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### Homework 1 Solutions W11

Course: EECS 203, Spring 2011
School: Michigan
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Word Count: 1043

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203: EECS Homework 1 Solutions Section 1.1 1. (E) 8bef b) You do not miss the final exam if and only if you pass the course. e) If you have the flu then you do not pass the course, or if you miss the final examination then you do not pass the course. f) You have the flu and you miss the final examination, or you do not miss the final examination and you pass the course. 2. (E) 10cde c) r p d) p q r e) (p q) r...

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203: EECS Homework 1 Solutions Section 1.1 1. (E) 8bef b) You do not miss the final exam if and only if you pass the course. e) If you have the flu then you do not pass the course, or if you miss the final examination then you do not pass the course. f) You have the flu and you miss the final examination, or you do not miss the final examination and you pass the course. 2. (E) 10cde c) r p d) p q r e) (p q) r 3. (E) 24abc a) Converse: Contrapositive: Inverse: Converse: Contrapositive: Inverse: Converse: Contrapositive: Inverse: If I stay at home, then it will snow tonight. If I do not stay at home, then it will not snow tonight. If it does not snow tonight, then I will not stay home. It is a sunny summer day whenever I go to the beach. It is not a sunny summer day whenever I do not go to the beach. I do not go to the beach whenever it is not a sunny summer day. When I sleep until noon, it is necessary that I stayed up late. When I do not sleep until noon, it is necessary that I did not stay up late. When I do not stay up late, it is necessary that I do not sleep until noon. b) c) 4. (E) 28ef and 32e p T T F F p T T F F p T T T T F F F F q T F T F q T F T F q T T F F T T F F p F F T T q F T F T r T F T F T F T F q p F T T T pq T F F T r F T F T F T F T pq T F F T p q F T T F (q p) (p q) F F F T (p q) (p q) T T T T 28 e) f) 32 e) pq T T T T T T F F (p q) r F T F T F T F F 5. (C) 65 (Show the entire solution table, not just who owns the zebra and who drinks water.) House Location 1 2 3 4 5 Section 1.2 6. (E) 8abd a) Kwame will neither take a job in industry nor go to graduate school. b) Yoshiko does not know Java or does not know calculus. d) Rita will move to neither Oregon nor Washington. 7. (M) 26 (Do not use truth tables. Use the rules in Tables 6, 7, and 8.) p (q r) p (q r) p (q r) q (p r) q (p r) because p q p q by the Double Negation Law by the Associative and Commutative Laws because p q p q Nationality Norwegian Italian English Spanish Japanese House Color Yellow Blue Red White Green Favorite Drink Mineral Water Tea Milk Orange Juice Coffee Job Diplomat Physician Photographer Violinist Painter Pet Fox Horse Snails Dog Zebra 8. (M) 30 (Do not use truth tables. Use the rules in Tables 6, 7, and 8.) (p q) (p r) (q r) ((p q) (p r)) (q r) (p q) (p r) (q r) (p q) (p r) (q r) (p q) (p r) (q r) (q (p q)) (r (p r)) ((q p) (q q)) ((r p) (r r)) ((q p) T) ((r p) T) (q p) (r p) (q r) p) (p (q r) T T because p q p q by De Morgan's Laws by De Morgan's Laws by the Double Negation Law by the Associative and Commutative Laws by the Distributive Laws by the Negation Laws by the Identity Laws by the Associative and Commutative Laws by the Negation Laws by the Domination Laws 9. (C) 44 (Hint: Read problems 42 and 43.) To show that {, } is functionally complete, we need to show that every compound proposition is logically equivalent to a compound proposition using only those two logical operators. In other words, given a compound proposition we need to find an equivalent proposition using only and . Begin by considering the truth table of the compound proposition. As in Problem 42, we construct a formula as follows: (p q r . . . ) (p q r . . . ) . . . We include one term of the form (p q r . . . ) for each line of the truth table for which the proposition is true (corresponding in this example to p = T, q = T, r = F, and so on). If the original proposition is true then one of these terms is true as well, making our constructed formula true. Furthermore, if our constructed formula is true then there must be some term that is true, from which we can conclude that the original proposition is true (since each term corresponds to a line in the truth table that makes the proposition true). Using De Morgan's Laws, we can construct an equivalent formula: ((p q r . . . ) (p q r . . . ) . . . ) This formula uses only the logical operators and . Therefore, this is a functionally complete set of logical operators. Section 1.3 10. (E) 38cde c) Some system is in an open state or some system is in a diagnostic state. d) Some system is not in an available state. e) All systems are not in a working state. 11. (M) 44 x(P (x) Q(x)) is not logically equivalent to xP (x) xQ(x). Suppose x has a domain of {a, b} such that P (a) is true, P (b) is false, Q(a) is false, and Q(b) is true. Then x(P (x) Q(x)) is false but xP (x) xQ(x) is true. 12. (M) 62 a) x(P (x) S(x)) b) x(R(x) S(x)) c) x(Q(x) P (x)) d) x(Q(x) R(x)) e) Yes, (d) follows from (a), (b), and (c). Section 1.4 13. (E) 10deh d) xyF (x, y) e) xyF (y, x) h) x(yF (y, x) z(yF (y, z) z = x)) 14. (E) 24bc b) A nonnegative real number minus a negative real number is positive. c) There exist two real numbers less than or equal to zero whose difference is positive. 15. (E) 28def d) False e) True f) False 16. (E) 30cde c) y(Q(y) xR(x, y)) d) y(xR(x, y) xS(x, y)) e) y(xzT (x, y, z) xzU (x, y, z))
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Michigan - EECS - 203
EECS 203: Homework 1 SolutionsSection 1.1 1. (E) 8bef b) You do not miss the final exam if and only if you pass the course. e) It is either the case that if you have the flu then you do not pass the course or the case that if you miss the final exam then
Michigan - EECS - 203
EECS 203: Homework 2 SolutionsSection 1.5 1. (E) 4bd b) Disjunctive syllogism. d) Addition. 2. (M) 14bd b) Let r(x) = &quot;r is one of the five roommates listed&quot;, d(x) = &quot;x has taken a course in discrete mathematics&quot;, a(x) = &quot;x can take a course in algorithm
Michigan - EECS - 203
EECS 203 Homework 2Section 1.31. (E) 6bef and 10cde 6a) Every student in school has visited North Dakota 6b) Not every student in the school has visited North Dakota 6c) No student in the school has visited North Dakota 10c) x(C(x) F (x) D(x) 10d) x(C(x
Michigan - EECS - 203
EECS 203: Homework 3 SolutionsSection 2.1 1. (E) 8defg d) True e) True f) True g) False, since cfw_, cfw_ = cfw_. 2. (E) 22 a) No, since a power set should contains at least the empty set ; note that P () = cfw_. b) Yes, P (cfw_a) = cfw_, cfw_a. c) No, s
Michigan - EECS - 203
EECS 203 Homework 3 SolutionsSection 1.6 6. and(an odd number) 24. Proof by contradiction: Assume that at most two days are selected from any month and still select as least 25 days. To select as many days as possible, we can select two from each month.
Michigan - EECS - 203
EECS 203: Homework 4 SolutionsSection 2.4 1. (E) 4cd c) a0 = 8, a1 = 11, a2 = 23, a3 = 71 d) a0 = 2, a1 = 0, a2 = 8, a3 = 0 2. (E) 10ef e) The general formula for this sequence is an = (3 an-1 ) + 2, given that a0 = 0 Therefore, this sequence from n=0 to
Michigan - EECS - 203
EECS 203 HOMEWORK 4 SOLUTIONS 1. (M) Prove that the union of all elements of a power set of a set is the set itself.Let S be the original set and P(S) be the power set and UPS be the union of all elements of a power set. To Prove: S = UPS We p
Michigan - EECS - 203
EECS 203: Homework 5 SolutionsSection 3.2 1. (E) 2bcd b) f (x) is O(x2 ) with C = 2 and k =21000.c) f (x) is O(x ) with C = 1 and k = 1. Note that log x &lt; x, so x log x &lt; x2 . d) f (x) is not O(x2 ). Suppose there exist C and k such that x4 /2 Cx2 for
Michigan - EECS - 203
EECS 203: Homework 5 SolutionsSection 2.4 1. (E) 16bc b) 3 j -2 j = 3 j - 2 j 3 -1 2 -1 - 3-1 2-1 =9841-511 =9330 =j=0 9 j=0 9 j=0 8 88c) 23 j32 j j=08=2 3 j3 2 jj=0 j=08823 9-1 329 -1 = 3-1 2-1 =298413511 =212152. (E)n 2i 3 ji=1 j=1 m n
Michigan - EECS - 203
EECS 203: Homework 6 SolutionsSection 4.1 1. (E) 4 a) 13 = (1(1 + 1)/2)2 b) Evaluating both sides 1=1. So True. c) The inductive hypothesis is the statement P (k) : 13 + 23 + . + k 3 =k(k+1) 2 2.(k+1)(k+2) 2 2d) We need to prove that for all k 1, giv
Michigan - EECS - 203
EECS 203: Homework 7 SolutionsSection 5.1 1. (E) 24 a) Since the strings do not contain the same digit, there are 10 ways to choose the first digit, 9 ways to choose the second and so on. Therefore the answer is 10 9 8 7 = 5040. b) There are 10 ways to c
Michigan - EECS - 203
Homework 7 SolutionsSection 4.2 4. a)b) Assume for all c) d) for some non-negative i, j e) We have shown that the statement is true for several basis cases ( ) and that in the inductive steps all other cases are reducible to smaller cases, which are the
Michigan - EECS - 203
EECS 203: Homework 8 SolutionsSection 5.4 1. (E) 8 By binomial theorem, the coefficient of the term x8 y 9 in the expansion of (x + y)17 is 17 = 24310. 8 When considering the term x8 y 9 in the expansion of (3x + 2y)17 , let x = 3x and y = 2y, meaning th
Michigan - EECS - 203
Homework 8 Solutions Section 5.3 30 a). There are C(16,5) ways to choose a committee if there are no restrictions. There are C(9,5) ways to select a committee from just the 9 men. Therefore there are C(16,5) C(9,5) = 4242 committees with at least one woma
Michigan - EECS - 203
EECS 203: Homework 9 SolutionsSection 8.1 1. (E) 4 a) antisymmetric, transitive b) reflexive, symmetric, transitive c) reflexive, symmetric, transitive d) reflexive, symmetric 2. (E) 10 Only the relation in part (a) is irreflexive. 3. (E) 20 An asymmetri
Michigan - EECS - 203
EECS 203: Homework 9 SolutionsSection 3.8 1. (E) 4b[4 -1 -7 6 -7 -5 8 5 4 0 7 3]2. (E) 12a Let A and B be m x k matrices and C be a k x n matrix. By the definition of matrix sum, A+B is an m x k matrix such that each (i,j) entry is equal to aij + bij
Michigan - EECS - 203
EECS 203: Homework 10 SolutionsSection 8.4 1. (E) 2 When we add the pairs (x, x) to the given relation we have all of Z Z; in other words, we have the relation that always holds. 2. (E) 6 We form the reflexive closure by taking the given directed graph a
Michigan - EECS - 203
Homework 10Section 8.11.(M) 48a) Since R and S are both having all pairs like (x,x) where x is an element of set A the union will also have it. Thus R S is also reflexive. 48b) R S will be reflexive as both R and S have pairs like (x,x). 48c) R S will b
Michigan - EECS - 203
EECS 203: Homework 11 SolutionsSection 8.5 1. (E) 14 In order to show that R is an equivalence relation, we need to show that it is reflexive, symmetric and transitive. It is easy to see that R is reflexive since x has the same nth character as x for any
Michigan - EECS - 203
Homework 11 SolutionsSection 8.4 Initial Matrix Using cfw_a as an interior vertex Using cfw_a,b as interior vertices Using cfw_a,b,c as interior vertices Using cfw_a,b,c,d as interior verticesUsing cfw_a,b,c,d,e as interior verticesFinal AnswerSection
Michigan - EECS - 203
EECS 203: Homework 12 SolutionsSection 9.3 1. (E) 6,8 (6) a 0 1 0 1 0 b 1 0 0 1 1 c 0 0 0 1 1 d 1 1 1 0 0 e 0 1 1 0 0a b c d e(8) a 0 1 0 1 0 b 1 0 1 0 0 c 0 1 1 0 1 d 1 1 0 0 0 e 0 1 0 1 1a b c d e2. (E) 14 a 0 3 0 1 b 3 0 1 0 c 0 1 0 3 d 1 0 3 0a
Michigan - EECS - 203
EECS 203: Homework 12Instructions: The questions below come from the course textbook and are rated as (E)asy, (M)edium or (C)hallenging. Answer them as completely, concisely, and legibly as possible and your answers will be scored and weighted according
Michigan - EECS - 203
EECS 203: Homework 13 SolutionsSection 4.3 1. (E) 6cd c) This is not valid because, for example, f (2) can have many different values and still be consistent with the recursive definition. d) This is not valid for the opposite reason. There are no possib
Michigan - EECS - 203
Homework 6Section 1.31. (E) 10. If x &gt; 1 then x3 &lt; x4 . So clearly x3 is O(x4 ) for C=1 and k=1. But if x4 is O(x3 ) then x4 &lt; C.x3 for some C and k. Dividing both sides by x3 we get x &lt; C. Clearly this condition will not hold for all large x. 2. (E) (n
Michigan - EECS - 203
EECS 203: Homework 7Instructions: The questions below come from the course textbook and are rated as (E)asy, (M)edium or (C)hallenging. Answer them as completely, concisely, and legibly as possible and your answers will be scored and weighted according t
Michigan - EECS - 203
Mail : Inbox: EECS 203 001 LEC Teaching Questionnaires Submission C.https:/web.mail.umich.edu/blue/imp/message.php?actionID=print_messa.Date: From: To: Subject:Mon, 13 Dec 2010 00:10:40 -0500 [12:10:40 AM EST] um_evaluations@umich.edu Undisclosed Recip
Michigan - EECS - 203
Mail : Inbox: EECS 203 014 DIS Teaching Questionnaires Submission C.https:/web.mail.umich.edu/blue/imp/message.php?actionID=print_messa.Date: From: To: Subject:Mon, 13 Dec 2010 00:14:04 -0500 [12:14:04 AM EST] um_evaluations@umich.edu Undisclosed Recip
Michigan - EECS - 203
EECS 203 Exam 1 Practice ProblemsThese are some problems to help you study for the exam. They are not meant to simulate an exam, although some of the problems are similar in scope and difficulty to actual exam problems. Other problems may be a little har
Michigan - EECS - 203
Michigan - EECS - 203
Michigan - EECS - 203
Michigan - EECS - 203
Michigan - EECS - 203
EECS 203 Midterm #1 Review Multiple Choice QuestionsDiscrete Math JeopardyPropositional Predicate Sets and Logic Logic Functions100 200 300 400 500100 200 300 400 500100 200 300 400 500Propositional Logic: 100Of tautology, contingency, and contradi
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Project1: How much car can you afford? EECS 280 Fall 2010 Due: Wednesday, January 19th, 11:59 PM Introduction This project will give you experience writing, compiling, and debugging a simple C+ program. You will gain experience with header files and
Michigan - EECS - 280
Project 2: Recursive Data Structures EECS 280 Winter 2011 Due: Tuesday, February 8th, 11:59 PM Introduction This project will give you experience writing recursive functions that operate on recursively-defined data structures and mathematical abstra
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CHAPTER 10Product Concepts iStockphoto.com/Nikolay TitovLearning OutcomesLO 1 Define the term product LO 2 Classify consumer products LO 3 Define the terms product item, product line, and product mix LO 4 Describe marketing uses of brandingLearning O
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Iowa State - SCM - 301
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Iowa State - SCM - 301
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Iowa State - SCM - 301
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Iowa State - SCM - 301
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