Let R be a relation defined on ℕ as a R b is 2a + 3b is a multiple...

Given:

$R = \left\{\left(a, b\right): 2a + 3bisdivisibleby5fora, b \in N\right\}$

Reflexive:

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

So, \left(a, a\right) \in R$</p><p><strong>Symmetric:</strong></p><p>Let \left(a, b\right) \in R$ i.e., $2a + 3b = 5\lambda$, where $\lambda \in {Z}^{+}$.

Now,

$5a + 5b = multipleof5$

$⇒ \left(2a + 3b\right) + \left(2b + 3a\right) = multipleof5$

$⇒ 5\lambda + \left(2b + 3a\right) = multipleof5$

$⇒ 2b + 3a = multipleof5 - 5\lambda$

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

Transitive:

Let \left(a, b\right), \left(b, c\right) \in R$, then</p><p>$2a + 3b = 5{\lambda}_{1}; {\lambda}_{1} \in {Z}^{+}$</p><p>$2b + 3c = 5{\lambda}_{2}; {\lambda}_{2} \in {Z}^{+}$</p><p>So,</p><p>$2a + 5b + 3c = 5\left({\lambda}_{1} + {\lambda}_{2}\right)$</p><p>$⇒ 2a + 3c = 5\left({\lambda}_{1} + {\lambda}_{2} - b\right)$</p><p>Hence,$2a + 3c$is divisible by$5$, so \left(a, c\right) \in R$

Hence, $R$ is an equivalence relation.