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1 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Introduction Chapter 1 - Introduction Applications of discrete mathematics: Formal Languages (computer languages) Compiler Design Data Structures Computability Automata Theory Algorithm Design Relational Database Theory Complexity Theory (counting) Example (counting): The Traveling Salesman Problem Important in circuit design many other CS problems Given: n cities c 1, c 2,..., c n distance between city i and j, d ij Find the shortest tour. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

2 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Chapter 1 - Introduction Assume a very fast PC: 1 flop = 1 nanosecond = 10-9 sec. = 1,000,000,000 ops/sec = 1 GHz. A tour requires n-1 additions. How many different tours? etc. Choose the first city n ways, the second city n-1 ways, the third city n-2 ways, # tours = n (n-1) (n-2)....(2) (1) = n! (Combinations) Total number of additions = (n-1) n! (Rule of Product) If n=8, T(n) = 7 8! = 282,240 flops < 1/3 second. HOWEVER If n=50, T(n) = 49 50! = Prepared by: David F. McAllister TP , 2007 McGraw-Hill

3 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications = seconds = minutes = hours = days = weeks = years....a long time. You ll be an old person before it s finished. There are some problems for which we do not know if efficient algorithms exist to solve them! Chapter 1 - Introduction Prepared by: David F. McAllister TP , 2007 McGraw-Hill

4 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 Section 1.1 Propositional Logic proposition : true = T (or 1) or false = F (or 0) (binary logic) the moon is made of green cheese go to town! X - imperative What time is it? X - interrogative propositional variables: P, Q, R, S,... New Propositions from old: calculus of propositions - relate new propositions to old using TRUTH TABLES logical operators: unary, binary Unary Negation not Symbol: Prepared by: David F. McAllister TP , 2007 McGraw-Hill

5 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 Example: P: I am going to town P: I am not going to town; It is not the case that I am going to town; I ain t goin. Truth Table: P P F (0) T (1) T (1) F (0) Binary Example: Conjunction and Symbol: P - I am going to town Q - It is going to rain Prepared by: David F. McAllister TP , 2007 McGraw-Hill

6 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 P Q: I am going to town and it is going to rain. Truth Table: P Q P Q Note: Both P and Q must be true!!!!! Example: Disjunction inclusive or Symbol: P - I am going to town Q - It is going to rain P Q: I am going to town or it is going to rain. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

7 Truth Table: Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 P Q P Q Note: Only one of P, Q need be true. Hence, the inclusive nature. Example: Exclusive OR Symbol: P - I am going to town Q - It is going to rain P Q: Either I am going to town or it is going to rain. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

8 Truth Table: Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 P Q P Q Note: Only one of P and Q must be true. Example: Implication If...then... Symbol: P - I am going to town Q - It is going to rain P Q: If I am going to town then it is going to rain. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

9 Truth Table: Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 Equivalent forms: P Q P Q If P, then Q P implies Q If P, Q P only if Q P is a sufficient condition for Q Q if P Q whenever P Q is a necessary condition for P Note: The implication is false only when P is true and Q is false! There is no causality implied here! If the moon is made of green cheese then I have more money than Bill Gates (T) If the moon is made of green cheese then I m on welfare (T) If 1+1=3 then your grandma wears combat boots (T) Prepared by: David F. McAllister TP , 2007 McGraw-Hill

10 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 If I m wealthy then the moon is not made of green cheese. (T) If I m not wealthy then the moon is not made of green cheese. (T) Terminology: P = premise, hypothesis, antecedent Q = conclusion, consequence More terminology: Q P is the CONVERSE of P Q Q P is the CONTRAPOSITIVE of P Q Example: Find the converse and contrapositive of the following statement: R: Raining tomorrow is a sufficient condition for my not going to town. Step 1: Assign propositional variables to component propositions P: It will rain tomorrow Q: I will not go to town Prepared by: David F. McAllister TP , 2007 McGraw-Hill

11 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 Step 2: Symbolize the assertion R: P Q Step 3: Symbolize the converse Q P Step 4: Convert the symbols back into words If I don t go to town then it will rain tomorrow or Raining tomorrow is a necessary condition for my not going to town. or My not going to town is a sufficient condition for it raining tomorrow. Biconditional if and only if, iff Symbol: Example: P - I am going to town, Q - It is going to rain P Q: I am going to town if and only if it is going to rain. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

12 Truth Table: Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 P Q P Q Note: Both P and Q must have the same truth value. Others: NAND ( ) Sheffer Stroke; NOR ( ) Peirce Arrow (see problems) Breaking assertions into component propositions - look for the logical operators! Example: If I go to Harry s or go to the country I will not go shopping. P: I go to Harry s Q: I go to the country R: I will go shopping If...P...or...Q...then...not...R Prepared by: David F. McAllister TP , 2007 McGraw-Hill

13 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 Constructing a truth table: (P Q) R - one column for each propositional variable - one for the compound proposition - count in binary - n propositional variables = 2 n rows You may find it easier to include columns for propositions which themselves are component propositions. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

14 Truth Table: Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.1 P Q R (P Q) R Question: How many different propositions can be constructed from n propositional variables? Prepared by: David F. McAllister TP , 2007 McGraw-Hill

15 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.2 Section 1.2 Propositional Equivalences A tautology is a proposition which is always true. Classic Example: P P A contradiction is a proposition which is always false. Classic Example: P P A contingency is a proposition which neither a tautology nor a contradiction. Example: (P Q) R Two propositions P and Q are logically equivalent if P Q is a tautology. We write P Q Example: (P Q) (Q P) (P Q) Prepared by: David F. McAllister TP , 2007 McGraw-Hill

16 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.2 Proof: The left side and the right side must have the same truth values independent of the truth value of the component propositions. To show a proposition is not a tautology: use an abbreviated truth table - try to find a counter example or to disprove the assertion. - search for a case where the proposition is false Case 1: Try left side false, right side true Left side false: only one of P Q or Q P need be false. 1a. Assume P Q = F. Then P = T, Q = F. But then right side P Q = F. Oops, wrong guess. 1b. Try Q P = F. Then Q = T, P = F. Then P Q = F. Another wrong guess. Case 2. Try left side true, right side false Prepared by: David F. McAllister TP , 2007 McGraw-Hill

17 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.2 If right side is false, P and Q cannot have the same truth value. 2a. Assume P =T, Q = F. Then P Q = F and the conjunction must be false so the left side cannot be true in this case. Another wrong guess. 2b. Assume Q = T, P = F. Again the left side cannot be true. We have exhausted all possibilities and not found a counterexample. The two propositions must be logically equivalent. Note: Because of this equivalence, if and only if or iff is also stated as is a necessary and sufficient condition for. Some famous logical equivalences: Logical Equivalences P T P Identity P F P P T T Domination P F F P P P Idempotency P P P ( P)) P Double negation P Q Q P Commutativity P Q Q P (P Q) R P (Q R) Associativity (P Q) R P (Q R) Prepared by: David F. McAllister TP , 2007 McGraw-Hill

18 P (Q R) Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.2 (P Q) (P R) P (Q R) (P Q) (P R) (P Q) P Q (P Q) P Q P Q P Q P P T P P F P T P Distributivity DeMorgan s laws Implication Tautology Contradiction P F P (P Q) (Q P) Equivalence (P Q) (P Q) (P Q) Absurdity P (P Q) ( Q P) Contrapositive P (P Q) P Absorption P (P Q) P (P Q) R Exportation P (Q R) Note: equivalent expressions can always be substituted for each other in a more complex expression - useful for simplification. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

19 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.2 Normal or Canonical Forms Unique representations of a proposition Examples: Construct a simple proposition of two variables which is true only when P is true and Q is false: P Q P is true and Q is true: P Q P is true and Q is false or P is true and Q is true: (P Q) (P Q) A disjunction of conjunctions where - every variable or its negation is represented once in each conjunction (a minterm) - each minterms appears only once Disjunctive Normal Form (DNF) Important in switching theory, simplification in the design of circuits. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

20 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.2 Method: To find the minterms of the DNF. Use the rows of the truth table where the proposition is 1 or True If a zero appears under a variable, use the negation of the propositional variable in the minterm Example: If a one appears, use the propositional variable. Find the DNF of (P Q) R P Q R (P Q) R There are 5 cases where the proposition is true, hence 5 minterms. Rows 1,2,3, 5 and 7 produce the following disjunction of minterms: Prepared by: David F. McAllister TP , 2007 McGraw-Hill

21 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.2 (P Q) R ( P Q R) ( P Q R) ( P Q R) (P Q R) (P Q R) Note that you get a Conjunctive Normal Form (CNF) if you negate a DNF and use DeMorgan s Laws. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

22 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 Section 1.3 Predicates and Quantifiers A generalization of propositions - propositional functions or predicates.: propositions which contain variables Predicates become propositions once every variable is bound- by U assigning it a value from the Universe of Discourse or quantifying it Examples: Let U = Z, the integers = {... -2, -1, 0, 1, 2, 3,...} P(x): x > 0 is the predicate. It has no truth value until the variable x is bound. Examples of propositions where x is assigned a value: P(-3) is false, P(0) is false, P(3) is true. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

23 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 The collection of integers for which P(x) is true are the positive integers. P(y) P(0) is not a proposition. The variable y has not been bound. However, P(3) P(0) is a proposition which is true. = z Let R be the three-variable predicate R(x, y z): x + y Find the truth value of R(2, -1, 5), R(3, 4, 7), R(x, 3, z) Prepared by: David F. McAllister TP , 2007 McGraw-Hill

24 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 Universal Quantifiers P(x) is true for every x in the universe of discourse. Notation: universal quantifier xp(x) For all x, P(x), For every x, P(x) The variable x is bound by the universal quantifier producing a proposition. Example: U={1,2,3} xp(x) P(1) P(2) P(3) Existential P(x) is true for some x in the universe of discourse. Notation: existential quantifier xp(x) Prepared by: David F. McAllister TP , 2007 McGraw-Hill

25 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 There is an x such that P(x), For some x, P(x), For at least one x, P(x), I can find an x such that P(x). Example: U={1,2,3} xp(x) P(1) P(2) P(3) Unique Existential P(x) is true for one and only one x in the universe of discourse. Notation: unique existential quantifier! xp(x) There is a unique x such that P(x), There is one and only one x such that P(x), One can find only one x such that P(x). Prepared by: David F. McAllister TP , 2007 McGraw-Hill

26 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 Example: U={1,2,3} Truth Table: P(1) P(2) P(3)!xP( x) How many minterms are in the DNF? Note: REMEMBER! A predicate is not a proposition until all variables have been bound either by quantification or assignment of a value! Equivalences involving the negation operator xp(x) x P(x) xp(x) x P(x) Prepared by: David F. McAllister TP , 2007 McGraw-Hill

27 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 Distributing a negation operator across a quantifier changes a universal to an existential and vice versa. Multiple Quantifiers: read left to right... Example: Let U = R, the real numbers, P(x,y): xy= 0 x yp(x, y) x yp(x, y) x yp(x, y) x yp(x, y) The only one that is false is the first one. Suppose P(x,y) is the predicate x/y=1? Example: Let U = {1,2,3}. Find an expression equivalent to x yp(x, y) where the variables are bound by substitution instead: Prepared by: David F. McAllister TP , 2007 McGraw-Hill

28 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 Expand from inside out or outside in. Outside in: (can be very difficult) Examples: yp(1, y) yp(2, y) yp(3, y) [P(1,1) P(1,2) P(1,3)] [P(2,1) P(2,2) P(2,3)] [P(3,1) P(3,2) P(3,3)] Converting from English F(x): x is a fleegle S(x): x is a snurd T(x): x is a thingamabob U={fleegles, snurds, thingamabobs} (Note: the equivalent form using the existential quantifier is also given) Everything is a fleegle xf(x) x F(x) Prepared by: David F. McAllister TP , 2007 McGraw-Hill

29 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 Nothing is a snurd. x S(x) xs( x) All fleegles are snurds. x[f(x) S(x)] x[ F(x) S(x)] x [F(x) S(x)] x[f(x) S(x)] Some fleegles are thingamabobs. x[f(x) T(x)] x[ F(x) T(x)] No snurd is a thingamabob. x[s(x) T(x)] x[s(x) T(x)] If any fleegle is a snurd then it's also a thingamabob x[(f(x) S(x)) T(x)] x[f(x) S(x) T(x)] Prepared by: David F. McAllister TP , 2007 McGraw-Hill

30 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 Extra Definitions: An assertion involving predicates is valid if it is true for every universe of discourse. An assertion involving predicates is satisfiable if there is a universe and an interpretation for which the assertion is true. Else it is unsatisfiable. The scope of a quantifier is the part of an assertion in which variables are bound by the quantifier Examples: Valid: x S(x) xs( x) Not valid but satisfiable: x[f(x) T(x)] Not satisfiable: x[f(x) F(x)] Scope: x[f(x) S(x)] vs. x[f(x)] x[s(x)] Dangerous situations: Commutativity of quantifiers Prepared by: David F. McAllister TP , 2007 McGraw-Hill

31 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.3 x yp(x, y) y xp(x, y)? YES! x yp(x, y) y xp(x, y)? NO! DIFFERENT MEANING! Distributivity of quantifiers over operators x[p(x) Q(x)] xp(x) xq(x)? YES! x[p(x) Q(x)] [ xp(x) xq(x)]? NO! Prepared by: David F. McAllister TP , 2007 McGraw-Hill

32 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.5 Section 1.5 Rules of Inference Definition: A theorem is a valid logical assertion which can be proved using other theorems axioms (statements which are given to be true) and rules of inference (logical rules which allow the deduction of conclusions from premises). A lemma (not a lemon ) is a 'pre-theorem' or a result which is needed to prove a theorem. A corollary is a 'post-theorem' or a result which follows directly from a theorem. Rules of Inference Many of the tautologies in Chapter 1 are rules of inference. They have the form where and H 1 H 2... H n C H i are called the hypotheses Prepared by: David F. McAllister TP , 2007 McGraw-Hill

33 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.5 C is the conclusion. As a rule of inference they take the symbolic form: H 1 H 2.. H n C where means 'therefore' or 'it follows that.' Examples: The tautology P (P Q) Q becomes P P Q Q This means that whenever P is true and P Q is true we can conclude logically that Q is true. This rule of inference is the most famous and has the name or modus ponens Prepared by: David F. McAllister TP , 2007 McGraw-Hill

34 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.5 the law of detachment. Other famous rules of inference: P P Q Addition P Q P Simplification Q P Q P Modus Tollens P Q Q R P R Hypothetical syllogism P Q P Q Disjunctive syllogism P Q P Q Conjunction Prepared by: David F. McAllister TP , 2007 McGraw-Hill

35 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.5 (P Q) (R S) P R Q S Constructive dilemma Rules of Inference for Quantifiers xp(x) Universal Instantiation (UI) P(c) P(x) Universal Generalization (UG) xp(x) P(c) Existential Generalization (EG) xp(x) xp( x) Existential Instantiation (EI) P(c) Note: In Universal Generalization, x must be arbitrary. In Universal Instantiation, c need not be arbitrary but often is assumed to be. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

36 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.5 In Existential Instantiation, c must be an element of the universe which makes P(x) true. Example: Every man has two legs. John Smith is a man. Therefore, John Smith has two legs. Define the predicates: The argument becomes The proof is Q. E. D. M(x): x is a man L(x): x has two legs J: John Smith, a member of the universe 1. x[ M(x) L(x)] 2.M(J) L(J) 1. x[ M(x) L(x)] Hypothesis 1 2.M(J) L(J) step 1 and UI 3.M( J ) Hypothesis 2 4. L(J) steps 2 and 3 and modus ponens Note: Using the rules of inference requires lots of practice. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

37 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.5 If the butler did it he has blood on his hands. The butler had blood on his hands. Therefore, the butler did it. This argument has the form Fallacies Fallacies are incorrect inferences. Some common fallacies: The Fallacy of Affirming the Consequent or P Q Q P [(P Q) Q] P which is not a tautology and therefore not a rule of inference! The Fallacy of Denying the Antecedent (or the hypothesis) If the butler is nervous, he did it. The butler is really mellow. Therefore, the butler didn't do it. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

38 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.5 This argument has the form P Q P Q or [(P Q) P] Q which is also not a tautology and hence not a rule of inference. Begging the question or circular reasoning This occurs when we use the truth of statement being proved (or something equivalent) in the proof itself. Example: Conjecture: if x 2 is even then x is even. Proof: If x2 is even then x2 = 2k for some k. Then x = 2l for some l. Hence, x must be even. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

39 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 Section 1.6 Introduction to Proofs Formal Proofs To prove an argument is valid or the conclusion follows logically from the hypotheses: Assume the hypotheses are true Use the rules of inference and logical equivalences to determine that the conclusion is true. Example: Consider the following logical argument: If horses fly or cows eat artichokes, then the mosquito is the national bird. If the mosquito is the national bird then peanut butter takes good on hot dogs. But peanut butter tastes terrible on hot dogs. Therefore, cows don't eat artichokes. Assign propositional variables to the component propositions in the argument: F A M P Horses fly Cows eat artichokes The mosquito is the national bird Peanut butter tastes good on hot dogs Prepared by: David F. McAllister TP , 2007 McGraw-Hill

40 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 Represent the formal argument using the variables 1.(F A) M 2.M P 3. P A Use the hypotheses 1., 2., and 3. and the above rules of inference and any logical equivalences to construct the proof. Q. E. D. Assertion Reasons 1.(F A) M Hypothesis 1. 2.M P Hypothesis 2. 3.(F A) P` steps 1 and 2 and hypothetical syll. 4. P Hypothesis (F A) steps 3 and 4 and modus tollens 6. F A step 5 and DeMorgan 7. A F step 6 and commutativity of 'and' 8. A step 7 and simplification Methods of Proof We wish to establish the truth of the 'theorem' P Q. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

41 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 P may be a conjunction of other hypotheses. P Q is a conjecture until a proof is produced. Example: Trivial proof If we know Q is true then P Q is true. If it's raining today then the void set is a subset of every set. The assertion is trivially true independent of the truth of P. Vacuous proof If we know one of the hypotheses in P is false then P Q is vacuously true. Example: If I am both rich and poor then hurricane Fran was a mild breeze. This is of the form Prepared by: David F. McAllister TP , 2007 McGraw-Hill

42 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 (P P) Q and the hypotheses form a contradiction. Hence Q follows from the hypotheses vacuously. Direct proof - assumes the hypotheses are true - uses the rules of inference, axioms and any logical equivalences to establish the truth of the conclusion. Example: the Cows don t eat artichokes proof above Indirect proof A direct proof of the contrapositive: is true) - assumes the conclusion of P Q is false ( Q - uses the rules of inference, axioms and any logical equivalences to establish the premise P is false. Note, in order to show that a conjunction of hypotheses is false is suffices to show just one of the hypotheses is false. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

43 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 Example: Theorem: If 6x + 9y = 101, then x or y is not an integer. Proof: (Direct) Assume 6x + 9y = 101 is true. Then from the rules of algebra 3(2x + 3y) = 101. But 101/3 is not an integer so it must be the case that one of 2x or 3y is not an integer (maybe both). Therefore, one of x or y must not be an integer. Q.E.D. Example: A perfect number is one which is the sum of all its divisors except itself. For example, 6 is perfect since = 6. So is 28. Theorem: A perfect number is not a prime. Proof: (Indirect). We assume the number p is a prime and show it is not perfect. But the only divisors of a prime are 1 and itself. Hence the sum of the divisors less than p is 1 which is not equal to p. Hence p cannot be perfect. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

44 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 Q. E. D. Proof by contradiction or reductio ad absurdum - assumes the conclusion Q is false - derives a contradiction, usually of the form P P which establishes Q 0. The contrapositive of this assertion is 1 Q from which it follows that Q must be true. Example: Theorem: There is no largest prime number. (Note that there are no formal hypotheses here.) We assume the conclusion 'there is no largest prime number' is false. There is a largest prime number. Call it p. Hence, the set of all primes lie between 1 and p. Form the product of these primes: r = p. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

45 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 But r + 1 is a prime larger than p. (Why?). This contradicts the assumption that there is a largest prime. Q.E.D. The formal structure of the above proof is as follows: Let P be the assertion that there is no largest prime. Let Q be the assertion that p is the largest prime. Assume P is true. Then (for some p) Q is true so P Q is true. We then construct a prime greater than p so Q Q. Applying hypothetical syllogism we get P Q. From two applications of modus ponens we conclude that Q is true and Q is true so by conjunction Q Q or a contradiction is true. Hence the assumption must be false and the theorem is true. Proof by Cases Break the premise of P Q into an equivalent disjunction of the form Prepared by: David F. McAllister TP , 2007 McGraw-Hill

46 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 Then use the tautology P 1 P 2... P n. [(P 1 Q) (P 2 Q)... (P n Q)] [(P 1 P 2... P n ) Q] Each of the implications P i Q is a case. You must Convince the reader that the cases are inclusive, i.e., they exhaust all possibilities Example: establish all implications Let be the operation 'max' on the set of integers: if a b then a b = max{a, b} = a = b a. Theorem: The operation is associative. For all a, b, c Proof: (a b) c = a (b c). Let a, b, c be arbitrary integers. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

47 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.6 Then one of the following 6 cases must hold (are exhaustive): Hence 1. a b c 2. a c b 3. b a c 4. b c a 5. c a b 6. c b a Case 1: a b = a, a c = a, and b c = b. (a b) c = a = a (b c). Therefore the equality holds for the first case. The proofs of the remaining cases are similar (and are left for the student). Q. E. D. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

48 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.7 Section 1.7 Proof Methods and Strategy Existence Proofs We wish to establish the truth of xp(x). Constructive existence proof: - Establish P(c) is true for some c in the universe. - Then xp(x) is true by Existential Generalization (EG). Example: Theorem: There exists an integer solution to the equation x2 + y2 = z2. Proof: Choose x = 3, y = 4, z = 5. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

49 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.7 Example: Theorem: There exists a bijection from A= [0,1] to B= [0, 2]. Proof: We build two injections and conclude there must be a bijection without ever exhibiting the bijection. Let f be the identity map from A to B. Then f is an injection (and we conclude that A B ). Define the function g from B to A as g(x) = x/4. Then g is an injection. Therefore, B A. We now apply a previous theorem which states that if A B and B A then A = B. Hence, there must be a bijection from A to B. (Note that we could have chosen g(x) = x/2 and obtained a bijection directly). Q. E. D. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

50 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.7 Nonconstructive existence proof. - Assume no c exists which makes P(c) true and derive a contradiction. Example: Theorem: There exists an irrational number. Proof: Assume there doesn t exist an irrational number. Then all numbers must be rational. Then the set of all numbers must be countable. Then the real numbers in the interval [0, 1] is a countable set. But we have already shown this set is not countable. Hence, we have a contradiction (The set [0,1] is countable and not countable). Therefore, there must exist an irrational number. Q. E. D. Note: we have not produced such a number! Prepared by: David F. McAllister TP , 2007 McGraw-Hill

51 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.7 Disproof by Counterexample: Recall that x P(x) xp(x). To establish that xp(x) is true (or xp(x) is false) construct a c such that P(c) is true or P(c) is false. In this case c is called a counterexample to the assertion xp(x) Nonexistence Proofs We wish to establish the truth of xp(x) (which is equivalent to x P(x)). Use a proof by contradiction by assuming there is a c which makes P(c) true. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

52 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.7 Universally Quantified Assertions We wish to establish the truth of xp(x). We assume that x is an arbitrary member of the universe and show P(x) must be true. Using UG it follows that xp(x). Example: Theorem: For the universe of integers, x is even iff x2 is even. Proof: The quantified assertion is We assume x is arbitrary. x[x is even x 2 is even] Recall that P Q is equivalent to (P Q) (Q P). Case 1. We show if x is even then x2 is even using a direct proof (the only if part or necessity). If x is even then x = 2k for some integer k. Hence, x2 = 4k2 = 2(2k2) which is even since it is an integer which is divisible by 2. This completes the proof of case 1. Prepared by: David F. McAllister TP , 2007 McGraw-Hill

53 Transparencies to accompany Rosen, Discrete Mathematics and Its Applications Section 1.7 Case 2. We show that if x 2 is even then x must be even (the if part or sufficiency). We use an indirect proof: Assume x is not even and show x2 is not even. If x is not even then it must be odd. So, x = 2k + 1 for some k. Then x2 = (2k + 1)2 = 4k2 + 4k + 1 = 2(2k2 + 2k) + 1 which is odd and hence not even. This completes the proof of the second case. Therefore we have shown x is even iff x2 is even. Since x was arbitrary, the result follows by UG. Q.E.D. Dear students: Learning how to construct proofs is probably one of the most difficult things you will face in life. Few of us are gifted enough to do it with ease. One only learns how to do it by practicing. Prepared by: David F. McAllister TP , 2007 McGraw-Hill