Question 26

By definition, $aRb$ if and only if:

1. $a \neq b$ ($a$ and $b$ are distinct), and
2. $\gcd(a, b) > 1$ ($a$ and $b$ share a common divisor other than 1).

• Reflexivity: For any positive integer $a$, $a$ is not distinct from itself ($a = a$), so $(a, a) \notin R$. Thus, $R$ is not reflexive.

• Symmetry: If $aRb$, then $a \neq b$ and $\gcd(a, b) > 1$. Because distinctness and greatest common divisor are commutative ($b \neq a$ and $\gcd(b, a) = \gcd(a, b) > 1$), $bRa$ holds. Thus, $R$ is symmetric.

• Transitivity: Consider $a = 2, b = 6, c = 9$:
  • $a \neq b$ and $\gcd(2, 6) = 2 > 1 \implies 2R6$.
  • $b \neq c$ and $\gcd(6, 9) = 3 > 1 \implies 6R9$.
  • However, $\gcd(2, 9) = 1$, so $(2, 9) \notin R$.
    Hence, $R$ is not transitive.

Therefore, $R$ is symmetric, but neither reflexive nor transitive (option D).