Let S be the set of all real numbers. A relat | vedclass

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

Question

(D) Given $a R b \Leftrightarrow |a-b| \leq 1$. $1$. Reflexivity: For any $a \in S$,$|a-a| = 0 \leq 1$. Thus,$a R a$ holds. $R$ is reflexive. $2$. Sym

Vedclass · Accepted Answer

reflexive and symmetric but not transitive

Vedclass · Answer

(D) Given $a R b \Leftrightarrow |a-b| \leq 1$. $1$. Reflexivity: For any $a \in S$,$|a-a| = 0 \leq 1$. Thus,$a R a$ holds. $R$ is reflexive. $2$. Symmetry: If $a R b$,then $|a-b| \leq 1$. Since $|a-b| = |b-a|$,it follows that $|b-a| \leq 1$,so $b R a$ holds. $R$ is symmetric. $3$. Transitivity: Consider $a = 1, b = 2, c = 3$. $|a-b| = |1-2| = 1 \leq 1$ (True,so $a R b$). $|b-c| = |2-3| = 1 \leq 1$ (True,so $b R c$). However,$|a-c| = |1-3| = 2 > 1$ (False,so $a$ is not related to $c$). Thus,$R$ is not transitive. Conclusion: $R$ is reflexive and symmetric but not transitive.

Let $S$ be the set of all real numbers. $A$ relation $R$ has been defined on $S$ by $a R b \Leftrightarrow |a-b| \leq 1$. Then $R$ is:

Similar Questions

If $R$ is an equivalence relation on a set $A$,then $R^{-1}$ is

Let $R$ be a transitive relation on a set $A$ and $I$ be the identity relation on $A$. Then:

Let $R = \{(x, y) \in N \times N : \log_e(x + y) \leq 2\}$. Then the minimum number of elements, required to be added in $R$ to make it a transitive relation, is . . . . . . .

Given the relation $R = \{(1, 2), (2, 3)\}$ on the set $A = \{1, 2, 3\}$,the minimum number of ordered pairs which when added to $R$ make it an equivalence relation is

The minimum number of elements that must be added to the relation $R = \{(a, b), (b, c)\}$ on the set $\{a, b, c\}$ so that it becomes symmetric and transitive is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module