Let R1 be a relation defined by R1-a-b-a- b-a-b- Then R1 is

Explanation for the correct option:Checking the relation:For any a∈ℝ, a≥ a therefore the relation is symmetricConsider a=2 b=1(a,b)∈R_1 because 2≥ 1But (b,a)∉R_ ...

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Standard XII

Mathematics

Reflexive Relations

Let R1 be a r...

Question

Let $R_{1}$ be a relation defined by $R_{1} = \{(a, b) | a \geq b, a, b \in ℝ\}$ . Then $R_{1}$ is

Reflexive, transitive and symmetric

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Reflexive, transitive but not symmetric

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Symmetric, transitive but not reflexive

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Neither transitive nor reflexive but symmetric

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Solution

The correct option is B
Reflexive, transitive but not symmetric

Explanation for the correct option:
Checking the relation:
For any $a \in ℝ, a \geq a$ therefore the relation is symmetric
Consider $a = 2 b = 1$
$(a, b) \in R_{1}$ because $2 \geq 1$
But $(b, a) \notin R_{1}$ since $1 \leq 2$
Therefore $R_{1}$ is not symmetric
Consider $(a, b) \in R_{1}$ and $(b, c) \in R_{1}$
Therefore $a \geq b$ and $b \geq c$
Which means $a \geq c$
Therefore $(a, c) \in R_{1}$
So whenever $(a, b) \in R_{1}$ and $(b, c) \in R_{1}$ it implies $(a, c) \in R_{1}$
$R_{1}$ is transitive
$R_{1}$ is reflexive, transitive but not symmetric
Hence, option (C) is the correct answer

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Q. Let

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is divisible by

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R_{1}

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Let R1 be a relation defined by R1=a,b|a≥b,a,b∈ℝ. Then R1 is

Let $R_{1}$ be a relation defined by $R_{1} = \{(a, b) | a \geq b, a, b \in ℝ\}$ . Then $R_{1}$ is