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: Understand the relation.
The set is \(\{1,2,\ldots,10\}\), and \((x,y) \in R\) exactly when \(xy\) is a perfect square. We check each property, reflexive, symmetric, transitive, and antisymmetric, one at a time using prime factorisation.

Step 2: Check reflexivity.
A relation is reflexive if \((x,x) \in R\) for every \(x\) in the set. Here \((x,x) \in R\) means \(x \cdot x = x^2\) is a perfect square. Since \(x^2\) is always a perfect square for any integer \(x\), this holds for every element of the set. So \(R\) is reflexive. Option (A) is satisfied.

Step 3: Check symmetry.
A relation is symmetric if \((x,y) \in R\) implies \((y,x) \in R\). If \(xy\) is a perfect square, then \(yx\) is the same product, since multiplication does not depend on order, so \(yx\) is a perfect square too. Hence \((y,x) \in R\) whenever \((x,y) \in R\). So \(R\) is symmetric. Option (B) is satisfied.

Step 4: Check transitivity using prime factorisation.
Suppose \((x,y) \in R\) and \((y,z) \in R\), meaning \(xy\) and \(yz\) are both perfect squares. For any prime \(p\), let \(v_p(n)\) denote the exponent of \(p\) in the prime factorisation of \(n\). A number \(n\) is a perfect square exactly when \(v_p(n)\) is even for every prime \(p\).
From \(xy\) being a square, \(v_p(x) + v_p(y)\) is even for every prime \(p\).
From \(yz\) being a square, \(v_p(y) + v_p(z)\) is even for every prime \(p\).
Subtracting these two even numbers gives another even number:
\[ \big(v_p(x)+v_p(y)\big) - \big(v_p(y)+v_p(z)\big) = v_p(x) - v_p(z) \]
So \(v_p(x) - v_p(z)\) is even for every prime \(p\). Adding \(2v_p(z)\), which is also even, gives
\[ v_p(x) + v_p(z) = \big(v_p(x)-v_p(z)\big) + 2v_p(z) \]
which is even as well, for every prime \(p\). This means \(xz\) is a perfect square, so \((x,z) \in R\). Hence \(R\) is transitive. Option (C) is satisfied.

Step 5: Check antisymmetry.
A relation is antisymmetric if \((x,y) \in R\) and \((y,x) \in R\) together force \(x = y\). Take \(x=1\) and \(y=4\). Then \(xy = 4 = 2^2\), a perfect square, so \((1,4) \in R\). Since \(R\) is symmetric, \((4,1) \in R\) as well. But \(1 \neq 4\), so this is a counterexample to antisymmetry. Option (D) is not satisfied.

Final Answer:
\(R\) is reflexive, symmetric, and transitive, but not antisymmetric.
\[ \boxed{\text{Reflexive, Symmetric, Transitive}} \]