TOPIC: REVIEW OF BASIC CONCEPTS OF SET
TOPIC: SETS
- Idea of a set, set notations, and applications.
- Disjoint sets, Venn diagram
IDEA OF SET, NOTATIONS, APPLICATIONS.
Definitions:
A set can be defined as a group or a collection of well defined objects or numbers e.g collection of books, cooking utensils.
A set is denoted by capital letters such as P, Q, and R e.t.c while small letters are used to denote the elements e.g. a, b, c
Elements of a set: These are the elements or members of a given set. The elements are separated by commas and enclosed by a curly bracket {}
e.g M ={ 1, 3 ,5, 7, 11}, 1 is an element of M.
Example: Write down the elements in each of the following sets.
A = {Odd numbers from 1 to 21}
F = {factors of 30}
M = {Multiples of 4 up to 40}
Solution:
A = { 1,3,5,7,9,11,13,15,17,19,21}
F = {1, 3, 5, 6, 10, 15, 30}
M = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40}
Cardinality of a set: This is the number of elements in a set.
Example: Given that µ= {all the days of the week}, B= {all days of the week whose letter begin with s}
- List all the elements of µ
- List the members of B
- What is n (µ) 4. What is n(µ) + n(B)
Solution:
1.µ = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
- B = {Sunday, Saturday}
- n (µ) = 7
- n (µ) + n(B) = 7 + 2= 9
Set notation: A set can be described algebraically using inequality and other symbols. E.g B = {x: -10≤x ≤ 3, x is an integer}
Example: List the members of the following sets
- A= {x: 5< x< 8} 2. B= {x: 0≤ x≤ 5}
Solution;
- A = {6, 7} 2. B = {0, 1, 2, 3, 4, 5}
GENERAL/REVISION EVALUATION: If µ= {all positive integers ≤ 30}, M= {all even number ≤ 20},
N = {all integers: 10≤ x≤ 30}
Find 1. n (µ) 2.n (N) 3. n (N) + n(s) 4. n (M) + n(N)
- Types of sets:
Finite and Infinite set: Finite set is a set in which all its members can be listed.
Infinite set: An infinite set is a set in which all its members cannot be listed.
Empty (Null) set: A set without any members. It is usually denoted by { } or Ø.
Subset and Supersets: If we have 2 sets A and B such that all the elements in A is contained in B, then A is a subset of B. Subset is denoted by C e.g. A C B. If there is at least one element in set B but not in A, then B is a superset of A.
Disjoint set: Two sets are disjoint when there is no common element between them. i.e no intersection.
A n B = Ø
A B
Universal set: This is a set that contains all the members under consideration for any given problem. It is denoted by µ or €.
Complementary Set: This is a set that contains the members in the universal set that are not in set A. It is denoted by Ac or A1.
Intersection of sets: This is the set which consists all the common elements in a given two or more sets. It is denoted by n.
Union of sets: This is the set of all members that belong to A or to B or to both A and B. It is denoted by u.
Example: If the universal set µ= {x: 1≤ x ≤ 12} and its subsets D, F and G are given as follows. D = {x: 2<x<8}, F={x: 4≤ x≤ 10}, G={x: 1< x ≤ 4}
Find (a) D U F (b) D n F (c) G1 (d) (D n G)1
Solution:
µ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
D = {3, 4, 5, 6, 7}
F = {4, 5, 6, 7, 8, 9, 10}
G = {2, 3, 4}
- D U F ={ 3, 4, 5, 6, 7,8, 9,10}
- DnF = {4, 5, 6, 7}
- G1 = { 1, 5, 6, 7, 8,9,10,11,12]
- (D n G)1
D n G = {3, 4}
(D n G)1 = {1, 2, 5, 6, 7,8, 9,10,11,12}
Relationship between union and intersection of sets
n(A or B) = n( A) + n( B) –n( A and B)\
or n(AUB) = n(A) + n( B) – n(A n B)
Example:
If n(A)=23, n(B)= 12, n( AUB) = 35, find n(AnB) and comment on set A and B.
Solution
n(AUB) = n( A) + n( B) – n(AnB)
- = 23 + 12 –n(AnB)
n(AnB) = 35- 35
n (AnB) = 0
Set A and B are disjoint.
Evaluation:
- A and B are two sets. The number of elements in AUB is 49, the number in A is 22 and the number in B is
34.How many elements are in AnB?
- The universal set µ ={ set of all integers}, p= {x:x≤ 2}, Q= { x: -7≤ x ≤15}R ={x: -2 ≤x ≤ 19}
Find 1. PnQ 2. P n (Q UR1)
Venn diagrams:
The Venn diagram is a geometric representation of sets using diagrams which shows different relationship between two or more sets. In order words, it is the diagrammatical representation of relationships between two or more sets. The operations of intersection, union and complementation of sets can be demonstrated by using Venn diagrams.
Venn diagram representation
E or U
The rectangle represents the universal set i.e E or U
A
The oval shape represents the subset A.
The shaded portion represents the complement of set P i.e p1 or Pc
The shaded portion shows the elements common to A and B i.e A∩B or A intersection B.
The shades portion shows P intersection Q| i.e P∩Q|
The shaded portion shows A Ʋ B i.e A union B
U or E
This shows that P and Q have no common element. i.e P and Q are disjoint sets i.e P∩Q= Ф
Q
P
P is a subset of Q i.e P C Q
U
P| ∩ Q| or (P Ʋ Q)|. This shows elements that are neither in P nor Q but are represented in the universal set
This shows the element common to set P,Q and R i.e the intersection of three sets P,Q and R i.e P∩Q∩R
This shows the elements in P only, but not in Q and R i.e P∩Q|∩R|
This shaded region shows the union of the three sets i.e PƲQƲ R
Use of Venn diagrams to solve problems involving two sets
Examples:
- Out of the 400 final year students in a secondary school, 300 are offering Biology and 190 are offering Chemistry. If only 70 students are offering neither Biology nor Chemistry. How many students are offering (i) both Biology and Chemistry? (ii) At least one of Biology or Chemistry?
Solution
n(E)= 400
Let the number of students who offered both Biology and Chemistry be X i.e (B∩C)= X. from the information given in the question
n(E)= 400
n(B)= 300
n(C)= 190
n(BƲC)|= 70
since the sum of the number of elements in all region is equal to the total number of elements in the universal sets, then:
300 – x + x +190 – x + 70 =400
560 – x= 400
-x= 400 – 560
X= 160
Number of students offer both Biology and Chemistry= 160
(ii)no of students offering at least one of biology and chemistry from the Venn diagram this includes those who offered biology only, chemistry only and those whose offered both i.e
300 – x + 190 – x + x= 490
490 – 160 (from (i) above)= 330
- In a youth club with 90 members, 60 likes modern music and 50 likes traditional music. The member of them who like both traditional and modern music are three times those who do not like any type of music. How many members like only one type of music
Solution
Let the members who do not like any type of music = X
Then,
n(TnM)= 3X
Also,
n(E)= 94
n(M)=60
n(T)= 50
n(MƲT)|= X
n(E)= 94
aa
Since the sum of the number of elements in all region is equal to the total number of elements in the universal set, then
60 – 3X + 3X + 50 – 3X = 94
110 – 2X= 94
16= 2X
Divide both sides by 2
16= 2X
2 2
X= 8
Therefore number of member who likes only one type of music are those who like modern music only + those who like traditional music only
60 -3x + 3X + 50 – 3X= 110
110 – 6 x 8 (from above)
= 110 – 48
= 62
Two Venn diagram;
µ
A B
1 2 3
4
Where 1 = AnBI , 2 = AnB, 3 = AInB, 4 = ( AUB)I
Therefore, µ = 1 + 2 + 3 + 4
µ = n( AnBI) + n( AnB) + n( AInB) + n( AUB)I
Example 1: In a class of 40 students, every student had to study French or Russian or both subjects. 25 students studied French and 20 studied Russian. Find the number of students who studied both languages.
Solution:
Let µ = {All the students}
F = {French students}, R = {Students studying Russian}
µ = 40, n(F)= 25, n(R) = 20
n( Fn R)= x
n(FnRI) = 25-x
n (Rn FI)= 20- x
µ = 25 –x +x + 20-x
40= 45 –x
x = 45- 40
x=5, n(FnR) = 5 students.
Evaluation
Two questions A and B were given to 50 students as class work 23 of them could answer question A but not B. 15 of them could answer B but not A. If 2x of them could answer none of the two questions and 2 could answer both questions.
- Represent the information in a Venn diagram. (b) Find the value of x
General evaluation
- In a senior secondary school, 90 students play hockey or football. The numbers that play football is 5 more than twice the number that play hockey. If 5 students play both games and every students in the school plays at least one of the game. Find:
- The number of students that play football
- The number of student that play football but not hockey
- The number of students that play hockey but not football
- A, B and C are subset of the universal set U such that
U={0,1,2,3,4………….12}
A={X: 0≤X7} B= {4,6,8,10,12} C= {1<y<8} where Y is a prime number.
- Draw a venn diagram to illustrate the information
- Find (i) BƲC (ii) AB∩C
Reading assignment: NGM bk1 pg 89 – 92 and Ex 8d number 11 pgs 91- 92
Weekend Assignment
- Given that µ= {-10≤ x ≤ 10}, p= { -10 < x< 10}, Q= { -5 < x ≤ 3}. Which of the following is correct? I PI n Q II P U Q =µ III PI C QI
- I and II only B. I and III only C II and III only
- P and Q are subsets of the µ={x is an integer and 1< x < 15}, P= { x is odd} and Q= { x is prime}, find n(PI n QI) A. 3 B. 4 C. 5
Use the information below to answer question 3 and 4, µ= {1, 2, 3… 10}, A= {2, 4, 6, 8, 10} B= {1, 3, 9} and C = {2, 5, 7}
- AI n C is A.{5, 7} B. { 1, 3, 4} C. { 6,7,8,9}
- BI U C A.{2,4,5,7,8,10} B.{2,4,5,6,7,8,10} C.{ 1, 2,3,4,5, 9}
- A set contains 7 members; find the number of subsets that can be obtained from it. A. 32 B. 64 C. 128
Theory
- 1. During one year in a school, 5/8 of the students had measles, ½ had chickenpox, and 1/8 had neither. What fraction of the school had both measles and chickenpox?
- In a class of 50 pupils, 24 like oranges, 23 like apples and 7 like the two fruits.
- How many do not like oranges and apples (b) What percentage of the class like apples only