Chapter v - Algebraic Structures II

8/4/2019 Chapter v - Algebraic Structures II

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Rings and fiels

Algebraic structures - Part II

NGUYEN CANH Nam1

1Faculty of Applied MathematicsDepartment of Applied Mathematics and Informatics

Hanoi University of Technologies

[email protected]

HUT - 2010

NGUYEN CANH Nam Mathematics I - Chapter 5
http://goforward/http://find/http://goback/


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Agenda

1 Rings and fiels

Rings

Fields

Ring of integers

Euclidean Algorithm

Presentation of integers



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Euclidean Algorithm


Agenda

1 Rings and fiels

Rings

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Euclidean Algorithm




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RingsDefinition

Definition

Let V be a set with two binary operators usually as addition andmultiplication (+ and ). Then V is a ring if

i) (V, +) is an Abelian group with the identity 0.

ii) (closure for multiplication) If x, y V then x y V.

iii) (associative law) For x, y, z V, (x y) z = x (y z),

iv) (distributive laws) For x, y, z V,x (y + z) = x y + x z, (x + y) z = x z + y z.

If V has the property : x y = y x for x, y V then V is called

commutative.

If V has the identity for the multiplication, V is called a ring with identity.NGUYEN CANH Nam Mathematics I - Chapter 5


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RingsExamples

Example

a) (ZZ, +, ), (OQ, +, ), (IR, +, ) are commutative rings with identity.

b) Let E be the set of even integers with usual addition andmultiplication. We have known that (E, +) is an Abelian group.Since the associativity, distributivity and commutativity hold forall integers and, therefore, are true whenever a, b, c are even.Consequently, E is a commutative ring. E does not have an

identity, however, because no even integer e has the propertythat ae = a= eafor every even integer.

c) The set of odd integers with the usual addition and multiplicationis not a ring. Because, as we have known, the set of oddintegers is not closed under addition.



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RingsExamples

Example

e) Let X be a non-empty set. Let R be the set of mappingsfrom X to IR. We introduce an addition and multiplication

into R as follows. Let f : X IR and g : IR. Define thesum f + g and the product f g as mapping from X to IR by,

(f + g)(x) = f(x) + g(x) (x X),

(f g)(x) = f(x)g(x) (x X).

We should note that on the right-hand side of the first

equation the plus sign + refers to addition in IR and on theright-hand side of the second equation the product is in IR.

R is a commutative ring.NGUYEN CANH Nam Mathematics I - Chapter 5

Ri


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Agenda

1 Rings and fiels

Rings

Fields

Ring of integers

Euclidean Algorithm



Rings


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FieldsDefinition

Definition

Let F be a ring. We say that F is a field if

i) F is a commutative ring with identity 1.

ii) For x= 0, y= 0, we have x y= 0

iii) For every x= 0, there exists the inverse element x1 suchthat x x1 = 1.

Remark

If F is a field, then F = F\ {0} is group under the multiplication


Rings


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Euclidean Algorithm


FieldsExamples

Example

a) (IR,+, ) is a field,

b) (OQ,+, ) is a field.

c) Z3 = {0, 1, 2}, where0 = {3n | n ZZ}, 1 = {3n+1 | n ZZ}, 2 = {3n+2 | n ZZ}.

Defind addition and multiplication operations as followsi + j = i + j, i j = i j. Then Z3 is a field.


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Euclidean Algorithm


CharacteristicDefinition

Let F be a field, x is an element of F the sum

x + x + + x (k terms )

is denoted by kx.

Definition

Let F be a field, e be the identity element of F. If p is the

smallest natural number such that pe = 0 then p is called thecharacteristic of the field F. If ke= 0 for every natural number kthen the characteristic is zero.


Rings


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CharacteristicExamples

Example

a ) Z3 = {0, 1, 2}, since 1 + 1 + 1 = 3.1 = 0 . Characteristicof Z3 is 3.

b) Field IR of real numbers has the characteristic zero.

c) Field OQ of rational numbers has the characteristic zero.

Theorem

Assume that p is the characteristic of a field. If p= 0 then p is aprime number.


Rings


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Euclidean Algorithm


Agenda

1 Rings and fiels

Rings

Fields

Ring of integers

Euclidean Algorithm



Rings


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g

Fields

Ring of integers

Euclidean Algorithm


DivisorsDefinition

Definition

Let m and n be integers with b being nonzero. We say that m

divides n and write m | n if there exists an integer k such that

n = km. We also say in this case m is a divisor of n and n is amultiple of m.

Notice that 0 is divisible by any integer b, b= 0, since 0 = b0and that any integer a= 0 has so-called trivial divisors 1,a.

Example

a) 3 is a divisor of 6, and 6 is a multiple of 3. We denote 3 | 6.

b) 7 has the trivial divisors 1,7. There are no otherdivisors.


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g

Fields

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Euclidean Algorithm


Binary operators

Definition

A strictly positive integer p= 1 is said to be a prime number or,simply, a prime if p has only trivial divisors.

Theorem

If a, d are natural numbers, d is a nonzero then there existunique unique integers q and r such that

a= qd + r, 0 r < d.

The number q is called the quotient and r is called the

remainder. If r = 0 then d is a divisor of aand a is a multiple ofd.


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Fields

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Euclidean Algorithm


Congruence

Definition

Let n be a given non zero integer. Two integers aand b are said to becongruent modulo n if a b is a multiple of n. We write

a b mod m.

Otherwise we say that a is not congruent to b modulo n if n does notdivide a b.

Remark

Two integers a, b are congruent modulo m if and only if they have the

same remainder after dividing by the modulo m.

Example

a) 3 8 mod 5,

b) 3 17 mod 10.NGUYEN CANH Nam Mathematics I - Chapter 5

Rings


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Fields

Ring of integers

Euclidean Algorithm


Congruence

Proposition

The congruence modulo m relation is an equivalence relation.

Proposition

If a b mod m and c d mod m then

i) (a+ c) (b+ d) mod m,

ii) (a c) (b d) mod m,

iii) ac bd mod m.


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Fields

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Euclidean Algorithm


Congruence

Definition

The set of integers congruent to an integer i modulo m is called

the congruence class of i modulo m. This class is denote by i.

The set of congruence class modulo m is denoted by ZZm orZZ/mZZ. Thus ZZm = {0, 1, 2, , m 1}.Now we define the addition and multiplication on ZZm as follows

a+ b = a+ b, a.b = ab

Proposition

ZZm with the above addition and multiplication becomes a

commutative ring.


Rings

Fi ld


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Fields

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Euclidean Algorithm


Agenda

1 Rings and fiels

Rings

Fields

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Euclidean Algorithm



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Euclidean Algorithm


gcdDefinitions

Definition

Let aand b be integers, not both 0. The greatest common

divisor (gcd) of aand b is the largest integer d that divides both

aand b. In other words, d is the gcd of aand b provided that

i) d | aand d | b,

ii) if c | aand c | b then c d.

Denote the greatest common divisor of aand b by GCD(a, b).

Definition

If GCD(a, b) = 1 then a, b are said to be coprime or relativelyprime.


Rings

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Euclidean Algorithm


gcdExamples

Example

a) GCD(6, 9) = 3,

b) GCD(180, 315) = 45,c) GCD(315, 143) = 1. So 315 and 143 are copime.

Proposition

Let a and b be integers, not both 0, and GCD(a, b) = d. Thereexists integers m, n (not necessarily unique) such thatam+ bn = d. Furthermore, d is the smallest positive integerthat can be written in the form am+ bn.


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Euclidean Algorithm


lcdDefinition

Definition

For two natural numbers a, b the lowest common denominator(lcd) is the least common multiple of aand b.

Example

a) LCD(6, 9) = 18,b) LCD(180, 315) = 1260.


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Euclidean Algorithm


gcd and lcdProperties

Proposition

For natural numbers a, b we have

ab = GCD(a, b).LCD(a, b).

Proposition

Suppose that natural numbers a, b, q, r satisfy the formula

a= bq+ r.

Then

GCD(a, b) = GCD(b, r).


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Euclidean Algorithm


Euclidean Algorithm.

Introduction

For any integer b, we know that b and b have the samedivisors. Consequently, the common divisors of aand b are the

same as the common divisors of aand b. Therefore, the

greatest common divisors must be the same, that is,GCD(a, b) = GCD(a,b). Using similar arguments, we seethat

GCD(a, b) = GCD(a,b) = GCD(a, b) = GCD(a,b).

So a method for finding the gcd of two positive integers can be

also used to find the gcd of any two integers.

Moreover, from the above proposition one can obtain a method

of finding GCD of aand b called Euclidean algorithm.


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Euclidean Algorithm



Algorithm

Let aand b be positive integers with a b. If b | a thenGCD(a, b) = b. Otherwise apply the division algorithmrepeatedly as follows:

a= q0b+ r0, 0 < r0 < b

b = q1r0 + r1, 0 r1 < r0r0 = q2r1 + r2, 0 r2 < r1

r1 = q3r2 + r3, 0 r3 < r2

...

This process ends when a remainder of 0 is obtained. This

must occur after a finite number of steps; that is, for someinteger t :

rt2 = qtrt1 + rt, 0 < rt < rt1

rt1 = qt+1rt + 0

Then rt, the last nonzero remainder, is the greatest common

divisor of aand b.NGUYEN CANH Nam Mathematics I - Chapter 5

RingsFields


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Euclidean Algorithm



Example

Example

Find gcd of 1071 and 1029.

Solution.

1071 = 1 1029 + 42,

1029 = 24 42 + 21,

42 = 2 21.

Hence the greatest common divisor of 1071 and 1029 is 21.


RingsFields


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Euclidean Algorithm


Agenda

1 Rings and fiels

RingsFields

Ring of integers

Euclidean Algorithm



RingsFields


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Euclidean Algorithm


Base 10

Introduction

We usually write integers as form 10-adic. For example,

2139 = 2.103 + 1.102 + 3.101 + 9.100.

Given a positive integer b. For a natural number n we have the

expression

n = akbk + ak1b

k1 + + a1.b+ a0, 0 aj < b, ak = 0. ()

Then the presentation () is said to be the expansion of n bybase b, denoted by n = (akak1...a1a0)b.If b = 2 the presentation () is called the binary expansion of n.


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Euclidean Algorithm


Bases

Examples

Example

a) The binary expansion of 35 is

35 = 1 25 + 0 24 + 0 23 + 0 22 + 1 21 + 120

35 = (100011)2.

b) The expansion of 135 by base 4 is

135 = 2 43 + 0 42 + 1 41 + 3 40

135 = (2013)4.


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Euclidean Algorithm


Algorithm for expansion of n by base b.

We now present an algorithm for finding the expansion of a

positive integer n by base b. The algorithm is proceeded

repeatedly as follows:

n = bq0 + r0,

q0 = bq1 + r1,

q1 = bq2 + r2

...

This process ends when a quotient of 0 is obtained. This mustoccur after a finite number of steps; that is, for some integer t :

qt1 = bqt + rt.

Then n = (rmrm1...r1r0)b.


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Euclidean Algorithm


Algorithm for expansion of n by base b.

Example

Example

Represent 1397 by base 8.

1397 = 8.174 + 5

174 = 8.21 + 6

21 = 8.2 + 5

2 = 8.0 + 2

Hence, 1397 = (2565)8.


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Midterm examination

Time : Monday, 25th October 2010Duration : 1h30m

Knowledge : From chapter 1 to chapter 5

Only dictionary (hard copy) is allowed


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Chapter v - Algebraic Structures II