Let A = {1, 2, 3,......, 9} and R be the relation in A × A defined by (*a*, *b*) R (*c*, *d*) if *a* + *d* = *b* + *c* for (*a*, *b*), (*c*, *d*) in A × A. Prove that R is an equivalence relation. Also, obtain the equivalence class [(2, 5)].

#### Solution

A = {1, 2, 3, ..., 9} ⊂ ℕ, the set of natural numbers

Let R be the relation in A × A defined by (*a*, *b*) R (*c*, *d*) if *a* + *d* = *b* + *c* for (*a*, *b*), (*c*, *d*) in A × A.

We have to show that R is an equivalence relation.

**Reflexivity**:

Let (*a*, *b*) be an arbitrary element of A × A. Then, we have:

(*a*, *b*) ∈ A × A

⇒ *a*, *b* ∈ A

⇒ *a* + *b* = *b* + *a* (by commutativity of addition on A ⊂ ℕ)

⇒ (*a*, *b*) R (*a*, *b*)

Thus, (*a*, *b*) R (*a*, *b*) for all (*a*, *b*) ∈ A × A.

So, R is reflexive.

**Symmetry**:

Let (*a*, *b*), (*c*, *d*) ∈ A × A such that (*a*, *b*) R (*c*, *d*).*a* + *d* = *b* + *c*

⇒ *b* +* c* = *a* + *d*

⇒ *c* + *b* = *d* + *a* (by commutativity of addition on A ⊂ ℕ)

⇒ (*c*, *d*) R (*a*, *b*)

Thus, (*a*, *b*) R (*c*, *d*) ⇒ (*c*, *d*) R (*a*, *b*) for all (*a*, *b*), (*c*, *d*) ∈ A × A.

So, R is symmetric

**Transitivity**:

Let (*a*, *b*), (*c*, *d*), (*e*, *f*) ∈ A × A such that (*a*, *b*) R (*c*, *d*) and (*c*, *d*) R (*e*, *f*). Then, we have:

(*a*, *b*) R (*c*, *d*)

⇒ *a* + *d* = *b* + *c* ... (1)

(*c*, *d*) R (*e*, *f*)

⇒ *c* + *f* = *d* + *e* ... (2)

Adding equations (1) and (2), we get:

(*a* + *d*) + (*c* + *f*) = (*b* + *c*) + (*d* + *e*)

⇒ *a* + *f* = *b* + *e*

⇒ (*a*, *b*) R (*e*, *f*)

Thus, (*a*, *b*) R (*c*, *d*) and (*c*, *d*) R (*e*, *f*) ⇒ (*a*, *b*) R (*e*, *f*) for all (*a*, *b*), (*c*, *d*), (*e*, *f*) ∈ A × A.

So, R is transitive on A × A.

Thus, R is reflexive, symmetric and transitive.

∴ R is an equivalence relation.

To write the equivalence class of [(2, 5)], we need to search all the elements of the type (*a*, *b*) such that 2 + *b* = 5 + *a.*

∴ Equivalence class of [(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}