Let W denote the words in the English dictionary. Define the relation R by $R = \{(x, y) \in W \times W \mid \text{the words x and y have at least one letter in common.}\}$ Then R is Options: (A) not reflexive, symmetric and transitive (B) reflexive, symmetric and not transitive (C) reflexive, symmetric and transitive (D) reflexive, not symmetric and transitive

### Analyzing the Setup We are examining a set $W$, which consists of all words in the English dictionary. We define a relation $R$ on $W$ such that $(x, y) \in R$ if and only if $x \cap y \neq \emptyset$. In this context, the intersection $x \cap y$ represents the set of common letters between word $x$ and word $y$. The condition $x \cap y \neq \emptyset$ implies that the two words share at least one common letter. ### The Mirror of Reflexivity To determine if the relation is reflexive, we must check if $(x, x) \in R$ for every word $x \in W$. By the definition of the relation, this requires $x \cap x \neq \emptyset$. Since every word $x$ in the dictionary contains at least one letter, the set of letters in $x$ is non-empty. Therefore, $x \cap x = x \neq \emptyset$. Because this condition holds for every word in the set $W$, the relation is **reflexive**. ### The Mutual Bond of Symmetry Next, we test for symmetry. We assume $(x, y) \in R$, which means $x \cap y \neq \emptyset$. Because the intersection of two sets is commutative, we know that: $$x \cap y = y \cap x$$ Since $x \cap y \neq \emptyset$, it follows that $y \cap x \neq \emptyset$. This satisfies the condition for $(y, x) \in R$. Thus, the relation is **symmetric**. ### The Broken Chain of Transitivity Finally, we investigate transitivity. We must determine if $(x, y) \in R$ and $(y, z) \in R$ necessarily implies $(x, z) \in R$. Consider the following counterexample: Let $x = \text{"cat"}$, $y = \text{"apple"}$, and $z = \text{"pen"}$. 1. $x \cap y = \{a\} \neq \emptyset$, so $(x, y) \in R$. 2. $y \cap z = \{p, e\} \neq \emptyset$, so $(y, z) \in R$. However, checking $x$ and $z$: $$x \cap z = \{c, a, t\} \cap \{p, e, n\} = \emptyset$$ Since $x \cap z = \emptyset$, it follows that $(x, z) \notin R$. Because the chain of connection is broken, the relation is **not transitive**. ### The Final Verdict We have evaluated the three properties of the relation $R$ on the set $W$. We have determined that the relation is **reflexive**, **symmetric**, and **not transitive**. This classification corresponds to **Option 2**. The logic confirms that while the relation captures the mutual nature of sharing letters, it fails to maintain that connection across a sequence of words.

JEE Main 2006

LEVELJEE Main

Animated Solution for Mathematics - Sets and Relations: Let W denote the words in the English dictionary. Define the relation R by $R = {(x, y) \in W \times W ∣ the words x and y have at least one letter in common.}$ Then R is

Select Answer:

Visualized Solution

Defining the Relation $R$ on Word Set $W$

Let $W$ be the set of all words in the English dictionary.
Relation $R$ is defined on $W$ such that two words $x$ and $y$ are related if they share at least one letter.
Mathematically, $(x, y) \in R ⟺ x \cap y \neq = \emptyset$ .

Testing for Reflexivity: Does $(x, x) \in R$ ?

A relation $R$ on a set $W$ is reflexive if every element is related to itself.
That is, for every word $x \in W$ , we must have $(x, x) \in R$ .
This means we need to check if a word $x$ shares at least one letter with itself.

Reflexivity Confirmed

Every word in the English dictionary contains at least one letter (no empty words exist in $W$ ).
Therefore, the set of letters in $x$ intersected with itself is just $x$ itself: $x \cap x = x \neq = \emptyset$ .
Since $x \cap x \neq = \emptyset$ , we have $(x, x) \in R$ for all $x \in W$ .
Thus, the relation $R$ is reflexive.

Testing for Symmetry: If $(x, y) \in R$ , is $(y, x) \in R$ ?

A relation $R$ is symmetric if whenever $(x, y) \in R$ , it must also be true that $(y, x) \in R$ .
In terms of our word relation, if word $x$ shares a letter with word $y$ , does word $y$ share a letter with word $x$ ?

Symmetry Confirmed

Assume $(x, y) \in R$ . This means $x \cap y \neq = \emptyset$ .
By the commutative property of set intersection, we know that $x \cap y = y \cap x$ .
Therefore, $y \cap x \neq = \emptyset$ , which directly implies $(y, x) \in R$ .
Thus, the relation $R$ is symmetric.

Testing for Transitivity: The Chain Rule

A relation $R$ is transitive if whenever $(x, y) \in R$ and $(y, z) \in R$ , it must follow that $(x, z) \in R$ .
For our relation, if word $x$ shares a letter with word $y$ , and word $y$ shares a letter with word $z$ , must word $x$ share a letter with word $z$ ?

Constructing a Counterexample

To disprove transitivity, we only need to find one single counterexample.
Let's choose three simple words:
- $x = cat$ (letters: ${c, a, t}$ )
- $y = apple$ (letters: ${a, p, l, e}$ )
- $z = pen$ (letters: ${p, e, n}$ )
Let's check the relations between these pairs.

Analyzing the Overlaps

$x = cat$ and $y = apple$ share the letter 'a'. Thus, $x \cap y = {a} \neq = \emptyset ⟹ (x, y) \in R$ .
$y = apple$ and $z = pen$ share the letters 'p' and 'e'. Thus, $y \cap z = {p, e} \neq = \emptyset ⟹ (y, z) \in R$ .
However, $x = cat$ and $z = pen$ share absolutely no letters: $x \cap z = \emptyset$ .
Therefore, $(x, z) \in / R$ .

Final Verdict on Relation $R$

We have established that the relation $R$ is:
1. Reflexive (every word shares letters with itself)
2. Symmetric (sharing letters is a mutual property)
3. Not Transitive (as shown by the cat-apple-pen counterexample)
Therefore, $R$ is reflexive, symmetric and not transitive.
This corresponds to Option 2.

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The Sigma Insight: Types of Relations

Solution Diagram

Analyzing the Setup

We are examining a set

W

, which consists of all words in the English dictionary. We define a relation

R

W

such that

(x, y) \in R

if and only if $x \cap y eq \emptyset$.

In this context, the intersection

x \cap y

represents the set of common letters between word

x

and word

y

. The condition $x \cap y eq \emptyset$ implies that the two words share at least one common letter.

The Mirror of Reflexivity

To determine if the relation is reflexive, we must check if

(x, x) \in R

for every word

x \in W

. By the definition of the relation, this requires $x \cap x eq \emptyset$.

Since every word

x

in the dictionary contains at least one letter, the set of letters in

x

is non-empty. Therefore, $x \cap x = x eq \emptyset$.

Because this condition holds for every word in the set

W

, the relation is reflexive.

The Mutual Bond of Symmetry

Next, we test for symmetry. We assume

(x, y) \in R

, which means $x \cap y eq \emptyset$.

Because the intersection of two sets is commutative, we know that:

x \cap y = y \cap x

Since $x \cap y eq \emptyset$, it follows that $y \cap x eq \emptyset$. This satisfies the condition for

(y, x) \in R

. Thus, the relation is symmetric.

The Broken Chain of Transitivity

Finally, we investigate transitivity. We must determine if

(x, y) \in R

and

(y, z) \in R

necessarily implies

(x, z) \in R

Consider the following counterexample: Let

x = "cat"

y = "apple"

, and

z = "pen"

1. $x \cap y = \{a\} eq \emptyset$, so

(x, y) \in R

. 2. $y \cap z = \{p, e\} eq \emptyset$, so

(y, z) \in R

However, checking

x

and

z

x \cap z = {c, a, t} \cap {p, e, n} = \emptyset

Since

x \cap z = \emptyset

, it follows that $(x, z) otin R$. Because the chain of connection is broken, the relation is not transitive.

The Final Verdict

We have evaluated the three properties of the relation

R

on the set

W

. We have determined that the relation is reflexive, symmetric, and not transitive.

This classification corresponds to Option 2. The logic confirms that while the relation captures the mutual nature of sharing letters, it fails to maintain that connection across a sequence of words.

Common Traps & FAQs

Wait, why do we assume the set of words W doesn't contain an empty word? What if it did?

That is a brilliant observation! In formal set theory, if the set

W

contained an empty string (a word with zero letters), the relation would fail to be reflexive because the empty set has no letters to share with itself. However, in the context of the 'English dictionary,' we define

W

as the set of actual words, all of which possess at least one character. Always check your domain definitions!

Why can't we just say that if they all share letters, they must be transitive?

It is tempting to think that 'sharing' is a transitive property, but it isn't! Transitivity requires a common thread that persists through the entire chain. If word

x

shares 'a' with

y

, and

y

shares 'p' with

z

, the letter 'a' is lost. The relation only requires *at least one* letter to be common, not the *same* letter.

Is there a trick to finding counterexamples for transitivity?

Yes! The best strategy is to look for 'disjoint' sets. Pick a word

x

and a word

z

that have absolutely nothing in common. Then, try to find a 'middle' word

y

that acts as a bridge by sharing a letter with

x

and a different letter with

z

. Once you find that bridge, you have successfully broken the transitivity chain.

Does the order of the words matter for symmetry?

Not at all! Because the relation is defined by the intersection of sets, and set intersection is commutative (

x \cap y = y \cap x

), the relationship is inherently mutual. If you are related to me, I am automatically related to you.

How do I know if I've checked enough cases for transitivity?

In mathematics, you don't need to check 'enough' cases to prove transitivity—you need a formal proof. But to *disprove* it, you only need one single counterexample. If you find one instance where the chain breaks, the entire relation fails the transitive test. One 'no' is enough to defeat a universal 'yes'!

Animated Solution for Mathematics - Sets and Relations: Let W denote the words in the English dictionary. Define the relation R by $R = {(x, y) \in W \times W ∣ the words x and y have at least one letter in common.}$ Then R is

Select Answer:

Visualized Solution

Defining the Relation $R$ on Word Set $W$

Testing for Reflexivity: Does $(x, x) \in R$ ?

Reflexivity Confirmed

Testing for Symmetry: If $(x, y) \in R$ , is $(y, x) \in R$ ?

Symmetry Confirmed

Testing for Transitivity: The Chain Rule

Constructing a Counterexample

Analyzing the Overlaps

Final Verdict on Relation $R$

The Sigma Insight: Types of Relations

Analyzing the Setup

The Mirror of Reflexivity

The Mutual Bond of Symmetry

The Broken Chain of Transitivity

The Final Verdict

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Visualized Solution

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

Testing for Reflexivity: Does .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x,x)∈R?

Reflexivity Confirmed

Testing for Symmetry: If .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (x,y)∈R, is .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (y,x)∈R?

Symmetry Confirmed

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

The Sigma Insight: Types of Relations

Analyzing the Setup

The Mirror of Reflexivity

The Mutual Bond of Symmetry

The Broken Chain of Transitivity

The Final Verdict

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