Question

Show that the relation R in the set N x N defined by (a, b}R(c, d) if a +d = b+Va, b, c, deN, is an equivalence relation.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

➢ R is a relation defined on the set N × N given by

(a, b) R (c, d) such that a + d = b + c, for every a, b, c, d ∈ N

Now,

➢ In order to prove that given relation is equivalence, we have to show that given relation R is reflexive, symmetric and transitive.

So

➢ Let we first prove R is Reflexive.

Let a, b ∈ N.

We know,

a + b = b + a

⇛ (a, b) R (a, b)

⇛ R is reflexive.

Now,

➢ We prove that R is symmetric.

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

⇛a + d = b + c

⇛b + c = a + d

⇛c + b = d + a

⇛(c, d) R (a, b).

⇛R is symmetric.

Now,

We prove that R is transitive.

Let a, b, c ∈ N such that (a, b) R (c, d) and (c, d) R (e, f).

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

⇛a + d = b + c ----(1)

Now

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

⇛c + f = d + e-----(2)

On adding equation (1) and (2), we get

⇛a + d + c + f = b + c + d + e

⇛a + f = b + e

⇛(a, b) R (e, f)

⇛R is transitive.

⇛ Thus, we prove that R is Reflexive, Symmetric and Transitive.

Hence,

R is an equivalence relation.

Result used :-

Relation R is reflexive if (a, b) R (a, b).

Relation R is symmetric if (a, b) R (c, d) then (c, d) R (a, b)

Relation R is transitive if (a, b) R (c, d) an d (c, d) R (e, f) then (a, b) R (e, f).