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Question

A relation R is defined on the set N of natural numbers as xRy ⇒ x 2– 4xy + 3y 2= 0. Then which one of the following is correct?

This question was previously asked in
NDA II 2019 GAT Previous Year Paper (17-Nov-2019)
The correct answer is

R is reflexive, but neither symmetric nor transitive

Analyzing the Relation R on Natural Numbers

The question asks us to determine the properties of a relation R defined on the set of natural numbers, \( \mathbb{N} \). The relation is given by \( xRy \iff x^2 - 4xy + 3y^2 = 0 \).

First, let's factor the given quadratic equation: \( x^2 - 4xy + 3y^2 = 0 \).

We can treat this as a quadratic in terms of \(x\) or \(y\), or factor it directly:

\( x^2 - xy - 3xy + 3y^2 = 0 \)
\( x(x - y) - 3y(x - y) = 0 \)
\( (x - y)(x - 3y) = 0 \)

So, the relation \( xRy \) holds if and only if \( x - y = 0 \) or \( x - 3y = 0 \). This means \( xRy \iff x = y \) or \( x = 3y \), where \(x, y \in \mathbb{N}\).

Checking for Reflexivity of Relation R

A relation R is reflexive if \( xRx \) for all \( x \) in the set. In this case, the set is \( \mathbb{N} \).

We need to check if \( xRx \) holds for all \( x \in \mathbb{N} \). According to the definition of R, \( xRx \) holds if \( x = x \) or \( x = 3x \).

The condition \( x = x \) is always true for any number, including all natural numbers.

Since \( x = x \) is always true for any \( x \in \mathbb{N} \), the condition \( x = y \) or \( x = 3y \) is satisfied when \( y=x \). Thus, \( xRx \) holds for all \( x \in \mathbb{N} \).

Therefore, the relation R is reflexive.

Checking for Symmetry of Relation R

A relation R is symmetric if whenever \( xRy \) holds, \( yRx \) also holds for all \( x, y \) in the set.

We know \( xRy \iff x = y \) or \( x = 3y \).

We know \( yRx \iff y = x \) or \( y = 3x \).

Let's consider an example. Let \( y = 1 \). If we choose \( x = 3 \), then \( x = 3y \), so \( 3R1 \) holds (since \( 3 = 3 \times 1 \)).

Now let's check if \( yRx \) holds, i.e., if \( 1R3 \) holds. According to the definition, \( 1R3 \) holds if \( 1 = 3 \) or \( 1 = 3 \times 3 \).

Is \( 1 = 3 \)? No.

Is \( 1 = 3 \times 3 \)? Is \( 1 = 9 \)? No.

Since neither condition holds, \( 1R3 \) does not hold.

We found a pair \((3, 1)\) such that \( 3R1 \) holds, but \( 1R3 \) does not hold. Therefore, the relation R is not symmetric.

Checking for Transitivity of Relation R

A relation R is transitive if whenever \( xRy \) holds and \( yRz \) holds, \( xRz \) also holds for all \( x, y, z \) in the set.

We know:

\( xRy \iff x = y \) or \( x = 3y \)

\( yRz \iff y = z \) or \( y = 3z \)

\( xRz \iff x = z \) or \( x = 3z \)

We need to check if \( (xRy \text{ and } yRz) \implies xRz \) for all \( x, y, z \in \mathbb{N} \).

Let's consider a case where both \( xRy \) and \( yRz \) hold, but \( xRz \) might not. Consider the scenario where \( x = 3y \) and \( y = 3z \).

Substitute the second equation into the first: \( x = 3(3z) = 9z \).

So, if \( x=9z \), then \( xRy \) (since \( x=3y \) becomes \( 9z=3y \), which means \( 3z=y \), but we assumed \( y=3z \) which is true) and \( yRz \) (since \( y=3z \)) hold.

Now we must check if \( xRz \) holds when \( x = 9z \). According to the definition of R, \( xRz \) holds if \( x = z \) or \( x = 3z \).

For \( x = 9z \) where \( z \in \mathbb{N} \) (so \( z \ge 1 \)):

Is \( 9z = z \)? This implies \( 8z = 0 \), which means \( z = 0 \). However, \( z \in \mathbb{N} \), so \( z \ge 1 \). Thus, \( 9z=z \) is false for \( z \in \mathbb{N} \).

Is \( 9z = 3z \)? This implies \( 6z = 0 \), which means \( z = 0 \). Again, this is false for \( z \in \mathbb{N} \).

Since neither condition \( x=z \) nor \( x=3z \) holds when \( x=9z \) (for \( z \in \mathbb{N} \)), \( xRz \) does not hold in this case.

Let's use a concrete example. Let \( z=1 \). Then \( y=3z=3 \). Then \( x=3y=3(3)=9 \). So we have \( x=9, y=3, z=1 \).

\( 9R3 \) holds because \( 9 = 3 \times 3 \).

\( 3R1 \) holds because \( 3 = 3 \times 1 \).

Does \( 9R1 \) hold? \( 9R1 \iff 9 = 1 \) or \( 9 = 3 \times 1 \). Neither is true.

Since \( 9R3 \) and \( 3R1 \) hold, but \( 9R1 \) does not hold, the relation R is not transitive.

Summary of Relation Properties

Based on our analysis:

Reflexivity: R is reflexive.

Symmetry: R is not symmetric.

Transitivity: R is not transitive.

Therefore, the relation R is reflexive, but neither symmetric nor transitive.

Revision Table: Relation Properties

Property Definition Check for Relation R (\(xRy \iff x=y\) or \(x=3y\)) Result
Reflexive \(xRx\) for all \(x\) in the set. Is \(x=x\) or \(x=3x\)? Yes, \(x=x\) is true. Yes
Symmetric If \(xRy\), then \(yRx\) for all \(x,y\). If (\(x=y\) or \(x=3y\)), is (\(y=x\) or \(y=3x\))? No. Example: \(3R1\) (\(3=3 \times 1\)) holds, but \(1R3\) (\(1=3\) or \(1=3 \times 3\)) does not. No
Transitive If \(xRy\) and \(yRz\), then \(xRz\) for all \(x,y,z\). If (\(x=y\) or \(x=3y\)) and (\(y=z\) or \(y=3z\)), is (\(x=z\) or \(x=3z\))? No. Example: \(9R3\) (\(9=3 \times 3\)) and \(3R1\) (\(3=3 \times 1\)) hold, but \(9R1\) (\(9=1\) or \(9=3 \times 1\)) does not (since \(x=9z\)). No

Additional Information on Types of Relations

Understanding the properties of relations is fundamental in set theory and discrete mathematics. A relation on a set A is simply a subset of \(A \times A\). Here, our set is \( \mathbb{N} \), and the relation R is the set of pairs \((x,y)\) such that \(x=y\) or \(x=3y\), where \(x, y \in \mathbb{N}\).

Let's briefly define the properties again:

Reflexive: Every element is related to itself. Think of it as looping back to the same element.

Symmetric: If there's a relationship from A to B, there's also one from B to A. It's a two-way relationship.

Transitive: If there's a relationship from A to B and from B to C, then there's automatically one from A to C. It's like a chain reaction.

Relations that have all three properties (reflexive, symmetric, and transitive) are called equivalence relations. Equivalence relations partition the set into disjoint subsets called equivalence classes. Our relation R is not an equivalence relation because it fails the symmetric and transitive properties.

Relations that are reflexive and transitive, but not necessarily symmetric, are important in defining orders, like partial orders (if also antisymmetric) or total orders.

The properties of a relation depend entirely on the specific definition of the relation and the set it is defined on.

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