Let N be the set of natural numbers and a relation R on N be defined by R = {(x, y) ∈ N × N : x³ - 3

Q: Wait, why can we just divide by $(x + y)$? Is that always allowed?

Great question! In general, dividing by a variable expression is dangerous because it might be zero. However, here we are working strictly within the set of natural numbers $\mathbb{N} = \{1, 2, 3, \dots\}$. Since the smallest value for $x$ and $y$ is $1$, the smallest value for $x + y$ is $1 + 1 = 2$. Because $x + y$ is always at least $2$, it is guaranteed to be non-zero. Thus, dividing by $(x + y)$ is perfectly safe and mathematically sound.

Q: Why is the relation not symmetric? If $(3, 1)$ is in the relation, shouldn't $(1, 3)$ be there too?

That is a classic trap! Symmetry requires that for *every* pair $(x, y)$ in the relation, the reversed pair $(y, x)$ must also be in the relation. While $(3, 1)$ satisfies $x = 3y$ (so it is in $R$), the pair $(1, 3)$ does not satisfy $x = y$ ($1 \neq 3$) nor does it satisfy $x = 3y$ ($1 \neq 3(3) = 9$). Since $(1, 3)$ fails both conditions, it is not in $R$. One counterexample is enough to break symmetry.

Q: I'm confused about transitivity. Why does the chain $(9, 3)$ and $(3, 1)$ fail?

Transitivity requires that if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z)$ must be in $R$. Here, $(9, 3)$ is in $R$ because $9 = 3(3)$, and $(3, 1)$ is in $R$ because $3 = 3(1)$. For transitivity to hold, $(9, 1)$ must be in $R$. But $9 \neq 1$ and $9 \neq 3(1)$. Since $(9, 1)$ is not in the relation, the chain is broken. Transitivity is a strict property—it must hold for *all* possible chains, and this one clearly fails.

Question

Let $\mathbb{N}$ be the set of natural numbers and a relation $R$ on $\mathbb{N}$ be defined by $R = \{(x, y) \in \mathbb{N} 	imes \mathbb{N} : x^3 - 3x^2y - xy^2 + 3y^3 = 0\}$. Then the relation $R$ is :

Options:

(A) symmetric but neither reflexive nor transitive

(B) reflexive but neither symmetric nor transitive

(C) reflexive and symmetric, but not transitive

(D) an equivalence relation

Sigma Percentile · Accepted Answer

### The Algebraic Surgery: Unmasking the Relation

Welcome, future engineers. Today, we are going to dissect a problem that, at first glance, looks like a chaotic cubic mess. We are given a relation $R$ on the set of natural numbers $\mathbb{N}$ defined by the equation:

$$x^3 - 3x^2y - xy^2 + 3y^3 = 0$$

When you see a high-degree polynomial like this, do not let the exponents intimidate you. Our first task is to perform some algebraic surgery to see what lies beneath.

We start by grouping the terms. Look at the first two terms: $x^3 - 3x^2y$. We can factor out an $x^2$, leaving us with $(x - 3y)$.

Now look at the last two terms: $-xy^2 + 3y^3$. If we factor out a $-y^2$, we are left with $(x - 3y)$. This is the moment of clarity!

Our equation becomes:

$$x^2(x - 3y) - y^2(x - 3y) = 0$$

Factoring out the common $(x - 3y)$, we get $(x^2 - y^2)(x - 3y) = 0$. Finally, recognizing the difference of squares, we arrive at the fully factored form:

$$(x - y)(x + y)(x - 3y) = 0$$

### The Domain Trap: Why Constraints Matter

In JEE Advanced, the domain is your best friend or your worst enemy. Here, we are told $x, y \in \mathbb{N}$. This is not just a label; it is a vital piece of information.

Because $x$ and $y$ are natural numbers, the smallest possible value for $x + y$ is $1 + 1 = 2$. This means $x + y$ can never be zero.

In the world of algebra, this is a gift. It allows us to divide both sides of our equation by $(x + y)$ without any fear of division by zero. Our condition simplifies beautifully to:

$$(x - y)(x - 3y) = 0$$

This implies that for any pair $(x, y)$ to be in our relation, we must have either $x = y$ or $x = 3y$. Geometrically, these are just two lines in the first quadrant of the coordinate plane.

### Testing the Properties: The Mirror, The Handshake, and The Chain

Now that we have our simplified conditions, $x = y$ and $x = 3y$, we can test the three pillars of relations: reflexivity, symmetry, and transitivity.

**Reflexivity (The Mirror):** A relation is reflexive if every element is related to itself. We need to check if $(x, x) \in R$ for all $x \in \mathbb{N}$.

Substituting $y = x$ into our condition, we get $x = x$ or $x = 3x$. Since $x = x$ is always true for any natural number, the condition is satisfied. Thus, the relation is **reflexive**.

**Symmetry (The Handshake):** Symmetry asks: if $(x, y) \in R$, is $(y, x)$ also in $R$? Let's test the point $(3, 1)$.

Since $3 = 3(1)$, the pair $(3, 1)$ is in $R$. Now, let's reverse it to $(1, 3)$. Does $1 = 3$ or $1 = 3(3)$? No.

Because we found a pair that is in the relation but whose reverse is not, the relation is **not symmetric**.

**Transitivity (The Chain):** Finally, we test transitivity. If $(x, y) \in R$ and $(y, z) \in R$, must $(x, z) \in R$?

Let's build a chain. We know $(9, 3) \in R$ because $9 = 3(3)$. We also know $(3, 1) \in R$ because $3 = 3(1)$.

For transitivity to hold, $(9, 1)$ must be in $R$. But $9 
eq 1$ and $9 
eq 3(1)$. The chain is broken. Therefore, the relation is **not transitive**.

### The Final Verdict

We have systematically dismantled the problem. We found that the relation is reflexive because every number is related to itself.

We proved it is not symmetric by finding a counterexample, and we proved it is not transitive by showing a broken chain.

Thus, the relation is **reflexive but neither symmetric nor transitive**. Keep this logical flow in your toolkit, and you will conquer any relation problem that comes your way.

Animated Solution for Mathematics - Sets and Relations: Let $N$ be the set of natural numbers and a relation $R$ on $N$ be defined by $R = {(x, y) \in N \times N : x^{3} - 3 x^{2} y - x y^{2} + 3 y^{3} = 0}$ . Then the relation $R$ is :

Select Answer:

Visualized Solution

The Given Relation

Factorizing the Equation

Expanding the Difference of Squares

Applying the Domain Constraint

The Simplified Relation

Checking Reflexivity

Reflexivity Confirmed

Checking Symmetry

Symmetry Counterexample

Checking Transitivity

Transitivity Counterexample

Final Conclusion

The Sigma Insight: Types of Relations

The Algebraic Surgery

The Domain Trap

Testing the Properties

The Final Verdict

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