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

Answer: (D)

For $(a,b),(c,d)\in N\times N$
$(a,b)R(c,d)$
$\Rightarrow ad(b+c)=bc(a+d)$
Reflexive Since, $ab(b+a)=ba(a+b),\forall ab\in N$
$\therefore (a,b)R(a,b)$
So, $R$ is reflexive.
Symmetric For $(a,b),(c,d)\in N\times N$,
Let $(a,b)R(c,d)$
$\therefore ad(b+c)=bc(a+d)$
$\Rightarrow bc(a+d)=ad(b+c)$
$\Rightarrow cd(d+a)=da(c+b)$
$\Rightarrow (c,d)R(a,b)$
So, $R$ is symmetric.
Transitive For $(a,b),(c,d),(e,f)\in N\times N$
Let $(a,b)R(c,d),(c,d)R(e,f)$
$\therefore ad(b+c),=bc(a+d),cf(d+e)=de(c+f)$
$\Rightarrow adb+adc=bca+bcd\dots$(i)
and $cfd+cfe=dec+def\dots$(ii)
On multiplying Eq. (i) by ef and Eq. (ii) by $ab$,
then we get $adbef+adcef+cfdab+cfeab$
$=bcuef+bcdef+decub+defub$
$\Rightarrow adcf(b+e)=bcde(a+f)$
$\Rightarrow af(b+e)=be(a+f)$
$\Rightarrow (a,b)R(e,f)$
So, $R$ is transitive.
Hence $R$ is an equivalence relation.