If R is a binary relation on a set A define

Solution:

3a + 2b = 15

$b=\frac{15-3 a}{2}$

$a=1$ è $b=6$

$a=3$ è $b=3$

$a=5$ è $b=0$

$R=\{(1,6),(3,3),(5,0)\}$

$R^{-1}=\{(6,1),(3,3),(0,5)\}$

The domain of R is the set of first co-ordinates of R

Dom(R) = {1, 3, 5}

The range of R is the set of second co-ordinates of R

Range(R) = {6, 3, 0}

The domain of $R^{-1}$ is the set of first co-ordinates of $R^{-1}$

$\operatorname{Dom}\left(R^{-1}\right)=\{6,3,0\}$

The range of $R^{-1}$ is the set of second co-ordinates of $R^{-1}$

Range $\left(R^{-1}\right)=\{1,3,5\}$

Thus,

$\operatorname{dom}(\mathrm{R})=\operatorname{range}\left(\mathrm{R}^{-1}\right)$

range $(\mathrm{R})=\operatorname{dom}\left(\mathrm{R}^{-1}\right)$