Question

Let N be the set of all natural numbers and R be a relation on N × N defined by (a, b) R (c, d) `⇔` ad = bc for all (a, b), (c, d) ∈ N × N. Show that R is an equivalence relation on N × N. Also, find the equivalence class of (2, 6), i.e., [(2, 6)].

Answer 1

Let (a, b) be an arbitrary element of N × N.

Then, (a, b) ∈ N × N and a, b ∈ N

We have, ab = ba; (As a, b ∈ N and multiplication is commutative on N)

`\implies` (a, b) R (a, b), according to the definition of the relation R on N × N

Thus (a, b) R (a, b), ∀ (a, b) ∈ N × N.

So, R is reflexive relation on N × N.

Let (a, b), (c, d) be arbitrary elements of N × N such that (a, b) R (c, d).

Then, (a, b) R (c, d) `\implies` ad = bc `\implies` bc = ad; (changing LHS and RHS)

`\implies` cb = da; (As, a, b, c, d ∈ N and multiplication is commutative on N)

`\implies` (c, d) R (a, b); according to the definition of the relation R on N × N

Thus (a, b) R (c, d) `\implies` (c, d) R (a, b)

So, R is symmetric relation on N × N.

Let (a, b), (c, d), (e, f) be arbitrary elements of N × N such that (a, b) R (c, d) and (c, d) R (e, f).

Then `{:((a, b) R (c, d) \implies ad = bc),((c, d) R (e, f) \implies cf = de):}} \implies` (ad) (cf) = (bc) (de) `\implies` af = be

`\implies` (a, b) R (e, f); (according to the definition of the relation R on N × N)

Thus (a, b) R (c, d) and (c, d) R (e, f) `\implies` (a, b) R (e, f)

So, R is transitive relation on N × N.

As the relation R is reflexive, symmetric and transitive so, it is equivalence relation on N × N.

[(2, 6)] = {(x, y) ∈ N × N : (x, y) R (2, 6)}

= {(x, y) ∈ N × N : 3x = y}

= {(x, 3x) : x ∈ N}

= {(1, 3), (2, 6), (3, 9),.........}