Question

For $\alpha \in N$, consider a relation $R$ on $N$ given by $R=\{(x,y):3x+\alpha y$ 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 $\Rightarrow xRx$
$\Rightarrow 3x+\alpha x=7x\Rightarrow (3+\alpha )x=7K$
$\Rightarrow 3+\alpha =7\lambda \Rightarrow \alpha =7\lambda -3=7N+4,K,\lambda ,N\in I$
$\therefore$ when $\alpha$ divided by 7 , remainder is 4 .
$R$ to be symmetric $xRy\Rightarrow yRx$
$3x+\alpha y=7N_1,3y+\alpha x=7N_2$
$\Rightarrow (3+\alpha )(x+y)=7\left(N_1+N_2\right)=7N_3$
Which holds when $3+\alpha$ is multiple of 7
$\therefore \alpha =7N+4\text{ (as did earlier) }$
$R$ to be transitive
$xRy\&yRz\Rightarrow xRz$.
$3x+\alpha y=7N_1\&3y+\alpha z=7N_2$ and $3x+\alpha z=7N_3$
$\therefore 3x+7N_2-3y=7N_3$
$\therefore 7N_1-\alpha y+7N_2-3y=7N_3$
$\therefore 7\left(N_1+N_2\right)-(3+\alpha )y=7N_3$
$\therefore (3+\alpha )y=7N$
Which is true again when $3+\alpha$ divisible by
7, i.e. when $\alpha$ divided by 7 , remainder is 4 .