Let ℕ be the set of all natural numbers and R be a relation on ℕ × ℕ…

The correct answer is Option (1) → $\{ (x, y) \in \mathbb{N} \times \mathbb{N} : y = 3x \}$ ##

Let $(a, b)$ be any arbitrary element of $\mathbb{N} \times \mathbb{N}$.

Then $(a, b) \in \mathbb{N} \times \mathbb{N}$ and $a, b \in \mathbb{N}$

We have, $ab = ba$; (As $a, b \in \mathbb{N}$ and multiplication is commutative on $\mathbb{N}$)

$\Rightarrow (a, b)R(a, b)$, according to the definition of the relation $R$ on $\mathbb{N} \times \mathbb{N}$

So, $R$ is reflexive relation on $\mathbb{N} \times \mathbb{N}$

Let $(a, b), (c, d)$ be any arbitrary elements of $\mathbb{N} \times \mathbb{N}$ such that $(a, b)R(c, d)$.

Then, $(a, b)R(c, d) \Rightarrow ad = bc \Rightarrow bc = ad$;

$\Rightarrow cb = da$; (As $a, b, c, d \in \mathbb{N}$ and multiplication is commutative on $\mathbb{N}$)

$\Rightarrow (c, d)R(a, b)$; according to the definition of the relation $R$ on $\mathbb{N} \times \mathbb{N}$

So, $R$ is symmetric relation on $\mathbb{N} \times \mathbb{N}$

Let $(a, b), (c, d), (e, f)$ be any arbitrary elements of $\mathbb{N} \times \mathbb{N}$ such that $(a, b)R(c, d)$ and $(c, d)R(e, f)$.

Then $\left. \begin{array}{l} (a, b)R(c, d) \Rightarrow ad = bc \\ (c, d)R(e, f) \Rightarrow cf = de \end{array} \right\} \Rightarrow (ad)(cf) = (bc)(de) \Rightarrow af = be$

$\Rightarrow (a, b)R(e, f)$; according to the definition of the relation $R$ on $\mathbb{N} \times \mathbb{N}$

So, $R$ is transitive relation on $\mathbb{N} \times \mathbb{N}$.

As the relation $R$ is reflexive, symmetric and transitive, so, it is an equivalence relation on $\mathbb{N} \times \mathbb{N}$.

Equivalence class of $(2, 6)$ i.e.,

$[(2, 6)] = \{(x, y) \in \mathbb{N} \times \mathbb{N} : (x, y)R(2, 6)\}$

$= \{(x, y) \in \mathbb{N} \times \mathbb{N} : 6x = 2y\} = \{(x, y) \in \mathbb{N} \times \mathbb{N} : 3x = y\}$

$= \{(x, 3x) : x \in \mathbb{N}\} = \{(1, 3), (2, 6), (3, 9), \dots \}$