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

Given:

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

Reflexive:

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

So, $\left(a,a\right)\in R$

Symmetric:

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

Now,

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

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

$\Rightarrow 5\lambda +\left(2b+3a\right)=\mathrm{multiple} \mathrm{of} 5$

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

So, $2b+3a$ is divisible by $5$

Transitive:

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

$2a+3b=5{\lambda }_1; {\lambda }_1\in Z^{+}$

$2b+3c=5{\lambda }_2; {\lambda }_2\in Z^{+}$

So,

$2a+5b+3c=5\left({\lambda }_1+{\lambda }_2\right)$

$\Rightarrow 2a+3c=5\left({\lambda }_1+{\lambda }_2-b\right)$

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

Hence, $R$ is an equivalence relation.