Question

A relation \(R\) on \(A = \{1, 2, 3\}\) is defined as \(R = \{(1, 1), (3, 3), (1, 2)\\). Is \(R\) a symmetric relation? Justify. Write the smallest relation set \(R_1\) such that \(R \cup R_1\) becomes an equivalence relation on the set \(\{1, 2, 3\}\).}

Hint:

To turn a relation into an equivalence relation, check the three rules in order: Reflexive, then Symmetric, then Transitive. Transitivity usually requires the most careful verification of "bridge" elements.

Answer 1

Concept:
• A relation is symmetric if \((a, b) \in R \implies (b, a) \in R\).
• An equivalence relation must be reflexive, symmetric, and transitive.

Step 1: Check for symmetry
In the given relation \(R = \{(1, 1), (3, 3), (1, 2)\}\): We see that \((1, 2) \in R\), but \((2, 1) \notin R\). Therefore, \(R\) is not a symmetric relation.

Step 2: Ensure reflexivity for the equivalence relation
The set is \(A = \{1, 2, 3\}\). For reflexivity, every element must relate to itself: \((1, 1), (2, 2), (3, 3)\). \(R\) already contains \((1, 1)\) and \((3, 3)\). We must add \((2, 2)\).

Step 3: Ensure symmetry for the equivalence relation
\(R\) contains \((1, 2)\). For symmetry, we must include \((2, 1)\). Now the set of pairs is \(\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\).

Step 4: Check for transitivity
Check all combinations: \((1, 2)\) and \((2, 1)\) are present, and \((1, 1)\) is present. Also \((2, 1)\) and \((1, 2)\) are present, and \((2, 2)\) is present. The set \(\{(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)\}\) is already transitive.

Step 5: Define the smallest set \(R_1\)
The elements we needed to add to \(R\) to reach this equivalence relation are \((2, 2)\) and \((2, 1)\). Thus, \(R_1 = \{(2, 2), (2, 1)\}\).