For α ∈ N , consider a relation R on N given by R = \ (x, y): 3x +...

Given, a relation $R$ on $N$ given by $R = \{\left(x, y\right): 3x + \alphay \text{ is a multiple of }7\}$,

Now for $R$ to be reflexive $⇒ xRx$

$⇒ 3x + \alphax = 7x$

$⇒ \left(3 + \alpha\right)x = 7K$

$⇒ 3 + \alpha = 7\lambda$

$⇒ \alpha = 7\lambda - 3 = 7N + 4, \left\{\text{where}K, \lambda, N \in I\right\}$

So, when $\alpha$ divided by $7$, remainder is $4$.

Now $R$ to be symmetric $xRy ⇒ yRx$

$3x + \alphay = 7{N}_{1}, 3y + \alphax = 7{N}_{2}$

$⇒ \left(3 + \alpha\right)\left(x + y\right) = 7\left({N}_{1} + {N}_{2}\right) = 7{N}_{3}$

Which holds when $3 + \alpha$ is multiple of $7$

So, $\alpha = 7N + 4$ (as did earlier)

Now, for $R$ to be transitive

$xRy \text{ and } yRz ⇒ xRz .</p><p>$⇒ 3x + \alphay = 7{N}_{1} . . . . \left(1\right)$</p><p>$⇒ 3y + \alphaz = 7{N}_{2} . . . . . . \left(2\right)

And $3x + \alphaz = 7{N}_{3} . . . . . \left(3\right)$

Now subtracting equation \left(3\right)$- \left(2\right)$ we get,

$3x + 7{N}_{2} - 3y = 7{N}_{3}$

Now putting the value of $3x$ from equation \left(1\right)$we get,$7{N}_{1} - \alphay + 7{N}_{2} - 3y = 7{N}_{3}$</p><p>$⇒ 7\left({N}_{1} + {N}_{2}\right) - \left(3 + \alpha\right)y = 7{N}_{3}$</p><p>$⇒ \left(3 + \alpha\right)y = 7N$</p><p>Which is true again when$3 + \alpha$divisible by$7$, i.e. when$\alpha$divided by$7$, remainder is$4 .