Chapter 1.6 - Intersection of Sets

In the previous section, we saw Venn diagrams and union of sets. In this section, we will see intersection of sets.

Intersection of sets

• This can be explained in 9 steps:
1. Let A and B be any two sets.
• We perform an operation called ‘intersection of sets’ between A and B.
2. The set which is formed as a result of that operation,
    ♦ will contain only those elements which are common to both A and B.
3. The symbol ‘∩’ is used to represent intersection.
• So the intersection of A and B can be represented as A ∩ B.
5. Let us see some examples:
A = {2, 4, 6, 8, 10} and B = {2, 4, 6, 12, 14}. Find A ∩ B.
Solution:
• A ∩ B must contain only those elements which are common to both A and B.
• The elements 2, 4 and 6 are the only elements that appear in both A and B. So A ∩ B will contain only those elements.
• Thus we get: A ∩ B = {2, 4, 6}
• We can represent this intersection using Venn diagrams also.
    ♦ Fig.1.9(a) below shows A and B before the intersection.
    ♦ Fig.1.9(b) shows the new set which is A ∩ B.

Fig.1.9

 6. Another example:
Let A = { a, e, i, o, u } and B = { a, i, u }. Show that A ∩ B = B
Solution:
• A ∪ B must contain only those elements which are common to both A and B.
• The elements a, i and u are the only elements that appear in both A and B. So A ∩ B will contain only those elements.
• Thus we get: A ∩ B = {a, i, u}
    ♦ We see that, A ∩ B is same as B.
• We can represent this intersection using Venn diagrams also.
    ♦ Fig.1.10(a) below shows A and B before the intersection.
    ♦ Fig.1.10(b) shows the new set which is A ∩ B.

Fig.1.10

• We see that, all elements of B are contained in A. So B ⊂ A.
    ♦ When the intersection takes place, the resulting set is same as the subset.
    ♦ We can write: B ⊂ A ⇒ A ∩ B = B
◼  Compare this result with that of the union: B ⊂ A ⇒ A ∪ B = A.
• We see that:
   ♦ When there is union between superset and subset, the result is same as the superset.
   ♦ When there is intersection between superset and subset, the result is same as the subset.
7. One more example:
Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in school hockey team. Let Y = {Geeta, David, Ashok} be the set of students from Class XI who are in the school football team. Find X ∩ Y and interpret the set.
Solution:
• We have: X ∩ Y = {Geeta}
• We see that:
The set X ∩ Y consists of members who are both in the hockey team and football team.
8. Thus we can write the definition:

Definition 7: The intersection of two sets A and B is the set C which consists of all those elements which belong to both A and B.

• In symbols, we write: A ∩ B = { x : x ∈ A and x ∈ B }
◼  Compare this result with that of the union: A ∩ B = { x : x ∈ A or x ∈ B}
• We see that,
   ♦ In the case of union, we have 'or' between x ∈ A and x ∈ B.
   ♦ In the case of intersection, we have 'and' between x ∈ A and x ∈ B.
9. Consider two sets: A = {2, 4, 6, 8} and B = {1, 3, 5, 7}
• There are no elements which are common to both sets. So if we perform the operation of intersection, there will not be any elements in A ∩ B.
   ♦ That means, A ∩ B will be a null set.
• Any two sets which give a null set when intersection is performed, are called disjoint sets.
• We can write: If A ∩ B = ɸ, then A and B are disjoint sets.
• Disjoint sets can be represented using Venn diagrams as shown in fig.1.11 below: 

Fig.1.11

• We see that, disjoint sets do not overlap in Venn diagram.

---

Some Properties of the Operation of Intersection

1. A ∩ B = B ∩ A (Commutative law)
• This is obvious from fig.1.9 above.
    ♦ Both A ∩ B and B ∩ A will give the same result in fig.1.9(b).
2. (A ∩ B) ∩ C = A ∩ (B ∩ C) (Associative law).
• This can be proved in 3 steps, using Venn diagrams.
(i) Fig.1.12(a) below shows the three sets A, B and C.

Fig.1.12

• Fig.b shows (A ∩ B) and C.
    ♦ Fig.c shows the intersection between (A ∩ B) and C.
(ii) Fig.1.13(a) below shows the three sets A, B and C.

Fig.1.13

• Fig.b shows A and (B ∩ C).
    ♦ Fig.c shows the intersection between A and (B ∩ C).
(iii) We see that
    ♦ Fig.1.12(c)
    ♦ is same as
    ♦ Fig.1.13(c)
3. ɸ ∩ A = ɸ (Law of ɸ)
• This is obvious because,
    ♦ There are no elements common to ɸ and A.
4. U ∩ A = A (Law of U).
• This is obvious because,
    ♦ 'Elements of A' are the only elements which are common to both U and A.
5. A ∩ A = A (Idempotent law).
• This is obvious because,
    ♦ 'Elements of A' are the only elements which are common to both A and A.
(6) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)  (Distributive law)
• This can be proved using Venn diagrams.
(i) The left side of the '=' sign is shown in fig.1.14 below:

Fig.1.14

• Fig.1.14(a) shows the three sets A, B and C.
• Fig.b shows A and (B ∪ C).
    ♦ Fig.c shows the intersection between A and (B ∪ C).
(ii) The right side of the '=' sign is shown in fig.1.15 below:

Fig.1.15

• Fig.1.15(a) shows A ∩ B.
• Fig.b shows A ∩ C.
    ♦ Fig.c shows the union between (A ∩ B) and (A ∩ C).
(iii) We see that
    ♦ Fig.1.14(c)
    ♦ is same as
    ♦ Fig.1.15(c)

---

In the next section, we will see difference of sets

Previous

Contents

Next

Copyright©2021 Higher secondary mathematics.blogspot.com

Chapter 1.3 - Subsets

In the previous section, we saw the empty set, finite set and equal sets. In this section, we will see subsets

Some basics about subsets can be written in 14 steps:
1. Consider the following two sets:
X = {x : x is a student in the school}
Y = {y : y is a student in class 11}
2. It is clear that, all elements in Y will be present in X.
• We say that: Y is a subset of X.
   ♦ Using symbols, this is written as: Y ⊂ X
   ♦ The symbol ‘⊂’ stands for ‘is a subset of’ or ‘is contained in’.
3. We can write the definition:

Definition 4 A set A is said to be a subset of set B if every element of A is contained in B.

4. We already know that:
If a is an element of A, we can write a ∈ A.
• But since A is a subset of B, the element a will be present in B also. So we can write: a ∈ B
5. That means,
If a ∈ A implies a ∈ B then, A ⊂ B.
• The word implies can be replaced by the symbol ‘⇒ ’. So we can write:
If a ∈ A ⇒ a ∈ B then, A ⊂ B.
6. There are two situations where A can be a subset of B
(i) Every element in A is present in B. Further, B has some extra elements also.
(ii) Every element in A is present in B. There are no extra elements in B.
   ♦ This means B has the exact same elements as A.
7. Let us consider the second situation carefully:
• If B has the exact same elements as A, we can write A ⊂ B alright.
   ♦ But we can write B ⊂ A also.
• That means, if two sets A and B have the exact same elements, we can write both:
A ⊂ B and B ⊂ A.
8. Recall that, if two sets have the exact same elements, they are equal sets.
• So we can write:
If A ⊂ B and B ⊂ A, the sets A and B are equal.
• Using symbols, we can write:
A ⊂ B and B ⊂ A ⇒ A = B
9. Conversely, we can write:
If two sets A and B are equal, then A ⊂ B and B ⊂ C.
10. A theorem and it’s converse can be written using the symbol ‘⇔’.
• This symbol indicates two way implication. So we get: A ⊂ B and B ⊂ A ⇔ A = B
   ♦ The forward direction indicates what we wrote in (8).
   ♦ The reverse direction indicates what we wrote in (9).
11. What we saw in steps (7) to (10) is the analysis of the situation in 6(ii).
• What if the situation is as in 6(i) ?
   ♦ In such a situation, we say two things:
         ✰ A is a proper subset of B.
         ✰ B is a superset of A.
12. Let us see some examples:
Example 1:
• Consider the following two sets:
A = {x : x is a divisor of 56}
B = {x : x is prime divisor of 56}
• Obviously, all elements in B will be present in A.
   ♦ So we can write: B ⊂ A
Example 2:
• Consider the following two sets:
A = {1, 3, 5}
B = {x : x is an odd number less than 6}
• The set B in roster form is {1, 3, 5}.
• So we can write:
   ♦ All elements of A are present in B.
         ✰ So A ⊂ B
   ♦ All elements of B are present in A.
         ✰ So B ⊂ A
• Since A ⊂ B and B ⊂ A, we can write A = B.
Example 3:
• Consider the following two sets:
A = {a, l, o, y}
B = {a, e, i, o, u}
• We see that, A is not a subset of B.
   ♦ We write: A ⊄ B
• We see that, B is not a subset of A.
   ♦ We write: B ⊄ A
13. From the above steps, it is clear that, every set A is a subset of itself
• Also, since the empty set ɸ has no elements, ɸ is a subset of every set
14. If a set A has only one element, we call it a singleton set
• For example, {a} is a singleton set

Solved example 1.18
Consider the sets
ɸ, A = { 1, 3 }, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}.
Insert the symbol ⊂ or ⊄ between each of the following pair of sets:
(i) ɸ . . . B  (ii) A . . . B  (iii) A . . . C  (iv) B . . . C
Solution:
(i) Since is a subset of every set, we can write: ɸ ⊂ B
(ii) 3 is not present in B. So we write: A ⊄ B
(iii) Both 1 and 3 are present in C. So we write: A ⊂ C
(iv) All three elements 1, 5 and 9 are present in C. So we write B ⊂ C

Solved example 1.19
Let A, B and C be three sets. If A ∈ B and B ⊂ C, is it true that A ⊂ C?. If not, give an example.
Solution:
1. Let A = {1, 2}
• Given that, set A is an element of B. So we can write: B = {3, {1, 2}, 4}
• Given that B is a subset of C. So we can write: C = {3, {1, 2}, 4, 5}
2. Here we are inclined to think that, A is a subset of C
• But if we write a subset of C, it will be: {{1, 2}}
   ♦ {{1, 2}} is different from {1, 2}
   ♦ That is: {{1, 2}} is different from A.
• So the answer is: No, A cannot be a subset of C.
3. Another way to analyze this situation can be written in 3 steps:
(i) If A is to be a subset of C, all elements of A should be present in C as separate elements.
(ii) But in C, the elements 1 and 2 are present together as a set. Not as separate elements.
(iii) So the answer is: No, A cannot be a subset of C.

---

In the next section, we will see subsets of the set R

Previous

Contents

Next

Copyright©2021 Higher secondary mathematics.blogspot.com

Chapter 1.2 - Empty, Finite and Equal Sets

In the previous section, we saw the roster form and set-builder form. In this section, we will see empty set, finite set and equal sets.

The empty set can be described in 6 steps:
1. Consider the set:
A = {x : x is a student of Class XI presently studying in a school}
• Let us analyze this set:
We can write the names of all students of Class XI in that school. So set A will contain a definite number of names.
2.Consider another set:
B = {x : x is a student of both Class X and Class XI presently studying in that school}
• Let us analyze this set:
We will not find any student who studies in both Class X and Class XI in the same school. So set B will not contain any element.

Definition 1: • A set which does not contain any element is called the empty set    ♦ Another name for empty set is null set    ♦ Yet another name for empty set is void set

3. In our present case,
   ♦ A is not an empty set.
   ♦ B is an empty set.
4. Representing empty set:  
   ♦ Empty set is denoted by the symbol { }
   ♦ Another symbol for empty set is Ø
5. In our present case, we can write:
   ♦ A ≠ { }
   ♦ B = { }
6. Let us see a few more examples:
(i) Let A = {x : 1 < x < 2, x is a natural number}.
• Here x should be a natural number. Also, it must lie between 1 and 2.
• There is no such natural number. So the set will not contain any elements.
• In other words, A is an empty set.

(ii) B = {x : x² – 2 = 0 and x is rational number}.
• Here, the only solution of (x² – 2 = 0) is +√2 and -√2
• Both +√2 and -√2 are irrational numbers. So B is an empty set.

(iii) C = {x : x is an even prime number greater than 2}.
• Here, x should be:
    ♦ A prime number.
    ♦ An even number.
    ♦ Greater than 2.
• All prime numbers greater than 2 are odd numbers. So C is an empty set.

(iv) D = { x : x² = 4, x is odd}.
Here, the only solution of (x² = 4) is +2 and -2
• Both +2 and -2 are even numbers. So D is an empty set.

Solved example 1.10
Which of the following are examples of the null set
(i) Set of odd natural numbers divisible by 2
(ii) Set of even prime numbers
(iii) { x : x is a natural numbers, x < 5 and x > 7}
(iv) { y : y is a point common to any two parallel lines}
Solution:
(i) None of the odd natural numbers are divisible by 2
• So it is a null set.
(ii) All prime numbers except 2 are odd numbers.
• That means, there is one even prime number, which is '2'. So this is not a null set.
(iii) No natural number can be less than 5 and greater than 7 at the same time. So it is a null set.
(iv) Two parallel lines will never have a common point. So it is a null set.

---

Finite and Infinite sets

This can be explained in 7 steps
1. Consider the following set:
A = {2, 4, 6, 8}
• This set A has four elements. We write: n(A) = 4
   ♦ 4 is a finite natural number.
   ♦ So n(A) is a finite natural number.
2. Consider the following set:
V = {a, e, i, o, u}
• This set V has five elements. We write: n(V) = 5
   ♦ 5 is a finite natural number.
   ♦ So n(V) is a finite natural number.
3. Consider the following set:
C = {x : x is a citizen of India}
• This set will contain the names of all citizen of India. The elements are names.
   ♦ The ‘number of elements’ will be very large. But it will be a finite natural number.
   ♦ That is., n(C) will be a very large but finite natural number.
4. Consider the following set:
D = {x : x is a positive even number}
• This set will contain the numbers 2, 4, 6, 8, . . . The elements are even numbers.
   ♦ The ‘number of elements’ will be very very large.
   ♦ We will not be able to count them.
   ♦ That is., n(D) will be infinite.
5. We can now write the definition

Definition 2 • The set A is said to be a finite set if n(A) is a finite number.    ♦ The set A is said to be a finite set even if it is an empty set. • The set A is said to be an infinite set if n(A) is not a finite number.

6. Let us see some more examples:
(i) Consider the set:
D = {x : x is a day of the week}
• Then D = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
• We get: n(D) = 7
   ♦ '7' is a finite natural number. So D is a finite set.
(ii) Consider the set:
S = {x : x² – 16 = 0}.
• Here, x can take two values only. They are 4 and -4. So S = {4, -4}
• We get: n(S) = 2
   ♦ '2' is a finite natural number. So S is a finite set.
(iii) Consider the set:
G = {x : x is a point on a line}.
• There will be infinite number of points on a line.
• So G will contain infinite number of elements.
   ♦ That means, n(G) is not a finite number.
   ♦ So G is an infinite set.
7. We know that, in roster form, we write all the elements and enclose them within braces.
• But if the set is an infinite set, we will not be able to write all the elements.
• If the set is infinite, we write a few elements followed by three dots.
• Those few elements that are written, should clearly indicate the structure of the set.
• For example,
   ♦ If we write {1, 3, 5, 9, . . .}, it clearly indicates odd numbers.
   ♦ If we write {2, 2², 2³, 2⁴, . . .}, it clearly indicates the powers of 2.
• Some times, the three dots may precede the elements.
• For example,
   ♦ If we write { . . . , -3, -2, -1, 0, 1, 2, 3, . . .}, it clearly indicates integers.
◼ This method of providing three dots, cannot be used to indicate real numbers. This is because, real numbers do not follow any particular pattern. For example, the distance (on the number line) between √2 and √3 is not the same as that between √3 and √4.

Solved example 1.11
Which of the following sets are finite or infinite
(i) The set of months of a year
(ii) {1, 2, 3, . . .}
(iii) {1, 2, 3, . . .99, 100}
(iv) The set of positive integers greater than 100
(v) The set of prime numbers less than 99
Solution:
(i) The set will contain the names of the 12 months. The number of elements is 12. So it is a finite set.
(ii) This set has the definite pattern of natural numbers. But the pattern is followed by 3 dots. The dots indicate that, all the natural numbers must be included in the set. So it is an infinite set.
(iii) This set has the definite pattern of natural numbers. The pattern is followed by 3 dots. The dots usually indicate that, the set is infinite. But here, the last two elements are also given. It is clear that, there will be exactly 100 elements in the set. So it is a finite set.
(iv) There are infinite number of positive integers which are greater than 100. So it is an infinite set.
(v) There is a finite number of prime numbers which are less than 99. So it is a finite set.

Solved example 1.12
State whether each of the following set is finite or infinite:
(i) The set of lines which are parallel to the x-axis
(ii) The set of letters in the English alphabet
(iii) The set of numbers which are multiple of 5
(iv) The set of animals living on the earth
(v) The set of circles passing through the origin (0,0)
Solution:
(i) There are infinite number of lines which are parallel to the x-axis. So it is an infinite set.
(ii) There is a finite number of letters in the English alphabet. So it is a finite set.
(iii) There are infinite numbers which are multiple of 5. So it is an infinite set.
(iv) There is a finite number of animals living on the earth. So it is a finite set.
(iii) There are infinite number of circles passing through the origin (0,0). So it is an infinite set.

---

Equal Sets

This can be explained in 5 steps:
1. Suppose that, we have two sets A and B. We will be able to say that the two sets are equal, if two conditions are satisfied:
(i) All elements in A are present in B.
(ii) All elements in B are present in A.
2. But if the above two conditions are satisfied, it is obvious that, the two sets will be having the same elements.
3. So we can write the definition:

Definition 3 • Two sets A and B are said to be equal if they have exactly the same elements.    ♦ The equality is expressed as A = B • Two sets A and B are said to be unequal if they do not have exactly the same elements.    ♦ The inequality is expressed as A ≠ B

4. Let us see some examples:
Example (i):
• Let A = {2, 2¹⁄₂, 3, 3¹⁄₂} and B = {2, 3, 2¹⁄₂, 3¹⁄₂}
• We see that,
   ♦ All elements in A are present in B.
   ♦ All elements in B are present in A.
• A and B have exactly the same elements. They are equal sets.
   ♦ We write: A = B

Example (ii):
• Consider the following set:
A = {x : x is a prime number less than 6}
• Consider another set:
B = {x : x prime factor of 30}
• In roster form, the above sets are: A = {2, 3, 5} and B = {2, 3, 5}
We see that,
   ♦ All elements in A are present in B.
   ♦ All elements in B are present in A.
• A and B have exactly the same elements. They are equal sets.
   ♦ We write: A = B
5. An interesting note:
• Consider the sets: A = {1, 2, 3} and B= {2, 2, 1, 3, 3}
• We see that:
   ♦ All elements in A are present in B.
   ♦ All elements in B are present in A.
   ♦ So A = B
◼ That means, it not necessary to write the repeating elements in B more than once. That is the reason why in general, we avoid writing repeating elements more than once.

Solved example 1.13
Find the pairs of equal sets, if any, give reasons:
A = {0},
B = {x : x > 15 and x < 5},
C = {x : x – 5 = 0 },
D = {x: x² = 25},
E = {x : x is an integral positive root of the equation x² – 2x –15 = 0}.
Solution:
1. Let us convert those last 5 sets which are in set-builder form into roster form:
• We have: B = {x : x > 15 and x < 5}
   ♦ Obviously, no number can be less than 5 and greater than 15 at the same time.
   ♦ It is a null set.
   ♦ We can write: B = {}
• We have: C = {x : x – 5 = 0 }
   ♦ We can write: C = {5}
• We have: D = {x: x² = 25}
   ♦ Solving (x² = 25), we can write: D = {5, -5}
• We have: E = {x : x is an integral positive root of the equation x² – 2x –15 = 0}
   ♦ Solving (x² – 2x –15 = 0), the solution is 5 and -3
   ♦ But the set must contain only integral positive roots
   ♦ So we can write: E = {5}
2. So the sets are:
A = {0}, B = {}, C = {5}, D = {5, -5} and E = {5}
3. Let us compare each set with the others
(i) Set A:
• The element '0' is not present in any other set. So we write:
A ≠ B, A ≠ C, A ≠ D, A ≠ E
(ii) Set B:
• None of the other sets is a null set. So we write:
B ≠ A, B ≠ C, B ≠ D, B ≠ E
(iii) Set C:
• The element '5' is present only in E. The other sets cannot be equal to C. So we write:
C ≠ A, C ≠ B, C ≠ D
• Since '5' is present in 'E', we must check further. We see that, E has only '5'
• So we can write:
   ♦ All elements in C are present in E
   ♦ All elements in E are present in C
   ♦ Thus we get: C = E
(iv) Set D:
• D has '5' and '-5'. No other sets have both these elements. So we write:
D ≠ A, D ≠ B, A ≠ C, D ≠ E
(v) Set E
   ♦ We have already seen that, A, B and D are not equal to E
   ♦ We have also seen that, C = E
(vi) So the only equal pair is: C and E

Solved example 1.14
Which of the following pairs of sets are equal ? Justify your answer.
(i) X, the set of letters in “ALLOY” and B, the set of letters in “LOYAL”.
(ii) A = {n : n ∈ Z and n 2 ≤ 4} and B = {x : x ∈ R and x 2 – 3x + 2 = 0}.
Solution:
(i) We have: X = {A, L, O, Y} and B = {L, O, Y, A}.
   ♦ All elements in X are present in B
   ♦ All elements in B are present in X
• So we write: X = B
(ii) We have: A = {–2, –1, 0, 1, 2} and B = {1, 2}.
• We see that, '0' is present only in A
   ♦ All elements in A are not present in B
   ♦ All elements in B are present in A
• Since both conditions are not satisfied, we can write: A ≠ B

Solved example 1.15
In the following, state whether A = B or not:
(i) A = { a, b, c, d }, B = { d, c, b, a }
(ii) A = { 4, 8, 12, 16 }, B = { 8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}, B = { x : x is positive even integer and x ≤ 10}
(iv) A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }
Solution:
(i) A = { a, b, c, d }, B = { d, c, b, a }
   ♦ All elements in A are present in B
   ♦ All elements in B are present in A
• So we write: A = B
(ii) A = { 4, 8, 12, 16 }, B = { 8, 4, 16, 18}
   ♦ All elements in A are not present in B
   ♦ All elements in B are not present in A
• So we write: A ≠ B
(iii) A = {2, 4, 6, 8, 10}, B = { x : x is positive even integer and x ≤ 10}
• Set B in roster form is: {2, 4, 6, 8, 10}
   ♦ All elements in A are present in B
   ♦ All elements in B are present in A
• So we write: A = B
(iv) A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }
• Set A in roster form is: {10, 20, 30, 40, . . .}
   ♦ All elements in A are present in B
   ♦ All elements in B are not present in A
• So we write: A ≠ B

Solved example 1.16
Are the following pair of sets equal ? Give reasons.
(i) A = {2, 3}, B = {x : x is solution of x2 + 5x + 6 = 0}
(ii) A = { x : x is a letter in the word FOLLOW}, B = { y : y is a letter in the word WOLF}
Solution:
(i) A = {2, 3}, B = {x : x is solution of x² + 5x + 6 = 0}
• Set B in roster form is: {-2, -3}
   ♦ All elements in A are not present in B
   ♦ All elements in B are not present in A
• So we write: A ≠ B
(ii) A = { x : x is a letter in the word FOLLOW}, B = { y : y is a letter in the word WOLF}
• Set A in roster form is: {F, O, L, W}
• Set B in roster form is: {W, O, L, F}
   ♦ All elements in A are present in B
   ♦ All elements in B are present in A
• So we write: A = B

Solved example 1.17
From the sets given below, select equal sets :
A = { 2, 4, 8, 12}, B = { 1, 2, 3, 4}, C = { 4, 8, 12, 14}, D = { 3, 1, 4, 2},
E = {–1, 1}, F = { 0, a}, G = {1, –1}, H = { 0, 1}
Solution:
(i) Consider set A
• Take the first element '2'
   ♦ C, E, F, G and H can be ruled out because, '2' is not present in them
   ♦ The remaining sets are B and D
• Take the second element '4'
   ♦ This element is present in B and D
• Take the third element 8
   ♦ B and D can be ruled out because, '8' is not present in them
• So A is not equal to any other set

(ii) Consider set B
• Take the first element '1'
   ♦ A, C and F can be ruled out because, '1' is not present in them
   ♦ The remaining sets are D, E, G and H
• Take the second element '2'
   ♦ E, G and H can be ruled out because, '2' is not present in them
   ♦ The remaining set is D
• Take the third element '3'
   ♦ This element is present in D
• Take the fourth element '4'
   ♦ This element is present in D
   ♦ All the elements in B are present in D
• Comparing B and D, we see that:
   ♦ All the elements in B are present in D
   ♦ All the elements in D are present in B
◼ So B and D are equal sets

(iii) Consider set C
• A and B can be ruled out because, we already compared them with C
• Take the first element '4'
   ♦ E, F, G and H can be ruled out because, '4' is not present in them
   ♦ The remaining set is D
• Take the second element '8'
   ♦ D can be ruled out because, '8' is not present in it
• So C is not equal to any other set

(iv) Consider set D
• We already saw that,
   ♦ D = B
   ♦ B is not equal to any other set
• So D cannot be equal to any set other than B

(v) Consider set E
• A, B, C and D can be ruled out because, we already compared them with E
• Take the first element '-1'
   ♦ F and H can be ruled out because, '-1' is not present in them
   ♦ The remaining set is G
• Take the second element '1'
   ♦ This element is present in G
   ♦ All the elements in E are present in G
• Comparing E and G, we see that:
   ♦ All the elements in E are present in G
   ♦ All the elements in G are present in E
◼  So E and G are equal sets

(vi) Consider set F
• A, B, C, D and E can be ruled out because, we already compared them with F
• Take the second element 'a'
   ♦ G and H can be ruled out because, 'A' is not present in them
   ♦ There are no remaining sets to compare with F
• So F is not equal to any other set

(vii) Consider set G
• We already saw that,
   ♦ E = G
   ♦ E is not equal to any other set
• So G cannot be equal to any set other than E

(viii) Consider set H
• All other sets can be ruled out because, we already compared them with H
• So H is not equal to any other set

◼  Thus we get the final answer:
• The equal sets are:
   ♦ From (ii): B = D
   ♦ From (v): E = G

---

In the next section, we will see subsets

Previous

Contents

Next

Copyright©2021 Higher secondary mathematics.blogspot.com