Problem 2
Question
(i) Let \(R\) be the relation on the set \(R\) of all real numbers defined by setting \(a R b\) if \(|a-b| \leq \frac{1}{2}\). Then \(R\) is (A) Reflexive and symmetric but not transitive (B) Symmetric and transitive but not reflexive (C) Transitive but neither reflexive nor symmetric (D) None of these
Step-by-Step Solution
Verified Answer
(A) Reflexive and symmetric but not transitive.
1Step 1: Checking Reflexivity
A relation is reflexive if every element relates to itself. In this case, for any real number \(a\), we need \(|a-a| \leq \frac{1}{2}\). Since \(|a-a| = 0\) for any \(a\), and \(0\leq \frac{1}{2}\) holds, the relation \(R\) is reflexive.
2Step 2: Checking Symmetry
A relation is symmetric if whenever \(aRb\), then \(bRa\). If \(|a-b| \leq \frac{1}{2}\), it follows that \(|b-a| = |a-b| \leq \frac{1}{2}\). Thus, \(R\) is symmetric.
3Step 3: Checking Transitivity
A relation is transitive if whenever \(aRb\) and \(bRc\), then \(aRc\). Assume \(|a-b| \leq \frac{1}{2}\) and \(|b-c| \leq \frac{1}{2}\). Applying the triangle inequality, \(|a-c| \leq |a-b| + |b-c| \leq \frac{1}{2} + \frac{1}{2} = 1\). Since this has not yielded \(|a-c| \leq \frac{1}{2}\) consistently, the relation \(R\) is not transitive.
4Step 4: Conclusion
Given our analysis, the relation \(R\) is reflexive and symmetric but not transitive. Hence, the correct answer is (A).
Key Concepts
Reflexive RelationSymmetric RelationTransitive Relation
Reflexive Relation
In mathematics, a **reflexive relation** is a fundamental concept in the study of binary relations. A relation is considered reflexive if every element is related to itself. This means that for a set with elements, each element must "mirror" itself in the relation.
Take, for instance, a set containing numbers. For this set to be reflexive, each number should satisfy the condition that relates it to itself. In our context of the relation on real numbers where the condition is \(|a-a| \leq \frac{1}{2}\), this simplifies to \(|0| \leq \frac{1}{2}\) as \(|a-a| = 0\) for any element \(a\).
Take, for instance, a set containing numbers. For this set to be reflexive, each number should satisfy the condition that relates it to itself. In our context of the relation on real numbers where the condition is \(|a-a| \leq \frac{1}{2}\), this simplifies to \(|0| \leq \frac{1}{2}\) as \(|a-a| = 0\) for any element \(a\).
- Every number can relate to itself, meeting the criteria for reflexivity.
- This characteristic ensures a stable "self-relationship" among the elements.
Symmetric Relation
Understanding a **symmetric relation** is key to grasping the nature of how elements interact within a set. A relation is symmetric if for any two elements, say \(a\) and \(b\), whenever \(a\) is related to \(b\), \(b\) is also related back to \(a\).
For the given relation where \(|a-b| \leq \frac{1}{2}\), this symmetry is verified by considering that \(|a-b|\) is equal to \(|b-a|\). Thus, if \(a\) is related to \(b\) under the set condition, \(b\) is automatically related to \(a\) as well.
For the given relation where \(|a-b| \leq \frac{1}{2}\), this symmetry is verified by considering that \(|a-b|\) is equal to \(|b-a|\). Thus, if \(a\) is related to \(b\) under the set condition, \(b\) is automatically related to \(a\) as well.
- The symmetric property means relationships are bidirectional.
- This quality reflects the mutual interaction between any two elements in the set.
Transitive Relation
A **transitive relation** is a slightly more intricate concept, crucial to understanding cascading relationships among elements in a set. A relation is deemed transitive if, whenever an element \(a\) is related to \(b\) and \(b\) is related to \(c\), \(a\) must necessarily be related to \(c\).
In our exercise, the transitive property fails because even when \(|a-b| \leq \frac{1}{2}\) and \(|b-c| \leq \frac{1}{2}\), the direct relationship \(|a-c|\) may not necessarily be \(\leq \frac{1}{2}\). Instead, it can be up to 1 due to the triangle inequality property, \(|a-c| \leq |a-b| + |b-c|\).
In our exercise, the transitive property fails because even when \(|a-b| \leq \frac{1}{2}\) and \(|b-c| \leq \frac{1}{2}\), the direct relationship \(|a-c|\) may not necessarily be \(\leq \frac{1}{2}\). Instead, it can be up to 1 due to the triangle inequality property, \(|a-c| \leq |a-b| + |b-c|\).
- Transitivity ensures a pathway through related elements.
- If broken, as in the current case, it prevents a seamless connection among all elements.
Other exercises in this chapter
Problem 1
Let \(F_{1}\) be the set of all parallelograms, \(F_{2}\) the set of rectangles, \(F_{3}\) the set of rhombuses, \(F_{4}\) the set of squares and \(F_{5}\) the
View solution Problem 3
\(n / m\) means that \(n\) is a factor of \(m\), then the relation \(\%\) 'is (A) reflexive and symmetric. (B) transitive and reflexive. (C) reflexive, transiti
View solution Problem 4
Set \(A\) and \(B\) have 3 and 6 elements respectively. What can be the minimum number of elements in \(A \cup B ?\) (A) 18 (B) 9 (C) 6 (D) 3
View solution Problem 5
Let \(R\) be a relation defined on the set of natural numbers \(N\) as \(R=[(x, y): x \in N, y \in N, 2 x+y=41]\). Then (A) Domain of \(R=\\{1,2,3, \ldots, 19,2
View solution