Let A = {0, 1, 2, ..., 9}. Let R be a relation on A defined by (x, y) ∈ R if and only if |x - y| is

Q: Wait, why do we partition the set into classes based on remainders?

Great question! The relation is defined by $|x - y|$ being a multiple of $3$, which is mathematically equivalent to $x \equiv y \pmod 3$. This means $x$ and $y$ must have the same remainder when divided by $3$. Elements with the same remainder are 'related' to each other, forming an equivalence class. This is the most efficient way to group elements that satisfy the relation.

Q: Why is the formula $n(R) = n_0^2 + n_1^2 + n_2^2$ used?

Think about the Cartesian product. If a class has $k$ elements, every element in that class is related to every other element in the same class (including itself). This creates $k \times k = k^2$ pairs for that specific class. Since the classes are disjoint, we simply sum these squares to get the total number of pairs in the relation.

Q: Is it possible for an element to belong to two different classes?

No, that is impossible! Every integer has a unique remainder when divided by $3$ (either $0, 1,$ or $2$). Therefore, every element in set $A$ belongs to exactly one equivalence class. This is a fundamental property of equivalence relations: they partition the set into disjoint, non-overlapping subsets.

Question

Let $A = \{0, 1, 2, \dots, 9\}$. Let $R$ be a relation on $A$ defined by $(x, y) \in R$ if and only if $|x - y|$ is a multiple of 3. Given below are two statements: Statement I: $n(R) = 36$. Statement II: $R$ is an equivalence relation. In the light of the above statements, choose the correct answer from the options given below

Options:

(A) (1) Statement I is incorrect but Statement II is correct

(B) (2) Statement I is correct but Statement II is incorrect

(C) (3) Both Statement I and Statement II are incorrect

(D) (4) Both Statement I and Statement II are correct

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a set of numbers, $A = \{0, 1, 2, \dots, 9\}$. At first glance, it is just a simple list.

When we define a relation $R$ where $(x, y) \in R$ if and only if $|x - y|$ is a multiple of $3$, we are not just looking at numbers; we are looking at a hidden structure. This is the world of equivalence relations, and it is more elegant than it appears.

### The Three Pillars of Equivalence

To prove that $R$ is an equivalence relation, we must verify three fundamental properties: reflexivity, symmetry, and transitivity.

First, **reflexivity**: for any $x \in A$, is $(x, x) \in R$? The condition is $|x - x| = 0$. Since $0 = 3 	imes 0$, $0$ is indeed a multiple of $3$. Thus, $R$ is reflexive.

Second, **symmetry**: if $(x, y) \in R$, then $|x - y| = 3k$. Since $|x - y| = |y - x|$, it follows that $|y - x| = 3k$, so $(y, x) \in R$. Symmetry holds.

Finally, **transitivity**: if $(x, y) \in R$ and $(y, z) \in R$, then $x - y = 3k$ and $y - z = 3m$. Adding these gives $x - z = 3(k + m)$, which is a multiple of $3$. Thus, $(x, z) \in R$.

With these three pillars standing tall, we have proven that $R$ is an equivalence relation. **Statement II is correct!**

### The Power of Partitioning

Now, let us tackle the counting problem. An equivalence relation acts like a sorter, partitioning our set $A$ into disjoint equivalence classes. We group elements by their remainders when divided by $3$.

Class $C_0$ contains numbers $x$ such that $x \equiv 0 \pmod 3$, which are $\{0, 3, 6, 9\}$. This gives us $n_0 = 4$ elements.

Class $C_1$ contains numbers $x$ such that $x \equiv 1 \pmod 3$, which are $\{1, 4, 7\}$, giving $n_1 = 3$ elements.

Class $C_2$ contains numbers $x$ such that $x \equiv 2 \pmod 3$, which are $\{2, 5, 8\}$, giving $n_2 = 3$ elements. Every element of $A$ is accounted for, and no element is left behind.

### The Final Calculation

To find the total number of ordered pairs $n(R)$, we use the formula:

$$n(R) = n_0^2 + n_1^2 + n_2^2$$

Substituting our values, we get:

$$n(R) = 4^2 + 3^2 + 3^2$$

This simplifies to $16 + 9 + 9$, which equals **$34$**.

Statement I claimed that $n(R) = 36$. Our rigorous calculation shows that $34 
eq 36$. Therefore, **Statement I is incorrect**.

We have navigated the logic, verified the properties, and performed the arithmetic with precision. The conclusion is clear: **Statement I is incorrect, but Statement II is correct.**

Select Answer:

Visualized Solution

Defining the Relation $R$

Checking Reflexivity

Checking Symmetry

Checking Transitivity

Statement II is Correct

Concept of Equivalence Classes

Finding Class $C_{0}$

Finding Classes $C_{1}$ and $C_{2}$

Calculating $n (R)$

Atomic Computation

Comparing with Statement I

Final Conclusion

The Sigma Insight: Equivalence Relations

Analyzing the Setup

The Three Pillars of Equivalence

The Power of Partitioning

The Final Calculation

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