Let N denote the set of all natural numbers. Define two binary relations on N as R1 = {(x, y) ∈ N ×

Q: Wait, why do we reject $x=5$ in $R_1$? Isn't 0 a natural number?

Ah, the classic trap! In the context of JEE mathematics, the set of natural numbers $\mathbb{N}$ is defined as $\{1, 2, 3, \dots\}$. Zero is a whole number, not a natural number. If you include zero, your entire relation set will be wrong. Always check the domain definition first!

Q: Why isn't $R_1$ symmetric? I thought if $2x+y=10$, then $2y+x=10$ would also work?

That's a common intuition, but look closely at the equation. $2x+y=10$ is not the same as $2y+x=10$. For symmetry, if $(1, 8)$ is in the relation, $(8, 1)$ must satisfy the *same* equation. Plugging $(8, 1)$ into $2x+y=10$ gives $2(8)+1 = 17$, which is not 10. The relation is asymmetric because the coefficients of $x$ and $y$ are different.

Q: How do I prove transitivity without checking every single pair?

You don't need to check every pair! You only need one counter-example. If you find $(a, b) \in R$ and $(b, c) \in R$, but $(a, c) \notin R$, the relation fails transitivity immediately. In our case, $(4, 2) \in R_1$ and $(2, 6) \in R_1$, but $(4, 6) \notin R_1$. One failure is enough to break the property.

Question

Let N denote the set of all natural numbers. Define two binary relations on N as $R_1 = \{(x, y) \in N 	imes N : 2x + y = 10\}$ and $R_2 = \{(x, y) \in N 	imes N : x + 2y = 10\}$. Then :

Options:

(A) Both $R_1$ and $R_2$ are symmetric relations

(B) Range of $R_1$ is $\{2, 4, 8\}$

(C) Both $R_1$ and $R_2$ are transitive relations

(D) Range of $R_2$ is $\{1, 2, 3, 4\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a problem; we are dissecting the very structure of binary relations.

Imagine you are standing on the infinite grid of natural numbers, $\mathbb{N} = \{1, 2, 3, \dots\}$. This is our domain.

We have two rules, $R_1$ and $R_2$, acting as filters on this grid. Our mission is to uncover their properties: symmetry, transitivity, and their ranges.

### Unpacking $R_1$: The Hunt for Pairs

To understand $R_1$, we must find the ordered pairs $(x, y)$ that satisfy the relation $2x + y = 10$. Let's isolate $y$ to make our calculations easier:

$$y = 10 - 2x$$

Now, we iterate through the natural numbers for $x$:

*   If $x=1$, $y=8$.
*   If $x=2$, $y=6$.
*   If $x=3$, $y=4$.
*   If $x=4$, $y=2$.

What happens if $x=5$? Then $y=0$. But wait! Zero is not a natural number, so we must stop our search here.

Our relation $R_1$ is the set $\{(1, 8), (2, 6), (3, 4), (4, 2)\}$. The range is the set of all second elements, which is $\{2, 4, 6, 8\}$.

### The Symmetry and Transitivity Trap

Now, let's test the properties. A relation is symmetric if $(x, y) \in R$ implies $(y, x) \in R$.

We have $(1, 8) \in R_1$. Is $(8, 1) \in R_1$? Testing this: $2(8) + 1 = 17$, which is not $10$. Thus, **$R_1$ is not symmetric**.

Next, we check for transitivity. A relation is transitive if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$.

We have $(4, 2) \in R_1$ and $(2, 6) \in R_1$. For transitivity, $(4, 6)$ must be in $R_1$. Checking this: $2(4) + 6 = 14$, which is not $10$. So, **$R_1$ is not transitive**.

### The Victory with $R_2$

Finally, we turn our attention to $R_2$, defined by $x + 2y = 10$. Rearranging for $x$, we get:

$$x = 10 - 2y$$

Iterating through $y$ values:

*   If $y=1$, $x=8$.
*   If $y=2$, $x=6$.
*   If $y=3$, $x=4$.
*   If $y=4$, $x=2$.

Again, $y=5$ gives $x=0$, which is invalid. The set $R_2$ is $\{(8, 1), (6, 2), (4, 3), (2, 4)\}$.

The range is the set of $y$-values: $\{1, 2, 3, 4\}$. This perfectly matches our final option. We have systematically dismantled the problem and arrived at the truth.

Animated Solution for Mathematics - Sets and Relations: Let N denote the set of all natural numbers. Define two binary relations on N as $R_{1} = {(x, y) \in N \times N : 2 x + y = 10}$ and $R_{2} = {(x, y) \in N \times N : x + 2 y = 10}$ . Then :

Select Answer:

Visualized Solution

Understanding the Set $N$

Analyzing Relation $R_{1}$

Finding Elements of $R_{1}$

Range of $R_{1}$

Checking Symmetry for $R_{1}$

Checking Transitivity for $R_{1}$

Analyzing Relation $R_{2}$

Finding Elements of $R_{2}$

Range of $R_{2}$

Final Conclusion

The Sigma Insight: Domain and Range of a Relation

Analyzing the Setup

Unpacking $R_{1}$

The Symmetry and Transitivity Trap

The Victory with $R_{2}$

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Select Answer:

Visualized Solution

Understanding the Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N

Analyzing Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Finding Elements of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Checking Symmetry for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Checking Transitivity for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Analyzing Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Finding Elements of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​

Range of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​