378 Relations Solutions for Chapter 16. Section 16.1 Exercises. 3. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses on A.

Size: px
Start display at page:

Download "378 Relations Solutions for Chapter 16. Section 16.1 Exercises. 3. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses on A."

Transcription

1 378 Reltions 16.7 Solutions for Chpter 16 Section 16.1 Exercises 1. Let A = {0,1,2,3,4,5}. Write out the reltion R tht expresses > on A. Then illustrte it with digrm. 2 1 R = { (5,4),(5,3),(5,2),(5,1),(5,0),(4,3),(4,2),(4,1), (4,0),(3,2),(3,1),(3,0),(2,1),(2,0),(1,0) } Let A = {0,1,2,3,4,5}. Write out the reltion R tht expresses on A. Then illustrte it with digrm. R = { 2 1 (5,5),(5,4),(5,3),(5,2),(5,1),(5,0), (4,4),(4,3),(4,2),(4,1),(4,0), 3 0 (3,3),(3,2),(3,1),(3,0), (2,2),(2,1),(2,0),(1,1),(1,0),(0,0) } The following digrm represents reltion R on set A. Write the sets A nd R. Answer: A = {0,1,2,3,4,5}; R = {(3,3),(4,3),(4,2),(1,2),(2,5),(5,0)} 7. Write the reltion < on the set A = Z s suset R of Z Z. This is n infinite set, so you will hve to use set-uilder nottion. Answer: R = {(x, y) Z Z : y x N} 9. How mny different reltions re there on the set A = { 1,2,3,4,5,6 }? Consider forming reltion R A A on A. For ech ordered pir (x, y) A A, we hve two choices: we cn either include (x, y) in R or not include it. There re 6 6 = 36 ordered pirs in A A. By the multipliction principle, there re thus 2 36 different susets R nd hence lso this mny reltions on A. 11. Answer: 2 ( A 2 ) 13. Answer: 15. Answer: (mod 3) Section 16.2 Exercises 1. Consider the reltion R = {(, ),(, ),(c, c),(d, d),(, ),(, )} on the set A = {,, c, d}. Which of the properties reflexive, symmetric nd trnsitive does R possess nd why? If property does not hold, sy why. This is reflexive ecuse (x, x) R (i.e., xrx )for every x A. It is symmetric ecuse it is impossile to find n (x, y) R for which (y, x) R. It is trnsitive ecuse (xr y yrz) xrz lwys holds. 3. Consider the reltion R = {(, ),(, c),(c, ),(, c)} on the set A = {,, c}. Which of the properties reflexive, symmetric nd trnsitive does R possess nd why? If property does not hold, sy why.

2 Solutions for Chpter This is not reflexive ecuse (, ) R (for exmple). It is not symmetric ecuse (, ) R ut (, ) R. It is not trnsitive ecuse cr nd Rc re true, ut crc is flse. 5. Consider the reltion R = { (0,0),( 2,0),(0, 2),( 2, 2) } on R. Sy whether this reltion is reflexive, symmetric nd trnsitive. If property does not hold, sy why. This is not reflexive ecuse (1,1) R (for exmple). It is symmetric ecuse it is impossile to find n (x, y) R for which (y, x) R. It is trnsitive ecuse (xr y yrz) xrz lwys holds. 7. There re 16 possile different reltions R on the set A = {, }. Descrie ll of them. (A picture for ech one will suffice, ut don t forget to lel the nodes.) Which ones re reflexive? Symmetric? Trnsitive? Only the four in the right column re reflexive. Only the eight in the first nd fourth rows re symmetric. All of them re trnsitive except the first three on the fourth row. 9. Define reltion on Z y declring xr y if nd only if x nd y hve the sme prity. Sy whether this reltion is reflexive, symmetric nd trnsitive. If property does not hold, sy why. Wht fmilir reltion is this? This is reflexive ecuse xrx since x lwys hs the sme prity s x. It is symmetric ecuse if x nd y hve the sme prity, then y nd x must hve the sme prity (tht is, xr y yrx). It is trnsitive ecuse if x nd y hve the sme prity nd y nd z hve the sme prity, then x nd z must hve the sme prity. (Tht is (xr y yr z) xr z lwys holds.) The reltion is congruence modulo Suppose A = {,, c, d} nd R = {(, ),(, ),(c, c),(d, d)}. Sy whether this reltion is reflexive, symmetric nd trnsitive. If property does not hold, sy why. This is reflexive ecuse (x, x) R for every x A. It is symmetric ecuse it is impossile to find n (x, y) R for which (y, x) R. It is trnsitive ecuse (xr y yrz) xrz lwys holds. (For exmple (R R) R is true, etc.) 13. Consider the reltion R = {(x, y) R R : x y Z} on R. Prove tht this reltion is reflexive nd symmetric, nd trnsitive.

3 380 Reltions Proof. In this reltion, xr y mens x y Z. To see tht R is reflexive, tke ny x R nd oserve tht x x = 0 Z, so xrx. Therefore R is reflexive. To see tht R is symmetric, we need to prove xr y yrx for ll x, y R. We use direct proof. Suppose xr y. This mens x y Z. Then it follows tht (x y) = y x is lso in Z. But y x Z mens yrx. We ve shown xr y implies yrx, so R is symmetric. To see tht R is trnsitive, we need to prove (xr y yrz) xrz is lwys true. We prove this conditionl sttement with direct proof. Suppose xr y nd yr z. Since xr y, we know x y Z. Since yrz, we know y z Z. Thus x y nd y z re oth integers; y dding these integers we get nother integer (x y) + (y z) = x z. Thus x z Z, nd this mens xrz. We ve now shown tht if xr y nd yrz, then xr z. Therefore R is trnsitive. 15. Prove or disprove: If reltion is symmetric nd trnsitive, then it is lso reflexive. This is flse. For counterexmple, consider the reltion R = {(, ),(, ),(, ),(, )} on the set A = {,, c}. This is symmetric nd trnsitive ut it is not reflexive. 17. Define reltion on Z s x y if nd only if x y 1. Sy whether is reflexive, symmetric nd trnsitive. This is reflexive ecuse x x = 0 1 for ll integers x. It is symmetric ecuse x y if nd only if x y 1, if nd only if y x 1, if nd only if y x. It is not trnsitive ecuse, for exmple, 0 1 nd 1 2, ut is not the cse tht 0 2. Section 16.3 Exercises 1. Let A = {1,2,3,4,5,6}, nd consider the following equivlence reltion on A: R = {(1,1),(2,2),(3,3),(4,4),(5,5),(6,6),(2,3),(3,2),(4,5),(5,4),(4,6),(6,4),(5,6),(6,5)}. List the equivlence clsses of R. The equivlence clsses re: [1] = {1}; [2] = [3] = {2,3}; [4] = [5] = [6] = {4,5,6}. 3. Let A = {,, c, d, e}. Suppose R is n equivlence reltion on A. Suppose R hs three equivlence clsses. Also Rd nd Rc. Write out R s set. Answer: R = {(, ),(, ),(c, c),(d, d),(e, e),(, d),(d, ),(, c),(c, )}. 5. There re two different equivlence reltions on the set A = {, }. Descrie them ll. Digrms will suffice. Answer: R = {(, ),(, )} nd R = {(, ),(, ),(, ),(, )} 7. Define reltion R on Z s xr y if nd only if 3x 5y is even. Prove R is n equivlence reltion. Descrie its equivlence clsses. To prove tht R is n equivlence reltion, we must show it s reflexive, symmetric nd trnsitive. The reltion R is reflexive for the following reson. If x Z, then 3x 5x = 2x is even. But then since 3x 5x is even, we hve xrx. Thus R is reflexive.

4 Solutions for Chpter To see tht R is symmetric, suppose xr y. We must show yrx. Since xr y, we know 3x 5y is even, so 3x 5y = 2 for some integer. Now reson s follows: 3x 5y = 2 3x 5y + 8y 8x = 2 + 8y 8x 3y 5x = 2( + 4y 4x). From this it follows tht 3y 5x is even, so yrx. We ve now shown xr y implies yrx, so R is symmetric. To prove tht R is trnsitive, ssume tht xr y nd yrz. (We will show tht this implies xrz.) Since xr y nd yrz, it follows tht 3x 5y nd 3y 5z re oth even, so 3x 5y = 2 nd 3y 5z = 2 for some integers nd. Adding these equtions, we get (3x 5y) + (3y 5z) = 2 + 2, nd this simplifies to 3x 5z = 2( + + y). Therefore 3x 5z is even, so xrz. We ve now shown tht if xr y nd yrz, then xrz, so R is trnsitive. We ve now shown tht R is reflexive, symmetric nd trnsitive, so it is n equivlence reltion. The completes the first prt of the prolem. Now we move on the second prt. To find the equivlence clsses, first note tht [0] = {x Z : xr0} = {x Z : 3x 5 0 is even} = {x Z : 3x is even} = {x Z : x is even}. Thus the equivlence clss [0] consists of ll even integers. Next, note tht [1] = {x Z : xr1} = {x Z : 3x 5 1 is even} = {x Z : 3x 5 is even} = { x Z : x is odd }. Thus the equivlence clss [1] consists of ll odd integers. Consequently there re just two equivlence clsses {..., 4, 2,0,2,4,...} nd {..., 3, 1,1,3,5,...}. 9. Define reltion R on Z s xr y if nd only if 4 (x+3y). Prove R is n equivlence reltion. Descrie its equivlence clsses. This is reflexive, ecuse for ny x Z we hve 4 (x + 3x), so xrx. To prove tht R is symmetric, suppose xr y. Then 4 (x+3y), so x+3y = 4 for some integer. Multiplying y 3, we get 3x +9y = 12, which ecomes y+3x = 12 8y. Then y + 3x = 4(3 2y), so 4 (y + 3x), hence yrx. Thus we ve shown xr y implies yrx, so R is symmetric. To prove trnsitivity, suppose xr y nd yrz. Then 4 (x + 3y) nd 4 (y + 3z), so x +3y = 4 nd y+3z = 4 for some integers nd. Adding these two equtions produces x + 4y + 3z = 4 + 4, or x + 3z = y = 4( + y). Consequently 4 (x + 3z), so xrz, nd R is trnsitive. As R is reflexive, symmetric nd trnsitive, it is n equivlence reltion. Now let s compute its equivlence clsses. [0] = {x Z : xr0} = {x Z : 4 (x + 3 0)} = {x Z : 4 x} = {... 4,0,4,8,12,16...}

5 382 Reltions [1] = {x Z : xr1} = {x Z : 4 (x + 3 1)} = {x Z : 4 (x + 3)} = {... 3,1,5,9,13,17...} [2] = {x Z : xr2} = {x Z : 4 (x + 3 2)} = {x Z : 4 (x + 6)} = {... 2,2,6,10,14,18...} [3] = {x Z : xr3} = {x Z : 4 (x + 3 3)} = {x Z : 4 (x + 9)} = {... 1,3,7,11,15,19...} 11. Prove or disprove: If R is n equivlence reltion on n infinite set A, then R hs infinitely mny equivlence clsses. This is Flse. Counterexmple: consider the reltion of congruence modulo 2. It is reltion on the infinite set Z, ut it hs only two equivlence clsses. 13. Answer: m A 15. Answer: 15 Section 16.4 Exercises 1. List ll the prtitions of the set A = {, }. Compre your nswer to the nswer to Exercise 5 of Section There re just two prtitions {{},{}} nd {{, }}. These correspond to the two equivlence reltions R 1 = {(, ),(, )} nd R 2 = {(, ),(, ),(, ),(, )}, respectively, on A. 3. Descrie the prtition of Z resulting from the equivlence reltion (mod 4). Answer: The prtition is {[0],[1],[2],[3]} = { {..., 4,0,4,8,12,...},{..., 3,1,5,9,13,...}, {..., 2,2,4,6,10,14,...}, {..., 1,3,7,11,15,...} } 5. Answer: Congruence modulo 2, or sme prity. Section 16.5 Exercises 1. Write the ddition nd multipliction tles for Z 2. + [0] [1] [0] [0] [1] [1] [1] [0] [0] [1] [0] [0] [0] [1] [0] [1] 3. Write the ddition nd multipliction tles for Z 4. + [0] [1] [2] [3] [0] [0] [1] [2] [3] [1] [1] [2] [3] [0] [2] [2] [3] [0] [1] [3] [3] [0] [1] [2] [0] [1] [2] [3] [0] [0] [0] [0] [0] [1] [0] [1] [2] [3] [2] [0] [2] [0] [2] [3] [0] [3] [2] [1] 5. Suppose [],[] Z 5 nd [] [] = [0]. Is it necessrily true tht either [] = [0] or [] = [0]?

6 Solutions for Chpter The multipliction tle for Z 5 is shown in Section In the ody of tht tle, the only plce tht [0] occurs is in the first row or the first column. Tht row nd column re oth heded y [0]. It follows tht if [] [] = [0], then either [] or [] must e [0]. 7. Do the following clcultions in Z 9, in ech cse expressing your nswer s [] with 0 8. () [8] + [8] = [7] () [24] + [11] = [8] (c) [21] [15] = [0] (d) [8] [8] = [1]

set is not closed under matrix [ multiplication, ] and does not form a group.

Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed

Lecture 3: Equivalence Relations

Mthcmp Crsh Course Instructor: Pdric Brtlett Lecture 3: Equivlence Reltions Week 1 Mthcmp 2014 In our lst three tlks of this clss, we shift the focus of our tlks from proof techniques to proof concepts

Coalgebra, Lecture 15: Equations for Deterministic Automata

Colger, Lecture 15: Equtions for Deterministic Automt Julin Slmnc (nd Jurrin Rot) Decemer 19, 2016 In this lecture, we will study the concept of equtions for deterministic utomt. The notes re self contined

Homework 4. 0 ε 0. (00) ε 0 ε 0 (00) (11) CS 341: Foundations of Computer Science II Prof. Marvin Nakayama

CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 4 1. UsetheproceduredescriedinLemm1.55toconverttheregulrexpression(((00) (11)) 01) into n NFA. Answer: 0 0 1 1 00 0 0 11 1 1 01 0 1 (00)

Minimal DFA. minimal DFA for L starting from any other

Miniml DFA Among the mny DFAs ccepting the sme regulr lnguge L, there is exctly one (up to renming of sttes) which hs the smllest possile numer of sttes. Moreover, it is possile to otin tht miniml DFA

Bases for Vector Spaces

Bses for Vector Spces 2-26-25 A set is independent if, roughly speking, there is no redundncy in the set: You cn t uild ny vector in the set s liner comintion of the others A set spns if you cn uild everything

Discrete Structures, Test 2 Monday, March 28, 2016 SOLUTIONS, VERSION α

Disrete Strutures, Test 2 Mondy, Mrh 28, 2016 SOLUTIONS, VERSION α α 1. (18 pts) Short nswer. Put your nswer in the ox. No prtil redit. () Consider the reltion R on {,,, d with mtrix digrph of R.. Drw

CS 310 (sec 20) - Winter Final Exam (solutions) SOLUTIONS

CS 310 (sec 20) - Winter 2003 - Finl Exm (solutions) SOLUTIONS 1. (Logic) Use truth tles to prove the following logicl equivlences: () p q (p p) (q q) () p q (p q) (p q) () p q p q p p q q (q q) (p p)

Matrix Algebra. Matrix Addition, Scalar Multiplication and Transposition. Linear Algebra I 24

Mtrix lger Mtrix ddition, Sclr Multipliction nd rnsposition Mtrix lger Section.. Mtrix ddition, Sclr Multipliction nd rnsposition rectngulr rry of numers is clled mtrix ( the plurl is mtrices ) nd the

Linear Inequalities. Work Sheet 1

Work Sheet 1 Liner Inequlities Rent--Hep, cr rentl compny,chrges \$ 15 per week plus \$ 0.0 per mile to rent one of their crs. Suppose you re limited y how much money you cn spend for the week : You cn spend

MATH 573 FINAL EXAM. May 30, 2007

MATH 573 FINAL EXAM My 30, 007 NAME: Solutions 1. This exm is due Wednesdy, June 6 efore the 1:30 pm. After 1:30 pm I will NOT ccept the exm.. This exm hs 1 pges including this cover. There re 10 prolems.

Multiplying integers EXERCISE 2B INDIVIDUAL PATHWAYS. -6 ì 4 = -6 ì 0 = 4 ì 0 = -6 ì 3 = -5 ì -3 = 4 ì 3 = 4 ì 2 = 4 ì 1 = -5 ì -2 = -6 ì 2 = -6 ì 1 =

EXERCISE B INDIVIDUAL PATHWAYS Activity -B- Integer multipliction doc-69 Activity -B- More integer multipliction doc-698 Activity -B- Advnced integer multipliction doc-699 Multiplying integers FLUENCY

Homework 3 Solutions

CS 341: Foundtions of Computer Science II Prof. Mrvin Nkym Homework 3 Solutions 1. Give NFAs with the specified numer of sttes recognizing ech of the following lnguges. In ll cses, the lphet is Σ = {,1}.

1. For each of the following theorems, give a two or three sentence sketch of how the proof goes or why it is not true.

York University CSE 2 Unit 3. DFA Clsses Converting etween DFA, NFA, Regulr Expressions, nd Extended Regulr Expressions Instructor: Jeff Edmonds Don t chet y looking t these nswers premturely.. For ech

CM10196 Topic 4: Functions and Relations

CM096 Topic 4: Functions nd Reltions Guy McCusker W. Functions nd reltions Perhps the most widely used notion in ll of mthemtics is tht of function. Informlly, function is n opertion which tkes n input

Regular Language. Nonregular Languages The Pumping Lemma. The pumping lemma. Regular Language. The pumping lemma. Infinitely long words 3/17/15

Regulr Lnguge Nonregulr Lnguges The Pumping Lemm Models of Comput=on Chpter 10 Recll, tht ny lnguge tht cn e descried y regulr expression is clled regulr lnguge In this lecture we will prove tht not ll

Math 4310 Solutions to homework 1 Due 9/1/16

Mth 4310 Solutions to homework 1 Due 9/1/16 1. Use the Eucliden lgorithm to find the following gretest common divisors. () gcd(252, 180) = 36 (b) gcd(513, 187) = 1 (c) gcd(7684, 4148) = 68 252 = 180 1

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

U.U.D.M. Project Report 07:4 Frey Frctions Rickrd Fernström Exmensrete i mtemtik, 5 hp Hledre: Andres Strömergsson Exmintor: Jörgen Östensson Juni 07 Deprtment of Mthemtics Uppsl University Frey Frctions

Convert the NFA into DFA

Convert the NF into F For ech NF we cn find F ccepting the sme lnguge. The numer of sttes of the F could e exponentil in the numer of sttes of the NF, ut in prctice this worst cse occurs rrely. lgorithm:

QUADRATIC EQUATIONS OBJECTIVE PROBLEMS +. The solution of the eqution will e (), () 0,, 5, 5. The roots of the given eqution ( p q) ( q r) ( r p) 0 + + re p q r p (), r p p q, q r p q (), (d), q r p q.

CMPSCI 250: Introduction to Computation. Lecture #31: What DFA s Can and Can t Do David Mix Barrington 9 April 2014

CMPSCI 250: Introduction to Computtion Lecture #31: Wht DFA s Cn nd Cn t Do Dvid Mix Brrington 9 April 2014 Wht DFA s Cn nd Cn t Do Deterministic Finite Automt Forml Definition of DFA s Exmples of DFA

Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.

Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one

CS 311 Homework 3 due 16:30, Thursday, 14 th October 2010

CS 311 Homework 3 due 16:30, Thursdy, 14 th Octoer 2010 Homework must e sumitted on pper, in clss. Question 1. [15 pts.; 5 pts. ech] Drw stte digrms for NFAs recognizing the following lnguges:. L = {w

Section 3.1: Exponent Properties

Section.1: Exponent Properties Ojective: Simplify expressions using the properties of exponents. Prolems with exponents cn often e simplied using few sic exponent properties. Exponents represent repeted

CSE : Exam 3-ANSWERS, Spring 2011 Time: 50 minutes

CSE 260-002: Exm 3-ANSWERS, Spring 20 ime: 50 minutes Nme: his exm hs 4 pges nd 0 prolems totling 00 points. his exm is closed ook nd closed notes.. Wrshll s lgorithm for trnsitive closure computtion is

Lecture 2e Orthogonal Complement (pages )

Lecture 2e Orthogonl Complement (pges -) We hve now seen tht n orthonorml sis is nice wy to descrie suspce, ut knowing tht we wnt n orthonorml sis doesn t mke one fll into our lp. In theory, the process

In this skill we review equations that involve percents. review the meaning of proportion.

6 MODULE 5. PERCENTS 5b Solving Equtions Mening of Proportion In this skill we review equtions tht involve percents. review the mening of proportion. Our first tsk is to Proportions. A proportion is sttement

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

CS 373, Spring Solutions to Mock midterm 1 (Based on first midterm in CS 273, Fall 2008.)

CS 373, Spring 29. Solutions to Mock midterm (sed on first midterm in CS 273, Fll 28.) Prolem : Short nswer (8 points) The nswers to these prolems should e short nd not complicted. () If n NF M ccepts

Formal Languages and Automata

Moile Computing nd Softwre Engineering p. 1/5 Forml Lnguges nd Automt Chpter 2 Finite Automt Chun-Ming Liu cmliu@csie.ntut.edu.tw Deprtment of Computer Science nd Informtion Engineering Ntionl Tipei University

7. Indefinite Integrals

7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find

MTH 505: Number Theory Spring 2017

MTH 505: Numer Theory Spring 207 Homework 2 Drew Armstrong The Froenius Coin Prolem. Consider the eqution x ` y c where,, c, x, y re nturl numers. We cn think of \$ nd \$ s two denomintions of coins nd \$c

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18

Infinite Geometric Series

Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to

Things to Memorize: A Partial List. January 27, 2017

Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved

AT100 - Introductory Algebra. Section 2.7: Inequalities. x a. x a. x < a

Section 2.7: Inequlities In this section, we will Determine if given vlue is solution to n inequlity Solve given inequlity or compound inequlity; give the solution in intervl nottion nd the solution 2.7

Chapter 1: Logarithmic functions and indices

Chpter : Logrithmic functions nd indices. You cn simplify epressions y using rules of indices m n m n m n m n ( m ) n mn m m m m n m m n Emple Simplify these epressions: 5 r r c 4 4 d 6 5 e ( ) f ( ) 4

The University of Nottingham SCHOOL OF COMPUTER SCIENCE A LEVEL 2 MODULE, SPRING SEMESTER LANGUAGES AND COMPUTATION ANSWERS

The University of Nottinghm SCHOOL OF COMPUTER SCIENCE LEVEL 2 MODULE, SPRING SEMESTER 2016 2017 LNGUGES ND COMPUTTION NSWERS Time llowed TWO hours Cndidtes my complete the front cover of their nswer ook

Answers and Solutions to (Some Even Numbered) Suggested Exercises in Chapter 11 of Grimaldi s Discrete and Combinatorial Mathematics

Answers n Solutions to (Some Even Numere) Suggeste Exercises in Chpter 11 o Grimli s Discrete n Comintoril Mthemtics Section 11.1 11.1.4. κ(g) = 2. Let V e = {v : v hs even numer o 1 s} n V o = {v : v

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.

State Minimization for DFAs

Stte Minimiztion for DFAs Red K & S 2.7 Do Homework 10. Consider: Stte Minimiztion 4 5 Is this miniml mchine? Step (1): Get rid of unrechle sttes. Stte Minimiztion 6, Stte is unrechle. Step (2): Get rid

Finite Automata-cont d

Automt Theory nd Forml Lnguges Professor Leslie Lnder Lecture # 6 Finite Automt-cont d The Pumping Lemm WEB SITE: http://ingwe.inghmton.edu/ ~lnder/cs573.html Septemer 18, 2000 Exmple 1 Consider L = {ww

Exponentials - Grade 10 [CAPS] *

OpenStx-CNX module: m859 Exponentils - Grde 0 [CAPS] * Free High School Science Texts Project Bsed on Exponentils by Rory Adms Free High School Science Texts Project Mrk Horner Hether Willims This work

Special Numbers, Factors and Multiples

Specil s, nd Student Book - Series H- + 3 + 5 = 9 = 3 Mthletics Instnt Workooks Copyright Student Book - Series H Contents Topics Topic - Odd, even, prime nd composite numers Topic - Divisiility tests

8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers.

8. Complex Numers The rel numer system is dequte for solving mny mthemticl prolems. But it is necessry to extend the rel numer system to solve numer of importnt prolems. Complex numers do not chnge the

Chapter Five: Nondeterministic Finite Automata. Formal Language, chapter 5, slide 1

Chpter Five: Nondeterministic Finite Automt Forml Lnguge, chpter 5, slide 1 1 A DFA hs exctly one trnsition from every stte on every symol in the lphet. By relxing this requirement we get relted ut more

HW3, Math 307. CSUF. Spring 2007.

HW, Mth 7. CSUF. Spring 7. Nsser M. Abbsi Spring 7 Compiled on November 5, 8 t 8:8m public Contents Section.6, problem Section.6, problem Section.6, problem 5 Section.6, problem 7 6 5 Section.6, problem

ad = cb (1) cf = ed (2) adf = cbf (3) cf b = edb (4)

10 Most proofs re left s reding exercises. Definition 10.1. Z = Z {0}. Definition 10.2. Let be the binry reltion defined on Z Z by, b c, d iff d = cb. Theorem 10.3. is n equivlence reltion on Z Z. Proof.

dx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω.

Chpter 8 Stility theory We discuss properties of solutions of first order two dimensionl system, nd stility theory for specil clss of liner systems. We denote the independent vrile y t in plce of x, nd

September 13 Homework Solutions

College of Engineering nd Computer Science Mechnicl Engineering Deprtment Mechnicl Engineering 5A Seminr in Engineering Anlysis Fll Ticket: 5966 Instructor: Lrry Cretto Septemer Homework Solutions. Are

2.4 Linear Inequalities and Interval Notation

.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or

Analytically, vectors will be represented by lowercase bold-face Latin letters, e.g. a, r, q.

1.1 Vector Alger 1.1.1 Sclrs A physicl quntity which is completely descried y single rel numer is clled sclr. Physiclly, it is something which hs mgnitude, nd is completely descried y this mgnitude. Exmples

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=

Consolidation Worksheet

Cmbridge Essentils Mthemtics Core 8 NConsolidtion Worksheet N Consolidtion Worksheet Work these out. 8 b 7 + 0 c 6 + 7 5 Use the number line to help. 2 Remember + 2 2 +2 2 2 + 2 Adding negtive number is

Myhill-Nerode Theorem

Overview Myhill-Nerode Theorem Correspondence etween DA s nd MN reltions Cnonicl DA for L Computing cnonicl DFA Myhill-Nerode Theorem Deepk D Souz Deprtment of Computer Science nd Automtion Indin Institute

Computing with finite semigroups: part I

Computing with finite semigroups: prt I J. D. Mitchell School of Mthemtics nd Sttistics, University of St Andrews Novemer 20th, 2015 J. D. Mitchell (St Andrews) Novemer 20th, 2015 1 / 34 Wht is this tlk

Introduction to Group Theory

Introduction to Group Theory Let G be n rbitrry set of elements, typiclly denoted s, b, c,, tht is, let G = {, b, c, }. A binry opertion in G is rule tht ssocites with ech ordered pir (,b) of elements

List all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.

Mth Anlysis CP WS 4.X- Section 4.-4.4 Review Complete ech question without the use of grphing clcultor.. Compre the mening of the words: roots, zeros nd fctors.. Determine whether - is root of 0. Show

Linear Systems with Constant Coefficients

Liner Systems with Constnt Coefficients 4-3-05 Here is system of n differentil equtions in n unknowns: x x + + n x n, x x + + n x n, x n n x + + nn x n This is constnt coefficient liner homogeneous system

Harvard University Computer Science 121 Midterm October 23, 2012

Hrvrd University Computer Science 121 Midterm Octoer 23, 2012 This is closed-ook exmintion. You my use ny result from lecture, Sipser, prolem sets, or section, s long s you quote it clerly. The lphet is

Nondeterminism and Nodeterministic Automata

Nondeterminism nd Nodeterministic Automt 61 Nondeterminism nd Nondeterministic Automt The computtionl mchine models tht we lerned in the clss re deterministic in the sense tht the next move is uniquely

38 Riemann sums and existence of the definite integral.

38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

On the diagram below the displacement is represented by the directed line segment OA.

Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel. Other exmples

STRAND J: TRANSFORMATIONS, VECTORS and MATRICES

Mthemtics SKE: STRN J STRN J: TRNSFORMTIONS, VETORS nd MTRIES J3 Vectors Text ontents Section J3.1 Vectors nd Sclrs * J3. Vectors nd Geometry Mthemtics SKE: STRN J J3 Vectors J3.1 Vectors nd Sclrs Vectors

Vectors , (0,0). 5. A vector is commonly denoted by putting an arrow above its symbol, as in the picture above. Here are some 3-dimensional vectors:

Vectors 1-23-2018 I ll look t vectors from n lgeric point of view nd geometric point of view. Algericlly, vector is n ordered list of (usully) rel numers. Here re some 2-dimensionl vectors: (2, 3), ( )

CS 330 Formal Methods and Models

CS 330 Forml Methods nd Models Dn Richrds, George Mson University, Spring 2017 Quiz Solutions Quiz 1, Propositionl Logic Dte: Ferury 2 1. Prove ((( p q) q) p) is tutology () (3pts) y truth tle. p q p q

Continuous Random Variables Class 5, Jeremy Orloff and Jonathan Bloom

Lerning Gols Continuous Rndom Vriles Clss 5, 8.05 Jeremy Orloff nd Jonthn Bloom. Know the definition of continuous rndom vrile. 2. Know the definition of the proility density function (pdf) nd cumultive

Intermediate Math Circles Wednesday, November 14, 2018 Finite Automata II. Nickolas Rollick a b b. a b 4

Intermedite Mth Circles Wednesdy, Novemer 14, 2018 Finite Automt II Nickols Rollick nrollick@uwterloo.c Regulr Lnguges Lst time, we were introduced to the ide of DFA (deterministic finite utomton), one

First Midterm Examination

24-25 Fll Semester First Midterm Exmintion ) Give the stte digrm of DFA tht recognizes the lnguge A over lphet Σ = {, } where A = {w w contins or } 2) The following DFA recognizes the lnguge B over lphet

HQPD - ALGEBRA I TEST Record your nswers on the nswer sheet. Choose the best nswer for ech. 1. If 7(2d ) = 5, then 14d 21 = 5 is justified by which property? A. ssocitive property B. commuttive property

Chapter 6 Techniques of Integration

MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

Chapter 8.2: The Integral

Chpter 8.: The Integrl You cn think of Clculus s doule-wide triler. In one width of it lives differentil clculus. In the other hlf lives wht is clled integrl clculus. We hve lredy eplored few rooms in

AQA Further Pure 1. Complex Numbers. Section 1: Introduction to Complex Numbers. The number system

Complex Numbers Section 1: Introduction to Complex Numbers Notes nd Exmples These notes contin subsections on The number system Adding nd subtrcting complex numbers Multiplying complex numbers Complex

Bridging the gap: GCSE AS Level

Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions

1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE

ELEMENTARY ALGEBRA nd GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE Directions: Study the exmples, work the prolems, then check your nswers t the end of ech topic. If you don t get the nswer given, check

The practical version

Roerto s Notes on Integrl Clculus Chpter 4: Definite integrls nd the FTC Section 7 The Fundmentl Theorem of Clculus: The prcticl version Wht you need to know lredy: The theoreticl version of the FTC. Wht

Qudrtic recirocity Frncisc Bozgn Los Angeles Mth Circle Octoer 8, 01 1 Qudrtic Recirocity nd Legendre Symol In the eginning of this lecture, we recll some sic knowledge out modulr rithmetic: Definition

AUTOMATA AND LANGUAGES. Definition 1.5: Finite Automaton

25. Finite Automt AUTOMATA AND LANGUAGES A system of computtion tht only hs finite numer of possile sttes cn e modeled using finite utomton A finite utomton is often illustrted s stte digrm d d d. d q

Regular expressions, Finite Automata, transition graphs are all the same!!

CSI 3104 /Winter 2011: Introduction to Forml Lnguges Chpter 7: Kleene s Theorem Chpter 7: Kleene s Theorem Regulr expressions, Finite Automt, trnsition grphs re ll the sme!! Dr. Neji Zgui CSI3104-W11 1

Section 3.2: Negative Exponents

Section 3.2: Negtive Exponents Objective: Simplify expressions with negtive exponents using the properties of exponents. There re few specil exponent properties tht del with exponents tht re not positive.

Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte

APPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line

APPENDIX D Preclculus Review APPENDIX D.1 Rel Numers n the Rel Numer Line Rel Numers n the Rel Numer Line Orer n Inequlities Asolute Vlue n Distnce Rel Numers n the Rel Numer Line Rel numers cn e represente

CS375: Logic and Theory of Computing

CS375: Logic nd Theory of Computing Fuhu (Frnk) Cheng Deprtment of Computer Science University of Kentucky 1 Tle of Contents: Week 1: Preliminries (set lger, reltions, functions) (red Chpters 1-4) Weeks

Here we study square linear systems and properties of their coefficient matrices as they relate to the solution set of the linear system.

Section 24 Nonsingulr Liner Systems Here we study squre liner systems nd properties of their coefficient mtrices s they relte to the solution set of the liner system Let A be n n Then we know from previous

First Midterm Examination

Çnky University Deprtment of Computer Engineering 203-204 Fll Semester First Midterm Exmintion ) Design DFA for ll strings over the lphet Σ = {,, c} in which there is no, no nd no cc. 2) Wht lnguge does

Lecture Solution of a System of Linear Equation

ChE Lecture Notes, Dept. of Chemicl Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updted /) Lecture 8- - Solution of System of Liner Eqution 8. Why is it importnt to e le to solve system of liner

A study of Pythagoras Theorem

CHAPTER 19 A study of Pythgors Theorem Reson is immortl, ll else mortl. Pythgors, Diogenes Lertius (Lives of Eminent Philosophers) Pythgors Theorem is proly the est-known mthemticl theorem. Even most nonmthemticins

STRUCTURE OF CONCURRENCY Ryszard Janicki. Department of Computing and Software McMaster University Hamilton, ON, L8S 4K1 Canada

STRUCTURE OF CONCURRENCY Ryszrd Jnicki Deprtment of Computing nd Softwre McMster University Hmilton, ON, L8S 4K1 Cnd jnicki@mcmster.c 1 Introduction Wht is concurrency? How it cn e modelled? Wht re the

Chapter 9 Definite Integrals

Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished

expression simply by forming an OR of the ANDs of all input variables for which the output is

2.4 Logic Minimiztion nd Krnugh Mps As we found ove, given truth tle, it is lwys possile to write down correct logic expression simply y forming n OR of the ANDs of ll input vriles for which the output

Finite Automata Theory and Formal Languages TMV027/DIT321 LP4 2018

Finite Automt Theory nd Forml Lnguges TMV027/DIT321 LP4 2018 Lecture 10 An Bove April 23rd 2018 Recp: Regulr Lnguges We cn convert between FA nd RE; Hence both FA nd RE ccept/generte regulr lnguges; More

Lecture 08: Feb. 08, 2019

4CS4-6:Theory of Computtion(Closure on Reg. Lngs., regex to NDFA, DFA to regex) Prof. K.R. Chowdhry Lecture 08: Fe. 08, 2019 : Professor of CS Disclimer: These notes hve not een sujected to the usul scrutiny

Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge

Parse trees, ambiguity, and Chomsky normal form

Prse trees, miguity, nd Chomsky norml form In this lecture we will discuss few importnt notions connected with contextfree grmmrs, including prse trees, miguity, nd specil form for context-free grmmrs

Chter 8 Qudrtic Residues 8. Qudrtic residues Let n>be given ositive integer, nd gcd, n. We sy tht Z n is qudrtic residue mod n if the congruence x mod n is solvble. Otherwise, is clled qudrtic nonresidue

CS 301. Lecture 04 Regular Expressions. Stephen Checkoway. January 29, 2018

CS 301 Lecture 04 Regulr Expressions Stephen Checkowy Jnury 29, 2018 1 / 35 Review from lst time NFA N = (Q, Σ, δ, q 0, F ) where δ Q Σ P (Q) mps stte nd n lphet symol (or ) to set of sttes We run n NFA

The Evaluation Theorem

These notes closely follow the presenttion of the mteril given in Jmes Stewrt s textook Clculus, Concepts nd Contexts (2nd edition) These notes re intended primrily for in-clss presenttion nd should not

Exercises with (Some) Solutions

Exercises with (Some) Solutions Techer: Luc Tesei Mster of Science in Computer Science - University of Cmerino Contents 1 Strong Bisimultion nd HML 2 2 Wek Bisimultion 31 3 Complete Lttices nd Fix Points