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


 Rose Hutchinson
 4 years ago
 Views:
Transcription
1 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 under mtrix multipliction, so it does not form group. To see tht it is not closed under mtrix multipliction, it is enough to consider the following exmple: [ ][ ] [ ] = [ ] [ ] 2 5 where the mtrix is not of the form. 3 7 d [ ] ii) Those of the form such tht c 2 c [ ] [ ] Answer: Agin, consider the two mtrices nd which re [ ] 7 6 of this form. Their product is, which is not in the correct form. So this 2 3 set is not closed under mtrix [ multipliction, ] nd does not form group. iii) Those of the form where c is not zero. 0 c Answer: These do form group. To show this, we need to check the following things. [ ] [ ] (1) The set is closed under multipliction: Suppose nd 0 c 0 c stisfy c, c 0. Then [ ][ ] [ ] 0 c 0 c = + c 0 c c where cc 0 ecuse c, c 0. Thus the product of two mtrices in this set is gin in the set, so it is closed under multipliction. [ ] 1 0 (2) The identity is in this set: This is true ecuse the mtrix is of 0 1 the correct form. [ ] (3) We hve inverses: Suppose the mtrix A = is in our set. Then 0 c consider the mtrix B = [ 1 c 1 0 c ]. Since c 0, the term c mkes sense, so B is welldefined. It is simple to check tht AB = BA = I, where I is the identity mtrix. Thus every mtrix in our set hs n inverse. (4) Associtivity: We hve this ecuse mtrix multipliction is ssocitive in generl. iv) The set of mtrices with nonzero determinnt nd integer entries. Answer: This set does not form [ group ecuse ] inverses my not e in the 2 0 set. For exmple, consider the mtrix. This mtrix is in our set. But its 0 2 inverse is the mtrix [ 1/ /2 ], which is not in our set. 1
2 2 Prolem 2.6 Show tht the collection of ll rottions of the plne out fixed point P forms group under composition of functions. Is the sme true of the set of ll reflections in lines which pss through P? Wht hppens if we tke ll the rottions nd ll the reflections? Answer: Fix point P in the plne. Let G ro e the set of ll rottions out P, let G re e the set of ll reflections out lines through P nd let G e the set of ll rottions nd reflections together. First, note tht ll of these sets stisfy the ssocitivity property. This is ecuse the composition of functions is lwys ssocitive. Suppose we hve three functions f,g nd h. We need to see tht for ech point x in their domin, f (g h)(x) = (f g) h(x). To show this tkes just it of mnipultion. Since (g h)(x) = g(h(x)), we hve f (g h)(x) = f(g(h(x))), nd likewise, (f g) h(x) = (f g)(h(x)), which is gin just f(g(h(x))). Therefore, composition of functions is lwys ssocitive, so composition of rottions nd/ or reflections is lso ssocitive. Now we show tht G ro is group. Let r θ G ro e the clockwise rottion y ngle θ out P. So if θ is negtive, we men counterclockwise rottion y θ. Then given two ngles θ nd φ, we clerly hve r θ r φ = r θ+φ. Thus the composition of two rottions is gin rottion, so G ro is closed under composition of functions. Now we hve to check the 3 group properties. (1) Associtivity: Composition of functions is ssocitive. (2) Identity: Clerly the identity is r 0, the rottion y ngle 0, since for ny ngle θ, r θ r 0 = r θ = r 0 r θ. (3) Inverses: Fix n ngle θ. Then the inverse of r θ is r θ since r θ r θ = r 0 = r θ r θ. Thus the set of rottions is group under function composition. Forthenextprtsoftheprolem,weusethefollowingfctoutthecomposition of two reflections. Property 1. Let R,R G re e reflections out lines nd through the point P. Suppose the ngle tht line sweeps out s it moves in clockwise direction to line is θ. Then R R = r θ/2, where R R mens we do R first nd then R. Proof. Choose some point Q in the plne, where Q P. Then s in the digrm, mesure the signed cute ngle etween the line segment QP nd line. We sy this ngle is positive if when you use QP to sweep out the ngle you go in clockwise direction. Otherwise, the ngle is negtive. Cll this signed ngle θ 1. In the digrm, the ngle is positive. Then the signed cute ngle etween the line segment R (Q),P nd is lso θ 1. Now let θ 2 e the signed cute ngle etween R (Q),P nd the line. Agin, the signed cute ngle etween R R (Q),P nd the line is lso θ 2. Note tht the distnce etween Q nd P is the sme s the distnce etween R R (Q) nd P. So if you rotte Q y n ngle of 2θ 1 +2θ 2 out P, we get to R R (Q). But θ 1 +θ 2 is just θ in the cse when t lest one of these two ngles is positive, nd θ 1 + θ 2 = θ π when oth of these ngles re negtive. So rottion y 2θ 1 +2θ 2 is either rottion y 2θ or rottion y 2(θ π). But the ltter is, in fct, the sme s rottion y θ since rotting y 2π is the sme s not rotting t ll. So we re done.
3 3 Q R (Q) θ 1 θ 1 P θ θ2θ 2 R (R (Q)) Figure 1. The point Q reflected first through nd then through By the ove property, we see tht the set of reflections is not closed under composition of functions, s two reflections give rottion. So G re is not group. On the other hnd, we clim tht G is group. Clim 1. The set of ll rottions nd reflections forms group. Proof.. It isn t hrd to see y the sme methods s ove tht if r θ is rottion nd R is reflection out line then r θ R = R r12 θ () tht is, reflecting out nd then rottion y ngle θ is the sme s reflecting out the line rotted y ngle 1 2θ. Likewise, R r θ = R r 1 2 θ () So the set G of rottions nd reflections is closed under composition of functions. The identity is in G, ecuse the identity is rottion y ngle 0. Inverses re in G ecuse ech reflection is its own inverse, nd the inverse of rottion y ngle θ is rottion y ngle θ. The set is ssocitive ecuse, gin, composition of functions is ssocitive. Therefore, G is group. Prolem 2.8 If x nd y re elements of group, prove tht (xy) 1 = y 1 x 1. Proof. The element (xy) 1 is the element s.t. (xy) 1 (xy) = 1 = (xy)(xy) 1. We show tht the element y 1 x 1 stisfies this property. (y 1 x 1 )(xy) = y 1 (x 1 (xy)) y ssocitivity = y 1 ((x 1 x)y) y ssocitivity = y 1 y y prop. of inverses = 1 y prop. of inverses so the first prt of tht eqution is stisfied. Now for the second prt. (xy)(y 1 x 1 ) = ((xy)y 1 )x 1 y ssocitivity So we hve shown tht (xy) 1 = y 1 x 1. = (x(yy 1 ))x 1 y ssocitivity = x 1 x y prop. of inverses = 1 y prop. of inverses
4 4 Prolem 3.2 Write Q( 2) for the set descried in Exercise 3.1 (iii). Given nonzero element + 2 express 1/(+ 2) in the form c+d 2 where c,d Q. Prove tht multipliction mkes Q( 2)\{0} into group. Proof. The set Q( 2) consists of ll rel numers of the form + 2 where nd re in Q. The numer 1/(+ 2) is the unique rel numer with the property tht (+ 2) 1/(+ 2) = 1. Let c = 2 2 nd let d = Then we 2 clim tht 1/(+ 2) = c+d 2. To see this, we clculte: Then nd (+ 2) c+d 2 = c+2d+(d+c) 2 2 c+2d = = 1 d+c = = 0. Thus, for the ove choices of c nd d, (+ 2) c+d 2 = 1, s promised. Now we show tht the set G = Q( 2)\{0} is group. The formul (+ 2) c+d 2 = c+2d+(d+c) 2 shows tht this set is closed under multipliction. Since G is suset of rel numers, it is ssocitive. The identity is in G since 1 = Finlly, the ove clcultion mens tht every nonzero element of Q( 2) hs multiplictive inverse lso in Q( 2). Therefore, Q( 2)\{0} is group. Prolem 3.3 Let n e positive integer nd let G consist of ll those complex numers z which stisfy z n = 1. Show tht G forms group under multipliction of complex numers. Proof. First we show tht G is closed under multipliction. Let z,z G. We need to show tht (zz ) n = 1. We hve tht (zz ) n = zz zz }{{} n times But multipliction in C is commuttive, so in fct, (zz ) n = z n z n. Since z,z G, this just mens tht (zz ) = 1. Thus G is closed under multipliction. Since C is group, nd G is suset of C, we know tht G is ssocitive. The identity is in G ecuse 1 n = 1. So the lst thing we need to show is tht G hs inverses. Let z G. Then z 1 is some complex numer (since ll complex numers hve inverses which re complex numers). We need to show tht (z 1 ) n = 1. But we know tht z n = 1. So multiply oth sides of this eqution y (z 1 ) n. This gives us tht (z 1 ) n z n = (z 1 ) n. But (z 1 ) n z n = 1 y the fct tht multipliction in C is commuttive, so this mens tht 1 = (z 1 ) n, s required. Thus G is group under multipliction. Prolem 3.5 Let n e positive integer. Prove tht (x ny) nz = x n(y nz).
5 5 Proof. By definition, (x n y) = xy kn where k Z is the numer s.t. xy kn [0,n 1]. Thus, (x n y) n z = (xy kn) n z = (xy kn)z k n = xyz k n where k is the numer s.t. (xy kn)z k n [0,n 1] nd so k = kz+k is the numer s.t. xyz k n [0,n 1]. With similr steps we get x n (y n z) = x n (yz ln) = x(yz ln) l n = xyz l n where gin l is the numer s.t. xyz l n [0,n 1]. We wnt to show tht xyz k n nd xyz l n re the sme numer. Note tht xyz k n (xyz l n) = (l k )n. So the difference etween these two numers is n integer multiple of n. On the other hnd, oth xyz k n nd xyz l n re etween 0 nd n 1. So (n 1) xyz k n (xyz l n) n 1. The only multiples of n inside these ounds is 0. Thus l k = 0 nd so xyz k n = xyz l n. Therefore, (x n y) n z = x n (y n z). Prolem 3.7 Which of the following sets form group under multipliction modulo 14? Answer: Note tht 5 5 = 11 mod 14. The sets {1,3,5} nd {1,3,5,7} contin 5 nd not 11, so they cnnot form groups. The set {1,7,13} cnnot form group mod 14 since 7 nd 14 shre common fctor, so 7 cnnot hve n inverse mod 14. In fct, suppose 7 14 x = 1. Then n = 2. But = 0, so 0 = 2 mod 14, which is impossile. Finlly, = 5, so the set {1,9,11,13} does not form group. So none of these sets form groups mod 14. Prolem 3.8 Show tht if suset of {1,2,...,21} contins n even numer, or contins the numer 11, then it cnnot form group under multipliction. Proof. Suppose n is n even integer etween 1 nd 21, inclusive. Then we cn write n = 2m for some other integer m. We clim tht n cnnot hve n inverse mod 22. Suppose not. Then n = 2m hs n inverse with the property tht 2m 22 n 1 = 1. Thus, m 22 n 1 = 11. By ssocitivity, I cn multiply 11 nd 2 first. Note tht = 0. Thus I get the sttement tht 0 22 n 1 = 11, i.e. 0 = 11. But this is not true mod 22. Therefore, n hs no inverse. Similrly, suppose n = 11. We clim tht 11 cnnot hve n inverse mod 22. Suppose not. Then n = 11 hs n inverse with the property tht n 1 = 1. Thus, n 1 = 2. By ssocitivity, I cn multiply 11 nd 2 first. Note tht = 0. Thus I get the sttement tht 0 22 n 1 = 2, i.e. 0 = 2. But this is not true mod 22. Therefore, 11 hs no inverse. So if we were given suset of {1,2,...,21} contining the numer 11 or ny even numer, then tht suset couldn t possily hve inverses for ll of its elements. Therefore it would not form group.
6 6 Prolem 4.7 Let G e the collection of ll rtionl numers x which stisfy 0 x < 1. Show tht the opertion { x+y if 0 x+y < 1 x+ G y = x+y 1 if 1 x+y < 2 mkes G into n infinite elin group ll of whose elements hve finite order. Proof. We will first show tht G is group. Agin, we do this in four steps. (1) G needs to e closed under the group opertion. Let x,y e in G. Then 0 x,y < 1 so 0 x + y < 2. Thus 0 x + G y < 1, so this is ok. Furthermore, oth x+y nd x+y 1 re rtionl numers. So x+ G y is rtionl numer etween 0 nd 1, s required. (2) Identity. The identity is 0, which is in the group. To see tht 0 is the identity, note tht if 0 x < 1, then 0 x+0 < 1, so x+ G 0 = x. (3) Inverses. If 0 x < 1 then 0 1 x < 1. Note tht x+(1 x) = 1, so x+ G (1 x) = 0. So for ny x G, 1 x is the inverse of x. Therefore the inverse of x is in G. (4) Associtivity. Let x,y nd z e in G. Ech time we dd two numers in G we sutrct 1 enough times to mke the result t lest 0, nd strictly less thn 1. Since 0 x+y +z < 3, we see tht { x+(y +G z) if 0 x+(y + x+ G (y + G z) = G z) < 1 x+(y + G z) 1 if 1 x+(y + G z) < 2 x+y + G z if 0 x+y +z < 1 = x+y +z 1 if 1 x+y +z < 2 x+y +z 2 if 2 x+y +z < 3 since y + G z is either y +z or y +z 1. Similrly, we get tht x+y + G z if 0 x+y +z < 1 (x+ G y)+ G z = x+y +z 1 if 1 x+y +z < 2 x+y +z 2 if 2 x+y +z < 3 Thus the group opertion is ssocitive. Therefore, G is group under this group opertion. There re infinitely mny rtionl numers etween 0 nd 1, so G is infinite. G is elin ecuse ddition of rtionl numers is elin. Finlly, we need to show tht elements of G hve finite order. Let e n element of G. Then define n G = + G + G }{{ } n times Since every time we dd on nother we sutrct either 0 or 1, we hve tht n G = n n m where m is n integer s.t. 0 m n nd 0 m < 1. Thus for n =, G = m where m is the integer s.t. 0 m < 1. But is lso n integer. So we must hve = m, nd so G = 0. But tht mens exctly tht hs order t most, since dding to itself times got us ck to the identity. So every element of G hs finite order.
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.
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),
More informationThings 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
More informationCoalgebra, 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
More informationVectors , (0,0). 5. A vector is commonly denoted by putting an arrow above its symbol, as in the picture above. Here are some 3dimensional vectors:
Vectors 1232018 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 2dimensionl vectors: (2, 3), ( )
More informationSUMMER 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
More informationMath 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
More informationMatrix 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
More informationQUADRATIC EQUATIONS OBJECTIVE PROBLEMS
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.
More informationpadic Egyptian Fractions
padic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Setup 3 4 pgreedy Algorithm 5 5 pegyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction
More informationBases for Vector Spaces
Bses for Vector Spces 22625 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
More informationHW3, 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
More informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
More information2.4 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
More informationLecture 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
More information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
More informationHow 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=
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationLecture 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
More informationFarey 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
More informationChapter 1: Fundamentals
Chpter 1: Fundmentls 1.1 Rel Numbers Types of Rel Numbers: Nturl Numbers: {1, 2, 3,...}; These re the counting numbers. Integers: {... 3, 2, 1, 0, 1, 2, 3,...}; These re ll the nturl numbers, their negtives,
More informationAnalytically, vectors will be represented by lowercase boldface 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
More informationSeptember 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
More informationLinear Inequalities. Work Sheet 1
Work Sheet 1 Liner Inequlities RentHep, 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
More information8. 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
More informationLecture 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
More informationParse 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 contextfree grmmrs
More informationIntroduction To Matrices MCV 4UI Assignment #1
Introduction To Mtrices MCV UI Assignment # INTRODUCTION: A mtrix plurl: mtrices) is rectngulr rry of numbers rrnged in rows nd columns Exmples: ) b) c) [ ] d) Ech number ppering in the rry is sid to be
More information2. VECTORS AND MATRICES IN 3 DIMENSIONS
2 VECTORS AND MATRICES IN 3 DIMENSIONS 21 Extending the Theory of 2dimensionl Vectors x A point in 3dimensionl spce cn e represented y column vector of the form y z zxis yxis z x y xxis Most of the
More informationGeometric Sequences. Geometric Sequence a sequence whose consecutive terms have a common ratio.
Geometric Sequences Geometric Sequence sequence whose consecutive terms hve common rtio. Geometric Sequence A sequence is geometric if the rtios of consecutive terms re the sme. 2 3 4... 2 3 The number
More informationIntroduction 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
More informationQuadratic Forms. Quadratic Forms
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
More information9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes
The Vector Product 9.4 Introduction In this section we descrie how to find the vector product of two vectors. Like the sclr product its definition my seem strnge when first met ut the definition is chosen
More informationInfinite 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
More informationLecture 2: Fields, Formally
Mth 08 Lecture 2: Fields, Formlly Professor: Pdric Brtlett Week UCSB 203 In our first lecture, we studied R, the rel numbers. In prticulr, we exmined how the rel numbers intercted with the opertions of
More informationSCHOOL OF ENGINEERING & BUILT ENVIRONMENT. Mathematics
SCHOOL OF ENGINEERING & BUIL ENVIRONMEN Mthemtics An Introduction to Mtrices Definition of Mtri Size of Mtri Rows nd Columns of Mtri Mtri Addition Sclr Multipliction of Mtri Mtri Multipliction 7 rnspose
More informationEECS 141 Due 04/19/02, 5pm, in 558 Cory
UIVERSITY OF CALIFORIA College of Engineering Deprtment of Electricl Engineering nd Computer Sciences Lst modified on April 8, 2002 y Tufn Krlr (tufn@eecs.erkeley.edu) Jn M. Rey, Andrei Vldemirescu Homework
More informationNatural examples of rings are the ring of integers, a ring of polynomials in one variable, the ring
More generlly, we define ring to be nonempty set R hving two binry opertions (we ll think of these s ddition nd multipliction) which is n Abelin group under + (we ll denote the dditive identity by 0),
More information4 VECTORS. 4.0 Introduction. Objectives. Activity 1
4 VECTRS Chpter 4 Vectors jectives fter studying this chpter you should understnd the difference etween vectors nd sclrs; e le to find the mgnitude nd direction of vector; e le to dd vectors, nd multiply
More informationHomework 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}.
More informationFirst Midterm Examination
2425 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
More informationDEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS
3 DEFINITION OF ASSOCIATIVE OR DIRECT PRODUCT AND ROTATION OF VECTORS This chpter summrizes few properties of Cli ord Algebr nd describe its usefulness in e ecting vector rottions. 3.1 De nition of Associtive
More informationad = 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.
More informationHow 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 bout solving systems of liner equtions. These re problems tht give couple of equtions with couple of unknowns, like: 6 2 3 7 4
More informationFirst Midterm Examination
Çnky University Deprtment of Computer Engineering 203204 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
More informationNumerical Linear Algebra Assignment 008
Numericl Liner Algebr Assignment 008 Nguyen Qun B Hong Students t Fculty of Mth nd Computer Science, Ho Chi Minh University of Science, Vietnm emil. nguyenqunbhong@gmil.com blog. http://hongnguyenqunb.wordpress.com
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationTorsion in Groups of Integral Triangles
Advnces in Pure Mthemtics, 01,, 11610 http://dxdoiorg/1046/pm011015 Pulished Online Jnury 01 (http://wwwscirporg/journl/pm) Torsion in Groups of Integrl Tringles Will Murry Deprtment of Mthemtics nd Sttistics,
More information5. (±±) Λ = fw j w is string of even lengthg [ 00 = f11,00g 7. (11 [ 00)± Λ = fw j w egins with either 11 or 00g 8. (0 [ ffl)1 Λ = 01 Λ [ 1 Λ 9.
Regulr Expressions, Pumping Lemm, Right Liner Grmmrs Ling 106 Mrch 25, 2002 1 Regulr Expressions A regulr expression descries or genertes lnguge: it is kind of shorthnd for listing the memers of lnguge.
More informationHomework 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)
More informationM A T H F A L L CORRECTION. Algebra I 2 1 / 0 9 / U N I V E R S I T Y O F T O R O N T O
M A T H 2 4 0 F A L L 2 0 1 4 HOMEWORK ASSIGNMENT #1 CORRECTION Alger I 2 1 / 0 9 / 2 0 1 4 U N I V E R S I T Y O F T O R O N T O 1. Suppose nd re nonzero elements of field F. Using only the field xioms,
More informationChapter 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
More informationAPPENDIX. 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
More informationUSA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year
1/1/21. Fill in the circles in the picture t right with the digits 18, one digit in ech circle with no digit repeted, so tht no two circles tht re connected by line segment contin consecutive digits.
More information(e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer
Divisibility In this note we introduce the notion of divisibility for two integers nd b then we discuss the division lgorithm. First we give forml definition nd note some properties of the division opertion.
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationMathCity.org Merging man and maths
MthCity.org Merging mn nd mths Exercise.8 (s) Pge 46 Textbook of Algebr nd Trigonometry for Clss XI Avilble online @ http://, Version: 3.0 Question # Opertion performed on the twomember set G = {0, is
More informationRudimentary Matrix Algebra
Rudimentry Mtrix Alger Mrk Sullivn Decemer 4, 217 i Contents 1 Preliminries 1 1.1 Why does this document exist?.................... 1 1.2 Why does nyone cre out mtrices?................ 1 1.3 Wht is mtrix?...........................
More informationBridging 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
More informationMATRIX DEFINITION A matrix is any doubly subscripted array of elements arranged in rows and columns.
4.5 THEORETICL SOIL MECHNICS Vector nd Mtrix lger Review MTRIX DEFINITION mtrix is ny douly suscripted rry of elements rrnged in rows nd columns. m  Column Revised /0 n Row m,,,,,, n n mn ij nd Order
More informationMTH 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
More informationReview: set theoretic definition of the numbers. Natural numbers:
Review: reltions A inry reltion on set A is suset R Ñ A ˆ A, where elements p, q re written s. Exmple: A Z nd R t pmod nqu. A inry reltion on set A is... (R) reflexive if for ll P A; (S) symmetric if implies
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationMATH 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.
More informationName Solutions to Test 3 November 8, 2017
Nme Solutions to Test 3 November 8, 07 This test consists of three prts. Plese note tht in prts II nd III, you cn skip one question of those offered. Some possibly useful formuls cn be found below. Brrier
More informationMatrices. Elementary Matrix Theory. Definition of a Matrix. Matrix Elements:
Mtrices Elementry Mtrix Theory It is often desirble to use mtrix nottion to simplify complex mthemticl expressions. The simplifying mtrix nottion usully mkes the equtions much esier to hndle nd mnipulte.
More information1. 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
More informationAQA 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
More information1 Nondeterministic Finite Automata
1 Nondeterministic Finite Automt Suppose in life, whenever you hd choice, you could try oth possiilities nd live your life. At the end, you would go ck nd choose the one tht worked out the est. Then you
More informationLinear Algebra 1A  solutions of ex.4
Liner Algebr A  solutions of ex.4 For ech of the following, nd the inverse mtrix (mtritz hofkhit if it exists  ( 6 6 A, B (, C 3, D, 4 4 ( E i, F (inverse over C for F. i Also, pick n invertible mtrix
More informationMultiplying 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 doc69 Activity B More integer multipliction doc698 Activity B Advnced integer multipliction doc699 Multiplying integers FLUENCY
More informationChapter 14. Matrix Representations of Linear Transformations
Chpter 4 Mtrix Representtions of Liner Trnsformtions When considering the Het Stte Evolution, we found tht we could describe this process using multipliction by mtrix. This ws nice becuse computers cn
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
2 The Prllel Circuit Electric Circuits: Figure 2 elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationDate Lesson Text TOPIC Homework. Solving for Obtuse Angles QUIZ ( ) More Trig Word Problems QUIZ ( )
UNIT 5 TRIGONOMETRI RTIOS Dte Lesson Text TOPI Homework pr. 4 5.1 (48) Trigonometry Review WS 5.1 # 3 5, 9 11, (1, 13)doso pr. 6 5. (49) Relted ngles omplete lesson shell & WS 5. pr. 30 5.3 (50) 5.3 5.4
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationCM10196 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
More informationPreSession Review. Part 1: Basic Algebra; Linear Functions and Graphs
PreSession Review Prt 1: Bsic Algebr; Liner Functions nd Grphs A. Generl Review nd Introduction to Algebr Hierrchy of Arithmetic Opertions Opertions in ny expression re performed in the following order:
More informationA 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 estknown mthemticl theorem. Even most nonmthemticins
More informationa a a a a a a a a a a a a a a a a a a a a a a a In this section, we introduce a general formula for computing determinants.
Section 9 The Lplce Expnsion In the lst section, we defined the determinnt of (3 3) mtrix A 12 to be 22 12 21 22 2231 22 12 21. In this section, we introduce generl formul for computing determinnts. Rewriting
More informationHarvard University Computer Science 121 Midterm October 23, 2012
Hrvrd University Computer Science 121 Midterm Octoer 23, 2012 This is closedook exmintion. You my use ny result from lecture, Sipser, prolem sets, or section, s long s you quote it clerly. The lphet is
More informationChapter 3 MATRIX. In this chapter: 3.1 MATRIX NOTATION AND TERMINOLOGY
Chpter 3 MTRIX In this chpter: Definition nd terms Specil Mtrices Mtrix Opertion: Trnspose, Equlity, Sum, Difference, Sclr Multipliction, Mtrix Multipliction, Determinnt, Inverse ppliction of Mtrix in
More informationQuadratic Residues. Chapter Quadratic residues
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
More informationReview of Gaussian Quadrature method
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
More informationOn 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
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationLecture 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
More information1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
1 PYTHAGORAS THEOREM 1 1 Pythgors Theorem In this setion we will present geometri proof of the fmous theorem of Pythgors. Given right ngled tringle, the squre of the hypotenuse is equl to the sum of the
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationRegular 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
More informationLecture Note 9: Orthogonal Reduction
MATH : Computtionl Methods of Liner Algebr 1 The Row Echelon Form Lecture Note 9: Orthogonl Reduction Our trget is to solve the norml eution: Xinyi Zeng Deprtment of Mthemticl Sciences, UTEP A t Ax = A
More informationMath Solutions to homework 1
Mth 75  Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationCIRCULAR COLOURING THE PLANE
CIRCULAR COLOURING THE PLANE MATT DEVOS, JAVAD EBRAHIMI, MOHAMMAD GHEBLEH, LUIS GODDYN, BOJAN MOHAR, AND REZA NASERASR Astrct. The unit distnce grph R is the grph with vertex set R 2 in which two vertices
More informationSection 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
More informationMATRICES AND VECTORS SPACE
MATRICES AND VECTORS SPACE MATRICES AND MATRIX OPERATIONS SYSTEM OF LINEAR EQUATIONS DETERMINANTS VECTORS IN SPACE AND SPACE GENERAL VECTOR SPACES INNER PRODUCT SPACES EIGENVALUES, EIGENVECTORS LINEAR
More informationAbsolute values of real numbers. Rational Numbers vs Real Numbers. 1. Definition. Absolute value α of a real
Rtionl Numbers vs Rel Numbers 1. Wht is? Answer. is rel number such tht ( ) =. R [ ( ) = ].. Prove tht (i) 1; (ii). Proof. (i) For ny rel numbers x, y, we hve x = y. This is necessry condition, but not
More informationQuadratic reciprocity
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
More informationHQPD  ALGEBRA I TEST Record your answers on the answer sheet.
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
More informationWhat else can you do?
Wht else cn you do? ngle sums The size of specil ngle types lernt erlier cn e used to find unknown ngles. tht form stright line dd to 180c. lculte the size of + M, if L is stright line M + L = 180c( stright
More informationHere 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
More informationMath 33A Discussion Example Austin Christian October 23, Example 1. Consider tiling the plane by equilateral triangles, as below.
Mth 33A Discussion Exmple Austin Christin October 3 6 Exmple Consider tiling the plne by equilterl tringles s below Let v nd w be the ornge nd green vectors in this figure respectively nd let {v w} be
More information