JEE Main 2010

LEVELJEE Main

Animated Solution for Mathematics - Sets and Relations: Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; S = {( $\frac{m}{n}, \frac{p}{q}$ ) | m, n, p and q are integers such that $n, q \neq = 0$ and qm = pn}. Then

Q: Wait, why is $R$ not symmetric? If $x = wy$, can't we just say $y = x/w$?

That is a common trap! While $y = x/w$ works when $w \neq 0$, the definition of $R$ requires $w$ to be a rational number. If $x=0$ and $y=1$, we have $0 = 0 \cdot 1$, so $w=0$. But for the symmetric pair $(1, 0)$, we would need $1 = w' \cdot 0$. No rational number $w'$ can satisfy this because $w' \cdot 0$ is always $0$. Thus, symmetry fails.

Select Answer:

Progressive Hints

Hint 1

To determine if a relation is an equivalence relation, you must verify three distinct properties: Reflexivity, Symmetry, and Transitivity. Think of these as the three pillars of a bridge; if one collapses, the entire structure fails. Start by testing each property one by one for both relations.

Hint 2

For Relation

R

, pay close attention to the Symmetry property. If

(x, y) \in R

, then

x = w y

for some rational

w

. To be symmetric, we need

(y, x) \in R

, meaning

y = w^{'} x

for some rational

w^{'}

. What happens if

x = 0

and

y = 1

? Does a rational

w^{'}

exist such that

1 = w^{'} \cdot 0

Hint 3

For Relation

S

, look at the condition

q m = p n

. If you divide both sides by

n q

(given

n, q \neq = 0

), what does the equation become? It reveals the true nature of the relation. Once you identify it as simple equality, the properties of reflexivity, symmetry, and transitivity become trivial to verify.

Visualized Solution

The Equivalence Trinity

A relation is an equivalence relation if it satisfies three properties:
Reflexive: $a \sim a$
Symmetric: $a \sim b ⟹ b \sim a$
Transitive: $a \sim b$ and $b \sim c ⟹ a \sim c$

Defining Relation $R$

Relation $R$ is defined for real numbers $x, y$ .
Condition: $x = w y$ for some rational number $w \in Q$ .

Reflexivity of $R$

For reflexivity, we need $(x, x) \in R$ for all $x \in R$ .
This means $x = w x$ for some $w \in Q$ .
If we choose $w = 1$ , then $x = 1 \cdot x$ is always true.
Since $1 \in Q$ , $R$ is Reflexive.

Symmetry of $R$ : The Setup

For symmetry, if $(x, y) \in R$ , then $(y, x)$ must also be in $R$ .
Let's pick a specific pair to test.
Let $x = 0$ and $y = 1$ .
Since $0 = 0 \cdot 1$ and $0 \in Q$ , the pair $(0, 1)$ is in $R$ .

Symmetry of $R$ : The Failure

Now we must check if $(1, 0) \in R$ .
We need $1 = w \cdot 0$ for some $w \in Q$ .
But $w \cdot 0 = 0$ for any $w$ .
So, $1 = 0$ , which is impossible!
Therefore, $(1, 0) \in / R$ .

Conclusion for Relation $R$

Since $(0, 1) \in R$ but $(1, 0) \in / R$ , symmetry fails.
If even one property fails, it is not an equivalence relation.

Defining Relation $S$

Relation $S$ is defined on pairs of rational numbers $(\frac{m}{n}, \frac{p}{q})$ .
Condition: $q \cdot m = p \cdot n$ , where $n, q \neq = 0$ .

Decoding Relation $S$

The condition $q m = p n$ can be rearranged.
Dividing both sides by $n q$ , we get $\frac{m}{n} = \frac{p}{q}$ .
This simply means the two rational numbers are equal.

Reflexivity of $S$

Is a rational number equal to itself?
Yes, $\frac{m}{n} = \frac{m}{n}$ is always true.
Thus, $S$ is Reflexive.

Symmetry of $S$

If $\frac{m}{n} = \frac{p}{q}$ , does $\frac{p}{q} = \frac{m}{n}$ ?
Yes, equality is symmetric.
Thus, $S$ is Symmetric.

Transitivity of $S$

If $\frac{m}{n} = \frac{p}{q}$ and $\frac{p}{q} = \frac{r}{s}$ .
Then logically, $\frac{m}{n} = \frac{r}{s}$ .
Thus, $S$ is Transitive.

Final Verdict

$S$ is Reflexive, Symmetric, and Transitive.
Therefore, $S$ is an equivalence relation.
Final Answer: $S$ is an equivalence relation but $R$ is not.

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The Sigma Insight: Types of Relations

Solution Diagram

The Equivalence Trinity

A Mathematical Journey

Welcome, future engineers! Today, we are going to demystify the concept of equivalence relations.

In the world of set theory, an equivalence relation is like a VIP club. To gain entry, a relation must satisfy three strict, non-negotiable rules: Reflexivity, Symmetry, and Transitivity.

If a relation fails even one of these, it is barred from the club. Let us analyze our two candidates,

R

and

S

, to see if they make the cut.

Analyzing Relation $R$

The Trap of Zero

We define

R

on real numbers

x

and

y

such that

x = w y

for some rational number

w

First, let us test for reflexivity. Does

x

relate to itself? We need

x = w x

for all

x \in R

If we choose

w = 1

, which is a rational number, the equation

x = 1 \cdot x

holds true for every real number. So,

R

is reflexive.

Now, for symmetry. If

(x, y) \in R

, then

(y, x)

must also be in

R

Let us test with

x = 0

and

y = 1

. Since

0 = 0 \cdot 1

and

0 \in Q

, the pair

(0, 1)

is in

R

But for symmetry, the pair

(1, 0)

must also be in

R

. This requires

1 = w \cdot 0

for some rational

w

As we know, any number multiplied by zero is zero, so

1 = 0

is impossible. Symmetry fails! Since symmetry is broken,

R

is not an equivalence relation.

Analyzing Relation $S$

The Hidden Equality

Now, let us look at

S

, defined on pairs of rational numbers

(\frac{m}{n}, \frac{p}{q})

with the condition

q m = p n

This looks intimidating, but let us simplify it. If we divide both sides by

n q

(given $n, q eq 0$), we get:

\frac{m}{n} = \frac{p}{q}

This is simply the definition of equality for fractions! Once we see that

S

is just equality, the rest is easy.

Equality is reflexive (

\frac{m}{n} = \frac{m}{n}

), symmetric (if

\frac{m}{n} = \frac{p}{q}

, then

\frac{p}{q} = \frac{m}{n}

), and transitive (if

\frac{m}{n} = \frac{p}{q}

and

\frac{p}{q} = \frac{r}{s}

, then

\frac{m}{n} = \frac{r}{s}

Thus, $S$ is an equivalence relation.

The Final Verdict

We have successfully navigated the logic.

R

failed the symmetry test, while

S

proved to be a standard equality relation.

Therefore, $S$ is an equivalence relation, but $R$ is not. Keep practicing this systematic approach, and you will master these concepts in no time!

Common Traps & FAQs

Wait, why is

R

not symmetric? If

x = w y

, can't we just say

y = x / w

That is a common trap! While

y = x / w

works when

w \neq = 0

, the definition of

R

requires

w

to be a rational number. If

x = 0

and

y = 1

, we have

0 = 0 \cdot 1

, so

w = 0

. But for the symmetric pair

(1, 0)

, we would need

1 = w^{'} \cdot 0

. No rational number

w^{'}

can satisfy this because

w^{'} \cdot 0

is always

0

. Thus, symmetry fails.

S

just equality? That seems too simple.

Exactly! In mathematics, we often define concepts in complex ways to test if you can see through the fog. The condition

q m = p n

is the fundamental definition of equality for fractions

\frac{m}{n}

and

\frac{p}{q}

. Once you strip away the notation, you are just checking if equality is an equivalence relation, which it always is.

Do I have to check all three properties for

R

if I find one that fails?

Great question! In a formal proof, yes, you should check all. But in a competitive exam like JEE, the moment you find a single counterexample for any of the three properties, you can immediately conclude that the relation is not an equivalence relation. Efficiency is key!

Animated Solution for Mathematics - Sets and Relations: Consider the following relations: R = {(x, y) | x, y are real numbers and x = wy for some rational number w}; S = {( $\frac{m}{n}, \frac{p}{q}$ ) | m, n, p and q are integers such that $n, q \neq = 0$ and qm = pn}. Then

Select Answer:

Visualized Solution

The Equivalence Trinity

Defining Relation $R$

Reflexivity of $R$

Symmetry of $R$ : The Setup

Symmetry of $R$ : The Failure

Conclusion for Relation $R$

Defining Relation $S$

Decoding Relation $S$

Reflexivity of $S$

Symmetry of $S$

Transitivity of $S$

Final Verdict

The Sigma Insight: Types of Relations

The Equivalence Trinity

Analyzing Relation $R$

Analyzing Relation $S$

The Final Verdict

Similar Questions

Let R be the set of real numbers. Statement-1: $A = {(x, y) \in R \times R : y - x is an integer}$ is an equivalence relation on R. Statement-2: $B = {(x, y) \in R \times R : x = α y for some rational number α}$ is an equivalence relation on R.

Let $R$ be the real line. Consider the following subsets of the plane $R \times R$ : $S = {(x, y) : y = x + 1 and 0 < x < 2}$ , $T = {(x, y) : x - y is an integer}$ . Which one of the following is true?

Let $R_{1} = {(a, b) \in N \times N : ∣ a - b ∣ \leq 13}$ and $R_{2} = {(a, b) \in N \times N : ∣ a - b ∣ \neq = 13}$ . Then on $N$ :

Let $R$ be a relation on $R$ , given by $R = {(a, b) : 3 a - 3 b + 7 is an irrational number}$ . Then $R$ is

Which of the following is not correct for relation $R$ on the set of real numbers?

Let $N$ be the set of natural numbers and a relation $R$ on $N$ be defined by $R = {(x, y) \in N \times N : x^{3} - 3 x^{2} y - x y^{2} + 3 y^{3} = 0}$ . Then the relation $R$ is :

Let a relation $R$ on $N \times N$ be defined as: $(x_{1}, y_{1}) R (x_{2}, y_{2})$ if and only if $x_{1} \leq x_{2}$ or $y_{1} \leq y_{2}$ . Consider the two statements: (I) $R$ is reflexive but not symmetric. (II) $R$ is transitive Then which one of the following is true?

Let $R_{1}$ and $R_{2}$ be two relation defined as follows: $R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}$ and $R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \in / Q}$ , where $Q$ is the set of all rational numbers. Then:

Let $A = {1, 2, 3, 4, 5, 6, 7}$ . Then the relation $R = {(x, y) \in A \times A : x + y = 7}$ is

Let $R$ be a relation on $N \times N$ defined by $(a, b) R (c, d)$ if and only if $a d (b - c) = b c (a - d)$ . Then $R$ is

Select Answer:

Visualized Solution

The Equivalence Trinity

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

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

Symmetry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R: The Setup

Symmetry of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R: The Failure

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

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

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

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

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

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

Final Verdict

The Sigma Insight: Types of Relations

The Equivalence Trinity

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

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

The Final Verdict

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R be the real line. Consider the following subsets of the plane R×R: S={(x,y):y=x+1 and 0<x<2}, T={(x,y):x−y is an integer}. Which one of the following is true?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R1​={(a,b)∈N×N:∣a−b∣≤13} and R2​={(a,b)∈N×N:∣a−b∣=13}. Then on N:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R be a relation on R, given by R={(a,b):3a−3b+7​ is an irrational number}. Then R is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Which of the following is *not* correct for relation R on the set of real numbers?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let N be the set of natural numbers and a relation R on N be defined by R={(x,y)∈N×N:x3−3x2y−xy2+3y3=0}. Then the relation R is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R1​ and R2​ be two relation defined as follows: R1​={(a,b)∈R2:a2+b2∈Q} and R2​={(a,b)∈R2:a2+b2∈/Q}, where Q is the set of all rational numbers. Then:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={1,2,3,4,5,6,7}. Then the relation R={(x,y)∈A×A:x+y=7} is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let R be a relation on N×N defined by (a,b)R(c,d) if and only if ad(b−c)=bc(a−d). Then R is

Defining Relation $R$

Reflexivity of $R$

Symmetry of $R$ : The Setup

Symmetry of $R$ : The Failure

Conclusion for Relation $R$

Defining Relation $S$

Decoding Relation $S$

Reflexivity of $S$

Symmetry of $S$

Transitivity of $S$

Analyzing Relation $R$

Analyzing Relation $S$

Let $R$ be the real line. Consider the following subsets of the plane $R \times R$ : $S = {(x, y) : y = x + 1 and 0 < x < 2}$ , $T = {(x, y) : x - y is an integer}$ . Which one of the following is true?

Let $R_{1} = {(a, b) \in N \times N : ∣ a - b ∣ \leq 13}$ and $R_{2} = {(a, b) \in N \times N : ∣ a - b ∣ \neq = 13}$ . Then on $N$ :

Let $R$ be a relation on $R$ , given by $R = {(a, b) : 3 a - 3 b + 7 is an irrational number}$ . Then $R$ is

Which of the following is not correct for relation $R$ on the set of real numbers?

Let $N$ be the set of natural numbers and a relation $R$ on $N$ be defined by $R = {(x, y) \in N \times N : x^{3} - 3 x^{2} y - x y^{2} + 3 y^{3} = 0}$ . Then the relation $R$ is :

Let $R_{1}$ and $R_{2}$ be two relation defined as follows: $R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}$ and $R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \in / Q}$ , where $Q$ is the set of all rational numbers. Then:

Let $A = {1, 2, 3, 4, 5, 6, 7}$ . Then the relation $R = {(x, y) \in A \times A : x + y = 7}$ is

Let $R$ be a relation on $N \times N$ defined by $(a, b) R (c, d)$ if and only if $a d (b - c) = b c (a - d)$ . Then $R$ is