P(x) and ~Q(x). Negate: 1. Every CS student studies CMSC Some CS students study CMSC203.

Section 2 - More Quantified Statements
Section 2 - More Quantified Statements Statements with multiple quantifiers; Negations of multiply quantified statements; Equivalent forms of universal conditionals.

Multiply Quantified Statements
Multiply Quantified Statements Consider the following statement: Given any real number, there is a smaller real number. This is equivalent to the formal statement: " xÎR, $ yÎR ' y < x. This is an example of a multiply quantified statement.

Examples The formal statement: $ xÎR+ ' " yÎR+, y < x can be interpreted informally as: There is a non-negative real number with the property that all other non-negative real numbers are smaller than this number; There is a non-negative real number that is larger than all other non-negative real numbers.

Another Example INFORMAL: Everybody loves somebody.
Another Example INFORMAL: Everybody loves somebody. FORMAL: " people x, $ a person y ' x loves y. INFORMAL: Somebody loves everybody. FORMAL: $ a person x ' " people y, x loves y.

Negation of Universal Existentials
Negation of Universal Existentials What is the negation of the statement: " people x, $ a person y such that x loves y? Recall this is “Everybody loves somebody,” so its negation would be the case of “Somebody who does not love anybody.” In formal terms: $ a person x ' " people y, x does not love y. Thus: ~ [" x, $ y ' P(x,y) ] º $ x ' " y, ~P(x,y)

Negation of Existential Universals
Negation of Existential Universals What is the negation of the statement: $ a person x such that " people y, x loves y? Recall this is “Somebody loves everybody,” so its negation would be the case of “Everybody has at least one person they do not love.” In formal terms: " people x, $ person y ' x does not love y. Thus: ~ [$ x ' " y, P(x,y) ] º " x, $ y ' ~P(x,y)

Equivalent Forms of Universal Conditionals
Equivalent Forms of Universal Conditionals Given the statement: " xÎD, if P(x), then Q(x) analogous to our definitions from propositional calculus, we can construct the following. Contrapositive: " xÎD, if ~Q(x), then ~P(x). Converse: " xÎD, if Q(x), then P(x). Inverse: " xÎD, if ~P(x), then ~Q(x). Negation: $ xÎD ' P(x), and ~Q(x).

Example Statement: " xÎR, if x > 2, then x2 > 4.
Example Statement: " xÎR, if x > 2, then x2 > 4. Converse: " xÎR, if x2 > 4, then x > 2. Inverse: " xÎR, if x £ 2, then x2 £ 4. Contrapositive: " xÎR, if x2 £ 4, then x £ 2. Negation: $ xÎR ' x > 2 and/but x2 £ 4.

Section 3 - Valid Arguments
Section 3 - Valid Arguments Argument Forms; Diagrams to Test for Validity; Quantified Converse and Inverse Errors; Abduction.

Universal Instantiation
Universal Instantiation Consider the following statement: All men are mortal Socrates is a man. Therefore, Socrates is mortal. This argument form is valid and is called universal instantiation. In summary, it states that if P(x) is true for all xÎD and if aÎD, then P(a) must be true.

Universal Modus Ponens
Universal Modus Ponens Formal Version: " xÎD, if P(x), then Q(x). P(a) for some aÎD \ Q(a). Informal Version: If x makes P(x) true, then x makes Q(x) true. a makes P(x) true. \ a makes Q(x) true. The first line is called the major premise and the second line is the minor premise.

Universal Modus Tollens
Universal Modus Tollens Formal Version: " xÎD, if P(x), then Q(x). ~Q(a) for some aÎD \ ~P(a). Informal Version: If x makes P(x) true, then x makes Q(x) true. a makes Q(x) false. \ a makes P(x) false.

Examples Universal Modus Ponens or Tollens???
Examples Universal Modus Ponens or Tollens??? If a number is even, then its square is even. 10 is even. Therefore, 100 is even. 25 is odd. Therefore, 5 is odd.

Using Diagrams to Show Validity
Using Diagrams to Show Validity Does this diagram portray the argument of the second slide? Mortals Men Socrates

Modus Ponens in Pictures
Modus Ponens in Pictures For all x, P(x) implies Q(x). P(a). Therefore, Q(a). {x | Q(x)} {x | P(x)} a

A Modus Tollens Example
A Modus Tollens Example All humans are mortal. Zeus is not mortal. Therefore, Zeus is not human. Zeus Mortals Humans

Modus Tollens in Pictures
Modus Tollens in Pictures For all x, P(x) implies Q(x). ~Q(a). Therefore, ~P(a). {x | Q(x)} a {x | P(x)}

Converse Error in Pictures
Converse Error in Pictures All humans are mortal. Felix the cat is mortal. Therefore, Felix the cat is human. Mortals Felix? Humans Felix?

Inverse Error in Pictures
Inverse Error in Pictures All humans are mortal. Felix the cat is not human. Therefore, Felix the cat is not mortal. Felix? Mortals Felix? Humans

Quantified Form of Converse and Inverse Errors
Quantified Form of Converse and Inverse Errors Converse Error: " x, P(x) implies Q(x). Q(a), for a particular a. \ P(a). Inverse Error: " x, P(x) implies Q(x). ~P(a), for a particular a. \ ~Q(a).

An Argument with “No” Major Premise: No Naturals are negative.
An Argument with “No” Major Premise: No Naturals are negative. Minor Premise: k is a negative number. Conclusion: k is not a Natural number. Negative numbers Natural numbers k

Abduction Major Premise: All thieves go to Paul’s Bar.
Abduction Major Premise: All thieves go to Paul’s Bar. Minor Premise: Tom goes to Paul’s Bar. Converse Error: Therefore, Tom is a thief. Although we can’t conclude decisively if Tom is a thief or not, if we have further information that 99 of the 100 people in Paul’s Bar are thieves, then the odds are that Tom is a thief and the converse error is actually valid here. This is called abduction by Artificial Intelligence researchers.

Chapter 3 - Elementary Number Theory and Proofs
Chapter 3 - Elementary Number Theory and Proofs Direct & Indirect Proofs; Properties of Primes, Integers, Rationals, and Reals; Divisibility (Unique Factorization Theorem); Modular Forms (Quotient-Remainder Theorem); The Division & Euclidean Algorithms.

Section 1 - Direct Proof and Counterexample
Section 1 - Direct Proof and Counterexample Mathematics is built on the Axiomatic Method. Start with Definitions and Axioms. Use these in valid arguments to demonstrate Theorems. Use all of the above to deduce NEW Theorems. Continue ad infinitum. Get paid! (or pass course!)

Even and Odd Integers Definition: An integer n is even provided there exists an integer k such that n = 2k. Definition: An integer n is odd provided there exists an integer k such that n = 2k + 1. 38 is even since 38 = 2(19) and 19 is an integer. 417 is odd since 417 = 2(208) + 1 and 208 Î Z. 417 is not even since 417 = 2(208.5) but Ï Z.

Prime and Composite Integers
Prime and Composite Integers Definition: An integer n is prime if, and only if, n > 1, and for all positive integers r and s, if n = r×s, then r = 1 or s = 1. Definition: An integer n is composite if, and only if, n > 1, and for all positive integers r and s, if n = r×s, then r ¹ 1 and s ¹ 1. Every natural number > 1 is either prime or composite. 2 is the only even prime number.

Proving Existential Statements
Proving Existential Statements To show: There exists an a such that P(a). Demonstrate an Example: Prove there is an even integer that can be written in two ways as the sum of two primes. Proof: 10 = = Construct an Example: Prove if r,s Î Z, then 4r + 6s is even. Proof: Let r,s Î Z. Thus 2r + 3s = k Î Z, and 4r + 6s = 2k, therefore (4r + 6s) is even.

Proving Universal Statements
Proving Universal Statements Most theorems are of the form: " xÎD, if P(x), then Q(x). If D is a finite set, we can just exhaust over each element n to verify that Q(n) holds. Example: Prove all n Î {4, 6, 8, 10, 12} can be written as the sum of two primes. Proof: 4 = = 3 + 3 8 = = 3 + 7 12 =

Generalizing from the Generic Particular
Generalizing from the Generic Particular When it is not feasible to exhaust over each element of the domain, we turn to the method of generalizing from the generic particular: To show that every element of a domain satisfies a certain property, suppose x is a particular, but arbitrarily chosen element of the domain, and show that x satisfies the property. This is the strategy we employ in the method of direct proof.

Method of Direct Proof Express the statement to be proved in the form: " xÎD, if P(x), then Q(x) if possible. (Often, this is done mentally) Start the proof by supposing that n is a particular but arbitrary element of D for which P(n) is true. (Suppose nÎD and P(n)) Show that the conclusion Q(n) follows from P(n) by using definitions, axioms, previously established results, and the rules for logical inference.

Theorem 3.1.1 Prove: If the sum of two integers is even, then so is their difference. Proof: Let m and n be any integers with (m + n) even. This means there is an integer k such that (m + n) = 2k. Now, (m - n) = (m + n) - 2n = 2k - 2n = 2 (k - n) = 2p, where k - n = p is an integer. Thus (m - n) is even. Also, (n - m) = -(m - n) = 2(-p), so (n - m) is also even. Therefore, the difference of m and n is even. QED

Directions for Writing Proofs
Directions for Writing Proofs Write the statement to be proved. Clearly mark the beginning of your proof with the word Proof. Make your proof self-contained: Identify each variable used in the body of the proof; Introduce only necessary variables and notation; Use Lemmas to show significant but related ideas. Write proofs in complete (English) sentences.

Common Mistakes Arguing from examples;
Common Mistakes Arguing from examples; Using the same letter to mean different things; Jumping to a conclusion; Begging the question (i.e. assuming true that which you want to prove); Using if when you mean since, hence, thus, therefore, hencely, thusly, hereforthwith, etc.

Section 2 - Rational Numbers
Section 2 - Rational Numbers Recall the definition of a Rational Number: A real number r is rational provided there exist integers a and b such that r = a/b and b ¹ 0. Theorem: Every integer is a rational number. Proof: Let a be an integer, then a = a/1. Moreover, 1 is an integer and 1 ¹ 0. Therefore a is a rational number. QED

Proving Properties of Rationals
Proving Properties of Rationals We will now look at some theorems and corollaries (theorems that follow essentially trivially from another theorem) about rational numbers. We will rely on the Closure Properties of the Integers under +, -, and ×: If a,b are integers, then (a+b), (a-b), (b-a), and a×b are also integers. We will also use their Zero-Product Property: If a,b Î Z, with a ¹ 0 and b ¹ 0, then a×b ¹ 0.

Closure of the Rationals Under +
Closure of the Rationals Under + Theorem: If r, s Î Q, then (r + s) Î Q. Proof: Let r, s Î Q. Thus $ a, b, c, d Î Z such that r = a/b with b ¹ 0 and s = c/d with d ¹ 0. Now, (r + s) = a/b + c/d = (ad + bc)/bd. Since a, b, c, d Î Z, we have that (ad + bc) Î Z and that bd Î Z. Moreover, since b ¹ 0 and d ¹ 0, we conclude that bd ¹ 0. Consequently, (r + s) is the quotient of integers with non-zero denominator. Therefore (r + s) Î Q. QED

A Corollary Corollary: Double a rational is rational.
A Corollary Corollary: Double a rational is rational. Proof: Let r = s in the previous theorem.

Section 3 - Divisibility
Section 3 - Divisibility Definition: If n and d are integers and d ¹ 0, then n is divisible by d provided n = d × k for some integer k. Alternatively, we say: n is a multiple of d d is a factor of n d is a divisor of n d divides n (denoted with d | n).

Properties of Divisibility
Properties of Divisibility Divisors of 0: If k is a non-zero integer, then k divides 0 since 0 = k × 0. Positive Divisors of a Positive Number: If a and b are positive integers and a | b, is a £ b? Yes. Since a | b, $ k Î Z,such that b = a × k. Moreover, 0 < k, since a and b are, so 1 £ k. Thus: a = a × 1 £ a × k = b. Therefore a £ b. Divisors of 1: The only divisors of 1 are 1 and -1.

Divisibility of Algebraic Terms
Divisibility of Algebraic Terms Let a and b be integers. Does 3 | (3a + 3b)? Yes, since (3a + 3b) = 3(a + b) and (a + b) Î Z. Does 5 | 10ab? Yes again, since 10ab = 5(2ab) and (2ab) Î Z. If m Î Z and m | (a + b), does m | a and m | b? No. 2 | 8 but 2 | 5 and 2 | 3.

Divisibility and Non-divisibility
Divisibility and Non-divisibility There is another way to test for divisibility: If d | n, there is integer k with n = dk, then k = (n/d). So, if (n/d) is an integer, then d | n. This leads to an easy way to test for non-divisibility: If (n/d) is not an integer, then d cannot divide n. Examples: 3 | 12 since 12/3 = 4 Î Z. 5 | 12 since 12/5 = 2.4 Ï Z.

Divisibility by a Prime
Divisibility by a Prime Theorem: Every positive integer greater than 1 is divisible by a prime number. Proof: Let n Î Z with n > 1. Then either n is prime or composite. If n is prime, it is divisible by itself, and we are done. Now, assume n is composite. Thus there are integers (greater than 1) a and b, such that n = ab. If a is prime, we are done. If not, factor a, .... Will we eventually get to a prime factor?

Standard Factored Form
Standard Factored Form Definition: Given any integer n > 1, the standard factored form of n is an expression of the form: n = (p1)e1 × (p2)e2 × (p3)e3...(pk)ek, where k is a positive integer ; p1,p2,...,pk are prime numbers with p1 < p2 < ... < pk; and e1,e2,...,ek are positive integers. Example: 3300 = 33 × 100 = 3 × 11 × = 22 × 3 × 52 × 11.

Unique Factorization Theorem
Unique Factorization Theorem Theorem: Given any integer n > 1, there exist positive integer k; prime numbers p1,p2,...,pk; and positive integers e1,e2,...,ek, with n = (p1)e1 × (p2)e2 × (p3)e3...(pk)ek, and any other expression of n as a product of prime numbers is identical to this except, perhaps, for the order in which the factors appears. This is also referred to as the Fundamental Theorem of Arithmetic.

Fundamental Theorem of Arithmetic
Fundamental Theorem of Arithmetic Theorem: Every positive integer greater than 1 has a unique factorization as the product of primes. Proof: (outline) 1. Apply the previous theorem to each composite factor encountered. 2. Sort the final listing to get the prime factors in increasing (decreasing?) numeric order. 3. Rewrite using exponents.

Section 4 - The Quotient Remainder Theorem
Section 4 - The Quotient Remainder Theorem The Quotient-Remainder Theorem; Modular Arithmetic (div and mod functions); Proofs Requiring Division into Cases; Representations of the Integers. The Parity Theorem

Quotient-Remainder Theorem
Quotient-Remainder Theorem Theorem: Given any integer n and a positive integer d, there exist unique integers q and r such that: n = d×q + r, and 0 £ r < d. Example: If n = 27 and d = 5, then consider: 27 = 0 × = 1 × = 2 × = 3 × = 4 × = 5 × here, r = 2 and q = = 6 × 5 + (-3)

div and mod Functions Definition: Given a nonnegative integer n and a positive integer d, n div d = the integer quotient obtained when n is divided by d; n mod d = the integer remainder obtained when n is divided by d. Symbolically, if n and d are positive integers: n div d = q and n mod d = r, where n, d, q, and r are as described in the Quotient-Remainder Theorem.

div and mod Examples Consider the previous example of n = 27 and d = 5. Since 27 = 5×5 + 2 yields q = 5 and r = 2, we have that: div 5 = 5; mod 5 = 2. More: 100 div 10 = mod 10 = 0 100 div 8 = mod 8 = 4 10 div 100 = mod 100 = 10 365 div 7 = mod 7 = 1

Representations of the Integers
Representations of the Integers Recall, we have claimed previously that every integer is either even or odd. Consider: Even: Odd: We note that all the evens are n = 2q = 2q + 0 and all the odds are n = 2q + 1. Moreover, each successive integer alternates parity (its mod 2 value).

More Representations of Integers
More Representations of Integers If we continue representing integers via the Quotient-Remainder Theorem, we observe: Modulus Forms 2 2n 2n + 1 3 3n 3n n + 2 4 4n 4n n n + 3 ... k kn kn kn kn + (k-1)

Division into Cases Sometimes when proving a theorem, the logical flow will fork into different directions, each of which need investigation. This is analogous to needing IF THEN ELSE instead of just IF THEN in programming flow. An example is the Parity Theorem. Theorem: Any two consecutive integers have opposite parity.

Division into Cases (cont’d.)
Division into Cases (cont’d.) Proof: Let m be an integer, so its successor is (m+1). Show m and (m+1) have opposite parity. Case 1 (m even): If m is even, there is an integer k such that m = 2k, hence (m+1) = 2k + 1, thus (m+1) is odd. So, m even implies (m+1) is odd. Case 2 (m odd): If m is odd, there is integer k such that m = 2k + 1. Hence: (m+1) = (2k + 1) + 1 = 2k + 2 = 2(k +1), and so (m+1) is even. So, m odd implies (m+1) is even. Therefore, consecutive integers have opposite parity. QED

The Square of an Odd Integer
The Square of an Odd Integer Theorem: If n is an odd integer, (n2 mod 8) = 1. Proof: Let n be an odd integer, so it has the representation modulo 4 of n = 4q+1 or 4q+3. Case 1: Let n = 4q+1. Thus n2 = (4q+1)2 = 16q2 + 8q + 1 = 8(2q2 + q) + 1. Case 2: Let n = 4q+3. Thus n2 = (4q+3)2 = 16q2 + 24q + 9 = 16q2 + 24q = 8(2q2 + 3q + 1) + 1. Therefore, in either case, (n2 mod 8) = 1. QED

Section 6 - Indirect Argument
Section 6 - Indirect Argument Method of Proof by Contradiction; Method of Proof by Contraposition; Examples of Each Method.

Proof by Contradiction
Proof by Contradiction Instead of the Universal Modus Ponens argument form: "x, [P(x) ® Q(x) AND P(x)] Þ Q(x), a Proof by Contradiction (reductio ad absurdum) follows the Universal Modus Tollens form: "x, [P(x) ® Q(x) AND ~Q(x)] Þ ~P(x). We obtain a contradiction when the conclusion of this form is combined with our standard assumption in a direct proof the P(x) holds. This differs marginally from the Method of Contraposition which proves directly the validity of the comtrapositive statement.

Method of Proof By Contradiction
Method of Proof By Contradiction Suppose the statement to be proved is FALSE; Show this supposition leads logically to a contradiction (either to the original hypotheses or to some other statement of fact); Conclude that the original statement to be proved is TRUE.

Example: No Greatest Integer
Example: No Greatest Integer Theorem: There is no greatest integer. Proof: (Contradiction) Suppose there is a greatest integer N. Thus for every integer k, k £ N. Now, since N is an integer, by closure, (N+1) is an integer. Thus: N + 1 £ N , hence £ 0.* Therefore, there is no greatest integer. QED

Sums of Rationals and Irrationals
Sums of Rationals and Irrationals Theorem: The sum of a rational and an irrational is irrational. Proof: (Contradiction) Let r be rational, s be irrational, and assume (r + s) is rational. Thus there exist a,b,c,d Î Z, with r = a/b, (r + s) = c/d and b,d ¹ 0. Now, s = (r + s) - r = c/d - a/b = (bc - ad)/bd. Since a,b,c,d Î Z and b,d ¹ 0, we have s Î Q.* Therefore (r + s) is irrational. QED

Argument by Contraposition
Argument by Contraposition Since we know that a statement and its contrapositive are logically equivalent, if we can pose our conjecture in the form of a conditional, we can work, equivalently, with its contrapositive form. We call this strategy, simply enough, Argument by Contraposition.

Method of Proof by Contraposition
Method of Proof by Contraposition Express the statement to be proved in the form "x, if P(x) then Q(x). Rewrite this as its contrapositive "x, if ~Q(x) then ~P(x). Prove the contrapositive form directly: Suppose x is such that Q(x) is FALSE. Show that P(x) is FALSE.

Example of Contraposition
Example of Contraposition Theorem: Given any integer n, if n2 is even, then n is even. (Contrapositive: If n is odd, then n2 is odd.) Proof: (Contraposition) Let n be an integer and assume that n is odd. Thus, there is an integer k such that n = 2k + 1. Show that n2 is odd. Now, n2 = (2k + 1)2 = 4k2 + 4k = 2(2k2 + 2k) + 1. Since k is an integer, (2k2 + 2k) is an integer. Therefore n2 is odd. QED

Section 7 - Two Classical Theorems
Section 7 - Two Classical Theorems Prove: The Ö2 is irrational. Proof: (Contradiction) Assume Ö2 is rational. Then, there exist p,q Î Z in “lowest terms” with Ö2 = p/q and q ¹ 0. Now, Ö2 = p/q implies 2q2 = p2, hence p2 is even. From a previous result, this yields that p is even, so we deduce $m Î Z with p = 2m. Thus 2q2 = (2m)2 = 4m2, so q2 = 2m2. This time, we get that q2 is even, hence q is even.* Therefore Ö2 is irrational.

A Corollary Corollary: (1 + 3Ö2) is irrational.
A Corollary Corollary: (1 + 3Ö2) is irrational. Proof: (Contradiction) Assume (1 + 3Ö2) is rational. Then, there exist p,q Î Z in “lowest terms” with (1 + 3Ö2) = p/q and q ¹ 0. Now, solving for Ö2, we get Ö2 = (p - q)/3q, which implies Ö2 is rational.* Therefore (1 + 3Ö2) is irrational.

The Infinitude of the Primes
The Infinitude of the Primes Theorem: The set of prime numbers is infinite. Lemma: If a Î Z and p is prime with p | a, then p | (a + 1). Proof: (Contradiction) Let a Î Z and p be prime with p | a. Assume that p | (a + 1). Then there are m,n Î Z such that a = mp and (a + 1) = np. Now, (a + 1) = np implies 1 = np - mp, so 1 = p(n - m), thus p | 1. However, this means p = 1,-1.* Therefore, p | (a + 1).

The Infinitude of the Primes
The Infinitude of the Primes Proof: (Contradiction) Assume there are a finite number of primes, say p1, p2, ..., pk. Now, consider a = p1p2...pk. Since each pi is an integer, a is likewise an integer, and so is (a+1). Thus, by a previous result, (a+1) is divisible by some prime, pj. Since pj is a prime, it must divide a, and by assumption, pj divides (a+1).* Therefore the collection of primes is infinite. QED

Section 8 - Two Algorithms
Section 8 - Two Algorithms We look at two algorithms to obtain values of use in number theory. The Division Algorithm is a process to calculate (n div d) and (n mod d) as specified in the Quotient-Remainder Theorem. The Euclidean Algorithm uses the Quotient-Remainder Theorem to calculate Greatest Common Divisors.

The Division Algorithm
The Division Algorithm Input: Integers n and d. Output: Integers q (= n div d) and r (= n mod d). Begin set r = n; set q = 0; while (r ³ d) set r = r - d; set q = q + 1; endwhile output q, r; End

Tracing the Division Algorithm
Tracing the Division Algorithm Calculate: (19 div 4) and (19 mod 4) Iteration Number n 19 d 4 r q Thus (19 div 4) = 4 and (19 mod 4) = 3.

Greatest Common Divisors
Greatest Common Divisors Definition: An integer d is a common divisor of integers m and n provided d | n and d | m. Example: Since 6 | 12 and 6 | 30, 6 is a common divisor of 12 and 30. Definition: If m,n Î Z not both 0, then d Î Z is the greatest common divisor of m and n when: 1. d is a common divisor of m and n; 2. if c is also a common divisor of m and n, then c £ d. We denote d = gcd(m,n).

Calculating GCDs Find gcd(72,63)
Calculating GCDs Find gcd(72,63) 72 = 9×8 = 3×3×2×4 = 2×3×3× = 9×7 = 3×3 ×7 hence gcd(72,63) = 3×3 = 9. Find gcd(1020,630) 1020 = (2×5)20 = 220 × = (2×3)30 = 230 × 330 = 220 × 210 × hence gcd(1020,630) = 220.

Two Lemmas Lemma1: If r is a positive integer, then gcd(r,0) = r.
Two Lemmas Lemma1: If r is a positive integer, then gcd(r,0) = r. Why? Everything divides 0 and r is the biggest thing to divide r. Lemma 2: If a,b Î Z, b ¹ 0 and q,r Î Z+ with a = b×q + r, then gcd(a,b) = gcd(b,r).

Proof of Lemma 2 Proof: Let a,bÎZ, b ¹ 0, and q,rÎZ+ with a = b×q+r.
Proof: Let a,bÎZ, b ¹ 0, and q,rÎZ+ with a = b×q+r. Case 1: Show gcd(a,b) £ gcd(b,r). Let x = gcd(a,b), so that x | a and x | b. Now, since a = b×q + r, we have r = a - b×q, thus x | r. But x | b, hence gcd(a,b) = x £ gcd(b,r). Case 2: Show gcd(a,b) ³ gcd(b,r). Let y = gcd(b,r), so that y | b and y | r. Now, since a = b×q + r, we see that y | a. But y | b, hence gcd(a,b) ³ y = gcd(b,r). Therefore gcd(a,b) = gcd(b,r). QED

Calculating GCDs a la Euclid
Calculating GCDs a la Euclid Find gcd(330,156). 330 = 156 × 156 = 18 × 18 = 12 × 1 + 6 12 = 6 × 2 + 0 Hence gcd(330,156) = gcd(156,18) = gcd(18,12) = gcd(12,6) = gcd(6,0) = 6.

The Euclidean Algorithm
The Euclidean Algorithm Input: Integers A and B. Output: gcd(A,B). Begin set a = A; set b = B; while (b ¹ 0) set r = a mod b; set a = b; set b = r; endwhile output a; End

Chapter 4 - Sequences and Mathematical Induction
Chapter 4 - Sequences and Mathematical Induction Sequences & Series (Summations); Principle of Mathematical Induction; Weak Form of Induction; Strong Form of Induction.

Section 1 - Sequences A sequence is a list of elements called terms.
Section 1 - Sequences A sequence is a list of elements called terms. We think of each term as occupying a specific position within the sequence, so we may use an index variable to denote specific but arbitrary terms in the sequence. For example, if we have the sequence: 1, 2, 4, 8, 16, 32, 64, 128, ... we can denote a0 = 1, a1 = 2, a2 = 4, a3 = 8, ... This is useful since we can now describe the sequence as {ai = 2i}

Explicit Formulas Define the sequences {ai} and {bj} as: ai = i / (i + 1) for i > 0; bj = (j - 1) / j for j > 1. Thus ai = 1/2, 2/3, 3/4, 4/5, 5/6, ... and bj = 1/2, 2/3, 3/4, 4/5, 5/6, ... Are these the same sequence? This indicates a natural way to check if two sequences are equal: Two sequences {ai} and {bi} are equal if ai = bi for each value of i.

Alternating Sequences
Consider the sequence defined as: ai = (-1)i, for all positive integers i. We get: a1 = a2 = 1 a3 = a4 = 1, etc. Such a sequence in which successive terms have opposite sign is called an alternating sequence.

Summation Notation Often, it is useful to sum up the terms of a sequence. This expression of summation is called a series. To save the tedium of continually writing a1 + a2 + a an, we will use the summation notation: n S ai = a1 + a2 + a an i=1

Working with Summations
Working with Summations n S ai = am + am+1 + am an i=m Given we call i the index of summation, m the lower limit, and n the upper limit. To compute a summation, we perform the substitutions and calculations called for by the formula: S i2 = = 54, S i2 = 92. i= i=9

Telescoping Series Consider: n S æ k _ k + 1ö k=1 è k + 1 k + 2ø
Telescoping Series Consider: n S æ k _ k + 1ö k=1 è k k + 2ø Plugging into the formula, we get: (1/2 - 2/3) + (2/3 - 3/4) + (3/4 - 4/5) [(n-1)/n - n/(n+1)] + [n/(n+1) - (n+1)/(n+2)] = 1/2 - (n+1)/(n+2). This type of series is called a telescoping series.

Changing Variables Rewrite: n S k / (k + 1) as a sum from 3 to (n+2). k =1 To do this, we use a change of variables from k to j using the transformation j = k + 2 (so that k = j - 2). Thus k = 1 becomes j = 3; k = n becomes j = n+2; and k + 1 becomes (j - 1). Consequently: n n S k / (k + 1) = S (j - 2) / (j - 1) k =1 j =1

Product Notation Instead of summing the terms of a sequence, if we want to multiply each term, we use the product notation: n P ak = a1×a2×a3× ... ×an k =1 For example: P 3k = 35×36×37× ... ×312 = k =5

Properties of Sums and Products
Properties of Sums and Products If {ak} and {bk} are sequences of real numbers, and c is a non-zero real number, then for m £ n: n n n S ak S bk = S (ak + bk) k = m k = m k = m n n S (c × ak)= c × S (ak) k = m k = m n n n P ak × P bk = P ak× bk k = m k = m k = m

Factorial Notation Although products of the terms of a sequence do not arise in problems as much as summations, one particular type of product does. Definition: For each positive integer n, the quantity n factorial, denoted n!, is defined to be the product of all integers from 1 to n. Thus, n! = n×(n-1)×(n-2)×...×3×2×1. For completeness, we define 0! = 1.

Computing with Factorials
Computing with Factorials 8!/7! = 8×7!/7! = 8. 5!/(2!×3!) = (5×4×3!)/2!×3! = 5×4/2×1 = 5×2 = 10. 1/(2!×4!) + 1/(3!×3!) = 1×3/(3×2!×4!) + 1×4/(3!×4×3!) = (3 + 4)/(3!×4!) = 7/(6×24) = 7/144. n!/(n - 3)! = [n×(n - 1)×(n - 2)×(n - 3)!]/(n - 3)! = n×(n - 1)×(n - 2) = n(n2 - 3n + 2) = n3 - 3n2 + 2n.

Section 2: Mathematical Induction
Section 2: Mathematical Induction Principle of Mathematical Induction; Formula for the Sum of the First n Integers; Formula for the Sum of a Geometric Series.

What Is Induction One of the more recently developed proof techniques;
What Is Induction One of the more recently developed proof techniques; Used to verify conjectures about processes that occur repeatedly, according to definite patterns; Used to prove statements indexed on the Natural Numbers (i.e. For all integers n ³ 0, ...)

Climbing Infinite Staircases
Climbing Infinite Staircases Consider the problem of trying to climb an infinitely tall staircase. How can I know that this staircase is climbable? First, show the staircase exists. (Basis) Second, show standing at any arbitrary step implies I can climb to the next step. (Induction) The basis shows there is an initial step. The induction validates a rule of motion from one step to another.

In Pictures To Infinity Induction Induction Induction Induction Basis
In Pictures To Infinity Induction Induction Induction Induction Basis

Principle of Mathematical Induction
Principle of Mathematical Induction Let P(n) be a predicate that is defined for integers n, and let a be a fixed integer. Suppose the following two statements are true: 1. P(a) is true; 2. For all integers k ³ a, if P(k) is true, then P(k + 1) is true. Then, for all integers n ³ a, P(n) is true.

Outline of an Inductive Proof
Outline of an Inductive Proof Basis step says to show P(initial value) is true. Inductive step says: Assume P(k) is true for an arbitrary k. Use this to show P(k + 1) is true. This assumption of P(k) is called the inductive hypothesis.

Tuppence and Nickels Show that any monetary value of 4 cents or more can be composed of tuppence (a two-cent coin) and nickels. BASIS: 4 cents = tuppence + tuppence. INDUCTION: Assume k-cents can be made up from tuppence and nickels (k = 2t + 5n). Show (k+1) can too.

Tuppence and Nickels (cont’d.)
Tuppence and Nickels (cont’d.) Case 1: (At least one of each coin: k = 2t + 5n) k + 1 = 2t + 5n + 1 = 2t + 5n = (2t + 6) + (5n - 5) = 2(t + 3) + 5(n - 1). Case 2: (No tuppence: k = 5n) k + 1 = 5n + 1 = 5n = 2(3) + 5(n - 1). Case 3: (No nickels and t > 2: k = 2t) k + 1 = 2t + 1 = 2t = 2(t - 2) + 5(1).

Sum of the First n Integers
Sum of the First n Integers Carl Friederich Gauss is bored in school. Teach instructs him to add together the numbers from 1 to 100. Zip! Carl is done. Sum is How? = ( ) + (2 + 99) + (3 + 98) ( ) = = 50(101) = 5050.

Sum of the First n Integers (cont’d.)
Sum of the First n Integers (cont’d.) Prove: n S k = n(n + 1) / k =1 Proof: (Induction) Basis: Show: S k = 1(2) / k =1 1 S k = 1 and 1(2)/2 = 2/2 = 1, so they are equal. k =1

Sum of the First n Integers (cont’d.)
Sum of the First n Integers (cont’d.) Inductive: Assume for some p: p = p(p + 1)/2. Show: ( p) + (p + 1) = (p + 1)(p + 2)/2. Now, ( p) + (p + 1) = p(p + 1)/2 + (p + 1) = (p + 1)[p/2 + 1] = (p + 1)[p/2 + 2/2] = (p + 1)(p + 2)/2. Therefore n = n(n+1)/2 for all integers n QED

The Geometric Series Any sum of the form: 1 + r + r2 + r rn is called a Geometric Series. Thus, n is a geometric series. To find the sum of this series, consider: S = 1 + r + r2 + r rn. So rS = -r - r2 - r r(n + 1). and (1 - r)S = 1 - r(n + 1). Therefore, 1 + r + r rn = 1 - r(n + 1) r

Proof of the Geometric Series
Prove: 1 + r + r rn = [r(n + 1) - 1] / (r - 1) Proof: (Induction) Basis: Show true for n = 0. LHS = 1. RHS = [r(0+1) - 1]/(r - 1) = (r-1)/(r-1) = 1. Therefore LHS = RHS. Induction: Assume 1+r+r2+...+rk = r(k+1)-1/r-1. Show 1+r+r2+...+rk+r(k+1) = [r(k+2) - 1]/(r - 1). Now, 1+r+r2+...+rk+r(k+1) = r(k+1)-1/r-1 + r(k+1) = [r(k+1) (r - 1)r(k+1) ]/(r - 1) = [r(k+1) r×r(k+1) - r(k+1) ]/(r - 1) = [r(k+2) - 1]/(r - 1). QED

Section 3: Mathematical Induction II
Section 3: Mathematical Induction II Prove: 22n - 1 is divisible by 3, for integers n > 0. Proof: (Induction) Basis: 22(1) - 1 = = = 3, which is clearly divisible by 3. Induction: Assume for some integer k, 22k - 1 is divisible by 3. Now, 22(k + 1) - 1 = 2(2k + 2) - 1 = 22k = 22k(4) - 1 = 22k(3 + 1) - 1 = 3(22k) + 22k = 3(22k) + (22k - 1). Since each term in parentheses is divisible by 3, we have therefore that 22(k + 1) -1 is also. QED

Proving An Inequality Prove: 2n + 1 < 2n, for all integers n ³ 3.
Proving An Inequality Prove: 2n + 1 < 2n, for all integers n ³ 3. Proof: (Induction) Basis: LHS = 2(3) + 1 = 7, and RHS = 23 = 8, so clearly 2n + 1 < 2n for n = 3. Induction: Assume for some integer k, 2k + 1 < 2k. Show 2(k + 1) + 1 < 2(k+1). Now, 2(k + 1) + 1 = (2k + 1) < 2k + 2 < 2k + 2k = 2(k+1). Therefore, 2n + 1 < 2n, for all integers n ³ 3.. QED

Number of Subsets Prove: A set with n elements has 2n subsets.
Number of Subsets Prove: A set with n elements has 2n subsets. Proof: (Induction) Basis: Since the empty set has 1 subset (itself), and 20 = 1, then a set with 0 elements has 20 subsets. Induction: Assume every k-element set has 2k subsets. Show every (k+1)-element set has 2(k+1) subsets. Now let A = {a1, a2, a3,..., ak, b}, so that A has (k+1) elements. We partition P(A) into two sub-collections where the first contains subsets of A which don’t have b in them and the second contains

Number of Subsets (cont’d.)
Number of Subsets (cont’d.) subsets of A which do have b in them. Thus: First Sub-collection Second Sub-collection {} {b} {a1} {a1, b} {a1, a2} {a1, a2, b} {a1, a2, ..., ak} {a1, a2, ..., ak, b} Clearly, the first collection is made up of all the subsets from the k-element set {a1, a2, ..., ak} so it has 2k entries.

Number of Subsets (cont’d.)
Number of Subsets (cont’d.) Now, by construction, it follows that the second collection must have the same number of entries as the first, so it too must have 2k entries. Since the collection of all subsets of A has been partitioned into these two sub-collections, we see that A must have 2k + 2k = 2(k+1) subsets. QED

Section 4: Strong Induction
Section 4: Strong Induction The Principle of Mathematical Induction asserts that P(k) being true implies P(k+1) is true. However, sometimes we need to “look” further back than 1 step to obtain P(k+1). That’s where the Strong Form of Mathematical Induction comes in useful.

Principle of Strong Mathematical Induction
Principle of Strong Mathematical Induction Let P(n) be a predicate defined over all integers n, and let a and b be fixed integers with a £ b. Suppose the following two statements are true: 1. P(a), P(a+1),..., P(b) are all true. (Basis step) 2. For any integer k > b, if P(i) is true for all integers i with a £ i < k, then P(k) is true (Inductive step) Then the statement P(n) is true for all integers n ³ a.

Divisibility by a Prime
Divisibility by a Prime To see, via strong induction, that every integer greater than 1 is divisible by a prime number, we note that the basis value of 2 is trivially divisible by a prime (itself). Now, if we assume the strong inductive hypothesis that every integer up to k is divisible by a prime, when we look at k itself, either it is prime (and hence divisible by itself), or can be factored as lesser integers, each of which having a prime factor. Thus k is divisible by a prime.

Binary Representation Theorem
Binary Representation Theorem Theorem: Every positive integer n can be expressed as: n = cr2r + cr-12r c222 + c12 + c0. Proof: (Strong Induction) Clearly n = 1 corresponds to just c0 = 1. Assume every integer up to k has a binary representation. Show k does too. If k is even, then k/2 is an integer which is less than k, so it has a binary expansion of the form: k/2 = cr2r + cr-12r c222 + c12 + c0. Therefore, k = cr2r+1 + cr-12r c223 + c122 + c02.

Binary Representation Theorem (cont’d.)
Binary Representation Theorem (cont’d.) In the case where k is odd, then (k-1)/2 is an integer which is less than k, so it has a binary expansion of the form: (k-1)/2 = cr2r + cr-12r c222 + c12 + c0. Solving for k, we have: k = cr2r+1 + cr-12r c223 + c122 + c Therefore k has a binary representation. QED

Property of a Sequence Consider the sequence a1, a2, a3, ... defined as: a1 = 1, a2 = 2, a3 = 3 and an = an-1 + an-2 + an-3. Show that an < 2n. Basis: a4 = a1 + a2 + a3 = = 6 < 16. Inductive: Assume ai < 2i for i = 5, 6, 7, ..., (k - 1). Show ak < 2k . Now, ak = ak-1 + ak-2 + ak-3, so applying the inductive hypothesis: ak-1 < 2k-1, ak-2 < 2k-2, and ak-3 < 2k-3, we get:

Property of a Sequence (cont’d.)
Property of a Sequence (cont’d.) ak = ak-1 + ak-2 + ak-3 < 2k-1 + 2k-2 + 2k-3 < ( )×2k-3 = 7×2k-3 < 8×2k-3 < 23×2k-3 = 2k. Therefore ak < 2k. QED

Why Use Strong Induction?
Why Use Strong Induction? In the first proof, we could not use weak induction since factoring a composite number yields values which are strictly less than the composite number minus 1. In the second theorem, we apply strong induction to a value (k/2 or (k - 1)/2) which is clearly much less than k - 1. In the sequence problem, we need to apply the inductive hypothesis to values at the three stages k - 1, k - 2, and k - 3.

Chapter 10: Relations Relations from one set to another;
Chapter 10: Relations Relations from one set to another; Relations on a set; Graphs of relations; Congruence Modulo n; Properties of Relations; Equivalence Relations and Classes; Partitions and Relations.

Section 1: Relations on Sets
Section 1: Relations on Sets Definition: Let A and B be sets. A (binary) relation R from A to B is a subset of A ´ B. If A = B, then we say R is a relation on A. Given an ordered pair (x,y) in A ´ B, x is related to y by R, written x R y, if and only if (x,y) Î R. Think of the usual relations we deal with and replace for R in the notation x R y: x = y, x £ y, x > y, x | y, etc.

A Binary Relation as a Subset
A Binary Relation as a Subset Let A = {1,2} and let B = {1,2,3} and define the binary relation from A to B as: R = {(x,y) | x Î A and y Î B and x - y is even}. What is R? (1,1) ® 0 is even þ (2,1) ® 1 is even ý (1,2) ® -1 is even ý (2,2) ® 0 is even þ (1,3) ® -2 is even þ (2,3) ® -1 is even ý Hence R = {(1,1), (2,2), (1,3)}.

Congruence Modulo 2 If we generalize the last example to relate Z to Z by: R = {(x,y) | x,y Î Z and x - y is even}, we get the relation congruence modulo 2: R = {(x,y) | x,y Î Z and x º y mod 2}. If we look hard at this relation, we see that: even R even, since 2m - 2n = 2(m - n), and odd R odd, since 2m+1 - (2n+1) = 2(m - n).

Relations on Strings Recall that if S is an alphabet, then Sn = {all strings over S of length n}. Now, let S = {0,1} and define the relation on S6 to be: R = {(s,t) | s,t Î S6 and first four characters of s = first four characters of t}. (110011,110011) Î R. (100000,100001) Î R and (100001,100000) Î R. (000000,000001), (000001,000011) and (000000,000011) are all in R.

Graphs of a Relation R = {(x,y) | x,y Î R and x2 + y2 = 1}
Graphs of a Relation R = {(x,y) | x,y Î R and x2 + y2 = 1} 1 x y Arrow Diagrams: R = {(1,a),(1,b),(2,b),(3,a)} 1 2 3 a b

Functions A function from A to B is a relation that satisfies the properties: 1. For each x in A, there is a y in B such that (x,y) is in the function; 2. If (x,y) and (x,z) are in the function, then y = z.

Examples of Functions (Not)
Examples of Functions (Not) To understand functions, it is easier to study what are not: Violates Property Violates Property 2 1 2 3 4 a b 1 2 3 4 a b

Inverse Relations Definition: If R is a relation from set A to set B, then the inverse relation of R, denoted R-1, is R-1 = {(y,x) | (x,y) Î R}. For example, if R = {(1,3), (2,1), (4,5), (6,6)}, then R-1 = {(3,1), (1,2), (5,4), (6,6)}. Could R = R-1? Sure - let R = {(1,2), (2,1), (2,3), (3,2), (4,4)}.

Directed Graph of a Relation
When R is a relation on a set A, we draw it using a directed graph. For example, if R = {(1,1), (2,4), (3,2), (4,1), (4,3)}, then its directed graph is: 2 3 1 4

Section 2: Reflexivity, Symmetry, and Transitivity
Definition: Let R be a binary relation on A. R is reflexive if for all x Î A, (x,x) Î R. (Equivalently, for all x e A, x R x.) R is symmetric if for all x,y Î A, (x,y) Î R implies (y,x) Î R. (Equivalently, for all x,y Î A, x R y implies that y R x.) R is transitive if for all x,y,z Î A, (x,y) Î R and (y,z) Î R implies (x,z) Î R. (Equivalently, for all x,y,z Î A, x R y and y R z implies x R z.)

Examples Reflexive: The relation R on {1,2,3} given by R = {(1,1), (2,2), (2,3), (3,3)} is reflexive. (All loops are present.) Symmetric: The relation R on {1,2,3} given by R = {(1,1), (1,2), (2,1), (1,3), (3,1)} is symmetric. (All paths are 2-way.) Transitive: The relation R on {1,2,3} given by R = {(1,1), (1,2), (2,1), (2,2), (2,3), (1,3)} is transitive. (If I can get from one point to another in 2 steps, then I can get there in 1 step.)

Violations of the Properties
Why is R = {(1,1), (2,2), (3,3)} not reflexive on {1,2,3,4}? Because (4,4) is missing. Why is R = {(1,2), (2,1), (3,1)} not symmetric? Because (1,3) is missing. Why is R = {(1,2), (2,3), (1,3), (2,1)} not transitive? Because (1,1) and (2,2) are missing. Is {(1,1), (2,2), (3,3)} symmetric? transitive? Yes! Yes!

The Transitive Closure
Definition: Let R be a binary relation on a set A. The transitive closure of R is the binary relation Rt on A satisfying the following three properties: 1. Rt is transitive; 2. R is a subset of Rt; 3. If S is any other transitive relation that contains R, then S contains Rt. In other words, the transitive closure of R is the smallest transitive relation containing R.

Example of the Transitive Closure
Example of the Transitive Closure Given the relation R on {1,2,3,4}, its transitive closure is: 1 2 4 3 1 2 4 3

Properties of Equality
Consider the Equality (=) relation on R: Equality is reflexive since for each x Î R, x = x. Equality is symmetric since for each x,y Î R, if x = y, then y = x. Equality is transitive since for each x,y,z Î R, if x = y and y = z, then x = z. As a graph, the relation contains only loops, so symmetry and transitivity are vacuously satisfied!

Properties of Congruence Mod p
Properties of Congruence Mod p Let p be an integer greater than 1, and consider the relation on Z given by: R = {(x,y) | x,y Î Z and x º y mod p}. When we say x º y mod p, this means (x - y) = kp for some integer k. Now, R is reflexive since (x - x) = 0 = 0p, for all integers x. Moreover, R is symmetric, since if x º y mod p, then (x - y) = kp, thus (y - x) = (-k)p, implying that y º x mod p.

Congruence Mod p (cont’d.)
Finally, R is transitive. Why? Let x º y mod p and y º z mod p. This means there are integers k and j such that (x - y) = kp and (y - z) = jp. Hence, (x - z) = (x - y) + (y - z) = kp + jp = (k + j)p. Therefore, x º z mod p.

Properties of Inequality
Consider the Inequality (< or >) relation on R: Inequality is not reflexive since for no x Î R is it true that x < x. Inequality is not symmetric since for each x,y Î R, if x < y is true, then y < x is false. Inequality is transitive since for each x,y,z Î R, if x < y and y < z, then x < z. Inequality is so pathelogically unsymmetric, that we define a special property to describe it.

The Anti-symmetry Property
Definition: A relation R on a set A is called anti-symmetric if (x,y) Î R and (y,x) Î R implies x = y. This is equivalent to requiring that if x ¹ y and (x,y) Î R, then (y,x) Ï R. (All streets are one-way.) Example: R = {(1,1), (1,2), (3,2), (3,3)} is anti-symmetric. Is every relation symmetric or anti-symmetric? No! Consider R = {(1,2), (2,1), (1,3)}.

Section 3: Equivalence Relations
Definition: Let R be a binary relation on A. R is an equivalence relation on A if R is reflexive, symmetric, and transitive. From the last section, we demonstrated that Equality on the Real Numbers and Congruence Modulo p on the Integers were reflexive, symmetric, and transitive, so we can describe them as equivalence relations.

Examples What is the “smallest” equivalence relation on a set A?
What is the “smallest” equivalence relation on a set A? R = {(a,a) | a Î A}, so that n(R) = n(A). What is the “largest” equivalence relation on a set A? R = A ´ A, so that n(R) = [n(A)]2.

Equivalence Classes Definition: If R is an equivalence relation on a set A, and a Î A, then the equivalence class of a is defined to be: [a] = {b Î A | (a,b) Î R}. In other words, [a] is the set of all elements which relate to a by R. For example: If R is congruence mod 5, then [3] = {..., -12, -7, -2, 3, 8, 13, 18, ...}. Another example: If R is equality on Q, then [2/3] = {2/3, 4/6, 6/9, 8/12, 10/15, ...}. Observation: If b Î [a], then [b] = [a].

A String Example Let S = {0,1} and denote L(s) = length of s, for any string s Î S*. Consider the relation: R = {(s,t) | s,t Î S* and L(s) = L(t)} R is an equivalence relation. Why? REF: For all s Î S*, L(s) = L(s); SYM: If L(s) = L(t), then L(t) = L(s); TRAN: If L(s) = L(t) and L(t) = L(u), L(s) = L(u). What are the equivalence classes of R? [q], [0], [00], [000], [0000], ... in other words, S0, S1, S2, S3, S4, ...

Relations and Partitions
Recall that a partition of a set is a collection of mutually disjoint subsets whose union is the original set. Equivalence relations and partitions are tied together by the following: Definition: Given a partition of a set A, the binary relation induced by the partition is R = {(x,y) | x and y are in the same partition set}. Theorem: If A is a set with a partition and R is the relation induced by the partition, then R is an equivalence relation.

Making Equivalence Relations
This example shows how to apply this theorem to create the induced equivalence relation. The collection {{1,2,3}, {4,5}, {6}} is a partition of {1,2,3,4,5,6}. To find the induced equivalence relation, observe: {1,2,3} ´ {1,2,3} = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3)} {4,5} ´ {4,5} = {(4,4), (4,5), (5,4), (5,5)} {6} ´ {6} = {(6,6)} R = {(1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1), (3,2), (3,3), (4,4), (4,5), (5,4), (5,5), (6,6)}

Formally Making Equivalence Relations
Theorem: Let A be a set partitioned by the collection {A1, A2, A3, ...}. Then the equivalence relation induced by the partition is given by: R = (A1 ´ A1) È (A2 ´ A2) È (A3 ´ A3) È ... From the last example, the collection {{1,2,3}, {4,5}, {6}} partitioned {1,2,3,4,5,6}, so the induced relation is: R = {1,2,3}´{1,2,3} È {4,5}´{4,5} È {6}´{6}

Equivalence Classes and Partitions
Theorem: Let A be a set partitioned by the collection {A1, A2, A3, ...}. If a1ÎA1, a2ÎA2, a3ÎA3, etc., then the equivalence relation induced by the partition is given by: R = ([a1]´[a1]) È ([a2]´[a2]) È ([a3]´[a3]) È ... From the last example, the collection {{1,2,3}, {4,5}, {6}} partitioned {1,2,3,4,5,6}, so the induced relation is: R = ([1] ´ [1]) È ([4] ´ [4]) È ([6] ´ [6])

Going the Other Way Theorem: Let A be a non-empty set and let R be an equivalence relation on A. Then the distinct equivalence classes of R partition A. For example, given the relation of congruence mod 5 on the Integers, we obtain the partition: Z = [0] È [1] È [2] È [3] È [4]. If S = {0,1}, what partition of S4 is induced by R = {(s,t) | s,t Î S4 and density(s) = density(t)}? S4 = [0000]È[0001]È[0011]È[0111]È[1111]

Functions, Relations & Partitions
Let f be a function defined on a set A, and consider the relation R={(a,b) | f(a) = f(b)}. Show R is an equivalence relation and describe the partition of A induced by R. REF: f(a) = f(a) for all a Î A; SYM: If f(a) = f(b), then f(b) = f(a); TRAN: If f(a) = f(b) and f(b) = f(c), f(a) = f(c). Each partition set contains those elements whose output from f is the same.

An Example Let f be the function on Z, given by f(x) = x4 + 1.
Let f be the function on Z, given by f(x) = x4 + 1. x: 0 ±1 ±2 ±3 ±4 ... f(x): , This function induces the partition: Z = {0} È {-1, 1} È {-2, 2} È {-3, 3} È {-4,4}...

Graphs & Partitions If we generate the directed graph of an equivalence relation just right, then the induced partition jumps out: 3 2 6 1 5 4

Chapter 7: Functions Functions defined on general sets;
Functions defined on general sets; One-to-one, onto, and inverse functions; Composition of functions; Cardinality and functions.

Section 1: Functions on General Sets
Section 1: Functions on General Sets Definitions: A function f from set X to set Y is a relation such that each x in X is related to a unique y in Y. We denote this action as f:X ® Y. We call the set X the domain of f, denoted dom(f ), and the set Y, the range of f, denoted ran(f ). The set of actual elements of Y that appear as outputs of f is called the image of f, denoted im(f ) or f(X). In summary, dom(f ) = X, ran(f ) = Y and im(f ) = f(X) = {y Î Y | "x Î X, f(x) = y}.

Arrow Diagrams of Functions
We use arrow diagrams to illustrate the behaviour of the function: X = {1,2,3,4} Y={a,b} f = {(1,a),(2,a),(3,b),(4,b)} X = {1,2,3} Y = {a,b} f(A) = {a} f = {(1,a),(2,a),(3,a)} 1 2 3 4 a b 1 2 3 a b

Equality of Functions If f:X ® Y and g:X ® Y are functions, then we say f = g provided f(x) = g(x) for each x Î X. Consider f:R ® R and g:R ® R given by f(x) = Ö(x2) and g(x) = |x|. In this instance, f = g. Consider f:R ® R and g:R ® R given by f(x) = x (the identity function) and g(x) = |x|. In this case, f and g are equal only for x ³ 0, but f(-1) = -1 and g(-1) = 1, hence f ¹ g.

Functions on Binary Strings
Let S = {0,1} represent the binary alphabet. We consider the following functions involving binary strings: Length: L:S* ® N defined as L(s) = length of s. Density: d:S* ® N defined as d(s) = # 1’s in s. Hamming Distance Function: H:Sn x Sn ® N defined as H(s,t) = # places where s and t disagree. For example, L(101100) = 6, d(101100) = 3, and H(101100,100110) = 2.

Hamming Distance Function
The Hamming Distance function is of fundamental importance in the world of Error Correcting Codes, to find the distance between binary codewords in digital communications. H(s,t) can be calculated in two steps: 1. Set w = s Å t; 2. Calculate d(w); In other words: H(s,t) = d(s Å t). Thus H(101100,100110) = d(001010) = 2.

Boolean Functions We can consider the boolean functions we studied in Chapter 1 in the context of our function definition. Notation: {0,1}n = {0,1} ´ {0,1} ´...´ {0,1} (n cross products) Definition: An (n-place) Boolean function is any f:{0,1}n ® {0,1}. Thus, we can think of the function f(x,y,z) = xy + z’ as a function f:{0,1}3 ® {0,1}

Boolean Function Example
In this case, the function f(x,y,z) = xy + z’ is: 111 110 101 100 011 010 001 000 1

Relations Which Are Not Functions
Relations Which Are Not Functions The following is an example of a relation that is not a function. Why? f:R ® R given by f(x) = Ö(-x2). Not defined for any value in the domain! Here is another relation that is not a function. Why? f:Q ® Z given by f(p/q) = p. 1/2 = 2/4 = ..., but f(1/2) = 1, f(2/4) = 2, etc.

Section 3: One-to-one, Onto, and Inverse Functions
Section 3: One-to-one, Onto, and Inverse Functions In this section, we will look at three special classes of functions and see how their properties lead us to the theory of counting. So far, we have the general notion of a function f:X ® Y, but in terms of the comparative sizes of the three sets involved (X, Y and f ), all we can say is that |f | = |X|. In this section, we compare |X| with |Y|.

One-to-one Functions Definition: A one-to-one (injective) function f from set X to set Y is a function such that each x in X is related to a different y in Y. More formally, we can restate this definition as either: f :X ® Y is 1-1 provided f(x1) = f(x2) implies x1 = x2, or f :X ® Y is 1-1 provided x1 ¹ x2 implies f(x1) ¹ f(x2).

Illustrative Examples
The function below is 1-1: This function is not: 1 2 3 4 a b c d e 1 2 3 a b

Proving Functions Are 1-1
If f:R ® R is given by f(x) = 3x + 7, prove it is one-to-one. Proof: Assume f(a) = f(b). Show a = b. Now f(a) = f(b) means 3a + 7 = 3b + 7, so 3a = 3b, therefore a = b. Why is f:R ® R given by f(x) = x2 not 1-1? Since 9 = f(3) = f(-3), but 3 ¹ -3, the definition is violated.

Onto Functions Definition: A function f:X ® Yis said to be onto (surjective) if for every y in Y, there is an x in X such that f(x) = y. This can be restated as: A function is onto when its image equals its range, i.e. f(X) = Y. Examples: ONTO NOT ONTO 1 2 3 4 1 2 3 1 2 3 1

Testing Onto For Infinite Functions
Show that f:R ® R given by f(x) = 5x - 7 is onto. Let y be in R. Then (y + 7) and (y + 7)/5 are also real numbers. Now f( (y + 7)/5 ) = 5[(y + 7)/5] - 7 = y, hence if y is in R, there exists an x in R such that f(x) = y. Is f:R ® R given by f(x) = 1/x onto? No! There is no x in R that has output = 0.

One-to-one Correspondences
Definition: A function is called a one-to-one correspondence (bijection) if it is one-to-one and onto. One-to-one correspondences define the theory of counting. Why? If f:X ® Y is one-to-one, then |X| £ |Y|, and if f is onto, then |X| ³ |Y|, so if f is both, |X| = |Y|. Hence, to count the elements of an unknown set, we create a 1-1 correspondence between the set and a set of known size. Simple!

Inverse Functions Recall that the inverse relation is created by inverting all the ordered pairs that comprise the original relation. When is the inverse of a function itself a function? onto both not onto not 1-1 (f -1 is a function) (f -1 not def.) (f -1 not well-def.) 1 2 3 4

Finding Inverse Functions
Theorem: If f:X ® Y is a one-to-one and onto, then f -1 is a one-to-one and onto function. Given f , how do we find f -1? Let f:R ® R be given by f(x) = 4x - 1 = y. Now, swap x and y and solve for y: 4y - 1 = x 4y = x y = x Thus f -1(x) = (x + 1)/4.

Section 5: Composition of Functions
Section 5: Composition of Functions Recall the Hamming distance function, which we calculated as the density of the XOR of two equal length binary strings. This functions is calculated by chaining or composing two functions in such a way that the output of the first becomes the input of the second. This chaining process demonstrates the composition of two functions.

Composition of Functions
Definition: Let f:X ® Y and g:Y ® Z be functions such that the image of f is a subset of the domain of g. Define the new function g ° f:X ® Z as (g ° f )(x) = g( f (x)) for all x in X. We call g ° f the composition of f and g, putting f first since it acts upon x first. x f (x) g ° f (x) X Z Y

Example on Finite Sets Let f = {(1,4),(2,3),(3,4),(4,5),(5,6)} and g = {(1,3),(2,5),(3,1),(4,2),(5,3),(6,4)}. Then g ° f = {(1,2),(2,1),(3,2),(4,3),(5,4)} 1 2 3 4 5 6 1 2 3 4 5 1 2 3 4 5 g f

Example on Infinite Sets
If f:R ® R is given by f (x) = 3x + 7, and g:R ® R given by g(x) = x2, then (g ° f )(x) = g(f (x)) = g(3x + 7) = (3x + 7)2, and (f ° g )(x) = f (g(x)) = f (x2) = 3x2 + 7. Note: this example illustrates that in general: (g ° f )(x) ¹ (f ° g )(x), since (3x + 7)2 ¹ 3x2 + 7

Identity Functions Definition: Given a set X, the identity function on X, iX:X ® X is the function iX(x) = x. In pictures: If X = {1,2,3,4} then When composing the identity function with a general function f:X ® Y, we see f ° iX = f and iY ° f = f. 1 2 3 4 iX

Inverse Functions Definition: Given a one-to-one and onto function f:X ® Y, the inverse function of f, is the function f -1:Y ® X such that f ° f -1 = iY and f -1 ° f = iX. In pictures: 1 2 3 4 f f -1

Composition of 1-1 Functions
Theorem: Given one-to-one functions f:X ® Y, and g:Y ® Z, then the composition g ° f: X ® Z is a one-to-one function. In pictures: 10 11 12 13 14 15 f g 1 2 3 4 5 6 7 8 9

Composition of Onto Functions
Theorem: Given onto functions f:X ® Y, and g:Y ® Z, then the composition g ° f: X ® Z is an onto function. In pictures: f g X Y Z

Chapter 6: Counting Counting and probability;
Counting and probability; Possibility trees & the Multiplication Rule; The Addition Rule; Inclusion/Exclusion Rule; Permutations and combinations; Generalized permutations and combinations; Pascal’s Triangle & Combinatorial Identities.

Section 1: Counting and Probablilty
Section 1: Counting and Probablilty Imagine tossing a coin and noting the results (Heads or Tails) repeatedly. You expect the outcome of each toss to be independent of any previous toss. You expect that either H or T is equally likely at any toss, so a given collection of outcomes occurred at random. A sample space is the set of all possible outcomes of a random process, and an event is a subset of the sample space.

Probability If S is a finite sample space in which all outcomes are equally likely and E is an event in S, then the probability of event E is P(E) = |E| / |S|. For example, if we toss our coin twice and record the outcomes, then S = {HH, HT, TH, TT}. Now if we ask what is the probability the two tosses match, then E = {HH, TT}, so P(E) = 2/4 = 0.5. The odds of an event are (# way for):(# ways not) so O(E) = |E|:(|S| - |E|). Thus the odds of flipping the coin twice to match is 2:(4 - 2) = 2:2 = 1:1.

Probabilities for Playing Cards
Probabilities for Playing Cards Many interesting counting problems involve the standard 52-card deck of playing cards: 2 red suits (Diamonds and Hearts), 2 black suits (Clubs and Spades) with 13 values (2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, Ace) in each suit. The Jack, Queen, King, and Ace cards are called face cards. P(Black face card) = 8/52 = 2/13. P(Ace) = 4/52 = 1/13. P(Royal Straight Flush) = ???

Probablilties for Dice
Consider the case of rolling two 6-sided dice: P(7) = 6/36 = 1/6 P(12) = 1/36 P(at least 1 die = 6) = 11/36

Counting Elements in a List
How many integers from 1 to 100, inclusive? 100 is the answer, but = 99. Need to correct for the missing endpoint. Theorem: If m and n are integers with m £ n, then there are (n - m + 1) integers from m to n. Hence = 100. From -5 to 18? (-5) + 1 = 24.

Section 2: The Multiplication Rule
Section 2: The Multiplication Rule Consider the game of tossing a coin, then rolling a die, then picking a card. One possible event would be (H, 2, 2clubs). How many events are in the sample space? One way to visualize this counting problem is to construct a possibility tree, where each intermediate leaf branches out successively according to the type of event generated at that stage.

Coin - Die - Card Possibility Tree
Coin - Die - Card Possibility Tree (Start) Heads Tail 2c...As 2c...As c...As

The Multiplication Rule
The total number of ways of generating output from this last example is obtained by counting the number of leaves at the end of the tree. Clearly in this case the number is 2×6×52 = 624 events. The Multiplication Rule: If event1 can be done e1 ways and event2 can be done e2 ways, then the number of ways to complete event1 AND event2 is e1×e2 ways. The key observations is that AND leads us to multiply the counts.

License Plates Suppose Maryland issued auto license plates using the following scheme: Letter Letter Letter Digit Digit Digit (where Letter = A,B,...,Z and Digit = 0,1,2,...,9). How many possible license plates could be issued without duplication? Answer: 26×26×26×10×10×10 = 263×103 How about Letter Digit Letter Digit Letter Digit? Answer: 26×10×26×10×26×10 = 263×103

Cartesian Products & Strings
If a set A has m elements and a set B has n elements, how many elements are in A ´ B? We view this as taking an element from A and then taking an element from B, the multiplication rule applies: |A ´ B| = |A|×|B| = m×n. If S is an alphabet, how many elements in S6? We model this as: __ __ __ __ __ __, where each slot can be filled in one of the elements of S. Therefore |S6| = |S|×|S|×|S|×|S|×|S|×|S| = |S|6. In general, we see that |Sn| = |S|n.

Permutations A permutation of a collection of objects is an ordering of the objects in a row. Equivalently, a permutation is a one-to-one and onto mapping of a set to itself. List: abcde ® acdbe a b c d e Bijective Function:

Counting Permutations
To count how many permutations (orderings) of n things there are, consider it in terms of filling in slots: n (n-1) (n-2) We denote this product n×(n-1)×(n-2)× ... ×3×2×1 as n! (read “n factorial”). Hence the number of different orderings of 6 distinct books on a shelf is 6! = 6×5×4×3×2×1 = 720. We define 0! = 1 because it makes things work.

More Counting Permutations
How many orderings of the letters of the word COMPUTER are there? C O M P U T E R Answer: 8! = 40,320 How many orderings of the letters of the word COMPUTER are there if the letters CO must stay together in this order? CO M P U T E R Answer: 7! = 5,040 What is the probablilty the CO will appear in a given permutation of the letters of COMPUTER? Answer: 7!/8! = 7!/(8×7!) = 1/8.

Circular Permutations
Permutations are a linear ordering. However, we can ask how many ways can we order things around a circle (or other closed polygon). In this case, the number of ways to arrange n things in a circle (square, hexagon, etc.) is (n-1)! since placing the first element does not matter. Consider ABCDE around a circle: A E B D C E D A C B D C E B A C B D A E B A C E D

Partial Permutations What happens if we don’t want to list out all n elements in a linear permutation. Suppose we only want the first r elements, where r < n. ______ ______ ______ ______ n n n n-(r-1) Total = n(n-1)(n-2)...(n-r+1) = n(n-1)(n-2)...(n-r+1)[(n-r)!/(n-r)!] = n!/(n-r)! We denote the partial permutation P(n,r). Theorem: P(n,r) = n(n-1)...(n-r+1) = n!/(n-r)!

Evaluating of Partial Permutations
Evaluating of Partial Permutations How many 5-letter strings can be formed from the letters of COMPUTER? Answer: P(8,5) = 8!/(8-5)! = 8!/3! = 8×7×6×5×4 How many 6-character license plates can be made if a character can be a letter or digit, but no character can repeat? Answer: P(36,6) = 36!/30! = 36×35×34×33×32×31 When do you use which formula: Use n!/(n-r)! when the form of the answer counts. Use the expanded product to actually calculate.

Summary Multiplication Rule: If there are m ways to do A and n ways to do B, then there are (m×n) ways to do A and B. (order matters & repetition allowed) Permutations (order matters & NO repetition): There are n! linear orderings of n things. There are n!/n = (n-1)! circular orderings of n things. There are P(n,r) = n!/(n-r)! orderings of r things from a total of n things.

Section 3: Counting Elements in Disjoint Sets
In the last section, we looked at counting events that branch from one state to the next. These events we modelled using possibility trees and obtained counts using the Multiplication Rule. In this section, we will learn how to count elements in the union, difference, or intersection of two sets using the Addition Rule. We will also look at counting when set operations are acting upon subset structures.

The Addition Rule Theorem: Suppose a finite set A is partitioned by the collection {A1, A2, ..., An}. Then: n(A) = n(A1) + n(A2) n(An). How many binary strings of length up to and including 4 are there? Answer: G4 = S0 È S1 È S2 È S3 È S4, and Si Ç Sj = Æ if i ¹ j, so |G4| = |S0| + |S1| + |S2| + |S3| + |S4| = = = 31.

Another Example Let D be set of 3-digit integers (from 100 to 999) which are divisible by 5. How big is D? Solution: We know an integer is divisible by 5 if it ends in a 0 or 5. Let’s call a typical 3-digit integer abc, where a is one of {1,2,3,4,5,6,7,8,9}, and b is one of {0,1,2,3,4,5,6,7,8,9}. Now, we partition D into those integers of the form ab0 and those of the form ab5. The first set has 9×10 elements and the second set has 9×10 elements, hence D has 180 elements.

License Plates We have already used the addition rule without realizing it in the last section. When we are counting the number of license plates with certain properties and allow a particular character to be either a digit or a letter, we have applied the addition rule: The slot drawing (Digit or Letter) (Digit or Letter) gives us the counts ( ) × ( )...

The Difference Rule Consider the following: so |A| = |A1| + |A2|. A1
Consider the following: so |A| = |A1| + |A2|. Now, in the addition rule, we know |A1| and |A2| and use this to find |A|. But if we know |A| and either |A1| or |A2|, then we get: Difference Rule: If A1 and A2 partition A, then |A2| = |A| - |A1| and |A1| = |A| - |A2|. A1 A2 A

Counting Things Without a Condition
Counting Things Without a Condition How many 3-digit integers are not divisible by 5? Before, we saw that there are digit integers which are divisible by 5, hence: {all 3-digit integers not divisible by 5} = {all 3-digit integers} - {all 3-digit integers divisible by 5}. Therefore there are ( ) = digit integers which are not divisible by 5.

Another Complement Example
Another Complement Example How many ways can 4 couples sit around a circular table if one couple cannot be seated next to each other? Solution: We see that {seatings w/ couple apart} = {all seatings} - {seating w/ couple together} Label the people abcdefgh and suppose couple ab cannot sit together. Thus we want to count how to seat Xcdefgh around a table, then repeat it since X represents both the ab and ba orderings. Answer = 7! - 6! - 6! = 7! - 2×6!.

A Probability Example Suppose there are 10,000 total ways to carry out an experiment with 2,123 of those experiments leading to success. What is the probability the experiment will fail? Solution: Starting from the sample space, we get: |{all experiments}| = |{successes}| + |{failures}|, Dividing boths sides by the LHS, we see that: 1 = P(success) + P(failure), so P(failure) = 1 - P(success) = = We find there is a 78.77% probability of failure.

Inclusion/Exclusion Rule
Inclusion/Exclusion Rule So far, our counting under the addition and subtraction rules have required that the files in question are disjoint. This is an unreasonable requirement in general. How do we account for the aggregrate knowing the parts that make it up? A A Ç B B A È B Inclusion/Exclusion Rule: For any sets A and B, |A È B| = |A| + |B| - |A Ç B|.

Inclusion/Exclusion Example
Inclusion/Exclusion Example How many 6-character license plates begin with “A” or end with “9”? Effectively, we want to count all plates of the forms Axxxxx or xxxxx9, but recognizing that these individual cases overlap for plates of the form Axxxx9. Now |{Axxxxx}| = 365, and |{xxxxx9}| = 365, and |{Axxxx9}| = 364, hence the total number of plates is 2×

What About For 3 Sets? Using the labeling in the above drawing, we see that A = 1 È 4 È 6 È 7, B = 2 È 4 È 5 È 7, and C = 3 È 5 È 6 È 7. 1 4 7 6 2 5 3 Thus |A È B È C| = |A| + |B| + |C| ( ) - |AÇB| - |AÇC| - |BÇC| ( ) + |AÇBÇC| (+7) =

Inclusion/Exclusion Rule for 3 Sets
Inclusion/Exclusion Rule for 3 Sets If A, B, and C are sets, then: |A È B È C| = |A|+|B|+|C|-|AÇB|-|AÇC|-|BÇC|+|AÇBÇC|. Example: The CS department teaches Algol, Basic, and C languages. Suppose: 22 students study Algol; 30 study Basic; 42 study C; 12 study Algol and Basic; 18 study Basic and C; 16 study Algol and C; 9 study all three. How many students study a language? Answer: 57 =

Section 4: Counting Subsets of a Set
In Section 2, we looked at counting events with or without repetition, but in either instance the order of the elements mattered. Now, we shall relax the order restriction to allow counting set structures where events are not distinguished by the order of elements, but by the mere clustering of elements together. This will lead to the last rule, the Division Rule (not in the text!).

The Division Rule Theorem: Suppose a set A has n elements and is partitioned by the collection {A1, A2, ..., Ap}, where each partition set has m elements. Then: p = n/m. In other words, if a set is partitioned into equal-sized partition sets, then the number of partition sets is the quotient of the size of the set with the size of any partition set. For example, if a set has 100 elements and is partitioned in 20-element subsets, then there must be 5 subsets (equivalence classes).

Counting Subsets How many 3-element subsets of a 4-element set are there: Let A = {1,2,3,4} then all 3-permutations are: 123, 132, 213, 231, 312, 321 ® {1,2,3} 124, 142, 214, 241, 412, 421 ® {1,2,4} 134, 143, 314, 341, 413, 431 ® {1,3,4} 234, 243, 324, 342, 423, 432 ® {2,3,4}. Hence # 3-element subsets = (# 3-permutations) / (# 3-orderings) = P(4,3) / 3! = 4! / (1!3!) = 4! / 3! = 4.

( ) = Combinations n! n k!(n - k)! k
What we have just counted is a combination. In this instance, it was a combination of 4 elements taken 3 at a time. We use the Division Rule to negate the order condition of the permutation counts. In general, C(n,k) = P(n,k) / k! Equivalently, we use the “choose” notation to get: n! k!(n - k)! n k ( ) =

More Counting Subsets How many subsets of a 10-element set have 3 elements? How many have 7 elements? Solution: C(10,3) = 10! / (3!7!) C(10,7) = 10! / (7!3!), the same! Note: Counting subsets containing 3 elements is the same as counting subsets NOT containing the other 7 elements! Theorem: C(n,k) = C(n,n-k). How many subsets have at least 8 elements? Solution: C(10,8) + C(10,9) + C(10,10)

Counting Binary Strings
How many 10-bit strings have three 1’s? Solution: We model this as requiring us to choose three of the ten slots to place a 1 then the other seven remaining slots will get a 0. Thus the number of 10-bit strings that have three 1’s is C(10,3) = 10! / (3!7!). In general, the number of n-bit binary strings with density k is C(n,k). As before, having k 1’s is the same thing as having (n - k) 0’s.

Counting Teams Suppose a group of 5 men and 7 women want to pick a 5-person team. How many teams can they make with 3 men and 2 women? C(5,3)C(7,2) = [5!/(3!2!)][7!/(5!2!)] = 7!/(3!2!2!). How many teams have at least 1 man? (All - teams w/no man) = C(12,5) - C(7,5) = [12!/(7!5!)] - [7!/(5!2!)] How many teams have at most 1 man? (teams w/no man) + (teams w/1 man) = C(7,5) + C(5,1)C(7,4) = 7!/(5!2!) + [5!/4!][7!/(3!4!)]

Generalized Permutations
In Section 2, we learned how to count the number of orderings of the letters of COMPUTER. What about the number of orderings of the letters of MISSISSIPPI? In this case, we note that not all the letters are distinct. In particular, MISSISSIPPI ® IIIISSSSPPM, so although we are still searching for an ordering structures, there are sub-unorderings present, induced by the repeated letters, for which we have to account.

Generalized Permutations Take 1
Generalized Permutations Take 1 Let’s apply the Division Rule to negate the effect of the unordering portions of the overall order problem. This leaves us with a total count of 11!/4!4!2!. Here, the first quotient of 4! “mods” out the effect of the unordered I’s, the second quotient of 4! “mods” out the effect of the unordered S’s, and the last quotient of 2! “mods” out the effect of the unordered P’s.

Generalized Permutations Take 2
Generalized Permutations Take 2 If we model this problem, purely as a combination, and not a permutation at all, we can reason the task as: 1. Choose 4 slots from 11 for the I’s; 2. Choose 4 slots from the remaining for the S’s; 3. Choose 2 slots from the remaining for the P’s; 4. Place the M (only 1 way remaining). This yields: C(11,4)C(7,4)C(3,2)C(1,1) = (11!7!3!)/(7!4!4!3!2!1!) = 11!/(4!4!2!).

Generalized Permutation Theorem
Generalized Permutation Theorem Theorem: Suppose a collection consists of n objects of which: n1 are of type 1, indistinguishable from each other; n2 are of type 2, indistinguishable from each other; ... nk are of type k, indistinguishable from each other; and n1 + n nk = n. Then the number of distinct permutations of the n objects is: C(n,n1)C(n-n1,n2)C(n-n1-n2,n3)...C(nk,nk) = n! / (n1!n2!n3!...nk!).

Section 5: Combinations with Repetition
In the last section, we saw how to count combinations, where order does not matter, based on permutation counts, and we saw how to count permutations where repetitions occur. Now, we shall consider the case where we don’t want order to matter, but we will allow repetitions to occur. This will complete the matrix of counting formulae, indexed by order and repetition.

A Motivating Example How many ways can I select 15 cans of soda from a cooler containing large quantities of Coke, Pepsi, Diet Coke, Root Beer and Sprite? We have to model this problem using the chart: Coke Pepsi Diet Coke Root Beer Sprite A: =15 B: =15 C: =15 Here, we set an order of the categories and just count how many from each category are chosen.

A Motivating Example (cont’d.)
A Motivating Example (cont’d.) Now, each event will contain fifteen 1’s, but we need to indicate where we transition from one category to the next. If we use 0 to mark our transitions, then the events become: A: B: C: Thus, associated with each event is a binary string with #1’s = #things to be chosen and #0’s = #transitions between categories.

Counting Generalized Combinations
Counting Generalized Combinations From this example we see that the number of ways to select 15 sodas from a collection of 5 types of soda is C(15 + 4,15) = C(19,15) = C(19,4). Note that #zeros = #transitions = #categories - 1. Theorem: The number of ways to fill r slots from n catgories with repetition allowed is: C(r + n - 1, r) = C(r + n - 1, n - 1). In words, the counts are: C(#slots + #transitions, #slots) or C(#slots + #transitions, #transitions).

Another Example How many ways can I fill a box holding 100 pieces of candy from 30 different types of candy? Solution: Here #slots = 100, #transitions = , so there are C(100+29,100) = 129!/(100!29!) different ways to fill the box. How many ways if I must have at least 1 piece of each type? Solution: Now, we are reducing the #slots to choose over to ( ) slots, so there are C(70+29,70) = 99!/70!29!

When to Use Generalized Combinations
When to Use Generalized Combinations Besides categorizing a problem based on its order and repetition requirements as a generalized combination, there are a couple of other characteristics which help us sort: In generalized combinations, having all the slots filled in by only selections from one category is allowed; It is possible to have more slots than categories.

Integer Solutions to Equations
Integer Solutions to Equations One other type of problem to be solved by the generalized combination formula is of the form: How many non-negative integer solutions are there to the equation a + b + c + d = 100. In this case, we could have 100 a’s or 99 a’s and 1 b, or 98 a’s and 2 d’s, etc. We see that the #slots = 100 and we are ranging over 4 categories, so #transitions = 3. Therefore, there are C(100+3,100) = 103!/100!3! non-negative solutions to a + b + c + d = 100.

Integer Solutions with Restrictions
Integer Solutions with Restrictions How many integer solutions are there to: a + b + c + d = 15, when a ³ 3, b ³ 0, c ³ 2 and d ³ 1? Now, solution “strings” are 111a0b011c01d, where the a,b,c,d are the remaining numbers of each category to fill in the remaining slots. However, the number of slots has effectively been reduced to 9 after accounting for a total of 6 restrictions. Thus there are C(9+3,9) = 12!/(9!3!) solutions.

More Integer Solutions & Restrictions
More Integer Solutions & Restrictions How many integer solutions are there to: a + b + c + d = 15, when a ³ -3, b ³ 0, c ³ -2 and d ³ -1? In this case, we alter the restrictions and equation so that the restrictions “go away.” To do this, we need each restriction ³ 0 and balance the number of slots accordingly. Hence a ³ -3+3, b ³ 0, c ³ -2+2 and d ³ -1+1, yields a + b + c + d = = 21 So, there are C(21+3,21) = 24!/(21!3!) solutions.

Summary Theorem: The number of integer solutions to: a1 + a2 + a an = r, when a1 ³ b1, a2 ³ b2, a3 ³ b3 , ..., an ³ bn is C(r+n-1-b1-b2-b3-...-bn , r-b1-b2-b3-...-bn). Theorem: The number of ways to select r things from n categories with b total restrictions on the r things is C(r + n b , r - b). Corollary: The number of ways to select r things from n categories with at least 1 thing from each category is C(r - 1 , r - n) (set b = n).

Section 6: The Algebra of Combinations
Section 6: The Algebra of Combinations In this section, we will look at some identities involving the choosery function. We will also resurrect Pascal’s Triangle and see how it relates to combinations. Three Simple Calculations: C(n,n) = n! / [n!(n-n)!] = n!/(n!0!) = n!/n! = 1. C(n,1) = n! / [1!(n-1)!] = [n(n-1)!] / (n-1)! = n. C(n,2) = n! / [2!(n-2)!] = [n(n-1)(n-2)!] / [2(n-2)!] = n(n-1)/2.

A Half-empty Set We have already seen and used the identity: C(n,r) = C(n,n-r). This identity is easy to demonstrate algebraically. However, we can reason it combinatorically. We use C(n,r) to count how many subset of size r an n-element set A has. But we can identify uniquely with any r-element subset B of A the (n-r)-element subset that is its complement, A-B. Moreover, this identification is a bijection. (Why?) Hence, the number of r-element subsets equals the number of (n-r)-element subsets.

Using Substitutions We have seen the identity C(n,2) = n(n-1)/2.
Using Substitutions We have seen the identity C(n,2) = n(n-1)/2. Combining this with C(n,2) = C(n,n-2), we see that C(n,n-2) = n(n-1)/2. We can now use this to get related identities by substituting “interesting” values for n: n ® n+1: C(n+1,n-1) = n(n+1)/2. n ® n-1: C(n-1,n-3) = (n-1)(n-2)/2. n ® n+2: C(n+2,n) = (n+2)(n+1)/2.

Pascal’s Triangle Recall the number array we call Pascal’s Triangle: 1
Pascal’s Triangle Recall the number array we call Pascal’s Triangle: 1 1 1 Rule of generation: T(n,r) = T(n-1,r-1) + T(n-1,r).

Using Pascal’s Triangle
Using Pascal’s Triangle One application of Pascal’s Triangle is to find the coefficients of the binomial expansion (a + b)n. For example, to expand (a + b)5 we look at the 6th row: 1,5,10,10,5,1 to get: 1a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + 1b5. However, the Binomial Theorem tells us each of these coefficients is of the form C(5,r). Thus each term of Pascal’s Triangle is actually a combination, that is T(n,r) = C(n,r).

Pascal’s Formula Now, if we replace the the T(n,r) terms of the rule of generation with the binomial coefficients, we get: C(n,r) = C(n-1,r-1) + C(n-1,r). Proof: C(n-1,r-1) + C(n-1,r) = (n-1)!/(r-1)!(n-1-r+1)! + (n-1)!/r!(n-r-1)! = (n-1)!/(r-1)!(n-r)! + (n-1)!/r!(n-r-1)! [want (r-1)! ® r! want (n-r-1)! ® (n-r)!] = (n-1)!r/r!(n-r)! + (n-1)!(n-r)/r!(n-r)! = [(n-1)!(r + n - r)] / r!(n-r)! = (n-1)!n / r(n-r)! = n!/r!(n-r)! = C(n,r).

Chapter 8: Recursion Recursively defined sequences;
Chapter 8: Recursion Recursively defined sequences; The Iteration Method to solve recurrence relations; Solve linear, homogeneous recurrence relations; Count Fibonacci’s bunnies.

Section 1: Recursively Defined Sequences
Section 1: Recursively Defined Sequences In Chapter 4, we looked at sequences, although most of them were generated by a function. In this section, we will study sequences where new terms are calculated based on the values of predecessor terms. Definition: A recurrence relation for a sequence a1, a2, ..., an, is a formula that calculates each term ak in terms of ak-1,ak-2,...,ak-i, for some integer i. The initial conditions for a recurrence relation specify values for a1, a2, ..., ak-1.

Examples of Recurrence Relations
Examples of Recurrence Relations To actually determine the sequence specified by a recurrence relation, you need: (1) the relation: an = an-1 + an-2, and (2) initial conditions: a0 = 1 and a1 = 3. This example yields the sequence: 1, 3, 4, 7, 11, 18, 29, ... If we keep the same recurrence relation, but change our initial conditions (say a0 = 2, a1 = -5), we get a completely different sequence: 2, -5, -3, -8, -11, -19, -30, ...

Different Representations
Different Representations We can specify a sequence using essentially the same recurrence relation, but in different ways if we get “unstuck” in time. Consider these two descriptions of the same sequence: (1) sk = (3sk-1 - 1), for all integers k ³ 1; (2) sk+1 = (3sk - 1), for all integers k ³ 0. If we let s1 = 1, (1) becomes 1, 2, 5, 14, 41, ... If we let s0 = 1, (2) becomes 1, 2, 5, 14, 41, ...

The Tower of Hanoi Here is a game that generates a recurrence relation. The Tower of Hanoi consist of a collection of disks with holes in the middle placed on a board with three vertical posts. The disks are such that they can be placed on a single post and from the bottom up, each successive disk has decreasing diameter.

The Tower of Hanoi (cont’d.)
The Tower of Hanoi (cont’d.) The game is to move the tower of disks from one post to another using the rules that you can only move 1 disk at a time and you cannot place a disk of larger diameter on top of a smaller disk. How many moves (mn) will it take to solve the n-disk Tower of Hanoi? Clearly m1 = 1. Now, claim m2 = 3. Why? (1,2)()() ® (2)(1)() ® ()(1)(2) ® ()()(1,2).

The Tower of Hanoi (cont’d.)
The Tower of Hanoi (cont’d.) How about the 3-disk Tower of Hanoi? (1,2,3)()() ® (2,3)(1)() ® (3)(1)(2) ® (3)()(1,2) ® ()(3)(1,2) ® (1)(3)(2) ® (1)(2,3)() ® ()(1,2,3)(). Thus, m3 = 7. This example illustrates the recursive nature of the game: To solve the 3-disk problem, we solve the 2-disk problem, then move disk 3, then solve the 2-disk problem again. So m3 = 2m2 + 1. Continuing this logic, we see mn= 2mn

Fibonacci Numbers Leonardo of Pisa, son of Bonacci (and hence, Fibonacci), posed the following in 1202: A single pair of rabbits (a male and a female) are born at the beginning of the year. Rabbit pairs are fertile one month after their birth, and produce one mixed pair each month thereafter. The rabbit population suffers no deaths during the course of the year. How many rabbits will there be at the end of the year? Clearly, at the end of any given month: #rabbits = #alive at the beginning + #babies.

Fibonacci Numbers (cont’d.)
Fibonacci Numbers (cont’d.) Suppose the number of pairs of rabbits alive at the beginning of month k is Fk. Then: (a) F0 = 1 (A) F1 = 1 (A,b) F2 = = 2 (A,B,c) F3 = = 3 = F2 + 1 = F2 + F1 (A,B,C,d,e) F4 = = = 5 = F3 + F2 (A,B,C,D,E,f,g,h) F5 = = F4 + F3. In general: Fk = Fk-1 + Fk-2 , so the sequence is: 1,1,2,3,5,8,13,21,34,55,89,144,233 = 1 year!

Solving Recurrence Relations
Solving Recurrence Relations The last example nicely illustrates why we want to “solve” recurrence relations; that is, obtain an expression for the general term (Fn) that only depends on n and not Fn-1, Fn-2, Fn-3, ... To find a term like F13, it requires us to find all the predecessor terms, a tedious and uninteresting task. What if we had to find F1,000,000????? From here on, we will develop methods to solve recurrence relations, so we won’t need to plug in values again and again.

Section 2: Solving Recurrence Relations by Iteration
Section 2: Solving Recurrence Relations by Iteration As we noted at the end of the last lecture, when analyzing recurrence relations, we want to rewrite the general term as a function of the index and independent of predecessor terms. This will allow us to compute any arbitrary term in the sequence without having to compute all the previous terms. In this section, we will look at one method for solving recurrence relations.

The Method of Iteration
The Method of Iteration Let {ai} be the sequence defined by: ak = ak with a0 = 1. Plugging values of k into the relation, we get: a1 = a0 + 2 = a2 = a1 + 2 = = 1 + 2(2) a3 = a2 + 2 = = 1 + 3(2) a4 = a3 + 2 = = 1 + 4(2) Continuing in this fashion reinforces the apparent pattern that an = 1 + n(2) = 1 + 2n. This brute force technique is the Method of Iteration.

The Tower of Hanoi Recall the Tower of Hanoi relation: mk = 2mk with m1 = 1. Plugging values of k into the relation, we get: m2 = 2m1 + 1 = m3 = 2m2 + 1 = 2(2 + 1) + 1 = m4 = 2m3 + 1 = 2( ) = m5 = 2m4 + 1 = 2( ) = Thus, we guess: mn = 2n-1 + 2n This is a Geometric Series, so mn = 2n - 1.

Another Example Let {ai} be the sequence given by: ak = ak-1 + k with a0 = 0. Solve this recurrence relation and find a100. Now, a1 = a0 + 1 = 1 + 0 a2 = a1 + 2 = a3 = a2 + 3 = a4 = a3 + 4 = Thus an = n + (n-1) + (n-2) so an = n(n+1)/2. Plugging in n = 100: a100 = 100(101)/2 = 5050.

A Geometric Sequence Let a and b be non-zero constants, and consider: sk = ask-1 with s0 = b. Thus: s1 = as0 = ab s2 = as1 = a(ab) = a2b s3 = as2 = a(a2b) = a3b s4 = as3 = a(a3b) = a4b. From this, we can make the conjecture that: sn = anb. Note: if b = 1, then sn = an.

A Perturbed Geometric Sequence
A Perturbed Geometric Sequence Let a be non-zero constants, and consider: sk = ask with s0 = 1. Thus: s1 = as0 + 1 = a + 1 s2 = as1 + 1 = a(a + 1) + 1 = a2 + a + 1 s3 = as2 + 1 = a(a2 + a + 1) = a3 + a2 + a + 1 s4 = as3 + 1 = a(a3 + a2 + a + 1) = a4 + a3 + a2 + a + 1. It appears that sn = an + an a2 + a + 1, so sn = (an+1 - 1) / (a - 1) .

Some Observations In solving these recurrence relations, we point out the following observations: 1. Each recurrence relation looks only 1 step back; that is each relation has been of the form sn = F(sn-1); 2. We have relied on luck to solve the relation, in that we have needed to observe a pattern of behavior and formulated the solution based on the pattern; 3. The initial condition has played a role in making this pattern evident; 4. Generating a formula from the generalization of the pattern looks back to our study of induction.

Verifying Solutions Once we “guess” the form of the solution for a recurrence relation, we need to verify it is, in fact, the solution. We use Mathematical Induction to do this. For example, in the Tower of Hanoi game, we conjecture that the solution is mn = 2n -1. Basis Step: m1 = 1 (by playing the game), and = = 1, therefore m1 = Inductive Step: Recall the recurrence relation is mk = 2mk Assume mk = 2k - 1.

Verifying Solutions (cont’d.)
Verifying Solutions (cont’d.) Inductive Step: Show mk+1 = 2k Now, mk+1 = 2mk = 2(2k - 1) = 2×2k = 2k Therefore mk = 2k - 1 is the solution to mk = 2mk-1 + 1, when m1 = 1. QED A similar verification process will work for all the other formulas we discovered in this section.

Section 3: Linear, Homogeneous Recurrence Relations
Section 3: Linear, Homogeneous Recurrence Relations So far, we have seen that certain simple recurrence relations can be solved merely by interative evaluation and keen observation. In this section, we seek a more methodical solution to recurrence relations. In particular, we shall introduce a general technique to solve a broad class of recurrence relations, which will encompass those of the last section as well as the tougher Fibonnaci relation.

Linear, Homogeneous Recurrence Relations with Constant Coefficients
Linear, Homogeneous Recurrence Relations with Constant Coefficients If A and B (¹ 0) are constants, then a recurrence relation of the form: ak = Aak-1 + Bak-2 is called a linear, homogeneous, second order, recurrence relation with constant coefficients. We will use the acronym LHSORRCC. Linear: All exponents of the ak’s are 1; Homogeneous: All the terms have the same exponent. Second order: ak depends on ak-1 and ak-2;

Higher Order Linear, Homogeneous Recurrence Relations
Higher Order Linear, Homogeneous Recurrence Relations If we let C1, C2, C3, ..., Cn be constants (Clast ¹ 0), we can create LHRRCC’s of arbitrary order. As we shall see, the techniques the book develops for second order relations generalizes nicely to higher order recurrence relations. Third order: ak = C1ak-1 + C2ak-2 + C3ak-3 Fourth order: ak = C1ak-1 + C2ak-2 + C3ak-3 + C4ak-4 nth order: ak = C1ak-1 + C2ak Cnak-n;

Solving LHSORRCC’s Let’s start with the second order case before we generalize to higher orders. Definition: Given ak = Aak-1 + Bak-2, the characteristic equation of the recurrence relation is x2 = Ax + B, and the characteristic polynomial of the relation is x2 - Ax - B. Theorem: Given ak = Aak-1 + Bak-2, if s,t,C,D are non-zero real numbers, with s ¹ t, and s,t satisfy the characteristic equation of the relation, then its General Solution is an = C(sn)+ D(tn).

An Example Let ak = 5ak-1 - 6ak-2. Find the general solution.
An Example Let ak = 5ak-1 - 6ak-2. Find the general solution. The relation has characteristic equation: x2 = 5x - 6, so x2 - 5x + 6 = 0 hence (x - 2)(x - 3) = 0 implying either (x - 2) = 0 or (x - 3) = 0 thus x = 2,3 General Solution is an = C(2n) + D(3n).

Finding Particular Solutions
Finding Particular Solutions Once we have found the general solution to a recurrence relation, if we have a sufficient number of initial conditions, we can find the particular solution. This means we find the values for the arbitrary constants C and D, so that the solution for the recurrence relation takes on those initial conditions. The required number of initial conditions is the same as the order of the relation.

An Example For the last example, we found the recurrence relation ak = 5ak-1 - 6ak-2 has general solution an = C(2n) + D(3n). Find the particular solution when a0 = 9 and a1 = 20. a0 = C(20)+ D(30) = C + D = 9 a1 = C(21)+ D(31) = 2C + 3D = 20, so 2C + 2D = 18 2C + 3D = 20, so D = 2 and C = 7. Therefore, the particular solution is: an = 7(2n) + 2(3n).

Generalizing These Methods
Generalizing These Methods We can extend these techniques to higher order LHRRCC quite naturally. Suppose we have a LHRRCC whose charateristic poylnomial has roots x = 2, -3,5,7,11, and 13. Then its general solution is: an = C2n + D(-3)n + E5n + F7n + G11n + H13n. Moreover, if we have initial conditions specified for a0, a1, a2, a3, a4, and a5, we can plug them into the general solution and get a 6´6 system of equations to solve for C, D, E, F, G, and H.

Solving The Fibonacci Relation
Solving The Fibonacci Relation Solve: an = an-1 + an-2 when a0 = 1 and a1 = 1. Solution: In this case, the characteristic polynomial is x2 - x - 1, which doesn’t factor nicely. We turn to the quadratic formula to find the roots. Quadratic Formula: If ax2 + bx + c = 0, then x = [-b ± Ö (b2 - 4ac)]/2a. In our case, we have a = 1, b = -1 and c = -1, so x = [-(-1) ± Ö((-1)2 - 4(1)(-1))]/2(1) = (1 ± Ö5)/2. Thus an = C[(1 + Ö5)/2]n + D[(1 - Ö5)/2]n.

Solving The Fibonacci Relation (cont’d.)
Solving The Fibonacci Relation (cont’d.) If we apply the initial conditions a0 = 1 and a1 = 1 to an = C[(1 + Ö5)/2]n + D[(1 - Ö5)/2]n, we get: a0 = C + D = 1 a1 = [(1 + Ö5)/2]C + [(1 - Ö5)/2]D = 1, yielding C = (1 + Ö5)/(2Ö5) and D = -(1 - Ö5)/(2Ö5). Therefore an = [(1 + Ö5)/(2Ö5)][(1 + Ö5)/2]n [-(1 - Ö5)/(2Ö5)][(1 - Ö5)/2]n This simplifies to an = (1/Ö5){[(1 + Ö5)/2]n+1 - [(1 - Ö5)/2]n+1

Single Root Case So far, our technique for solving LHSORRCCs has depended on the fact that the two roots of the characteristic polynomial are distinct. This is not always the case, however. We can find that a polynomial has only one root, s, whenever the polynomial factors as (x - s)2. In this case, our solution takes on a special variant to ensure “linear independence” of the solutions. Theorem: If an LHSORRCC has a repeated root s, then the general solution is an = (A + Bn)sn.

Single Root Case Example
Single Root Case Example Find the general solution of an - 6an-1 + 9an-2 = 0. This LHSORRCC has a characteristic polynomial equation of x2 - 6x + 9 = 0, so (x - 3)2 = 0, which yields the sole root x = 3. Therefore, the general solution is an = (A + Bn)3n. If we add initial conditions a0 = 2 and a1 = 21, we get: a0 = (A + B(0))30 = A = 2, and a1 = (A + B(1))31 = 3(A + B) = 3(2 + B) = 21, so 2 + B = 7, hence B = 5. Therefore the particular solution is an = (2 + 5n)3n

Higher Order Repeated Root Case
Higher Order Repeated Root Case This method of building up the “coefficient” when the variable part degenerates because of repeated roots extends nicely to higher order problems as well. Example: If a LHRRCC has characteristic polynomial with roots x = 7,7,7,7,7,7,7,9,9,9 then its general solution is: an = (A + Bn + Cn2 + Dn3 + En4 + Fn5 + Gn6)7n + (H + In + Jn2)9n. How many IC are needed for a particular solution?

Summary Our general technique for solving LHRRCCs is a two-step process. Step 1: Find the roots of the characteristic polynomial and use them to develop the general solution. How do I find roots of polynomials? Step 2: Use the initial conditions to make and solve a system of linear equations that determine the arbitrary constants in the general solution to get the particular solution. How do I solve systems of linear equations?

Board Example #1 Given the recurrence relation an = 4an-1 - 3an-2, find a999 when a0 = 5 and a1 = 7.

Board Example #2 Given the recurrence relation an = 4an-1 - 4an-2, find a999 when a0 = 5 and a1 = 7.

Board Example #3 What is the general solution for the LHRRCC whose characteristic polynomial is: (x + 5)6(x - 3)4(x + 8)2

Board Example #4 Given the LHRRCC an = 2an-1 + 5an-2 - 6an-3, find a999 when a0 = 17, a1 = 14, and a2 = 110.

Validity of the General Solution I
Prove: If Aan + Ban-1 + Can-2 = 0, and s ¹ t satisfy Ax2 + Bx + C = 0, then ak = Msk + Ntk satisfies the relation. Proof: Let Aan + Ban-1 + Can-2 = 0, and s ¹ t satisfy Ax2 + Bx + C = 0. Thus: As2 + Bs + C = At2 + Bt + C = 0. Now, an = Msn + Ntn, an-1 = Msn-1 + Ntn-1, and an-2 = Msn-2 + Ntn-2 hence Aan + Ban-1 + Can-2 = A(Msn + Ntn) + B(Msn-1 + Ntn-1) + C(Msn-2 + Ntn-2) = M(Asn + Bsn-1 + Csn-2) + N(Atn + Btn-1 + Ctn-2) = Msn-2(As2 + Bs + C) + Ntn-2(At2 + Btn-1 + C) = 0.QED

Validity of the General Solution II
Prove: If Aan + Ban-1 + Can-2 = 0, and s is the only solution of Ax2 + Bx + C = 0, then ak = (P + Qk)sk satisfies the relation. Proof: Let Aan + Ban-1 + Can-2 = 0, and s be the only solution of Ax2 + Bx + C = 0, so As2 + Bs + C = 0. Now, an = (P + Qn)sn, an-1 = [P + Q(n-1)]sn-1, and an-2 = [P + Q(n-2)]sn-2 hence Aan + Ban-1 + Can-2 = A(P + Qn)sn + B[P + Q(n-1)]sn C[P + Q(n-2)]sn-2 = P(Asn + Bsn-1 + Csn-2) + Q[Ansn + B(n-1)sn-1 + C(n-2)sn-2]

Validity of the General Solution II
Thus, Aan + Ban-1 + Can-2 = Psn-2(As2 + Bs + C) + Q(Ansn + Bnsn-1 - Bsn Cnsn-2 - 2Csn-2) = Qnsn-2(As2 + Bs + C ) + Qsn-2(-Bs - 2C) = Qsn-2(-Bs - 2C) = 0????? However, since s is the only root of the characteristic polynomial, from the Quadratic Formula, we have that (B2 - 4AC) = 0 and s = -B/2A. Thus (-Bs - 2C) = -B(-B/2A) - 2C = B2/2A - 2C(2A/2A) = (B2 - 4AC)/2A = 0. QED

2. Empty Set, Partitions, Power Set

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2. Empty Set, Partitions, Power Set

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