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

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

Question

(A) $1$. Reflexivity: For any $a \in \mathbb{R}$,$|a - a| = 0 \le 1$. Thus,$a \, R \, a$ holds for all $a \in \mathbb{R}$. So,$R$ is reflexive. $2$. S

Vedclass · Accepted Answer

Reflexive and symmetric

Vedclass · Answer

(A) $1$. Reflexivity: For any $a \in \mathbb{R}$,$|a - a| = 0 \le 1$. Thus,$a \, R \, a$ holds for all $a \in \mathbb{R}$. So,$R$ is reflexive. $2$. Symmetry: If $a \, R \, b$,then $|a - b| \le 1$. Since $|a - b| = |b - a|$,it follows that $|b - a| \le 1$,which means $b \, R \, a$. So,$R$ is symmetric. $3$. Transitivity: Consider $a = 1, b = 1.5, c = 2$. Here,$|1 - 1.5| = 0.5 \le 1$ $(1 \, R \, 1.5)$ and $|1.5 - 2| = 0.5 \le 1$ $(1.5 \, R \, 2)$. However,$|1 - 2| = 1 \le 1$ holds,but if we take $a = 1, b = 1.8, c = 2.6$,then $|1 - 1.8| = 0.8 \le 1$ and $|1.8 - 2.6| = 0.8 \le 1$,but $|1 - 2.6| = 1.6 > 1$. Thus,$R$ is not transitive. Conclusion: $R$ is reflexive and symmetric.

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:

Similar Questions

Let the relation $R$ on the set $M = \{1, 2, 3, \dots, 16\}$ be given by $R = \{(x, y) : 4y = 5x - 3, x, y \in M\}$. Then the minimum number of elements required to be added in $R$, in order to make the relation symmetric, is equal to

The void relation on a set $A$ is

Let ${R_1}$ be a relation defined by ${R_1} = \{ (a, b) | a \ge b, a, b \in R \}$. Then ${R_1}$ is

Let $R$ and $S$ be equivalence relations on a set $A$. Then,

Let $R$ be a relation defined on the set $\{1,2,3,4\} \times \{1,2,3,4\}$ by $R = \{((a,b), (c,d)) : 2a + 3b = 3c + 4d\}$. Then the number of elements in $R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module