(i) R= {(1, 2), (1, 3), (2, 3), (3, 2), (5, 6)}
(ii) R= {(x, y) : x, y ∈ N; x + 2y = 8}
Solution:
(i) It is given that:
$ R=\left\{ \left( 1,\text{ }2 \right),\text{ }\left( 1,\text{ }3 \right),\text{ }\left( 2,\text{ }3 \right),\text{ }\left( 3,\text{ }2 \right),\text{ }\left( 5,\text{ }6 \right) \right\} $
So, R‑1 is given as follows:
$ {{R}^{-1}}=\left\{ \left( 2,\text{ }1 \right),\text{ }\left( 3,\text{ }1 \right),\text{ }\left( 3,\text{ }2 \right),\text{ }\left( 2,\text{ }3 \right),\text{ }\left( 6,\text{ }5 \right) \right\} $
(ii) It is given that : R= {(x, y): x, y ∈ N; x + 2y = 8}
Here, we have: x + 2y = 8
Or, x = 8 – 2y
As y ∈ N, Put the values of y = 1, 2, 3,…… till x ∈ N
When, y = 1, we get:
x = 8 – 2(1) = 8 – 2
x = 6
When, y = 2, we get
x = 8 – 2(2) = 8 – 4
x = 4
When, y = 3, we get
x = 8 – 2(3) = 8 – 6
x = 2
When, y = 4, we get
x = 8 – 2(4) = 8 – 8
x = 0
Now, y cannot have the value 4 because when y = 4, x becomes zero which is clearly not a natural number.
Therefore, we can write:
$ R\text{ }=\text{ }\left\{ \left( 2,\text{ }3 \right),\text{ }\left( 4,\text{ }2 \right),\text{ }\left( 6,\text{ }1 \right) \right\} $
$ {{R}^{-1}}~=\text{ }\left\{ \left( 3,\text{ }2 \right),\text{ }\left( 2,\text{ }4 \right),\text{ }\left( 1,\text{ }6 \right) \right\} $