Given a relation $R$, what is the $R^{-1}$, and what is the complement of $R$ on $A$?

Question

For example if I have the following group $A$ and a relation $R$: $$A= {1,2,3}$$ $$R= \{(1,1)(2,2)(3,3)(1,2)(3,2)\}$$ $$R^{-1} = ?$$ $$R^C = \;?$$

Thanks.

Answer 1

$$ R^{-1} = \{(x,y) \mid (y,x) \in R\}$$ $$ R^C = \{ (x,y) \in A \times A \mid (x,y) \notin R\}$$ so with

$$ R = \{(1,1),(2,2),(3,3),(1,2),(3,2)\}$$

we get

$$R^{-1} = \{(1,1),(2,2),(3,3),(2,1),(2,3)\}$$ $$R^C = \{(2,1),(2,3),(1,3),(3,1)\}$$

Also, there was no significance of calling $A$ a group. Calling it the set $$ A = \{1,2,3\}$$ is enough for this problem.

Answer 2

$$\begin{align} A\times A &= \{(x, y)\mid (x\in A) \land (y\in A)\} \\ &= \{(1, 1), (1,2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3. 2), (3, 3)\}\end{align}$$

$$R\subset A\times A = \{(1,1),(2,2),(3,3),(1,2),(3,2)\}\tag{given}$$

$$R^{-1} = \{(y, x) \mid (x, y) \in R \} = \{(1, 1), (2, 2), (3, 3), (2, 1), (2, 3)\}\tag{inverse of $R$}$$

$$\begin{align} R^C \subset A\times A &= \{(x, y) \in A\times A \mid (x, y)\notin R\}\\ &= \{(1, 3), (2, 1), (2, 3), (3, 1)\}\tag{complement of $R$}\end{align}$$

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Note that $R^{-1}$ represents the inverse relation of $R$ on $A$.

Also, $R^C \subset A\times A$ represents the complement of the relation $R$ on $A$, and that $R \cup R^{C} = A\times A$.