Question

please answer it fast​

Answer 1

We know that a relation R is an equivalence relation, if it is reflexive, symmetric and transitive.

For reflexive:

∣a−a∣=0⇔(a,a)∈R, ∀a∈A

For symmetric:

(a,b)∈R⇔∣a−b∣=4k=∣b−a∣, k∈Z⇔(b,a)∈R, ∀(a,b)∈R

For transitive:

Let (a,b)∈R and (b,c)∈R

Thus, we have

(a,b)∈R⇔∣a−b∣=4k1, k1∈Z and

(b,c)∈R⇔∣b−c∣=4k2, k2∈Z

Since a, b and c are integers, we have

∣a−c∣=∣a−b+b−c∣=∣a−b∣±∣b−c∣=4(k1±k2)=4m, m∈Z⇔(a,c)∈R

So, we have shown that the relation R is reflexive, symmetric and transitive. Therefore, the relation is an equivalence relation.

Let x be the element of A such that (x,1)∈R  

∣x−1∣ is a multiple of 4

⟹∣x−1∣=0,4,8,12

x−1=0,4,8,12