Question

If R and R₁ are equivalence relations on a set A, then so are the relations

• A. R-1
• B. R∪R₁
• C. R∩R1
• D. All of these

Answer 1

The Correct Option is A, C

The correct answer is/are option(s):
(A): R^-1 
(C): R ∩ R₁

Given, R and R₁ are equivalence relations on set A.

A relation is called an equivalence relation if it is:

• Reflexive: \( (a, a) \in R \) for all \( a \in A \)
• Symmetric: if \( (a, b) \in R \), then \( (b, a) \in R \)
• Transitive: if \( (a, b) \in R \) and \( (b, c) \in R \), then \( (a, c) \in R \)

 

Since R₁ is an equivalence relation, it satisfies:

• Reflexive: \( (a, a) \in R_1 \) for all \( a \in A \)
• Symmetric: if \( (a, b) \in R_1 \), then \( (b, a) \in R_1 \)
• Transitive: if \( (a, b) \in R_1 \), \( (b, c) \in R_1 \), then \( (a, c) \in R_1 \)

 

Similarly, R is also an equivalence relation and hence satisfies the same three properties.

We now prove that \( R \cap R_1 \) is an equivalence relation:

• Reflexive:
  Since \( (a, a) \in R \) and \( (a, a) \in R_1 \) for all \( a \in A \),
  \( (a, a) \in R \cap R_1 \).
  So, \( R \cap R_1 \) is reflexive.
• Symmetric:
  Let \( (a, b) \in R \cap R_1 \).
  Then \( (a, b) \in R \) and \( (a, b) \in R_1 \).
  Since both are symmetric,
  \( (b, a) \in R \) and \( (b, a) \in R_1 \).
  Thus, \( (b, a) \in R \cap R_1 \).
  So, \( R \cap R_1 \) is symmetric.
• Transitive:
  Let \( (a, b) \in R \cap R_1 \) and \( (b, c) \in R \cap R_1 \).
  Then \( (a, b), (b, c) \in R \) and \( (a, b), (b, c) \in R_1 \).
  Since both R and R₁ are transitive,
  \( (a, c) \in R \) and \( (a, c) \in R_1 \).
  So, \( (a, c) \in R \cap R_1 \).
  Hence, \( R \cap R_1 \) is transitive.

Conclusion: Since \( R \cap R_1 \) is reflexive, symmetric, and transitive, it is an equivalence relation.

Also, if R is an equivalence relation, its inverse \( R^{-1} \) is also an equivalence relation.
So both (A) and (C) are correct.