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

Solution

Maths sir · 4 steps

Reflexivity alone fixes α: put y = x and see what α must be modulo 7.

1. Reflexive first. For (x, x) ∈ R we need 3x + α x = (3 + α)x to be a multiple of 7 for every natural x. Taking x = 1 forces 7 | (3 + α), so α ≡ -3 ≡ 4 (mod 7).
2. Rewrite the relation. With α ≡ -3 (mod 7), the condition becomes 3x + α y ≡ 3x - 3y = 3(x - y) (mod 7). Since 7 does not divide 3, this is a multiple of 7 exactly when 7 | (x - y).

   (x, y) ∈ R ⇔ x ≡ y (mod 7)

3. Check the rest. Congruence modulo 7 is reflexive, symmetric and transitive, so R is an equivalence relation. If α ≡ 4 (mod 7), then (1, 1) ∉ R and reflexivity already fails.
4. So. R is an equivalence relation if and only if α leaves remainder 4 when divided by 7, which is option (d).

The idea to keep. For a relation defined by a divisibility condition, test reflexivity with y = x first. It usually pins down the parameter, and then the relation often turns into a familiar congruence.

Answer from the NTA final answer key. Solution by PrepWiser.