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Chapter 2: Relations – Exercise 2.1

Relations – Exercise 2.1 – Q.1

By the definition of equality of ordered pairs

Relations – Exercise 2.1 – Q.1

By the definition of equality or ordered pairs

(x + 1, 1) = (3, y - 2)

⟹ x + 1 = 3 and 1 = y - 2

⟹ x = 3 - 1 and 1 + 2 = y

⟹ x = 2 and 3 = y

⟹ x = 2 and y = 3

 

Relations – Exercise 2.1 – Q.2

We have,

(x, -1) ϵ {(a,b): b = 2a - 3} and , (5, y) ϵ {(a, b): b = 2a - 3}

⟹ -1 = 2 × x - 3 and y = 2 × 5 - 3

⟹ -1 = 2x - 3 and y = 10 - 3

⟹ 3 - 1 = 2x and y = 7

⟹ 2 = 2x and y = 7

⟹ x = 1 and y = 7

 

Relations – Exercise 2.1 – Q.3

We have,

a + b = 5

⟹ a = 5 - b

∴ b = 0 ⟹ a = 5 - 0 = 5,

b = 3 ⟹ a = 5 - 3 = 2,

b = 6 ⟹ a = 5 - 6 = -1,

Hence, the required set of ordered pairs (a, b) is {(-1, 6), (2, 3), (5, 0)}

 

Relations – Exercise 2.1 – Q.4

We have,

a ϵ {2, 4, 6, 9} and,

b ϵ {4, 6, 18, 27}

Now, a/b stands for 'a divides b'. For the elements of the given sets, we find that 2/4, 2/6, 2/18, 9/18 and 9/27

∴ {(2, 4), (2, 6), (2,18), (6,18), (9,18), (9, 27)} are the required set of ordered pairs (a, b).

 

Relations – Exercise 2.1 – Q.5

We have,

A = {1,2} and B = {1,3}

Now, A × B = {1,2} × {1, 3}

= {(1,1), (1,3), (2,1), (2, 3)}

and, B × A = {1, 3} × {1, 2}

= {(1,1), (1, 2), (3,1), (3, 2)}

 

Relations – Exercise 2.1 – Q.6

We have,

A = {1, 2, 3} and B = {3, 4}
∴ A×B = {1, 2, 3} × {3, 4}

= {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}

In order to represent A × B graphically, we follow the following steps:

(a) Draw two mutually perpendicular line one horizontal and other vertical.

(b) On the horizontal line represent the element of set A and on the vertical line represent the elements of set B.

(c) Draw vertical dotted lines through points representing elements of A on horizontal line and horizontal

line and horizontal lines through points representing elements of B on the vertical line points of intersection of these lines will represent A × B graphically.

Vertical dotted lines

 

Relations – Exercise 2.1 – Q.7

We have,

A = {1, 2, 3} and B = {2, 4}

∴ A × B = {1, 2, 3} × {2, 4}

= {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)},

B × A = {2, 4} × {1, 2, 3}

= {(2,1), (2, 2), (2, 3), (4, 1), (4, 2), (4, 3)},

A × A = {1, 2, 3} × {1, 2, 3}

= {(1, 1), (1, 2), (1, 3), (2,1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)},

B × B = {2, 4} × {2, 4}

= {(2, 2), (2, 4), (4, 2), (4, 4)},

and {A × B} ∩ {B × A}

= {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)} ∩ {(2,1), (2, 2), (2, 3), (4,1), (4, 2), (4, 3)}

= {(2, 2)}

⟹ (A × B) ∩ (B × A) = {(2, 2)}.

 

Relations – Exercise 2.1 – Q.8

We have,

n(A) = 5 and n(B) = 4

We know that, if A and B are two finite sets, then n(A × B) = n(A) × n(B)

∴ n(A × B) = 5 × 4 = 20

Now, n[(A × B) ∩ (B × A)] = 3 × 3 = 9 [∵ A and B have 3 elements in common]

 

Relations – Exercise 2.1 – Q.9

Let (a, b) be an arbitrary element of (A × B) ∩ (B × A). Then,

(a, b) ϵ (A × B) ∩ (B × A)

⟺ (a, b) ϵ A × B and (a, b) ϵ B × A

⟺ (a ϵ A and b ϵ B) and (a ϵ B and b ϵ A)

⟺ (a ϵ A and a ϵ B) and (b ϵ A and b ϵ B)

⟺ a ϵ A ∩ B and b ϵ A ∩ B

Hence, the sets A × B and B × A have and element in common if the sets A and B have an element in common.

 

Relations – Exercise 2.1 – Q.10

Since (x, 1), (y,2), (z,1) are elements of A × B. Therefore, x, y, z ϵ A and 1, 2 ϵ B

It is given that n(A) = 3 and n(B) = 2

∴ x, y, z ϵ A and n (A) = 3

⟹ A = {x, y, z}

1, 2 ϵ B and n(B) = 2

⟹ B = {1, 2}.

 

Relations – Exercise 2.1 – Q.11

We have,

A = {1, 2, 3, 4}

and R = {(a, b) = a ϵ A, b ϵ A, a divides b}

Now,

a/b stands fro 'a divides b'. For the elements of the given sets, we find that 1/1, 1/2, 1/3, 1/4, 2/2, 3/3 and 4/4

∴ R = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 4), (3, 3), (4, 4)}

 

Relations – Exercise 2.1 – Q.12

We have,

A = {-1, 1}

∴ A × A = {-1, 1} × {-1, 1}

= {(-1,-1), (-1,1), (1,-1), (1,1)}

∴ A × A × A = {-1, 1} × {(-1, -1, 1), (-1, 1), (1, -1), (1,1)}

= {(-1, -1, -1), (-1, -1, 1), (-1, 1, -1) (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}

 

Relations – Exercise 2.1 – Q.13

(i) False,

If P = {m, n} and Q = (n, M)

Then,

P × Q = {(m, n), (m, m), (n, n), (n, m)}

(ii) False,

If A and B are non-empty sets, then AB is a non-empty set of ordered pairs (x, y) such that x ϵ A and y ϵ B.

(iii) True

 

Relations – Exercise 2.1 – Q.14

We have,

A = {1, 2}

∴ A × A = {1, 2} × {1, 2}

= {(1, 1), (1, 2), (2, 1), (2, 2)}

∴ A × A × A = {1, 2} × {(1, 1), (1, 2), (2, 1), (2, 2)}

= {(1, 1, 1), (1,1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)}


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