Question -
Answer -
(i) Given:
R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
So, R 1 = {(2, 1), (3, 1), (3, 2), (2, 3), (6, 5)}
(ii) Given,
R= {(x, y): x, y ∈ N; x + 2y = 8}
Here, x + 2y = 8
x = 8 – 2y
As y ∈ N, Put the values of y = 1, 2, 3,…… till x ∈ N
When, y = 1, x = 8 – 2(1) = 8 – 2 = 6
When, y = 2, x = 8 – 2(2) = 8 – 4 = 4
When, y = 3, x = 8 – 2(3) = 8 – 6 = 2
When, y = 4, x = 8 – 2(4) = 8 – 8 = 0
Now, y cannot hold value 4 because x = 0 for y = 4 which is not a natural number.
∴ R = {(2, 3), (4, 2), (6, 1)}
R 1 = {(3, 2), (2, 4), (1, 6)}
(iii) Given,
R is a relation from {11, 12, 13} to (8, 10, 12} defined by y = x – 3
Here,
x = {11, 12, 13} and y = (8, 10, 12}
y = x – 3
When, x = 11, y = 11 – 3 = 8 ∈ (8, 10, 12}
When, x = 12, y = 12 – 3 = 9 ∉ (8, 10, 12}
When, x = 13, y = 13 – 3 = 10 ∈ (8, 10, 12}
∴ R = {(11, 8), (13, 10)}
R 1 = {(8, 11), (10, 13)}