GATE 2026 CS2 Q.26: solution and answer

Official answer: A, B, CSource: Official answer key, GATE 2026 CS2 (IIT Guwahati)

1. $(x,y)\in R$ exactly when $xy$ is the square of an integer.

2. Reflexive: $x\cdot x=x^2$ is a perfect square for every $x$, so $(x,x)\in R$ for all $x\in \{1,\dots ,10\}$. True.

3. Symmetric: $xy=yx$, so if $xy$ is a perfect square then so is $yx$. True.

4. Transitive: let $xy=a^2$ and $yz=b^2$ with $a,b$ integers. Then $(xy)(yz)=y^2\cdot xz=(ab{)}^2$, so $xz={\left(\frac{ab}{y}\right)}^2$. Thus $xz$ is an integer and the square of a rational number. A rational whose square is an integer must itself be an integer, so $xz$ is the square of an integer and $(x,z)\in R$. True.

5. Equivalently, $xy$ is a square exactly when $x$ and $y$ have the same square-free part, so $R$ is an equivalence relation. For example, on $\{1,\dots ,10\}$ the classes are $\{1,4,9\}$, $\{2,8\}$, $\{3\}$, $\{5\}$, $\{6\}$, $\{7\}$, $\{10\}$.

6. Antisymmetric: $(1,4)\in R$ since $1\cdot 4=2^2$, and $(4,1)\in R$, but $1\ne 4$. So $R$ is not antisymmetric. False.

7. Therefore $R$ is reflexive, symmetric and transitive: options A, B and C.