R={(x,y):3x+αy is multiple of 7}, now R to be an equivalence relation (1) R should be reflexive : (a,a)∈R∀a∈N ∴3a+aα=7k ∴(3+α)a=7k ∴3+α=7k1⇒α=7k1−3 =7k1+4 (2) R should be symmetric : aRb⇔bRa aRb:3a+(7k−3)b=7m ⇒3(a−b)+7kb=7m ⇒3(b−a)+7ka=7m So, aRb⇒bRa ∴ R will be symmetric for a=7k1−3 (3) Transitive : Let (a,b)∈R,(b,c)∈R ⇒3a+(7k−3)b=7k1 and 3b+(7k2−3)c=7k3 Adding 3a+7kb+(7k2−3)c=7(k1+k3)
3a+(7k2−3)c=7m ∴(a,c)∈R ∴R is transitive ∴α=7k−3=7k+4