Mathematics - Sets and Relations Question with Solution | TestHub

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

Now for $R$ to be reflexive $\Rightarrow xRx$

$\Rightarrow 3x+\alpha x=7x$

$\Rightarrow \left(3+\alpha \right)x=7K$

$\Rightarrow 3+\alpha =7\lambda$

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

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

Now $R$ to be symmetric $xRy\Rightarrow yRx$

$3x+\alpha y=7N_1,3y+\alpha x=7N_2$

$\Rightarrow \left(3+\alpha \right)\left(x+y\right)=7\left(N_1+N_2\right)=7N_3$

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

So, $\alpha =7N+4$ (as did earlier)

Now, for $R$ to be transitive

$xRy \& yRz\Rightarrow xRz.$

$\Rightarrow 3x+\alpha y=7N_1 ....\left(1\right)$

$\Rightarrow 3y+\alpha z=7N_2 ......\left(2\right)$

And  $3x+\alpha z=7N_3 .....\left(3\right)$

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

$3x+7N_2-3y=7N_3$

Now putting the value of $3x$ from equation $\left(1\right)$ we get, $7N_1-\alpha y+7N_2-3y=7N_3$

$\Rightarrow 7\left(N_1+N_2\right)-\left(3+\alpha \right)y=7N_3$

$\Rightarrow \left(3+\alpha \right)y=7N$

Which is true again when $3+\alpha$ divisible by $7$, i.e. when $\alpha$ divided by $7$, remainder is $4$ .