Let $R$ be a relation defined on $N \times N$ by $(a, b) R(c, d) \Leftrightarrow a(b + d) = c(b + d)$ is incorrect,the correct relation is $(a, b) R(c, d) \Leftrightarrow ad = bc$. Given the relation $(a, b) R(c, d) \Leftrightarrow a(b + d) = c(a + d)$ is not standard,let us analyze the relation $(a, b) R(c, d) \Leftrightarrow ad = bc$. Then $R$ is:

  • A
    reflexive,symmetric
  • B
    symmetric,transitive
  • C
    transitive only
  • D
    equivalence

Explore More

Similar Questions

Let $A = \{1, 2, 3, 4, 6\}$. Let $R$ be the relation on $A$ defined by $R = \{(a, b) : a, b \in A, b \text{ is exactly divisible by } a\}$. Find the domain of $R$.

Let $M$ denote the set of all $3 \times 3$ non-singular matrices. Define the relation $R$ by $R = \{ (A,B) \in M \times M : AB = BA \}$. Then $R$ is-

The relation $R = \{(a, b) : \operatorname{gcd}(a, b) = 1, 2a \neq b, a, b \in \mathbb{Z}\}$ is:

Let $R$ be a relation on $N \times N$ defined by $(a, b) R (c, d)$ if and only if $ad(b-c) = bc(a-d)$. Then $R$ is

Show that the relation $R$ in the set $\{1, 2, 3\}$ given by $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)\}$ is reflexive but neither symmetric nor transitive.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo