On the set of real numbers R, a relation rho | vedclass

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

Question

(C) $1$. Reflexivity: For any $x \in R$, $x - x = 0$. Since $0$ is allowed, $x \rho x$ holds. Thus, $\rho$ is reflexive. $2$. Symmetry: If $x \rho y$,

Vedclass · Accepted Answer

$\rho$ is reflexive and symmetric but not transitive

Vedclass · Answer

(C) $1$. Reflexivity: For any $x \in R$, $x - x = 0$. Since $0$ is allowed, $x \rho x$ holds. Thus, $\rho$ is reflexive. $2$. Symmetry: If $x \rho y$, then $x - y$ is $0$ or irrational. Since $y - x = -(x - y)$, if $x - y$ is $0$, $y - x$ is $0$. If $x - y$ is irrational, $y - x$ is also irrational. Thus, $y \rho x$ holds. $\rho$ is symmetric. $3$. Transitivity: Let $x = \sqrt{2} + 1$, $y = \sqrt{2}$, and $z = 0$. Here, $x - y = 1$ (rational, not zero or irrational), so $x \rho y$ is false. Let us test $x = \sqrt{2}$, $y = 0$, $z = -\sqrt{2}$. Then $x - y = \sqrt{2}$ (irrational) and $y - z = \sqrt{2}$ (irrational). However, $x - z = 2\sqrt{2}$ (irrational). Consider $x = \sqrt{2} + 1$, $y = \sqrt{2}$, $z = 1$. $x - y = 1$ (not allowed). Consider $x = 1 + \sqrt{2}$, $y = 1$, $z = \sqrt{2}$. $x - y = \sqrt{2}$ (irrational) and $y - z = 1 - \sqrt{2}$ (irrational). But $x - z = 1$ (rational, not zero). Thus, $x \rho y$ and $y \rho z$ hold, but $x \rho z$ does not hold. Therefore, $\rho$ is not transitive.

On the set of real numbers $R$, a relation $\rho$ is defined by $x \rho y$ if and only if $x-y$ is zero or an irrational number. Then:

Similar Questions

Given a non-empty set $X$,consider $P(X)$ which is the set of all subsets of $X$. Define the relation $R$ in $P(X)$ as follows: For subsets $A, B$ in $P(X)$,$ARB$ if and only if $A \subset B$. Is $R$ an equivalence relation on $P(X)$? Justify your answer.

Let $R$ be an equivalence relation on a finite set $A$ having $n$ elements. Then the number of ordered pairs in $R$ is

On $R$, the set of real numbers, a relation $\rho$ is defined as $a \rho b$ if and only if $1+a b > 0$. Then,

Show that the relation $R$ in the set of real numbers $\mathbb{R}$,defined as $R = \{(a, b) : a \leq b^2\}$,is neither reflexive,nor symmetric,nor transitive.

$A$ relation $R$ is defined on the set of natural numbers such that $m$ is related to $n$ if $m$ is a multiple of $n$. Then the relation is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module