Question

For $\alpha \in N$, consider a relation $R$ on $N$ given by $R=\{(x, y): 3 x+\alpha y$ is a multiple of 7$\}$. The relation $R$ is an equivalence relation if and only if :

• A. $\alpha=14$
• B. $\alpha$ is a multiple of 4
• C. 4 is the remainder when $\alpha$ is divided by 10
• D. 4 is the remainder when $\alpha$ is divided by 7

Answer 1

Answer: (D)

For $R$ to be reflexive $\Rightarrow x R x$
$\Rightarrow 3 x +\alpha x =7 x \Rightarrow(3+\alpha) x =7 K$
$\Rightarrow 3+\alpha=7 \lambda \Rightarrow \alpha=7 \lambda-3=7 N +4, K , \lambda, N \in I$
$\therefore$ when $\alpha$ divided by 7 , remainder is 4 .
$R$ to be symmetric $x R y \Rightarrow y R x$
$3 x +\alpha y =7 N _1, 3 y +\alpha x =7 N _2$
$\Rightarrow(3+\alpha)( x + y )=7\left( N _1+ N _2\right)=7 N _3$
Which holds when $3+\alpha$ is multiple of 7
$\therefore \alpha=7 N +4 \text { (as did earlier) }$
$R$ to be transitive
$x R y \& y R z \Rightarrow x R z$.
$3 x +\alpha y =7 N _1 \& 3 y +\alpha z =7 N _2$ and $3 x +\alpha z =7 N _3$
$\therefore 3 x +7 N _2-3 y =7 N _3$
$\therefore 7 N _1-\alpha y +7 N _2-3 y =7 N _3$
$\therefore 7\left( N _1+ N _2\right)-(3+\alpha) y =7 N _3$
$\therefore(3+\alpha) y =7 N$
Which is true again when $3+\alpha$ divisible by
7, i.e. when $\alpha$ divided by 7 , remainder is 4 .