Question
Find the inverse relation $R^{-1}$ in the following case: $\text{R}=\{(\text{x, y}):\text{x},\text{ y}\in\text{N},\text{ x}+2\text{y}=8\}$

Answer

We have, $\text{R}=\{(\text{x, y}):\text{x},\text{ y}\in\text{N},\text{ x}+2\text{y}=8\}$ Now, x + 2y = 8 ⇒ x = 8 - 2y Putting y = 1, 2, 3 we get x = 6, 4, 2 respectively. For y = 4, we get $\text{x}=0\notin\text{N}$ Also for $\text{y}>4,\text{ x}\notin\text{N}$ $\therefore\ \text{R}=\{(6,1),(4,2),(2,3)\}$ Thus, $\text{R}^{-1}=\{(1,6),(2,4),(3,2)\}$ $\Rightarrow\text{R}^{-1}=\{(3,2),(2,4),(1,6)\}$

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