Problem 6 - Calculus for JEE ADVANCED

Question

(1) \(|a-a|=0<1 \therefore a R a \forall a \in R\) Therefore, \(R\) is reflexive. Again \(a R b \Rightarrow|a-b| \leq \mid\) and \(|b-a| \leq \mid \Rightarrow b R a\) Therefore, \(R\) is symmetric. Therefore, \(R\) is not anti-symmetric. Further, \(1 R 2\) and \(2 R 3\) but \(1 \not R 3,[\because|1-3|=2>1]\) Therefore, \(R\) is not transitive.

Step-by-Step Solution

Answer

The relation is reflexive and symmetric, but not anti-symmetric or transitive.

Step 1: Understanding Reflexivity

A relation \( R \) is reflexive if each element is related to itself. For all \( a \in \mathbb{R} \), we check \( |a-a| = 0 < 1 \). This condition is satisfied for all real numbers \( a \), hence \( a R a \) for all \( a \), making the relation reflexive.

Step 2: Checking Symmetry

A relation is symmetric if for all \( a, b \), whenever \( a R b \), then \( b R a \). Given \( |a-b|<1 \), we have \( |b-a|<1 \) as the absolute value is the same for \( a-b \) and \( b-a \). Thus, if \( a R b \), it follows that \( b R a \), making the relation symmetric.

Step 3: Assessing Anti-symmetry

A relation is anti-symmetric if, for all \( a, b \), whenever \( a R b \) and \( b R a \), then \( a = b \). In this case, even if \( a R b \) and \( b R a \), \( a \) and \( b \) do not have to be equal, demonstrating that the relation is not anti-symmetric.

Step 4: Evaluating Transitivity

A relation is transitive if for all \( a, b, c \), whenever \( a R b \) and \( b R c \), then \( a R c \). However, for example \( 1 R 2 \) and \( 2 R 3 \), since \( |1-2|<1 \) and \( |2-3|<1 \). But \( |1-3| = 2 > 1 \), so \( 1 ot R 3 \), indicating the relation is not transitive.

Key Concepts

Reflexivity

A relation is reflexive if every element is related to itself. Think of a mirror where every element sees its own reflection. For a particular element to satisfy reflexivity, it must hold that it is in a relation with itself. In more mathematical terms, for any element \( a \), if \( a R a \), then the relationship \( R \) has reflexivity. For example, when you have an expression like \( |a-a| = 0 < 1 \), it shows that every number \( a \) aligns with itself because the absolute difference is zero, and zero is clearly less than one. Thus, the relation stands as reflexive for all \( a \) in the set of real numbers.

Symmetry

Symmetry in a relation is akin to having a two-way street; if one element can reach another, then the reverse should also be true. For a relation to be symmetric, whenever you have a situation where \( a R b \), it must also imply that \( b R a \). In the case of checking relations using absolute values, consider \( |a-b| < 1 \). If this condition holds true, the absolute value of \( |b-a| \) will similarly be less than one because the absolute function is symmetric—\( |a-b| \) equals \( |b-a| \). That's why, in this context, if \( a \) is related to \( b \), then \( b \) is necessarily related to \( a \) as well, thereby confirming the symmetry of the relation.

Transitivity

A relation has transitivity if it follows a kind of domino effect. That means, if \( a \) is related to \( b \) and \( b \) is related to \( c \), then \( a \) should also be related to \( c \). In a perfectly transitive setup, you can string together relations without missing a connection. However, our example outlines a case where this is not respected. Though \( 1 R 2 \) and \( 2 R 3 \) might hold true (given that \( |1-2| < 1 \) and \( |2-3| < 1 \)), the gap between 1 and 3 breaks the deal. That's because \( |1-3| = 2 \), which is greater than 1, meaning \( 1 ot R 3 \). Thus, the relation fails the test for transitivity.

Anti-symmetry

In anti-symmetry, things take a peculiar turn compared to symmetry. For a relation to be anti-symmetric, if two different elements are mutually related, then they must actually turn out to be the same. Essentially, if both \( a R b \) and \( b R a \) are true, then \( a \) should equal \( b \). In the given examples, if \( a R b \) and \( b R a \) do not force \( a = b \), then the relation violates the rule of anti-symmetry. Here, our relation shows that even with bidirectional linking allowed by \( |a-b| < 1 \) and \( |b-a| < 1 \), \( a \) and \( b \) don't end up being equal, thereby confirming a lack of anti-symmetry.