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Chapter 2: Relations – Exercise 2.3

Relations – Exercise 2.3 – Q.1

(i) We have,

A = {1, 2, 3} and B = {4, 5, 6}

{(1, 6), (3, 4), (5, 2)} is not a relation from A to 8 as it is not a subset of A × B.

(ii) We have,

A = {1, 2, 3} and B = {4, 5, 6}

{(1, 5), (2, 6), (3, 4), (3, 6} is a subset of A × B, so it is a relation from A to B.

(iii) We have,

A = {1, 2, 3} and B = {4, 5, 6}

{(4, 2), (4, 3), (5,1)} is not a relation from A to B as it is not a subset of A × B.

(iv) We have,

A = {1, 2, 3} and B = {4, 5, 6)

A × B is a relation from A to B.

 

Relations – Exercise 2.3 – Q.2

We have,

A = {2, 3, 4, 5} and B = {3 ,6, 7, 10}

It is given that (x, y) ϵ R ⟺ x is relatively prime to y

∴ (2, 3) ϵ R, (2, 7) ϵ R, (3, 7) ϵ R, (3, 10) ϵ R, (4, 3) ϵ R, (4, 7) ϵ R, (5, 3) ϵ R, and (5, 7) ϵ R

Thus,

R = {(2, 3), (2, 7), (3, 7), (3,10), (4, 3), (4, 7), (5, 3), (5, 7)}

Clearly, Domain (R) = {2, 3, 4, 5} and Range = {3, 7,10}.

 

Relations – Exercise 2.3 – Q.3

We have,

A = {1, 2, 3, 4, 5}    [∵A is the set of first five natural number]

It is given that R be a relation on A defined as (x, y) ϵ R ⟺ x ≤ y

For the elements of the given sets A and A, we find that

1 = 1, 1 < 2, 1 < 3, 1 < 4, 1 < 5, 2 = 2, 2 < 3, 2 < 4, 2 < 5, 3 = 3, 3 < 4, 3 < 5, 4 = 4, 4 < 5, and 5 = 5

∴ (1, 1) ϵ R, (1, 2) ϵ R, (1, 3) ϵ R, (1, 4) ϵ R, (1, 5) ϵ R, (2, 2) ϵ R, (2, 3) ϵ R, (2, 4) ϵ R, (2, 5) ϵ (3, 3) ϵ R, (3, 4) ϵ R, (3, 5) ϵ R, (4, 4) ϵ R, (4, 5) ϵ R, and (5, 5) ϵ R

Thus,

R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 5)}

Also,

R-1 = {(1,1), (2,1), (3,1), (4,1), (5,1), (2, 2), (3, 2), (4, 2), (5, 2), (3, 3), (4, 3), (5, 3), (4, 4), (5, 4), (5, 5)}

(i) Domain(R-1) = {1, 2, 3, 4, 5}

(ii) Range(R) = {1, 2, 3, 4, 5}

 

Relations – Exercise 2.3 – Q.4

(i) we have,

R = {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}

⟹ R-1 = {(2,1), (3,1), (3, 2), (2, 3), (6, 5)}

(ii) We have,

R = {(x, y) : x, y ϵ N, x +2y = 8}

Now,

x +2y = 8

⟹ x = 8 – 2y

Putting y = 1, 2, 3 we get x = 6,4,2 respectively.

For y - 4, we get x = 0 ∉ N. Also for y > 4, x ∉ N

∴ R = {(6, 1), (4, 2), (2, 3)}

Thus,

R-1 = {(1, 6), (2, 4), (3, 2)}

⟹ R-1 = {(3, 2) , (2, 4), (1,6)}

(iii) We have,

R is a relation from {11, 12, 13,} to {8, 10, 12} defined by y = x – 3

Now,

y = x – 3

Putting x = 11, 12, 13 we get y = 8, 9, 10 respectively

⟹ (11, 8) ϵ R (12, 9) ∉ R and (13, 10) ϵ R

Thus,

R = {(11, 8), (13,10)}

⟹ R-1 = {(8, 11), (10, 13)}

 

Relations – Exercise 2.3 – Q.5

(i) We have,

x - 2y

Putting y = 1, 2, 3 we get x = 2, 4, 6 respectively.

∴ R = {(2, 1), (4, 2), (6, 3)}

(ii) We have,

It is given that relation R on the set {1, 2, 3, 4, 5, 6, 7) defined by (x, y) ϵ R ⟺ x is relatively prime to y.

∴ (2, 3) ϵ R, (2, 5) ϵ R, (2, 7) ϵ R , (3, 2) ϵ R, (3, 4) ϵ R, (3, 5) ϵ R, (3, 7) ϵ R, (4, 3) ϵ R, (4, 5) ϵ R, (4, 7) ϵ R, (5, 2) ϵ R, (5, 3) ϵ R, (5, 4) ϵ R, (5, 6) ϵ R, (5, 7) ϵ R, (6, 5) ϵ R, (6, 7) ϵ R, (7, 2) ϵ R, (7, 3) ϵ R, (7, 4) ϵ R, (7, 5) ϵ R and (7, 6) ϵ R.

Thus,

R=

{(2, 3), (2, 5), (2, 7), (3, 2), (3, 4), (3, 5), (3, 7), (4, 3), (4, 5), (4, 7), (5, 2) (5, 3), (5, 4), (5, 6), (5, 7), (6, 5), (6, 7), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6)}

(iii) We have,

2x +3y - 12

⟹ 2x = 12 - 3y

relation

Putting y - 0, 2, 4 we get x = 6, 3, 0 respectively.

For y = 1, 3, 5, 6, 7, 8, 9, 10, x ∉ given set

∴ R = {(6, 0), (3, 2), (0, 4)}

= {(0, 4), (3, 2), (6, 0)}

(iv) We have, A = (5, 6, 7, 8) and B = (10, 12, 15, 16, 18)

Now,

Thus,

a/b stands for 'a divides b'. For the elements of the given set A and B, we find that 5/10, 5/15, 6/12, 6/18 and 8/16

∴ (5, 10) ϵ R, (5,15) ϵ R, (6, 12) ϵ R, (6, 18) ϵ R, and (8, 16) ϵ R

Thus,

R = {(5, 10), (5, 15), (6, 12), (6, 18), (8, 16)}

 

Relations – Exercise 2.3 – Q.6

We have,

(x, y) ϵ R ⟺ x + 2y = 8

Now

x + 2y = 8

⟹ x = 8 – 2y

Putting y = 1, 2, 3 we get x = 6, 4, 2 respectively

For y = 4, we get x = 0 ∉ N

Also, for y > 4, x ∉ N

∴ R = {(6, 1), (4, 2), (2, 3)}

Thus,

R-1 = {(1, 6), (2, 4), (3, 2)}

R-1 = {(3, 2), (2, 4), (1, 6)}

 

Relations – Exercise 2.3 – Q.7

We have,

A = {3, 5}, B = {7, 11}

and, R = {(a, b): a ϵ  A, b ϵ B, a - b is odd}

For the elements of the given sets A and B, we find that

3-7 = – 4, 3 – 11 = – 8, 5 – 7 = – 2 and 5 –11 = 6

∴ (3, 7) ∉ R, (3,11) ∉ R, (5, 7) ∉ R and (5,11) ∉ R,

Thus, R is an Empty relation from , A into B.

 

Relations – Exercise 2.3 – Q.8

We have,

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

∴ n (A) = 2 and n(B) = 2

⟹ n(A) × n(B) = 2 × 2 = 4

⟹ n(A × B) = 4 [∵ n(A × B) =  n(A) × n(B)]

So, there are 24 = 16 relations from A to B. [∵ n(x) = a, n(y) = b ⟹ Total number of relations = 2ab]

 

Relations – Exercise 2.3 – Q.9

(i) We have,

R = {(x, x + 5) : x ϵ (0, 1, 2, 3, 4, 5)}

For the elements of the given sets, we find that

R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}

Clearly, Domain (R) = {0, 1, 2, 3,  4, 5} and Range(R) = {5, 6, 7, 6, 4, 10}

(ii) We have,

R = {(x, x3)} : x is a prime number less than 10}

For the elements of the given sets, we find that

x = 2, 3, 5,7

∴ (2,13) ϵ R, (3,27) ϵ R, (5,125) ϵ R and (7,343) ϵ R

⟹ R = {(2, 8), (3, 27) , (5,125), (7, 343)}

Clearly, Domain (R) = {2, 3, 5, 7} and Range(R) = {8, 27,125, 343}


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