Question

Let \(R\) be a binary relation on the set \(\{1, 2, \ldots, 10\}\), where \((x,y) \in R\) if the product of \(x\) and \(y\) is square of an integer. Which of the following properties is/are satisfied by \(R\)?

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. Antisymmetric

Hint:

Use prime factorisation or the squarefree-part idea: two numbers multiply to a perfect square exactly when they share the same squarefree part.

Answer 1

The Correct Option is A, B, C

Step 1: Restate the rule with a useful trick.
Every positive integer $n$ can be written as $n = s \cdot k^2$, where $s$ is the squarefree part of $n$ (what is left after pulling out every square factor). Two numbers multiply to a perfect square exactly when they share the same squarefree part. We use this trick instead of a full prime by prime proof.

Step 2: Reflexive check.
Take any $x$ in the set. Then $x \cdot x = x^2$, which is obviously a perfect square. So every element is related to itself, and $R$ is reflexive.

Step 3: Symmetric check.
If $xy$ is a perfect square, $x$ and $y$ have the same squarefree part. Since "having the same squarefree part" does not care about order, $y$ and $x$ also share that squarefree part, so $yx$ is a perfect square too. This makes $R$ symmetric.

Step 4: Transitive check.
Say $x$ and $y$ share squarefree part $s_1$, and $y$ and $z$ share squarefree part $s_1$ again, since $y$ only has one squarefree part. Then $x$ and $z$ both share that same squarefree part $s_1$, which means $xz$ is a perfect square. So $R$ passes the transitive test as well.

Step 5: Antisymmetric check with a concrete example.
Try $x = 2$ and $y = 8$. Both have squarefree part $2$, since $2 = 2 \cdot 1^2$ and $8 = 2 \cdot 2^2$, so $xy = 16$, a perfect square, and $(2,8) \in R$. Also $(8,2) \in R$ by symmetry. But $2 \neq 8$, so antisymmetry fails.

Step 6: Collect the results.
$R$ is reflexive, symmetric and transitive, but it is not antisymmetric.
\[ \boxed{\text{Reflexive, Symmetric, Transitive}} \]