Question

If the relation \( R \) is given by \[ R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\}, \] then find \( R^{-1} \circ R^{-1} \).

Hint:

To find the composition \(S \circ T\), think of it as a path. An element (a, c) is in the composition if you can go from 'a' to 'b' using relation T, and then from 'b' to 'c' using relation S. In this problem, both T and S are \(R^{-1}\).

Answer 1

Step 1: Understanding the Concept:
First, we need to find the inverse relation, \(R^{-1}\), which is obtained by swapping the first and second elements of each ordered pair in R. Second, we need to find the composition \(R^{-1} \circ R^{-1}\). An ordered pair (a, c) belongs to this composition if there exists an element 'b' such that (a, b) is in \(R^{-1}\) and (b, c) is in \(R^{-1}\).
Step 2: Key Formula or Approach:
1. Find \(R^{-1}\) from R. 2. For each pair \((a, b) \in R^{-1}\), search for all pairs \((b, c) \in R^{-1}\). 3. The resulting pairs \((a, c)\) form the set \(R^{-1} \circ R^{-1}\).
Step 3: Detailed Explanation:
The given relation is: \[ R = \{(4, 5), (1, 4), (4, 6), (7, 6), (3, 7)\} \] First, find the inverse relation \(R^{-1}\) by swapping the elements in each pair: \[ R^{-1} = \{(5, 4), (4, 1), (6, 4), (6, 7), (7, 3)\} \] Now we find the composition \(R^{-1} \circ R^{-1}\) by finding "chains" of length two within \(R^{-1}\).
We have \((5, 4) \in R^{-1}\) and \((4, 1) \in R^{-1}\). This gives us the pair \(\mathbf{(5, 1)}\).
We have \((4, 1) \in R^{-1}\), but there are no pairs in \(R^{-1}\) that start with 1. So this chain ends.
We have \((6, 4) \in R^{-1}\) and \((4, 1) \in R^{-1}\). This gives us the pair \(\mathbf{(6, 1)}\).
We have \((6, 7) \in R^{-1}\) and \((7, 3) \in R^{-1}\). This gives us the pair \(\mathbf{(6, 3)}\).
We have \((7, 3) \in R^{-1}\), but there are no pairs in \(R^{-1}\) that start with 3. So this chain ends.
Combining the resulting pairs, we get the final relation.
Step 4: Final Answer:
The composition \(R^{-1} \circ R^{-1}\) is \(\{(5, 1), (6, 1), (6, 3)\}\).