Problem 123 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Consider the following relations: \(R=\\{(x, y) \mid x, y\) are real numbers and \(x=w y\) for some rational number \(w\\} ;\) \(S=\left\\{\begin{array}{l}\left(\frac{m}{p}, \frac{p}{q}\right) m, n, p \text { and } q \in \mathbf{Z} \\ \text { such that } n, q \neq 0 \text { and } q m=p n\end{array}\right\\}\) Then [2010] (A) neither \(R\) nor \(S\) is an equivalence relation (B) \(S\) is an equivalence relation but \(R\) is not an equivalence relation (C) \(R\) and \(S\) both are equivalence relations (D) \(R\) is an equivalence relation but \(S\) is not an equivalence relation

Step-by-Step Solution

Answer

Option (D) is correct: \( R \) is an equivalence relation but \( S \) is not.

Step 1: Check if R is Reflexive

For a relation to be reflexive, for every element \( x \), the pair \( (x, x) \) must exist in \( R \). For \( (x, x) \) to be in \( R \), \( x = wx \) must hold for some rational number \( w \). This is possible when \( w = 1 \), which is a rational number. Hence, \( R \) is reflexive.

Step 2: Check if R is Symmetric

For a relation to be symmetric, if \( (x, y) \in R \), then \( (y, x) \) must also be in \( R \). If \( x = wy \) for some rational \( w \), then \( y = \frac{x}{w} \). Since the inverse of a rational number is also rational, \( y = x/z \) where \( z \) is rational, making \( (y,x) \in R \). Hence, \( R \) is symmetric.

Step 3: Check if R is Transitive

For a relation to be transitive, if \( (x, y) \in R \) and \( (y, z) \in R \), then \( (x, z) \) must also be in \( R \). If \( x = wy \) and \( y = uz \), then \( x = wuz \). Since the product of two rationals \( w \) and \( u \) is also rational, \( (x, z) \) is in \( R \), making \( R \) transitive.

Step 4: Determine if R is an Equivalence Relation

Since \( R \) is reflexive, symmetric, and transitive, \( R \) is an equivalence relation.

Step 5: Check if S is Reflexive

To be reflexive, for every pair \( (a, a) \), \( (a, a) \) must be in \( S \). If \( a = \frac{m}{p} \), then we need \( q m = p n \). Not all \( a = \frac{m}{p} \) satisfy this with just \((a, a)\). Therefore, \( S \) is not reflexive.

Step 6: Check if S is Symmetric

For \( S \) to be symmetric, if \( (\frac{m}{p}, \frac{p}{q}) \in\ S \), \( (\frac{p}{q}, \frac{m}{p}) \) should also be in \( S \). With \( qm = pn \), reversing roles leads to \( qn = mp \), which is not guaranteed, thus \( S \) is not symmetric.

Step 7: Determine if S is an Equivalence Relation

Since \( S \) is neither reflexive nor symmetric, \( S \) is not an equivalence relation.

Step 8: Identify the Correct Option

Based on previous steps, \( R \) is an equivalence relation and \( S \) is not. Therefore, the correct answer is option (D) \( R \) is an equivalence relation but \( S \) is not an equivalence relation.

Key Concepts

Reflexive Relation

A reflexive relation is a type of relation that every element must have a loop back to itself. This means for a set to be reflexive, every element \( x \) in a set is related to itself, represented as \( (x, x) \).

• For instance, consider a relation \( R \) on a set where a relation is defined as \( x = w x \) for some rational number \( w \).
• To find if \( R \) is reflexive, check if \( (x, x) \) is always present. This occurs if \( w \) can equal 1, which maintains \( x = x \).
• So, for every \( x \), there exists a rational \( w \) (in this case, 1) making \( R \) reflexive as every element is indeed related to itself.

This ensures that reflexivity is a simple way of ensuring each element is individually centered in the set, covering its own relation.

Symmetric Relation

A symmetric relation is a straightforward concept where if one element is related to another, the reverse must also hold true. In simpler terms, for a relation \( R \) to be symmetric, if \( (x, y) \) is in \( R \), then \( (y, x) \) must also be in \( R \).

• Take, for example, numbers \( x \) and \( y \) connected by \( x = w y \) for a rational number \( w \).
• If you flip the pair as \( (y, x) \), this condition leads to \( y = \frac{x}{w} \).
• The key here is remembering that the inverse of a rational number (like \( \frac{1}{w} \)) is also rational. Hence, the pair \( (y, x) \) is maintained in \( R \).

Thus, symmetry in a relation ensures that if you can go one way around an element pair, you can also go back the other way.

Transitive Relation

A transitive relation holds when a direct connection between elements implies an indirect connection. This might sound more complex, but it is just a chain rule. For a relation \( R \) to be transitive, if \( (x, y) \) and \( (y, z) \) are in \( R \), then \( (x, z) \) must also be in \( R \).

• Consider that \( x \) related to \( y \) and \( y \) related to \( z \) implies \( x = w y \) and \( y = u z \).
• The multiplication of two rational numbers \( w \) and \( u \) results in another rational number, let’s denote this as \( w u \).
• Thus, \( x = w u z \) satisfies the condition, and \( (x, z) \) appears in \( R \).

Transitivity binds elements into a coherent structure, closing gaps between chains of relationships within the set, ensuring a continuous link among all elements linked indirectly.