Sigma Percentile
JEE Main 2023 (08 Apr Shift 2)
LEVELBoard

Animated Solution for Mathematics - Sets and Relations: Let . Then the relation is

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Visualized Solution

The Set

  • Set :
  • We are working within the Cartesian product .

Defining the Relation

  • Relation
  • Condition: The sum of the two numbers must be exactly .

Finding Elements of

  • Let's find pairs that sum to .
  • If ,
  • If ,

All Elements of

  • Continuing this, we get:
  • Notice is not included because .

Checking Reflexivity

  • A relation is reflexive if for all .
  • Geometrically, all points on the line must be in .

Is Reflexive?

  • For , we need .
  • Since , no such pair exists.
  • Therefore, is not reflexive.

Checking Symmetry

  • A relation is symmetric if .
  • If a point is in , its reflection across must also be in .

Is Symmetric?

  • We know .
  • By commutative property of addition, .
  • Thus, .
  • Example: and .
  • Therefore, is symmetric.

Checking Transitivity

  • A relation is transitive if and .
  • Let's test with an example from our set.

Is Transitive?

  • We have (since ).
  • We also have (since ).
  • For transitivity, we must have .

Transitivity Fails

  • But .
  • So, .
  • Therefore, is not transitive.

Final Conclusion

  • Reflexive: No
  • Symmetric: Yes
  • Transitive: No
  • Conclusion: is symmetric but neither reflexive nor transitive.

The Sigma Insight: Types of Relations

Solution Diagram

The Universe of Relations

Welcome, fellow traveler on the path of mathematics. Today, we are going to explore the elegant world of relations.
Imagine you are standing in a vast, discrete landscape defined by the set . This is our universe, where every point is an integer between and .
We are interested in the Cartesian product , a grid of all possible pairs where both and are from our set. We are looking for a specific subset, a relation , defined by the constraint .

The Geometry of the Relation

To understand , we must list its members systematically. We seek pairs such that their sum is exactly .
If , then . If , then . Continuing this, we find the pairs , , , and .
If , then would have to be . However, $0 otin A$, so the pair is forbidden. Our relation is defined as:

The Three Pillars of Relations

Now, we test the three fundamental pillars of relations: Reflexivity, Symmetry, and Transitivity.
Reflexivity: A relation is reflexive if every element satisfies . This requires:
Since $3.5 otin A$, there is no element related to itself. The pillar of reflexivity crumbles; is not reflexive.
Symmetry: A relation is symmetric if implies .
Looking at our set , we observe that for every pair , the mirror image is also present. Algebraically, because is equivalent to , symmetry is guaranteed. The pillar of symmetry stands tall; is symmetric.
Transitivity: A relation is transitive if and implies .
Consider the pairs and . For transitivity to hold, we would require . Since $1 + 1 = 2 eq 7$, the pair is missing. The pillar of transitivity falls; is not transitive.

The Verdict

We have journeyed through the set , mapped the relation , and tested the three pillars.
We found that is symmetric, but it lacks both reflexivity and transitivity. It is a beautiful, specific structure that teaches us that in mathematics, properties are not guaranteed—they must be earned through rigorous testing.

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