Given:
R=(a,b):2a+3bisdivisibleby5fora,b∈N
Reflexive:
2a+3a=5a is always divisible by 5.
So, (a,a)∈R
Symmetric:
Let (a,b)∈R i.e., 2a+3b=5λ, where λ∈Z+.
Now,
5a+5b=multipleof5
⇒(2a+3b)+(2b+3a)=multipleof5
⇒5λ+(2b+3a)=multipleof5
⇒2b+3a=multipleof5−5λ
So, 2b+3a is divisible by 5
Transitive:
Let (a,b),(b,c)∈R, then
2a+3b=5λ1;λ1∈Z+
2b+3c=5λ2;λ2∈Z+
So,
2a+5b+3c=5(λ1+λ2)
⇒2a+3c=5(λ1+λ2−b)
Hence, 2a+3c is divisible by 5, so (a,c)∈R
Hence, R is an equivalence relation.