Question

If $N$ 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 $(a,b),(c,d)ϵN\times N$
$(a,b)R(c,d)$
$\Rightarrow ad(b+c)=bc(a+d)$
Reflexive: $ab(b+a)=ba(a+b),\forall abϵN$
$\therefore (a,b)R(a,b)$
So, $R$ is reflexive,
Symmetric: $ad(b+c)=bc(a+d)$
$\Rightarrow bc(a+d)=ad(b+c)$
$\Rightarrow cd(d+a)=da(c+b)$
$\Rightarrow (c,d)R(a,b)$
So, $R$ is symmetric.
Transitive: For $(a,b),(c,d),(e,f)ϵN\times N$
Let $(a,b)R(c,d),(c,d)R(e,f)$
$\therefore ad(b+c)=bc(a+d)andcf(d+e)=de(c+f)$
$\Rightarrow adb+adc=bca+bcd....\left(i\right)$
and $cfd+cfe=dec+def...\left(ii\right)$
On multiplying eq. (i) by $ef$ and eq. (ii) by $ab$ and then adding, we have
$adbef+adcef+cfdab+cfeab$
$=bcaef+bcdef+decab+defab$
$\Rightarrow adcf(b+e)=bcde(a+f)$
$\Rightarrow af(b+e)=be(a+f)$
$\Rightarrow (a,b)R(e,f)$
So, $R$ is transitive.
Hence, $R$ is an equivalence relation.