Let R1 and R2 be two relations defined as follows-R1 - a-b -2- a2 - b2 - and R2 - a-b -2- a2 - b2 - Then

R_1 = {(a,b) ∈ℝ^2: a^2 + b^2 ∈ℚ} Let nbsp; a = 2 + √(3) and b = 2 √(3) ⇒ a^2 + b^2 = 14 ∈ℚ Let c = nbsp;(1 + 2√(3)) ⇒ b^2 + c^2 = 20 ∈ℚ But a^2 + c^2 = ...

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Standard XI

Mathematics

Transitive Relations

Let R1 and R2...

Question

Let $R_{1}$ and $R_{2}$ be two relations defined as follows:
$R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}$ and $R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q} .$ Then

Neither

R_{1}

nor

R_{2}

is a transitive relation

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R_{1}

and

R_{2}

both are transitive relations

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R_{1}

is transitive but

R_{2}

is not a transitive relation

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R_{2}

is transitive but

R_{1}

is not a transitive relation

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Solution

The correct option is A Neither $R_{1}$ nor $R_{2}$ is a transitive relation
$R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}$

Let $a = 2 + \sqrt{3}$ and $b = 2 - \sqrt{3}$
$\Rightarrow a^{2} + b^{2} = 14 \in Q$
Let $c = (1 + 2 \sqrt{3})$
$\Rightarrow b^{2} + c^{2} = 20 \in Q$
But $a^{2} + c^{2} = (2 + \sqrt{3})^{2} + (1 + 2 \sqrt{3})^{2} \notin Q$

Since $(2 + \sqrt{3}, 2 - \sqrt{3})$ and $(2 - \sqrt{3}, 1 + 2 \sqrt{3}) \in R_{1}$ but $(2 + \sqrt{3}, 1 + 2 \sqrt{3}) \notin R_{1} .$
So, $R_{1}$ is not a transitive relation $\dots (1)$

$R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q} .$
Let $a^{2} = 1, b^{2} = \sqrt{3}$ and $c^{2} = 2$
$\Rightarrow a^{2} + b^{2} \notin Q$ and $b^{2} + c^{2} \notin Q$
But $a^{2} + c^{2} \in Q$
Since $(1, \sqrt{\sqrt{3}})$ and $(\sqrt{\sqrt{3}}, \sqrt{2}) \notin R_{2}$ but $(1, \sqrt{2}) \in R_{2} .$
So, $R_{2}$ is not a transitive relation $\dots (2)$

$∴$ From $(1)$ and $(2)$ we can say that Neither $R_{1}$ nor $R_{2}$ is a transitive relation.

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Similar questions

Q. Let

R_{1}

and

R_{2}

be two relations defined as follows:

R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}

and

R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q} .

Then

Q. Let

R_{1}

and

R_{2}

be two relations defined as follows:

R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}

and

R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q}

, where

Q

is the set of all rational numbers. Then :

Q. Let

R_{1}

and

R_{2}

be two relations defined on a non empty set A. Which of the following statements is false? Give reasons in support of your answer.
(a) If

R_{1}

and

R_{2}

are reflexive, then so is

R_{1} \cap R_{2}

(b) If

R_{1}

and

R_{2}

are symmetric, then so is

R_{1} \cup R_{2}

R_{1}

and

R_{2}

are transitive, then so is

R_{1} \cup R_{2}

(d) If

R_{1}

and

R_{2}

are transitive, then so is

R_{1} \cap R_{2}

(e) If

R_{1}

and

R_{2}

are symmetric, then so is

R_{1} \cap R_{2}

Q. A circle of radius

r (< a)

is concentric with ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, (a > b)

, then slope of the common tangents to ellipse and circle is

Q. Test whether the following relations R₁, R₂, and R₃ are (i) reflexive (ii) symmetric and (iii) transitive:
(i) R₁ on Q₀ defined by (a, b) ∈ R₁ ⇔ a = 1/b.
(ii) R₂ on Z defined by (a, b) ∈ R₂ ⇔ |a – b| ≤ 5
(iii) R₃on R defined by (a, b) ∈ R₃ ⇔ a²– 4ab + 3b²= 0.

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Transitive Relations

Standard XI Mathematics

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Let R1 and R2 be two relations defined as follows: R1={(a,b) ∈R2:a2+b2∈Q} and R2={(a,b) ∈R2:a2+b2∉Q}. Then

Let $R_{1}$ and $R_{2}$ be two relations defined as follows:
$R_{1} = {(a, b) \in R^{2} : a^{2} + b^{2} \in Q}$ and $R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q} .$ Then