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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
22
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
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 = 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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