Let r be a relation from R (set of real numbe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The relation $r$ on $R$ is defined as $r = \{(a,b) \in R 	imes R \mid a - b + \sqrt{3} \in R \setminus Q \}$.$1$. Reflexivity: For any $a \in R$,

Vedclass · Accepted Answer

reflexive only

Vedclass · Answer

(B) The relation $r$ on $R$ is defined as $r = \{(a,b) \in R 	imes R \mid a - b + \sqrt{3} \in R \setminus Q \}$.$1$. Reflexivity: For any $a \in R$,$aRa \iff a - a + \sqrt{3} = \sqrt{3}$. Since $\sqrt{3}$ is an irrational number,$aRa$ holds for all $a \in R$. Thus,$r$ is reflexive.$2$. Symmetry: For $r$ to be symmetric,$aRb \implies bRa$. Let $a = \sqrt{3}$ and $b = 0$. Then $a - b + \sqrt{3} = \sqrt{3} - 0 + \sqrt{3} = 2\sqrt{3}$,which is irrational. So,$(\sqrt{3}, 0) \in r$. However,$b - a + \sqrt{3} = 0 - \sqrt{3} + \sqrt{3} = 0$,which is a rational number. So,$(0, \sqrt{3}) 
otin r$. Thus,$r$ is not symmetric.$3$. Transitivity: For $r$ to be transitive,$aRb$ and $bRc \implies aRc$. Consider $a = \sqrt{3}$,$b = 0$,and $c = 2\sqrt{3}$. $aRb \implies \sqrt{3} - 0 + \sqrt{3} = 2\sqrt{3}$ (irrational).$bRc \implies 0 - 2\sqrt{3} + \sqrt{3} = -\sqrt{3}$ (irrational).$aRc \implies \sqrt{3} - 2\sqrt{3} + \sqrt{3} = 0$ (rational).Since $aRc$ is not irrational,$r$ is not transitive.Therefore,$r$ is reflexive only.

Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = \{(a,b) \mid a,b \in R \text{ and } a - b + \sqrt{3} \text{ is an irrational number} \}$. The relation $r$ is

Similar Questions

Let $R$ and $S$ be two relations on a set $A$. Then which of the following is true?

The number of relations, defined on the set ${a, b, c, d}$, which are both reflexive and symmetric, is equal to:

The relation "congruence modulo $m$" is:

On the set of all real numbers,a relation $R$ is defined as $a \, R \, b$ if and only if $|a - b| \le 1$. Then $R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module