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OM Swastiastu
    CopyRigth@Microsoft Algebraic
GROUP 6
I Putu Budiana       (1013011067)
Ni Luh Suciari                (1113011082)
Desak Made Pindari Dwi Putra (1113011091)
Gusti Ayu Ari Primandani      (1113011099)
 Ni Wayan Eka Sentana Wati    (1113011105)
Algebraic Set
Introduction of Algebraic set

The Fundamental law of set algebra
 The principle of Duality
Some additional laws for union and
 intersection
The algebra of insclusion
The algebra of relative complements
 FIELDS
                EXERCISE
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


    Closure Properties
     A1 : a + b is any number
     M1 : a b is any number

    Asosiative Properties
     A2 : ( a + b ) + c = a + ( b + c )
     M2 : (a b) c = a ( b c )
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




• 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 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
Inclusion-Exclusion Principle
|A1     A2| = |A1| + |A2| - |A1      A2|

 A      B = A + B –2 A            B




                                       22
• 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
Let :
          3 or 5?

   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
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 =             Ai –              Ai   Aj +
                                i              1 i j r

                                        Ai   Aj          Ak + … +
                            1 i j k r

                             (-1)r-1 A1           A2       …    Ar

                                                                      26
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.
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?




                                                31
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 :      A   B   =?

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

To Find :

            A   B   =U– A     B = 500 – 175 = 325




                                                                 32
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 oleh
  5, 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!
• answer
  Suppose 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
OM SANTIH SANTIH
   SANTIH OM
DISKUSI

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Pdm presentation

  • 1. OM Swastiastu CopyRigth@Microsoft Algebraic
  • 2. GROUP 6 I Putu Budiana (1013011067) Ni Luh Suciari (1113011082) Desak Made Pindari Dwi Putra (1113011091) Gusti Ayu Ari Primandani (1113011099) Ni Wayan Eka Sentana Wati (1113011105)
  • 4. Introduction of Algebraic set The Fundamental law of set algebra The principle of Duality Some additional laws for union and intersection The algebra of insclusion The algebra of relative complements FIELDS EXERCISE
  • 5. 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 Closure Properties A1 : a + b is any number M1 : a b is any number Asosiative Properties A2 : ( a + b ) + c = a + ( b + c ) M2 : (a b) c = a ( b c )
  • 6. 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
  • 7. 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 )
  • 8. 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’ =
  • 9. The algebra of sets • 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
  • 10. The fundamental laws of set algebra
  • 11.
  • 12.
  • 13. The principle of duality Duality principle : two different concepts are interchangeable but still give the correct answer.
  • 14. 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
  • 15. (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.
  • 16. Exercise. Dualty form of (A B) (A B ) = A IS (A B) (A B ) = A.
  • 17. Some additional laws for unions and intersections The following proposition states six more important laws of set algebra, involving unions and intersections
  • 18.
  • 19. Some additional laws for complements The following proposition states five more important laws of set algebra, involving complements
  • 20.
  • 21.
  • 22. Inclusion-Exclusion Principle |A1 A2| = |A1| + |A2| - |A1 A2| A B = A + B –2 A B 22
  • 23. • 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.
  • 24. Example . How many integers between 1 and 100 are divisible by Let : 3 or 5? 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.
  • 25. For three sets A, B, and C A B C = A + B + C – A B – A C – B C + A B C
  • 26. 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 = Ai – Ai Aj + i 1 i j r Ai Aj Ak + … + 1 i j k r (-1)r-1 A1 A2 … Ar 26
  • 27. 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}
  • 28. 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}
  • 29. 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
  • 30. 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? 31
  • 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 : A B =? Calculate A B = A + B –2 A B = 125 + 100 – 50 = 175 To Find : A B =U– A B = 500 – 175 = 325 32
  • 33.
  • 34.
  • 35.
  • 36.
  • 37. 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
  • 38. 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
  • 39. 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)
  • 41. 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?
  • 42. 2.Berapa banyak bilangan bulat positif yang tidak melampaui 1000 yang habis dibagi oleh 5, 7 atau 11 ?
  • 44.
  • 46. 3. We want to count the number of integers between 1 and 100 is divisible by 3 or 5!
  • 47. • answer Suppose 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
  • 48. 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.
  • 49. 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. . . .
  • 50. 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
  • 52. OM SANTIH SANTIH SANTIH OM