Let A = {0, 1, 2, 3, 4, 5}. Let R be a relation on A defined by (x, y) ∈ R if and only if max{x, y}

Q: Why is the total number of elements 16 and not 18? Did I miss some pairs?

It is a common trap to double-count or miscalculate the L-shape. For $\max\{x, y\} = 3$, we have pairs where $x=3$ (4 pairs: $(3,0), (3,1), (3,2), (3,3)$) and $y=3$ (3 pairs: $(0,3), (1,3), (2,3)$). Total 7. Similarly for 4, we have 9. $7+9=16$. Stay systematic!

Q: Why is the relation not reflexive? Shouldn't every element be related to itself?

Reflexivity is a strict condition: $(x, x) \in R$ must hold for ALL $x \in A$. Here, for $x=0$, $\max\{0, 0\} = 0$, which is not in $\{3, 4\}$. Thus, $(0, 0) \notin R$. One failure is enough to break reflexivity.

Q: How do I prove it's not transitive without checking every combination?

Look for a counter-example! We found $(0, 3) \in R$ and $(3, 0) \in R$. If it were transitive, $(0, 0)$ would have to be in $R$. Since it isn't, the chain is broken. A single counter-example is the most powerful tool in your arsenal.

Question

Let $A = \{0, 1, 2, 3, 4, 5\}$. Let $R$ be a relation on $A$ defined by $(x, y) \in R$ if and only if $\max\{x, y\} \in \{3, 4\}$. Then among the statements $(S1)$ : The number of elements in $R$ is 18, and $(S2)$ : The relation $R$ is symmetric but neither reflexive nor transitive

Options:

(A) both are true

(B) both are false

(C) only (S2) is true

(D) only (S1) is true

Sigma Percentile · Accepted Answer

### The Geometry of Relations: Unlocking the L-Shape

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a journey into the heart of set theory and relations.

Often, when we see a definition like $\max\{x, y\} \in \{3, 4\}$, our minds might race to calculate, but I want you to pause. I want you to visualize.

Imagine a $6 	imes 6$ grid representing the Cartesian product $A 	imes A$, where $A = \{0, 1, 2, 3, 4, 5\}$. Every point $(x, y)$ on this grid is a potential candidate for our relation $R$.

The condition $\max\{x, y\} \in \{3, 4\}$ is not just an algebraic constraint; it is a geometric signature. It carves out a specific shape on our grid.

### Phase 1: The L-Shape of Maximums

Let us dissect the condition $\max\{x, y\} = 3$. This means either $x=3$ (and $y \le 3$) or $y=3$ (and $x \le 3$).

If you plot this on your grid, you will see a vertical line segment at $x=3$ and a horizontal line segment at $y=3$. They meet at $(3, 3)$.

Counting these points, we have $(3, 0), (3, 1), (3, 2), (3, 3)$ and $(0, 3), (1, 3), (2, 3)$. That is exactly $4 + 3 = 7$ points. This is our first L-shape.

Now, let us turn our attention to $\max\{x, y\} = 4$. Following the same logic, we have $x=4$ (with $y \le 4$) and $y=4$ (with $x \le 3$).

This gives us $(4, 0), (4, 1), (4, 2), (4, 3), (4, 4)$ and $(0, 4), (1, 4), (2, 4), (3, 4)$. Counting these, we find $5 + 4 = 9$ points.

The total number of elements in $R$ is $7 + 9 = 16$. Statement (S1) claims the count is 18. We have just proven, with absolute clarity, that (S1) is **false**.

### Phase 2: The Anatomy of Properties

Now, let us investigate the properties of this relation. Is it symmetric?

Symmetry requires that if $(x, y) \in R$, then $(y, x) \in R$. Our condition is $\max\{x, y\} \in \{3, 4\}$.

Since the maximum function is commutative—meaning $\max\{x, y\} = \max\{y, x\}$—the condition is inherently symmetric. If $(x, y)$ satisfies the condition, $(y, x)$ must satisfy it as well. Thus, $R$ is **symmetric**.

Next, reflexivity. This is where many students stumble. Reflexivity demands that for every $x \in A$, the pair $(x, x)$ must be in $R$.

Let us test $x=0$. The pair is $(0, 0)$. Is $\max\{0, 0\} \in \{3, 4\}$? No, $\max\{0, 0\} = 0$.

Since $0 
otin \{3, 4\}$, the pair $(0, 0)$ is not in $R$. Because this fails for $x=0$, the relation is **not reflexive**. One single failure is all it takes to break the property.

Finally, transitivity. This is the most subtle of the three. Transitivity requires that if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.

Let us look for a counter-example. We know $(0, 3) \in R$ because $\max\{0, 3\} = 3$. We also know $(3, 0) \in R$ because $\max\{3, 0\} = 3$.

If the relation were transitive, then $(0, 0)$ would have to be in $R$. But we already established that $(0, 0) 
otin R$. The chain is broken. Therefore, the relation is **not transitive**.

### Conclusion

We have systematically dismantled the problem. We found the count to be 16, not 18, making (S1) false.

We confirmed the relation is symmetric but fails both reflexivity and transitivity, making (S2) true.

The beauty of this problem lies in the transition from abstract definitions to concrete geometric visualization. You have mastered the L-shape, the counter-example, and the logical rigor required for JEE Advanced. Keep this clarity with you, and no relation will ever be too complex to solve.

Select Answer:

Visualized Solution

Understanding the Set $A$ and Relation $R$

Case 1: $max {x, y} = 3$

Counting Pairs for Case 1

Case 2: $max {x, y} = 4$

Counting Pairs for Case 2

Verifying Statement (S1)

Checking Symmetry in (S2)

Checking Reflexivity in (S2)

Checking Transitivity in (S2)

Final Conclusion

The Sigma Insight: Types of Relations

The Geometry of Relations

Phase 1

Phase 2

Conclusion

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