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(D) $1$. Reflexivity: For any $x \in N$,the divisors of $x$ include at least $1$ and $x$ itself (since $x > 100$). Thus,$x$ and $x$ have at least two

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Reflexive,Symmetric and $NOT$ transitive relation

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(D) $1$. Reflexivity: For any $x \in N$,the divisors of $x$ include at least $1$ and $x$ itself (since $x > 100$). Thus,$x$ and $x$ have at least two common divisors ($1$ and $x$). So,$(x, x) \in R$. $R$ is reflexive.$2$. Symmetry: If $(x, y) \in R$,then $x$ and $y$ have at least two common divisors. This implies $y$ and $x$ also have at least two common divisors. So,$(y, x) \in R$. $R$ is symmetric.$3$. Transitivity: Let $(x, y) \in R$ and $(y, z) \in R$. This means $x, y$ share at least two divisors and $y, z$ share at least two divisors. Consider $x = 102, y = 105, z = 110$. Divisors of $102$ are ${1, 2, 3, 6, 17, 34, 51, 102}$. Divisors of $105$ are ${1, 3, 5, 7, 15, 21, 35, 105}$. Common divisors of $102$ and $105$ are ${1, 3}$ (at least two). Divisors of $110$ are ${1, 2, 5, 10, 11, 22, 55, 110}$. Common divisors of $105$ and $110$ are ${1, 5}$ (at least two). However,common divisors of $102$ and $110$ are ${1, 2}$ (at least two). Let's check another case: $x = 105 (3 	imes 5 	imes 7), y = 110 (2 	imes 5 	imes 11), z = 143 (11 	imes 13)$. $(105, 110)$ share ${1, 5}$. $(110, 143)$ share ${1, 11}$. But $(105, 143)$ only share ${1}$. Thus,$(105, 143) 
otin R$. $R$ is not transitive.

Let $N$ be the set of natural numbers greater than $100$. Define the relation $R$ on $N$ by: $R = \{(x, y) \in N \times N : \text{the numbers } x \text{ and } y \text{ have at least two common divisors}\}.$ Then $R$ is-

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