Question

Show that the relation R on N × N, defined by (a, b) R (c, d) a + d = b + c.

Is an equivalent relation.

Answer 1

In order to show R is an equivalence relation we need to show R is Reflexive, Symmetric and Transitive.

Given that, R be the relation in N × N defined by (a, b) R (c, d) if a + d = b + c for (a, b), (c, d) in N × N.

R is Reflexive if (a, b) R (a, b) for (a, b) in N × N

Let (a, b) R (a, b)

⇒ a + b = b + a

which is true since addition is commutative on N.

⇒ R is reflexive.

R is Symmetric if (a, b) R (c, d) ⇒ (c, d) R (a, b) for (a, b), (c, d) in N × N

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

⇒ a + d = b + c

⇒ b + c = a + d

⇒ c + b = d + a [since addition is commutative on N]

⇒ (c, d) R (a, b)

⇒ R is symmetric.

R is Transitive if (a, b) R (c, d) and (c, d) R (e, f) ⇒ (a, b) R (e, f) for (a, b), (c, d),(e, f) in N × N

Let (a, b) R (c, d) and (c, d) R (e, f)

⇒ a + d = b + c and c + f = d + e

⇒ (a + d) – (d + e) = (b + c) – (c + f)

⇒ a - e = b - f

⇒ a + f = b + e

⇒ (a, b) R (e, f)

⇒ R is transitive.

Hence, R is an equivalence relation.

Answer 2

Reflexivity-

Consider (a, b) as an arbitrary element of N × N

Here, (a, b) ∈ N × N where (a, b) ∈ N

It can be written as

a + b = b + a

We get (a, b) R (a, b) for all (a, b) ∈ N × N

Hence, R is reflexive on N × N

Symmetry-

Consider (a, b), (c, d) ∈ N × N such that (a, b) R (c, d)

It can be written as

a + d = b + c and c + b = d + a

We get

(c, d) R (a, b)

(a, b) R (c, d) => (c, d) R (a, b) for all (a, b), (c, d) ∈ N × N

Hence, R is symmetric on N × N.

Transitive-

Consider (a, b), (c, d), (r, f) ∈ N × N such that (a, b) R (c, d) and (c, d) R (e, f)

It can be written as

a + d = b + c and c + f = d + e

By adding both

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

On further calculation

a + f = b + e where (a, b) R (e, f)

So (a, b) R (c, d) and (c, d) R (e, f) we get (a, b) R (e, f) for all (a, b), (c, d), (e, f) ∈ N × N

Hence, R is transitive on N × N.

Therefore, R is reflexive, symmetric and transitive is an equivalence relation on N × N.