Question

anyone to solve this only for brilliant people Q 29​

Answer 1

★Question:-

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

★Proof:-

We need to know:

✦If the relation R is reflexive,symmetric&transitive, is it an equivalence relation.

=>If R is a relation on a set A,R is:

Reflexive

• For all x∈A,xRx

Symmetric

• For all x,y∈A, xRy & yRx

Transitive

• For all x,y,z∈A,xRy & yRz & xRz

Case - 1:-

Let (a,b)∈N×N

Then,

⇒ a²+b²=a²+b²

→ (a,b)R(a,b)

Hence, R is reflexive

Case-2:-

Let (a,b),(c,d)∈N×N

Then,

→ (a,b)R(c,d)

⇒ a²+d²=b²+c²

⇒ c²+b²=d²+a²

→ (c,d)R(a,b)

Hence, R is symmetric

Case - 3:-

Let (a,b),(c,d),(e,f)∈N×N

Then,

→ (a,b)R(c,d) , (c,d)R(e,f)

⇒ a²+d²=b²+c²........(1)

⇒ c²+f²=d²+e².........(2)

Adding eqn (1)&(2):

⇒ a²+d²+c²+f²=b²+c²+d²+e²

[c² and d² gets cancelled]

⇒ a²+f²=b²+e²

→ (a,b)R(e,f)

Hence,R is transitive

We can see , R is reflexive,symmetric & transitive.

Hence,

R is an equivalence relation.

_______________

Answer 2

Answer:

Required Answer -

Let (a, b) € N×N

then,

${a}^{2} + {b}^{2} = {a}^{2} + {b}^{2}$

(a, b) R (a, b)

Hence, R is reflexive.

Let (a, b), (c, d) € N×N be such that

(a, b) R (c, d)

${a}^{2} + {d}^{2} = {b}^{2} + {c}^{2}$

${c}^{2} + {b}^{2} = {d}^{2} + {a}^{2}$

(c, d) R (a, b)

Hence, R is symmetric

Let (a, b), (c, d), (e, f) € N×N be such that

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

${a}^{2} + {d}^{2} = {b}^{2} + {c}^{2} .............1)$

${c}^{2} + {f}^{2} = {d}^{2} + {e}^{2} ...................2)$

Adding eq. 1) and 2)

${a}^{2} + {d}^{2} + {c}^{2} + {f}^{2} = {b}^{2} + {c}^{2} + {d}^{2} + {e}^{2}$

${a}^{2} + {f}^{2} = {b}^{2} + {e}^{2}$

(a, b) R (e, f)

Hence, R is transitive

Since, R is reflexive, symmetric and transitive

Therefore, R is an equivalence relation.

Thank You