Question

Let R₁ and R₂ be two relations defined on ℝ by a R₁b ⇔ ab ≥ 0 and aR₂b ⇔ a ≥ b. Then,

• A. R₁ is an equivalence relation but not R₂
• B. R₂ is an equivalence relation but not R₁
• C. Both R₁ and R₂ are equivalence relations
• D. Neither R₁ nor R₂ is an equivalence relation

Answer 1

The Correct Option is D

To determine whether \( R_1 \) and \( R_2 \) are equivalence relations, we need to check each of them for reflexivity, symmetry, and transitivity.

Checking \( R_1 \) 

Reflexivity: A relation \( R \) is reflexive if for every element \( a \in \mathbb{R} \), \( aRa \) holds. For \( R_1 \), this means \( aa \ge 0 \), which is true for all \( a \in \mathbb{R} \). Thus, \( R_1 \) is reflexive.

Symmetry: A relation \( R \) is symmetric if for any \( a, b \in \mathbb{R} \), whenever \( aRb \), it implies \( bRa \). In \( R_1 \), if \( ab \ge 0 \), then it implies \( ba \ge 0 \). Since multiplication is commutative, \( ab = ba \). Therefore, \( R_1 \) is symmetric.

Transitivity: A relation \( R \) is transitive if for any \( a, b, c \in \mathbb{R} \), whenever \( aRb \) and \( bRc \), it implies \( aRc \). Suppose \( ab \ge 0 \) and \( bc \ge 0 \). This implies different cases:

• Both \( a \) and \( b \), and both \( b \) and \( c \) could be either positive or negative, leading to \( ac \ge 0 \).
• However, if \( a = 1 \), \( b = 0 \), and \( c = -1 \), we see that \( ab = 0 \ge 0 \) and \( bc = 0 \ge 0 \), but \( ac = -1 \not\ge 0 \).

This counterexample shows that \( R_1 \) is not transitive. Therefore, \( R_1 \) is not an equivalence relation.

Checking \( R_2 \)

Reflexivity: For \( R_2 \), we check if \( a \ge a \) holds for all \( a \in \mathbb{R} \). This is true, so \( R_2 \) is reflexive.

Symmetry: We need \( a \ge b \) to imply \( b \ge a \). This is not generally true unless \( a = b \). For example, if \( a = 2 \) and \( b = 1 \), then \( 2 \ge 1 \), but \( 1 \not\ge 2 \). Therefore, \( R_2 \) is not symmetric.

Transitivity: For transitivity, suppose \( a \ge b \) and \( b \ge c \). Then \( a \ge c \) must be true, which holds under the assumption. Therefore, \( R_2 \) is transitive.

Since \( R_2 \) is not symmetric, it is not an equivalence relation.

Conclusion

Neither \( R_1 \) nor \( R_2 \) qualifies as an equivalence relation because \( R_1 \) lacks transitivity and \( R_2 \) lacks symmetry. Therefore, the correct answer is: Neither \( R_1 \) nor \( R_2 \) is an equivalence relation.