Class CBSE Class 12 Mathematics Relations and Functions Q #1329
KNOWLEDGE BASED
UNDERSTAND
5 Marks 2024 AISSCE(Board Exam) LA
Check whether the relation S in the set of real numbers R defined by $S=\{(a,b)$: where $a-b+\sqrt{2}$ is an irrational number is reflexive, symmetric or transitive.
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Detailed Solution

Step 1: Check for Reflexivity

For a relation to be reflexive, $(a, a) \in S$ for all $a \in R$. In this case, we need to check if $a - a + \sqrt{2}$ is an irrational number for all $a \in R$. $a - a + \sqrt{2} = \sqrt{2}$, which is an irrational number. Therefore, $(a, a) \in S$ for all $a \in R$, and the relation is reflexive.

Step 2: Check for Symmetry

For a relation to be symmetric, if $(a, b) \in S$, then $(b, a) \in S$. This means if $a - b + \sqrt{2}$ is irrational, then $b - a + \sqrt{2}$ must also be irrational. Let's consider a counterexample: Let $a = \sqrt{2}$ and $b = 2\sqrt{2}$. Then $a - b + \sqrt{2} = \sqrt{2} - 2\sqrt{2} + \sqrt{2} = 0$, which is a rational number. So, $(a,b) \notin S$. However, if we consider $a = 0$ and $b = \sqrt{2}$, then $a - b + \sqrt{2} = 0 - \sqrt{2} + \sqrt{2} = 0$, which is rational. So $(0, \sqrt{2}) \notin S$. Let's consider $a = 1$ and $b = 1 + \sqrt{2}$. Then $a - b + \sqrt{2} = 1 - (1 + \sqrt{2}) + \sqrt{2} = 1 - 1 - \sqrt{2} + \sqrt{2} = 0$, which is rational. So $(1, 1+\sqrt{2}) \notin S$. Let's consider $a = \sqrt{2}$ and $b = 1$. Then $a - b + \sqrt{2} = \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$, which is irrational. So $(\sqrt{2}, 1) \in S$. Now, let's check if $(1, \sqrt{2}) \in S$. $b - a + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1$, which is rational. So $(1, \sqrt{2}) \notin S$. Since $(\sqrt{2}, 1) \in S$ but $(1, \sqrt{2}) \notin S$, the relation is not symmetric.

Step 3: Check for Transitivity

For a relation to be transitive, if $(a, b) \in S$ and $(b, c) \in S$, then $(a, c) \in S$. This means if $a - b + \sqrt{2}$ is irrational and $b - c + \sqrt{2}$ is irrational, then $a - c + \sqrt{2}$ must also be irrational. Let's consider a counterexample: Let $a = 1$, $b = \sqrt{2}$, and $c = 1 + \sqrt{2}$. Then $a - b + \sqrt{2} = 1 - \sqrt{2} + \sqrt{2} = 1$, which is rational. So $(1, \sqrt{2}) \notin S$. Let $a = \sqrt{2}$, $b = 1$, and $c = 2\sqrt{2}$. $a - b + \sqrt{2} = \sqrt{2} - 1 + \sqrt{2} = 2\sqrt{2} - 1$, which is irrational. So $(\sqrt{2}, 1) \in S$. $b - c + \sqrt{2} = 1 - 2\sqrt{2} + \sqrt{2} = 1 - \sqrt{2}$, which is irrational. So $(1, 2\sqrt{2}) \in S$. Now, let's check if $(a, c) \in S$. $a - c + \sqrt{2} = \sqrt{2} - 2\sqrt{2} + \sqrt{2} = 0$, which is rational. So $(\sqrt{2}, 2\sqrt{2}) \notin S$. Since $(\sqrt{2}, 1) \in S$ and $(1, 2\sqrt{2}) \in S$ but $(\sqrt{2}, 2\sqrt{2}) \notin S$, the relation is not transitive.

Step 4: Conclusion

The relation S is reflexive, but it is neither symmetric nor transitive.

Final Answer: Reflexive only. Not symmetric and not transitive.

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Pedagogical Audit
Bloom's Analysis: This is an UNDERSTAND question because it requires the student to understand the definitions of reflexive, symmetric, and transitive relations and apply them to a specific relation defined on the set of real numbers. The student needs to check the given condition for each property.
Knowledge Dimension: CONCEPTUAL
Justification: The question tests the understanding of the concepts of relations, specifically reflexive, symmetric, and transitive relations. It requires applying these concepts to a specific relation defined using irrational numbers.
Syllabus Audit: In the context of CBSE Class 12, this is classified as KNOWLEDGE. The question directly assesses the student's understanding of relations and their properties, which is a core topic in the Relations and Functions chapter. The question requires applying the definitions of reflexive, symmetric, and transitive relations.

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