Unit Set
1
1.1 Set - Looking back
Classroom - Exercise
1. Let’s tell and write the elements of these sets.
a) A = {first five multiples of 3 }, A = {....................................................}
b) B = { common factors of 10 and 20}, B = {....................................................}
c) C = { x : x ≤ 15, x ∈ prime numbers }, C = {....................................................}
2. Let’s rewrite these sets in set-builder forms.
a) M = {2, 4, 6, 8}, M = ........................................................................................
b) N = {3, 6, 9, 12}, N = .........................................................................................
3. a) Let’s write all possible subsets of the set S = (1, 4, 9)
............................ ............................ ............................ ............................
............................ ............................ ............................ ............................
b) Among these subsets, let’s write an improper subset .............................................
4. Let’s tell and write the members of the following set operations from the
Venn-diagram given alongside.
a) P ∪ Q = {.........................................................} PQ
b) P ∩ Q = {.........................................................}
c) P – Q = {.........................................................} 2
13
5 11
7
9 13
d) Q – P = {.........................................................}
5. If X = {1, 3, 5, 7, 9} and Y = {2, 3, 5, 7, 11}, tell and write the elements of these
set operations.
a) X ∪ Y = {.....................................} b) X ∩ Y = {.....................................}
c) X – Y = {.....................................} d) Y – X = {.....................................}
Let’s take a collection of even numbers less than 10. The members of this collection
are definitely 2, 4, 6, 8. These members are distinct objects when considered
separately. However, when they are considered collectively, they form a single set
of size four, written {2, 4, 6, 8}. It is a set of even numbers less than 10.
Here, any even number less than 10 is definitely the member of the set.
Therefore, a set is a collection of ‘well-defined objects’.
5 Vedanta Excel in Mathematics - Book 8
Set
1.2 Notation of set
We usually denote sets by capital letters. The members or elements of a set are
enclosed inside the braces { } and the members are separated with commas. The
table given below shows a summary of the symbols which are used in the notation
of sets.
Symbol Name Example Explanation
{} Set W = {0, 1, 2, 3, 4} The members of the sets
are enclosed inside braces
∈ Membership O = {1, 3, 5, 7, 9} { } and separated with
∉ Non- commas.
⊂ 1 ∈ W, 4 ∈ W, The symbol '∈' denotes the
⊆ membership membership of an element
⊃ Proper 5 ∈ O, 9 ∈ O of the given set.
subset 5 ∉ W, 6 ∉ W, The symbol '∉' denotes
the non-membership of an
Improper 2 ∉ 0, 4 ∉ 0 element to the given set.
subset {0, 3} ⊂ W, {1, 2, 4} ⊂ W A set which is contained in
{1} ⊂ O, {1, 5, 7, 9} ⊂ O another set.
Super set {0, 1, 2, 3, 4} ⊆ W, It A set which is contained in
means {0, 1, 2, 3, 4} ⊂ W or equal to another set.
and {0, 1, 2, 3, 4} = W
W ⊃ {0, 1, 2} Set W includes {0, 1, 2,}
and set O includes {5, 7, 9}
O ⊃ {5, 7, 9}
1.3 Methods of describing sets
We usually use three methods to describe a set. These methods are description,
listing (or roster) and set-builder (or rule) methods.
Method Example Explanation
Description
Listing M is a set of multiples of The common properties of elements
(or roster) 3 less than 15. of a set are described by words.
Set-builder
(or rule) M = {3, 6, 9, 12} The elements of a set are listed inside
braces (or curly brackets) { }.
M = {x : x ∈ multiples of A variable such as ‘x’ is used to
3, x < 15} describe the common properties of
the elements of set by using symbols.
1.4 Cardinal number of sets
Let’s consider a set A = {2, 3, 5, 7}. Here, the cardinal number of the set A
represented by n(A) is 4. Thus, the number of elements contained by a set is called
its cardinal number.
Similarly, if B = {0, 1, 2, 3, 4, 5}, then n(B) = 6 and so on.
Vedanta Excel in Mathematics - Book 8 6
Set
1.5 Types of sets
On the basis of the number of elements of sets, there are four types of sets: null or
empty set, unit or singleton set, finite set and infinite set.
Types of sets Examples Explanation
Null or empty The set of odd numbers It does not contain any element.
set between 5 and 7. It is denoted by empty braces { }
Unit or or by φ (phi)
singleton set A = { } or A = φ
Finite set The set of prime numbers It contains only one element.
between 5 and 9.
Infinite set P = {7} It contains finite numbers of
The set of natural numbers elements. It means counting of
less than 50. elements can be ended.
N = {1, 2, 3, 4, 5, … 49} It contains infinite number of
elements. It means, counting of
The set of natural numbers. elements is never ended.
N = {1, 2, 3, 4, 5, …}
On the basis of the types of elements contained by two or more sets, the types of
their relationship can be defined in the following ways.
Types of Examples Explanation
relationship
Equal sets A = {a, e, i, o, u} Equal sets have exactly the
Equivalent B = {o, i, u, a, e} same elements.
sets ∴A=B
Equivalent sets have the
Overlapping P = {2, 3, 5, 7, 11} equal cardinal numbers,
sets Q = {2, 4, 6, 8, 10} i.e. they have the equal
n(P) = n(Q) = 5 number of elements.
Disjoint sets ∴P~Q ‘~’ is the symbol used to
denote equivalent sets.
C = {1, 3, 5, 7, 9}
D = {1, 2, 3, 4, 5} elements is Overlapping sets have at
Set of common least one element common.
{1, 3, 5}
∴ C and D are overlapping sets
M = {s, v, u, 3, ª}
N = {c, cf, O, O}{ Disjoint sets do not have
any element common.
There are no element common.
∴ M and N are disjoint sets.
7 Vedanta Excel in Mathematics - Book 8
Set
1.6 Subset, proper and improper subsets and universal set
Subset
Let’s consider any two sets A = {1, 2, 3, ... 10} and B = {1, 3, 5, 7}. Here, every element
of B is contained by A. In other words, the set B is contained by the set A. So, set B is
called the subset of set A.
Thus, between two sets A and B, the set B is said to be a subset of A if every element of
B is contained by A. It is denoted as B ⊂ A. Here, A is said to be the super set of B and
it is denoted as A ⊃ B.
Proper subset
Let's consider any two sets A = {1, 2, 3, 4, 5, … 10} and B = {2, 4, 6, 8}, where B is a
subset of A. Here, B is called a proper subset of the set A.
Thus, between any two sets A and it's subset B, the set B is said to be a proper subset of
A, if it contains at least one element less than set A. It is denoted as B ⊂ A
Improper subset
Let B be a subset of A. Then, B is said to be the improper subset of A when B is equal to
A i.e. every element of A are contained by B. For example,
If A = {s, v, u, 3, ª} and B = {s, v, u, 3, ª}, then B is the improper subset of A. It is
denoted as B ⊆ A.
Universal set
Let's consider a set of whole numbers less than 21, U = {0, 1, 2, 3, … 20}.
From this set, we can make many other subsets such as:
A = {1, 2, 3, 4, 5}, set of natural numbers less than 6.
B = {2, 4, 6, 8}, set of even numbers less than 10.
C = {1, 4, 9, 16}, set of square numbers less than 20, and so on.
Thus, in a given situation, the set of all the elements being considered from which many
other subsets can be formed is called a universal set.
A set of students of a school is also a universal set. From this universal set, the subsets
like the set of girls, the set of boys, the set of 8 class students, etc. can be formed.
A universal set is denoted by U or ξ (pxi).
1.7 Number of subsets of a given set
If n be the cardinal number of a given set, then, the number of subsets of a given set = 2n.
For example,
Let A = {x, y, z} be a given set, where n = 3.
Here, the number of possible subsets of A = 2n = 23 = 8
These subsets are {x}, {y}, {z}, (x, y), {x, z}, (y, z), (x, y, z) and { } or φ.
Remember that an empty (or null) set φ is always a subset of every given set.
Vedanta Excel in Mathematics - Book 8 8
Set
EXERCISE 1.1
General Section – Classwork
1. If A = { 2, 4, 6, 8, ...} , let's tell and write ‘true’ or ‘false’ as quickly as possible.
a) 6 ∈ A ......................... b) 12 ∈ A ......................... c) 5 ∈ A .........................
d) 10 ∉ A ......................... e) 15 ∉ A ......................... f) 18 ∉ A .........................
2. Let's tell and write whether the following sets are empty, unit, finite or infinite.
a) A = { 1, 3, 5, 7, ...} ...............................
b) B = { 2, 4, 6, 8, ... 100} ...............................
c) C = { x : x ∈ natural numbers and x < 1 } ...............................
d) D = { composite numbers between 7 and 10} ...............................
3. Let's tell and tick the correct answers as quickly as possible.
a) W = { 0, 1, 2, 3, 4} and A = { whole numbers less than 5 } sets W and A are
(i) Equal sets (ii) Equivalent sets
b) P = { 2, 3, 5, 7} and N = { x : x is a natural number, x < 5} sets P and N are
(i) Equal sets (ii) Equivalent sets
c) A = { 3, 6, 9, 12} and B = { 5, 10, 15} sets A and B are
(i) Overlapping sets (ii) Disjoint sets
d) C = { 1, 2, 3, 4, 6, 12} and D = {1, 2, 5, 10} sets C and D are
(i) Overlapping sets (ii) Disjoint sets
4. Let's tell and write which one the universal set is and the other is its subset.
a) A = {teachers of a school}, B = {maths teachers of the school}
Universal set is ...................................... and its subset is ......................................
b) N = {natural numbers less than 100} , W = {whole numbers less than 100}
Universal set is ...................................... and its subset is ......................................
Creative section - A
5. a) A set of composite numbers less than 12. Express it in listing and set-builder
methods.
b) B = {the first five multiples of 2}. Express it in roster method and rewrite in
set-builder method.
9 Vedanta Excel in Mathematics - Book 8
Set
c) Z = { x : x ∈ integers – 2 < x < 2}. List the elements of this set and also express
it in description method.
d) P = {2, 3, 5, 7}. Express it in description method and in rule method.
e) If A = {x : x is a letter in the word ‘SCHOOL’}, list the members and find n (A).
f) If B = {p : p is a factor of 12}, list the elements of this set and find n (B).
6. How many subsets are possible from the following sets? Also write the subsets.
a) A = {7} b) B = {s, v} c) C = { a, e, i} d) D = {1, 3, 5, 7}
7. What can be the universal sets from which the following subsets can be formed?
a) The set of cricket players of class 8.
b) The set of cricket players of a school.
c) The set of odd numbers less than 10.
Creative section - B
8. Let's answer the following questions.
a) Is A = {1, 4, 9, 16} a subset of B = {1, 2, 3, … 15}? Why?
b) Is P = {a, e, i, o, u} a subset of Q = {a, b, c, d, e, ... z}? Why?
c) Is X = {1, 2, 4, 5, 10, 20} a proper or improper subset of
Y = {x : x ∈ factors of 20}? Why?
d) Are M = {x : x ∈ multiplies of 2, x < 10} and N = {y : y ∈ multiples of 3, y < 10}
overlapping sets? Why?
e) Are A = {p : p ∈ factors of 15, p > 1} and B = {q : q ∈ factors of 16, q > 1},
disjoint sets? Why?
It's your time - Project work!
9. a) Let's write a pair of sets of your own in each of the following cases.
(i) Equal sets (ii) Equivalent sets (iii) Overlapping sets (iv) Disjoint sets
b) Let's write a universal set of your own and write as many possible subsets of the
universal set. What is the improper subset of this universal set?
10. a) Make different sets from the object which you find in your classroom. Such as
students, furniture, stationeries, maths teacher of your class, etc. Then classify
them by the number of elements they have.
b) Make a group of your friends and write a set with the names of members of the
group. Then make different subsets from the set. For example: A {students who
like football}, B = {students who like maths}, C = {girls students}, and so on.
1.8 Venn – diagrams
We can represent sets and set operations by using different types of diagrams.
We can use different types of diagrams to represent sets and operations of sets. Usually,
a rectangle is used to represent a universal set and circles (or oval shapes) are used to
represent the subsets of the universal set. The concept was first introduced by Swiss
Mathematician, Euler and it was further developed by the British mathematician John
Venn. So, the diagrams are famous as Venn Euler diagrams or simply Venn diagrams.
Vedanta Excel in Mathematics - Book 8 10
Set
Let's study the following illustrations and learn to represent various set relations by
using Venn-diagrams.
U UU
A A BA B
A ⊂ U (A is a subset of U.) A and B are overlapping sets. A and B are disjoints.
U U U
A, B and C are overlapping A and B are overlapping, A and B are overlapping,
sets. A and C are overlapping, B and C are overlapping, but
but B and C are disjoint A and C are disjoint sets.
sets.
1.9 Set operations with Venn-diagrams
Sets can be combined in a number of different ways to produce another set. It is known
as set operation. Here, we discuss about four basic set operations. These four basic
operations are:
(i) Union of sets (ii) Intersection of sets
(iii) Difference of sets (iv) Complement of a set
(i) Union of sets
Let's take any two sets, A = {1, 2, 3, 4, 5} and B = {2, 4, 6, 8, 10}
Then, the union of A and B, denoted by A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10}.
When the elements of two or more sets are combined and listed A 2 6B
together in a single set, the operation is said to be the union of sets. 1
The shaded region of the diagram shows A ∪ B. 3 4 8
5 10
A∪B
Thus, the union of two sets A and B is the set which contains all the
elements that belong to A or B (or both). It is denoted by A ∪ B and read as
A union B. The symbol ∪ (cup) is used denote the union of sets.
In set builder form, the union of sets A and B can be defined as follows.
A ∪ B = {x : x ∈ A or x ∈ B}
Remember that in the case of overlapping sets, the common elements are mentioned
only once while making the union of the given sets.
11 Vedanta Excel in Mathematics - Book 8
Set
Let's study the following illustrations of the union of sets by using Venn-diagrams.
UU U
The shaded region contains The shaded region contains The shaded region contains
the elements of A ∪ B. Here, the elements of A ∪ B. Here, the elements of A ∪ B ∪ C.
A and B are disjoint sets. A and B are overlapping sets. A, B and C are disjoint sets.
U U U
The shaded region contains The shaded region contains The shaded region contains
the elements of A ∪ B ∪ C. the elements of A ∪ B ∪ C. the elements of A ∪ B ∪ C.
A, B and C are overlapping A and B, A and C are overlapping A and B, B and C are overlapping
sets. but B and C are disjoint sets. but A and C are disjoint sets.
(ii) Intersection of sets
Let's take any two sets A = {1, 2, 3, 6} and B = {1, 2, 4, 8}.
Then, the intersection of A and B denoted by A ∩ B = {1, 2} AB
When the common elements of two or more sets are taken and listed in 3 14
a separate set, the operation is said to be the intersection of sets. 6 28
The shaded region of the diagram shows A ∩ B.
Thus, the intersection of sets A and B is the set which contains all the elements that
belong to A and B. It is denoted by A ∩ B and read as A intersection B. The symbol
∩ (cap) is used to denote the intersection of sets.
In set-builder form, the intersection of sets A and B can be defined as follows.
A ∩ B = {x : x ∈ A and x ∈ B}
Study the following illustrations of the intersection of sets by using Venn-diagrams.
U UU
AB
The shaded region contains Sets A and B are the disjoint C
sets. So, they do not have The shaded region contains
the elements of A ∩ B. any common elements.
the elements of A ∩ B ∩ C
U U
U
AB AB
AB
C C C
The shaded region contains The shaded region contains The shaded region contains the
the elements of only (A ∩ B). the elements of only (A ∩ C). elements of only (B ∩ C).
Vedanta Excel in Mathematics - Book 8 12
Set
(iii) Difference of two sets
Let's take any two sets A = {2, 3, 5, 7, 11} and B = {1, 3, 5, 7, 9}.
Then, the difference of A and B denoted by A – B = {2, 11}
The difference of B and A denoted by B – A = {1, 9}
Between two sets, when the elements of one set which do not belong to another set, are
listed in a separate set, the operation is said to be the A BA B
difference of two sets. 2 3 1 2 3 1
11 75 9 11 57 9
The shaded regions of the diagrams show A – B and
B – A. A–B B–A
Thus, the difference of two sets A and B denoted by A – B is the set of all elements
contained only by A but not by B.
Similarly, the difference of two sets B and A denoted by B – A is the set of all elements
contained only by B but not by A.
In set-builder form, the difference of A and B or B and A can be defined as follows.
A – B = {x : x ∈ A, but x ∉ B} and B – A = {x : x ∈ B, but x ∉ A}
Study the following illustrations of the difference of sets by using Venn-diagrams.
UU UU
The shaded region The shaded region The shaded region The shaded region
contains the contains the contains the contains the
elements of A – B. elements of A – B. elements of B – A. elements of B – A.
(iv) Complement of a set U
Let's take a universal set U = {1, 2, 3, ... 10} and it's subset A1
A = {1, 4, 9}.
Then, the complement of A denoted by 2 49 10
A = U – A = {2, 3, 5, 6, 7, 8, 10} 3 8
When the elements of a universal set, which do not belong to 567
its given subset are listed in a separate set, the operation is said
to be the complement of the given subset. The shaded region contains
the elements of A
Thus, if a set A is the subset of a universal set U, then its complement denoted by A is
the set which is formed from the difference of U and A, i.e. U – A.
In set builder form, the complement of a subset A of the universal set U can be defined
in the following ways.
A= {x : x ∈ U, but x ∉ A}
13 Vedanta Excel in Mathematics - Book 8
Set
Worked-out examples
Example 1: Let A, B and C are any three sets. Write the set operations represented
by the shaded regions. Define each operation by set-builder form.
a) b) C c) A B d)
A BB
C
Solution:
a) The shaded region represents A ∩ B.
A ∩ B = {x : x ∈ A and x ∈ B}
b) The shaded region represents B – C.
B – C = {x : x ∈ B, but x ∉ C}
c) The shaded region represents A ∩ B ∩ C.
A ∩ B ∩ C = {x : x ∈ A, x ∈ B and x ∈ C}
d) The shaded region represents A ∪ B.
A ∪ B = {x : A or x ∈ B}
Example 2: If A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7, 9} and C = {2, 3, 5, 7, 11}, find
Solution: (i) A ∪ B (ii) A ∪ B ∪ C (iii) A ∩ B (iv) A ∩ B ∩ C (v) A – B
(vi) C – B. Illustrate the results in Venn-diagrams.
Here, A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7, 9} and C = {2, 3, 5, 7, 11} 2 1
(i) A ∪ B = {1, 2, 3, 4, 5, 7, 9} 9
The shaded region contains the elements of A ∪ B. 9
2
(ii) A ∪ B ∪ C = {1, 2, 3, 4, 5, 7, 9, 11}
The shaded region contains the elements of A ∪ B ∪ C. 11
(iii) A ∩ B = {1, 3, 5} 21
The shaded region contains the elements of A ∩ B. 9
(iv) A ∩ B ∩ C = {3, 5}
The shaded region contains the elements of A ∩ B ∩ C.
11
Vedanta Excel in Mathematics - Book 8 14
Set
(v) A – B = {2, 4} 9
The shaded region contains the elements of A – B.
12
(vi) C – B = {2, 11} 11
The shaded region contains the elements of C – B.
Example 3: A, B and C are the subsets of a universal set U. If U = {1, 2, 3, ... 20},
A ={1, 2, 3, 4, 5, 6, 7}, B ={2, 4, 6, 8, 10, 12} and C ={3, 6, 9, 12, 15, 18},
perform the following set operations and illustrate them in Venn-
diagrams.
(i) (A ∪ B) ∩ C (ii) A ∩ (B ∪ C) (iii) A ∪ B (iv) A ∪ B ∪ C
(v) A ∩ C (vi) A ∩ B ∩ C (vii) (A – C) ∩ B
Solution:
Here, U = {1, 2, 3, ... 20}, A = {1, 2, 3, 4, 5, 6, 7}, B = {2, 4, 6, 8, 10, 12} and
C = {3, 6, 9, 12, 15, 18}
(i) A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10, 12} U
∴(A ∪ B) ∩ C = {3, 6, 12}
The shaded region contains the elements of 18 20
(A ∪ B) ∩ C. 16
(ii) B ∪ C = {2, 3, 4, 6, 8, 9, 10, 12, 15, 18} 19
17
U
∴A ∩ (B ∪ C) = {2, 3, 4, 6}
The shaded region contains the elements of
A ∩ (B ∪ C). 18 19 20
(iii) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 10, 12} 16 17
U
∴ (A ∪ B) = U – (A ∪ B) 20
19
= {9, 11, 13, 14, 15, 16, 17, 18, 19, 20} 9 18
11 13 14 15 16 17
The shaded region contains the elements of (A ∪ B).
U
(iv) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 18}
∴ A ∪ B ∪ C = U – (A ∪ B ∪ C) 14 16 18 17 19 20
= {11, 13, 14, 16, 17, 19, 20}
15 Vedanta Excel in Mathematics - Book 8
Set
(v) A ∩ C = {3, 6} U
∴ A ∩ C = U – (A ∩ C)
20
={1, 2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} 18 19
The shaded region contains the elements of A ∩ C.
(vi) A ∩ B ∩ C = {6} 16 17
U
∴ A ∩ B ∩ C = U – (A ∩ B ∩ ∩)
={1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20} 16 18 20
The shaded region contains the elements of A ∩ B ∩ C. 17 19
U
(vii) (A – C) = {1, 2, 4, 5, 7}
∴ A – C ={3, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20}
∴ A – C ∩ B={6, 8, 10, 12} 16 18 17 19 20
The shaded region contains the elements of A – C ∩ B.
Example 4: From the given Venn-diagram, list the elements of the following sets. U
(i) A ∪ B (ii) B ∪ C (iii) A ∪ B ∪ C
(iv) A ∩ B (v) A ∩ C (vi) A ∩ B ∩ C
(vii) A ∪ B ∪ C (viii) A ∩ B ∩ C (ix) B 13
(x) A (xi) (A ∪ B) – C (xii) (A ∩ B) ∩ C 14 15
Solution:
Here, U = {1, 2, 3, … 15}, A = {2, 3, 5, 7, 11, 13}, B = {1, 3, 5, 7, 9}, C = {1, 2, 3, 4, 5, 6}
(i) A ∪ B = {1, 2, 3, 5, 7, 9, 11, 13}
(ii) B ∪ C = {1, 2, 3, 4, 5, 6, 7, 9}
(iii) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 9, 11, 13}
(iv) A ∩ B = {3, 5, 7}
(v) A ∩ C = {2, 3, 5}
(vi) A ∩ B ∩ C = {3, 5}
(vii) A ∪ B ∪ C = {8, 10, 12, 14, 15}
(viii)A ∩ B ∩ C = {1, 2, 4, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
(ix) B = {2, 4, 6, 8, 10, 11, 12, 13, 14, 15} ∴ B = {1, 3, 5, 7, 9} It shows that B = B
(x) A = {1, 4, 6, 8, 9, 10, 12, 14, 15} ∴ A = {2, 3, 5, 7, 11, 13} It shows that A = A
(xi) (A ∪ B) – C = {7, 9, 11, 13}
(xii) (A ∪ B) ∩ C = {1, 2, 3, 5}
Vedanta Excel in Mathematics - Book 8 16
Set
Example 5: If A = {a, e, i, o, u}, B = {a, b, c, d, e} and C = {e, f, g, h}, show that
Solution: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Here, A = {a, e, i, o, u}, B = {a, b, c, d, e}, C = {e, f, g, h}
Now, B ∩ C = {e} and A ∩ (B ∩ C) = {a, e, i, o, u}
Again, A ∪ B = {a, b, c, d, e, i, o, u}, A ∪ C = {a, e, f, g, h, i, o, u}
Then, (A ∪ B) ∩ (A ∪ C) = {a, e, i, o, u}
Thus, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) proved.
EXERCISE 1.2
General Section - Classwork
1. Let's tell and write the answers as quickly as possible.
a) If A = {1, 3, 5, 7, 9} and B = { 2, 3, 5, 7, 11}, then
(i) A ∪ B = ..................................................................................................
(ii) A ∩ B = ..................................................................................................
(iii) A – B = ..................................................................................................
(iv) B – A = ..................................................................................................
b) A universal set U = { 1, 2, 3, ... 10}. Its subsets A = {2, 4, 6, 8} and
B = {1, 2, 3, 4, 5}. Let's tell and write
(i) A = ................................................ and A = ................................................
(ii) B = ................................................ and B = ...............................................
2. Let's tell and list the elements as quickly as possible.
a) P ∪ Q = ............................................................. PQ
b) P ∩ Q = .............................................................
c) P – Q = ............................................................. r aect m
d) Q – P = ............................................................. i
hs
Creative Section - A
3. Let A, B and C are any three sets. Let's write the set operations represented by
the shaded regions. Define each operation by set-builder form.
a) A B b) B C c) A C
17 Vedanta Excel in Mathematics - Book 8
Set
d) A B e) A B A B
f)
C
C
4. P and Q are two overlapping subsets of a universal set U. Draw Venn-diagrams
and shade the regions that contain the element of the following set operations.
a) P ∪ Q b) P ∩ Q c) P – Q d) Q – P e) P f) Q
5. If A = {a, b, c, d, e}, B = {d, e, f, g, h, i} and C = {a, e, i, o, u}, find the elements
of the following set operations and illustrate them in Venn-diagrams.
a) A ∪ B b) B ∪ C c) A ∩ B d) A ∪ B ∪ C e) A ∩ B ∩ C
f) B – C g) C – A h) A ∪ (B ∩ C) i) (A ∪ B) ∩ C j) A – (B ∩ C)
6. A, B, C are the subsets of universal set U. If U = {1, 2, 3, ... 15},
A = {2, 4, 6, 8, 10, 12}, B = {3, 6, 9, 12, 15} and C = {1, 2, 3, 4, 5, 6, 7}, find the
elements of the following set operations and illustrate them in Venn-diagrams.
a) (A ∪ B) ∩ C b) (A ∩ B) ∪ C c) A ∪ B ∪ C d) (A ∪ C) ∩ B
e) A – (B ∩ C) f) (A ∪ B) – (B ∩ C) g) (A – B) ∪ (B – C) h) (A – C) ∪ B
7. From the adjoining Venn-diagram, list the elements of the following sets.
a) P ∪ Q b) Q ∪ R c) P ∪ R P U
d) P ∩ Q e) Q ∩ R f) P ∪ Q ∪ R 8
g) P ∪ Q ∪ R h) P ∩ Q ∩ R i) P ∩ Q ∩ R 12 Q
j) P ∪ (Q ∩ R) k) (Q ∪ R) ∩ P l) P – (Q ∩ R) 10 2 9
m) P n) Q o) R 4
63 R
1
5
7
11 13 14 15
Creative Section - B
8. a) If A = {1, 2, 3, 4, 5}, B = {1, 3, 5, 7} and C = {2, 3, 5}, show that:
(i) A ∪ (B ∪ C) = (A ∪ B) ∪ C (ii) (A ∩ B) ∩ C = A ∩ (B ∩ C)
b) If A = {a, b, c, d, e, f}, B = {b, c, d, f, g} and C = (a, e, i, o, u), show that
i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
9. a) A and B are the subsets of a universal set U. If U = {1, 2, 3, ... 10},
A = {1, 3, 5, 7, 9} and B = {1, 2, 3, 6}, show that:
(i) A ∪ B = A ∩ B (ii) A ∩ B = A ∪ B
b) P and Q are the subsets of a universal set U. If U = {x : x is a whole number, x≤9},
A = {y : y is a multiple of 2} and B = {z : z is a factor of 12}, verify that:
(i) P ∪ Q = P ∩ Q (ii) P ∩ Q = P ∪ Q
Vedanta Excel in Mathematics - Book 8 18
Set
It's your time - Project work
10. a) Make a group of your 10 friends and collect the data to complete the table given
below.
Name of Friends who like Friends who Friends who like Mathematics
friends Mathematics like Science and Science both
Now, draw Venn-diagram and shade each diagram to show each of the following
sets.
(i) Set of friends who like Mathematics or Science or both
(ii) Set of friends who like Mathematics and Science both
(iii) Set of friends who like only Mathematics
(iv) Set of friends who like none of these two subjects.
b) Let's write any three non-empty and overlapping sets A, B and C. Then verify the
following operations.
(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) (ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
1.10 Cardinality relations of sets
The cardinal number of a set is called its cardinality. Let's study the following relations
which are generalised by taking the cardinalities of different sets.
(i) Cardinality relation of union of two disjoint sets
Let A = {a, b, c, d} and B = {e, f, g, h, i}, are two disjoint subsets of a universal set
U = {a, b, c, d, e, f, g, h, i, j}
Here, n (A) = 4, n (B) = 5 and n (U) = 10
Now, A ∪ B = {a, b, c, d, e, f, g, h, i}
∴ n (A ∪ B) = 9 = 4 + 5 = n (A) + n (B)
Thus, n (A ∪ B) = n (A) + n (B), where A and B are two disjoint sets.
Also, n (A ∪ B) = n (U) – n (A ∪ B)
(ii) Cardinality relation of union of two overlapping sets
Let A = {a, b, c, d, e} and B = {d, e, f, g, h, i}, are two overlapping subsets of a
universal set U = {a, b, c, d, e, f, g, h, i, j}.
Here, n (A) = 5, n (B) = 6 and n (U) = 10. U
Now, A ∪ B = {a, b, c, d, e, f, g, h, i} a d f h
∴ n (A ∪ B) = 9 g
Also, A ∩ B = {d, e} b ei
∴ n (A ∩ B) = 2 c
j
Again, n (A ∪ B) = 9 = 5 + 6 – 2 = n (A) + n (B) – n (A ∩ B)
19 Vedanta Excel in Mathematics - Book 8
Set
It’s clear that in n (A) + n (B), the number of common elements [n (A ∩ B)] appears
two times. So, to correct it n (A ∩ B) is subtracted from n (A) + n (B).
Thus, n (A ∪ B) = n (A) + n (B) – n (A ∩ B) where A and B are overlapping sets.
Also, n (A ∪ B) = n (U) – n (A ∪ B)
Let’s learn a few more relations about the cardinality relation of the operations of
two overlapping sets from the following Venn-diagrams.
U UU U
A BA B A B A B
The shaded region The shaded region The shaded region It is n (only B) or no (B)
represents n (A ∪ B). represents n(A ∩ B) or n(B – A)
represents n (only A) ∴ no (B) = n (B) – n (A ∩ B)
∴ornnoo(A(A))=ornn((AA) – B). ∩ B).
– n(A
Worked-out examples
Example 1: A and B are the subsets of a universal set U, where n (U) = 100, n (A) = 70,
Solution: n (B) = 50 and n (A ∩ B) = 30.
(i) Illustrate this information in Venn-diagram.
(ii) Find n (A ∪ B) and (A ∪ B).
Here, n (U) = 100, n (A) = 70, n (B) = 50 and n(A ∩ B) = 30
(i) Venn-diagram (ii) From the Venn-diagram
U n (A ∪ B) = 40 + 30 + 20 = 90
A ∴ n (A ∪ B) = n (U) – n (A ∪ B) = 100 – 90 = 10
B
Alternative process
40 30 20 n (A ∪ B) = n(A) + n (B) – n(A ∩ B)
10 = 70 + 50 – 30 = 90
∴ n (A ∪ B) = n (U) – n (A ∪ B) = 100 – 90 = 10.
Example 2: P and Q are the subsets of a universal set U. If n (U) = 85, n (P) = 45,
n (Q) = 55 and n (P ∪ Q) = 65, find
(i) n (P ∩ Q) (ii) no (P) or (P – Q) (iii) no (Q) or (Q – P)
Also illustrate the information in Venn-diagram.
Solution:
Here, n (U) = 85, n (P) = 45, n (Q) = 55 and n (P ∪ Q) = 65 U
(i) Now, n (P ∪ Q) = n (P) + n (Q) – n (P ∩ Q) P
or, 65 = 45 + 55 – n(P ∩ Q) Q
or, n(P ∩ Q) = 100 – 65 = 35.
10 35 20
(ii) no (P) or (P – Q) = n (P) – n (P ∩ Q) = 45 – 35 = 10 20
(iii) no (Q) or (Q – P) = n (Q) – n (P ∩ Q) = 55 – 35 = 20
Vedanta Excel in Mathematics - Book 8 20
Set
Example 3: There are 45 students in a class. 24 of them like cricket, 30 like football
and 14 like cricket as well as football.
(i) Illustrate this information in Venn-diagram
(ii) How many students like both games?
(iii) How many students do not like both games?
(iv) How many of them like only cricket?
(v) How many of them like only football?
Solution:
Let, C and F denote the sets of students who like cricket and football respectively.
Here, n (U) = 45, n (C) = 24, n (F) = 30 and n (C ∩ F) = 14
(i) Venn diagram: (ii) n (C ∪ F) = n (C) + n (F) – n (C ∩ F)
U
(iii) n (C ∪ F) = 24 + 30 – 14 = 40
CF (iv) n (only C) = n (U) – n (C ∪ F) = 45 – 40 = 5
10 14 16 (v) n (only F) = n (C) – n (C ∩ F) = 24 – 14 = 10
= n (F) – n (C ∩ F) = 30 – 14 = 16
5
Example 4: In a survey of 5,000 Chinese tourists who recently visited Nepal during
'Visit Nepal 2020', it was found that 3,500 tourists visited Pokhara, 4,000
visited Lumbini and 500 tourists did not visit both places.
(i) How many tourists visited Pokhara or Lumbini?
(ii) How many tourists visited Pokhara and Lumbini?
(iii) Represent the above information in a Venn-diagram.
Solution:
Let the set of Chinese tourist who visited Pokhara be P and Lumbini be L.
Here, n(U) = 5,000, n(P) = 3,500, n(L) = 4,000 and n (P ∪ L) = 500
(i) Now, n(P ∪ L) = n(U) – n (P ∪ L) = 5,000 – 500 = 4,500
∴ 4,500 tourists visited Pokhara or Lumbini. (iii) 3000 U
(ii) Also, n(P ∪ L) = n(P) + n(L) – n(P ∩ L)
P L
or, 4,500 = 3,500 + 4,000 – n(P ∩ L) 500 1000
or, n(P ∩ L) = 7,500 – 4,500 = 3,000
∴ 3,000 tourists visited Pokhara and Lumbini. 500
Example 5: In a group of 175 people, everybody can speak at least one of the two
languages, Nepali or Maithili. 125 people can speak Nepali and 105
people can speak Maithili.
(i) Illustrate the above information in a Venn-diagram.
(ii) How many people can speak both Nepali and Maithili languages?
(iii) Find the number of people who can speak only one language.
Solution:
Let N and M denote the sets of people who can speak Nepali and Maithili language
respectively.
Here, n(U) = n(N ∪ M) = 175, n(N) = 125, and n(M) = 105
21 Vedanta Excel in Mathematics - Book 8
Set
Let, the number of people who can speak both languages, n(N ∩ M) = x.
(i) Illustration in Venn-diagram N U
M
125–x x 105–x
(ii) From the Venn-diagram,
125 – x + x + 105 – x = 175 Alternative process
or, 230 – x = 175 n (N ∪ M) = n (N) + n (M) – n (N ∩ M)
or, 175 = 125 + 105 – n (N ∩ M)
or, x = 230 – 175 = 55 or, n (N ∩ M) = 230 – 175 = 55
∴ 55 people can speak both languages.
(iii) Again, n(only N) = 125 – x = 125 – 55 = 70 Alternative process
n(only M) = 105 – x = 105 – 55 = 50 n (only N) = n (N) – n (N ∩ M)
= 125 – 55 = 70
Number of people who can speak n (only M) = n (M) – n (N ∩ M)
only one language = 70 + 50 = 120
= 105 – 55 = 50
EXERCISE 1.3
General Section - Classwork
1. Let's tell and write the correct answers as quickly as possible.
a) A and B are two disjoint subsets of a universal set U . If n (U) = 20, n (A) = 9
and (B) = 8, find,
(i) n (A ∪ B) = ......................... (ii) n (A ∪ B) = .........................
b) P and Q are two overlapping subsets of a universal set U. If n (U) = 50,
n (P) = 25, n (Q) = 30 and n (P ∩ Q) = 15, find,
(i) n (P ∪ Q) = .............. (ii) n (P ∪ Q) = ..............
2. In the given Venn-diagram, T and D denote the sets of students who want to
be teacher and doctor respectively. Let's tell and write the correct answers as
quickly as possible.
a) n(T ∪ D) = ................ b) n (T ∩ D) = ................ U
c) n (T ∪ D) = ................ d) n (T – D ) = ................ TD
578
10
e) n (D – T) = ................ f) n(U) = ................
Vedanta Excel in Mathematics - Book 8 22
Set
Creative Section-A
3. In the given Venn – diagram, A and B are the sets of students U
who like apple and banana respectively. If n (U) = 48 and AB
n (A ∪ B) = 36, find:
27–x x 18–x
b) n (A ∪ B) c) no (A)
a) n (A ∩ B) d) n (B – A)
4. a) If n (U) = 50, n (P) = 27, n (Q) = 25 and n (P ∩ Q) = 12, find
(i) n (P ∪ Q) (ii) n (P ∪ Q) (iii) n (only P) (iv) n (only Q)
Also illustrate the above information in a Venn-diagram.
b) A and B are the subsets of a universal set U. If n (U) = 100, n (A) = 45,
n (B) = 60 and n (A ∪ B) = 95, find
(i) n (A ∩ B) (ii) n (A ∪ B) (iii) no (A) (iv) no (B)
Also illustrate the above information in a Venn-diagram.
c) A and B are the subsets of a universal set U. If n (U) = 75, n (A) = 40,
n (B) = 35 and n (A ∪ B) = 10, illustrate this information in a Venn-diagram
and find:
(i) n (A ∪ B) (ii) n (A ∩ B) (iii) n (A – B) (iv) n (only B)
d) X and Y are two subsets of a universal set U in which there are n (U) = 54,
n (X) = 30, n (Y) = 20 and n (X ∩ Y) = 9.
(i) Draw a Venn-diagram to illustrate the above information.
(ii) Find the value of n (X ∪ Y) .
e) A and B are the subsets of a universal set U. If n (A) = 63, n (B) = 72,
n (A ∪ B) = 115, n (A ∪ B) = 20, find
(i) n (U) (ii) n (A ∩ B) (iii) no (A) (iv) no (B)
Also illustrate the above information in Venn-diagram.
Creative Section-B
5. a) Among 72 students, 24 are selected to participate in music only, 26 are selected
to participate in dance only and none of them participated in both activities.
The rest of the students are selected in other activities.
(i) How many students are participating in music or in dance?
(ii) How many students are participating in other than music or dance?
(iii) Illustrate this information in a Venn-digram.
b) In a survey of 110 people, 50 use internet, 75 use cellular data and 25 use
internet as well as cellular data.
23 Vedanta Excel in Mathematics - Book 8
Set
(i) Draw a Venn-diagram to illustrate the above information.
(ii) Find the number of people who use either internet or cellular data.
(iii) Find the number of people who use neither internet nor cellular data.
(iv) Find the number of people who use only one type of Wi-Fi connection.
c) In a survey of some students, 55% of the students like basketball, 65% like
cricket and 35% like basketball as well as cricket.
(i) Show the above information in a Venn-diagram.
(ii) How many percent of students do not like basketball as well as cricket?
6. a) In a survey of 250 people, 180 like farming, 120 like civil service and every
people like at least one occupation. Draw a Venn-diagram and find the number
of people who like farming as well as civil service. Also find how many people
like only civil service?
b) In a survey of 486 tourists who visited Nepal during 'Visit Nepal 2020', it was
found that 275 visited Rara, 300 visited Bardiya and 56 tourists did not visit
both places.
(i) How many tourists were there who visited Rara as well as Bardiya?
(ii) Represent the above information in a Venn-diagram.
c) In a school's Annual Day, 60 students involved in Science Exhibition, 75
involved in Mathematics Exhibition, 20 involved in both Exhibitions and the
rest of 285 students involved in many other activities but not in the Exhibitions.
(i) Illustrate the above information in a Venn-diagram
(ii) How many students took part in the School's Annual Day?
(iii) How many students involved in only one exhibition?
It's your time - Project work
7. a) Let's conduct a survey inside your classroom and collect data about how many
of your friends like singing, dancing and singing as well as dancing. Then find
the following numbers by using cardinality relation of the sets.
(i) Number of friends who like sing and dancing.
(ii) Number of friends who like singing only.
(iii) Number of friends who like dancing only.
(iv) Number of friends who do not like singing and dancing.
(v) Also draw Venn-diagrams to show these numbers.
b) Let's conduct a survey among the teachers of your school and collect the data
about their favourite fruits between apple and orange. Then, perform as many
set operations as you can by using the cardinality properties of sets. Illustrate
each operation in Venn-diagrams.
Vedanta Excel in Mathematics - Book 8 24
Unit Number System in Different Bases
2
2.1 Denary, binary and quinary numbers - Looking back
Classroom - Exercise
1. Let's tell and write the short forms of the given bases of numbers.
a) The decimal number of 4 × 103 + 6 × 101 + 8 × 10° is ...........................
b) The binary number of 1 × 24 + 1 × 22 + 1 × 21 + 1 × 2° is ...........................
c) The quinary number of 3 × 53 + 4 × 52 + 2 × 51 + 1 × 5° is ...........................
2. Let's tell and write the answers as quickly as possible.
a) The digits in denary system are ......................................................................
e) The digits in binary system are ......................................................................
f) The digits in quinary system are .........................................................................
3. Let's tell and tick the correct value.
a) Value of 3 in binary number is (i) 102 (ii) 112 (iii) 1012
b) Value of 9 in quinary number is (i) 145 (ii) 1045 (iii) 415
c) Value of 1012 in denary number is (i) 3 (ii) 6 (iii) 5
d) Value of 135 in denary number is (i) 7 (ii) 8 (iii) 9
2.2 Natural numbers and whole numbers - Looking back
N = {1, 2, 3, 4, 5, ...} is the set of natural numbers which are also called the counting
numbers. The operations of addition and multiplication of any two natural numbers
always give natural numbers. For example,
3 + 4 = 7 (7 is also a natural number)
3 × 4 = 12 (12 is also a natural number)
Similarly, in 4 – 3 = 1, 1 is a natural number. However, in 3 – 3 = 0, 0 is not a natural
number. It demands another set of numbers that should include 0 (zero) also. So, 0 is
included in the set of natural numbers to develop a new set of numbers that we call the
set of whole numbers. It is denoted by W.
Thus, W = {0, 1, 2, 3, 4, 5, ...} is the set of whole numbers.
25 Vedanta Excel in Mathematics - Book 8
Number System in different Bases
2.3 Decimal numeration system
Let’s take 18 blocks of cubes and regroup them into the group of 10 blocks.
18 = 10 + 8 = 1 × 101 + 8 × 10°
Let’s take 36 pencils and regroup them into the group of 10 blocks.
36 = 30 + 6 = 3 × 101 + 6 × 10°
Similarly, 342 = 300 + 40 + 2 = 3 × 102 + 4 × 101 + 2 × 100.
In this way, whole numbers can be regrouped into the base of 10 with some power of 10.
It is called the decimal numeration system or denary system.
Example 1: Express 472510 in the expanded form.
Solution:
472510 = 4 × 1000 + 7 × 100 + 2 × 10 + 5
= 4× 103 + 7 ×102 + 2 × 101 + 5 × 100.
2.4 Binary number system
Decimal or denary number system is base 10 system. Besides this number system, there
are four alternative base systems that are most useful in computer applications. These
are binary (base two), quinary (base five), octal (base eight), and hexadecimal (base
sixteen) systems.
Computers and handheld calculators use the binary system for their internal calculations.
The system consists of only two digits 0 and 1. So, all numbers can then be represented
by electronic switches of one kind or another, where “on” indicates 1 and “off”
indicates 0.
The octal system is used extensively by programmers who work with internal computer
codes. In a computer, the CPU often uses the hexadecimal system to communicate with
a printer or other output devices.
2.5 Formation of binary number system
Again, let’s take 15 blocks of cubes and regroup them into the group of 2 blocks.
7 pairs blocks of cube and 1 cube
26
Vedanta Excel in Mathematics - Book 8
Number System in different Bases
Now, let’s arrange the groups of 2 blocks into the base of 2 with the maximum possible
powers.
8 4 21
23 22 21 20
So, 15 = 1 × 23 + 1 × 22 + 1 × 21 + 1 × 2° = 11112
In this way, the denary number 15 can be expressed in binary number as 11112.
The system of numeration in base 2 with some power of 2 is called the binary numeration
system. In this system, we use only two digits 0 and 1.
2.6 Conversion of decimal numbers to binary numbers
We convert a decimal number into binary number, dividing the given number successively
by 2 until the quotient becomes 0. The remainder obtained in each successive division
is listed in a separate column. The remainders arranging in the reverse order is the
required binary number. For example,
Example 2: Convert 185 into binary system.
Solution Arranging the remainders from
the bottom up: 111011012
2 185 Remainder 1×28 1×27 1×26 1×25 1×24 1×23 1×22 1×21 1×20
256 128 64 32 16 8 4 2 1
2 92 1
185 = 128 + 0 + 32 + 16 + 8 + 0 + 0 + 1
2 46 0 = 1×27 + 0 + 1×25 + 1×24 + 1×23 + 0 + 0 + 1×10
= 101110012
2 23 0
2 11 1
25 1
22 1
10
∴ 185 = 101110012
2.7 Conversion of binary numbers to decimal numbers
We convert a binary number into decimal number, just by expanding the given binary
number in the power of 2. Then, by simplifying the expanded form of the binary number,
we obtain the required decimal number. For example:
Example 3: Convert 1101102 into decimal system.
Solution:
1101102 = 1 × 25 + 1 × 24 + 0 × 23 + 1 × 22 + 1 × 21 + 0 × 20
= 32 + 16 + 4 + 2 + 0 = 54
27 Vedanta Excel in Mathematics - Book 8
Number System in different Bases
EXERCISE 2.1
General Section - Classwork
1. Let's tell and tick the short form of these numbers as quickly as possible.
a) 3 × 102 + 7 × 101 + 5 × 100 (i) 3075 (ii) 375 (iii) 370
b) 2 × 104 + 8 × 102 + 4 × 101 + 9 × 100 (i) 20849 (ii) 28049 (iii) 2849
c) 1 × 24 + 1 ×23 + 1 × 22 + 1 × 21 (i) 110112 (ii) 111012 (iii) 111102
d) 1 × 25 + 1 × 23 + 1 × 21 + 1 × 20 (i) 1010112 (ii) 1100112 (iii) 1010102
2. Let's tell and write the values of these decimal numbers in binary numbers as
quickly as possible.
a) 1= ............. b) 2 = ............. c) 3 = ............. d) 4 = .............
e) 5 = ............. f) 6 = ............. g) 7 = ............. h) 8 = .............
3. Let's tell and write the values of these binary numbers in decimal numbers as
quickly as possible.
a) 12 = ............. b) 102 = ............. c) 112 = ............. d) 1002 = .............
e) 1012 = ............. f) 1102 = ............. g) 1112 = ............. h) 10002= .............
Creative Section
4. Expand the following decimal numbers in power of 10s.
a) 276 b) 3816 c) 54027 d) 809005 e) 7000409
5. Convert the following decimal numbers into binary numbers.
a) 9 b) 10 c) 11 d) 12 e) 18
f) 55 g) 124 h) 216 i) 361 j) 490
6. Convert the following binary numbers into decimal numbers.
a) 10012 b) 10102 c) 101112 d) 1010102 e) 1111012
f) 10101112 g) 11001112 h) 110010112 i) 111011102 j) 1101110112
It's your time - Project work
7. a) Let's take a few number of matchsticks and represent the following binary
numbers by pairing the required number of matchsticks.
(i) 102 (ii) 112 (iii) 1012 (iv) 1112 (v) 1102 (vi) 11102 (vii) 11112
b) Let's draw the required number of pairs of dots in the table to convert the given
denary numbers into binary numbers.
Denary numbers 24 23 22 21 20
5
13
18
Vedanta Excel in Mathematics - Book 8 28
Number System in different Bases
2.8 Quinary number system
We have learned that denary is base ten and binary is base two number systems.
Similarly, quinary is the base five number system. In the quinary place system, five
digits 0, 1, 2, 3 and 4 are used to represent any real number. The numbers in the quinary
system can be expressed as the product of digits and powers of 5. For example:
245 = 2 × 51 + 4 × 5°, 4325 = 4 × 52 + 3 × 51 + 2 × 5° and so on.
2.9 Conversion of decimal numbers to quinary numbers
To convert a decimal number into quinary number, we should divide the given number
successively by 5 until the quotient becomes zero. The remainder obtained in each
successive division is listed in a separate column. Then, the remainders arranging in the
reverse order is the required quinary number. For example:
Example 1: Convert 728 into quinary number.
Solution:
5 728 Remainder Arranging the remainders 54 53 52 51 50
from, the bottom 625 125 25 5 1
5 145 3 up, 104035 728 = 625 + 0 + 4 × 25 + 0 + 3
5 29 0 = 1×54 + 0 + 4×52 + 0 + 3×50
55 4 = 104035
51 0
01
∴ 728 = 104035
2.10 Conversion of quinary numbers to decimal numbers
To convert a quinary number into decimal number, it is expanded in the power of 5.
Then, by simplifying the expanded form of the quinary number, we get the required
decimal number. For example,
Example 2 : Convert 31045 into decimal number.
Solution:
31045 = 3 × 53 + 1 × 52 + 0 × 51 + 4 × 5°
= 3 × 125 + 1 × 25 + 4
= 375 + 25 + 4
= 404
EXERCISE 2.2
General Section
1. Let's tell and and tick the short form of the quinary numbers.
a) 2 × 53 + 4 × 51 + 1 × 50 (i) 2415 (ii) 24105 (iii) 20415
(i) 423105 (ii) 423015 (iii) 42315
b) 4 × 54 + 2 × 53 + 3 × 52 + 1 × 51 (i) 31205 (ii) 301025 (iii) 301205
c) 3 × 54 + 1 × 52 + 2 × 50
29 Vedanta Excel in Mathematics - Book 8
Number System in different Bases
2. Let's tell and write the quinary numbers as quickly as possible.
a) 5 = ............ b) 6 = ............ c) 7 = ............ d) 8 = ...........
e) 9 = ............ f) 10 =............ g) 15 = ............ h) 25 = ............
3. Let's tell and write the decimal numbers as quickly as possible.
a) 105 = ............ b) 115 = ............ c) 205 = ............ d) 235 = ............
e) 305 = ............ f) 405 = ............ g) 1005 = ............ h) 1105 = ............
Creative section - A
4. Let's convert the following decimal numbers into quinary numbers.
a) 8 b) 10 c) 27 d) 72 e) 136
f) 257 g) 444 h) 516 i) 817 j) 1539
5. Let's convert the following quinary numbers into decimal numbers.
a) 145 b) 235 c) 435 d) 2145 e) 4315
f) 14235 g) 21305 h) 43215 i) 120435 j) 302425
Creative section - B
6. Let's convert binary into quinary or quinary into binary.
a) 11102 b) 110112 c) 1111012 d) 135 e) 2415 f) 30245
7. a) If the value of a number in decimal system is 955, find its value in quinary
system and in binary system.
b) Which one is greater between 11110112 and 1435 ? Also, find their difference in
decimal system.
c) Which one is greater 110112 or 3225? Also, find their difference in denary
system.
8. Three students Ramila, Shashwat and Arabi were asked to convert 2215 into binary
number. Ramila got the answer 1110112, Shashwat got the answer 1111012 and
Arabi got the answer 1101102. Who got the correct answer? Find by calculation.
It's your time - Project work
9. a) Write the number of your family members and convert it into binary as well as
quinary systems.
b) Take the total number of students of your class. Then convert it into binary as
well to as to quinary systems.
10. Let's draw the required number of groups of 5 dots in the table to convert given
denary numbers into quinary numbers.
Denary numbers 52 51 50
6
9
12
16
27
Vedanta Excel in Mathematics - Book 8 30
Unit Integers
3
3.1 Integers - Looking back
Classroom - Exercise
1. Let's tell and write 'true' or 'false' for the following statements.
a) Zero is the least whole number. ................
b) Set of whole number is the subset of natural number. ................
c) {..., –1, 0, 1, …} is the set of integers. ................
d) Set of integers is the subset of whole numbers. ................
e) The least integer is infinite. ................
f) 0 is less than –1. ................
g) The difference of any two whole numbers is always a whole number. ................
h) The difference of any two integers is always an integer. ................
2 Let's tell and write the following sets of integers.
a) integers between - 3 and 3 .................................................................
b) integers between –5 and 1 .................................................................
3. Let's tell and write the answer as quickly as possible.
a) 3 + 6 = ............. 6 – 3 = ............. 3 – 6 = ............. –3 – 6 = ...........
b) 3 × 6 = ........... , (–3) × (6) = ..........., 3 × (–6) = ..........., (–3) × (–6) = ...........
c) 6 ÷ 3 = ..........., (–6) ÷ 3 = ..........., 6 ÷ (–3) = ..........., (–6) ÷ (–3) = ...........
Now, let’s take any two whole numbers 4 and 5. Tell and write the answers as quickly
as possible.
4 + 5 = 9, is 9 a whole number? .........................................................
5 – 4 = 1, is 1 a whole number? .........................................................
5 × 4 = 20, is 20 a whole number? .........................................................
4 – 5 = –1, is –1 a whole number? .........................................................
We know that –1 is not the member of the set of whole numbers. It must be the member
of another set of numbers that we call the set of integers. So, –1 is an integer.
31 Vedanta Excel in Mathematics - Book 8
Integers
Thus, the set of all numbers both positive and negative numbers including zero (0) is
called the set of integers. The set of integers is denoted by the letter ‘Z’.
Z = {..., –3, –2, –1, 0, 1, 2, 3, ...} is the set of integers.
Z+ = {+1, +2, +3, ...} is the set of positive integers.
Z– = {–1, –2, –3, ...} is the set of negative integers.
The integers can be shown in the number line.
decreasing order zero increasing order
– 1 0 – 9 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10
The positive integers are written right to the zero in increasing order whereas the
negative integers are written left to the zero in decreasing order.
3.2 Laws of addition of integers
There are some properties of addition of integers. These properties are called the laws
of addition of integers.
(i) Closure law
The sum of two or more integers is also an integer. It is called closure law of
addition of integers.
For example,
4 + 3 = 7, 9 + (–5) = 4, –6 + 4 + (–3) = –5 and so on.
Here, the sums 7, 4 and –5 are also the integers.
Thus, if a, b, c are any three integers and Z is the set of integers, then
a + b ∈ Z, a + b + c ∈ Z, b + c ∈ Z, a + c ∈ Z
(ii)Commutative law
The sum of two integers remains unchanged if their places are interchanged. It
is called the commutative law of addition of integers.
For example,
5+4=4+5=9 –7 + 3 = 3 + (–7) = –4 and so on.
Thus, if a and b are any two integers, then a + b = b + a.
Vedanta Excel in Mathematics - Book 8 32
Integers
(iii) Associative law
The sum of three integers remains unchanged if the order in which they are
grouped is altered. It is called the associative law of addition of integers.
For example,
(3 + 4) + 2 = 3 + (4 + 2) = (3 + 2) + 4 = 9
(–6 + 5) + (–4) = –6 + [5 + (–4)] = [–6 + (–4)] + 5 = –5
Thus, if a, b and c are any three integers, then
(a + b) + c = a + (b + c) = (a + c) + b
(iv) Identity law
If zero (0) is added to any integer, the sum is equal to the integer itself. It is called
the identity law of addition of integers.
For example,
9 + 0 = 9, –3 + 0 = –3, 0 + (–7) = –7
Thus, if a is any integer, then a + 0 = a.
Here, zero is known as the identity element of addition.
(v) Inverse law
For every integer, there exists an integer such that their sum is zero (0). It is
called the inverse law of addition of integers.
For example,
5 + (–5) = 0, –8 + 8 = 0
Thus, if a is any integer, then a + (–a) = 0. Here, each integer is said to be the
additive inverse of the other. So, (+a) is the additive inverse of (–a) and (–a) is the
additive inverse of (+a).
3.3 Sign rules of addition of integers
(i) The positive integers are always added and the sum holds the positive (+) sign.
For example,
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
(+2) + (3) = 2 + 3 = + 5 or 5
(ii) The negative integers are always added and the sum holds the negative (–) sign.
For example,
(–2) + (–5) = –2 – 5 = – 7 -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
(iii) The positive and negative integers are always subtracted and the difference holds
the sign of the bigger digit.
For example, -10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
(+6) + (–4) = 6 – 4 = 2 [Positive digit (+6) is bigger]
33 Vedanta Excel in Mathematics - Book 8
Integers
(+2) + (–7) = 2 – 7 = –5 [Negative digit (–7) is bigger]
-10 -9 -8 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 10
Thus, the integers with the same sign are always added and the integers with the
different sign are always subtracted.
3.4 Law of multiplication of integers
(i) Closure law
The product of two integers is always an integer. It is called the closure law of
multiplication of integers.
For example,
4 × 2 = 8, –3 × 4 = –12, (–5) × (–4) = 20 and so on.
Thus, if a and b are any two integers and z is the set of integers, then a × b ∈ Z.
(ii) Commutative law
The product of two integers remains unchanged if their places are interchanged.
It is called the commutative law of multiplication of integers.
For example,
3 × 4 = 4 × 3 = 12, (–2) × 3 = 3 × (–2) = –6 and so on.
Thus, if a and b are any two integers, then a × b = b × a.
(iii) Associative law
The product of three integers remains unchanged if the order in which they are
grouped is altered. It is called the associative law of multiplication of integers.
For example,
(2 × 4) × 3 = 2 × (4 × 3) = (2 × 3) × 4 = 24, and so on.
Thus, if a, b and c are any three integers, then
(a × b) × c = a × (b × c) = (a × c) × b.
(iv) Distributive law
The distributive law of multiplication of integers states that multiplying
an integer by a group of integers added together is the same as doing each
multiplication separately.
For example,
3 × (2 + 4) = 3 × 2 + 3 × 4, and so on.
Thus, if a, b and c are any three integers, then,
a × (b + c) = a × b + a × c.
(v) Identity law
Identify law of multiplication states that the product of 1 and any integer is the
integer itself.
For example,
2 × 1 = 2, 3 × 1 = 3, –4 × 1 = –4, and so on.
Thus, if a is any integer then a × 1 = a.
Vedanta Excel in Mathematics - Book 8 34
Integers
3.5 Sign rules of multiplication and division of integers
(i) The product or quotient of two positive integers is always positive.
For example,
(+4) × (+2) = 4 × 2 = + 4, (+8) ÷ (+4) = 8 ÷ 4 = +2, and so on.
(ii) The product or quotient of two negative integers is always positive. For example,
(–3) × (–2) = + 6, (–12) ÷ (–4) = + 3, and so on.
Let’s investigate the idea of this fact from the following pattern of operations.
2 × (–4) = –8
multiplicand 1 × (–4) = –4 products are
are decreasing 0 × (–4) = 0 increasing
(–1) × (–4) = 4
(–2) × (–4) = 8
(iii) The product or quotient of two positive and negative integers is always negative.
For example,
(+3) × (–2) = –6, (–4) × (+5) = –20, (+18) ÷ (–3) = –6, and so on.
3.6 Simplification of integers
We should follow the order to operations while simplifying the mixed operations
involving addition, subtraction, multiplication and division. In the order, the first
operation is division (÷), then multiplication (×), then addition (+) and subtraction
(–) at the last.
Worked-out examples
Example 1: Simplify 27 + 72 ÷ 8 – 4 × 7
Solution:
27 + 72 ÷ 8 – 4 × 7 = 27 + 9 – 4 × 7 [1st is division: 72 ÷ 8 = 9.]
= 27 + 9 – 28 [Then, multiplication: 4 × 7 = 28]
= 36 – 28 [Then, addition: 27 + 9 = 36]
= 8 [Then, subtraction: 36 – 28 = 8]
In the case of simplification of mixed operations inside brackets, we should start
the order of operation from small brackets ( ), then middle brackets { }, then the
bigger or box bracket [ ]. If there is parenthesis (–––) inside the brackets, it should be
removed at first.
35 Vedanta Excel in Mathematics - Book 8
Integers
Example 2: Simplify –98 ÷ 14 + [7 × {10 ÷ (18 ÷ 11 – 2)}]
Solution:
–98 ÷ 14 + [7 × {10 ÷ (18 ÷ 11 – 2)}]
= –98 ÷ 14 + [7 × {10 ÷ (18 ÷ 9}] → Operation in parenthesis inside small brackets
= –98 ÷ 14 + [7 × {10 ÷ 2}] → Operation inside ( )
= – 98 ÷ 14 + [7 × 5] → Operation inside { }
= –98 ÷ 14 + 35 → Operation inside [ ]
= –7 + 35
= 28
Example 3: 9 is subtracted from 5 times the sum of 4 and 3. Simplify by making a
mathematical expression.
Solution:
5 times the sum of 4 and 3 = 5 × (4 + 3)
∴ The expression = 5 × (4 + 3) – 9
=5×7–9
= 35 – 9
= 26
Example 4: The product of 8 and the difference of 17 and 12 is divided by 10. Simplify
by making mathematical expression.
Solution:
The product of 8 and the difference of 17 and 12 = 8 × (17 – 12)
∴ The expression = {8 × (17 – 12)} ÷ 10
= {8 × 5} ÷ 10
= 40 ÷ 10
=4
Example 5: A man had Rs 2,000. He purchased 2 kg of apples at Rs 150 per kg,
3 kg of sugar at Rs 80 per kg and he donated Rs 1,000 to a charity. If he
divided the remaining sum between his daughter and a son equally,
find the share of the daughter and the son.
Solution:
[2000 – {(2 × 150) + (3 × 80) + 1000}] ÷ 2
= [2000 – {300 + 240 + 1000}] ÷ 2
= [2000 – 1540] ÷ 2
= 460 ÷ 2
= 230
Hence, so, the share of each daughter is Rs 230.
Vedanta Excel in Mathematics - Book 8 36
Integers
EXERCISE 3.1
General Section - Classwork
1. Let's, tell and write the correct answers as quickly as possible.
a) If x and y are any two integers and Z is the set of integers, then write the following
laws of addition and multiplication of integers in mathematical expressions.
(i) closure law of addition of integers ..................................................
(ii) commutative law of addition of integers ..................................................
(iii) closure law of multiplication of integers ..................................................
(iv) commutative law of multiplication of integers ..................................................
b) If x, y and z are any three integers, then write the following laws in mathematical
expressions.
(i) associative law of addition of integers .....................................................
(ii) associative law of multiplication of integers .......................................................
c) If p is any integer, then write mathematical expressions to show the following laws.
(i) identity law of addition of integer ..............................................................
(ii) identity law of multiplication of integer ..............................................................
d) If x, y and z are any three integers, then according to the distributive law of
multiplication of integers .......................................................................................
2. a) What is the identity element of addition of integers? ............................................
b) What is the identity element of multiplication of integers? ...................................
c) What is the additive inverse of + 6? ...............................
d) What is the additive inverse of –3? ...............................
3. Let's simplify as quickly as possible.
a) 3 + 6 = ............ 6 – 3 = ............ 3 – 6 = ............ –3 – 6 = ............
b) 2 × 4 = ............ –2 × 4 = ............ 2 × (–4) = ............ (–2) × (–4) = ............
c) 6 ÷ 2 = ............ –6 ÷ 2 = ............ 6 ÷ (–2) = ............ (–6) ÷ (–2) = ............
d) –3 + 7 + 5 = ............ e) 2 + 4 × 3 – 5 = ............
f) 9 – 4 × 12 ÷ 6 + 1 = ............ g) 20 ÷ (3 + 2) × 10 = ............
37 Vedanta Excel in Mathematics - Book 8
Integers
Creative Section - A b) –5 × 3 + 7 × 4 – 54 ÷ 9 + 2
d) 60 ÷ 3 × (–4 + 12 – 6) × 2
4. Let's simplify.
a) 18 – 3 × 5 + 32 ÷ 4
c) 24 – (5 – 9 + 18) ÷ 7
5. Let's simplify.
a) 48 – {30 – (17 + 5)} ÷ 2 b) 27 ÷ {12 – (3 + 6)} × 5
c) 37 + {35 ÷ (4 – 7 + 8)} – 10 d) 48 – 15 ÷ {3 × (–2 – 4 + 11)} + 12
e) 7 × 50 ÷ {30 – (5 × 12 ÷ 6 + 5) + 10}
f) 5 × 12 ÷ [–12 ÷ {15+ (4 – 13)}] g) 10 [25 – {8 – 6(16 – 13)}] ÷ 5
h) –19 + [27 – {14 + (5 – 2) × 4 ÷ 2}] i) 56 ÷ 7 – 3 [4 + {8 – 4 (4 + 5 – 3)}]
j) –72 ÷ 9 + [11 × {10 ÷ (15 ÷17 – 14)}]
k) 100 ÷ 2 [500 ÷ 5 {6 + (21 + 7 – 24)}]
l) 300 ÷ 25 [56 ÷ 7 {24 – 2(6 – 5 – 9)}]
6. Let's simplify by making mathematical expressions.
a) 4 times the sum of 3 and 5 is subtracted from 35.
b) 9 is subtracted from 3 times the sum of 4 and 2.
c) 3 is subtracted from 9 times 2 and 5 is added.
d) 5 is subtracted from one-fourth part of the product of 12 and 3 and multiplied
by 2.
e) 7 is subtracted from the quotient of 48 divided by the sum of 5 and difference
of 11 and 8.
f) The sum of 35 and one-fifth part of itself is added to the sum of one-seventh
part of itself and 3.
g) The sum of 48 and itself, its half and half of the half is added to 18.
Creative Section - B
7. Let's make the mathematical expressions. Simplify them and find the correct answers.
a) You bought 3 gel pens at Rs 25 each, 6 pencils at Rs 10 each and 2 boxes at
Rs 80 each. If you gave a Rs 500 note to the shopkeeper, what changes did the
shopkeeper return you?
b) Sunayana had 50 sweets. She ate 10 sweets and gave 8 sweets to her brother.
If she divided the rest of them among her 8 friends equally, how many sweets
would each friend get?
c) Mr. Hamal had Rs 4,400. He purchased 6 kg of rice at Rs 75 per kg, 2 packets
of oil at Rs 125 per packet and he gave Rs 3,300 to his wife. If he divided the
remaining sum between his son and daughter equally, find the share of each of
them.
It's your time - Project work
8. a) Let's use 3 eights with proper signs (+, –, ×, ÷) and brackets and make the
separate expressions to get the result 2, 7, 8 and 9.
b) Let's make any two mixed operations using all four signs (+, –, ×, ÷) using middle
{ } and small ( ) brackets. Then simplify them and get the correct answers.
Vedanta Excel in Mathematics - Book 8 38
Unit Real Number System
4
4.1 Rational numbers – Looking back
Classroom - Exercise
1. Let's tell and write ‘True’ or ‘False’ for the following statements.
a) Set of integers is a subset of the set of rational numbers. ............................
b) Every rational number is an integer. ............................
c) The product of any two integers is always an integer. ............................
d) When an integer is divided by another integer, ............................
the quotient is always an integer.
e) When an integer is divided by another integer, ............................
the quotient is always a rational number.
f) In 9 ÷ 3 = 3, 3 is an integer. ............................
............................
g) In 3 ÷ 9 = 31, 1 is an integer. ............................
3
h) In –10 ÷ 2 = –5, –5 is a rational number. ............................
i) In 5 ÷ (–10) = – 1 , – 1 , is a rational number.
2 2
j) 5 is an irrational number.
k) 5 is a rational number. ............................
2
Let's study the following sets of numbers.
N = {1, 2, 3, ...} is the set of natural numbers.
W = { 0, 1, 2, 3, ...} is the set of whole numbers.
Z = {... –3, –2, – 1, 0, 1, 2, 3, ...} is the set of integers.
Thus, every natural number is a whole number and every whole number is an integer.
Therefore, sets of natural numbers and whole numbers are the subsets of the set of
integers.
Now, let's discuss the answers of these questions.
In 3 + 6 = 9, is 9 an integer?
In 3 – 6 = – 3, is –3 an integer?
In 3 × 6 = 18, is 18 an integer?
In 6 ÷ 3 = 2, is 2 an integer?
In 3÷6 = 21, is 1 an integer?
Of course, 21, is 2
not an integer. Thus, when an integer is divided by another integer, the
quotient is not always an integer. Therefore, there must be another set of numbers that
includes such quotients which are not integers. The set of such numbers is called the
set of rational numbers.
39 Vedanta Excel in Mathematics - Book 8
Real Number System
Any numbers which can be expressed in the form a , where a and b are integers and
b ≠ 0 are called rational numbers. b
The set of rational number is denoted by ‘Q’.
5 3 1 3 5
∴ Q = {... – 3, – 2 , – 2, – 2 , – 1, – 2 , 0, , 1, 2 , 2, 2 , 3, ...}
In this way, the sets of natural numbers, whole numbers and integers are the subsets of
the set of rational numbers. It means every natural number, whole number and integer
is a rational number. For example:
0 = 0, 7 = 7, – 9 = –9, …, where 0, 7, –9, … are all rational numbers.
1 1 1
4.2 Terminating and non-terminating recurring decimals
When we express a rational number into decimal, it may be terminating or
non-terminating decimal. If the decimal is non-terminating, a digit or block of digits in
the decimal part repeat after a certain interval. It is called non-terminating recurring
decimal. For example,
1 = 0.5 (Terminating decimal) 1 = 0.3 ....... (Non-terminating decimal)
2 (Terminating decimal) 3
2 = 0.4 5 = 1.83 ....... (Non-terminating decimal)
5 6
3 = 0.75 (Terminating decimal) 8 = 0.72 ....... (Non-terminating decimal)
4 11
The non-terminating recurring decimals can be indicated by putting dots just above the
beginning and end of the repeated digit or block of digits.
Worked-out examples
Example 1: Multiply a) 0.•3 b) 0.6•3• c) 0.5•9•2 by 10, 100 and 1000.
Solution:
a) 0.•3 × 10 = 0.3•3 × 10 = 3.•3
0.•3 × 100 = 0.333• × 100 = 33.•3
0.•3 × 1000 = 0.333•3 × 1000 = 333.•3
b) 0.6•3• × 10 = 0.636•3• × 10 = 6.36•3•
0.6•3• × 100 = 0.636•3• × 100 = 63.6•3•
0.6•3• × 1000 = 0.636•3• × 1000 = 636.36•3•
c) 0.•59•2 × 10 = 0.5925•9•2 × 10 = 5.925•9•2
0.•59•2 × 100 = 0.5925•9•2 × 100 = 59.25•9•2
0.•59•2 × 1000 = 0.5925•9•2 × 1000 = 592.5•9•2
Vedanta Excel in Mathematics - Book 8 40
Real Number System
Example 2: Convert the following non-terminating recurring decimals into fractions.
a) 0.•2 b) 0.53• c) 0.2•9•6
Solution:
a) Let x = 0.•2
or, x = 0.22 .................. (i) 2 is the non-terminating recurring digit.
Multiplying equation (i) by 10,
10x =2.22 ....................(ii)
Subtracting (i) from (ii)
10x – x = 2.22 – 0.22
or, 9x = 2
or, x = 2
∴ 0.•2 9
b) 2
Let, x = 09.53• = 0.5333 3 is the non-terminating recurring digit.
=
or, 10x = 5.333 ............... (i) Multiplying both sides by 10.
or, 100x = 53.333 ...............(ii) Multiplying both sides by 100.
Subtracting (i) from (ii),
100x – 10x = 53.333 – 5.333
or, 90x = 48
or, x = 48 = 8
90 15
0.53• 8
∴ = 15
c) Let, x = 0.2•9•6 = 0.296296 .................(i) 296 is the non-terminating recurring
block of digits.
or, 1000x = 296.296296 ..........................(ii) Multiplying both sides by 1000.
Vedanta Excel in Mathematics - Book 8
Subtracting (i) from (ii),
1000x – x = 296.296296 – 0.296296
or, 999x = 296
or, x = 296
999
0.2•9•6
∴ = 296
999
41
Real Number System
4.3 Irrational numbers
Let’s take a square of unit length (say 1 cm). C 1 unit B
OB is the diagonal of the square. O 1 unit
Δ OAB is the right-angled triangle right-angled at O.
So, OA is the base, AB is the perpendicular and OB is the A
hypotenuse of the right-angled triangle OAB. By using
Pythagorus theorem to find the length of OB,
OB = OA2 + AB2 = 12 + 12 = 2 unit
Here, 2 does not belong to the set of rational numbers.
Therefore, besides rational numbers, another set of numbers also must exist in the
number system and that we call the set of irrational numbers.
Now, the question is whether it is possible to show 2 in the number line.
Let’s take a point P of the coordinates (1, 1) in the coordinate axes of the graph paper and
join it with the origin O (0, 0).
P (1, 1)
2
O 1 Q2
2
Now, by using distance formula between the points O (0, 0) and P (1, 1),
OP = (x2 – x1)2 + (y2 – y1)2 = (1 – 0)2 + (1 – 0)2 = 12 + 12 = 2 units
Let’s draw a circle with the radius OP and centre at O. The circumference of the circle
intersect OX (the number line) at Q.
Here, radius of the circle OP = OQ = 2 units.
In this way, 2 can also be shown in the number line. It means irrational numbers can
also be shown in the number line.
Again, let’s consider the number 2 . Here, 2 = 1.414213562...
In this way, when 2 is expressed into decimal, it is non-terminating and non-recurring.
So, it is not a rational number. It is an irrational number. 3 , 5 , 6 , 7 , 8 , etc. are
a few examples of irrational numbers. On the other hand, 4 is not an irrational number
because the square root of 4 is 2 which is a rational number.
Vedanta Excel in Mathematics - Book 8 42
Real Number System
4.4 Real numbers
We have already learned the following facts about the number system.
(i) Set of natural numbers is the subset of the set of whole numbers.
(ii) Set of whole numbers is the subset of the set of integers.
(iii) Set of integers is the subset of the set of rational numbers.
(iv) The set of numbers which are not the rational numbers are the irrational numbers.
Now, the set of numbers that contains both the rational and irrational numbers and can
be shown in the number line is called the set of real numbers. It is denoted by R.
∴ R = {x : x ∈ Q and x ∈ Q }
Where Q is the set of rational numbers and Q is the set of irrational numbers.
Study the following diagram to know the relationship of different sets of numbers to the
real number system.
Real numbers
Rational numbers Irrational numbers
0, 1, 2, –1, –5, 21, 23, … 2, 3, 5, …
Fractional numbers Integers
21, 23, 14, 75, … –2, –1, 0, 1, 2, 3, …
Terminating Non-terminating Positive integers Negative integers
decimals recurring decimals 0, 1, 2, 3, 4, … –4, –3, –2, –1, …
21, 41, 53, 87, … 31, 65, 76, …
Natural numbers Whole number
1, 2, 3, 4, … 0, 1, 2, 3, 4, …
EXERCISE 4.1
General Section - Classwork
1. Let's list the rational and irrational numbers separately.
2, 2 5, 22 , π, 6.1428, Rational numbers Irrational numbers
7
9 , 49 , – 4, 3 27, 27
16 50
2. Let's list the terminating and non-terminating recurring decimal numbers
separately.
12, 13, 34, 27, 53, Terminating Decimal numbers Non-terminating Decimal numbers
5 , 34, 58, 190, 4
6 9
43 Vedanta Excel in Mathematics - Book 8
Real Number System
3. Let's read the statements carefully and write ‘T' for the true statements and ‘F' for
the false statements.
a) Set of whole number is the universal set of natural numbers.
b) Set of integers is the subset of whole numbers.
c) The set of rational numbers is the universal set of integers.
d) The set of rational numbers is the improper subset of
the set of real numbers.
e) The set of irrational numbers is the subset of the set of real numbers.
f) If R is the set of real numbers, Q is the set of rational numbers
and Q is the set of irrational numbers, then R = Q ∪ Q.
Creative Section
4. aM)u0l.t•2iply the bfo)l0lo.•3wing non-tce)r0m.8in3•ating recudrr)i0n.g•54•decimals bey)100.•2, 11•600 and 1000.
5. aC)o0n.v•3ert the fbo)ll0o.w•2ing non-tecr)m0.i•4n1•ating recurdri)n0g.•2d4•ecimals into 0fr.a•3c2t•4ions.
e)
f) 0.•13•2 g) 1.•57• h) 1.•43•2 i) 0.83• j) 0.26•
6. a) Draw a Venn-diagram to show the relation between the sets of real
numbers (R), rational numbers (Q) and irrational numbers (Q).
b) Draw a Venn-diagram to show the relation between the sets of real numbers
(R), rational numbers (Q), integers (Z), whole numbers (W), natural numbers
(N) and irrational numbers (Q).
It's your time -Project work
7. a) Let's write a few rational numbers with the denominators 2, 4, 5, 8 and 10.
Identify whether they are terminating or non-terminating recurring decimals.
b) Let's write a few rational numbers with the denominators 3, 6, 7 and 9. Identify
whether they are terminating or non-terminating recurring decimals.
c) Let's discuss in the class, what types of numbers in the denominators of rational
number always produce terminating decimals while dividing any numerator?
8. Let's draw a diagram showing all possible subsets of the set of real number on a
chart paper.
4.5 Scientific notation of numbers
It is somehow tedious to read, write and compute with very large or a very small
numbers. In such cases, we express a very large or a very small number as the product
of digit (or digits) and some power of 10. It is known as scientific notation of numbers.
For example:
4500 = 4.5 × 103, 299000 = 2.99 × 105, 693000000 km = 6.93 × 108 km,
0.0072 = 7.2 × 10–3, 0.00000000954 kg = 9.54 × 10–9 kg and so on
Vedanta Excel in Mathematics - Book 8 44
Real Number System
Now, let's study the following rules and learn to express numbers in scientific notation.
Rule 1: The whole number part in scientific notation of a number should be of one digit
number. For example,
59400000000 = 5.94 × 1010 (But 59.4 × 109 is incorrect scientific notation)
0.0000000826 = 8.26 × 10–8 (But 82.6 × 10–9 is incorrect scientific notation)
Rule 2: In the case of a large number, after separating whole number part and decimal
part by using decimal point, count the number of digits after the decimal point
and express as the product of the same number of power of 10. For example:
Insert decimal point here.
3275000000 = 3.275 × 109 There are 9 digits after the decimal
point. So the power of 10 will be 109.
123456789
9 digits
Rule 3: In the case of a very small number, after separating whole number part and
decimal part by using decimal point, count the number of digits before the
decimal point and express as the product of the same number of negative power
of 10. For example:
0.0000000653 = 6.53 × 10–8 Decimal point is shifted after eight
number of digits to make a one digit whole
12345678 number. So, the power of 10 will be 10–8.
8 digits
Worked-out examples
Example 1: Write these numbers in scientific notation.
a) 1700000 b) 2745.8 c) 0.000054
Solution: The whole number part in scientific notation
a) 1700000 = 1.7 × 106 should have one digit. So, after the decimal,
there are six digits, so 106 is written.
b) 2745.8 = 2.7458 × 103 2745.8 = 274.58 × 101
= 2.75 × 103
274.58 = 27.458 × 102
27.458 = 2.7458 × 103
= 2.75 × 103
0.000054 = 0.00054 × 10–1
0.00054 = 0.0054 × 10–2
c) 0.000054 = 5.4 × 10–5 0.0054 = 0.054 × 10–3
0.054 = 0.54 × 10–4
0.54 = 5.4 × 10–5
45 Vedanta Excel in Mathematics - Book 8
Real Number System
Example 2: Write the decimal values of the following scientific notation of numbers.
a) 2.52 × 106 b) 5.723 × 102 c) 9.81 × 10–7
Solution:
a) 2.52 × 106 = 2.52 × 1000000 = 2520000
b) 5.723 × 102 = 5.723 × 100 = 572.3
c) 9.81 × 10–7 = 9.81 = 0.000000981
10000000
Example 3: Simplify a) 3.6 × 104 + 4.5 × 105 b) 2.8 × 10–2 – 1.2 × 10–3
Solution:
a) 3.6 × 104 + 4.5 × 105 = 0.36 × 105 + 4.5 × 105 = 4.86 × 105
b) 2.8 × 10–2 – 1.2 × 10–3 = 2.8 × 10–2 – 0.12 × 10–2 = 2.68 × 10–2
Example 4: Simplify a) (2.4 × 103) × (6.2 × 104) b) 4.5 × 106 + 2.55 × 107
Solution: 1.4 × 103 + 1.1 × 103
a) (2.4 × 103) × (6.2 × 104) = (2.4 × 6.2) × 103 + 4 = 14.88 × 107 = 1.488 × 108
b) 4.5 × 106 + 2.55 × 107 = 0.45 × 107 + 2.65 × 107
1.4 × 103 + 1.1 × 103 2.5 × 103
= 3 × 107 = 1.2 × 107 – 3 = 1.2 × 104
2.5 × 103
EXERCISE 4.2
Section A – Classwork
1. Let's tell the scientific notation of these numbers and write as quickly as possible.
a) 16000 = .................... b) 3250000 = .................... c) 517000000 = ....................
d) 0.294 = .................... e) 0.0058 = .................... f) 0.0000063 = ....................
g) 84.6 = .................... h) 145.3 = .................... i) 9137.6 = ....................
2. Let's tell and write the values of these numbers in decimal system.
a) 3.2 × 10 = ................. b) 5.19 × 102 = ................. c) 7.4629 × 104 = .................
Creative Section - A
3. Let's express the following numbers in scientific notations.
a) 324000000 b) 612000000000 c) 74835000000
d) 0.000003 e) 0.000000054 f) 0.000000000936
4. a) 1 square kilometres = 10000000000 square centimetres. Express it in scientific
notation.
b) If the volume of 1 litre of water is 1000 c.c., find the volume of 2000 litres of
water and rewrite it in scientific notation.
Vedanta Excel in Mathematics - Book 8 46
Real Number System
c) 1 square centimetres = 1 = 0.0001 square metres. Express it in scientific
notation. 10000
d) 1 square centimetres = 1 square kilometres. Express it in scientific
notation. 10000000000
e) 1 cubic centimetres = 1 cubic metres. Rewrite it in scientific notation.
1000000
f) How many seconds are there in a month of 30 days? Express it in scientific
notation.
5. a) The radius of the earth is 6.371 × 103 km. Write the value of this scientific
notation and express the radius in words.
b) The distance between the sun and the earth is 148000000 km. Rewrite it in
scientific notation.
c) The distance travelled by light in 1 year is called a light year. The value of 1 light
year is about 9460000000000 km. Express it in scientific notation.
6. Let's express the value of the following scientific notation of numbers in general
number system.
a) 2.7 × 103 b) 4.5 × 104 c) 7.56 × 105 d) 8.45 × 109
e) 2.5 × 10–2 f) 5.6 × 10–3 g) 4.95 × 10–5 h) 7.83 × 10–8
7. Let's simplify.
a) 2.6 × 103 + 1.8 × 103 b) 3.5 × 104 + 5.7 × 104
c) 4.3 × 105 + 6.7 × 105 d) 6.4 × 109 + 9.7 × 109
e) 3.4 × 102 + 2.6 × 103 f) 2 × 103 + 4 × 102
g) 6 × 105 + 3 × 104 h) 1.5 × 107 – 9.5 × 105
i) 1.3 × 105 – 4.9 × 104 j) 3.3 × 10– 4 – 5.2 × 10– 5
k) 4.8 × 10–5 – 3.5 × 10–6 l) 5.4 × 10–3 – 8.6 × 10–5
8. Let's simplify. b) (2.7 × 103) × (3.9 × 104)
a) (3.4 × 102) × (1.6 × 103) d) (3.6 × 10–8) × (5.4 × 10–5)
c) (4.5 × 10–3) × (6.7 × 105)
Creative Section - B
9. Let's simplify.
a) 1.5 × 104 × 4.5 × 103 b) 6.4 × 10 – 5 × 3.6 × 109
2.5 × 102 1.6 × 1010 × 1.8 × 10–3
c) 6.7 × 10–11 × 2.5 × 1020 × 3.6 × 1015 d) 7.8 × 107 + 6.5 × 107
(3.35 × 106)2 1.3 × 103
e) 3.2 × 103 + 4.8 × 104 f) 9.6 × 106 – 7.2 × 105
1.6 × 102 2.4 × 103
47 Vedanta Excel in Mathematics - Book 8
Real Number System
10. a) A petrol tank has 3.6 × 103 litres of petrol. If 1.5 × 104 litres of petrol is added
b) into it how much petrol does it have now?
c) An underground water tank contains 1.8 × 104 litres of water. When
d) 1.8 × 103 litres of water is used up how much water is left in the tank?
e) Sound travels 3.43 × 102 metres in 1 second. Calculate the distance travelled
by sound in 3.6 × 103 seconds.
How many petrol tanks of capacity 1.2 × 104 litres each are required to empty
4.8 × 105 litres of petrol?
The speed of light in Vacuum is 2.99 × 105 km in 1 second. Calculate the
distance travelled by light in 3.6 × 103 seconds.
It's your time - Project work
11. a) Let's visit to the available website and search the distance of 7 planets from
b) the earth. Express the distances in scientific notation.
What is your body weight? Write it in grams and milligrams, then express in
12. a) scientific notation.
b) Write any two 7-digit and 9-digit numerals and express them in scientific
notation.
Write any two decimal numbers with 6 and 7 decimal places. Express them in
scientific notation.
4.6 Introduction to surd
Numbers which are written in the radical sign ( ) and cannot be simplified to remove
the radical sign are surds. For example, 2 , 3 , 5 , etc. are the surds. Surds
Surds have a non-terminating and non-recurring decimals places. So, they are actually
irrational numbers. For example:
2 = 1.4142135…, 3 = 1.7320508…, 5 = 2.236068… and so on.
4 = 2, 9 = 3, 25 = 5, … are not surds. Because 4 , 9 , 25, etc. can be simplified
to get rational numbers.
4.7 Addition and subtraction of irrational numbers
Let's study the following illustrations and learn the addition and subtraction of the same
irrational numbers.
i) 2 + 2 = 2 2 It is like x + x = 2x (ii) 2 3 + 3 3 = 5 3 It is like 2y + 3y = 5y
(iii) 7 5 – 4 5 = 3 5 It is like 7a – 4a = 3a (iv) 5 2 – 3 2 = 2 2 It is like 5p – 3p = 2p
Thus, while adding or subtracting the same irrational numbers, we should simply add or
subtract their rational coefficients as like in the case of addition and subtraction of like
algebraic terms.
Vedanta Excel in Mathematics - Book 8 48
Real Number System
4.8 Multiplication and division of irrational numbers
In the case of multiplication and division of irrational numbers, we multiply or divide
their rational coefficients as well as the irrational numbers. For example,
(i) 2 2 × 2 = (2 × 1) 2 × 2 = 2 4 = 2 × 2 = 4 We got it !
(ii) 2 2 × 5 3 = (2 × 5) 2 × 3 = 10 6
(iii) 9 10 ÷ 3 2 = (9 ÷ 3) 10 ÷ 2 = 3 5 We should simply multiply or
divide rational numbers and
irrational numbers separately.
(iv) 4 15 ÷ 2 5 = (4 ÷ 2) 15 ÷ 5 = 2 3
4.9 Rationalisation
Let’s consider an irrational number 5 .
When it is multiplied by 5 , the product = 5 × 5 = 25 = 5 and it is a rational
number. The process of changing an irrational number into a rational number is called
rationalisation. Study the following examples.
2 × 2 = 4 =2 2 is the rationalising factor of 2 .
2 3 × 3 =2 9 =2×3 3 is the rationalising factor of 2 3 .
Thus, if the product of two irrational numbers is a rational number, each of them is
called the rationalising factor of the other.
4.10 Conjugate
Let’s consider a binomial irrational number 3 + 2 .
Here, ( 3 + 2 ) × ( 3 – 2 ) = ( 3 )2 – ( 2 )2 (a + b) (a – b) = a2 – b2
=3–2=1
Thus, when 3 + 2 is multiplied by 3 – 2 , the product is a rational number. So,
3 – 2 is the rationalising factor of 3 + 2 . In this case, the rationalising factor
3 – 2 is called the conjugate of 3 + 2 or 3 + 2 is the conjugate of 3 – 2 .
Study a few more examples given below.
Conjugate of 5 + 3 is 5 – 3 , conjugate of 7 – 2 is 7 + 2 and so on.
Therefore, in a + b , it's conjugate is a – b or vice versa.
Worked-out examples
Example 1: Simplify a) 3 3 + 5 3 – 4 3 b) 5 8 – 2 18 + 50
Solution:
a) 3 3 + 5 3 – 4 3 = 8 3 – 4 3 = 4 3
49 Vedanta Excel in Mathematics - Book 8
Real Number System
b) 5 8 – 2 18 + 50 8 = 4 × 2 → 4 is a perfect square
= 5 4 × 2 – 2 9 × 2 + 25 × 2 18 = 9 × 2 → 9 is a perfect square
=5×2 2 –2×3 2 +5 2
= 10 2 – 6 2 + 2 50 = 25 × 2 → 25 is a perfect square
= 11 2 – 6 2
=5 2 Thus, one factor is a perfect square and
another is 2 in each case.
Example 2: Simplify a) 2 3 × 3 6 b) 15 24 ÷ 3 8
Solution:
I understood !
a) 2 3 × 3 6 = 2 × 3 3 × 6 In 18, I should factorise 18 in
= 6 18 such a way that one of the factors
=6 9×2 should be a perfect square.
= 6 × 3 2 = 18 2
b) 15 24 ÷ 3 8 = (15 ÷ 3) 24 ÷ 8 or 15 24 =5 3
3 8
=5 3
Example 3: Simplify ( 5 – 2 ) ( 5 + 2 ).
Solution:
( 5 – 2 ) ( 5 + 2 ) = ( 5 )2 – ( 2 )2 = 5 – 2 = 3
Example 4: Rationalise the denominator of a) 3 b) 10 6 c) 2
3 35 5+ 3
Solution:
a) Multiplying the numerator and denominator by the rationalising factor 3 .
3 = 3× 3 = 3 3 = 3
3 3× 3 3
b) Multiplying the numerator and the denominator by the rationalising factor 5 .
10 6 = 10 6 × 5 = 10 30 = 10 30 =2 30
35 3 5× 5 3×5 15 3
c) Multiplying the numerator and denominator by the rationalising factor 5 – 3 .
2= 2 ( 5 – 3)
5+ 3 ( 5+ 3) ( 5 – 3)
= 2( 5 – 3 ) = 2( 5 – 3 ) =2 5– 3) = 5– 3
( 5 )2 – ( 3 )2 5–3 2
Vedanta Excel in Mathematics - Book 8 50
Real Number System
Example 5: Simplify a) 3 – 1 b) 2 – 6 + 3 18 – 32
23 28
Solution:
a) 3 – 1 = 3 – 1× 3 = 3 – 3 = 3 3 –2 3) = 3
2 3 2 3× 3 2 3 6 6
b) 2 – 6 + 3 18 – 32 = 22 – 6 + 3 9×2– 16 × 2
2 8 2× 2 4×2
= 22 – 6 +9 2 –4 2
2 22
= 2– 3 +5 2
2
=6 2– 3× 2 =6 2 – 32 = 12 2 –3 2) = 92
2× 2 2 2 2
EXERCISE 4.3
General Section - Classwork
1. Let's tell and write the rationalising factors of the following surds.
Surds Rationalising factors Surds Rationalising factors
a) 2
b) 2 3 e) 3 + 5
c) 3 5 f) 3 – 2
g) 1
d) 1
2 3+ 2
h) 1
5 –2
2. Let's simplify the following operations on surds.
a) 3 + 3 = ................... b) 5 + 5 = ..................
c) 2 2 + 3 2 = ................. d) 2 3 – 3 = ................
e) 5 6 – 2 6 = .................. f) 2 × 3 = ..................
g) 7 × 2 = ................... h) 2 3 × 3 5 = ...................
i) 6 ÷ 2 = ................... j) 6 15 ÷ 2 3 = ...................
51 Vedanta Excel in Mathematics - Book 8
Real Number System
3. Let's rationalise the denominators.
a) 1 = ................... b) 1 = ................... c) 1 = ...................
2 3 5
Creative Section - A
4. Let's simplify.
a) 3 2 + 4 2 b) 7 3 – 5 3 c) 3 + 3 3 + 2 3
d) 4 7 – 2 7 + 7 e) 9 6 – 3 6 – 6 f) 2 + 8
g) 2 2 + 3 18 h) 4 3 – 27 i) 4 8 + 3 18 + 32
j) 6 3 + 5 75 – 2 48 k) 9 32 – 6 50 – 2 2 l) 3 20 – 2 45 + 80 – 5
5. Let's simplify.
a) 2 2 × 3 b) 3 5 × 2 c) 2 2 × 4 5
d) 4 3 × 3 5 e) 2 × 6 f) 6 × 3
g) 2 × 2 × 2 h) 10 ÷ 2 i) 12 30 ÷ 6 6
j) 8 27 ÷ 3 12 k) 9 24 ÷ 18 l) 12 35 ÷ 2 28
6. Let's simplify.
a) 5 × ( 2 + 8 ) b) 4 3 × (2 12 – 3 3 )
c) ( 3 + 2 ) ( 3 – 2 ) d) ( 5 – 2 ) ( 5 + 2 )
e) ( 7 + 3 ) ( 7 – 3 ) f) (2 2 – 3 ) (2 2 + 3 )
g) (3 3 + 5 ) (3 3 – 5 ) h (4 6 – 2 7 ) (4 6 + 2 7 )
7. Let's rationalise the denominators of the following irrational numbers and simplify.
2 3 c) 3 d) 3 2 e) 5
a) 2 b) 3 22 23 56
g) 2
f) 1 h) 7 i) 2 5 3 2 j) 3 2
2 +1 3 –1 3+ 2 3 – 2 6 +3 2
Creative Section - B
8. Let's rationalise the denominators as per necessary and simplify.
a) 3 + 1 b) 5 – 1 c) 4 + 6 d) 21 – 5 7
2 3 2 5 3 72
4 f) 3 + 5 g) 32 – 6 +5 2 h) 6 –2 5 – 53
e) 3 2 – 2 52 2 3 4
Vedanta Excel in Mathematics - Book 8 52
Real Number System
9. Simplify:
a) 42 + 60 –2 20 + 2 175 b) 16 – 24 +2 50 –3 8
28 45 2 32
6+
c) 98 – 40 – 48 + 2 72 d) 2 24 + 2 46 + 3 12
50 128 12
e) 15 – 30 – 18 + 20 f) 72 – 48 – 45 +2 98
5 45 27 50 128
10. a) Find the perimeter of a rectangle with width 18 cm and length 50 cm.
b) Find the perimeter of a triangle whose sides are 75 cm, 48 cm and 27 cm.
c) Find the area of rectangular ground of length 25 3 m and breadth 20 6 m.
d) The base of a triangular field is 30 2 m and it's height (h) is 24 2 m. Find it's
area using the formula, area of triangle = 1 × b × h.
2
11. a) Find the perimeter and area of each of the following rectangles.
p = 27cm(i) (ii) (2 + 5) cm (iii) (7 + 2) m
( 3 + 3) cm
( 3 + 5) cm (4 + 5 cm (10 + 8 m
b) Find the area of each triangle. (Area of triangle =21 × base (b) × perpendicular (p))
(i) (ii) (iii) b = 5 5 cm
p = 4 18cm p = 3 20cm
b = 12 cm b = 6 8 cm
It's your time - Project work!
12. a) Let's write any four fractions with irrational denominators. Then rationalise
each denominator and simplify.
b) Let's write any four pairs of irrational numbers that give the following sums or
difference after addition or subtraction.
(i) sum is 9 2 (ii) sum is 12 3 (iii) difference is 5 (iv) difference is 2 7
c) Let's write any four pairs of irrational numbers that give the following products
or quotients after multiplication or division.
(i) product is 24 (ii) product is 30 (iii) quotient is 2 (iv) quotient is 3
53 Vedanta Excel in Mathematics - Book 8
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