Question

Let N denote the set of all natural numbers and R be the relation on $N \times N$ defined by (a, b) R (c, d) if

$a d ( b + c ) = b c ( a + d )$ then R is

• A. Symmetric only
• B. Reflexive only
• C. Transitive only
• D. An equivalence relation

Answer 1

Answer: (D)

An equivalence relation

Answer 2

For (a, b), (c, d) ∈ N × N

$( a , b ) R ( c , d ) \Rightarrow a d ( b + c ) = b c ( a + d )$Reflexive: Since $a b ( b + a )$

= $b a ( a + b ) \forall a b \in N$ ,

$\therefore$ $\therefore$R is reflexive.

Symmetric: For $( a , b ) R ( c , d )$

$\therefore$ $a d ( b + c ) = b c ( a + d )$ ⇒ $b c ( a + d ) = a d ( b + c )$

⇒ $c b ( d + a ) = d a ( c + b )$ ⇒ $( c , d ) R ( a , b )$

$\therefore$ R is symmetric

Transitive: For $( a , b ) , ( c , d ) , ( e , f ) \in N \times N$

Let $( a , b ) R ( c , d ) , ( c , d ) R ( e , f )$

$\therefore$ $a d ( b + c ) = b c ( a + d )$ , $c f ( d + e ) = d e ( c + f )$

⇒ $a d b + a d c = b c a + b c d$ .....(i)

and $c f d + c f e = d e c + d e f$ .......(ii)

(i) × $e f +$ (ii) × $a b$ gives,

= $a d c f ( b + e ) = b c d e ( a + f )$

⇒ $a f ( b + e ) = b e ( a + f )$⇒ $( a , b ) R ( e , f )$.

$\therefore$ R is transitive. Hence R is an equivalence relation.