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 :
Solutions · 1
Official solution
\[R = \{ (x,y):3x + \alpha y\] is multiple of 7\[\} \], now R to be an equivalence relation
(1) R should be reflexive : \[(a,a) \in R\,\forall \,a \in N\]
\[\therefore\] \[3a + a\alpha = 7k\]
\[\therefore\] \[(3 + \alpha )a = 7k\]
\[\therefore\] \[3 + \alpha = 7{k_1} \Rightarrow \alpha = 7{k_1} - 3\]
=7k1+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=7k1−3
(3) Transitive : Let \[(a,b) \in R,\,(b,c) \in R\]
\[ \Rightarrow 3a + (7k - 3)b = 7{k_1}\] and
3b+(7k2−3)c=7k3
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
\[\therefore\] \[\alpha = 7k - 3 = 7k + 4\]