Question

Let 𝑅 be a binary relation on the set {1, 2, … , 10}, where (𝑥, 𝑦) ∈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:

Think of each number as (square-free part) times (a perfect square); $x$ and $y$ relate exactly when their square-free parts match, which forms an equivalence relation - reflexive, symmetric and transitive, but not antisymmetric.

Answer 1

The Correct Option is A, B, C

We are given a relation \(R\) on \(\{1,2,\ldots,10\}\) where \((x,y)\in R\) if \(xy\) is a perfect square.

Step 1: Key idea - square-free part. Every positive integer \(n\) can be written as \(n = s \cdot k^2\), where \(s\) is the square-free part of \(n\) (what is left after removing all square factors). Two numbers \(x\) and \(y\) satisfy \(xy = \text{perfect square}\) exactly when they share the same square-free part.

Step 2: Check Reflexive. For any \(x\), \(x \cdot x = x^2\), always a perfect square. So \((x,x) \in R\) for every \(x\). Hence \(R\) is reflexive.

Step 3: Check Symmetric. If \((x,y) \in R\), then \(xy\) is a perfect square. Since \(xy = yx\), \((y,x) \in R\) too. Hence \(R\) is symmetric.

Step 4: Check Transitive. Suppose \((x,y) \in R\) and \((y,z) \in R\). Then \(x,y\) share a square-free part, and \(y,z\) share a square-free part, so \(x,z\) share that same square-free part, meaning \(xz\) is a perfect square. Example: \(x=2, y=8, z=18\): \(2\times8=16\), \(8\times18=144\), \(2\times18=36\) - all perfect squares. Hence \(R\) is transitive.

Step 5: Check Antisymmetric. Take \(x=2, y=8\). Both \(2\times8=16\) and \(8\times2=16\) are perfect squares, so \((2,8)\in R\) and \((8,2)\in R\), yet \(2 \neq 8\). This violates antisymmetry, so \(R\) is not antisymmetric.

Final Answer: \(\boxed{\text{Reflexive, Symmetric, Transitive (options A, B, C)}}\)