Homework 1/Solutions. Graded Exercises


 Jared Paul
 3 years ago
 Views:
Transcription
1 MTH Abstract Algebra I and Number Theory S18 Homework 1/Solutions Graded Exercises Exercise 1. Below are parts of the addition table and parts of the multiplication table of a ring. Complete both tables. + w x y z w w x y z y z w x z w y w x y z w x y y z Note that w + w = w and so the Additive Identity Law implies w = 0 R. Thus Ax 4 implies that (I) w + a = a and a + w = a for all a in the ring. By the given part of the addition table: (II) z + x = w and y + z = x Using Ax 3 we compute (III) x + z Ax 3 = z + x (II) = w and z + y Ax 3 = y + z (II) = x By (I),(II) and (III) we conclude that the addition table is ( ) + w x y z w w x y z x x y z w y y z w x z z w x y By the given part of the multiplication table we have We compute xx = y (1) Since w = 0 R, Theorem 1.2.9(c) gives xy (*) =x(x + x) Ax 8 = xx + xx (1) = y + y (*) = w (2) yx (*) =(x + x)x Ax 8 = xx + xx (1) = y + y (*) = w (3) xz (*) =x(x + y) Ax 8 = xx + xy (1),(2) = y + w (*) = y (4) zx (*) =(x + y)x Ax 8 = xx + yx (1),(3) = y + w (*) = y (5) yy (*) =y(x + x) Ax 8 = yx + yx (3) = w + w (*) = w (6) yz (*) =y(x + y) Ax 8 = yx + yy (3),(6) = w + w (*) = w (7) zy (*) =z(x + x) Ax 8 = zx + zx (5) = y + y (*) = w (8) zz (*) =z(x + y) Ax 8 = zx + zy (5),(8) = y + w (*) = y (9) wa = 0 R a = 0 R = w and aw = a0 R = 0 R = w (10)
2 for all a in R. Thus the multiplication table is: w x y z w w w w w x w y w y y w w w w z w y w y Exercise 2. Let R be a ring and a, b, c, d R. Prove that (a b)(c d) = ( (ac ad) + bd ) bc In each step of your proof, quote exactly one Axiom, Definition or Theorem. (a b)(c d) = a(c d) b(c d) Theorem 1.2.9j = (ac ad) (bc bd) Theorem 1.2.9j, twice = (ac ad) + ( (bc bd)) Definition of, see = (ac ad) + (bd bc) Theorem 1.2.9h = (ac ad) + (bd + ( bc)) Definition of = ( (ac ad) + bd ) + ( bc)) Ax 2 = ( (ac ad) + bd ) bc Definition of Exercise 3. Prove or give a counterexample: If R is a ring with identity, then 1 R 0 R. The ring R = {0 R } in Example (f) is a counterexample. Exercise 4. Let R be a ring such that a a = 0 R for all a R. Show that ab = (ba) for all a, b R. By hypothesis we have ( ) x x = 0 R for all x R Let a, b R. We compute 0 R = (a + b)(a + b) a + b R by Ax 1 and (*) for x = a + b = a(a + b) + b(a + b) Ax 8 = (aa + ab) + (ba + bb)) Ax 8, twice = (0 R + ab) + (ba + 0 R ) (*) for x = a and x = b = ab + ba Ax 4, twice We proved that ab + ba = 0 R. Thus ab = (ba) by the Additive Inverse Law. Not Graded Exercises Exercise 5. Let R = {0, e, b, c} with addition and multiplication defined by the following tables. Assume associativity and distributivity and show that R is a ring with identity. Is R commutative? + 0 e b c 0 0 e b c e e 0 c b b b c 0 e c c b e 0 0 e b c e 0 e b c b 0 b b 0 c 0 c 0 c (Ax1) (closure of additions) All entrees in the addition table are from R. So a + b R for all a, b R.
3 (Ax2) (associative addition) Luckily we are allowed to assume this without proof. (Ax3) (commutative addition) The addition table is symmetric along the diagonal through +. Note that this means that a + b = b + a for all a, b R. (Ax4) (additive identity) The second row of the addition table equals the first. Thus 0 + r = r for all r R. The second column equals the first. So r + 0 = r for all r R. Hence 0 R := 0 is an additive identity. (Ax5) (additive inverse) All entrees in the diagonal through + equal 0. So r + r = 0 = 0 = 0 R for all r R. Thus every element has an additive inverse, namely itself. (Ax6) (closure of multiplication) All entrees in the multiplication table are from R. a, b R. So ab R for all (Ax7) (associative multiplication) Given to us for granted. (Ax8) (distributive laws) Given to us for granted. (Ax9) (commutative multiplication) The multiplication table is symmetric along the diagonal through. Note that this means that ab = ba for all a, b R. (Ax10) (multiplicative identity) The third row of the multiplication table equals the first. Thus e r = r for all r R. The third column equals the first. So r e = r for all r R. Hence 1 R := e is a multiplicative identity. Since Axiom 18 hold, R is a ring. Since Axiom 9 holds, R is commutative and since Axiom 10 holds, R has an identity. Exercise 6. Prove the following Theorem: Theorem Let R and S be rings. Define an addition and multiplication on R S by for all r, r R and s, s S. Then (a) R S is a ring. (b) 0 R S = (0 R, 0 S ). (c) (r, s) = ( r, s) for all r R, s S. (r, s) + (r, s ) = (r + r, s + s ) (r, s)(r, s ) = (rr, ss ) (d) If R and S are both commutative, then so is R S. (e) If both R and S have an identity, then R S has an identity and 1 R S = (1 R, 1 S ). (a): We need to verify all the eight axioms of a ring. Let a, b, c R and d, e, f S. (Ax1) (closure of addition) By Ax1 for R and S, a + b R and d + e S and so (a + b, d + e) R S. By definition of addition, (a, d) + (b, e) = (a + b, d + e) and so (a, d) + (b, e) R S. (Ax2) (associative addition) (a, d) + ((b, e) + (c, f)) = (a, d) + (b + c, e + f) (by the definition of addition) = (a + (b + c), d + (e + f) (by the definition of addition) = ((a + b) + c, (d + e) + f) ((Ax2) for R and S) = (a + b, d + e) + (c, f) (by the definition of addition) = ((a, d) + (b, e)) + (c, f) (by the definition of addition). So addition in R S is associative.
4 (Ax3) (commutative addition) (a, d) + (b, e) = (a + b, d + e) (by the definition of addition) = (b + a, e + d) ((Ax3) for R and S) = (a, d) + (b, e) (by the definition of addition). So the addition in R S is commutative. (Ax4) (additive identity) Define ( ) 0 R S := (0 R, 0 S ). Then (a, d) + 0 R S = (a, d) + (0 R, 0 S ) (by the definition of 0 R S ) = (a + 0 R, d + 0 S ) (by the definition of addition) = (a, d) ((Ax4) for R and S) Since the addition in R S is commutative, we also get 0 R S + (a, d) = (a, d) and so 0 R S is an additive identity. (Ax5) (additive inverse) Define ( ) (a, d) := ( a, d) Then (a, d) + ( (a, d)) = (a, d) + ( a, d) (by the definition of (a, d)) = (a + ( a), d + ( d)) (by the definition of addition) = (0 R, 0 S ) ((Ax5) for R and S) So (a, d) is an additive inverse of (a, d). (Ax6) (closure of multiplication) By Ax6 for R and S, ab R and de S and so (ab, de) R S. By definition of multiplication, (a, d)(b, e) = (ab, de) and so (a, d)(b, e) R S. (Ax7) (associative multiplication) (a, d)((b, e)(c, f)) = (a, d)(bc, ef) (by the definition of multiplication) = (a(bc), d(ef) (by the definition of multiplication) = ((ab)c, (de)f) ((Ax7) for R and S) = (ab, de)(c, f) (by the definition of multiplication) = ((a, d)(b, e))(c, f) (by the definition of multiplication). So multiplication in R S is associative.
5 (Ax8) (distributive laws) (a, d)((b, e) + (c, f)) = (a, d)(b + c, e + f) (by the definition of addition) = (a(b + c), d(e + f) (by the definition of multiplication) = (ab + ac, de + df) ((Ax7) for R and S) = (ab, de) + (ac, df) (by the definition of addition) = (a, b)(b, e) + (a, b)(d, f) (by the definition of multiplication) and ((a, d) + (b, e))(c, f) = (a + b, d + e)(c, f) (by the definition of addition) = ((a + b)c, (d + e)f) (by the definition of multiplication) = (ac + bc, df + ef) ((Ax7) for R and S) = (ac, df) + (bc, ef) (by the definition of addition) = (a, d)(c, f) + (b, e)(c, f) (by the definition of multiplication). So the distributive laws hold in R S. We verified all the eight axioms and so R S is a ring. (b) By definition of 0 R S, see (*), we have 0 R S = (0 R, 0 S ). (c) By definition of (r, s), see (**), we have (r, s) = ( r, s). (d) Suppose that both R and S are commutative. So (Ax9) holds in R and S. (a, d)(b, e) = (ab, de) (by the definition of multiplication) = (ba, ed) ((Ax9) for R and S) = (b, e)(a, d) (by the definition of multiplication). So the multiplication in R S is commutative. (e) Suppose that R has an identity. Put 1 R S := (1 R, 1 S ). Then (a, d)1 R S = (a, d)(1 R, 1 S ) (by the definition of 1 R S ) = (a1 R, d1 S ) (by the definition of multiplication ) = (a, d) ((Ax10) for R and S) and 1 R S (a, d) = (1 R, 1 S )(a, d) (by the definition of 1 R S ) = (1 R a, 1 S d) (by the definition of multiplication ) = (a, d) ((Ax10) for R and S) So 1 R S is an multiplicative identity in R S, and 1 R S = (1 R, 1 S ).
6 Exercise 7. Let R be a ring such that a a = a for all a R. Show that (a) a + a = 0 R for all a R (b) R is commutative. By hypothesis we have ( ) xx = x for all x R (a) Let a R. Then a = aa (*) for x = a = ( a)( a) Theorem 1.2.9(i) = a (*) for x = a We proved that ( ) a = a for all a R Thus a + a = a + ( a) = 0 R, by Axiom 5. (b) Let a, b R. Then a + b = (a + b)(a + b) a + b R by Ax 1 and (*) for x = a + b = a(a + b) + b(a + b) Ax 8 = (aa + ab) + (ba + bb) Ax 8, twice = (aa + ab) + (bb + ba) Ax 3 = ( (aa + ab) + bb ) + ba Ax 2 = ( aa + (ab + bb) ) + ba Ax 2 = ( aa + (bb + ab) ) + ba Ax 3 = ( (aa + bb) + ab) ) + ba Ax 2 = (aa + bb) + (ab + ba) Ax 2 = (a + b) + (ab + ba) (*) for x = a and x = b We proved that a + b = (a + b) + (ab + ba) and so the Additive Identity Law shows that ab + ba = 0 R. Hence the Additive Inverse Law implies that ab = (ba). By (**) applied with ba in place of a, we have (ba) = ba. As ab = (ba), this gives ab = ba. Hence Ax 9 holds and so R is commutative.
Homework 3/ Solutions
MTH 3103 Abstract Algebra I and Number Theory S17 Homework 3/ Solutions Exercise 1. Prove the following Theorem: Theorem Let R and S be rings. Define an addition and multiplication on R S by for all r,
More informationHomework Solution #1. Chapter 2 2.6, 2.17, 2.24, 2.30, 2.39, 2.42, Grading policy is as follows:  Total 100
Homework Solution #1 Chapter 2 2.6, 2.17, 2.24, 2.30, 2.39, 2.42, 2.48 Grading policy is as follows:  Total 100 Exercise 2.6 (total 10)  Column 3 : 5 point  Column 4 : 5 point Exercise 2.17 (total 25)
More informationCHAPTER 2 BOOLEAN ALGEBRA
CHAPTER 2 BOOLEAN ALGEBRA This chapter in the book includes: Objectives Study Guide 2.1 Introduction 2.2 Basic Operations 2.3 Boolean Expressions and Truth Tables 2.4 Basic Theorems 2.5 Commutative, Associative,
More informationA DARK GREY P O N T, with a Switch Tail, and a small Star on the Forehead. Any
Y Y Y X X «/ YY Y Y ««Y x ) & \ & & } # Y \#$& / Y Y X» \\ / X X X x & Y Y X «q «z \x» = q Y # % \ & [ & Z \ & { + % ) / / «q zy» / & / / / & x x X / % % ) Y x X Y $ Z % Y Y x x } / % «] «] # z» & Y X»
More informationAlgebraic Expressions
Algebraic Expressions 1. Expressions are formed from variables and constants. 2. Terms are added to form expressions. Terms themselves are formed as product of factors. 3. Expressions that contain exactly
More informationAxioms of Kleene Algebra
Introduction to Kleene Algebra Lecture 2 CS786 Spring 2004 January 28, 2004 Axioms of Kleene Algebra In this lecture we give the formal definition of a Kleene algebra and derive some basic consequences.
More information( ) Chapter 6 ( ) ( ) ( ) ( ) Exercise Set The greatest common factor is x + 3.
Chapter 6 Exercise Set 6.1 1. A prime number is an integer greater than 1 that has exactly two factors, itself and 1. 3. To factor an expression means to write the expression as the product of factors.
More informationECE 238L Boolean Algebra  Part I
ECE 238L Boolean Algebra  Part I August 29, 2008 Typeset by FoilTEX Understand basic Boolean Algebra Boolean Algebra Objectives Relate Boolean Algebra to Logic Networks Prove Laws using Truth Tables Understand
More informationCHAPTER 3 BOOLEAN ALGEBRA
CHAPTER 3 BOOLEAN ALGEBRA (continued) This chapter in the book includes: Objectives Study Guide 3.1 Multiplying Out and Factoring Expressions 3.2 ExclusiveOR and Equivalence Operations 3.3 The Consensus
More informationKevin James. MTHSC 412 Section 3.1 Definition and Examples of Rings
MTHSC 412 Section 3.1 Definition and Examples of Rings A ring R is a nonempty set R together with two binary operations (usually written as addition and multiplication) that satisfy the following axioms.
More informationTHE RING OF POLYNOMIALS. Special Products and Factoring
THE RING OF POLYNOMIALS Special Products and Factoring Special Products and Factoring Upon completion, you should be able to Find special products Factor a polynomial completely Special Products  rules
More informationAlgebraic Expressions and Identities
ALGEBRAIC EXPRESSIONS AND IDENTITIES 137 Algebraic Expressions and Identities CHAPTER 9 9.1 What are Expressions? In earlier classes, we have already become familiar with what algebraic expressions (or
More informationGrade 8 Factorisation
ID : ae8factorisation [1] Grade 8 Factorisation For more such worksheets visit www.edugain.com Answer the questions (1) Find factors of following polynomial A) y 22xy + 3y  6x B) 3y 212xy  2y + 8x
More information[3] (b) Find a reduced rowechelon matrix rowequivalent to ,1 2 2
MATH Key for sample nal exam, August 998 []. (a) Dene the term \reduced rowechelon matrix". A matrix is reduced rowechelon if the following conditions are satised. every zero row lies below every nonzero
More informationLOWELL WEEKLY JOURNAL
Y » $ 5 Y 7 Y Y Y Q x Q» 75»»/ q } # ]»\   $ { Q» / X x»» 3 q $ 9 ) Y q  5 5 3 3 3 7 Q q   Q _»»/Q Y  9    ) [ X 7» »  )»? / /? Q Y»» # X Q»  ?» Q ) Q \ Q    3? 7» ? #»»» 7  / Q
More informationName: Class: IM8 Block:
Name: Block: Class: IM8 Investigation 1: Mathematical Properties and Order of Operations Mathematical Properties 2 Practice Level 1: Write the name of the property shown in the examples below. 1. 4 + 5
More informationDIPOLES III. q const. The voltage produced by such a charge distribution is given by. r r'
DIPOLES III We now consider a particularly important charge configuration a dipole. This consists of two equal but opposite charges separated by a small distance. We define the dipole moment as p lim q
More informationChapter y. 8. n cd (x y) 14. (2a b) 15. (a) 3(x 2y) = 3x 3(2y) = 3x 6y. 16. (a)
Chapter 6 Chapter 6 opener A. B. C. D. 6 E. 5 F. 8 G. H. I. J.. 7. 8 5. 6 6. 7. y 8. n 9. w z. 5cd.. xy z 5r s t. (x y). (a b) 5. (a) (x y) = x (y) = x 6y x 6y = x (y) = (x y) 6. (a) a (5 a+ b) = a (5
More informationExpanding brackets and factorising
CHAPTER 8 Epanding brackets and factorising 8 CHAPTER Epanding brackets and factorising 8.1 Epanding brackets There are three rows. Each row has n students. The number of students is 3 n 3n. Two students
More informationSubrings and Ideals 2.1 INTRODUCTION 2.2 SUBRING
Subrings and Ideals Chapter 2 2.1 INTRODUCTION In this chapter, we discuss, subrings, sub fields. Ideals and quotient ring. We begin our study by defining a subring. If (R, +, ) is a ring and S is a nonempty
More informationNew Coding System of Grid Squares in the Republic of Indonesia
September14, 2006 New Coding System of Grid Squares in the Republic of Indonesia Current coding system of grid squares in the Republic of Indonesia is based on similar
More informationNeatest and Promptest Manner. E d i t u r ami rul)lihher. FOIt THE CIIILDIIES'. Trifles.
» ~ $ ) 7 x X ) / ( 8 2 X 39 ««x» ««! «! / x? \» «({? «» q «(? (?? x! «? 8? ( z x x q? ) «q q q ) x z x 69 7( X X ( 3»«! ( ~«x ««x ) (» «8 4 X «4 «4 «8 X «x «(» X) ()»» «X «97 X X X 4 ( 86) x) ( ) z z
More informationMATH 422, CSUSM. SPRING AITKEN
CHAPTER 3 SUMMARY: THE INTEGERS Z (PART I) MATH 422, CSUSM. SPRING 2009. AITKEN 1. Introduction This is a summary of Chapter 3 from Number Systems (Math 378). The integers Z included the natural numbers
More informationLinear Algebra. Chapter 8: Eigenvalues: Further Applications and Computations Section 8.2. Applications to Geometry Proofs of Theorems.
Linear Algebra Chapter 8: Eigenvalues: Further Applications and Computations Section 8.2. Applications to Geometry Proofs of Theorems May 1, 2018 () Linear Algebra May 1, 2018 1 / 8 Table of contents 1
More informationreview To find the coefficient of all the terms in 15ab + 60bc 17ca: Coefficient of ab = 15 Coefficient of bc = 60 Coefficient of ca = 17
1. Revision Recall basic terms of algebraic expressions like Variable, Constant, Term, Coefficient, Polynomial etc. The coefficients of the terms in 4x 2 5xy + 6y 2 are Coefficient of 4x 2 is 4 Coefficient
More informationChapter 2: Boolean Algebra and Logic Gates
Chapter 2: Boolean Algebra and Logic Gates Mathematical methods that simplify binary logics or circuits rely primarily on Boolean algebra. Boolean algebra: a set of elements, a set of operators, and a
More informationAlgebra 1. Predicting Patterns & Examining Experiments. Unit 5: Changing on a Plane Section 4: Try Without Angles
Section 4 Examines triangles in the coordinate plane, we will mention slope, but not angles (we will visit angles in Unit 6). Students will need to know the definition of collinear, isosceles, and congruent...
More informationAMC Lecture Notes. Justin Stevens. Cybermath Academy
AMC Lecture Notes Justin Stevens Cybermath Academy Contents 1 Systems of Equations 1 1 Systems of Equations 1.1 Substitution Example 1.1. Solve the system below for x and y: x y = 4, 2x + y = 29. Solution.
More informationGROUPS. Chapter1 EXAMPLES 1.1. INTRODUCTION 1.2. BINARY OPERATION
Chapter1 GROUPS 1.1. INTRODUCTION The theory of groups arose from the theory of equations, during the nineteenth century. Originally, groups consisted only of transformations. The group of transformations
More informationTwo Posts to Fill On School Board
Y Y 9 86 4 4 qz 86 x : ( ) z 7 854 Y x 4 z z x x 4 87 88 Y 5 x q x 8 Y 8 x x : 6 ; : 5 x ; 4 ( z ; ( ) ) x ; z 94 ; x 3 3 3 5 94 ; ; ; ; 3 x : 5 89 q ; ; x ; x ; ; x : ; ; ; ; ; ; 87 47% : () : / : 83
More informationPart 1. For any Amodule, let M[x] denote the set of all polynomials in x with coefficients in M, that is to say expressions of the form
Commutative Algebra Homework 3 David Nichols Part 1 Exercise 2.6 For any Amodule, let M[x] denote the set of all polynomials in x with coefficients in M, that is to say expressions of the form m 0 + m
More informationNozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Ismailia Road Branch
Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep. Nozha Language Schools Sheet Ismailia Road Branch Sheet ( 1) 1Complete 1. in the parallelogram, each two opposite
More informationLOWELL JOURNAL. MUST APOLOGIZE. such communication with the shore as Is m i Boimhle, noewwary and proper for the comfort
 7 7 Z 8 q ) V x  X > q  < Y Y X V  z     V  V  q \  q q <  V    x   V q > x  x q  x q  x    7 »     6 q x  >   x    x   q q  V  x   ( Y q Y7  >»>  x Y  ] [
More informationSYMBOL NAME DESCRIPTION EXAMPLES. called positive integers) negatives, and 0. represented as a b, where
EXERCISE A1 Things to remember: 1. THE SET OF REAL NUMBERS SYMBOL NAME DESCRIPTION EXAMPLES N Natural numbers Counting numbers (also 1, 2, 3,... called positive integers) Z Integers Natural numbers, their
More informationMath 10  Unit 5 Final Review  Polynomials
Class: Date: Math 10  Unit 5 Final Review  Polynomials Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Factor the binomial 44a + 99a 2. a. a(44 + 99a)
More informationChapter 2 : Boolean Algebra and Logic Gates
Chapter 2 : Boolean Algebra and Logic Gates By Electrical Engineering Department College of Engineering King Saud University 14311432 2.1. Basic Definitions 2.2. Basic Theorems and Properties of Boolean
More information5200: Similarity of figures. Define: Lemma: proof:
5200: Similarity of figures. We understand pretty well figures with the same shape and size. Next we study figures with the same shape but different sizes, called similar figures. The most important ones
More information2 Application of Boolean Algebra Theorems (15 Points  graded for completion only)
CSE140 HW1 Solution (100 Points) 1 Introduction The purpose of this assignment is threefold. First, it aims to help you practice the application of Boolean Algebra theorems to transform and reduce Boolean
More information' Liberty and Umou Ono and Inseparablo "
3 5? #< q 8 2 / / ) 9 ) 2 ) > < _ / ] > ) 2 ) ) 5 > x > [ < > < ) > _ ] ]? <
More informationMath 1060 Linear Algebra Homework Exercises 1 1. Find the complete solutions (if any!) to each of the following systems of simultaneous equations:
Homework Exercises 1 1 Find the complete solutions (if any!) to each of the following systems of simultaneous equations: (i) x 4y + 3z = 2 3x 11y + 13z = 3 2x 9y + 2z = 7 x 2y + 6z = 2 (ii) x 4y + 3z =
More informationProblem Set 2 Due Thursday, October 1, & & & & # % (b) Construct a representation using five d orbitals that sit on the origin as a basis:
Problem Set 2 Due Thursday, October 1, 29 Problems from Cotton: Chapter 4: 4.6, 4.7; Chapter 6: 6.2, 6.4, 6.5 Additional problems: (1) Consider the D 3h point group and use a coordinate system wherein
More informationMath 2331 Linear Algebra
2.2 The Inverse of a Matrix Math 2331 Linear Algebra 2.2 The Inverse of a Matrix ShangHuan Chiu Department of Mathematics, University of Houston schiu@math.uh.edu math.uh.edu/ schiu/ ShangHuan Chiu,
More informationLESSON 7.1 FACTORING POLYNOMIALS I
LESSON 7.1 FACTORING POLYNOMIALS I LESSON 7.1 FACTORING POLYNOMIALS I 293 OVERVIEW Here s what you ll learn in this lesson: Greatest Common Factor a. Finding the greatest common factor (GCF) of a set of
More informationCourse 2BA1: Hilary Term 2007 Section 8: Quaternions and Rotations
Course BA1: Hilary Term 007 Section 8: Quaternions and Rotations David R. Wilkins Copyright c David R. Wilkins 005 Contents 8 Quaternions and Rotations 1 8.1 Quaternions............................ 1 8.
More informationCM2104: Computational Mathematics General Maths: 2. Algebra  Factorisation
CM204: Computational Mathematics General Maths: 2. Algebra  Factorisation Prof. David Marshall School of Computer Science & Informatics Factorisation Factorisation is a way of simplifying algebraic expressions.
More informationPolynomial Functions
Polynomial Functions NOTE: Some problems in this file are used with permission from the engageny.org website of the New York State Department of Education. Various files. Internet. Available from https://www.engageny.org/ccsslibrary.
More informationUNIT 3 BOOLEAN ALGEBRA (CONT D)
UNIT 3 BOOLEAN ALGEBRA (CONT D) Spring 2011 Boolean Algebra (cont d) 2 Contents Multiplying out and factoring expressions ExclusiveOR and ExclusiveNOR operations The consensus theorem Summary of algebraic
More informationChapter 2. Boolean Algebra and Logic Gates
Chapter 2 Boolean Algebra and Logic Gates Basic Definitions A binary operator defined on a set S of elements is a rule that assigns, to each pair of elements from S, a unique element from S. The most common
More informationMTH Abstract Algebra I and Number Theory S17. Homework 6/Solutions
MTH 3103 Abstract Algebra I and Number Theory S17 Homework 6/Solutions Exercise 1. Let a, b and e be integers. Suppose a and b are not both zero and that e is a positive common divisor of a and b. Put
More informationMathematics. Exercise 6.4. (Chapter 6) (Triangles) (Class X) Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2.
() Exercise 6.4 Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2. If EF = 15.4 cm, find BC. Answer 1: 1 () Question 2: Diagonals of a trapezium ABCD with AB DC intersect each other
More informationUNIVERSITY OF MICHIGAN UNDERGRADUATE MATH COMPETITION 28 APRIL 7, 2011
UNIVERSITY OF MICHIGAN UNDERGRADUATE MATH COMPETITION 28 APRIL 7, 20 Instructions. Write on the front of your blue book your student ID number. Do not write your name anywhere on your blue book. Each question
More informationJUST THE MATHS UNIT NUMBER 1.5. ALGEBRA 5 (Manipulation of algebraic expressions) A.J.Hobson
JUST THE MATHS UNIT NUMBER 1.5 ALGEBRA 5 (Manipulation of algebraic expressions) by A.J.Hobson 1.5.1 Simplification of expressions 1.5.2 Factorisation 1.5.3 Completing the square in a quadratic expression
More information( ) Chapter 7 ( ) ( ) ( ) ( ) Exercise Set The greatest common factor is x + 3.
Chapter 7 Exercise Set 7.1 1. A prime number is an integer greater than 1 that has exactly two factors, itself and 1. 3. To factor an expression means to write the expression as the product of factors.
More informationA B CDE F B FD D A C AF DC A F
International Journal of Arts & Sciences, CDROM. ISSN: 19446934 :: 4(20):121 131 (2011) Copyright c 2011 by InternationalJournal.org A B CDE F B FD D A C A BC D EF C CE C A D ABC DEF B B C A E E C A
More informationChapter2 BOOLEAN ALGEBRA
Chapter2 BOOLEAN ALGEBRA Introduction: An algebra that deals with binary number system is called Boolean Algebra. It is very power in designing logic circuits used by the processor of computer system.
More informationMath 203A  Solution Set 1
Math 203A  Solution Set 1 Problem 1. Show that the Zariski topology on A 2 is not the product of the Zariski topologies on A 1 A 1. Answer: Clearly, the diagonal Z = {(x, y) : x y = 0} A 2 is closed in
More informationMATH 31BH Homework 5 Solutions
MATH 3BH Homework 5 Solutions February 4, 204 Problem.8.2 (a) Let x t f y = x 2 + y 2 + 2z 2 and g(t) = t 2. z t 3 Then by the chain rule a a a D(g f) b = Dg f b Df b c c c = [Dg(a 2 + b 2 + 2c 2 )] [
More informationAnswers and Solutions to Selected Homework Problems From Section 2.5 S. F. Ellermeyer. and B =. 0 2
Answers and Solutions to Selected Homework Problems From Section 2.5 S. F. Ellermeyer 5. Since gcd (2; 4) 6, then 2 is a zero divisor (and not a unit) in Z 4. In fact, we see that 2 2 0 in Z 4. Thus 2x
More informationIOAN ŞERDEAN, DANIEL SITARU
Romanian Mathematical Magazine Web: http://www.ssmrmh.ro The Author: This article is published with open access. TRIGONOMETRIC SUBSTITUTIONS IN PROBLEM SOLVING PART IOAN ŞERDEAN, DANIEL SITARU Abstract.
More informationProblem Set 2 Due Tuesday, September 27, ; p : 0. (b) Construct a representation using five d orbitals that sit on the origin as a basis: 1
Problem Set 2 Due Tuesday, September 27, 211 Problems from Carter: Chapter 2: 2ad,g,h,j 2.6, 2.9; Chapter 3: 1ad,f,g 3.3, 3.6, 3.7 Additional problems: (1) Consider the D 4 point group and use a coordinate
More informationANSWERS. CLASS: VIII TERM  1 SUBJECT: Mathematics. Exercise: 1(A) Exercise: 1(B)
ANSWERS CLASS: VIII TERM  1 SUBJECT: Mathematics TOPIC: 1. Rational Numbers Exercise: 1(A) 1. Fill in the blanks: (i) 21/24 (ii) 4/7 < 4/11 (iii)16/19 (iv)11/13 and 11/13 (v) 0 2. Answer True or False:
More informationMANY BILLS OF CONCERN TO PUBLIC
 6 8 96 8 9 6 9 XXX 4 > ?  8 9 x 4 z )  ! x  x   X      x 00      x z    x x  x      ) x       0 >  00090   4 0 x 00  ? z 8 & x   8? > 9     64 49 9 x  
More informationOWELL WEEKLY JOURNAL
Y \»<  } Y Y Y & #»»» q ] q»»»>) &    } ) x (  { Y» & ( x  (» & )<  Y X  & Q Q» 3  x Q Y 6 \Y > Y Y X 3 39 33 x   /  » 
More informationMultivariable Calculus and Matrix AlgebraSummer 2017
Multivariable Calculus and Matrix AlgebraSummer 017 Homework 4 Solutions Note that the solutions below are for the latest version of the problems posted. For those of you who worked on an earlier version
More information7.5 Proportionality Relationships
www.ck12.org Chapter 7. Similarity 7.5 Proportionality Relationships Learning Objectives Identify proportional segments when two sides of a triangle are cut by a segment parallel to the third side. Extend
More informationAlgebra I. Book 2. Powered by...
Algebra I Book 2 Powered by... ALGEBRA I Units 47 by The Algebra I Development Team ALGEBRA I UNIT 4 POWERS AND POLYNOMIALS......... 1 4.0 Review................ 2 4.1 Properties of Exponents..........
More informationProblem Set Mash 1. a2 b 2 0 c 2. and. a1 a
Problem Set Mash 1 Section 1.2 15. Find a set of generators and relations for Z/nZ. h 1 1 n 0i Z/nZ. Section 1.4 a b 10. Let G 0 c a, b, c 2 R,a6 0,c6 0. a1 b (a) Compute the product of 1 a2 b and 2 0
More informationMTH 250 Graded Assignment 4
MTH 250 Graded Assignment 4 Measurement Material from Kay, sections 2.4, 3.2, 2.5, 2.6 Q1: Suppose that in a certain metric geometry* satisfying axioms D1 through D3 [Kay, p78], points A, B, C and D are
More informationAPPM 3310 Problem Set 4 Solutions
APPM 33 Problem Set 4 Solutions. Problem.. Note: Since these are nonstandard definitions of addition and scalar multiplication, be sure to show that they satisfy all of the vector space axioms. Solution:
More informationSolutions to Assignment 3
Solutions to Assignment 3 Question 1. [Exercises 3.1 # 2] Let R = {0 e b c} with addition multiplication defined by the following tables. Assume associativity distributivity show that R is a ring with
More informationTRIANGLES CHAPTER 7. (A) Main Concepts and Results. (B) Multiple Choice Questions
CHAPTER 7 TRIANGLES (A) Main Concepts and Results Triangles and their parts, Congruence of triangles, Congruence and correspondence of vertices, Criteria for Congruence of triangles: (i) SAS (ii) ASA (iii)
More informationSection 19 Integral domains
Section 19 Integral domains Instructor: Yifan Yang Spring 2007 Observation and motivation There are rings in which ab = 0 implies a = 0 or b = 0 For examples, Z, Q, R, C, and Z[x] are all such rings There
More informationSect Least Common Denominator
4 Sect.3  Least Common Denominator Concept #1 Writing Equivalent Rational Expressions Two fractions are equivalent if they are equal. In other words, they are equivalent if they both reduce to the same
More informationEXERCISE SET 5.1. = (kx + kx + k, ky + ky + k ) = (kx + kx + 1, ky + ky + 1) = ((k + )x + 1, (k + )y + 1)
EXERCISE SET 5. 6. The pair (, 2) is in the set but the pair ( )(, 2) = (, 2) is not because the first component is negative; hence Axiom 6 fails. Axiom 5 also fails. 8. Axioms, 2, 3, 6, 9, and are easily
More informationTo Find the Product of Monomials. ax m bx n abx m n. Let s look at an example in which we multiply two monomials. (3x 2 y)(2x 3 y 5 )
5.4 E x a m p l e 1 362SECTION 5.4 OBJECTIVES 1. Find the product of a monomial and a polynomial 2. Find the product of two polynomials 3. Square a polynomial 4. Find the product of two binomials that
More informationLINEAR ALGEBRA WITH APPLICATIONS
SEVENTH EDITION LINEAR ALGEBRA WITH APPLICATIONS Instructor s Solutions Manual Steven J. Leon PREFACE This solutions manual is designed to accompany the seventh edition of Linear Algebra with Applications
More informationNozha Directorate of Education Form : 2 nd Prep
Cairo Governorate Department : Maths Nozha Directorate of Education Form : 2 nd Prep Nozha Language Schools Geometry Revision Sheet Ismailia Road Branch Sheet ( 1) 1Complete 1. In the parallelogram, each
More informationCH 37 DOUBLE DISTRIBUTING
CH 37 DOUBLE DISTRIBUTING 343 The Double Distributive Property W hat we need now is a way to multiply two binomials together, a skill absolutely necessary for success in this class. For example, how do
More informationA. H. Hall, 33, 35 &37, Lendoi
7 X x >  z Z  »»x  % x x» [> Q  ) < %   7» Q 9 Q # 5  z > Q x > z» ~»  x " < z Q q»» > X»? Q ~   % % <  <   7  x X   6 97 9
More informationUnit 3 Factors & Products
1 Unit 3 Factors & Products General Outcome: Develop algebraic reasoning and number sense. Specific Outcomes: 3.1 Demonstrate an understanding of factors of whole number by determining the: o prime factors
More informationMidterm1 Review. Jan 24 Armita
Midterm1 Review Jan 24 Armita Outline Boolean Algebra Axioms closure, Identity elements, complements, commutativity, distributivity theorems Associativity, Duality, De Morgan, Consensus theorem Shannon
More informationChapter 2 Combinational Logic Circuits
Logic and Computer Design Fundamentals Chapter 2 Combinational Logic Circuits Part 1 Gate Circuits and Boolean Equations Chapter 2  Part 1 2 Chapter 2  Part 1 3 Chapter 2  Part 1 4 Chapter 2  Part
More informationLesson 7: Algebraic Expressions The Commutative and Associative Properties
: Algebraic Expressions The Commutative and Associative Properties Four Properties of Arithmetic: The Commutative Property of Addition: If a and b are real numbers, then a + b = b + a. The Associative
More informationExpanding brackets and factorising
Chapter 7 Expanding brackets and factorising This chapter will show you how to expand and simplify expressions with brackets solve equations and inequalities involving brackets factorise by removing a
More informationLesson 13: More Factoring Strategies for Quadratic Equations & Expressions
: More Factoring Strategies for Quadratic Equations & Expressions Opening Exploration Looking for Signs In the last lesson, we focused on quadratic equations where all the terms were positive. Juan s examples
More informationChapter 1: Foundations for Algebra
Chapter 1: Foundations for Algebra 1 Unit 1: Vocabulary 1) Natural Numbers 2) Whole Numbers 3) Integers 4) Rational Numbers 5) Irrational Numbers 6) Real Numbers 7) Terminating Decimal 8) Repeating Decimal
More informationREVIEW Chapter 1 The Real Number System
REVIEW Chapter The Real Number System In class work: Complete all statements. Solve all exercises. (Section.4) A set is a collection of objects (elements). The Set of Natural Numbers N N = {,,, 4, 5, }
More informationChapter 3. Boolean Algebra. (continued)
Chapter 3. Boolean Algebra (continued) Algebraic structure consisting of: set of elements B binary operations {+, } unary operation {'} Boolean Algebra such that the following axioms hold:. B contains
More informationBoolean Algebra & Logic Gates. By : Ali Mustafa
Boolean Algebra & Logic Gates By : Ali Mustafa Digital Logic Gates There are three fundamental logical operations, from which all other functions, no matter how complex, can be derived. These Basic functions
More informationOn a class of threevariable inequalities. Vo Quoc Ba Can
On a class of threevariable inequalities Vo Quoc Ba Can 1 Theem Let a, b, c be real numbers satisfying a + b + c = 1 By the AM  GM inequality, we have ab + bc + ca 1, therefe setting ab + bc + ca = 1
More informationMatrix Algebra. Matrix Algebra. Chapter 8  S&B
Chapter 8  S&B Algebraic operations Matrix: The size of a matrix is indicated by the number of its rows and the number of its columns. A matrix with k rows and n columns is called a k n matrix. The number
More informationPage 1 of 11 Name: 1) Which figure always has exactly four lines of reflection that map the figure onto itself? A) rectangle B) square C) regular octagon D) equilateral triangle ee4caab3  Page 1 2) In
More information12 16 = (12)(16) = 0.
Homework Assignment 5 Homework 5. Due day: 11/6/06 (5A) Do each of the following. (i) Compute the multiplication: (12)(16) in Z 24. (ii) Determine the set of units in Z 5. Can we extend our conclusion
More informationA Note on UNAR LASemigroup
Punjab University Journal of Mathematics (ISSN 10162526) Vol. 50(3)(2018) pp. 113121 A Note on UNAR LASemigroup Muhammad Rashad Department of Mathematics, University of Malakand, Pakistan, Email: rashad@uom.edu.pk
More informationchapter 1 vector geometry solutions V Consider the parallelogram shown alongside. Which of the following statements are true?
chapter vector geometry solutions V. Exercise A. For the shape shown, find a single vector which is equal to a)!!! " AB + BC AC b)! AD!!! " + DB AB c)! AC + CD AD d)! BC + CD!!! " + DA BA e) CD!!! " "
More informationCSE 167: Introduction to Computer Graphics Lecture #2: Linear Algebra Primer
CSE 167: Introduction to Computer Graphics Lecture #2: Linear Algebra Primer Jürgen P. Schulze, Ph.D. University of California, San Diego Spring Quarter 2016 Announcements Project 1 due next Friday at
More informationLesson 7: Algebraic Expressions The Commutative, Associative and Distributive Properties
Hart Interactive Algebra 1 Algebraic Expressions The Commutative, Associative and Distributive Properties Exploratory Activity In elementary school you learned that 3 + 4 gives the same answer as 4 + 3
More informationChapter 2: Switching Algebra and Logic Circuits
Chapter 2: Switching Algebra and Logic Circuits Formal Foundation of Digital Design In 1854 George Boole published An investigation into the Laws of Thoughts Algebraic system with two values 0 and 1 Used
More informationSTRAND J: TRANSFORMATIONS, VECTORS and MATRICES
Mathematics SKE, Strand J STRAND J: TRANSFORMATIONS, VECTORS and MATRICES J4 Matrices Text Contents * * * * Section J4. Matrices: Addition and Subtraction J4.2 Matrices: Multiplication J4.3 Inverse Matrices:
More information