Question

Q39. Show that the relation R in the set N × N defined by (a, b) R (c, d) if a² + d² = b² + c² ∀ a, b, c, d ∈ N, is an equivalence relation.​

Answer 1

Basic Concept :-

Let R be a relation defined on sample space S, then

• 1. R is Reflexive if (a, b) R (a, b) where (a, b) ∈ S

• 2. R is Symmetric if (a, b) R (c, d) then (c, d) R (a, b) where (a, b), (c, d) ∈ S.

• 3. R is Transitive if (a, b) R (c, d), (c, d) R (e, f) then (a, b) R (e, f) where (a, b), (c, d), (e, f) ∈ S.

If R is Reflexive, Symmetric and Transitive, then R is an equivalence relation.

Let's solve the problem now!!

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

Reflexive :-

• Let (a, b) ∈ N × N

Then,

We know,

• a² + b² = b² + a²

• ⇛ (a, b) R (a, b)

⇛ R is Reflexive.

Symmetric :-

• Let (a, b), (c, d) ∈ N × N

such that

• (a, b) R (c, d)

• ⇛ a² + d² = b² + c²

• ⇛ c² + b² = d² + a²

• ⇛ (c, d) R (a, b)

Hence, R is symmetric.

Transitive :-

• Let (a, b), (c, d) and (e, f) ∈ N × N

such that

• (a, b) R (c, d)

• ⇛ a² + d² = b² + c² -------(1)

Also,

• (c, d) R (e, f)

• ⇛ c² + f² = d² + e²------(2)

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

• a² + f² = b² + e²

• ⇛ (a, b) R (e, f)

Hence, R is Transitive.

Since,

• R is Reflexive, Symmetric and Transitive,

Hence,

• R is an equivalence relation.

Answer 2

Answer:

hope its help full!

Step-by-step explanation:

Basic Concept :-

Let R be a relation defined on sample space S, then

1. R is Reflexive if (a, b) R (a, b) where (a, b) ∈ S

2. R is Symmetric if (a, b) R (c, d) then (c, d) R (a, b) where (a, b), (c, d) ∈ S.

3. R is Transitive if (a, b) R (c, d), (c, d) R (e, f) then (a, b) R (e, f) where (a, b), (c, d), (e, f) ∈ S.

If R is Reflexive, Symmetric and Transitive, then R is an equivalence relation.

Let's solve the problem now!!

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

Solution−

Reflexive :-

Let (a, b) ∈ N × N

Then,

We know,

a² + b² = b² + a²

⇛ (a, b) R (a, b)

⇛ R is Reflexive.

Symmetric :-

Let (a, b), (c, d) ∈ N × N

such that

(a, b) R (c, d)

⇛ a² + d² = b² + c²

⇛ c² + b² = d² + a²

⇛ (c, d) R (a, b)

Hence, R is symmetric.

Transitive :-

Let (a, b), (c, d) and (e, f) ∈ N × N

such that

(a, b) R (c, d)

⇛ a² + d² = b² + c² -------(1)

Also,

(c, d) R (e, f)

⇛ c² + f² = d² + e²------(2)

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

a² + f² = b² + e²

⇛ (a, b) R (e, f)

Hence, R is Transitive.

Since,

R is Reflexive, Symmetric and Transitive,

Hence,

R is an equivalence relation.