Question

If \(R = \{(1,1), (2,2), (1,2), (2,1), (3,3)\}\) and \(S = \{(1,1), (2,2), (2,3), (3,2), (3,3)\}\) are two relations on the set \(X = \{1, 2, 3\}\), the incorrect statement is:

• A. R and S are both equivalence relations
• B. \(R \cap S\) is an equivalence relation
• C. \(R^{-1} \cap S^{-1}\) is an equivalence relation
• D. \(R \cup S\) is an equivalence relation

Hint:

Check reflexive, symmetric and transitive for each combination separately; the union of two equivalence relations is not always transitive.

Answer 1

The Correct Option is D

Step 1: Recall what makes a relation an equivalence relation.
A relation on a set is an equivalence relation only when it is reflexive (every element relates to itself), symmetric (if a relates to b, then b relates to a), and transitive (if a relates to b, and b relates to c, then a must relate to c).
We check each of the four statements against these three rules.

Step 2: Check that R is an equivalence relation.
R contains \((1,1),(2,2),(3,3)\), so it is reflexive on \(X=\{1,2,3\}\).
It contains \((1,2)\) and \((2,1)\) together, so it is symmetric.
Every combination, such as \((1,2)\) with \((2,1)\) giving \((1,1)\), and \((2,1)\) with \((1,2)\) giving \((2,2)\), lands back inside R, so R is transitive. R is an equivalence relation, grouping the elements as \(\{1,2\}\) and \(\{3\}\).

Step 3: Check that S is an equivalence relation.
S contains \((1,1),(2,2),(3,3)\), so it is reflexive.
It contains \((2,3)\) and \((3,2)\) together, so it is symmetric.
Combining \((2,3)\) with \((3,2)\) gives \((2,2)\), and combining \((3,2)\) with \((2,3)\) gives \((3,3)\), both already inside S, so S is transitive. S is also an equivalence relation, grouping the elements as \(\{1\}\) and \(\{2,3\}\).

Step 4: Check \(R \cap S\) and \(R^{-1} \cap S^{-1}\).
The only pairs common to both R and S are \((1,1),(2,2),(3,3)\), so \(R \cap S = \{(1,1),(2,2),(3,3)\}\), which is just the identity relation, trivially reflexive, symmetric and transitive.
Since R and S are each symmetric, \(R^{-1}=R\) and \(S^{-1}=S\), so \(R^{-1}\cap S^{-1}\) is the same set \(\{(1,1),(2,2),(3,3)\}\), also an equivalence relation.

Step 5: Check \(R \cup S\).
Taking the union gives \(R\cup S = \{(1,1),(2,2),(3,3),(1,2),(2,1),(2,3),(3,2)\}\).
This union relates 1 to 2 (through \((1,2)\)) and relates 2 to 3 (through \((2,3)\)). For transitivity, we would then need 1 to relate to 3, that is, \((1,3)\) should be in the set.
But \((1,3)\) is not in \(R \cup S\), so the union fails the transitive test and is NOT an equivalence relation.

Final Answer:
R, S, \(R \cap S\) and \(R^{-1}\cap S^{-1}\) are all genuine equivalence relations, but \(R \cup S\) is not, since it is not transitive. So the incorrect statement is the one claiming \(R \cup S\) is an equivalence relation. \[ \boxed{R \cup S \text{ is not an equivalence relation}} \]