Quiz 1-a (Math 13): Negations and Quantifiers. 1. (2 pts.) Compute the negation of: a Z n, p(a) (q 1 (a) q 2 (a)).

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1 Quiz 1-a (Math 13): Negations and Quantifiers. 1. (2 pts.) Compute the negation of: a Z n, p(a) (q 1 (a) q 2 (a)). 2. (6 pts.) Let G = {1, 9, 12}, and H = {0, 2, 4}. For each of the following statements, determine whether it is true (T) or false (F): (a) ( ) x G, x is even or x > 5. (b) ( ) x G, y G, z G s.t. x + y + z = 20. (c) ( (d) ( ) a G, b H s.t. a + b is even. ) a G s.t. b H, a + b is even. (e) ( ) h H, if h is odd then h G. (f) ( ) All the odd numbers that belong to the set H are positive. 3. (2 pts.) Prove the following statement: B C X (B X C X c ). 4. (1 pts.) Bonus: Let A = Z [1, 10], and B = {(x, y) A A : x < y}. Find the cardinality of the set B.

2 Name: Quiz 2 -version b (for practice only) (Math 13): Implications. Id #: Thursday, October 10, UCI. Do not use calculators, notes or textbooks. 1. (4 pts.) Let (X, Y, Z) = ({0}, {1}, {0, a, 5}) (where a is some unknown integer!). For each of the following statements, determine whether it is true (T), false (F), or we would need additional informacion (?): (a) ( ) If X Y, then (if 2 X then 7 Y ). (b) ( (c) ( ) X Y Z. ) X Z = Z and X Y = X. (d) ( ) X = 4 if and only if Y = (6 pts.) For each of the following statements, write its contrapositive and then prove it: (a) If the integer x 3 is odd, then x must be odd as well. (b) If A B X, then A X and B X. (c) If A B X, then A X or B X. 3. (1 pts.) Bonus: Let Σ = {1, 2, 3}, and Γ = {3, 4}. Find the cardinality of the following sets: (i) P(Σ Γ), (ii) P(Σ) P(Γ). Can you elaborate a general conjecture about the sets P(Σ Γ) and P(Σ) P(Γ)? Try to prove it!

3 Quiz 1-a (Discrete Math): Negation, Contrapositive. 1. (4 pts.) Consider the sentence P, given by: if Kim has at least one brother, then the following is true: Kim does not live in a castle and every friend of Kim is a dragon. (i) Compute the negation of P. (ii) Compute the contrapositive of P. [Note: Make sure you use the rules of logic (De Morgan, negate quantifiers, etc) as much as possible, as done in your Homework and during examples in class.] 2. (4 pts.) Prove/disprove: y : (y is even) ( x Z : y 2 = 2x 10). 3. (2 pts.) Explain in full detail: If the set E contains the set A B, can we be certain that E contains A? [Answer in your own words, writing a convincing argument]. 4. (1 pts.) Bonus: If P(A) = 4 and P(B) = 16, find P(B A).

4 Massachusetts Institute of Technology 6.042J/18.062J, Fall 02: Mathematics for Computer Science Professor Albert Meyer and Dr. Radhika Nagpal Quiz 1 Appendix Appendix Contents 1 Induction 2 2 Relations Equivalence Partial Order Operations Closure Graphs Common graphs Graph coloring Trees State Machines Well-founded Partial Orders Copyright 2002, Prof. Albert R. Meyer and Dr. Radhika Nagpal.

5 Quiz 1 Appendix: Appendix 2 1 Induction Axiom (Induction). Axiom (Strong Induction). P (0), m N P (m) P (m + 1) n N P (n). P (0), n N m n P (m) P (m + 1) n N P (n) Axiom (Least Number Principle). Every nonempty subset, S N, has a smallest element. 2 Relations 2.1 Equivalence A binary relation, R, on a set A is reflexive if for every a A, ara. symmetric if for every a, b A, arb implies bra. transitive if for every a, b, c A, arb and brc implies arc. R is an equivalence relation iff it is reflexive, symmetric and transitive. 2.2 Partial Order A binary relation, R, on a set A is areflexive if for every a A, (ara). asymmetric if for every a, b A, arb implies (bra). anti-symmetric if for every a, b A arb and bra implies a = b. a partial order if it is reflexive, transitive and anti-symmetric. a strict partial order if it is areflexive and transitive. a total order if it is a partial order and for every a b A either arb or bra.

6 Quiz 1 Appendix: Appendix If R is a partial order, then the set A is called a partially ordered set. with the partial order, R, is called a poset, (A, R). In this case, 3 The set, A, together a is a minimal element of A if b A[bRa b a] a is a maximal element of A if b A[aRb b a] a subset of A is a chain iff it is totally ordered by R. elements a 1, a 2 A are incomparable iff neither a 1 Ra 2 nor a 2 Ra 1 holds. a subset of A is an anti-chain iff its elements are pairwise incomparable. A topological sort of a finite poset (A, R) is a total ordering of all the elements of A, a 1, a 2,, a n in such a way that for all i < j, either a i Ra j or a i and a j are incomparable. Theorem. Given any finite poset (A, R) for which the longest chain has length t, A can be partitioned into t antichains. Theorem (Dilworth). For all t, every poset with n elements must have either a chain of size greater than t or an antichain of size at least n/t. 2.3 Operations If R is a relation on A B, then the inverse of R is the relation on B A given by R 1 = {(b, a) (a, b) R}. The composition of relations R 1 A B and R 2 B C is the relation R 2 R 1 = {(a, c) ( b)((a, b) R 1 ) ((b, c) R 2 )}. A path in a relation R is a sequence a 0,..., a k with k 0 such that (a i, a i+1 ) R for every i < k. We call k the length of the path. Let R be a relation on the set A and n N. Then { R if n = 1 R n = R n 1 R if n > 1 Lemma. R n = {(a, b) there is a length n path from a to b in R}.

7 Quiz 1 Appendix: Appendix Closure The closure of relation R with respect to property P is the relation S that 1. contains R, 2. has property P, and 3. is contained in any relation satisfying (1) and (2). That is, S is the smallest relation satisfying (1) and (2). Lemma. Let R be a relation on the set A. The reflexive closure of R is S = R {(a, a), a A}. The symmetric closure of R is S = R R 1. The transitive closure of a relation R is the set S = {(a, b) A A given there is a path from a to b in R}. The reflexive and transitive closure (also called the connectivity relation) of R is R = {(a, b) A A there exists a path from a to b in R}. or, equivalently, R = n=1 Rn. Lemma. If R is a reflexive relation on A then R = R A. 3 Graphs A simple graph is a pair of sets (V, E) called vertices and edges, respectively. An edge {a, b} E is a set where a, b V and a b. A directed graph or digraph is a pair of sets of (V, E) where E V V. In other words, an edge (a, b) E is an ordered pair of vertices. Theorem. The sum of the degrees of the vertices in a simple graph equals twice the number of edges. Theorem (Handshake). In every graph, there are an even number of vertices of odd degree. An Euler tour of an undirected graph G is a circuit that traverses every edge of G exactly once. Theorem. If undirected graph G has Euler tour, then G is connected and every vertex has even degree. A Hamiltonian path of an undirected graph G is simple path that traverses every vertex exactly once.

8 Quiz 1 Appendix: Appendix Common graphs The empty graph or anticlique on n vertices, is the graph A n = ({v 1, v 2,..., v n }, ). The line graph on n vertices is the graph L n = ({v 1, v 2,..., v n }, {{v 1, v 2 }, {v 2, v 3 },..., {v n 1, v n }}). The cycle on n vertices C n is the L n plus the edge {v n, v 1 }. The wheel on n 4 vertices W n is the graph consisting of C n 1 plus an additional vertex connected to every other vertex. The complete graph or clique on n vertices, K n has an edge between every pair of vertices. 3.2 Graph coloring The minimum number of colors needed to color the vertices of a graph G so that no two adjacent vertices are the same is called the chromatic number and is written χ(g). Lemma. A graph with maximum degree d max can be colored with d max + 1 colors. 3.3 Trees A tree is any simple graph G = (V, E), such that any of these (equivalent) conditions hold: G is connected and E = V 1. G is connected, but removing any edge from G leaves a disconnected graph. G is connected and acyclic. There is a unique simple path between any two distinct vertices of G. A rooted tree is called finite-path iff it has no infinite paths away from the root. 4 State Machines A state machine has three parts: 1. a nonempty set, Q, whose elements are called states, 2. a nonempty subset Q 0 Q, called the set of start states, 3. a binary relation, δ, on Q, called the transition relation. An invariant for a state machine is a predicate, P, on states, such that whenever P (q) is true of a state, q, and q r for some state, r, then P (r) holds.

9 Quiz 1 Appendix: Appendix 6 Theorem (Invariant Theorem). Let P be an invariant predicate for a state machine. If P holds for all start states, then P holds for all reachable states. A derived variable f : Q R is strictly decreasing iff q q implies f (q ) < f (q). Theorem. If f : Q N is a strictly decreasing derived variable of a state machine, then the length of any execution starting at a start state q is at most f (q). 4.1 Well-founded Partial Orders If (P 1, 1 ) and (P 2, 2 ) are posets, then the lexicographic partial order, lex, on P 1 P 2 is defined by the condition that (p 1, p 2 ) lex (q 1, q 2 ) ::= p 1 1 q 1 or (p 1 = q 1 p 2 2 q 2 ). The coordinatewise partial order, c, on P 1 P 2 is defined by the condition that (p 1, p 2 ) c (q 1, q 2 ) ::= (p 1 1 q 1 p 2 2 q 2 ). A poset (P, ) is well-founded iff every nonempty subset S P has a minimal element. Lemma. Suppose (P 1, 1 ) and (P 2, 2 ) are posets. Then 1. so are (P 1 P 2, lex ) and (P 1 P 2, c ). Moreover, 2. if (P 1, 1 ) and (P 2, 2 ) are both well-founded, then so are (P 1 P 2, lex ) and (P 1 P 2, c ). 3. if (P 1, 1 ) and (P 2, 2 ) are both totally ordered, then so is (P 1 P 2, lex ). Lemma. A poset is well-founded iff it has no infinite decreasing chain. Theorem. Fundamental Theorem for two-person games of perfect information: For games in which every play is finite and ends in win or lose, there is a winning strategy for one of the players.

10 Massachusetts Institute of Technology 6.042J/18.062J, Fall 02: Mathematics for Computer Science Professor Albert Meyer and Dr. Radhika Nagpal Quiz 1 Your name: Circle the name of your Tutorial Instructor: Adrian Georgi Josh Karen Lee Min Nikos Tina This quiz is closed book. There is an Appendix with standard definitions. There are five (5) problems totaling 100 points. Total time is 110 minutes. Put your name on the top of every page these pages may be separated for grading. Write your solutions in the space provided. If you need more space, write on the back of the sheet containing the problem. Please keep your entire answer to a problem on that problem s page. You may assume any of the results presented in class or in the lecture notes. Be neat and write legibly. You will be graded not only on the correctness of your answer, but also on the clarity with which you express it. GOOD LUCK! DO NOT WRITE BELOW THIS LINE Problem Points Grade Grader Total 100 Copyright 2002, Prof. Albert R. Meyer and Dr. Radhika Nagpal.

11 Quiz 1 Your name: 2 Problem 1 (20 points). Suppose S(n) is a predicate on natural numbers, n, and suppose k N S(k) S(k + 2). (1) If (1) holds, some of the assertions below must always (A) hold, some can (C) hold but not always, and some can never (N) hold. Indicate which case applies for each of the assertions by circling the correct letter. (a) (2 points) A C N n 0 S(n) (b) (2 points) A C N S(0) n 1 S(n) (c) (2 points) A C N n 0 S(n) (d) (2 points) A C N ( n 100 S(n)) ( n > 100 S(n)) (e) (2 points) A C N ( n 100 S(n)) ( n > 100 S(n)) (f) (2 points) A C N S(0) n S(n + 2) (g) (2 points) A C N S(1) n S(2n + 1) (h) (2 points) A C N [ n S(2n)] n S(2n + 2) (i) (2 points) A C N n m > n [S(2n) S(2m)] (j) (2 points) A C N [ n S(n)] n m > n S(m)

12 Quiz 1 Your name: 3 Problem 2 (15 points). Attila the Hun is planning another excursion into a Roman village. This requires a number of tasks, each of which takes him one minute to complete. The prerequisites associated with these tasks are listed below. ABBRV. TASK PREREQUISITES C Assemble the barbarians N Plunder the village B D Get shots for his own cat, also named Emilios B B Blow the trumpet T Sell T-shirts: I got A LOT more than just this lousy t-shirt. N Q Grade the quiz S G Cook a feast D,N S Burn the village C,N (a) (4 points) Draw the Hasse diagram for the tasks and their prerequisites. (b) (2 points) Attila has decided that since his barbarians are quite smart and he has so many, he can get as many tasks done at a time as he wishes. What is the minimum amount of time required for him to finish the excursion? It turns out that Attila s graph is actually far more complicated than the one above. In fact, he doesn t know what the actual diagram because his Scribe forgot to tell him. All Attila has been told is that he must complete n tasks, each of which takes 1 minute, and that the minimum amount of time required to finish is t minutes. Without knowing anything more about the actual graph, Attila is trying to figure out how many barbarians to recruit. A barbarian can only complete one task in 1 minute, but has the stamina to work for days on end. Let n and t be fixed, and n > t > 1. (c) (4 points) Write a simple formula in n and t for the smallest number of barbarians Attila can recruit in order to be guaranteed to finish in t minutes.

13 Quiz 1 Your name: 4 (d) (5 points) Write a simple formula in n and t for the smallest number of barbarians he could recruit and still possibly finish the job in t minutes. (That is, if he recruits fewer barbarians, he will never be able to finish in the minimum number, t, of minutes, no matter how favorable graph turns out to be.)

14 Quiz 1 Your name: 5 Problem 3 (15 points). In this problem, let R be a binary relation on a set A, and S be a binary relation from A to a set B. Indicate whether each of the following statements is True or False. For the false ones, describe a counterexample 1. (a) (3 points) If R = R, then R is a transitive relation. (b) (3 points) If R = R, then R is an equivalence relation. (c) (3 points) If R is an equivalence relation, then R = R. (d) (3 points) If A = B, then S R = R S. (e) (3 points) If R = R 1 and R is nonempty, then R is reflexive. 1 No explanation required for statements that are True.

15 Quiz 1 Your name: 6 Problem 4 (25 points). Recall that a k-coloring of a simple graph is an assignment of a color to each vertex such that no two adjacent vertices have the same color, and no more than k colors are used. In class we proved Lemma. If a graph has maximum degree k, then it is k + 1-colorable, Consider the following variation of this Lemma: False Claim. A graph with maximum degree k that also has a vertex of degree less than k, is k-colorable. (a) (5 points) Give a counterexample to the False Claim when k = 2. Consider the following proof of the False Claim: False proof. Proof by induction on the number n of vertices: Induction hypothesis: P (n)::= Every graph with n vertices and maximum degree k that also has a vertex of degree less then k, is k-colorable. Base case: (n=1) The graph has only one vertex of degree zero, so P (1) holds vacuously. Inductive step: We may assume P (n). To prove P (n + 1), let G n+1 be a graph with maximum degree k that has a vertex, v, of degree less than k. Remove the vertex v to produce a graph G n. Removing v reduces the degree of all vertices adjacent to v by 1. Therefore G n must contain at least one vertex with degree less than k. Also the maximum degree of G n is at most k. If the maximum degree of G n is less than k, then by the Lemma above, G n is k-colorable. Otherwise G n has maximum degree k and a vertex of degree less than k, so by our induction hypothesis, G n is k-colorable. So in any case, G n is k-colorable. Now a k-coloring of G n gives a coloring of all the vertices of G n+1, except for v. Since v has degree less than k, there will be fewer than k colors assigned to the nodes adjacent to v. So there will be a color left over among the k colors that can be assigned to v. Hence G n+1 is k-colorable.

16 Quiz 1 Your name: 7 (b) (5 points) Identify the exact sentence where the proof goes wrong (underlining or circling the sentence is sufficient). (c) (5 points) Briefly explain why the sentence is incorrect. (d) (5 points) The False Claim can be made true by adding an additional assumption about the graph. Which of the following is the most general asumption that will make the False Claim true? 2 1. The graph is a line graph. 2. The graph has only even length cycles. 3. The graph is connected. 4. The graph does not contain a complete graph on k vertices. 5. The graph has no node of degree zero. 6. The graph has a Hamiltonian cycle. 7. k < 2. 2 By most general, we mean it is implied by all the other assumptions that verify the False Claim. For example, the assumption that G is a line graph is more general than the assumption that G is a line graph with 3 vertices.

17 Quiz 1 Your name: 8 (e) (5 points) Assuming that G n+1 satisifes the additional assumption from part (d), both the False Claim and the sentence that was incorrect from part (b) become correct. But now another sentence in the proof becomes incorrect and requires fixing. Indicate the new incorrect sentence and briefly explain what s wrong. (You are not expected to suggest a fix.)

18 Quiz 1 Your name: 9 Problem 5 (25 points). One of the three monks working on the famed Towers of Hanoi project recently rubbed his pained back and burst out, Yo! What are we doing? This is for chumps! Let s punt! But before wandering off to start up fast food joints, they must evenly divide the monastery s collection of prayer beads. Initially, monk A has 5 beads, monk B has 3 beads, and monk C has 4 beads. The monastic order has strict rules regarding the exchange of prayer beads. Only the following transactions are allowed. 1. Monk B may give a bead to monk A at any time. 2. If C has an odd number of beads, then monk A may give a bead to monk B. 3. If C has an even number of beads, then monk C may give or take a bead from either monk A or monk B. 4. If monk A has at least two more beads than monk B, then monk C may give or take a bead from either monk A or monk B. (a) (10 points) Model the situation with a state machine. Define the set of states, the set of start states, and the set of transitions.

19 Quiz 1 Your name: 10 (b) (5 points) Describe a sequence of steps (transitions) leading to a state where monk A has 0 beads, monk B has 9 beads, and monk C has 3 beads. (c) (4 points) A clever TA tells you that the following predicate is an invariant. (C has an odd number of beads) (A has more beads than B) Assuming she is correct, prove that the monks can not reach the state where every monk has 4 beads.

20 Quiz 1 Your name: 11 (d) (6 points) is an invariant. Now prove that the TA was correct in her assumption that the predicate

21 Quiz 1 (Discrete Math): Implication. 4/10/2014. Name and id: Recall: 0 is not a positive integer. 1. (3 pts.) What does it mean that No French clown is both fast and furious is false? 2. (5 pts.) Prove/Disprove the following: Conjecture: The sum of a multiple of 8 and a multiple of 20, must be a multiple of (2 pts.) Which of the following gives a list of all positive integers x such that x 8 and the implication If x then x 2 is odd is true? Select exactly one. No justification needed. ( ) 5, 7. ( ) 5, 6, 7, 8. ( ) 1, 3, 5, 7. ( ) 1, 2, 3, 5, 7. ( ) 1, 2, 3. ( ) 1, 2, 3, 4, 6, 8. ( ) 1, 9, 25. ( ) Other.

22 Solutions 1. (3 pts.) What does it mean that No French clown is both fast and furious is false? Solution: There is a french clown that is slow or is not furious. Using logical symbols: x : ( x is french x is a clown ) ( x is not fast x is not furious ). 2. (5 pts.) Prove/Disprove the following: Conjecture: The sum of a multiple of 8 and a multiple of 20, must be a multiple of 4. Solution 1: TRUE. Proof: Let x be a multiple of 8 and let y be a multiple of 20. WTS: x + y is a multiple of 4. By definition, there are p, q Z such that x = 8p and y = 20q. Then x + y = 4(2p + 5q). Since 2p + 5q Z, we conclude that x + y is a multiple of 4. The proof is complete. Solution 2 (Using Lemmas): First prove two Lemmas: (a) Divisibility is transitive (a b and b c implies a c), (b) Divisibility is nice with sum (a b and a c implies a b + c). [You would need to prove the Lemmas aside in the Quiz. Easy proofs!]. Proof of exercise: Let x, y Z such that 8 x and 20 y. Since 4 8 and 4 20, then by Lemma (a), 4 x and 4 y, so by Lemma (b), 4 x + y, as needed. 3. (2 pts.) Which of the following gives a list of all positive integers x such that x 8 and the implication If x then x 2 is odd is true? Select exactly one. No justification needed. Answer: 1, 2, 3, 5, 7. We give two solutions (note that the universe is U = {1, 2, 3, 4, 5, 6, 7, 8}, so we choose the values from there). Solution 1: Complement method: when does the implication fail? by logic, this would be: P Q, i.e., x x 2 is even, Values: {4, 5, 6, 7, 8} {2, 4, 6, 8} = {4, 6, 8}. So the solution set is: {1, 2, 3, 4, 5, 6, 7, 8} {4, 6, 8} = {1, 2, 3, 5, 7}. So answer as list: 1, 2, 3, 5, 7. Solution 2: Using reduction: P Q is equivalent to the disjunction P Q (Important: the negation only applies to P. So if you want extra parenthesis: ( P ) (Q)). P Q gives: x + 1 < 5 x 2 is odd, Values: {1, 2, 3} {1, 3, 5, 7} = {1, 2, 3, 5, 7}.

23 Massachusetts Institute of Technology 6.042J/18.062J, Fall 05: Mathematics for Computer Science September 21 Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld revised September 21, 2005, 1076 minutes Solutions to Problem Set 1 Problem 1. A real number r is called sensible if there exist positive integers a and b such that a/b = r. For example, setting a = 2 and b = 1 shows that 2 is sensible. Prove that 3 2 is not sensible. (Consider only positive real roots in this problem) 3 Solution. The proof is by contradiction. Assume for the purpose of contradiction that 2 3 is sensible. Then there exist positive integers a and b such that a/b = 2. Squaring both 3 3 sides of this equation gives a/b = 4, which implies that 4 is rational. 3 Since 4 is rational, we can write it as a fraction x/y in lowest terms, where x is an integer and y is a positive integer. Therefore, we have: 3 4 = x/y 4 = x 3 /y 3 4y 3 = x 3 In the last equation, the left side is even, and so the right side must be even. Since x 3 is even, x itself must be even. This implies that the right side is actually divisible by 8, and so the left side must also be divisible by 8. Therefore, y 3 is even, and so y itself must be even. The fact that both x and y are even contradicts the fact that x/y is a fraction in lowest 3 terms. Therefore, 2 is not sensible. Problem 2. Translate the following sentence into a predicate formula: There is a student who has e mailed exactly two other people in the class, besides possibly herself. Copyright 2005, Prof. Albert R. Meyer and Prof. Ronitt Rubinfeld.

24 Solutions to Problem Set 1 2 The domain of discourse should be the set of students in the class; in addition, the only predicates that you may use are equality and E(x, y), meaning that x has sent e mail to y. Solution. A good way to begin tackling this problem is by trying to translate parts of the sentance. Begin by trying to assert that student x has ed students y and z: E(x, y) E(x, z). Now we want to say that y and z not the same student, and neither of them is x either: where x = y abbreviates = (x = y). x = y x = z y = z, Now, we must think of a way to say that the only people x might have e mailed are x, y and z: s, E(x, s) s = x s = y s = z. Finally, we can say there is some student who ed exactly two other two students by existentially quantifying x, y and z. So the complete translation is x y z. E(x, y) E(x, z) (1) x = y x = z y = z (2) s, E(x, s) s = x s = y s = z. (3) Problem 3. Express each of the following predicates and propositions in formal logic notation. The domain of discourse is the nonnegative integers, N. In addition to the propositional operators, variables and quantifiers, you may define predicates using addition, multiplication, and equality symbols, but no constants (like 0, 1,... ). For example, the proposition n is an even number could be written m. (m + m = n). (a) n is the sum of three perfect squares. Solution. x y z. (x x + y y + z z = n)

25 Solutions to Problem Set 1 3 Since the constant 0 is not allowed to appear explicitly, the predicate x = 0 can t be written directly, but note that it could be expressed in a simple way as: Then the predicate x > y could be expressed x + x = x. w. (y + w = x) (w = 0). Note that we ve used w = 0 in this formula, even though it s technically not allowed. But since w = 0 is equivalent to the allowed formula (w + w = w), we can use w = 0 with the understanding that it abbreviates the real thing. And now that we ve shown how to express x > y, it s ok to use it too. (b) x > 1. Solution. The straightforward approach is to define x = 1 as y. xy = y and then express x > 1 as y. (y = 1) (x > y). (c) n is a prime number. Solution. IS PRIME(n) ::= (n >1) ( x y. (x > 1) (y > 1) (x y = n)) (d) n is a product of two distinct primes. Solution. x y. (x = y) (n = x y) IS PRIME(x) IS PRIME(y). (e) There is no largest prime number. Solution. Of course this is a true statement, so it could be expressed by the logically equivalent formula 1 = 1, but even if we didn t know this, we could transcribe the statement directly as: ( p. IS PRIME(p) ( q. IS PRIME(q) p q))

26 Solutions to Problem Set 1 4 (f) (Goldbach Conjecture) Every even natural number n > 2 can be expressed as the sum of two primes. Solution. We can define n > 2 with the formula y. (y = 1) (x > y+ y). Likewise, n = 2k can be expressed as n = k + k. Then we can express the Conjecture with: n. ((n > 2 k. n = 2k) p q. IS PRIME(p) IS PRIME(q) (n = p + q))) (g) (Bertrand s Postulate) If n > 1, then there is always at least one prime p such that n < p < 2n. Solution. n. ((n > 1) ( p. IS PRIME(p) (n < p) (p < 2n))) Problem 4. If a set, A, is finite, then A < 2 A = P (A), and so there is no surjection from set A to its powerset. Show that this is still true if A is infinite. Hint: Remember Russell s paradox and consider { x A x / f(x)} where f is such a surjection. Solution. We prove there is no surjection by contradiction: suppose there was a surjection f : A P (A) for some set A. Let W ::= { x A x / f(x)}. So by definition, (x W ) (x / f(x)) (4) for all x A. But W A by definition and hence is a member of P (A). This means W = f(a) for some a A, since f is a surjection to P (A). So we have from (4), that (x f(a)) (x / f(x)) (5) for all x A. Substituting a for x in (5) yields a contradiction, proving that there cannot be such an f. Problem 5. (a) Prove that z. [P (z) Q(z)] [ x. P (x) y. Q(y)] (6) is valid. (Use the proof in the subsection on Validity in Week 2 Notes as a guide to writing your own proof here.)

27 Solutions to Problem Set 1 5 Solution. Proof. Assume z. [P (z) Q(z)]. (7) That is, P (z) Q(z) holds for some element, z, of the domain. Let c be this element; that is, we have P (c) Q(c). In particular, P (c) holds by itself. So we conclude (by Existential Generalization) x P (x). We conclude y Q(y) similarly. Hence, holds. x. P (x) y. Q(y) (8) This shows that (8) holds in any interpretation in which (7) holds. Therefore, (7) implies (8) in all interpretations, that is, (6) is valid. (b) Prove that the converse of (6) is not valid by describing a counter model as in Week 2 Notes. Solution. Proof. We describe a counter model in which, (8) is true and (7) is false. Namely, let the domain, D, be {π, e}, P (x) mean x = π, and Q(y) mean y = e. Then, x. P (x) is true (let x be π) and likewise y. Q(y) is true (let y be e), so (8) holds. On the other hand, Q(π) is not true, so P (π) Q(π) is not true. Likewise P (e) Q(e) is not true. Since these are the only elements of D, it is not true that there is an element, z, of D, such that P (z) Q(z), That is, (7) is not true. Problem 6. (a) Give an example where the following result fails: False Theorem. For sets A, B, C, and D, let L ::= (A C) (B D), R ::= (A B) (C D). Then L = R. Solution. If A = D = and B and C are both nonempty, then L = C B =, but R =. (b) Identify the mistake in the following proof of the False Theorem.

28 Solutions to Problem Set 1 6 Bogus proof. Since L and R are both sets of pairs, it s sufficient to prove that (x, y) L (x, y) R for all x, y. The proof will be a chain of iff implications: (x, y) L iff x A C and y B D, iff either x A or x C, and either y B or y D, iff (x A and y B) or else (x C and y D), iff (x, y) A B, or (x, y) C D, iff (x, y) (A B) (C D) = R. Solution. The mistake is in the third iff. If [x A or x C, and either y B or y D], it does not necessarily follow that (x, y) (A B) (C D). It might be that (x, y) is in A D instead. This happens, for example, if A = {1}, B = {2}, C = {3}, D = {4}, and (x, y) = (1, 4). (c) Fix the proof to show that R L. Solution. Replacing the third iff by which will be true when, yields a correct proof that (x, y) L will be true when (x, y) R, which implies that R L.

29 Massachusetts Institute of Technology 6.042J/18.062J, Spring 02: Mathematics for Computer Science March 1 Professor Albert Meyer and Dr. Radhika Nagpal revised April 28, 2002, 858 minutes Quiz 1 Your name: Circle the name of your Tutorial Instructor: Ashish Carole Christos Eric George Jack Nick Tina This quiz is closed book. There is an Appendix giving the definitions of standard properties of relations. There are four (4) problems totaling 100 points. Problems are labeled with their point values. Put your name on the top of every page these pages may be separated for grading. Write your solutions in the space provided. If you need more space, write on the back of the sheet containing the problem. Please keep your entire answer to a problem on that problem s page. You may assume any of the results presented in class or in the lecture notes. Be neat and write legibly. You will be graded not only on the correctness of your answer, but also on the clarity with which you express it. GOOD LUCK! DO NOT WRITE BELOW THIS LINE Problem Points Grade Grader Total 100 Copyright c 2002, Prof. Albert R. Meyer and Dr. Radhika Nagpal.

30 Quiz 1 Your name: 2 Problem 1 (25 points). Consider the following system specifications The system is in multiuser state iff it is operating normally. 2. If the system is operating normally, then the kernel is functioning. 3. The kernel is not functioning or the system is in interrupt mode. 4. If the system is not in multiuser state, then it is in interrupt mode. 5. The system is not in interrupt mode. (a) (0 points) To make sense of these confusing conditions, let s introduce four Boolean variables. M ::= in Multiuser state (1) N ::= operating Normally (2) K ::= Kernel is functioning (3) I ::= in Interrupt mode (4) Translate the five statements in the specification into propositional logic notation:,,,, Rosen, Exercise

31 Quiz 1 Your name: 3 (b) (0 points) Are these system specifications consistent?. Prove it!

32 Quiz 1 Your name: 4 Problem 2 (20 points). For each of the following logical formulas with domain of discourse the natural numbers, N, indicate whether it is a possible formulation of I: the Induction Axiom, S: the Strong Induction Axiom, L: the Least Number Principle (also known as Well ordering), or N: None of these. For example, the ordinary Induction Axiom could be expressed by the following formula, so it gets labelled I. (P (0) [ k P (k) P (k + 1)]) k P (k) I This is a multiple choice problem: do not explain your answer. (a) (0 points) (P (b) [ k b P (k) P (k + 1)]) k b P (k) (b) (0 points) (P (b) [ k b P (k) P (k + 1)]) k b P (k) (c) (0 points) [ b ( k < b P (k)) P (b)] k P (k) (d) (0 points) ( n P (n)) n k < n P (k) (e) (0 points) n [P (n) ( n P (n) k < n P (k))] Problem 3 (20 points). Classify each of the following binary relations as E: An equivalence relation. T: A Total order, P: A Partial order that is not total. S: A Symmetric relation that is not transitive. N: None of the above. This is a multiple choice problem: do not explain your answer. (a) (0 points) The relation xry between times of day such that x and y are at most twenty minutes apart.

33 Quiz 1 Your name: 5 (b) (0 points) The relation xry between times of day such that x is more than twenty minutes later than y. (c) (0 points) The relation xry over all words in this sentence such that x does not appear after y. (Consider x, y, and xry to be words.) (d) (0 points) The relation xry over all words in this sentence such that word x does not appear after word y. (e) (0 points) The relation xry over all words in this sentence such that the final appearance of y occurs after x. Problem 4 (35 points). To encourage collaborative study, the staff is considering assigning each student to a study group with two or three other students. Prove that as long as the enrollment is large enough, the class can always be divided into such study groups.

34 Quiz 1 Appendix A binary relation, R, on a set A is reflexive iff xrx, symmetric iff xry yrx, Your name: 6 anti symmetric iff xry yrx x = y, transitive iff xry yrz xrz, for all x, y, z A. R is an equivalence relation iff it is reflexive, symmetric and transitive. R is a partial order iff it is transitive and anti symmetric. R is a total order iff it is a partial order and for all x = y A either xry or yrx.

35 Quiz 1-a (Math 13): Implications. 1. (4 pts.) Let (X, Y, Z) = ({0}, {1}, {0, a, 5}) (where a is some unknown integer!). For each of the following statements, determine whether it is true (T), false (F), or we need additional informacion (?) to determine its truth value: (a) ( (b) ( (c) ( (d) ( ) If X = Y, then Y = Z. ) X Y Z. ) Z X Y and 3 Z. ) X Y Z if and only if 1 Z. 2. (6 pts.) For each of the following statements, write its contrapositive 1 and then prove it: (a) If the integer x 3 is odd, then x must be odd as well. (b) If A B X, then A X. 3. (1 pts.) Bonus: Let Σ = {1, 2, 3}, and Γ = {3, 4}. Find the cardinality of the following sets: (i) P(Σ Γ), (ii) P(Σ) P(Γ). 1 Recall that the contrapositive of p q is q p.