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Algebraic Set

The Fundamental law of set algebra

The principle of Duality

FIELDS

Introduction of Algebraic set

The algebra of insclusion

Some additional laws for union and

intersection

The algebra of relative complements

EXERCISE

Closure PropertiesA1 : a + b is any number

M1 : a b is any number

Asosiative PropertiesA2 : ( a + b ) + c = a + ( b + c )

M2 : (a b) c = a ( b c )

5

Algebraic set has an analogy with the algebraic properties of

arithmetic. Arithmetic operation on the algebra is the addition (+)

and multiplication( )

The properties of mathematical operations on the algebra, ie a, b,

c, is any number

Identity Properties

A3 : There is a unique number that is zero (0) such that for all numbers that apply :

a + 0 = 0 + a = a

M3 : There is a unique number that is 1 such that for all numbers that apply:

a 1 = 1 a = a

Inverse Properties

A4 : For every number there is a unique number (-a) such that the applicable :

a + (-a) = (-a) + a = 0

M4 : For any numbers a ≠ 0, there is a unique number (1/a) such that the applicable a 1/a = 1/a a = 1

commutative Properties

A5 : a + b = b + a

M6 : a b = b a

distributive

D1 : a ( b + c ) = ( a b ) + ( a c )

D2 : (a + b) c = ( a c ) + ( b c )

The properties of algebra apply also in the set where there is a change

Addition operator (+) is replaced by symmetric difference operator(Δ),

Multiplication operator ( ) is replaced by intersection operator ( ),

Unique nature of the M4 numbers zero (0) replaced the set , a unique number of the universe replaced the set S,

Unique Numbers A4 (-a) is replaced by A ', such that the applicable,

A A’ = S A A’ =

• The algebra of sets develops and describes the basic properties and laws of sets, the set-theoretic operations of union, intersection, and complementation and the relations of set equality and set inclusion

The algebra of sets

The fundamental laws of set algebra

The principle of duality

Duality principle : two different concepts are interchangeable but still give the correct answer.

Example: US steering of the car in the front left

England (also Indonesia) Steering of the cara in the front right

Rules:

(a) United State,

- Cars must run on the right side of the road,

- On the road a lot side, the left lane to pass,

- When the red light, the car can turn right directly

(b) England(UK)

- Cars must run on the left side Of the road,

- On the road a lot side, the right to pass

- When the red light, the car can turn left directly

Principle of duality:

The concept of left and right can be interchanged in the two countries so

that the regulations in force in the United States to be valid also in the UK

(Duality principle of set )Suppose S is a similarity (identity) and which involves the set of operations such as Union, intersection, and complement. If S * is obtained from S by replacing

while the complement is left as before, then the similarity S * is also true and is called the dual of the similarity of S.

Exercise. Dualty form of (A B) (A B) = A IS

(A B) (A B) = A.

Some additional laws for unions and intersections

The following proposition states six more important laws of set algebra, involving unions and intersections

Some additional laws for complements

The following proposition states five more important laws of set algebra, involving complements

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Inclusion-Exclusion Principle

|A1 A2| = |A1| + |A2| - |A1 A2|

A B = A + B – 2 A B

• How to calculate the number of members in the union of two sets of up to?

• We can use Inclusion-Exclusion Principle to calculate the number of member in the union of two sets of up to.

Example . How many integers between 1 and 100 are divisible by 3 or 5?

Let :

A = set of integers is divisible by 3,

B = the set of integers divisible by 5,

A B = set of integers is divisible by 3 and 5 (i.e the set of integers is

divisible by LCD (least Common Division- of 3 and 5, namely 15),

Find A B .

A = 100/3 =33 A ={3,6,9,12,15,18,….,99}

B = 100/5 =20 B ={5,10,15,20,25,…,100}

A B = 100/15 =6 A B = {15,30,45,60,75,90}

A B = A + B – A B = 33 + 20 – 6 = 47

So, there are 47 numbers is divisible by 3 or 5.

For three sets A, B, and C

A B C = A + B + C – A B –

A C – B C + A B C

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For three sets A, B, and C, applies

A B C = A + B + C – A B –

A C – B C + A B C

For any sets A1, A2, …, Ar, Applies:

A1 A2 … Ar = i

Ai – rji1

Ai Aj +

rkji1

Ai Aj Ak + … +

(-1)r-1 A1 A2 … Ar

Example. How many positive integers not exceeding 1000 are divisible by 5, 7 or 11?

Answer:

Let

P = { Positive integers not exceeding 1000 are divisible by 5}

Q = { Positive integers not exceeding 1000 are divisible by 7}

R = { Positive integers not exceeding 1000 are divisible by 11}

Thus

P ∪ Q ∪ R = {Positive integers not exceeding 1000 are divisible

by 5 or 7 or 11}

P ∩ Q ∩ R = { Positive integers not exceeding 1000 are divisible

by 5, 7 and 11}

P ∩ Q = { Positive integers not exceeding 1000 are divisible by 5 and 7}

P ∩ R = { Positive integers not exceeding 1000 are divisible by 5 and 11}

Q ∩ R = { Positive integers not exceeding 1000 are divisible by 7 and 11}

A = 1000/5 = 200

B = 1000/7 = 142

C = 1000/11 = 90

A B = 1000/5.7 = 28

B C = 1000/7.11 = 17

A B C = 1000/5.7.11 =12

P ∪ Q ∪ R = 200 + 142 + 90 – 28 – 18 – 12 + 2

= 376.

31

exercise:Among the integers between 101-600 (including 101 and 600 itself), how many numbers are not divisible by 4 or 5 but not both?

32

Answer :

U = 600

A = 600/4 – 100/4 = 150 – 25 = 125

B = 600/5 – 100/5 = 120 – 20 = 100

A B = 600/20 – 100/20 = 30 – 5 = 25

FIND : BA = ?

Calculate

A B = A + B – 2 A B = 125 + 100 – 50 = 175

To Find :

BA = U – A B = 500 – 175 = 325

Prove Using Algebraic Sets

Example. Let A and B are Sets.

Prove that :

(A B) (A B) = A

Prove :

(A B) (A B) = A (B B)

= A U

= A

1. Let A and B are sets. Prove that : A (B – A) = A B

proof:

A (B – A) = A (B A)

= (A B) (A A)

= (A B) U

= A B

2. Prove that for any set A and B, that

(i) A ( A B) = A B and

(ii) A ( A B) = A B

Proof:

(i) A ( A B) = ( A A) (A B) (Distributive law)

= U (A B) (Complement law)

= A B (Identities law)

(ii) Duality form of (i)

A ( A B) = (A A) (A B) (Distributive law)

= (A B) (Complement law)

= A B (Identities law)

EXERCISE

1.In a classroom there are 25 students who love the discrete mathematics, 13 students like linear algebra and 8 of them liked the discrete mathematics and linear algebra. How many students are in class?

2.Berapa banyak bilangan bulat positif yang tidak melampaui 1000 yang habis dibagi oleh5, 7 atau 11 ?

Petunjuk untuk soal no 2

Kesimpuan soal no 2

3. We want to count the number of integers between 1 and 100 is divisible by 3 or 5!

• answerSuppose that:

A = the set of integers divisible by 3

B = the set of integers divisible by 5

A B = set of integers is divisible by 3 x 5 in question is A B

A = {3,6,9,………….,99} n(A)= 33

B = {5,10,15,……….,100} n(B)= 20

A B = {15,30,45,……,90} n(A B) = 6

to obtain:

A ᴜ B = |A | + | B | - | A B | = 33+20-6 = 47

So, there are 47 fruit number is divisible by 3 and 5.

4. In the selection of scholarship recipients,

each student must pass the math and language

tests. Of the 180 participants there were 103

people passed the math test and 142 passed

the language test. Many students who passed

the scholarship recipients there. . . .

Discussion

• n(S) = 180 persons

• n(M) = 103 persons

• n(B) = 142 persons

• n(M B ) = x persons

• n(S) = n( M B ) = n(M) + n(B) – n( M B)

180 = 103 + 142 - X

X = 245 – 180 = 65

• So the pass is 65 persons

That’s ALL

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