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

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$R = \{ (x,y):3x + \alpha y$ is multiple of 7$\}$, now R to be an equivalence relation

(1) R should be reflexive : $(a,a) \in R\,\forall \,a \in N$

$\therefore$ $3a + a\alpha = 7k$

$\therefore$ $(3 + \alpha )a = 7k$

$\therefore$ $3 + \alpha = 7{k_1} \Rightarrow \alpha = 7{k_1} - 3$

$= 7{k_1} + 4$

(2) R should be symmetric : $aRb \Leftrightarrow bRa$

$aRb:3a + (7k - 3)b = 7\,m$

$\Rightarrow 3(a - b) + 7kb = 7\,m$

$\Rightarrow 3(b - a) + 7ka = 7\,m$

So, $aRb \Rightarrow bRa$

$\therefore$ R will be symmetric for $a = 7{k_1} - 3$

(3) Transitive : Let $(a,b) \in R,\,(b,c) \in R$

$\Rightarrow 3a + (7k - 3)b = 7{k_1}$ and

$3b + (7{k_2} - 3)c = 7{k_3}$

Adding $3a + 7kb + (7{k_2} - 3)\,c = 7({k_1} + {k_3})$

$3a + (7{k_2} - 3)\,c = 7\,m$

$\therefore$ $(a,c) \in R$

$\therefore$ R is transitive

$\therefore$ $\alpha = 7k - 3 = 7k + 4$

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

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

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