Question
Choose the correct answer from the given four options.
Let A = {1, 2, 3} and consider the relation R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}. Then R is:
  1. Reflexive but not symmetric.
  2. Reflexive but not transitive.
  3. Symmetric and transitive.
  4. Neither symmetric, nor transitive.

Answer

  1. Reflexive but not symmetric.
Solution:
Given that, A = {1, 2, 3}
and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}
$\because\ (1,1), (2,2),(3,3)\in\text{R}$
Hence, R is reflexive.
$(1,2)\in\text{R}$ but $(2,1)\notin\text{R}$
Hence, R is not symmetric.
$(1,2)\in\text{R}$ and $(2,3)\in\text{R}$
$\Rightarrow\ (1,3)\in\text{R}$
Hence, R is transitive.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $A=\left[\begin{array}{ll}-2 & 4 \\ -1 & 2\end{array}\right]$, then $A^2$ is
The position vectors of points $P$ and $Q$ are $\vec{p}$ and $\vec{q}$ respectively. The point $R$ divides line segment $P Q$ in the ratio $3: 1$ and $S$ is the mid-point of line segment $P R$. The position vector of $S$ is
$l = m = n = 1$ represents the direction cosines of:
Choose the correct answer from the given four options:
The area of the region bounded by the y-axis, $\text{y}=\cos\text{x}$ and $\text{y}=\sin\text{x},0\leq\text{x}\leq\frac{\pi}{2}$ is:
  1. $\sqrt{2}\text{ sq. units}$
  2. $\big(\sqrt{2}+1)\text{ sq. units}$
  3. $\big(\sqrt{2}-1)\text{ sq. units}$
  4. $\big(2\sqrt{2}-1)\text{ sq. units}$
If $\cos ^{-1} \alpha+\cos ^{-1} \beta+\cos ^{-1} \gamma=3 \pi$, then $\alpha(\beta+\gamma)+\beta(\gamma+\alpha)+\gamma(\alpha+\beta)$ equals
If $A=\left[\begin{array}{cc}\sin ^2 \theta & \sec ^2 \theta \\ \operatorname{cosec}^2 \theta & \frac{1}{2}\end{array}\right]$ and $B=\left[\begin{array}{cc}\cos ^2 \theta & -\tan ^2 \theta \\ -\cot ^2 \theta & \frac{1}{2}\end{array}\right]$ the value of $A + B$ will be :
If the direction ratios of two lines are given by 3lm - 4ln + mn = 0 and l + 2m + 3n = 0, then the angle between the lines is:
  1. $\frac{\pi}{6}$
  2. $\frac{\pi}{4}$
  3. $\frac{\pi}{3}$
  4. $\frac{\pi}{2}$
The equation of a line passing through point (2, -1, 0) and parallel to the line $\frac{ x }{1}=\frac{ y -1}{2}=\frac{2- z }{2}$ is:
Evaluate $\begin{bmatrix}2&5\\-1&-1\end{bmatrix}$
  1. 3
  2. -7
  3. 5
  4. -2
Let $\vec{\text{a}}$ and $\vec{\text{b}}$ be two unit vectors and a be the angle between them. Then, $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if:
  1. $\text{a}=\frac{\pi}{4}$
  2. $\text{a}=\frac{\pi}{3}$
  3. $\text{a}=\frac{2\pi}{3}$
  4. $\text{a}=\frac{\pi}{2}$