Correct : a,b,c

The correct answer is Options A, B, and C — Reflexive, Symmetric, and Transitive. R is an equivalence relation.
Key insight: (x, y) ∈ R iff x·y is a perfect square. This happens exactly when x and y have the same square-free part — the product of prime factors appearing an odd number of times. Numbers 1–10 grouped by square-free part: {1,4,9}, {2,8}, {3}, {5}, {6}, {7}, {10}.
Reflexive: For any x, x·x = x² is always a perfect square. So (x,x) ∈ R for all x ∈ {1,...,10}. TRUE.
Symmetric: If x·y is a perfect square, then y·x = x·y is also a perfect square. So (x,y) ∈ R implies (y,x) ∈ R. TRUE.
Transitive: If (x,y) ∈ R and (y,z) ∈ R, then sqf(x) = sqf(y) and sqf(y) = sqf(z), so sqf(x) = sqf(z), meaning x·z is a perfect square → (x,z) ∈ R. TRUE.
Antisymmetric: Requires (x,y) ∈ R and (y,x) ∈ R to imply x = y. But (1,4) ∈ R since 1×4 = 4 = 2², and (4,1) ∈ R, yet 1 ≠ 4. FALSE.
Since R is reflexive, symmetric, and transitive, it is an equivalence relation — it partitions {1,...,10} into classes of numbers sharing the same square-free part.