Let $T$ be the set of all triangles in a plane with $R$ a relation in $T$ given by $R = \{(T_1, T_2) : T_1 \text{ is congruent to } T_2\}$. Show that $R$ is an equivalence relation.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
$R$ is reflexive,since every triangle is congruent to itself.
Further,$(T_1, T_2) \in R \implies T_1 \text{ is congruent to } T_2 \implies T_2 \text{ is congruent to } T_1 \implies (T_2, T_1) \in R$.
Hence,$R$ is symmetric.
Moreover,$(T_1, T_2) \in R$ and $(T_2, T_3) \in R \implies T_1 \text{ is congruent to } T_2$ and $T_2 \text{ is congruent to } T_3 \implies T_1 \text{ is congruent to } T_3 \implies (T_1, T_3) \in R$.
Therefore,$R$ is an equivalence relation.

Explore More

Similar Questions

Let $P = \{ (x, y) | x^2 + y^2 = 1, x, y \in \mathbb{R} \}$. Then $P$ is:

The number of reflexive relations on a set with $4$ elements is equal to

Let $R = \{( P , Q ) \mid P \text{ and } Q \text{ are at the same distance from the origin} \}$ be a relation. Then the equivalence class of $(1, -1)$ is the set:

If $A = \{x \in Z^+ : x < 10\}$ and $x$ is a multiple of $3$ or $4$,where $Z^+$ is the set of positive integers,then the total number of symmetric relations on $A$ is

Let $R_{1} = \{(a, b) \in N \times N : |a - b| \leq 13\}$ and $R_{2} = \{(a, b) \in N \times N : |a - b| \neq 13\}$. Then on $N$:

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