basi

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)

3. Algebraic Set

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

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

19. Some additional laws for
complements
The following proposition states five more
important laws of set algebra, involving
complements

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

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)

40. EXERCISE

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 ?

43. Petunjuk untuk soal no 2

45. Kesimpuan soal no 2

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

51. That’s ALL

52. OM SANTIH SANTIH
SANTIH OM

53. DISKUSI