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(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,
{(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,
{(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.
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}.
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
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}
(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)}
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)}
R-1 = {(1, 6), (2, 4), (3, 2)}
⟹ R-1 = {(3, 2) , (2, 4), (1,6)}
R is a relation from {11, 12, 13,} to {8, 10, 12} defined by y = x – 3
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
R = {(11, 8), (13,10)}
⟹ R-1 = {(8, 11), (10, 13)}
x - 2y
Putting y = 1, 2, 3 we get x = 2, 4, 6 respectively.
∴ R = {(2, 1), (4, 2), (6, 3)}
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.
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)}
2x +3y - 12
⟹ 2x = 12 - 3y
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)
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
R = {(5, 10), (5, 15), (6, 12), (6, 18), (8, 16)}
(x, y) ϵ R ⟺ x + 2y = 8
Now
x + 2y = 8
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-1 = {(3, 2), (2, 4), (1, 6)}
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.
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]
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}
R = {(x, x3)} : x is a prime number less than 10}
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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Chapter 2: Relations – Exercise 2.2...
Chapter 2: Relations – Exercise 2.1...