GATE CS 2026 Set 2 Q16 — Sets, Relations and Functions

Let $S = \{1, 2, \ldots, 10\}$. The relation $R$ is defined on $S$ as $(x, y) \in R \iff xy = k^2$ for some integer $k$. Let us analyze the properties of $R$:

1. Reflexive: For any $x \in S$, $x \cdot x = x^2$, which is always a perfect square. Thus, $(x, x) \in R$ for all $x \in S$. Hence, $R$ is reflexive.
2. Symmetric: If $(x, y) \in R$, then $xy = k^2$ for some integer $k$. Since multiplication is commutative, $yx = xy = k^2$, which means $(y, x) \in R$. Hence, $R$ is symmetric.
3. Transitive: Any positive integer $n$ can be uniquely written in its square-free form as $n = p \cdot u^2$, where $p$ is a square-free integer. The product of two numbers $x = p_x u^2$ and $y = p_y v^2$ is a perfect square if and only if their square-free parts are equal ($p_x = p_y$).
   If $(x, y) \in R$ and $(y, z) \in R$, then $p_x = p_y$ and $p_y = p_z$, which implies $p_x = p_z$. Thus, $xz$ is also a perfect square, meaning $(x, z) \in R$. Hence, $R$ is transitive.
4. Antisymmetric: For $x = 1$ and $y = 4$, we have $1 \cdot 4 = 4 = 2^2$, so $(1, 4) \in R$ and $(4, 1) \in R$, but $1 \neq 4$. Thus, $R$ is not antisymmetric.

Since $R$ is reflexive, symmetric, and transitive, options A, B, and C are correct.