Question

A relation \( R \) is defined on the set \( \mathbb{Q}^* \) of non-zero rational numbers as follows: \[ a \, R \, b \quad \text{if} \quad a = \frac{1}{b}. \] Then on \( \mathbb{Q}^* \), this relation \( R \) is:

• A. reflexive, but not symmetric and transitive
• B. symmetric, but not reflexive and transitive
• C. transitive, but not reflexive and symmetric
• D. reflexive and transitive, but not symmetric \bigskip

Hint:

When proving properties of relations, check each property (reflexivity, symmetry, transitivity) by using the definitions.

Answer 1

The Correct Option is A

Step 1: A relation is reflexive if for every element \( a \in \mathbb{Q}^* \), \( a \, R \, a \). In this case, since \( a = \frac{1}{a} \), the relation is reflexive.

Step 2: A relation is symmetric if \( a \, R \, b \) implies \( b \, R \, a \). In this case, if \( a = \frac{1}{b} \), then \( b = \frac{1}{a} \), so the relation is symmetric.

Step 3: A relation is transitive if \( a \, R \, b \) and \( b \, R \, c \) implies \( a \, R \, c \). In this case, it holds that if \( a = \frac{1}{b} \) and \( b = \frac{1}{c} \), then \( a = \frac{1}{c} \), so the relation is transitive.

Thus, the relation is reflexive, symmetric, and transitive, so the correct answer is (A). \bigskip