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

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

Answer 1

The correct option is D An equivalence relation

For $(a,b),(c,d)\in N\times N$

$(a,b)R(c,d)\Rightarrow ad(b+c)=bc(a+d)$

Reflexive: Since $ab(b+a)=ba(a+b)\forall ab\in N$

$\therefore (a,b)R(a,b)\Rightarrow R$ is reflexive.

Symmetric: For $(a,b),(c,d)\in N\times N$, let $(a,b)R(c,d)$

$\therefore ad(b+c)=bc(a+d)\Rightarrow bc(a+d)=ad(b+c)$

$\Rightarrow cb(d+a)=da(c+b)\Rightarrow (c,d)R(a,b)$

$\therefore$ $R$ is symmetric

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

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

$\therefore ad(b+c)=bc(a+d),cf(d+e)=de(c+f)$

$\Rightarrow adb+adc=bca+bcd......\left(i\right)$

and $cfd+cfe=dec+def.......\left(ii\right)$

$\left(i\right)\times ef+\left(ii\right)\times ab$ gives,

$adbef+adcef+cfdab+cefab$

$=bcaef+bcdef+decab+defab$

$\Rightarrow adcf(b+e)=bcde(a+f)$

$\Rightarrow af(b+e)=be(a+f)$

$\Rightarrow (a,b)R(e,f)$

$\therefore R$ is transitive.

Hence $R$ is an equivalence relation