Q1473: Let R be a relation defined on a set N of natural numbers su...

Class CBSE Class 12 Mathematics Relations and Functions Q #1473

KNOWLEDGE BASED

REMEMBER

3 Marks 2025 AISSCE(Board Exam) SA

Let R be a relation defined on a set N of natural numbers such that $R=\{(x,y): xy \text{ is a square of a natural number, } x, y\in N\}$. Determine if the relation R is an equivalence relation.

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Detailed Solution

Step 1: Check for Reflexivity

For R to be reflexive, we need $(x, x) \in R$ for all $x \in N$. This means $x \cdot x = x^2$ must be a square of a natural number. Since $x^2$ is always a square for any natural number $x$, R is reflexive.

Step 2: Check for Symmetry

For R to be symmetric, if $(x, y) \in R$, then $(y, x) \in R$. If $xy$ is a square of a natural number, then $yx$ is also a square of a natural number (since multiplication is commutative). Therefore, if $(x, y) \in R$, then $(y, x) \in R$, and R is symmetric.

Step 3: Check for Transitivity

For R to be transitive, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$. This means if $xy$ and $yz$ are squares of natural numbers, then $xz$ must also be a square of a natural number. Let $xy = m^2$ and $yz = n^2$ for some natural numbers $m$ and $n$. Then, $x = \frac{m^2}{y}$ and $z = \frac{n^2}{y}$. So, $xz = \frac{m^2}{y} \cdot \frac{n^2}{y} = \frac{m^2n^2}{y^2} = (\frac{mn}{y})^2$. For $xz$ to be a square, $\frac{mn}{y}$ must be a natural number. Consider $x=2, y=8, z=32$. Then $xy = 2*8 = 16 = 4^2$, $yz = 8*32 = 256 = 16^2$. So $(2,8) \in R$ and $(8,32) \in R$. Now, $xz = 2*32 = 64 = 8^2$. So $(2,32) \in R$. However, consider $x=1, y=4, z=9$. $xy = 4 = 2^2$ and $yz = 36 = 6^2$. Then $xz = 9 = 3^2$. Consider $x=2, y=2, z=8$. $xy = 4 = 2^2$ and $yz = 16 = 4^2$. Then $xz = 16 = 4^2$. Consider $x=a^2, y=1, z=b^2$. $xy = a^2$ and $yz = b^2$. Then $xz = a^2b^2 = (ab)^2$. Since $xy = m^2$ and $yz = n^2$, we have $x = m^2/y$ and $z = n^2/y$. $xz = (mn/y)^2$. For $xz$ to be a perfect square, $mn/y$ must be an integer. Multiplying the two equations, we get $xy^2z = m^2n^2$, so $y^2 = \frac{m^2n^2}{xz}$. $y = \frac{mn}{\sqrt{xz}}$. If $xz$ is a perfect square, then $\sqrt{xz}$ is an integer. So, $xz$ must be a square of a natural number. Therefore, R is transitive.

Step 4: Conclusion

Since R is reflexive, symmetric, and transitive, R is an equivalence relation.

Final Answer: R is an equivalence relation.

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Pedagogical Audit

Bloom's Analysis: This is an REMEMBER question because the student needs to recall the definitions of reflexive, symmetric, and transitive relations to determine if the given relation is an equivalence relation.

Knowledge Dimension: CONCEPTUAL

Justification: The question requires understanding the concepts of relations, specifically reflexive, symmetric, transitive, and equivalence relations. It involves applying these concepts to a specific relation defined on natural numbers.

Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly tests the student's understanding of relations and equivalence relations, which is a core concept in the Relations and Functions chapter.