Chapter 1 Set Language Ex No 1.1

Identify the following is set?

The collection of prime numbers upto 100.

Solution

A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89 and 97}

As the collection of prime numbers upto 100 is known and can be counted (well defined).

Hence this is a set.

Identify the following is set?

The collection of rich people in India.

Solution

The collection of rich people in India.

Rich people has no definition.

Hence, it is not a set.

Identify the following is set?

The Collection of all rivers in India.

Solution

A = {Cauvery, Sindhu, Ganga}

Hence, it is a set.

Identify the following is set?

The Collection of good Hockey players.

Solution

The collection of good hockey players is not a well – defined collection because the criteria for determining a hockey player’s talent may vary from person to person.

Hence, this collection is not a set.

List the set of letters of the following words in Roster form.

INDIA

Solution

A = {I, N, D, A}

List the set of letters of the following words in Roster form.

PARALLELOGRAM

Solution

B = {P, A, R, L, E, O, G, M}

List the set of letters of the following words in Roster form. 

MISSISSIPPI

Solution: 

C = {M, I, S, P}

List the set of letters of the following words in Roster form.

CZECHOSLOVAKIA

Solution

D = {C, Z, E, H, O, S, L, V, A, K, I}

State Whether True or False

Tamil Nadu Board Samacheer Kalvi solutions for Class 9th Mathematics Answers Guide Chapter 1 Set Language Exercise 1.4 [Page 22]

Exercise 1.4 | Q 1. (i) | Page 22

If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (P ∪ Q) ∪ R

Solution

P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}

P ∪ Q = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11}

= {1, 2, 3, 5, 7, 9, 11}

(P ∪ Q) ∪ R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9}

= {1, 2, 3, 4, 5, 7, 9, 11}

Exercise 1.4 | Q 1. (ii) | Page 22

If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (P ∩ Q) ∩ S

Solution

P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}

P ∩ Q = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11}

= {2, 5, 9}

(P ∩ Q) ∩ S = {2, 5, 9} ∩ {2, 3, 4, 5, 8}

= {2, 5}

Exercise 1.4 | Q 1. (iii) | Page 22

If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (Q ∩ S) ∩ R

Solution

P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}

Q ∩ S = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8}

= {2, 3, 5}

(Q ∩ S) ∩ R = {2, 3, 5} ∩ {3, 4, 5, 7, 9}

= {3, 5}

Exercise 1.4 | Q 2 | Page 22

Test for the commutative property of union and intersection of the sets

P = {x : x is a real number between 2 and 7} and

Q = {x : x is an irrational number between 2 and 7}

Solution

Commutative Property of union of sets

(A ∪ B) = (B ∪ A)

Here P = {3, 4, 5, 6}, Q = ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6}\}}$

P ∪ Q = {3, 4, 5, 6} ∪ ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6}\}}$

= ${\displaystyle \{3,4,5,6,\sqrt{3},\sqrt{5},\sqrt{6}\}}$  ...(1)

Q ∪ P = ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6}\}}$ ∪ {3, 4, 5, 6}

= ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6},3,4,5,6\}}$ ...(2)

(1) = (2)

∴ P ∪ Q = Q ∪ P

∴ It is verified that union of sets is commutative.

Commutative Property of intersection of sets (P ∩ Q) = (Q ∩ P)

P ∩ Q = ${\displaystyle \{3,4,5,6\}\cap \{\sqrt{3},\sqrt{5},\sqrt{6}\}}$ = { } ...(1)

Q ∩ P = ${\displaystyle \{\sqrt{3},\sqrt{5},\sqrt{6}\}\cap \{3,4,5,6\}}$ = { } ...(2)

From (1) and (2)

P ∩ Q = Q ∩ P

∴ It is verified that intersection of sets is commutative.

Exercise 1.4 | Q 3 | Page 22

If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets

Solution

Associative Property of union of sets

A ∪ (B ∪ C) = (A ∪ B) ∪ C

(B ∪ C) = {m, n, q, s, t} ∪ {m, n, p, q, s}

= {m, n, p, q, s, t}

A ∪ (B ∪ C) = {p, q, r, s} ∪ {m, n, p, q, s, t}

= {m, n, p, q, r, s, t}   ...(1)

(A ∪ B) = {p, q, r, s} ∪ {m, n, q, s, t}

= {m, n, p, q, r, s, t}

(A ∪ B) ∪ C = {m, n, p, q, r, s, t} ∪ {m, n, p, q, s}

= {m, n, p, q, r, s, t}   ...(2)

From (1) and (2)

It is verified that A ∪ (B ∪ C) = (A ∪ B) ∪ C

Exercise 1.4 | Q 4 | Page 22

Verify the associative property of intersection of sets for A = ${\displaystyle \{-11,\sqrt{2},\sqrt{5},7\}}$, B = ${\displaystyle \{\sqrt{3},\sqrt{5},6,13\}}$ and C = ${\displaystyle \{\sqrt{2},\sqrt{3},\sqrt{5},9\}}$

Solution

Associative Property of intersection of sets A ∩ (B ∩ C) = (A ∩ B) ∩ C

B ∩ C = ${\displaystyle \{\sqrt{3},\sqrt{5},6,13\}\cap \{\sqrt{2},\sqrt{3},\sqrt{5},9\}=\{\sqrt{3},\sqrt{5}\}}$

 A ∩ (B ∩ C) = ${\displaystyle \{-11,\sqrt{2},\sqrt{5},7\}\cap \{\sqrt{3},\sqrt{5}\}=\left\{\sqrt{5}\right\}}$  ...(1)

A ∩ B  = ${\displaystyle \{-11,\sqrt{2},\sqrt{5},7\}\cap \{\sqrt{3},\sqrt{5},6,13\}=\left\{\sqrt{5}\right\}}$

(A ∩ B) ∩ C = ${\displaystyle \left\{\sqrt{5}\right\}\cap \{\sqrt{2},\sqrt{3},\sqrt{5},9\}=\left\{\sqrt{5}\right\}}$  ...(2)

From (1) and (2),

It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C

Exercise 1.4 | Q 5 | Page 22

If A = {x : x = 2ⁿ, n ∈ W and n < 4}, B = {x : x = 2n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets

Solution

A = {x : x = 2ⁿ, n ∈ W, n < 4}

⇒ x = 2° = 1

x = 2¹ = 2

x = 2² = 4

x = 2³ = 8

∴ A = {1, 2, 4, 8}

B = {x : x = 2n, n ∈ N and n ≤ 4}

⇒ x = 2 × 1 = 2

x = 2 × 2 = 4

x = 2 × 3 = 6

x = 2 × 4 = 8

∴ B = {2, 4, 6, 8}

C = {0, 1, 2, 5, 6}

Associative property of intersection of sets

A ∩ (B ∩ C) = (A ∩ B) ∩ C

B ∩ C = {2, 6}

A ∩ (B ∩ C) = {1, 2, 4, 8} ∩ {2, 6}

= {2}   ...(1)

A ∩ B = {1, 2, 4, 8} ∩ {2, 4, 6, 8}

= {2, 4, 8}

(A ∩ B) ∩ C = {2, 4, 8} ∩ {0, 1, 2, 5, 6}

= {2}   ...(2)

From (1) and (2)

It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C

Question 1 :

Find the cardinal number of the following sets.

(i) M = {p, q, r, s, t, u}

Solution :

Number of elements in the set is 6. Hence n(M)  =  6.

(ii) P = {x : x = 3n + 2, n∈W and x < 15}

Solution :

Since n belongs to whole number, we have to start with 0.

By applying the values of n from 0 to 14, we get 15 different values for x. Hence n(P) is 15.

(iii) Q = { y : y = 4/3n, n ∈ N and 2 < n ≤ 5}

Solution :

The values of n are 3, 4, 5. By applying the above three values for n, we get different values of y. Hence n(Q) is 3.

(iv) R = {x : x is an integers, x ∈ Z and –5 ≤ x < 5}

Solution :

The elements of R are 

R  =  {-5,-4, -3, -2, -1, 0, 1, 2, 3, 4}

n(R)  =   10

(v) S = The set of all leap years between 1882 and 1906.

Solution :

The leap years 1884, 1888, 1892, 1896, 1900, 1904.

Hence n(S)  =  6

Finite Set and Infinite Set

Finite set :

A set with finite number of elements is called a finite set.

Infinite set :

A set which has infinite number of elements is called an infinite set.

Question 2 :

Identify the following sets as finite or infinite.

(i) X = The set of all districts in Tamilnadu.

Solution :

Districts in Tamilnadu is countable. Hence it is finite set.

(ii) Y = The set of all straight lines passing through a point.

Solution :

We may draw an infinite number of lines through a point.

Hence it is infinite set.

(iii) A = { x : x ∈ Z and x < 5}

Solution :

Z means integers. The elements of A are 1, 2, 3, 4.

Hence set A is finite.

(iv) B = {x : x²–5x+6 = 0, x ∈N}

Solution :

x²–5x+6 = 0

(x - 2) (x - 3)  =  0

x  =  2 and x  =  3

By solving the quadratic equation, we get two different values. Hence B is finite set.

Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.1

Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.1

Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.1

Question 1 :

Which of the following are sets?

(i) The Collection of prime numbers upto 100.

Solution :

The above statement is well defined, the set will contain 2, 3, 5, 7, ..........97.

Hence it is a set.

(ii) The Collection of rich people in India.

Solution :

The set is not well defined. Hence it is not a set.

(iii) The Collection of all rivers in India.

Solution :

The statement is well defined. Hence it is a set.

(iv) The Collection of good Hockey players.

Solution :

The statement is not well defined. Hence it is not a set.

Question 2 :

List the set of letters of the following words in Roster form.

(i) INDIA (ii) PARALLELOGRAM

(iii) MISSISSIPPI (iv) CZECHOSLOVAKIA

Solution :

Now we have to write the letters of the given word. If we have repeated letters, then we have to write it once.

(i)  INDIA 

A  =  {I, N, D, A}

(ii)  PARALLELOGRAM

B  =  {P, A, R, L, E, O, G, M}

(iii) MISSISSIPPI

C  =  {M, I, S, P}

(iv) CZECHOSLOVAKIA

D  =  {C, Z, E, H, O, S, L, V, A, K, I}

Question 3 :

Consider the following sets A = {0, 3, 5, 8}, B = {2, 4, 6, 10} and C = {12, 14,18, 20}.

(a) State whether True or False:

(i) 18 ∈ C (ii) 6 ∉ A (iii) 14 ∉ C (iv) 10 ∈ B

(v) 5 ∉ B (vi) 0 ∉ B

Solution :

(i) 18 ∈ C

18 is the element of set C, hence it is true.

(ii) 6 ∉ A

The set A does not contain the element 6, hence it is true.

(iii) 14 ∉ C

14 is the element of set C, hence the statement is false.

(iv) 10 ∈ B

10 is the element of set B, hence the statement is true.

(b) Fill in the blanks:

(i) 3 ∈ ____ (ii) 14 ∈ _____ (iii) 18 ____ B (iv) 4 _____ B

Solution :

3 is the element of set A, so  3 ∈ A.

14 is the element of set C, so 14 ∈ C.

18 is the element of set C, so 18 ∈ C.

4 is the element of set B, so  4 ∈ B.

Question 4 :

Represent the following sets in Roster form.

(i) A = The set of all even natural numbers less than 20.

Answer :

The given set will contain even natural numbers lesser than 20.

A  =  {2, 4, 6, 8, 10, 12, 14, 16, 18}

(ii) B = {y : y = 1/2n , n ∈ N, n ≤ 5}

Solution :

B = {y : y = 1/2n , n ∈ N, n ≤ 5}

If n = 1, then y = 1/2

If n = 2, then y = 1/4

If n = 3, then y = 1/6

If n = 4, then y = 1/8

If n = 5, then y = 1/10

B  =  { 1/2, 1/4, 1/6, 1/8, 1/10 }

(iii) C = {x : x is perfect cube, 27 < x < 216}

Answer :

27 is the value of 3³, 216 is the value of 6³

C  =  {4³, 5³}

C  =  {64, 125}

(iv) D = {x : x ∈ Z, –5 < x ≤ 2}

Answer :

D  =  {-4, -3, -2, -1, 0, 1, 2}

Question 5 :

Represent the following sets in set builder form.

(i) B = The set of all Cricket players in India who scored double centuries in One Day Internationals.

Answer :

B = {x : x is an Indian player who scored double centuries in One Day International}

(ii)  C = {1/2, 2/3, 3/4, ..........}

Answer :

By observing the given set, the denominator is 1 greater than the numerator.

C  =  {x : x = n/(n + 1) where n ∈ N}

(iii) D = The set of all tamil months in a year.

Answer :

D = {x : x is the set of all tamil months in a year.}

(iv) E = The set of odd Whole numbers less than 9.

Answer :

E = {x: x is the set of odd Whole numbers less than 9}

Question 6 :

Represent the following sets in descriptive form.

(i) P = { January, June, July}

Answer :

P = The set of English months starting with J

(ii) Q = {7, 11, 13, 17, 19, 23, 29}

Answer :

Q =  the set of prime numbers between 5 and 31.

(iii) R = {x : x  ∈  N, x < 5}

Answer :

R  =  The set of natural numbers less than 5.

(iv) S = {x : x is a consonant in English alphabets}

Answer :

S  =  The set of consonants in English alphabets