Question

 Let \( R \) be a relation on \( \mathbb{Z} \times \mathbb{Z} \) defined by \((a, b) R (c, d)\) if and only if \(ad - bc\) is divisible by 5.
Then \( R \) is:

• A. Reflexive but neither symmetric nor transitive
• B. Reflexive and symmetric but not transitive
• C. Reflexive, symmetric and transitive
• D. Reflexive and transitive but not symmetric

Answer 1

The Correct Option is B

Reflexive : for \((a, b) R (a, b) \)
\( ⇒ ab – ab = 0\) is divisible by 5.  
So \((a, b) R(a, b) ∀ a, b ∈ Z  \)
∴ R is reflexive  Symmetric : For \((a, b) R(c, d)  \)
If \(ad – bc\) is divisible by 5.  
Then \(bc – ad\) is also divisible by 5. 
\(⇒ (c, d) R(a, b) ∀ a, b, c, d ∈ Z  \)
∴ R is symmetric Transitive : If \((a, b) R(c, d) \) 
\(⇒ ad – bc\) divisible by 5 and \((c, d) R (e, f) \)
\(⇒ cf – de\) divisible by 5 
\(ad – bc = 5k_1\)   \( k_1\) and \(k_2\) are integers  
\(cf – de = 5k_2\) 
 \(afd – bcf = 5k_1f  \)
\(bcf – bde = 5k_2b  \)
\(afd – bde = 5(k_1f + k_2b)  \)
\(d(af – be) = 5 (k_1f + k_2b) \) 
\(⇒ af – be\) is not divisible by 5 for every a, b, c, d, e, f ∈ Z. 
\( ∴\) R is not transitive  
For e.g., take \(a = 1, b = 2, c = 5, d = 5, e = 2, f = 2\)

The correct option is (B): Reflexive and symmetric but not transitive