Question:medium

Let 𝑅 be a binary relation on the set {1, 2, … , 10}, where (π‘₯, 𝑦) βˆˆπ‘… if the product
of π‘₯ and 𝑦 is square of an integer. Which of the following properties is/are satisfied
by 𝑅?

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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.
Updated On: Aug 3, 2026
  • Reflexive
  • Symmetric
  • Transitive
  • Antisymmetric
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The Correct Option is A, B, C

Solution and Explanation

A quicker way to analyze this relation is to treat it as a partition problem instead of checking each property directly from the definition.

Step 1: Group the numbers 1 to 10 by their square-free part (strip out the largest square factor from each): 1 = 1 (square-free 1), 2 = 2, 3 = 3, 4 = \(2^2\) (square-free 1), 5 = 5, 6 = 6, 7 = 7, 8 = \(2 \times 2^2\) (square-free 2), 9 = \(3^2\) (square-free 1), 10 = 10.

Step 2: So the groups by square-free value are: {1,4,9} share square-free 1, {2,8} share square-free 2, while 3, 5, 6, 7, 10 are singleton groups.

Step 3: \((x,y) \in R\) exactly when \(x\) and \(y\) fall in the same group - this is precisely how an equivalence relation induced by a partition behaves.

Step 4: Every equivalence relation is automatically reflexive, symmetric and transitive by construction, since it partitions the set into disjoint classes and only relates elements within the same class.

Step 5: Antisymmetry fails whenever a group has more than one element, since two distinct elements of the same group relate to each other in both directions. Groups {1,4,9} and {2,8} both have more than one element, so antisymmetry breaks down.

Conclusion: The relation is reflexive, symmetric and transitive, but not antisymmetric - final answer A, B, C.

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