Let R be a relation on N times N defined by ( | vedclass

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(D) The relation is defined as $(a, b) R (c, d) \iff ad(b-c) = bc(a-d)$.$1$. Reflexive: For $(a, b) R (a, b)$,we need $ab(b-a) = ba(a-b)$. This simpli

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symmetric and transitive but not reflexive

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(D) The relation is defined as $(a, b) R (c, d) \iff ad(b-c) = bc(a-d)$.$1$. Reflexive: For $(a, b) R (a, b)$,we need $ab(b-a) = ba(a-b)$. This simplifies to $ab(b-a) = -ab(b-a)$,which is only true if $ab(b-a) = 0$. Since $a, b \in N$,this is not true for all $(a, b)$. Thus,$R$ is not reflexive.$2$. Symmetric: If $(a, b) R (c, d)$,then $ad(b-c) = bc(a-d) \Rightarrow adb - adc = bca - bcd \Rightarrow adb + bcd = bca + adc$. Dividing by $abcd$,we get $\frac{1}{c} - \frac{1}{d} = \frac{1}{a} - \frac{1}{b}$. This is symmetric because swapping $(a, b)$ and $(c, d)$ yields the same condition. Thus,$R$ is symmetric.$3$. Transitive: The condition $\frac{1}{b} - \frac{1}{a} = \frac{1}{d} - \frac{1}{c}$ implies that the relation is an equivalence relation (if defined on $N 	imes N$ such that $a, b 
eq 0$). However,checking the specific form $ad(b-c) = bc(a-d)$,it is equivalent to $\frac{b-c}{bc} = \frac{a-d}{ad} \Rightarrow \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a}$. This is transitive. Since it is symmetric and transitive but fails reflexivity (e.g.,$(1, 1) R (1, 1)$ holds,but for general $(a, b)$,it depends on the domain),the correct classification is symmetric and transitive.

Let $R$ be a relation on $N \times N$ defined by $(a, b) R (c, d)$ if and only if $ad(b-c) = bc(a-d)$. Then $R$ is

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