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Animated Solution for Mathematics - Sets and Relations: The relation $R = {(x, y) : x, y \in Z and x + y is even}$ is:

Select Answer:

Progressive Hints

Hint 1

Hint 1: The Spark. To understand this relation, visualize the set of integers

Z

as being partitioned into two distinct 'worlds': the world of even numbers and the world of odd numbers. If

x + y

is even, what does that imply about the parity of

x

and

y

? Do they have to be the same or different?

Hint 2

Hint 2: The Arsenal. Recall the formal definitions required for an equivalence relation. Reflexivity requires

(x, x) \in R

for all

x \in Z

. Symmetry requires that if

(x, y) \in R

, then

(y, x) \in R

. Transitivity requires that if

(x, y) \in R

and

(y, z) \in R

, then

(x, z) \in R

. Use these definitions as your checklist.

Hint 3

Hint 3: The Execution. For the transitivity proof, express the conditions as

x + y = 2 k_{1}

and

y + z = 2 k_{2}

, where

k_{1}

and

k_{2}

are integers. Your goal is to show

x + z

is even. Try adding these two equations and isolating

x + z

. How does the term

2 y

affect the parity of the result?

Visualized Solution

Introduction to the Relation $R$

Given set: $Z$ (Integers)
Relation: $R = {(x, y) : x, y \in Z and x + y is even}$
Goal: Determine if $R$ is reflexive, symmetric, and transitive.

Defining Reflexivity

Reflexive Property: A relation $R$ on set $A$ is reflexive if $(x, x) \in R$ for all $x \in A$ .
For our relation, we must check if $x + x$ is even for all $x \in Z$ .

Testing Reflexivity

For any $x \in Z$ :
$x + x = 2 x$
Since $2 x$ is always a multiple of $2$ , it is always even.
$\Rightarrow (x, x) \in R$ . Thus, $R$ is reflexive.

Defining Symmetry

Symmetric Property: A relation $R$ is symmetric if $(x, y) \in R \Rightarrow (y, x) \in R$ .
Assume $(x, y) \in R$ , which means $x + y$ is even.

Testing Symmetry

Since addition is commutative: $x + y = y + x$ .
If $x + y$ is even, then $y + x$ is also even.
$\Rightarrow (y, x) \in R$ . Thus, $R$ is symmetric.

Defining Transitivity

Transitive Property: A relation $R$ is transitive if $(x, y) \in R$ and $(y, z) \in R \Rightarrow (x, z) \in R$ .
Assume $(x, y) \in R$ and $(y, z) \in R$ .

Setting up Transitivity

Let $x + y = 2 k_{1}$ for some $k_{1} \in Z$ .
Let $y + z = 2 k_{2}$ for some $k_{2} \in Z$ .

Computing the Sum

Adding the equations: $(x + y) + (y + z) = 2 k_{1} + 2 k_{2}$
$x + 2 y + z = 2 (k_{1} + k_{2})$

Isolating $x + z$

$x + z = 2 (k_{1} + k_{2}) - 2 y$
$x + z = 2 (k_{1} + k_{2} - y)$

Finalizing Transitivity

Since $k_{1}, k_{2}, y$ are integers, $k_{1} + k_{2} - y$ is an integer.
$\Rightarrow x + z$ is a multiple of $2$ , so it is even.
$\Rightarrow (x, z) \in R$ . Thus, $R$ is transitive.

Conclusion: Equivalence Relation

The relation $R$ is Reflexive, Symmetric, and Transitive.
Therefore, $R$ is an equivalence relation.
Correct Option: (2)

00:00 / 00:00

The Sigma Insight: Equivalence Relations

Solution Diagram

Analyzing the Reflexive Property

A relation

R

is reflexive if every element relates to itself. For any integer

x \in Z

, we must determine if

(x, x) \in R

By definition, this requires checking if the sum

x + x

is even. Since

x + x = 2 x

, and any integer multiplied by

2

results in an even number, the condition holds.

Thus, the relation is reflexive.

Analyzing the Symmetric Property

The symmetric property requires that if

x

is related to

y

, then

y

must be related to

x

. Suppose

(x, y) \in R

, which implies that

x + y

is even.

Because addition is commutative, we know that:

x + y = y + x

Since

x + y

is even,

y + x

must also be even. Therefore,

(y, x) \in R

, and the relation is symmetric.

Analyzing the Transitive Property

Transitivity requires that if

(x, y) \in R

and

(y, z) \in R

, then

(x, z) \in R

. We start with the following equations for some integers

k_{1}

and

k_{2}

x + y = 2 k_{1}

y + z = 2 k_{2}

Adding these two equations together, we obtain:

(x + y) + (y + z) = 2 k_{1} + 2 k_{2}

x + 2 y + z = 2 (k_{1} + k_{2})

To determine if

x + z

is even, we isolate the term:

x + z = 2 (k_{1} + k_{2}) - 2 y

x + z = 2 (k_{1} + k_{2} - y)

Since

k_{1}

k_{2}

, and

y

are integers, their combination

K = (k_{1} + k_{2} - y)

is also an integer. This confirms that

x + z = 2 K

, meaning

x + z

is even.

Thus,

(x, z) \in R

, and the relation is transitive.

Conclusion

Because the relation

R

is reflexive, symmetric, and transitive, it is formally classified as an equivalence relation.

This relation effectively partitions the set of integers

Z

into two distinct equivalence classes: the set of even integers and the set of odd integers.

Common Traps & FAQs

Wait, why do we need to prove reflexivity for all

x

? Can't I just pick a few numbers like 2 and 4 to check?

That is a fantastic intuition-building exercise! Testing specific numbers is a great way to see the pattern, but in the rigorous world of JEE Advanced, we must prove the property for the entire set

Z

. By using the variable

x

, we are proving it for every single integer simultaneously. It is the difference between checking a few bricks and proving the entire wall is solid.

Is there a faster way to see this without all the

2 k_{1}

and

2 k_{2}

algebra?

Yes, there is! You can use parity logic: the sum of two numbers is even if and only if both numbers have the same parity (both even or both odd). Since 'having the same parity' is an equivalence relation, this relation must be one too. However, the algebraic method using

2 k_{1}

and

2 k_{2}

is the gold standard for formal proofs because it leaves absolutely no room for ambiguity.

Why does the

2 y

term not ruin the transitivity proof?

That is the most common trap! When we add the equations, we get

(x + y) + (y + z) = 2 k_{1} + 2 k_{2}

, which simplifies to

x + z + 2 y = 2 (k_{1} + k_{2})

. When we move

2 y

to the other side, we get

x + z = 2 (k_{1} + k_{2} - y)

. Since

k_{1}

k_{2}

, and

y

are all integers, their combination is also an integer. The factor of

2

remains, which is all we need to prove the sum is even!

Animated Solution for Mathematics - Sets and Relations: The relation $R = {(x, y) : x, y \in Z and x + y is even}$ is:

Select Answer:

Visualized Solution

Introduction to the Relation $R$

Defining Reflexivity

Testing Reflexivity

Defining Symmetry

Testing Symmetry

Defining Transitivity

Setting up Transitivity

Computing the Sum

Isolating $x + z$

Finalizing Transitivity

Conclusion: Equivalence Relation

The Sigma Insight: Equivalence Relations

Analyzing the Reflexive Property

Analyzing the Symmetric Property

Analyzing the Transitive Property

Conclusion

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Animated Solution for Mathematics - Sets and Relations: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The relation R={(x,y):x,y∈Z and x+y is even } is:

Select Answer:

Visualized Solution

Introduction to the Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Defining Reflexivity

Testing Reflexivity

Defining Symmetry

Testing Symmetry

Defining Transitivity

Setting up Transitivity

Computing the Sum

Isolating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x+z

Finalizing Transitivity

Conclusion: Equivalence Relation

The Sigma Insight: Equivalence Relations

Analyzing the Reflexive Property

Analyzing the Symmetric Property

Analyzing the Transitive Property

Conclusion

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Animated Solution for Mathematics - Sets and Relations: The relation $R = {(x, y) : x, y \in Z and x + y is even}$ is:

Introduction to the Relation $R$

Isolating $x + z$

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Define a relation R over a class of $n \times n$ real matrices A and B as "ARB iff there exists a non-singular matrix P such that $P A P^{- 1} = B$ ". Then which of the following is true?

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