Question

A relation R is defined on the set $z$ of integers as follows. $(x,y)\in R\iff x^2+y^2=25$. Express $R$ and $R^{-1}$ as the set of ordered pairs and hence find their respective domains.

Answer 1

We have

$(x,y)\in R\iff x^2+y^2=25$

$\iff y=\pm \sqrt{25-x^2}$

We observe that $x=0\Rightarrow y=\pm 5$

$\therefore (0,5)\in R$ and $(0,-5)\in R$

$x=\pm 3\Rightarrow y=\pm \sqrt{25-9}=\pm 4$

$\therefore (3,4)\in R,(-3,4)\in R,(3,-4)\in R$ and $(-3,-4)\in R$

$x=\pm 4\Rightarrow y=\pm \sqrt{25-16}=\pm 3$

$\therefore (4,3)\in R,(-4,3)\in R,(4,-3)\in R$ and $(-4,-3)\in R$

$x=\pm 5\Rightarrow y=\pm \sqrt{25-25}=0$

$\therefore (5,0)\in R$ and $(-5,0)\in R$

We also notice that for any other integral value of $x,y$ is not an integer.

$\therefore R=\{(0,5),(0,-5),(3,4),(-3,4),(3,-4),(-3,-4),(4,3),(-4,3),(4,-3),(-4,-3),(5,0),(-5,0)\}$

and

$R^{-1}=\{(5,0),(-5,0),(4,3),(4,-3),(-4,3),(-4,-3),(3,4),(3,-4),(-3,4),(-3,-4),(0,5),(0,-5)\}$

Clearly domain $\left[R\right]=\{0,3,-3,4,-4,5,-5\}=$ domain$\left(R^{-1}\right)$