Problem 26 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(W\) denote the words in the English dictionary. Define the relation \(R\) by: \([2006]\) \(R=\\{(x, y) \in W \times W \mid\) the words \(x\) and \(y\) have at least one letter in common \(\\}\). Then \(R\) is (A) not reflexive, symmetric and transitive (B) reflexive, symmetric and not transitive (C) reflexive, symmetric and transitive (D) reflexive, not symmetric and transitive

Step-by-Step Solution

Answer

The relation \(R\) is reflexive, symmetric, and not transitive. Answer: (B)

Step 1: Understanding Reflexivity

A relation is reflexive if every element is related to itself. For any word \(x\), the word shares at least one letter with itself, since they are the same word. Therefore, \((x, x) \in R\) for all words \(x\). Thus, the relation \(R\) is reflexive.

Step 2: Checking Symmetry

A relation is symmetric if for every \( (x, y) \in R \), \( (y, x) \in R \) also holds. If words \(x\) and \(y\) have at least one letter in common, then \(y\) and \(x\) must also have this letter in common. Thus, \(R\) is symmetric.

Step 3: Assessing Transitivity

A relation is transitive if whenever \((x, y) \in R\) and \((y, z) \in R\), then \((x, z) \in R\). Consider words where \((x, y)\) have a letter 'a' in common, and \((y, z)\) have a letter 'b' in common. If 'a' and 'b' are different and \(x\) and \(z\) have no letter in common, \((x, z)\) is not in \(R\). Therefore, \(R\) is not transitive.

Key Concepts

Reflexive Relations

A reflexive relation is one where every element is related to itself.
In the case of the relation \( R \) involving words from the English dictionary, we consider each word as an element.
To determine if \( R \) is reflexive, we need to check if each word \( x \) has at least one letter in common with itself.

This condition is naturally satisfied because any word \( x \) will obviously contain all its letters.
Thus, for any word \( x \), the pair \((x, x)\) will always be part of the relation \( R \). Therefore, the relation \( R \) is reflexive.

This means that no matter what word we pick, it will always relate to itself in this context.
This property is quite intuitive because a reflection bears the same identity.

Symmetric Relations

A symmetric relation is one where if \((x, y)\) is in the relation, then \((y, x)\) must also be in the relation.
This means the relationship works both ways between any two elements.
In the context of the English dictionary words relation \( R \), if two words \( x \) and \( y \) share at least one letter, then swapping the words should not affect the relation.

Let’s say \( x \) and \( y \) share the letter 'e'.
If \((x, y)\) is a pair in the relation because they share 'e', then flipping it to \((y, x)\) doesn't change the fact that they still share the letter 'e'.
This fulfills the symmetric condition, making \( R \) symmetric.

In simpler terms, the bond of sharing a common letter is mutual, no matter the order of the words.

Transitive Relations

Transitive relations involve a little more complexity.
A relation \( R \) is transitive if whenever \((x, y)\) and \((y, z)\) are in \( R \), then \((x, z)\) must also be in \( R \).

For our word relation \( R \), consider this scenario: if word \( x \) and word \( y \) share, say, the letter 'p', and word \( y \) and word \( z \) share the letter 'q', \((x, y)\) and \((y, z)\) are both part of \( R \).

However, unless word \( x \) and word \( z \) share at least one letter themselves, \((x, z)\) will not be in \( R \).
This breaks the transitive condition.
Such scenarios make the relation \( R \) non-transitive as there's no guarantee of a connecting shared letter between all involved words.

This explanation shows that while words might connect in a "friend of a friend" way sharing letters through a mutual contact, this does not establish a direct connection unless a common letter is proved between the starting and ending word.