If `R` be a relation `<` from `A = {1, 2, 3,4) to B = (1,3,5)` that is `(a, b) in R iff a lt b`, then `R o R^-1` is

Options

A. a) `{(1,3),(1,5),(2,3),(2,5),(3,5),(4,5)}`
B. b) `{(3,1),(5,1),(3,2),(5,2),(5,3),(5,4)}`
C. c)
`{(3,3),(3,5),(5,3),(5,5)}`
D. d) `{(3,3),(3,4),(4,5)}`

Answer

To solve the problem, we need to find the composition of the relation \( R \) and its inverse \( R^{-1} \). Let's go through the steps systematically.

### Step 1: Define the Sets
We are given two sets:
- \( A = \{1, 2, 3, 4\} \)
- \( B = \{1, 3, 5\} \)

### Step 2: Define the Relation \( R \)
The relation \( R \) is defined as \( (a, b) \in R \) if \( a < b \) where \( a \in A \) and \( b \in B \).

Now, let's find all pairs \( (a, b) \) that satisfy this condition:
- For \( a = 1 \): \( b \) can be \( 3 \) and \( 5 \) → pairs: \( (1, 3), (1, 5) \)
- For \( a = 2 \): \( b \) can be \( 3 \) and \( 5 \) → pairs: \( (2, 3), (2, 5) \)
- For \( a = 3 \): \( b \) can only be \( 5 \) → pair: \( (3, 5) \)
- For \( a = 4 \): \( b \) can only be \( 5 \) → pair: \( (4, 5) \)

Thus, the relation \( R \) is:
\[ R = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\} \]

### Step 3: Find the Inverse Relation \( R^{-1} \)
The inverse relation \( R^{-1} \) is obtained by swapping the elements in each pair of \( R \):
\[ R^{-1} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\} \]

### Step 4: Compute the Composition \( R \circ R^{-1} \)
The composition \( R \circ R^{-1} \) consists of pairs \( (a, c) \) such that there exists a \( b \) where \( (a, b) \in R \) and \( (b, c) \in R^{-1} \).

We will check each pair in \( R^{-1} \) and find corresponding pairs in \( R \):
1. For \( (3, 1) \):
- \( b = 1 \) gives \( (1, 3) \) and \( (1, 5) \) → pairs: \( (3, 3), (3, 5) \)

2. For \( (5, 1) \):
- \( b = 1 \) gives \( (1, 3) \) and \( (1, 5) \) → pairs: \( (5, 3), (5, 5) \)

3. For \( (3, 2) \):
- \( b = 2 \) gives \( (2, 3) \) and \( (2, 5) \) → pairs: \( (3, 3), (3, 5) \) (already counted)

4. For \( (5, 2) \):
- \( b = 2 \) gives \( (2, 3) \) and \( (2, 5) \) → pairs: \( (5, 3), (5, 5) \) (already counted)

5. For \( (5, 3) \):
- \( b = 3 \) gives \( (3, 5) \) → pairs: \( (5, 5) \) (already counted)

6. For \( (5, 4) \):
- \( b = 4 \) gives \( (4, 5) \) → pairs: \( (5, 5) \) (already counted)

Combining all unique pairs, we have:
\[ R \circ R^{-1} = \{(3, 3), (3, 5), (5, 3), (5, 5)\} \]

### Final Answer
Thus, the result of \( R \circ R^{-1} \) is:
\[ \{(3, 3), (3, 5), (5, 3), (5, 5)\} \]