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(D) The relation $R$ is defined on $N 	imes N$ as $(a, b) R (c, d) \iff ad(b + c) = bc(a + d)$.$1$. Reflexive: For any $(a, b) \in N 	imes N$,we hav

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An equivalence relation

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(D) The relation $R$ is defined on $N 	imes N$ as $(a, b) R (c, d) \iff ad(b + c) = bc(a + d)$.$1$. Reflexive: For any $(a, b) \in N 	imes N$,we have $ab(b + a) = ba(a + b)$,which is true. Thus,$(a, b) R (a, b)$. So,$R$ is reflexive.$2$. Symmetric: Let $(a, b) R (c, d)$. Then $ad(b + c) = bc(a + d)$. This implies $bc(a + d) = ad(b + c)$,which can be rewritten as $cb(d + a) = da(c + b)$. Thus,$(c, d) R (a, b)$. So,$R$ is symmetric.$3$. Transitive: Let $(a, b) R (c, d)$ and $(c, d) R (e, f)$. Then $ad(b + c) = bc(a + d)$ and $cf(d + e) = de(c + f)$.Dividing by $abcd$ and $cdef$ respectively,we get $\frac{b+c}{bc} = \frac{a+d}{ad} \implies \frac{1}{c} + \frac{1}{b} = \frac{1}{d} + \frac{1}{a} \implies \frac{1}{a} - \frac{1}{b} = \frac{1}{c} - \frac{1}{d}$.Similarly,$\frac{1}{c} - \frac{1}{d} = \frac{1}{e} - \frac{1}{f}$.Therefore,$\frac{1}{a} - \frac{1}{b} = \frac{1}{e} - \frac{1}{f} \implies \frac{1}{a} + \frac{1}{f} = \frac{1}{e} + \frac{1}{b} \implies \frac{a+f}{af} = \frac{b+e}{be} \implies af(b + e) = be(a + f)$.Thus,$(a, b) R (e, f)$. So,$R$ is transitive.Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.

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

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