Question

Let A = \{1, 2, 3\}. Then, the number of relations containing (1, 2) and (1, 3), which are reflexive and symmetric but not transitive, is

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When asked to find the number of relations with certain properties, start by building the smallest possible relation that includes all the mandatory elements and satisfies the given properties (like reflexivity and symmetry). Then, check if this minimal relation meets the final condition (e.g., not transitive). Finally, see if adding any other allowed elements still keeps the properties valid.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to construct a relation R on the set A = \{1, 2, 3\} that satisfies four conditions: 1. It contains (1, 2) and (1, 3). 2. It is reflexive. 3. It is symmetric. 4. It is not transitive. We then need to count how many such unique relations exist.
Step 3: Detailed Explanation:
Let R be the relation we are looking for.
Condition 1 \& 2 (Reflexivity): For R to be reflexive on A, it must contain all pairs (a, a) for a $\in$ A.
So, R must contain \{(1, 1), (2, 2), (3, 3)\}.
Condition 1 \& 3 (Symmetry): The relation must contain (1, 2) and (1, 3). For R to be symmetric, if (a, b) $\in$ R, then (b, a) must also be in R.
Since (1, 2) $\in$ R, we must have (2, 1) $\in$ R.
Since (1, 3) $\in$ R, we must have (3, 1) $\in$ R.
Combining these, the smallest possible relation R that satisfies the first three properties is:
\[ R_{min} = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1)\} \] Condition 4 (Not Transitive): Now we must check if R\textsubscript{min} is transitive. A relation is transitive if (a, b) $\in$ R and (b, c) $\in$ R implies (a, c) $\in$ R. Let's look for a counterexample.
Consider (2, 1) $\in$ R\textsubscript{min} and (1, 3) $\in$ R\textsubscript{min}. For transitivity, the pair (2, 3) must be in R\textsubscript{min}. However, (2, 3) $\notin$ R\textsubscript{min}.
Similarly, consider (3, 1) $\in$ R\textsubscript{min} and (1, 2) $\in$ R\textsubscript{min}. For transitivity, (3, 2) must be in R\textsubscript{min}. But (3, 2) $\notin$ R\textsubscript{min}.
Since we found a case where transitivity fails, R\textsubscript{min} is not transitive.
So, R\textsubscript{min} is one such relation. Are there any others?
The only pairs not in R\textsubscript{min} are (2, 3) and (3, 2). If we add these to R\textsubscript{min}, we must add both to maintain symmetry.
Let's define a new relation R' = R\textsubscript{min} $\cup$ \{(2, 3), (3, 2)\}.
R' = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)\}. This is the universal relation A $\times$ A.
Is R' transitive? Let's check the pair that failed before: (2, 1) $\in$ R' and (1, 3) $\in$ R'. Is (2, 3) $\in$ R'? Yes, it is now. The universal relation on a set is always an equivalence relation, meaning it is reflexive, symmetric, and transitive. So R' is transitive.
This means we cannot add the pair \{(2, 3), (3, 2)\} because it makes the relation transitive.
Step 4: Final Answer:
The only relation that satisfies all the given conditions is R\textsubscript{min}. Therefore, there is only 1 such relation.