Let R1 and R2 be two relations defined as follows-R1- a-b - R2-a2-b2- andR2- a- b- R2-a2-b2 - where - is the set of all rational numbers- Then -

For nbsp;R_1 Let a=1+√(2),b=1 √(2),c=8^1/4 aR_1b⇒ a^2+b^2=6∈ℚ bR_1c⇒ b^2+c^2=3∈ℚ aR_1c⇒ a^2+c^2=3+4√(2)∉ℚ ∴ R_1 is not transitive. For R_2 Let a=1+√(2),b=√(2 ...

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Byju's Answer

Standard XII

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
$R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q}$ , where $Q$ is the set of all rational numbers. Then :

R_{1}

is transitive but

R_{2}

is not transitive.

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R_{1}

and

R_{2}

are both transitive.

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R_{2}

is transitive but

R_{1}

is not transitive.

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Neither

R_{1}

nor

R_{2}

is transitive.

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Solution

The correct option is D Neither $R_{1}$ nor $R_{2}$ is transitive.
For $R_{1}$
Let $a = 1 + \sqrt{2}, b = 1 - \sqrt{2}, c = 8^{\frac{1}{4}}$
$a R_{1} b \Rightarrow a^{2} + b^{2} = 6 \in Q$
$b R_{1} c \Rightarrow b^{2} + c^{2} = 3 \in Q$
$a R_{1} c \Rightarrow a^{2} + c^{2} = 3 + 4 \sqrt{2} \notin Q$
$∴ R_{1}$ is not transitive.

For $R_{2}$
Let $a = 1 + \sqrt{2}, b = \sqrt{2}, c = 1 - \sqrt{2}$
$a R_{2} b \Rightarrow a^{2} + b^{2} = 5 + 2 \sqrt{2} \notin Q$
$b R_{2} c \Rightarrow b^{2} + c^{2} = 5 - 2 \sqrt{2} \notin Q$
$a R_{2} c \Rightarrow a^{2} + c^{2} = 6 \in Q$
$∴ R_{2}$ is not transitive.

Suggest Corrections

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. 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. Let

R

be a relation from

Q

Q

defined by

R = {(a, b) : a, b \in Q and a - b \in Z}

.
Show that:

(i) (a, a) \in R

for all

a \in Q

(i i) (a, b) \in R

implies that

(b, a) \in R

(i i i) (a, b) \in R and (b, c) \in R

implies that

(a, c) \in R

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}

(a,b) is the mid point of the chord $¯ A B$ of the circle $x^{2} + y^{2} = r^{2}$ . The tangent at A,B meet a C. then area of $Δ A B C$

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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}, where Q is the set of all rational numbers. 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
$R_{2} = {(a, b) \in R^{2} : a^{2} + b^{2} \notin Q}$ , where $Q$ is the set of all rational numbers. Then :