Answer to Question #127924 in Discrete Mathematics for jaya

Question

(b) (i) Let A = {2,4,6} and R = {(2,2), (2,4), (2,6), (4,4), (6,6)} be a relation on A.
Find R−1
.
(ii) Let A = {6,8,10,15} and B = {2,3,4}. Define a relation R from A to B by
(x,y) ∈ R iff x-y is divisible by 2. Find R−1

Answer 1

(i) Given $A = \{2,4,6\}$ and $R = \{(2,2), (2,4), (2,6), (4,4), (6,6)\}$ is a relation on A.

We have $R^{-1} = \{(y,x)|(x,y) \in R\}$. Therefore, $R^{-1} = \{(2,2),(4,2),(6,2),(4,4),(6,6)\}$

(ii) Given $A = \{6,8,10,15\}$ and $B = \{2,3,4\}$. Then,

$A \times B = \left\{(6,2),(6,3),(6,4),(8,2),(8,3),(8,4),\right.\\ \hspace{1in}\left.(10,2),(10,3),(10,4),(15,2),(15,3),(15,4)\right\}$

Let R be a relation from A to B defined by (x,y) ∈ R iff x-y is divisible by 2. That is,

$R =\{(x,y)\in A \times B|x-y\equiv 0(\text{mod}~2)\}$. Therefore,

$R = \{(6,2),(6,4),(8,2),(8,4),(10,2),(10,4),(15,3)\}$.

Hence, $R^{-1} = \{(2,6),(4,6),(2,8),(4,8),(2,10),(4,10),(3,15)\}$.