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

To determine if the given relation \( R \) on \( \mathbb{Z} \times \mathbb{Z} \) is reflexive, symmetric, and/or transitive, we analyze each property in the context of the definition of \( R \): \((a, b) R (c, d)\) if and only if \(ad - bc\) is divisible by 5.

1. Reflexivity:
   • A relation \( R \) is reflexive if \((a, b) R (a, b)\) for all \((a, b) \in \mathbb{Z} \times \mathbb{Z}\).
   • Substitute \((a, b)\) for \((c, d)\): we get \(ad - bc = ab - ba = 0\).
   • Since 0 is divisible by 5, the relation is reflexive.
2. Symmetry:
   • A relation \( R \) is symmetric if whenever \((a, b) R (c, d)\), then \((c, d) R (a, b)\).
   • If \(ad - bc\) is divisible by 5, \(-ad + bc\) is also divisible by 5 (because \(-1 \times\) a number divisible by 5 is also divisible by 5).
   • Hence, the relation is symmetric.
3. Transitivity:
   • A relation \( R \) is transitive if whenever \((a, b) R (c, d)\) and \((c, d) R (e, f)\), then \((a, b) R (e, f)\).
   • Let \(x = ad - bc\) and \(y = cf - de\), both divisible by 5. For transitivity, \(af - be\) should be divisible by 5.
   • However, knowing \(x\) and \(y\) are divisible by 5 doesn't ensure \(af - be\) will also be divisible by 5.
   • For example, if \(a = 1, b = 0, c = 0, d = 1, e = 1, f = 0\), then \(ad - bc = 1\), \(cf - de = 1\), but \(af - be = 0\), which is divisible by 5.
   • Instead, if \( b = 1 \) and other choices remain 0, then it's not guaranteed that \(af - be = 0 - 1\) is not divisible by 5, specifically.
   • Thus, the relation may not be transitive.

Considering these observations, the correct answer is: Reflexive and symmetric but not transitive.

Answer 2

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