Math 4107: Abstract Algebra I Fall Webwork Assignment1-Groups (5 parts/problems) Solutions are on Webwork.
|
|
- Sibyl Griffin
- 5 years ago
- Views:
Transcription
1 Math 4107: Abstract Algebra I Fall 2017 Assigmet 1 Solutios 1. Webwork Assigmet1-Groups 5 parts/problems) Solutios are o Webwork. 2. Webwork Assigmet1-Subgroups 5 parts/problems) Solutios are o Webwork. 3. Problem 2.1 # 22. Note that you must verify that is, you ca ot assume) the two relatios i Example 10 give i the setece Oe ca verify.... Proof. Let > 2 be a iteger. Cosider S {x, y) x, y R} ad f, h AS) where fx, y) x, y) ad hx, y) is rotatio of the plae about the origi through a agle of / i the couterclockwise directio. I matrix form, we have that fx, y) 1 0 ) ) x cos ad hx, y) y si si cos ) ) x y Let i deote the idetity mappig i AS). We see that f 2 x, y) ffx, y)) f x, y) x, y) so f 2 i. Likewise, h is rotatio of the plae about the origi through a agle of /) so h i. I the matrix form, we ca also cofirm that f ) ) ad that h i. To see the secod relatio, we give a quick proof by iductio. Let P k) be the statemet that h k cos k si k si k ) cos k for all k N. We kow that h 1 is true. For our iductio hypothesis, we assume that P i) is true for some i 1. We cosider h i+1 h i h. Sice h i cos i h cos si i si cos i si si i cos ) si i cos cos i si si i si cos i cos ad siu ± v) si u cos v ± cosu si v, cosu ± v) cos u cos v si u si v, we have that h i+1 cosi + 1) sii + 1) ) sii + 1) cosi + 1). Sice our assumptio that P i) is true for some i 1 implies that P i + 1) is true, by the priciple of iductio we kow that P k) is true for all k N. I particular, ) h cos si si cos ) ) 1 0. Now cosider cos fh)fh) si si cos ) 2 ) 1 0
2 sice cos 2 + si2 1. Because fh)fh) i, we have that fh) fh) 1 h 1 f 1 h 1 f by the defiitio of iverses ad our previous idetities. Let G {f k h j k 0, 1; j 0, 1,..., 1}. Cosider f i h j, f s h t G with 0 i, s 1 ad 0 j, t 1. If s 0, the f i h j )f s h t ) f i h t+j f i h t+j) mod. If s 0, the s 1 ad f i h j )f s h t ) f i+s h t j f i+s) mod 2 h t j) mod. Note that this is true eve whe j 0. I either case, we have that f i h j )f s h t ) f a h b for some a {0, 1}, b {0,..., 1}. Hece, G is closed uder the product of compositio of fuctios. We kow that compositio of fuctios is associative i AS). Hece, sice G AS), the product i G is associative. We have the idetity mappig i f 2 h i G. Let f s h t G. If s 0, the h t h t mod h t mod h t h 0 i. If s 1, the fh t )fh t ) f 2 h t t) mod i. Hece, every elemet of G has a iverse i G. Sice G satisfies the four coditios of closure, associativity, existece of a idetity, ad existece of iverses, the G is a group. We see that G is oabelia sice si si ad cos fh si si cos ) cos si si ) hf. cos By defiitio, G has order at most 2. Sice h is a rotatio by a agle of, we kow that h j i for 1 i 1. Hece the elemets h i for 0 i i are all distict. Likewise, the elemets fh i for 0 i i are all distict. If ot, the there would be some fh i fh t for 0 i < t 1. However, cacelig by f o the left ad h o the right we have that h t i i for 1 i t < 1, which cotradicts our previous assertio. Thus, there are 2 distict elemets i G. Hece, G is a oabelia group of order Problem 2.1 # 24. Proof. Let G be the dihedral group of order 2 where > 2 as above. a) Let be odd. Suppose there is a G such that ab ba for all b G. We claim that a e where e is the idetity i G. Let a f i h j for 0 i 1 ad 0 j 1. Suppose that i 1 ad cosider b h 2 G sice 3. We have that ab fh j )h 2 ) fh 2+j) mod ad ba h 2 )fh j ) fh j 2) mod. Sice ab ba, we have fh 2+j) mod fh j 2) mod. By cacellatios, we arrive at the equatio 2 + j) j 2) mod. But this is true if ad oly if 4 0 mod. Sice is odd, though, it s ot possible that 4. Hece i 0. We have that a h j for some 0 j 1. Now let b f. By assumptio ab h j f fh j ba. Sice h j f fh j mod we have that j j mod. But the 2j 0 mod. Sice is odd, it must be that j. But sice j <, this ca be true oly if j 0. Hece a h 0 e ad our claim is proved.
3 b) Let be eve ad let a G where a h /2. We claim that ab ba for all b G. Let b f i h j G for 0 i 1, 0 j 1. Suppose i 0. The ab h j+/2 h /2+j ba. Suppose i 1. The ab)a h /2 fh j ))h /2 fh j /2) mod )h /2 fh j b. Sice aba b ad a 2 h e, we have that ba aba)a ab)e ab so our claim is true. c) Let be eve. Let a G such that ab ba for all b G. Let a f i h j for 0 i 1, 0 j 1. ad let b f s h t for 0 s 1, 0 t 1. Suppose i 0. The ab h j f s h t ) ad ba f s h t )h j. If s 0, the ab h j+t ba for j. Suppose s 1. The ab fh t j) mod ad ab ba implies that fh t j) mod fh t+j) mod. This is true if ad oly if t+j t j) mod which holds exactly whe 2j 0 mod, that is whe j 0, /2. Suppose that i 1. The ab fh j )f s h t ) ad ba f s h t )fh j ). If s 0, the ab fh j+t) mod ad ba fh j t) mod. Assumig that ab ba implies that j+t) j t) mod ad that 2t 0 mod. Let t 1. Sice > 2, we have that 2 ad 2t 0 mod. Hece for b h, we have that ab fh j+1 fh j 1 ba. Thus, there is o a f i h j G with i Problem 2.1 # 28. Proof. Let G be a set with a operatio such that: a) G is closed uder. b) is associative. c) e G such that e x x x G. d) x G y G such that y x e. We claim that G is a group. We kow that G satisfies the first two requiremets of a group, closure ad associativity. We must show that the left idetity of G is also a right idetity, ad that left iverses are also right iverses. For all x, z G, we write x z as xz. Let e G such that ex x for all x G. Let x G ad y G such that yx e. Cosider xy G. The xy)xy) xyx)y) xey)) xy by repeated applicatio of associativity ad the fact that e is a left idetity i G ad y is a left iverse of x i G. There exists z G such that zxy) e. Hece zxy)xy)) zxy))xy) exy) xy. We also have that zxy)xy)) zxy)) e. Thus xy e ad y is also a right iverse of x i G. Now cosider xe. The xe xyx) xy)x ex x sice we have show that xy e. Thus, e is also a right idetity for ay a G. Sice G has a elemet which is both a left ad right idetity ad for every x G there exists some y G which is both a left ad a right iverse, we kow that G is a group. 6. Problem 2.2 # 2.
4 Proof. We are give G as a set, closed uder a associative operatio, such that, for all a, x, y, u, w G, ax ay implies x y ad ua wa implies u w. Assume the operatio is called. By defiitio, G is closed uder, ad is associative. Hece, we eed oly prove two thigs: the existece of a idetity elemet, ad the existece of iverse elemets. Suppose G. To fid the idetity elemet, fix a G ad defie the map f : G G defied by fx) ax for all x G. For all x, y G, ax ay implies x y. Therefore, f is a ijective map. Because f is ijective, the elemets i the domai G map to distict elemets i the rage G. Hece, by the pigeohole priciple, f is also surjective. This implies that, for all z G, there exists x G such that z ax. Similarly, if we defie g : G G such that gu) ua for all u G, for all u, w G, ua wa implies u w. Thus, g is ijective, ad by the same pigeohole argumet above, g must also be bijective. Hece, for all z G, there exists u G such that gu) ua z. Therefore, there exists elemet b G such that fb) ab a ad elemet c G such that gc) ca a. Applyig the elemets b ad c, for all z G, zb uab ua z ad cz cax ax z. Hece, c is a left-idetity elemet, ad b is a right-idetity elemet. Sice the above equatios apply to all z G, by applyig c z i the first equatio, ad b z i the secod, we have cb c ad cb b. Hece b c. Reamig this elemet as e, it satisfies ze ez z for all z G. Therefore, G has a idetity elemet uder. For each elemet z G, we ca use the defiitios of f ad g above as templates to defie surjective maps f z x) xz ad g z u) zu. Let z be fixed. Because f z ad g z are bijective, there exist elemets s, t G such that e f z s) zs ad e g z t) tz. The s es tz)s tzs) te t. This implies s t, ad thus zs sz e. Thus, s is the iverse of z, z 1. Sice this argumet holds for all z, every elemet of G has a iverse. We are give that G is closed uder the operatio, with associative. As we have show, G has a idetity elemet ad each elemet has a iverse. Therefore, G is a group. Additioally, give a example to that this result is false if G is ifiite. Proof. Let G be the group of atural umbers, G N {x Z x > 0}, ad let the operatio be stadard additio, +. The set of atural umbers is a ifiite set. We already kow that additio is associative o Z: for all x, y, z Z, x + y) + z x + y + z). Therefore, + is associative o G Z. If x, y G, the x, y Z, ad sice Z is closed, x + y Z. Further, sice x > 0 ad y > 0, x + y > 0. Hece, x + y G. Therefore, G is also closed uder +. Now let a, x, y, u, w G, such that a + x a + y ad u + a w + a. By applyig the properties of additio, we kow x y ad u w. Therefore, G is a ifiite set with the operatio + satisfyig all the other properties listed i Problem 2.2 # 2.
5 However, G does ot cotai a idetity elemet. The equatio a + e a is valid if ad oly if e 0, but 0 / G. Similarly, G does ot cotai ay of the iverses of its elemets. For all a G, a + b e is oly valid if b a, but a < 0, which meas b a / G. Therefore, G N with the operatio + is ot a group.
Different kinds of Mathematical Induction
Differet ids of Mathematical Iductio () Mathematical Iductio Give A N, [ A (a A a A)] A N () (First) Priciple of Mathematical Iductio Let P() be a propositio (ope setece), if we put A { : N p() is true}
More informationBasic Sets. Functions. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4. (e) 2 S.
Basic Sets Example 1. Let S = {1, {2, 3}, 4}. Idicate whether each statemet is true or false. (a) S = 4 (b) {1} S (c) {2, 3} S (d) {1, 4} S (e) 2 S. (f) S = {1, 4, {2, 3}} (g) S Example 2. Compute the
More informationSolutions to Math 347 Practice Problems for the final
Solutios to Math 347 Practice Problems for the fial 1) True or False: a) There exist itegers x,y such that 50x + 76y = 6. True: the gcd of 50 ad 76 is, ad 6 is a multiple of. b) The ifiimum of a set is
More informationChapter 0. Review of set theory. 0.1 Sets
Chapter 0 Review of set theory Set theory plays a cetral role i the theory of probability. Thus, we will ope this course with a quick review of those otios of set theory which will be used repeatedly.
More informationMath Solutions to homework 6
Math 175 - Solutios to homework 6 Cédric De Groote November 16, 2017 Problem 1 (8.11 i the book): Let K be a compact Hermitia operator o a Hilbert space H ad let the kerel of K be {0}. Show that there
More informationHomework 1 Solutions. The exercises are from Foundations of Mathematical Analysis by Richard Johnsonbaugh and W.E. Pfaffenberger.
Homewor 1 Solutios Math 171, Sprig 2010 Hery Adams The exercises are from Foudatios of Mathematical Aalysis by Richard Johsobaugh ad W.E. Pfaffeberger. 2.2. Let h : X Y, g : Y Z, ad f : Z W. Prove that
More informationSOLVED EXAMPLES
Prelimiaries Chapter PELIMINAIES Cocept of Divisibility: A o-zero iteger t is said to be a divisor of a iteger s if there is a iteger u such that s tu I this case we write t s (i) 6 as ca be writte as
More informationSeunghee Ye Ma 8: Week 5 Oct 28
Week 5 Summary I Sectio, we go over the Mea Value Theorem ad its applicatios. I Sectio 2, we will recap what we have covered so far this term. Topics Page Mea Value Theorem. Applicatios of the Mea Value
More informationModern Algebra. Previous year Questions from 2017 to Ramanasri
Moder Algebra Previous year Questios from 017 to 199 Ramaasri 017 S H O P NO- 4, 1 S T F L O O R, N E A R R A P I D F L O U R M I L L S, O L D R A J E N D E R N A G A R, N E W D E L H I. W E B S I T E
More informationM 340L CS Homew ork Set 6 Solutions
1. Suppose P is ivertible ad M 34L CS Homew ork Set 6 Solutios A PBP 1. Solve for B i terms of P ad A. Sice A PBP 1, w e have 1 1 1 B P PBP P P AP ( ).. Suppose ( B C) D, w here B ad C are m matrices ad
More informationM 340L CS Homew ork Set 6 Solutions
. Suppose P is ivertible ad M 4L CS Homew ork Set 6 Solutios A PBP. Solve for B i terms of P ad A. Sice A PBP, w e have B P PBP P P AP ( ).. Suppose ( B C) D, w here B ad C are m matrices ad D is ivertible.
More informationPROBLEMS ON ABSTRACT ALGEBRA
PROBLEMS ON ABSTRACT ALGEBRA 1 (Putam 197 A). Let S be a set ad let be a biary operatio o S satisfyig the laws x (x y) = y for all x, y i S, (y x) x = y for all x, y i S. Show that is commutative but ot
More informationChapter 2. Periodic points of toral. automorphisms. 2.1 General introduction
Chapter 2 Periodic poits of toral automorphisms 2.1 Geeral itroductio The automorphisms of the two-dimesioal torus are rich mathematical objects possessig iterestig geometric, algebraic, topological ad
More informationThe Structure of Z p when p is Prime
LECTURE 13 The Structure of Z p whe p is Prime Theorem 131 If p > 1 is a iteger, the the followig properties are equivalet (1) p is prime (2) For ay [0] p i Z p, the equatio X = [1] p has a solutio i Z
More informationIt is often useful to approximate complicated functions using simpler ones. We consider the task of approximating a function by a polynomial.
Taylor Polyomials ad Taylor Series It is ofte useful to approximate complicated fuctios usig simpler oes We cosider the task of approximatig a fuctio by a polyomial If f is at least -times differetiable
More informationMAS111 Convergence and Continuity
MAS Covergece ad Cotiuity Key Objectives At the ed of the course, studets should kow the followig topics ad be able to apply the basic priciples ad theorems therei to solvig various problems cocerig covergece
More informationMath 140A Elementary Analysis Homework Questions 1
Math 14A Elemetary Aalysis Homewor Questios 1 1 Itroductio 1.1 The Set N of Natural Numbers 1 Prove that 1 2 2 2 2 1 ( 1(2 1 for all atural umbers. 2 Prove that 3 11 (8 5 4 2 for all N. 4 (a Guess a formula
More informationM A T H F A L L CORRECTION. Algebra I 1 4 / 1 0 / 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 #4 CORRECTION Algebra I 1 4 / 1 0 / 2 0 1 4 U N I V E R S I T Y O F T O R O N T O P r o f e s s o r : D r o r B a r - N a t a Correctio Homework Assigmet
More information6 Integers Modulo n. integer k can be written as k = qn + r, with q,r, 0 r b. So any integer.
6 Itegers Modulo I Example 2.3(e), we have defied the cogruece of two itegers a,b with respect to a modulus. Let us recall that a b (mod ) meas a b. We have proved that cogruece is a equivalece relatio
More informationPRELIM PROBLEM SOLUTIONS
PRELIM PROBLEM SOLUTIONS THE GRAD STUDENTS + KEN Cotets. Complex Aalysis Practice Problems 2. 2. Real Aalysis Practice Problems 2. 4 3. Algebra Practice Problems 2. 8. Complex Aalysis Practice Problems
More informationDIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS
DIVISIBILITY PROPERTIES OF GENERALIZED FIBONACCI POLYNOMIALS VERNER E. HOGGATT, JR. Sa Jose State Uiversity, Sa Jose, Califoria 95192 ad CALVIN T. LONG Washigto State Uiversity, Pullma, Washigto 99163
More informationMath F215: Induction April 7, 2013
Math F25: Iductio April 7, 203 Iductio is used to prove that a collectio of statemets P(k) depedig o k N are all true. A statemet is simply a mathematical phrase that must be either true or false. Here
More informationIn number theory we will generally be working with integers, though occasionally fractions and irrationals will come into play.
Number Theory Math 5840 otes. Sectio 1: Axioms. I umber theory we will geerally be workig with itegers, though occasioally fractios ad irratioals will come ito play. Notatio: Z deotes the set of all itegers
More informationSolutions for May. 3 x + 7 = 4 x x +
Solutios for May 493. Prove that there is a atural umber with the followig characteristics: a) it is a multiple of 007; b) the first four digits i its decimal represetatio are 009; c) the last four digits
More information1. By using truth tables prove that, for all statements P and Q, the statement
Author: Satiago Salazar Problems I: Mathematical Statemets ad Proofs. By usig truth tables prove that, for all statemets P ad Q, the statemet P Q ad its cotrapositive ot Q (ot P) are equivalet. I example.2.3
More informationMATH 324 Summer 2006 Elementary Number Theory Solutions to Assignment 2 Due: Thursday July 27, 2006
MATH 34 Summer 006 Elemetary Number Theory Solutios to Assigmet Due: Thursday July 7, 006 Departmet of Mathematical ad Statistical Scieces Uiversity of Alberta Questio [p 74 #6] Show that o iteger of the
More information1+x 1 + α+x. x = 2(α x2 ) 1+x
Math 2030 Homework 6 Solutios # [Problem 5] For coveiece we let α lim sup a ad β lim sup b. Without loss of geerality let us assume that α β. If α the by assumptio β < so i this case α + β. By Theorem
More informationHomework 2 January 19, 2006 Math 522. Direction: This homework is due on January 26, In order to receive full credit, answer
Homework 2 Jauary 9, 26 Math 522 Directio: This homework is due o Jauary 26, 26. I order to receive full credit, aswer each problem completely ad must show all work.. What is the set of the uits (that
More informationHoggatt and King [lo] defined a complete sequence of natural numbers
REPRESENTATIONS OF N AS A SUM OF DISTINCT ELEMENTS FROM SPECIAL SEQUENCES DAVID A. KLARNER, Uiversity of Alberta, Edmoto, Caada 1. INTRODUCTION Let a, I deote a sequece of atural umbers which satisfies
More informationExam 2 CMSC 203 Fall 2009 Name SOLUTION KEY Show All Work! 1. (16 points) Circle T if the corresponding statement is True or F if it is False.
1 (1 poits) Circle T if the correspodig statemet is True or F if it is False T F For ay positive iteger,, GCD(, 1) = 1 T F Every positive iteger is either prime or composite T F If a b mod p, the (a/p)
More informationBasic Sets. MTH299 - Examples. Example 1. Let S = {1, {2, 3}, 4}. Indicate whether each statement is true or false. (a) S = 4
Basic Sets Example 1. Let S = {1, {, 3}, 4}. Idicate whether each statemet is true or false. (a) S = 4 False. Note that the elemets of S are 1, the set {, 3}, ad 4. Thus S = 3. (b) {1} S False. While 1
More informationand each factor on the right is clearly greater than 1. which is a contradiction, so n must be prime.
MATH 324 Summer 200 Elemetary Number Theory Solutios to Assigmet 2 Due: Wedesday July 2, 200 Questio [p 74 #6] Show that o iteger of the form 3 + is a prime, other tha 2 = 3 + Solutio: If 3 + is a prime,
More informationLecture 7: Properties of Random Samples
Lecture 7: Properties of Radom Samples 1 Cotiued From Last Class Theorem 1.1. Let X 1, X,...X be a radom sample from a populatio with mea µ ad variace σ
More informationMath 299 Supplement: Real Analysis Nov 2013
Math 299 Supplemet: Real Aalysis Nov 203 Algebra Axioms. I Real Aalysis, we work withi the axiomatic system of real umbers: the set R alog with the additio ad multiplicatio operatios +,, ad the iequality
More informationANSWERS TO MIDTERM EXAM # 2
MATH 03, FALL 003 ANSWERS TO MIDTERM EXAM # PENN STATE UNIVERSITY Problem 1 (18 pts). State ad prove the Itermediate Value Theorem. Solutio See class otes or Theorem 5.6.1 from our textbook. Problem (18
More informationFind a formula for the exponential function whose graph is given , 1 2,16 1, 6
Math 4 Activity (Due by EOC Apr. ) Graph the followig epoetial fuctios by modifyig the graph of f. Fid the rage of each fuctio.. g. g. g 4. g. g 6. g Fid a formula for the epoetial fuctio whose graph is
More informationLecture 10: Mathematical Preliminaries
Lecture : Mathematical Prelimiaries Obective: Reviewig mathematical cocepts ad tools that are frequetly used i the aalysis of algorithms. Lecture # Slide # I this
More informationi is the prime factorization of n as a product of powers of distinct primes, then: i=1 pm i
Lecture 3. Group Actios PCMI Summer 2015 Udergraduate Lectures o Flag Varieties Lecture 3. The category of groups is discussed, ad the importat otio of a group actio is explored. Defiitio 3.1. A group
More informationA brief introduction to linear algebra
CHAPTER 6 A brief itroductio to liear algebra 1. Vector spaces ad liear maps I what follows, fix K 2{Q, R, C}. More geerally, K ca be ay field. 1.1. Vector spaces. Motivated by our ituitio of addig ad
More informationCardinality Homework Solutions
Cardiality Homework Solutios April 16, 014 Problem 1. I the followig problems, fid a bijectio from A to B (you eed ot prove that the fuctio you list is a bijectio): (a) A = ( 3, 3), B = (7, 1). (b) A =
More informationLecture Notes for Analysis Class
Lecture Notes for Aalysis Class Topological Spaces A topology for a set X is a collectio T of subsets of X such that: (a) X ad the empty set are i T (b) Uios of elemets of T are i T (c) Fiite itersectios
More informationTEACHER CERTIFICATION STUDY GUIDE
COMPETENCY 1. ALGEBRA SKILL 1.1 1.1a. ALGEBRAIC STRUCTURES Kow why the real ad complex umbers are each a field, ad that particular rigs are ot fields (e.g., itegers, polyomial rigs, matrix rigs) Algebra
More informationPAPER : IIT-JAM 2010
MATHEMATICS-MA (CODE A) Q.-Q.5: Oly oe optio is correct for each questio. Each questio carries (+6) marks for correct aswer ad ( ) marks for icorrect aswer.. Which of the followig coditios does NOT esure
More informationChapter 6 Infinite Series
Chapter 6 Ifiite Series I the previous chapter we cosidered itegrals which were improper i the sese that the iterval of itegratio was ubouded. I this chapter we are goig to discuss a topic which is somewhat
More informationMATH 304: MIDTERM EXAM SOLUTIONS
MATH 304: MIDTERM EXAM SOLUTIONS [The problems are each worth five poits, except for problem 8, which is worth 8 poits. Thus there are 43 possible poits.] 1. Use the Euclidea algorithm to fid the greatest
More informationExercises 1 Sets and functions
Exercises 1 Sets ad fuctios HU Wei September 6, 018 1 Basics Set theory ca be made much more rigorous ad built upo a set of Axioms. But we will cover oly some heuristic ideas. For those iterested studets,
More informationApply change-of-basis formula to rewrite x as a linear combination of eigenvectors v j.
Eigevalue-Eigevector Istructor: Nam Su Wag eigemcd Ay vector i real Euclidea space of dimesio ca be uiquely epressed as a liear combiatio of liearly idepedet vectors (ie, basis) g j, j,,, α g α g α g α
More informationn=1 a n is the sequence (s n ) n 1 n=1 a n converges to s. We write a n = s, n=1 n=1 a n
Series. Defiitios ad first properties A series is a ifiite sum a + a + a +..., deoted i short by a. The sequece of partial sums of the series a is the sequece s ) defied by s = a k = a +... + a,. k= Defiitio
More information11. FINITE FIELDS. Example 1: The following tables define addition and multiplication for a field of order 4.
11. FINITE FIELDS 11.1. A Field With 4 Elemets Probably the oly fiite fields which you ll kow about at this stage are the fields of itegers modulo a prime p, deoted by Z p. But there are others. Now although
More informationHomework 4. x n x X = f(x n x) +
Homework 4 1. Let X ad Y be ormed spaces, T B(X, Y ) ad {x } a sequece i X. If x x weakly, show that T x T x weakly. Solutio: We eed to show that g(t x) g(t x) g Y. It suffices to do this whe g Y = 1.
More informationMath 25 Solutions to practice problems
Math 5: Advaced Calculus UC Davis, Sprig 0 Math 5 Solutios to practice problems Questio For = 0,,, 3,... ad 0 k defie umbers C k C k =! k!( k)! (for k = 0 ad k = we defie C 0 = C = ). by = ( )... ( k +
More informationSecond day August 2, Problems and Solutions
FOURTH INTERNATIONAL COMPETITION FOR UNIVERSITY STUDENTS IN MATHEMATICS July 30 August 4, 1997, Plovdiv, BULGARIA Secod day August, 1997 Problems ad Solutios Let Problem 1. Let f be a C 3 (R) o-egative
More informationObjective Mathematics
. If sum of '' terms of a sequece is give by S Tr ( )( ), the 4 5 67 r (d) 4 9 r is equal to : T. Let a, b, c be distict o-zero real umbers such that a, b, c are i harmoic progressio ad a, b, c are i arithmetic
More informationAssignment 1 : Real Numbers, Sequences. for n 1. Show that (x n ) converges. Further, by observing that x n+2 + x n+1
Assigmet : Real Numbers, Sequeces. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a upper boud of A for every N. 2. Let y (, ) ad x (, ). Evaluate
More informationChapter 4. Fourier Series
Chapter 4. Fourier Series At this poit we are ready to ow cosider the caoical equatios. Cosider, for eample the heat equatio u t = u, < (4.) subject to u(, ) = si, u(, t) = u(, t) =. (4.) Here,
More informationRead carefully the instructions on the answer book and make sure that the particulars required are entered on each answer book.
THE UNIVERSITY OF WARWICK FIRST YEAR EXAMINATION: Jauary 2009 Aalysis I Time Allowed:.5 hours Read carefully the istructios o the aswer book ad make sure that the particulars required are etered o each
More informationName of the Student:
SUBJECT NAME : Discrete Mathematics SUBJECT CODE : MA 65 MATERIAL NAME : Problem Material MATERIAL CODE : JM08ADM010 (Sca the above QR code for the direct dowload of this material) Name of the Studet:
More informationLesson 10: Limits and Continuity
www.scimsacademy.com Lesso 10: Limits ad Cotiuity SCIMS Academy 1 Limit of a fuctio The cocept of limit of a fuctio is cetral to all other cocepts i calculus (like cotiuity, derivative, defiite itegrals
More information4 Mathematical Induction
4 Mathematical Iductio Examie the propositios for all, 1 + + ( + 1) + = for all, + 1 ( + 1) for all How do we prove them? They are statemets ivolvig a variable ruig through the ifiite set Strictly speakig,
More informationWe are mainly going to be concerned with power series in x, such as. (x)} converges - that is, lims N n
Review of Power Series, Power Series Solutios A power series i x - a is a ifiite series of the form c (x a) =c +c (x a)+(x a) +... We also call this a power series cetered at a. Ex. (x+) is cetered at
More informationHomework 3. = k 1. Let S be a set of n elements, and let a, b, c be distinct elements of S. The number of k-subsets of S is
Homewor 3 Chapter 5 pp53: 3 40 45 Chapter 6 p85: 4 6 4 30 Use combiatorial reasoig to prove the idetity 3 3 Proof Let S be a set of elemets ad let a b c be distict elemets of S The umber of -subsets of
More informationFermat s Little Theorem. mod 13 = 0, = }{{} mod 13 = 0. = a a a }{{} mod 13 = a 12 mod 13 = 1, mod 13 = a 13 mod 13 = a.
Departmet of Mathematical Scieces Istructor: Daiva Puciskaite Discrete Mathematics Fermat s Little Theorem 43.. For all a Z 3, calculate a 2 ad a 3. Case a = 0. 0 0 2-times Case a 0. 0 0 3-times a a 2-times
More informationMath 4400/6400 Homework #7 solutions
MATH 4400 problems. Math 4400/6400 Homewor #7 solutios 1. Let p be a prime umber. Show that the order of 1 + p modulo p 2 is exactly p. Hit: Expad (1 + p) p by the biomial theorem, ad recall from MATH
More informationSequences and Series of Functions
Chapter 6 Sequeces ad Series of Fuctios 6.1. Covergece of a Sequece of Fuctios Poitwise Covergece. Defiitio 6.1. Let, for each N, fuctio f : A R be defied. If, for each x A, the sequece (f (x)) coverges
More informationDefinition 4.2. (a) A sequence {x n } in a Banach space X is a basis for X if. unique scalars a n (x) such that x = n. a n (x) x n. (4.
4. BASES I BAACH SPACES 39 4. BASES I BAACH SPACES Sice a Baach space X is a vector space, it must possess a Hamel, or vector space, basis, i.e., a subset {x γ } γ Γ whose fiite liear spa is all of X ad
More informationMATH 205 HOMEWORK #2 OFFICIAL SOLUTION. (f + g)(x) = f(x) + g(x) = f( x) g( x) = (f + g)( x)
MATH 205 HOMEWORK #2 OFFICIAL SOLUTION Problem 2: Do problems 7-9 o page 40 of Hoffma & Kuze. (7) We will prove this by cotradictio. Suppose that W 1 is ot cotaied i W 2 ad W 2 is ot cotaied i W 1. The
More informationMTH Assignment 1 : Real Numbers, Sequences
MTH -26 Assigmet : Real Numbers, Sequeces. Fid the supremum of the set { m m+ : N, m Z}. 2. Let A be a o-empty subset of R ad α R. Show that α = supa if ad oly if α is ot a upper boud of A but α + is a
More informationIt is always the case that unions, intersections, complements, and set differences are preserved by the inverse image of a function.
MATH 532 Measurable Fuctios Dr. Neal, WKU Throughout, let ( X, F, µ) be a measure space ad let (!, F, P ) deote the special case of a probability space. We shall ow begi to study real-valued fuctios defied
More information2.1. The Algebraic and Order Properties of R Definition. A binary operation on a set F is a function B : F F! F.
CHAPTER 2 The Real Numbers 2.. The Algebraic ad Order Properties of R Defiitio. A biary operatio o a set F is a fuctio B : F F! F. For the biary operatios of + ad, we replace B(a, b) by a + b ad a b, respectively.
More informationSolutions to Homework 1
Solutios to Homework MATH 36. Describe geometrically the sets of poits z i the complex plae defied by the followig relatios /z = z () Re(az + b) >, where a, b (2) Im(z) = c, with c (3) () = = z z = z 2.
More informationInternational Baccalaureate LECTURE NOTES MATHEMATICS HL FURTHER MATHEMATICS HL Christos Nikolaidis TOPIC NUMBER THEORY
Iteratioal Baccalaureate LECTURE NOTES MATHEMATICS HL FURTHER MATHEMATICS HL TOPIC NUMBER THEORY METHODS OF PROOF. Couterexample - Cotradictio - Pigeohole Priciple Strog mathematical iductio 2 DIVISIBILITY....
More informationlim za n n = z lim a n n.
Lecture 6 Sequeces ad Series Defiitio 1 By a sequece i a set A, we mea a mappig f : N A. It is customary to deote a sequece f by {s } where, s := f(). A sequece {z } of (complex) umbers is said to be coverget
More informationBertrand s Postulate
Bertrad s Postulate Lola Thompso Ross Program July 3, 2009 Lola Thompso (Ross Program Bertrad s Postulate July 3, 2009 1 / 33 Bertrad s Postulate I ve said it oce ad I ll say it agai: There s always a
More informationSection 4.3. Boolean functions
Sectio 4.3. Boolea fuctios Let us take aother look at the simplest o-trivial Boolea algebra, ({0}), the power-set algebra based o a oe-elemet set, chose here as {0}. This has two elemets, the empty set,
More information1. ARITHMETIC OPERATIONS IN OBSERVER'S MATHEMATICS
1. ARITHMETIC OPERATIONS IN OBSERVER'S MATHEMATICS We cosider a ite well-ordered system of observers, where each observer sees the real umbers as the set of all iite decimal fractios. The observers are
More informationDe Moivre s Theorem - ALL
De Moivre s Theorem - ALL. Let x ad y be real umbers, ad be oe of the complex solutios of the equatio =. Evaluate: (a) + + ; (b) ( x + y)( x + y). [6]. (a) Sice is a complex umber which satisfies = 0,.
More informationResolution Proofs of Generalized Pigeonhole Principles
Resolutio Proofs of Geeralized Pigeohole Priciples Samuel R. Buss Departmet of Mathematics Uiversity of Califoria, Berkeley Győrgy Turá Departmet of Mathematics, Statistics, ad Computer Sciece Uiversity
More information5 Birkhoff s Ergodic Theorem
5 Birkhoff s Ergodic Theorem Amog the most useful of the various geeralizatios of KolmogorovâĂŹs strog law of large umbers are the ergodic theorems of Birkhoff ad Kigma, which exted the validity of the
More informationBeurling Integers: Part 2
Beurlig Itegers: Part 2 Isomorphisms Devi Platt July 11, 2015 1 Prime Factorizatio Sequeces I the last article we itroduced the Beurlig geeralized itegers, which ca be represeted as a sequece of real umbers
More information[ 11 ] z of degree 2 as both degree 2 each. The degree of a polynomial in n variables is the maximum of the degrees of its terms.
[ 11 ] 1 1.1 Polyomial Fuctios 1 Algebra Ay fuctio f ( x) ax a1x... a1x a0 is a polyomial fuctio if ai ( i 0,1,,,..., ) is a costat which belogs to the set of real umbers ad the idices,, 1,...,1 are atural
More information, then cv V. Differential Equations Elements of Lineaer Algebra Name: Consider the differential equation. and y2 cos( kx)
Cosider the differetial equatio y '' k y 0 has particular solutios y1 si( kx) ad y cos( kx) I geeral, ay liear combiatio of y1 ad y, cy 1 1 cy where c1, c is also a solutio to the equatio above The reaso
More informationSOME TRIBONACCI IDENTITIES
Mathematics Today Vol.7(Dec-011) 1-9 ISSN 0976-38 Abstract: SOME TRIBONACCI IDENTITIES Shah Devbhadra V. Sir P.T.Sarvajaik College of Sciece, Athwalies, Surat 395001. e-mail : drdvshah@yahoo.com The sequece
More informationComplex Numbers Solutions
Complex Numbers Solutios Joseph Zoller February 7, 06 Solutios. (009 AIME I Problem ) There is a complex umber with imagiary part 64 ad a positive iteger such that Fid. [Solutio: 697] 4i + + 4i. 4i 4i
More informationAPPENDIX F Complex Numbers
APPENDIX F Complex Numbers Operatios with Complex Numbers Complex Solutios of Quadratic Equatios Polar Form of a Complex Number Powers ad Roots of Complex Numbers Operatios with Complex Numbers Some equatios
More informationCHAPTER I: Vector Spaces
CHAPTER I: Vector Spaces Sectio 1: Itroductio ad Examples This first chapter is largely a review of topics you probably saw i your liear algebra course. So why cover it? (1) Not everyoe remembers everythig
More informationNAME: ALGEBRA 350 BLOCK 7. Simplifying Radicals Packet PART 1: ROOTS
NAME: ALGEBRA 50 BLOCK 7 DATE: Simplifyig Radicals Packet PART 1: ROOTS READ: A square root of a umber b is a solutio of the equatio x = b. Every positive umber b has two square roots, deoted b ad b or
More informationMathematical Foundation. CSE 6331 Algorithms Steve Lai
Mathematical Foudatio CSE 6331 Algorithms Steve Lai Complexity of Algorithms Aalysis of algorithm: to predict the ruig time required by a algorithm. Elemetary operatios: arithmetic & boolea operatios:
More informationCSE 1400 Applied Discrete Mathematics Number Theory and Proofs
CSE 1400 Applied Discrete Mathematics Number Theory ad Proofs Departmet of Computer Scieces College of Egieerig Florida Tech Sprig 01 Problems for Number Theory Backgroud Number theory is the brach of
More informationCALCULUS BASIC SUMMER REVIEW
CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=
More informationCS / MCS 401 Homework 3 grader solutions
CS / MCS 401 Homework 3 grader solutios assigmet due July 6, 016 writte by Jāis Lazovskis maximum poits: 33 Some questios from CLRS. Questios marked with a asterisk were ot graded. 1 Use the defiitio of
More information} is said to be a Cauchy sequence provided the following condition is true.
Math 4200, Fial Exam Review I. Itroductio to Proofs 1. Prove the Pythagorea theorem. 2. Show that 43 is a irratioal umber. II. Itroductio to Logic 1. Costruct a truth table for the statemet ( p ad ~ r
More informationProduct measures, Tonelli s and Fubini s theorems For use in MAT3400/4400, autumn 2014 Nadia S. Larsen. Version of 13 October 2014.
Product measures, Toelli s ad Fubii s theorems For use i MAT3400/4400, autum 2014 Nadia S. Larse Versio of 13 October 2014. 1. Costructio of the product measure The purpose of these otes is to preset the
More informationMath 140A Elementary Analysis Homework Questions 3-1
Math 0A Elemetary Aalysis Homework Questios -.9 Limits Theorems for Sequeces Suppose that lim x =, lim y = 7 ad that all y are o-zero. Detarime the followig limits: (a) lim(x + y ) (b) lim y x y Let s
More informationMathematical Induction
Mathematical Iductio Itroductio Mathematical iductio, or just iductio, is a proof techique. Suppose that for every atural umber, P() is a statemet. We wish to show that all statemets P() are true. I a
More information2 Geometric interpretation of complex numbers
2 Geometric iterpretatio of complex umbers 2.1 Defiitio I will start fially with a precise defiitio, assumig that such mathematical object as vector space R 2 is well familiar to the studets. Recall that
More information(A sequence also can be thought of as the list of function values attained for a function f :ℵ X, where f (n) = x n for n 1.) x 1 x N +k x N +4 x 3
MATH 337 Sequeces Dr. Neal, WKU Let X be a metric space with distace fuctio d. We shall defie the geeral cocept of sequece ad limit i a metric space, the apply the results i particular to some special
More informationRegn. No. North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: ,
. Sectio-A cotais 30 Multiple Choice Questios (MCQ). Each questio has 4 choices (a), (b), (c) ad (d), for its aswer, out of which ONLY ONE is correct. From Q. to Q.0 carries Marks ad Q. to Q.30 carries
More informationConvergence of random variables. (telegram style notes) P.J.C. Spreij
Covergece of radom variables (telegram style otes).j.c. Spreij this versio: September 6, 2005 Itroductio As we kow, radom variables are by defiitio measurable fuctios o some uderlyig measurable space
More informationMath 110 Assignment #6 Due: Monday, February 10
Math Assigmet #6 Due: Moday, February Justify your aswers. Show all steps i your computatios. Please idicate your fial aswer by puttig a box aroud it. Please write eatly ad legibly. Illegible aswers will
More information