Problem 24 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let \(R=\\{(1,3),(4,2),(2,4),(2,3),(3,1)\\}\) be a relation on the set \(A=\\{1,2,3,4\\} .\) The relation \(R\) is \(\quad\) [2004] (A) a function (B) reflexive (C) not symmetric (D) transitive

Step-by-Step Solution

Answer

The relation is not symmetric.

Step 1: Understanding the Relation

The relation \( R \) is given by \( R = \{ (1,3), (4,2), (2,4), (2,3), (3,1) \} \). A relation is a set of ordered pairs from set \( A \), where \( A = \{ 1, 2, 3, 4 \} \). Each ordered pair consists of an element from the domain and an element from the codomain.

Step 2: Checking if R is a Function

A relation is a function if every element in the domain has exactly one image in the codomain. Here, the domain \( \{ 1, 2, 3, 4 \} \) associates with multiple images: both \( (2,4) \) and \( (2,3) \) include the element 2. Therefore, \( R \) is not a function.

Step 3: Checking if R is Reflexive

A relation is reflexive on set \( A \) if every element in \( A \) is related to itself, i.e., \( (1,1), (2,2), (3,3), (4,4) \) must all be in \( R \). None of these pairs are in \( R \), so \( R \) is not reflexive.

Step 4: Checking if R is Symmetric

A relation \( R \) is symmetric if for every \( (a,b) \in R \), there exists \( (b,a) \in R \). Check the pairs: \( (1,3) \) is present, but \( (3,1) \) is also present. However, \( (2,4) \) and \( (4,2) \) are both present. Thus, \( R \) is not symmetric.

Step 5: Checking if R is Transitive

A relation is transitive if whenever \( (a,b) \in R \) and \( (b,c) \in R \), then \( (a,c) \) must also be in \( R \). Checking the pairs, we find that for \( (1,3) \) and \( (3,1) \), \( (1,1) \) does not have to be present as it's not included. But, \( (2,4) \) and \( (4,2) \) imply \( (2,2) \), which is not there, so the relation is not transitive in all cases.

Key Concepts

Reflexive Relation

In set theory, a relation is termed reflexive when each element is related to itself. Let's take a closer look at what that means:

A reflexive relation on a set \( A \) must fulfill a specific condition: for every element \( x \) in \( A \), the pair \( (x, x) \) must be part of the relation \( R \). In simpler words, every element must loop back to itself.

For example, if our set is \( A = \{1, 2, 3, 4\} \), then for the relation \( R \) to be reflexive, it needs to include the pairs:

• \((1, 1)\)
• \((2, 2)\)
• \((3, 3)\)
• \((4, 4)\)

Without these, the relation isn't reflexive. In our exercise, because none of these pairs are present in \( R \), it is clear that \( R \) is not reflexive.

Symmetric Relation

A symmetric relation holds a fun twist. For every pair \((a, b)\) in the relation, there must also be the pair \((b, a)\). Let's break this down further.

If a relation is symmetric, it must show mutual pairs. Check for a reflection: if \((1, 3)\) is part of \( R \), then you should also find \((3, 1)\) in \( R \) as proof of symmetry.

In our example relation \( R = \{(1, 3), (4, 2), (2, 4), (2, 3), (3, 1)\} \), we can check:

• \((1, 3)\) is present and \((3, 1)\) is, interestingly, present too.
• \((2, 4)\), but no matching \((4, 2)\).

Although some pairs have reflections, not all do. As a result, \( R \) is determined to be not symmetric by these few exceptions.

Transitive Relation

Moving on to transitive relations, let’s see what makes a relation transitive.

A relation is transitive if whenever it includes triples \((a, b)\) and \((b, c)\), it also contains \((a, c)\). In essence, transitivity connects dots between related elements broadly.

Here’s how this concept works in \( R \):

Given \((2, 4)\) and \((4, 2)\) in \( R \), for \( R \) to be transitive, it should also include \((2, 2)\). Unfortunately, this is not the case here.

Another check with the pair \((1, 3)\) and pair \((3, 1)\) could intuitively require \((1, 1)\), but \( R \) lacks such a link as well.

In conclusion, our relation lacks the consistency required for transitivity, so it is not transitive.