Let R be a relation defined on N as a R b is 2a + 3b is a multiple of 5, a, b ∈ N. Then R is (1) not

Q: Wait, why is $2a + 3a = 5a$ enough to prove reflexivity?

Great question! Reflexivity requires that $aRa$ holds for every $a \in \mathbb{N}$. The condition $aRa$ translates to $2a + 3a$ being a multiple of $5$. Since $2a + 3a = 5a$, and $5a$ is clearly $5$ times the integer $a$, it satisfies the definition of being a multiple of $5$ for any natural number $a$. Thus, the relation is reflexive.

Q: I tried to prove symmetry by just swapping $a$ and $b$ in $2a + 3b = 5k$, but I got $2b + 3a = 5k$. Is that wrong?

Yes, that is a common trap! You cannot simply swap variables in the equation. You must start with the assumption $2a + 3b = 5\alpha$ and then manipulate it to show that $2b + 3a$ must also be a multiple of $5$. The coefficients $2$ and $3$ are different, so $2a + 3b$ is not the same as $2b + 3a$. You need the algebraic trick of adding the two expressions to bridge the gap.

Q: Is it always an equivalence relation if it is reflexive and symmetric?

Not necessarily! An equivalence relation must satisfy all three properties: reflexivity, symmetry, AND transitivity. If you only prove the first two, you haven't finished the job. In this problem, we had to perform the extra step of proving transitivity to confirm it is an equivalence relation.

Question

Let $R$ be a relation defined on $\mathbb{N}$ as $a R b$ is $2a + 3b$ is a multiple of 5, $a, b \in \mathbb{N}$. Then $R$ is

Options:

(A) not reflexive

(B) transitive but not symmetric

(C) symmetric but not transitive

(D) an equivalence relation

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the threshold of a mathematical structure, looking at a set of natural numbers $\mathbb{N}$. We are given a rule, a relation $R$, that acts like a secret handshake between two numbers $a$ and $b$.

The rule is simple yet profound: $aRb$ if and only if $2a + 3b$ is a multiple of $5$. In the world of JEE Advanced, these problems are about understanding the underlying symmetry of numbers.

### Phase 1: The Identity (Reflexivity)

First, we must test the mirror. Does every number relate to itself? We check for reflexivity by asking: Is $aRa$ true for all $a \in \mathbb{N}$?

We substitute $b = a$ into our condition $2a + 3b$:

$$2a + 3a = 5a$$

Since $a$ is a natural number, $5a$ is obviously a multiple of $5$. The condition holds, and the relation is **reflexive**.

### Phase 2: The Mirror (Symmetry)

Now, we test the symmetry. If $a$ is related to $b$, does $b$ relate back to $a$? We assume $aRb$ is true, which means:

$$2a + 3b = 5\alpha \quad 	ext{for some integer } \alpha$$

Our goal is to prove $bRa$, which means $2b + 3a$ must be a multiple of $5$. Consider the sum of our known expression and our target expression:

$$(2a + 3b) + (2b + 3a) = 5a + 5b = 5(a + b)$$

We can now isolate our target:

$$2b + 3a = 5(a + b) - (2a + 3b)$$

Substituting our assumption $2a + 3b = 5\alpha$, we get:

$$2b + 3a = 5(a + b) - 5\alpha = 5(a + b - \alpha)$$

Since $a, b,$ and $\alpha$ are integers, the term $(a + b - \alpha)$ is an integer. Thus, $2b + 3a$ is a multiple of $5$, and **symmetry is confirmed**.

### Phase 3: The Chain (Transitivity)

Finally, we reach the most rigorous test: transitivity. If $aRb$ and $bRc$, must $aRc$ be true? We have two equations:

$$2a + 3b = 5\alpha$$
$$2b + 3c = 5\beta$$

We want to prove $2a + 3c$ is a multiple of $5$. Let us add these two equations:

$$(2a + 3b) + (2b + 3c) = 5\alpha + 5\beta$$

This simplifies to:

$$2a + 5b + 3c = 5(\alpha + \beta)$$

Now, we isolate our target $2a + 3c$:

$$2a + 3c = 5(\alpha + \beta) - 5b = 5(\alpha + \beta - b)$$

Since $\alpha, \beta,$ and $b$ are integers, the expression $(\alpha + \beta - b)$ is an integer. Therefore, $2a + 3c$ is a multiple of $5$, and the chain is complete.

### The Final Verdict

We have walked through the three gates of logic. We found the relation to be reflexive, symmetric, and transitive.

Because it satisfies all three, we conclude that **$R$ is an equivalence relation**. This problem teaches us that even abstract relations follow a beautiful, logical order.

Animated Solution for Mathematics - Sets and Relations: Let $R$ be a relation defined on $N$ as $a R b$ is $2 a + 3 b$ is a multiple of 5, $a, b \in N$ . Then $R$ is

Select Answer:

Visualized Solution

Understanding the Relation $R$

Conditions for Equivalence

Checking Reflexivity

Reflexivity Conclusion

Checking Symmetry

Symmetry: The Clever Trick

Symmetry Conclusion

Checking Transitivity

Transitivity: Algebraic Manipulation

Transitivity Conclusion

Final Verdict

The Sigma Insight: Equivalence Relations

Analyzing the Setup

Phase 1

Phase 2

Phase 3

The Final Verdict

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