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Q.
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
For $(a, b),(c, d) \in N \times N$
$(a, b) R(c, d)$
$\Rightarrow a d(b+c)=b c(a+d)$
Reflexive Since, $a b(b+a)=b a(a+b), \forall a b \in N$
$\therefore (a, b) R(a, b)$
So, $R$ is reflexive.
Symmetric For $(a, b),(c, d) \in N \times N$,
Let $ (a, b) R(c, d)$
$\therefore a d(b+c)=b c(a+d)$
$\Rightarrow b c(a+d)=a d(b+c)$
$\Rightarrow c d(d+a)=d a(c+b)$
$\Rightarrow (c, d) R(a, b)$
So, $R$ is symmetric.
Transitive For $(a, b),(c, d),(e, f) \in N \times N$
Let $(a, b) R(c, d),(c, d) R(e, f)$
$\therefore a d(b+c),=b c(a+d), c f(d+e)=d e(c+f)$
$\Rightarrow a d b+a d c=b c a+b c d \ldots$(i)
and $c f d+c f e=d e c+d e f \ldots$(ii)
On multiplying Eq. (i) by ef and Eq. (ii) by $a b$,
then we get $adbef + a d c e f + c f d a b+c feab$
$= bcuef +b c d e f + decub + d e f u b $
$ \Rightarrow adcf (b+e)=bcde (a+f) $
$ \Rightarrow af (b+e)=b e(a+f) $
$ \Rightarrow (a, b) R(e, f) $
So, $R$ is transitive.
Hence $R$ is an equivalence relation.