For α ∈ N , consider a relation R on N given by R = { x , y : 3 x + α y is a multiple of 7 }…
For , consider a relation on given by { is a multiple of }. The relation is an equivalence relation if and only if
is a multiple of
is the remainder when is divided by
is the remainder when is divided by
Solution
Given, a relation $R$ on $N$ given by $R=\{(x,y):3x+\alpha y \text{ is a multiple of } 7\}$,
Now for $R$ to be reflexive $\Rightarrow xRx$
$\Rightarrow 3x+\alpha x=7x$
$\Rightarrow (3+\alpha)x=7k$
$\Rightarrow 3+\alpha=7\lambda$
$\Rightarrow \alpha=7\lambda-3=7N+4$, where $K, \lambda, N \in I$
So, when $\alpha$ divided by $7$, remainder is $4$.
Now $R$ to be symmetric $xRy \Rightarrow yRx$
$3x+\alpha y=7N_1, 3y+\alpha x=7N_2$
$\Rightarrow (3+\alpha)(x+y)=7(N_1+N_2)=7N_3$
Which holds when $3+\alpha$ is multiple of $7$
So, $\alpha=7N+4$ (as did earlier)
Now, for $R$ to be transitive
$xRy \& yRz \Rightarrow xRz$.
$\Rightarrow 3x+\alpha y=7N_1 ...1$
$\Rightarrow 3y+\alpha z=7N_2 ...2$
And $3x+\alpha z=7N_3 ...3$
Now subtracting equation $3-2$ we get,
$3x+7N_2-3y=7N_3$
Now putting the value of $3x$ from equation $1$ we get, $7N_1-\alpha y+7N_2-3y=7N_3$
$\Rightarrow 7(N_1+N_2)-(3+\alpha)y=7N_3$
$\Rightarrow (3+\alpha)y=7N$
Which is true again when $3+\alpha$ divisible by $7$, i.e. when $\alpha$ divided by $7$, remainder is $4$.