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

1.	Show that the relation R on N × N, defined by (a, b) R (c, d) a + d = b + c.Is an equivalent relation.
Answer» 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.

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

Answer»

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.