Understanding the Problem: Relations and Their Composition

The question asks us to find the composite relation \(R \circ R^{-1}\) for a given relation \(R\) defined from set \(A\) to set \(B\). We are provided with the sets \(A = \{1, 2, 3, 4\}\) and \(B = \{1, 3, 5\}\), and the relation \(R\) is defined by the condition \(R = \{(a, b) : a < b, \text{ where } a \in A \text{ and } b \in B\}\).

To solve this, we need to follow these steps:

1. Determine the explicit set of ordered pairs in relation \(R\).
2. Determine the inverse relation \(R^{-1}\).
3. Determine the composite relation \(R \circ R^{-1}\).

Step 1: Finding Relation R

The relation \(R\) consists of pairs \((a, b)\) where \(a \in A\), \(b \in B\), and \(a < b\). Let's list the elements of \(A\) and \(B\) and find the pairs satisfying the condition:

• For \(a = 1 \in A\), find \(b \in B\) such that \(1 < b\). The elements in \(B\) greater than 1 are 3 and 5. So, pairs are \((1, 3), (1, 5)\).
• For \(a = 2 \in A\), find \(b \in B\) such that \(2 < b\). The elements in \(B\) greater than 2 are 3 and 5. So, pairs are \((2, 3), (2, 5)\).
• For \(a = 3 \in A\), find \(b \in B\) such that \(3 < b\). The elements in \(B\) greater than 3 is 5. So, the pair is \((3, 5)\).
• For \(a = 4 \in A\), find \(b \in B\) such that \(4 < b\). The elements in \(B\) greater than 4 is 5. So, the pair is \((4, 5)\).

Thus, the relation \(R\) is:

\(R = \{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}\)

Step 2: Finding the Inverse Relation R^-1

The inverse relation \(R^{-1}\) is obtained by swapping the elements in each ordered pair of \(R\). If \((a, b) \in R\), then \((b, a) \in R^{-1}\). The relation \(R\) is from A to B, so \(R^{-1}\) is from B to A.

Using the pairs in \(R\), we find the pairs in \(R^{-1}\):

• \((1, 3) \in R \implies (3, 1) \in R^{-1}\)
• \((1, 5) \in R \implies (5, 1) \in R^{-1}\)
• \((2, 3) \in R \implies (3, 2) \in R^{-1}\)
• \((2, 5) \in R \implies (5, 2) \in R^{-1}\)
• \((3, 5) \in R \implies (5, 3) \in R^{-1}\)
• \((4, 5) \in R \implies (5, 4) \in R^{-1}\)

Thus, the inverse relation \(R^{-1}\) is:

\(R^{-1} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}\)

Step 3: Finding the Composite Relation R ˆ R^-1

The composite relation \(R \circ R^{-1}\) is defined such that an ordered pair \((x, y)\) is in \(R \circ R^{-1}\) if and only if there exists an element \(z\) such that \((x, z) \in R^{-1}\) and \((z, y) \in R\).

Since \(R^{-1}\) is a relation from \(B\) to \(A\), the first element \(x\) of the pair \((x, z)\) must be from set \(B\). Since \(R\) is a relation from \(A\) to \(B\), the second element \(y\) of the pair \((z, y)\) must be from set \(B\). The intermediate element \(z\) must be from set \(A\).

So, \(R \circ R^{-1}\) is a relation from \(B\) to \(B\). We need to find pairs \((x, y)\) where \(x, y \in B = \{1, 3, 5\}\) such that there exists \(z \in A = \{1, 2, 3, 4\}\) with \((x, z) \in R^{-1}\) and \((z, y) \in R\).

Let's examine the possible values for \(x\) from \(B\) and \(y\) from \(B\), and check for a connecting \(z\) from \(A\):

• Check pairs starting with x = 1 (from B):

  Are there pairs \((1, z)\) in \(R^{-1}\) for any \(z \in A\)? Looking at \(R^{-1} = \{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}\), there are no pairs starting with 1. So, no pairs \((1, y)\) will be in \(R \circ R^{-1}\).

• Check pairs starting with x = 3 (from B):

  Pairs in \(R^{-1}\) starting with 3 are \((3, 1)\) and \((3, 2)\). The connecting element \(z\) can be 1 or 2.

  • If \(z = 1\): We have \((3, 1) \in R^{-1}\). Now check for pairs \((1, y)\) in \(R\). Pairs in \(R\) starting with 1 are \((1, 3)\) and \((1, 5)\).
    • For \(y = 3\), \((1, 3) \in R\). Since \((3, 1) \in R^{-1}\) and \((1, 3) \in R\), we have \((3, 3) \in R \circ R^{-1}\).
    • For \(y = 5\), \((1, 5) \in R\). Since \((3, 1) \in R^{-1}\) and \((1, 5) \in R\), we have \((3, 5) \in R \circ R^{-1}\).
  • If \(z = 2\): We have \((3, 2) \in R^{-1}\). Now check for pairs \((2, y)\) in \(R\). Pairs in \(R\) starting with 2 are \((2, 3)\) and \((2, 5)\).
    • For \(y = 3\), \((2, 3) \in R\). Since \((3, 2) \in R^{-1}\) and \((2, 3) \in R\), we have \((3, 3) \in R \circ R^{-1}\). (Already found)
    • For \(y = 5\), \((2, 5) \in R\). Since \((3, 2) \in R^{-1}\) and \((2, 5) \in R\), we have \((3, 5) \in R \circ R^{-1}\). (Already found)

  So, pairs starting with 3 in \(R \circ R^{-1}\) are \((3, 3)\) and \((3, 5)\).

• Check pairs starting with x = 5 (from B):

  Pairs in \(R^{-1}\) starting with 5 are \((5, 1), (5, 2), (5, 3), (5, 4)\). The connecting element \(z\) can be 1, 2, 3, or 4.

  • If \(z = 1\): We have \((5, 1) \in R^{-1}\). Now check for pairs \((1, y)\) in \(R\). Pairs in \(R\) starting with 1 are \((1, 3)\) and \((1, 5)\).
    • For \(y = 3\), \((1, 3) \in R\). Since \((5, 1) \in R^{-1}\) and \((1, 3) \in R\), we have \((5, 3) \in R \circ R^{-1}\).
    • For \(y = 5\), \((1, 5) \in R\). Since \((5, 1) \in R^{-1}\) and \((1, 5) \in R\), we have \((5, 5) \in R \circ R^{-1}\).
  • If \(z = 2\): We have \((5, 2) \in R^{-1}\). Now check for pairs \((2, y)\) in \(R\). Pairs in \(R\) starting with 2 are \((2, 3)\) and \((2, 5)\).
    • For \(y = 3\), \((2, 3) \in R\). Since \((5, 2) \in R^{-1}\) and \((2, 3) \in R\), we have \((5, 3) \in R \circ R^{-1}\). (Already found)
    • For \(y = 5\), \((2, 5) \in R\). Since \((5, 2) \in R^{-1}\) and \((2, 5) \in R\), we have \((5, 5) \in R \circ R^{-1}\). (Already found)
  • If \(z = 3\): We have \((5, 3) \in R^{-1}\). Now check for pairs \((3, y)\) in \(R\). Pairs in \(R\) starting with 3 is \((3, 5)\).
    • For \(y = 5\), \((3, 5) \in R\). Since \((5, 3) \in R^{-1}\) and \((3, 5) \in R\), we have \((5, 5) \in R \circ R^{-1}\). (Already found)
  • If \(z = 4\): We have \((5, 4) \in R^{-1}\). Now check for pairs \((4, y)\) in \(R\). Pairs in \(R\) starting with 4 is \((4, 5)\).
    • For \(y = 5\), \((4, 5) \in R\). Since \((5, 4) \in R^{-1}\) and \((4, 5) \in R\), we have \((5, 5) \in R \circ R^{-1}\). (Already found)

  So, pairs starting with 5 in \(R \circ R^{-1}\) are \((5, 3)\) and \((5, 5)\).

Combining all the pairs found, the composite relation \(R \circ R^{-1}\) is:

\(R \circ R^{-1} = \{(3, 3), (3, 5), (5, 3), (5, 5)\}\)

Comparing with Options

Let's compare our result with the given options:

1. \(\{(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)\}\) - This is \(R\).
2. \(\{(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)\}\) - This is \(R^{-1}\).
3. \(\{(3, 3), (3, 5), (5, 3), (5, 5)\}\) - This matches our calculated \(R \circ R^{-1}\).
4. \(\{(3, 3), (3, 4), (4, 5)\}\) - This does not match.

The calculated composite relation \(R \circ R^{-1}\) matches option 3.

Conclusion on Relation Composition

Based on the detailed steps of finding \(R\), \(R^{-1}\), and their composition \(R \circ R^{-1}\), the composite relation is indeed \(\{(3, 3), (3, 5), (5, 3), (5, 5)\}\).

Revision Table: Key Concepts in Relations

Additional Information on Composite Relations

When dealing with composite relations like \(R \circ S\), it's important to remember the order of composition and the domains and codomains involved. If \(S\) is a relation from set X to set Y, and \(R\) is a relation from set Y to set Z, then \(R \circ S\) is a relation from X to Z. The composition \(R \circ S\) means that we first follow a connection from X to Y via relation \(S\), and then from Y to Z via relation \(R\).

In this problem, \(R\) is from A to B, and \(R^{-1}\) is from B to A. Therefore, \(R \circ R^{-1}\) means apply \(R^{-1}\) first (from B to A) and then \(R\) (from A to B). The resulting relation \(R \circ R^{-1}\) is a relation from B to B.

Similarly, \(R^{-1} \circ R\) would mean apply \(R\) first (from A to B) and then \(R^{-1}\) (from B to A). The resulting relation \(R^{-1} \circ R\) is a relation from A to A. In general, \(R \circ R^{-1}\) and \(R^{-1} \circ R\) are not the same relation unless the original relation \(R\) has specific properties (e.g., being a bijection if A and B are the same set and R is a function). In our case, \(R\) is not a function (e.g., 1 maps to both 3 and 5), so \(R \circ R^{-1}\) and \(R^{-1} \circ R\) will be different.