(i) the domain of R‑1
(ii) The Range of R.
Solution:
According to the question, A is the set of first five natural numbers. So, we can write:
\[A=\left\{ 1,\text{ }2,\text{ }3,\text{ }4,\text{ }5 \right\}\]
We are given that: (x, y) R x ≤ y
The number one is less than the numbers two, three, four, and five. Two is less than three, four, and five. Three is less than four and five. 4 is a lower number than 5. And 5 is not lower than any number A. Therefore,we can write:
$ R\text{ }=\text{ }\!\!\{\!\!\text{ }\left( 1,\text{ }1 \right),\text{ }\left( 1,\text{ }2 \right),\text{ }\left( 1,\text{ }3 \right),\text{ }\left( 1,\text{ }4 \right),\text{ }\left( 1,\text{ }5 \right),\text{ }\left( 2,\text{ }2 \right),\text{ }\left( 2,\text{ }3 \right),\text{ }\left( 2,\text{ }4 \right), $
$ \left( 2,\text{ }5 \right),\text{ }\left( 3,\text{ }3 \right),\text{ }\left( 3,\text{ }4 \right),\text{ }\left( 3,\text{ }5 \right),\text{ }\left( 4,\text{ }4 \right),\text{ }\left( 4,\text{ }5 \right),\text{ }\left( 5,\text{ }5 \right)\} $
The set of ordered pairs created by swapping the first and second elements of each pair in the original relation is known as an inverse relation. If a point (a, b) appears on the graph of a function, the graph of the inverse relation of that function also contains that point (b, a). Therefore, we can write:
$ {{R}^{-1}}~=\text{ }\!\!\{\!\!\text{ }\left( 1,\text{ }1 \right),\text{ }\left( 2,\text{ }1 \right),\text{ }\left( 3,\text{ }1 \right),\text{ }\left( 4,\text{ }1 \right),\text{ }\left( 5,\text{ }1 \right),\text{ }\left( 2,\text{ }2 \right),\text{ }\left( 3,\text{ }2 \right),\text{ }\left( 4,\text{ }2 \right), $
$ \left( 5,\text{ }2 \right),\text{ }\left( 3,\text{ }3 \right),\text{ }\left( 4,\text{ }3 \right),\text{ }\left( 5,\text{ }3 \right),\text{ }\left( 4,\text{ }4 \right),\text{ }\left( 5,\text{ }4 \right)\text{ }\left( 5,\text{ }5 \right)\} $
(i) Domain of R‑1 is {1, 2, 3, 4, 5}
(ii) Range of R is {1, 2, 3, 4, 5}