Students can go through AP Inter 1st Year Maths Notes 1st Lesson Sets will help students in revising the entire concepts quickly.
Sets Class 11 Notes AP Inter 1st Year Maths 1st Lesson
Sets:
A set is a well-defined collection of objects.
- Objects, elements, and members of a set are synonymous terms.
- Sets are usually denoted by capital letters A, B, C; X, Y, Z, etc.
- The elements of a set are represented by small letters a, b, c, x, y, z, etc.
Examples:
- N: the set of all natural numbers
- Z: the set of all integers
- Q: the set of all rational numbers
- R: the set of real numbers
- The rivers of India
- The vowels in the English alphabet, namely, a, e, i, o, u
- Various kinds of triangles
Roster or Tabular Form:
In roster form, all the elements of a set are listed; the elements are separated by commas and are enclosed within braces { }.
In roster form, the order in which the elements are listed is immaterial. Thus, the above set can also be represented as {1, 3, 7, 21, 2, 6, 14, 42}.
It may be noted that while writing the set in roster form, an element is not generally repeated, i.e., all the elements are taken as distinct.
For example, the set of letters forming the word ‘SCHOOL’ is {S, C, H, O, L} or {H, O, L, C, S}. Here, the order of listing elements has no relevance.
Examples:
- The set of all natural numbers that divide 42 is {1, 2, 3, 6, 7, 14, 21, 42}.
- The set of all vowels in the English alphabet is {a, e, i, o, u}.
- The set of odd natural numbers is represented by {1, 3, 5,…}.
The dots tell us that the list of odd numbers continues indefinitely.
Set-builder form:
In set-builder form, all the elements of a set possess a single common property that is not possessed by any element outside the set.
Examples:
A = {x: x is a natural number and 3 < x < 10} is read as “the set of all x such that x is a natural number and x lies between 3 and 10”.
Hence, the numbers 4, 5, 6, 7, 8, and 9 are the elements of the set A.
If we denote the sets described in (a), (b), and (c) above in roster form by A, B, and C, respectively, then A, B, and C can also be represented in set-builder form as follows:
A = {x: x is a natural number that divides 42}
B = {y: y is a vowel in the English alphabet}
C = {z: z is an odd natural number}
The Empty Set:
A set that does not contain any element is called the empty set or the null set, or the void set.
Examples:
- Let A = {x: 1 < x < 2, x is a natural number}. Then A is the empty set, because there is no natural number between 1 and 2.
- Let B = {x: x2 – 2 = 0 and x is rational number}. Then B is the empty set because the equation x2 – 2 = 0 is not satisfied by any rational value of x.
- Let C = {x: x is an even prime number greater than 2}. Then C is the empty set, because 2 is the only even prime number.
Infinite Sets:
A set that is empty or consists of a definite number of elements is called finite; otherwise, the set is called infinite.
Examples:
- Let W be the set of the days of the week. Then W is finite.
- Let S be the set of solutions of the equation x² – 16 = 0. Then S is finite.
- Let G be the set of points on a line. Then G is infinite.
Equal Sets:
Two sets A and B are said to be equal if they have the same elements, and we write A = B. Otherwise, the sets are said to be unequal, and we write A ≠ B.
Examples:
- Let A = {1, 2, 3, 4} and B = {3, 1, 4, 2}. Then A = B.
- Let A be the set of prime numbers less than 6 and P the set of prime factors of 30. Then A and P are equal, since 2, 3, and 5 are the only prime factors of 30, and also these are less than 6.
Subsets:
A set A is said to be a subset of a set B if every element of A is also an element of B.
In other words, A ⊂ B if whenever a ∈ A, then a ∈ B. It is often convenient to use the symbol “⇒” which means implies.
Using this symbol, we can write the definition of a subset as follows:
A ⊂ B if a ∈ A ⇒ a ∈ B
We read the above statement as “A is a subset of B if a is an element of A implies that a is also an element of B”.
If A is not a subset of B, we write A ⊄ B.
Intervals as Subsets of R:
Let a, b ∈ R and a < b. Then the set of real numbers {y: a < y < b} is called an open interval and is denoted by (a, b).
All the points between a and b belong to the open interval (a, b), but a, b themselves do not belong to this interval.
The interval that contains the endpoints is also called a closed interval and is denoted by [a, b]. Thus [a, b] = {x : a ≤ x ≤ b}
We can also have intervals closed at one end and open at the other, i.e.,
[a, b) = {x: a ≤ x ≤ b} is an open interval from a to b, including a but excluding b.
(a, b] = {x: a < x ≤ b} is an open interval from a to b, including b but excluding a.
Complement of a Set:
Let U be the universal set and A a subset of U. Then the complement of A is the set of all elements of U that are not the elements of A.
Symbolically, we write A’ to denote the complement of A with respect to U.
Thus, A’ = {x: x ∈ U and x ∉ A}.
Obviously A’ = U – A
We note that the complement of a set A can be looked upon, alternatively, as the difference between a universal set U and the set A.
Properties of the Operation of Union:
- A ∪ B = B ∪ A (Commutative law)
- (A ∪ B) ∪ C = A ∪ (B ∪ C) (Associative law)
- A ∪ φ = A (Law of identity element, φ is the identity of U)
- A ∪ A = A(Idempotent law)
- U ∪ A = U (Law of U)
Properties of the Operation of Intersection:
- A ∩ B = B ∩ A (Commutative law)
- (A ∩ B) ∩ C = A ∩ (B ∩ C) (Associative law)
- φ ∩ A = φ, U ∩ A = A (Law of φ and U)
- A ∩ A = A (Idempotent law)
- A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) (Distributive law)
i.e., ∩ distributes over ∪
Properties of Complement Sets:
Complement laws:
- A ∪ A’ = U
- A ∩ A’ = φ
De Morgan’s law:
- (A ∪ B)’ = A’ ∩ B.’
- (A ∩ B)’ = A’ ∪ B.’
Law of double complementation: (A’)’ = A
Laws of the empty set and the universal set: φ’ = U and U’ = φ.
These laws can be verified by using Venn diagrams.
Example Problems
Question 1.
Write the solution set of the equation x2 + x – 2 = 0 in roster form.
Solution:
The given equation can be written as (x – 1) (x + 2) = 0, i.e., x = 1, -2
Therefore, the solution set of the given equation can be written in roster form as {1, -2}.
Question 2.
Write the set {x: x is a positive integer and x² < 40} in roster form.
Solution:
The required numbers are 1, 2, 3, 4, 5, 6.
So, the given set in the roster form is {1, 2, 3, 4, 5, 6}.
Question 3.
Write the set A = {1, 4, 9, 16, 25,…} in set-builder form.
Solution:
We may write the set A as A = {x: x is the square of a natural number}
Alternatively, we cam write A = {x: x = n2, where n ∈ N}
Question 4.
Write the set \(\left\{\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}\right\}\) in the set-builder form.
Solution:
We observe that each member in the given set has a numerator one less than its denominator.
Additionally, the numerator starts at 1 and does not exceed 6.
Hence, in the set-builder form the given set is {x: x = \(\frac{\mathrm{n}}{\mathrm{n}+1}\), where n is a natural number and 1 ≤ n ≤ 6}.
Question 5.
Match each of the sets on the left described in the roster form with the same set on the right described in the set-builder form:
(i) {P, R, I, N, C, A, L} – (a){x: x is a positive integer and is a divisor of 18}.
(ii) {0} – (b) {x: x is an integer and x2 – 9 = 0}
(iii) {1, 2, 3, 6, 9, 18} – (c) {x: x is an integer and x + 1 = 1}
(iv) {3, -3} – (d) {x: x is a letter of the word PRINCIPAL}
Solution:
Since in (d), there are 9 letters in the word PRINCIPAL and two letters P and I are repeated, so (i) matches (d).
Similarly, (ii) matches (c) as x + 1 = 1 implies x = 0.
Also, 1, 2, 3, 6, 9, 18 are all divisors of 18, and so (iii) matches (a).
Finally, x2 – 9 = 0 implies x = 3, -,3 and so (iv) matches (b).
Question 6.
State which of the following sets are finite or infinite:
(i) {x: x ∈ N and (x – 1) (x – 2) = 0}
(ii) {x: x ∈ N and x2 = 4}
(iii) {x: x ∈ N and 2x – 1 = 0}
(iv) {x: x ∈ N and x is prime}
(v) {x: x ∈ N and x is odd}
Solution:
(i) Given set = {1, 2}. Hence, it is finite.
(ii) Given set = {2}. Hence, it is finite.
(iii) Given set = φ. Hence, it is finite.
(iv) The given set is the set of all prime numbers, and the set of prime numbers is infinite. Hence, the given set is infinite.
(v) Since there is are infinite number of odd numbers, hence, the given set is infinite.
Question 7.
Find the pairs of equal sets, if any, and 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:
Since 0 ∈ A and 0 does not belong to any of the sets B, C, D, and E, it follows that A ≠ B, A ≠ C, A ≠ D, and A ≠ E.
Since B = φ, but none of the other sets are empty.
Therefore, B ≠ C, B ≠ D, and B ≠ E.
Also, C = {5} but -5 ∈ D, hence C ≠ D.
Since E = {5}, C = E.
Further, D = {-5, 5} and E = {5}, we find that D ≠ E.
Thus, the only pair of equal sets is C and E.
Question 8.
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 n2 ≤ 4} and B = {x: x ∈ R and x2 – 3x + 2 = 0}.
Solution:
(i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}.
Then X and B are equal sets, as repetition of elements in a set does not change a set.
Thus, X = {A, L, O, Y} = B.
(ii) A = {-2, -1, 0, 1, 2}, B = {1, 2}.
Since 0 ∈ A and 0 ∉ B, A and B are not equal sets.
Question 9.
Consider the sets φ, A = {1, 3}, B = {1, 5, 9}, C = {1, 3, 5, 7, 9}. Insert the symbol ⊂ or ⊄ between each of the following pairs of sets:
(i) φ…B
(ii) A…B
(iii) A…C
(iv) B…C
Solution:
(i) φ ⊂ B as φ is a subset of every set.
(ii) A ⊄ B as 3 ∈ A and 3 ∉ B.
(iii) A ⊂ C, as 1, 3 ∈ A also belong to C.
(iv) B ⊂ C as each element of B is also an element of C.
Question 10.
Let A = {a, e, i, o, u} and B = {a, b, c, d}.
Is A a subset of B? No. (Why?).
Is B a subset of A? No. (Why?).
Solution:
A = {a, e, i, o, u} and B = {a, b, c, d}
⇒ a ∈ B and e, i, o, u ∉ B
Since not all elements of A are in B,
No, A is not a subset of B.
Reason: Elements like e, i, o, u are not in B
⇒ a ∈ A and b, c, d ∉ A
No, B is not a subset of A
Reason: Elements like b, c, and d are not in A
∴ A ⊆ B? No, because e, i, o, u ∉ B.
∴ B ⊆ A? No, because b, c, d ∉ A.
Question 11.
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:
No. Let A = {1}, B = {{1}, 2} and C = {{1}, 2, 3}.
Here, A ∈ B as A = {1} and B ⊂ C.
But A ⊄ C as 1 ∈ A and 1 ∉ C.
Note that an element of a set can never be a subset of itself.
Question 12.
Let A = {2, 4, 6, 8} and B = {6, 8, 10, 12}. Find A ∪ B.
Solution:
We have A ∪ B = {2, 4, 6, 8, 10, 12}
Note that the common elements 6 and 8 have been taken only once while writing A ∪ B.
Question 13.
Let A = {a, e, i, o, u} and B = {a, i, u}. Show that A ∪ B = A.
Solution:
We have A ∪ B = {a, e, i, o, u} = A.
This example illustrates that the union of sets A and its subset B is the set A itself,
i.e., if B ⊂ A, then A ∪ B = A.
Question 14.
Let X = {Ram, Geeta, Akbar} be the set of students of Class XI, who are in the 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 = {Ram, Geeta, Akbar, David, Ashok}.
This is the set of students from Class XI who are in the hockey team or the football team or both.
Question 15.
Consider the sets A and B of Question 12. Find A ∩ B.
Solution:
We see that 6, 8 are the only elements that are common to both A and B.
Hence A ∩ B = {6, 8}
Question 16.
Consider the sets X and Y of Question 14. Find X ∩ Y.
Solution:
We see that the element Geeta is the only element common to both.
Hence, X ∩ Y = {Geeta}
Question 17.
Let A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and B = {2, 3, 5, 7}. Find A ∩ B and hence show that A ∩ B = B.
Solution:
We have A ∩ B = {2, 3, 5, 7} = B.
We note that B ⊂ A and that A ∩ B = B.
Question 18.
Let A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6, 8}. Find A – B and B – A.
Solution:
We have, A – B = {1, 3, 5},
Since the elements 1, 3, 5 belong to A but not to B, and B – A = { 8 },
Since element 8 belongs to Band, not to A.
We note that A – B ≠ B – A.
Question 19.
Let V = {a, e, i, o, u} and B = {a, i, k, u}. Find V – B and B – V.
Solution:
We have, V – B = {e, o},
since the elements e, o belong to V but not to B, and B-V = {k},
since the element k belongs to B but not to V.
We note that V – B ≠ B – V.
Using the set builder notation, we can rewrite the definition of difference as
A – B = {x: x ∈ A and x ∉ B}
The difference between two sets A and B can be represented by a Venn diagram as shown in the Figure.
The shaded portion represents the difference between the two sets, A and B.
Remark: The sets A – B, A ∩ B and B – A are mutually disjoint.
i.e., the intersection of any of these two sets is the null set as shown in the Figure.
Question 20.
Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A’.
Solution:
We note that 2, 4, 6, 8, 10 are the only elements of U that do not belong to A.
Hence A’ = {2, 4, 6, 8, 10}
Question 21.
Let U be the universal set of all the students of Class XI of a coeducational school, and A be the Set of all girls in Class XI. Find A’.
Solution:
Since A is the set of all girls, A’ is clearly the set of all boys in the class.
If A is a subset of the universal set U, then its complement A’ is also a subset of U.
Again in Question 20 above, we have A’ = {2, 4, 6, 8, 10}
Hence (A’)’ = {x: x ∈ U and x ∈ A’} = {1, 3, 5, 7, 9} = A
It is clear from the definition of the complement that for any subset of the universal set U,
We have (A’)’ = A
Now, we want to find the results for (A ∪ B)’ and A’ ∩ B’ in the following example.
Question 22.
Let U = {1, 2, 3, 4, 5, 6}, A = {2, 3}, and B = {3, 4, 5}. Find A’, B’, A’ ∩ B’, A ∪ B, and hence show that (A ∪ B)’ = A’ ∩ B’.
Solution:
Clearly A’ = {1, 4, 5, 6}, B’ = {1, 2, 6}
Hence A’ ∩ B’ = {1, 6}
Also, A ∪ B = {2, 3, 4, 5}
so that (A ∪ B)’ = {1, 6}
(A ∪ B)’ = {1, 6} = A’ ∩ B’
It can be shown that the above result is true in general.
If A and B are any two subsets of the universal set U, then (A ∪ B)’ = A’ ∩ B’.
Similarly, (A ∪ B)’ = A’ ∪ B’.
These two results are stated in words as follows:
The complement of the union of two sets is the intersection of their complements, and the complement of the intersection of two sets is the union of their complements. These are called De Morgan’s laws. These are named after the mathematician De Morgan.
The complement A’ of a set A can be represented by a Venn diagram as shown in the Figure.
The shaded portion represents the complement of the set A.
Question 23.
Show that the set of letters needed to spell “CATARACT” and the set of letters needed to spell “TRACT” are equal.
Solution:
Let X be the set of letters in “CATARACT”.
Then X = {C, A, T, R}
Let Y be the set of letters in “TRACT”.
Then, Y = {T, R, A, C, T} = {T, R, A, C}
Since every element in X is in Y, and every element in Y is in X.
It follows that X = Y.
Question 24.
List all the subsets of the set {-1, 0, 1}.
Solution:
Let A = {-1, 0, 1}.
The subset of A having no element is the empty set φ.
The subsets of A having one element are {-1}, {0}, and {1}.
The subsets of A having two elements are {-1, 0}, {-1, 1}, and {0, 1}.
The subset of A having three elements of A is A itself.
So, all the subsets of A are φ, {-1}, {0}, {1}, {-1, 0}, {-1, 1}, {0, 1} and {-1, 0, 1}.
Question 25.
Show that A ∪ B = A ∩ B implies A = B.
Solution:
Let a ∈ A Then a ∈ A ∪ B.
Since A ∪ B = A ∩ B, a ∈ A ∩ B.
So a ∈ B. Therefore, A ⊂ B.
Similarly, if b ∈ B, then b ∈ A ∪ B.
Since A ∪ B = A ∩ B, b ∈ A ∩ B.
So, b ∈ A.
Therefore, B ⊂ A.
Thus, A = B.
Multiple Choice Questions
Question 1.
List of elements of the set {x: x is an integer, x3 ≤ 50}
(1) {2, 3, 4}
(2) {1, 3, 4}
(3) {1, 2, 3}
(4) {1, 2, 4}
Answer:
(3) {1, 2, 3}
Question 2.
The set builder form of {1, 2, 3, 4} is
(1) {x: x is a positive integer, x2 < 30}
(2) {x: x is a positive integer, x2 + 1 < 20}
(3) {x: x is a positive integer, x2 – 1 < 15}
(4) {x: x is a positive integer, 2x < 20}
Answer:
(2) {x: x is a positive integer, x2 + 1 < 20}
Question 3.
Which of the following statements is true?
(1) {a, b} ⊂ {x: x is a vowel in the English alphabet}
(2) {1, 2, 3, 4} ⊂ {1, 2, 3}
(3) {5} ∈ {4, 5, 6}
(4) {2, 3} ∈ {1, {2, 3}, 4, 5}
Answer:
(4) {2, 3} ∈ {1, {2, 3}, 4, 5}
Question 4.
The set builder form interval [-4, 5) is
(1) {x: x is an integer -4 ≤ x < 5}
(2) {x: x is an integer -4 < x < 5}
(3) {x: x is an integer -4 ≤ x ≤ 5}
(4) {x: x is an integer -4 < x ≤ 5}
Answer:
(1) {x: x is an integer -4 ≤ x < 5}
Question 5.
A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, then (A – B) ∪ (B – A)
(1) {1, 2, 3, 4, 5, 6, 7, 8}
(2) {1, 2, 3, 6, 7, 8}
(3) {1, 2, 4, 5, 6}
(4) {2, 4, 6, 8}
Answer:
(2) {1, 2, 3, 6, 7, 8}
Question 6.
A = {1, 2, 3, 4}, B = {2, 3, 4}, C = {3, 4, 5, 6} then A ∩ (B ∪ C)
(1) {2, 3, 4}
(2) {1, 2, 3, 4}
(3) {3, 4, 5, 6}
(4) {1, 6}
Answer:
(1) {2, 3, 4}
Question 7.
Which of the following is false?
(1) (A’)’ = A
(2) A ∪ A’ = φ
(3) (A ∪ B)’ = A’ ∩ B.’
(4) φ = U (U is the universal set)
Answer:
(2) A ∪ A’ = φ
Question 8.
If U = {1, 2, 3, 4, 5} and A = {2, 3}, then φ’ ∩ A = __________
(1) φ
(2) U
(3) A
(4) {1, 4, 5}
Answer:
(3) A
Question 9.
U = {x: x is a natural number 1 < x < 15 } and A = {x: x is a prime number 1 < x < 15} then A’ = __________
(1) {2, 3, 5, 7, 11, 13}
(2) {4, 6, 8, 9, 10, 12, 14}
(3) {2, 3, 4, 5, 7, 11, 13}
(4) {2, 4, 6, 8, 10, 12, 14}
Answer:
(2) {4, 6, 8, 9, 10, 12, 14}
Question 10.
Which of the following is not a subset of {a, b, c, d}?
(1) φ
(2) {a, b}
(3) {d, e}
(4) {b, d}
Answer:
(3) {d, e}