- Given A= U where U is the Universal set.
Let U ={ 1,2,3,4,5,6,7} and A ={ 2,4,6}.
First draw a rectangle to represent the Universal set U.
Second draw a circle (or closed bounded figure) inside the rectangle.
Third write the elements of A inside the closed bounded figure and the remaining elements of U outside it, in the rectangle as shown in the figure below:
Clearly the shaded region representing Ā is i.e. Ā = {1, 3, 5, 7}
1, 3, 5, 7
U
Ā 2, 4 , 6
2.Two intersecting subsets A and B of U.
Let U = {1, 2, 3, 4, 5, 6, 7, 8} and A= {1, 3, 4, 5} and B= {2, 4, 5, 6}.
First draw a rectangle to represent U. Since A and B are overlapping sets, draw two overlapping closed bounded figures inside the rectangle to represent A and B. Thirdly, in the common portion of closed bounded figures, write the elements common to both A and B i.e. A B and A B. The fourth step, in the remaining portion of A, write those elements of A which are not in B, i.e. A-B, the fifth step, in the remaining portion of B write those elements of B which are not in A, i.e. B-A. Lastly, the remaining elements of X are written in the rectangle, outside these closed bonded figures as shown here under:
U
1, 4 , 2
7 3 5 6
8
A B
3. Two intersecting subsets A and B of U representing regions.
From the figure below, (i) regions 1, 2, 3 and 4 together represent the Universal set U.
(ii) Regions 1 and 2 represent set A (iii) regions 2 and 3 represent set B (iv) region 2 represents sets A B (v) regions 1, 2 and 3 represent sets A B, (vi) regions 3 and 4 represent set Ā (Á) and (vii) region 1 represent A-B (complement of B with respect to A), and consists of all elements that are in A and not in B).
U
A B
1 2 3
4
Note that the numbers 1,2,3,4 used here in the Venn diagrams do represent regions and not single elements (members) of sets A and B.
Prove that (A B)' =A' B'.
By observing the Venn-diagram in 3 above, (A B) = {1, 2, 3}. So (A B)' = {4}. Similarly, A' {3, 4}; B' = {1, 4}, and A’ B' = {4}. Since both sides of the equation represent the same region {4} i.e. LHS=RHS of the equation. Hence (A B)' =A' B' (Q.E.D).
- Given: Two disjoint subsets of A and B=A B (mutually exclusive) of the Universal set U.
If we let U= {4, 5, 6, 7, 8, 9, 10, 11, 12}, A= {5, 7, 9} and B= {4, 6, 8}.
10, 11 12 U
5, 7 4 6
9 8
A B
Draw a rectangle to represent the universal set U, inside the rectangle, draw two disjoint closed figures to represent the sets A and B. Then write the members of A and B inside their respective regions and remaining members of the universal set U in the rectangle as shown in the figure above.
- Given: B A U.
Let the universal set U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A= {2, 4, 6, 8, 10} and B= {2, 4, 6}.
1, 3 5 U
7 9 2, 4 6
8
A B
First draw a rectangle to represent the universal set U and then inside the rectangle, draw a closed bounded figure to represent set B. Since B A, then draw a larger closed bonded figure containing the figure representing B. Write the elements (members) of set B inside the smaller closed bounded region and the remaining elements of set A ,in between the remaining two figures. Finally, write the remaining elements of the universal set U in the rectangle, outside these closed bounded figures.
4. By shading appropriate areas, all set combinations can be represented pictorially as shown below:
- A B: (ii) A B:
U U
A B
A B
(iii)A B=Φ (Disjoint sets): (iv) (A B') or A-B:
U U
A B
Mutually exclusive sets A B
Non mutually exclusive sets
(v)(B A') or (B-A): (vi) Ā (A'):
U U
A B A B
(vii)B’ (complement of B):
U (viii)(A B
U
A B A B
(ix)(A B)':
U
A B
By using the Venn –Euler diagram given below find the following sets:
(i)A B (ii) A B (iii) A-B (iv) B-A (v) B' (vi) A’ (vii) (A B)' (viii) (A B)'
4 3 1 5
6 2 10
7 A B
9 8
Solution: We have in:
(i)A B = {elements which are set A or set B or in both sets A and B} = {3, 6, 1, 2, 5, 10}.
(ii)A B = {elements common to both sets A and B} = {1, 2}.
(iii)A-B = {elements which are in set A and not in set B} = 3, 3}.
(iv) B-A = {elements which are in set A and not in set A = {5, 10}.
(v) A' = {members which are in the universal set U and not in set A} = {4, 7, 8, 9, 5, 10}.
(vi) B' = {members which are in the universal set U and not in set B} = {4, 7, 8, 9, 3, 6}.
(vii) (A B)' = {members of the universal set U which are not in sets A B} = {4, 7, 8, 9, 3, 6, 5, 10}.
(viii) (A B)' = {elements of the universal set U which are not in sets A B} =4, 7, 8 9}.
Using the Venn diagram given below, find the following set combinations:
9 10 11 12
1 3 2 4
A 5 7 B 6 8
(i)A B (ii) A B (iii) A-B (iv) B-A (v) A’ (vi) B' (vii) (A B)’ (viii) (A B)'
Solution: We have in:
(i)A B= {elements that are common to both sets A and B = Φ or {} =void set, null set or empty set.
(ii)A B = {elements which are in set A or in set B or both in set A and B} = {1, 3, 5, 7, 2, 4, 6, 8}.
(iii) A-B = {elements in set A which are not in set B} = {1, 3, 5, 7}.
(iv) B-A = {members in set B which are not in set A} = {2, 4, 6 8}.
(v)A' = {members in the universal set which are not in set A} = {9, 10, 11, 12, 2, 4, 6, 8}.
(vi) B' = {members in the universal set which are not in set B} = {9, 10, 11, 12, 1, 3, 5, 7}.
(vii) (A B)' = {members in the universal set which are not in set A B} = {9, 10, 11, 12}.
(viii) (A B)'= {members in the universal set which are not in set A B=Φ'=U= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}.
Using the given Venn diagram below find the following combinations of sets:
U
6 10 11
4 3 9
8 5 7
2
B
A
(i)A B (ii) A B (iii) A-B (iv) B-A (v) A' (vi) B’ (vii) (A B)’ (viii) (A B)'.
Solutions: We have in:
(i)A B = {members common in both sets A and B} = {2, 3, 5}.
(ii)A B = {members which are in set A or in set B or in both set A and B} = {2, 3, 5, 7, 9, 11}.
(iii)A-B= {members which are in set A and not in set B} = {7, 9, 11}.
(iv)B-A = {members which are in set B and not in set A} =Φ =null set, void set, or empty set.
(v) A'= {members which are in the universal set U which are not in set A} = {4, 6, 8, 10}.
(vi) B' = {members which are in the universal set U which are no in B} = {4, 6, 8, 10, 7, 9, 11).
(vii) (A B)' = {elements which are in the universal set which are no in set A B} = {4, 6, 8, 10, 7, 9, 11}.
(viii) (A B)' = {elements which are in the universal set which do not belong to set A B} =4, 6, 8, 10}.
Exercises
- From the given Venn diagram below, find the following:
r s t U
i a p
e u o q
B
A
(i) A (ii) B (iii) A B (iv) A B (v) A-B (vi) B-A (vii) Á (viii) complement B (ix) (A B)' (x) ( A B)'
2. Given the Venn diagram below, find the combination of the following sets:
8 10 16 20 U
3 6
9 12 7 14
15 21 28
S T
(i)S T (ii) S T (iii) T-S (iv) S T' (v) S’ (vi) (S T)' (vii) is (S T)'=S' T'?
4. What set is being represented by the shaded portion in each of the following Venn diagram
U U
- (ii)
A B A B
(iii) U (iv) U
A B A B
(v)
U
A B
5. Let the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16) and set A={x: x is a factor of 12}, set B={x: x is a factor of 16}. Draw a Venn diagram to show the relationship between the sets, and use the diagram the find the combination of the following sets:
(i)A B (ii) A B (iii) (A-B) (B-A) (iv) (A B)' (v) (A B)'
6. Draw Venn diagrams to show the relationship between the following pairs of sets:
(i) Set T= {people who like tea}, set C= {people who like tea}. Condition: it is being given that T C≠Φ.
(ii) Set A= {children who like apples}, set B= {children who like oranges}. Condition: it is being given that A B≠Φ.
(iii) Set P= {prime factors of 105}, set Q= {all factors of 105}.
(iv) Set G= {letters of the word ‘Integers’}, set H= {letters of the word ‘teenagers’}.
(v) Set S= {all schools in Tanzania}, set T= {all Veta schools in Tanzania}
7. Draw Venn diagrams to represent the following situations of sets:
(i) Q P universal set U (ii) A B≠Φ (iii) R S=Φ.
4.2The Number of Elements in a Set
The Notation n (A) defines the number of elements within a set A. For example, if set A = {2, 3, 4, 6,} then the number of elements in set A denoted as n (A) =4.
5.0Cardinal properties of sets
Let set A and set B be two non-empty intersecting sets as shown in the diagram below:
A B
A-B A B B-A
In counting the elements (members) in A B, the elements of A B are counted twice, once in the counting of elements of A and second time in the counting the elements of B. This, thus, introduces a useful rule in calculating the number of elements in the ‘ ’, the union of two sets A and B. Therefore, the rule is: n (A B) =n (A) +n (B) – n (A B).
Note: If A B=Φ, then n (A B) =n (A) +n (B)
It is also clear from the Venn diagram that:
- n (A-B) =n (A) –n(A B) and (ii) n(B-A) = n(B)-n(A B).
The Rule can be easily proved by using Venn diagram as shown below:
A B
1 2 3
4
Note that: 1, 2, 3 and 4 in the Venn diagram are regions and not elements.
The Proof: Set A= {1, 2}, and set B= {2, 3}. The number of elements in set A =n (A) =n ({1, 2}) and the number of elements in set B=n (B) =n ({2, 3}).
But the intersection of set A and B, A B=2, and the n (A B) =n ({2}). Now, n(A)+n(B)-n(A B)=n({1,2})+n({2,3})-n({2})=n({1, 2, 3})= n(A B).
6.0Application of Set theory to Arithmetic and Logical problems
Certain arithmetic problems can often be quickly solved by use of a Venn –Euler diagram. This can be illustrated by the following examples:
1. In a meeting of 20 people, 10 are farmers, 15 are soldiers and 6 are both. (i) How many people are farmers and soldiers? (ii) How many are neither?
Solution:
First draw the Venn diagram to show the relationships of the set interpreted from the problem posed.
20
10 6 15
F S
From the diagram let set F= {farmers}, set S= {soldiers}.
Then n(F) =10, n(S)=15 and n(F S)=6, BUT, n(F S)=n(F)+n(S)-n(F S), then we substitute 10+15 -6=19.
(i)Therefore, there are 19 people who are either farmers or soldiers.
(ii) n (F S)'= n(U) –n(F S)= 20 -19=1 (is a person who is neither a farmer nor a soldier).
2. In the class of 25 pupils, 17 study physics, 10 study chemistry and physics and 6 take neither chemistry nor physics. How many take chemistry?
Solution:
First interpret and represent the information of the problem in the Venn diagram as
Here below:
25
6 10 17
C P
From the diagram let set P= {physics}, set C= {chemistry}.
Then, n (U) =25, n (P) =17, n (P C) =10 n (P C)'=6.
From the n(C) is the number of pupils who take chemistry, but n(C) = n (U) –n (P C)'-n (P) + n (P C) =25-6-17+10=35-23=12.
Those pupils who study chemistry alone are 12.
3. In a class of 45 students, each one plays either crickets or hockey. If 30 of them play cricket, and 27 play hockey, find (i) How many play both the games (ii) How many play crickets only? (iii) How many play hockeys only?
Solution: We first interpret the information posed in the problem and then represent it in the Venn diagram as follows:
C-H C H H-C
C H
From the diagram, let cricket students be set {C} and hockey students be set {H}.
Then C-H = {students who play crickets only} and H-C = {students who play hockey only}. From the information given n(C) =30, n (H) =27 and n(C H) =45.
(i)Applying the rule, n (A B) = [n (A) +n (B) - n (A B)], then n(C H) is the number of students who both play the games. Therefore n(C H) =n(C) +n (H) –n(C H) =30+27-45=57-45=12. (The number of students who play cricket and hockey)
(ii) n(C-H) =n(C)-n(C H), applying n(A-B)=[n(A)-n(A B)]
=30-12=18(the number of students who play cricket only)
(iii)n (H-C)= n(H)-n(C H) =27 -12=15 (the number of students who play hockey only)
Alternative method:
Represent the information in the Venn diagram as here under:
Let set C= {crickets students) and set H= {hockey students) and x = (the number of who play both games. Then the numbers of students who play cricket only are (30-x) and those who play hockey only are (27-x).
45
30-x x 27-x
C H
From the figure and the information given we then form an equation for solving the value of ‘x’. Thus, 30-x +x+27-x =45
57-x=45 -x=45-57=-12 x=12(students who play both games)
Then students who play cricket only are 30-12=18, and students who play hockey only are 27-12=15.
4. An investigation of sporting interest of 100 players yielded the following data: 50 played badminton, 43 played squash, 45 played tennis, and 12 players played badminton and squash; 13 played squash and tennis; 15 played tennis and badminton; 5 played all the three sports. Show if the data is consistent or not.
Solution:
Represent the three games as sets in the Venn diagram as follows:
Let B= {set of badminton playing students}, S= {set of squash playing students} and T= {set of tennis playing students)
U
B S
T
From the information given we have: n(U) =100, n(S) =43, n(T)=45, n(B S)=12, n(S T)=13, n(T B)=15, n(B S T) =5 (all play the three games, they are common to all the circles in the diagram).
Also, since, n (B S T') =n (B S)-n (B S T). We have n (B S T') =12-5=7; similarly, we have n(S T B') = n(S T)-n (B’) 13-5=8; and, n (T B S’) =n(T B)-n(S’ ) 15-5=10. From this information we can draw another Venn diagram to represent it in the corresponding regions.
7
B 5 S
10 T 8
From the figure n (B S’ T’) =n (B)-7-5-10=50=22=28.
Also, n (S T’ B’) =n(S)-7-5-8=43-20=23; and n (T B’ S’) =n (T)-8-5-10=45-23=22. Again these numerical figures can be represented in the Venn diagram as here under
28 7 23
B 5
10 8
T 22 S
Hence, the total number of players is: 28+7+5+10+23+8+22= 103.
Therefore, the data is not consistent since the number of players given is 100.
5.In a battle involving 100 men, 42 were shot in the head, 43 in the arm, 32 in the leg, 5 in the head and arm, 8 in the arm and leg, 6 in the head and leg. How many were shot in all three places?
Solution:
Let set H = {the men shot in the head}, L= {men shot in the leg}and A={men shot in a arm}.
Then we know that, n(H)=42, n(A)=43, n(L)=32, n(H A)=5, n(A L)=8 and n(L H)=6.
From the question, we are asked n (H A L), the number of men who were shot all the three places. If let the number of men shot all the three places be x= n (H A L), then are able to represent this data (figures) in their respective regions as shown in the Venn diagram here under:
From the diagram, (i) 42-(5-x)-x-(6-x) =42-5+x-x-6+x=42-11+x=31+x.
- 43-(5-x)-x-(8-x) =43-5+x-x-8+x=43-13+x=30+x.
- 32-(6-x)-x-(8-x)=32-6+x-x-8+x=32-14+x=18+x
Since we know that the universal set U=100men, the 42+38+18+x=100.
98+x=100 x=2 (men who were shot all the three places:
H
31+X 5-X
X 30+X
6-X 8-X
L 18+X A
Brainstorming: Suppose we add (i) + (ii) + (iii) above and solve for x, do we get the same answer, i.e. x=2? Explain why?
6. In a class of 50 students, 20 have offered painting, but not crafts, and 32 have offered painting. (i) How many have offered crafts but not painting? (ii) How many have offered both?
Solution:
Let set P= {students who have offered painting}, set C= {students who have offered crafts}. Then U =50 and n (P) =32. Therefore, (i) n (50)-n (32) =18(students who offered crafts but not painting).
(ii)Now, n(P-C) =20 , thus, n(P-C)=n(P)-n(P C), but n(P C)=n(P)-n(P-C).
Then we have 30-20=12=n (P C) –students who have offered both.
7. It is found that out of 100 students, 25 can drive neither a scooter nor a car, while 15 can drive both these, and 52 of them can drive a scooter. How many can drive a car?
Solution:
Let set S represent students who can drive a scooter, set C represent students who can drive a car.
Then, n(S) =52, n(S C) =15. The number of students who can drive a scooter or a car or both is: 100=25 =75, thus n(S C) =75.
From the rule: n(S C) =n(S) +n(C)-n(S C); and n(C) =n(S C) +n(S C)-n(S).
Therefore, n(C) =75+15-52=90-52=38(the number of students who can drive a car).
n(S C)
S C
8.In a group of 50 people, 31 like taking tea, 20 like taking coffee, and 6 like both taking tea and coffee.
(i)How many like tea only?
(ii) How many like coffee only?
(iii)How many like none tea and coffee?
Solution:
The given data in the question can be represented in a Venn diagram as follows:
T-C T C C-T
From the figure, let T be the set of people who take tea, and C be the set of people who take coffee. Then, T C= {people who take both tea and coffee}; T-C= {people who take tea only} and C-T = {people who take coffee only}.
From the information given, n (T) =31, n(C) =20 and n (T C) =6; n (U) =50.
- n(T-C)=n(T)-n(T C) applying [n(A-B)=n(A)-n(A B)]
Thus, 31-6 =25 (people who take tea only)
- n(C-T) =n(C)-n(C T) applying [n(B-A)=n(B)-n(B A)]
20-6=14 (people who take coffee only)
- n (T C)= n(T) +n(C)-n(T C)=31+20-6=51-6=45(people who like tea or coffee or both). Then, people who like none tea and coffee will be:
n (U)-n(T C)= n(T' C')=50-45=5.
Exercises:
1. By use of Venn diagrams verify the relationship A (B C)=(A B) (A C)
2. Let F and G be two set such that n(F)=52, n(G)=60 and n(F G)=16. Find:
(i)n(F G) (ii)n(F-G) (iii) n(G-F) (iv) Draw the corresponding Venn diagram for each in (i)-(iv).
3. There is a group of 50 persons who can speak English or Kiswahili or both. Out of these 37 can speak English and 30 can speak Kiswahili.
(i)How many can speak both languages?
(ii)How many can speak English only?
(iii)How many can speak Kiswahili only?
4. In a group of 100 lecturers, 48 drank coffee, 48 drank lemonade, 36 drank tea, 15 drank both coffee and lemonade, 12 drank both lemonade and tea, 10 drank both tea and coffee, and 7 drank coffee, lemonade and tea. Verify if this data is consistent.
7.0Cartesian Products of two ordered pairs
The Cartesian Products is a unique among the mathematical set (product set or simply product) from multiple sets’ operations in that produces new elements that are not members of the Universal set. That is sets A and B, the Cartesian products A x B is the set of all Ordered Pairs (a, b) where a ϵ A and b ϵ B, products can be specified using a set builder notation. An ordered pair of objects where one element is considered first and the other element is considered the second. For example, a pair of socks is not an ordered pair, a pair of shoes not an ordered pair and etc.
A table can be created by taking the Cartesian product of a set of rows and a set of columns. The Cartesian product rows x columns are taken; the cells of the table contain ordered pairs of the form (row value, column value). More generally, Cartesian product of n-sets (known as n-fold Cartesian products) can be represented by an array of n-dimensions, where each element in an n-tuple. An ordered pair is a 2-tuple.
Mathematically, an ordered pair of elements consists of ‘a’ and ‘b’ are denoted as (a, b) letting us that element ‘a’ is considered first, and element, ‘b’ is considered the second. Other examples are: Given set A= {Ali, Juma} and set B= {p, q) find A cross B.
Solution:
First of all we must pair the elements in set A, Ali and Juma with elements in set B, the p and q. Ali may be paired with p or q, hence: the possible ordered pairs with Ali are (Ali, p); (Ali, q). We do the same to Juma, and the possible ordered pairs are (Juma, p) and (Juma, q).Therefore, A cross B= {(Ali, p), (Ali, q), (Juma, p) and (Juma, q)}. Note however that A cross B gives us the set where the elements are ordered pairs. These are not members of the Universal set, since the Universal set consists of single elements such p, q , Ali and Juma.
7.1Definition: Cartesian products is named after Rene’ Descartes, where formulation of analytic geometry gave rise to the concept, which further generalized in terms of direct products. Such products examples are: a deck of card, a two dimensional coordinate system, set theory (most implementation-in intersections, union, subsets, cardinality), non-cummulativity and non- associativity, graph theory and etc.
Example
A deck of cards: The standard playing cards ranks { A, K, Q, J, 10,9, 8, 7, 6, 5, 4, 3, 3, 2} form a 13 element set . The card suits (Ace, Spade, Diamond, and Heart) form a 4 element set. The Cartesian product of these sets returns a 52 element set consisting of 52 ordered pairs which correspond to all 52 possible playing cards. Rank x Suits returns a set form and the Suits x Ranks returns a set form, both sets are distinct, even disjoint.
Example
A two dimensional coordinate system: The historically main example of 2-dimensional system is the Cartesian plane in analytic geometry. In order to represent geometrical shapes in a numerical way and extract numerical information from the shapes ’numerical representations we assign to each point in a plane a pair of real numbers (denoted as ‘R’) called its coordinates. Usually, such pair’s first and second component is called x and y coordinates respectively (simply picture). The set of all such pairs, i.e. the Cartesian product R x R, is thus assigned to the set of all points in the plane.
Example
Set theory: A formal definition of Cartesian products from set theoretical principle follows from a definition of ordered pairs. The most common definition of ordered pairs, the Kuratowski definition, is: (x, y) =\ {\{x\}, \{x, y\}\}.
7.2Intersection
The Cartesian products behaves nicely with respect to intersections,
Given that two sets A, B, C and D such that set A={x ϵ A | 2≤x≤5}, set B={x ϵ B | 3≤x≤7}, set C= {y ϵ C| 1≤y≤3}, and set D ={y ϵ D| 2≤y≤4}, this demonstrates that (A B) x (C D) = (A x C) (B x D).
7.3Union
The Cartesian products with respect to union of the above sets behave as: (A B) x (C D) ≠ (A x C) (B x D).
Given the sets A={y ϵ A | 1≤y≤4}, set B ={x ϵ B| 2≤x≤5} and set C={x ϵ C| 4≤x≤7}, this also demonstrates that A X (B C)= (A x B) x (A C); A x (B C)=(A x B) (A x C) and A x (B\C)=(A x B)\ (A x C). For the set differences we can have the following distributivity with other operators:
- A x (B C)=( A x B) (A x C)
- A x ( B C)=(A x B) (A x C)
- A x (B\ C)=(A x B)\ (A x C)
- (A x B)ͨ =(Aᶜ x Bᶜ) (Aᶜ x B) (A x Bᶜ)
7.4Operation with Subsets
If set A B, then A x C B x C; if both A, B ≠Φ then A x B C x D.
Example
Non-cummulativity and non- associativity
(i)Non-commutative
Let A, B, C, and D be the sets. The Cartesian product A x B products of the sets A and B, denoted by AxB read as A cross B, is the set of all possible ordered pairs such that the first element of the ordered pair is an element of A and the second element of the ordered pair is an element of B and is not commutative, since A x B ≠ B x A, because the ordered pairs are reversed except if at least one the following conditions is satisfied:
- Set A is equal set B OR (ii)A or B is the empty set (Φ)
Given that set A= {4, 8} and set B= {a, b, c} find A cross B.
Solution:
The first ordered pairs must come from set A. Therefore element 4 is paired with element a then 4 is paired with b, and then 4 paired with c. The same procedure is done for element 8. Hence, the results of A cross B= {(4, a), (4, b), (4, c), (8, a), (8, b) (8, c)}.
Given set A= {4, 8} and set B= {a, b, c} find B cross B.
Solution:
We do the same as in the previous example except that the sets have been interchanged. Therefore, B cross A= {(a, 4), (a, 8), (b, 4), (b, 8), (c, 4), (c, 8)}. By comparing the results of A cross B and B cross A we can conclude that the two results are not equal i.e. A x B ≠ B x A, since the two set contain different ordered pairs.
Note that in example 1, the number of elements in A x B =4 i.e. n (A x B) and in example 2 the number of elements in A x B =6 i.e. n (A x B), then we conclude that n(A x B) ≠ n(A x B).
Example
Given that set A ={1, 2} , set B={3, 4} , then A x B= {1, 2}x {3, 4}={(1,3),(1, 4),(2, 3), (2, 4); and B x A ={3, 4}x {1, 2} ={(3, 1), (3, 2), (4,1), (4, 2)}.
If set A =set B ={1, 2} , then A x B = B x A= {1, 2} x {1,2} ={(1, 1),(1, 2), (2, 1), (2, 2)}.
(ii)Non-associative
If set A= {1, 2} and set B=Φ, then A x B = {1, 2} x Φ=Φ, and B x A= {Φ} x {1, 2} =Φ.
Strictly speaking, the Cartesian product is not associative (unless one of the involved sets is empty). (A x B) x C ≠ A x (B x C).
If for example, set A ={1}, then (A x A )x A= {(1, 1),1)} ≠{(1, (1,1)}= A x (A x A).
8.0Cardinality
8.1 Determining the number of Products
In the Cartesian product A x B, the ordinality of sets is the number of elements of a set. If we define two seta A= {a, b} and B= {5, 6}, then both sets A and B consist of two elements each. Their Cartesian product, we normally multiply the number of elements in set A and multiply it by the number of elements in set B. If set A has m elements in it, and set B has n elements in it then the number of elements in A x B is equal to m x n; in other words, n (A x B) = n (A) x n (B). Each of the elements of A is paired with element of B. Each pair makes up one element of the output set. The number of values in each pair is equal to the number of sets whose Cartesian product being taken, 2 in each case. The cardinality of the output set is equal to the product of the cardinalities of all inputs sets. That is |AxB|=|A|.|B|. Similarly, |A x B x C|=|A|.|B|.|C| and etc. The set A x B is infinite if either A or B infinite and the other set is not the empty set.
Example:
Find A x A given the set A = {1, 2, 3}.
Solution:
A x A ={1, 2, 3} x { 1, 2, 3} ={(1, 1), (1, 2), (1 ,3), (2, 1), (2, 2), (2, 3), (3, 1), 3, 2), (3, 3)}. We see that since there are three elements in each of the sets A, there are 3x3 =9 elements in A x A.
8.2Graph Theory
In graph theory, the Cartesian product of two graphs G and H is the graph denoted by G x H whose vertex set is the ordinary Cartesian product V(G) x V(H) and each of the two vertices (u, v) and (u', v') are adjacent in G x H if and only if u=u' and v is adjacent with v' in H or V=v' and u is adjacent with u' in G. The Cartesian product of graphs is not a product in the sense of category theory. Instead, the categorically product is known as tensor products of graphs.
8.3Cartesian products of two sets
The Cartesian products of two sets A x B gives a set of ordered pairs. This set of ordered pairs may be pictured by means of an ARRAY or Lattice (pattern, regular arrangement of objects
Consider the following example.
Find A x B if set A= {a, b, c) and set B= {d, e}
Solution:
There are 3 elements in set A and two in set B, then there are 6 ordered pairs of Cartesian products ,i.e. A x B ={a, b, c} x { d, e} ={(a, d), (a, e), (b, d), (b, e), ( c, d) , (c ,e)}. Now representing the results in a pictured form of an array or lattice, this can be shown as here under:
(Second) B
e (a, e) (b, e) (c, e)
(a, d) (b, d) (c, d)
A (first)
This is an array of ordered pairs in the lattice each point representing an ordered pair. It is traditional to use a horizontal axis to represent the first element in an ordered pair ‘
We can as well picture the Cartesian product in another way altogether. By using a Tree diagram consisting of a number of branches to illustrate the possible pairings of A x B.
Given that set A= {a, b, c} and set B = {d, e).
Set B A x B
Set A d (a, d)
a
e (a, e)
d (b, d)
b
e (b, e)
d (c, d)
c
e (c, e)
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