Let R be a relation defined on mathbbN as a R b is 2a+3b is a multiple of 5,a,b∈ mathbbN. Then R is

Given:

$$R={(a,b):2a+3b\mathrm{is}\mathrm{divisible}\mathrm{by}5\mathrm{for}a,b\in N}$$R=(a,b):2a+3bisdivisibleby5fora,b∈N

Reflexive:

$2a+3a=5a$2a+3a=5a is always divisible by $5$5.

So, $(a,a)\in R$(a,a)∈R

Symmetric:

Let $(a,b)\in R$(a,b)∈R i.e., $2a+3b=5\lambda$2a+3b=5λ, where $\lambda \in {Z}^{+}$λ∈Z+.

Now,

$$5a+5b=\mathrm{multiple}\mathrm{of}5$$5a+5b=multipleof5

$$\Rightarrow (2a+3b)+(2b+3a)=\mathrm{multiple}\mathrm{of}5$$⇒(2a+3b)+(2b+3a)=multipleof5

$$\Rightarrow 5\lambda +(2b+3a)=\mathrm{multiple}\mathrm{of}5$$⇒5λ+(2b+3a)=multipleof5

$$\Rightarrow 2b+3a=\mathrm{multiple}\mathrm{of}5-5\lambda$$⇒2b+3a=multipleof5−5λ

So, $2b+3a$2b+3a is divisible by $5$5

Transitive:

Let $(a,b),(b,c)\in R$(a,b),(b,c)∈R, then

$$2a+3b=5{\lambda }_{1};{\lambda }_{1}\in {Z}^{+}$$2a+3b=5λ1​;λ1​∈Z+

$$2b+3c=5{\lambda }_{2};{\lambda }_{2}\in {Z}^{+}$$2b+3c=5λ2​;λ2​∈Z+

So,

$$2a+5b+3c=5({\lambda }_{1}+{\lambda }_{2})$$2a+5b+3c=5(λ1​+λ2​)

$$\Rightarrow 2a+3c=5({\lambda }_{1}+{\lambda }_{2}-b)$$⇒2a+3c=5(λ1​+λ2​−b)

Hence, $2a+3c$2a+3c is divisible by $5$5, so $(a,c)\in R$(a,c)∈R

Hence, $R$R is an equivalence relation.