Question:

If a relation R on the set {1, 2, 3} be defined by R = {(1, 1)}, then R is

Updated On: Apr 9, 2025
  • Reflexive and symmetric
  • Reflexive and transitive
  • symmetric and transitive
  • Only symmetric
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The Correct Option is C

Approach Solution - 1

Given a relation R on the set {1, 2, 3} defined by R = {(1, 1)}, we need to determine the properties of R.

Reflexive: A relation R is reflexive if for all a in the set, (a, a) belongs to R. In this case, the set is {1, 2, 3}. For R to be reflexive, (1, 1), (2, 2), and (3, 3) must be in R. However, R = {(1, 1)} does not contain (2, 2) and (3, 3), so R is not reflexive.

Symmetric: A relation R is symmetric if for every (a, b) in R, (b, a) is also in R. In this case, R = {(1, 1)}. Since (1, 1) is in R, and (1, 1) is equal to (1, 1), R is symmetric.

Transitive: A relation R is transitive if for every (a, b) and (b, c) in R, (a, c) is also in R. In this case, R = {(1, 1)}. If (1, 1) and (1, 1) are in R, then (1, 1) must be in R, which it is. Thus, R is transitive.

Since R is symmetric and transitive, but not reflexive, the correct option is (C) Symmetric and transitive.

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Approach Solution -2

The given relation \( R = \{(1, 1)\} \) is on the set \( \{1, 2, 3\} \). 

1. Reflexive: A relation \( R \) is reflexive if for every element \( x \) in the set, \( (x, x) \) is in \( R \). - Here, the set is \( \{1, 2, 3\} \). For the relation to be reflexive, we need \( (1, 1) \), \( (2, 2) \), and \( (3, 3) \) to be in \( R \). - Since \( (2, 2) \) and \( (3, 3) \) are not in \( R \), the relation is not reflexive

2. Symmetric: A relation \( R \) is symmetric if whenever \( (x, y) \) is in \( R \), \( (y, x) \) is also in \( R \). - The only pair in \( R \) is \( (1, 1) \), and its reverse, \( (1, 1) \), is also in \( R \). - Thus, the relation is symmetric

3. Transitive: A relation \( R \) is transitive if whenever \( (x, y) \) and \( (y, z) \) are in \( R \), \( (x, z) \) must also be in \( R \). - The only pair in \( R \) is \( (1, 1) \), and since it is a self-loop, transitivity holds trivially. 

- Therefore, the relation is transitive

The relation \( R = \{(1, 1)\} \) is symmetric and transitive, but not reflexive.

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