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Bijection Functions - Discrete Mathematics - Solved Exam, Exams of Discrete Mathematics

During the study of discrete mathematics, I found this course very informative and applicable.The main points in these lecture slides are:Bijection Functions, Propositions, Tautology, Power Set, Subsets of Set, Set Identities, Expresses Negation, Inverse Function, Predicate Symbols, Appropriate Translation, Real Number, Inference Rule

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2012/2013

Uploaded on 04/27/2013

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CS 173: Midterm Exam 1
Spring 2005
Name: SOLUTION
NetID:
Lecture Section:
General Directions
1. Make sure your name is on every page.
2. There are 11 pages, including a sheet of scratch paper. Make sure that you answer all 14 questions.
3. Remember to write clearly and legibly. Unreadable answers will receive no credit.
4. This is a closed book exam. No notes of any kind are allowed.
5. Remember to time yourself.
Question Points Out of
1 5
2 5
3 5
4 5
5 5
6 5
7 5
85
910
10 10
11 10
12 10
13 10
14 10
Total 100
Page 1 of 12
pf3
pf4
pf5
pf8
pf9
pfa

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CS 173: Midterm Exam 1

Spring 2005

Name: SOLUTION

NetID:

Lecture Section:

General Directions

  1. Make sure your name is on every page.
  2. There are 11 pages, including a sheet of scratch paper. Make sure that you answer all 14 questions.
  3. Remember to write clearly and legibly. Unreadable answers will receive no credit.
  4. This is a closed book exam. No notes of any kind are allowed.
  5. Remember to time yourself.

Question Points^ Out of 1 5 2 5 3 5 4 5 5 5 6 5 7 5 8 5 9 10 10 10 11 10 12 10 13 10 14 10 Total 100

Multiple Choice

Problem 1 (5pts)

Which one of the following functions is a bijection from ^ to ^?

a)    b)   c) ^ ^ ! " d)  (^) ##%$

Solution

&

Problem 2 (5pts)

Which one of the following propositions is NOT a tautology?

a) ('),+-). ' 0/ b) 3'4,+5 /63'7),. c) 9.+:);(',/<+5 = />. ' d) 3'/>+5 ?)@+A/215  B/C3'/

Solution

D

Problem 3 (5pts)

Let EF HGIKJLGMI?NOJLGMIPJQGIPNNRN .Which one of the following statements is true?

a) GMIKJLGGMIPNNN is in the power set of E. b) EHS;GMI?NT I. c) Let U HGMIPJQGIPNN. Then UWV,E and UWXYE. d) E[Z\GRGIPNN] HGIKJLGMI?NOJLGRGIPNNOJLGIKJQGIPNNRN

Solution

&

Problem 7 (5pts)

Given the predicate symbols

x  is “  is a dog.” y  is “ is a rabbit.”

^zKJ{| (^) is “  chases {.”

Which one of the following is an appropriate translation of the English statement “Only dogs chase rabbits”?

a) ~} {| L€m Q y {| ?)^zPJ={|  B/ x  = b) 9m{| L~}  L y {| ?)^zKJ{|  B/ x   c) ‚} {| Q‚}  L= y {| ƒ)^zKJ{|  B/ x   d) 9m{| L€m Q y {| P)^zKJ{|  B/ x  

Solution

c

Problem 8 (5pts)

Which one of the following arguments is NOT valid?

a) If „ is a real number with „@ f† , then „  F‡. Suppose that „ vˆ ‡. Then „ ˆ †. b) If „ is a real number with „@ ‰ , then „Kv І. Suppose that „ ˆ ‰. Then „P ˆ †. c) All students in this class love CS173. Xavier is a student in this class. Therefore, Xavier loves CS173. d) Every computer science major takes CS173. Natasha does not take CS173. Therefore, Natasha is not a computer science major.

Solution

b

Short Answer Problems

Problem 9 (10pts)

Prove the inference rule known as ”simplification” without using truth tables.

Solution

'),+F ?‹Œ'F.w3'),+5 ?4c'e€. 'c4.+5 ?4c'e3'74,. ' ƒ4,+vq

Comments

Each mistake drops two points.

Problem 11 (10pts)

Prove or disprove the following statements.

a) For any integers l and j, l”“z–•—A FjH“z–•—A/<l–˜“z–•,—™ Fjš›“z–•o—. b) For any integers l and j, l– “7–•o—A jL›“z–•—™/>l‘“z–•o—A jH“z–•o—.

Solution

a) We first prove that lœ“7–•—™ "1A je“7–•o—A/<l  “z–•o—v Fj  “7–•o—.

Let l7 F—v d1 , and j: F—QžT d1 , with 1 cVŸG ¡JQJ%¢£¢¢J†|N. (Any integer can be written in this form.) Then l  ‰R—  % 515A [1 , and j ‰R—Qž  " 51Lž] [1 . Thus, lmk —¡9—R–B f‰515| P [1T Y1M¤“7–•o—. Similarly, jLk Y1M¤“zm•—. b) We disprove that l– “z–•o—v Fjš “7–•o—A/<lœ“z–•,—™ FjH“z–•,—.

All that is needed is a counterexample. Let lz "¥ , jw ‰. Then, l| “zm•o—™ jš “z–•—™ "†. However, l\¦j.

Comments

a) 5 points. 1 point for lz —A [. 1 point for j: —%ž] [. 1 point for squaring. 1 point for factoring. 1 point for the argument. b) 5 points. 1 point for recognizing that the claim is false. 4 points for the counterexample.

Problem 12 (10pts)

Prove that if _fXq , then _oZaq" q.

Solution

The proof can be done directly or indirectly, but similar intricacies exist in each. Assume ŠXq. First, show ZqFXŠq. Choose oVoZq. Then, \Vo or oVq by definition of union. Case 1, \V_. Then oVq since _[Xq. Case 2, oVq.

Second, show q§XY_Zq. Choose oV,q. Then oV_,Zq by addition.

Comments

10 points. 2 points for a reasonable attempt. 1 point each for writing down correct definitions of X , and Z. 1 point for assuming _fXŠq. 2 points each for _oZaq§XŠq , and vice versa. 1 point for cases.

Solution

a)   R  b)   c) ¨ZqbZ,E ¨ ¨ ¨ ¨q ¨ ¨ E

_

oZaq ¨ ¨q[ZE ¨ ¨,Z,E ¨ ¨qb,E ¨   © %  R w % R  R kd‰Tb‰Td‰©  R% RªR—

Comments

a) 3 points This one is saying that set _ (students that are sick) is a subset of q and q is a subset of E. So _Z7q;Z!E is just E. Therefore,

_ZqdZE

E

.

b) 3 points Pairwise disjoint means the every pair of sets does not have overlapping part. Therefore,

_Zdq§ZfE

_

q

E

.

c) 4 points This is a standard problem of set operations. First, we add up ¨_ ¨, ¨q ¨, and ¨E ¨. Since the green parts (i.e. ¨_`q ¨, ¨qŠoE ¨, and ¨_;oE ¨) are added twice

and the red part (¨ ¨_,q[,E ¨) is added three times, we next subtract duplicate ones. After subtracting ¨oq ¨, qfoE ¨, and ¨;oE ¨from the sum, we find that ¨_;oqfoE ¨is actually subtracted three times. So we finally add ¨_qdE ¨once. Therefore, the overall process looks like: ¨_ZqbZ,E ¨ ¨_ ¨ ¨q ¨ ¨E ¨ ¨_Zq ¨ ¨q[Z,E ¨ ¨_oZE ¨ ¨_qd`,E ¨

Problem 14 (10pts)

Let «! , ¬! , and y  be the statements ” is a duck,” ”  is annoying,” and ” is a dancer,” respectively. Express each of these statements using quantifiers, logical connectives, and «! , ¬ , and y .

a) All ducks are annoying. b) Some dancers are not annoying. c) Some dancers are not ducks. d) Does c) follow from a) and b)? Prove your response.