Question

Let T be the set of all triangles in the Euclidean plane, and let a relation R and T be defined as a R b if a is congruent to b for all a, b $\in $ T. Then R is
(a) Reflexive but not symmetric
(b) Transitive but not symmetric
(c) Equivalence
(d) None of these

Answer 1

Hint: For solving this problem, we consider all options individually. By using the necessary conditions for a set to be reflexive, symmetric and transitive, we proceed for solving the question. If any of the options fails to satisfy the condition, it would be rejected.

Complete step-by-step answer:
The conditions for a set to be reflexive, transitive and symmetric are:
1)For a relation to be reflexive, $\left( a,a \right)\in R$.
2)For a relation to be symmetric, $\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R$.
3)For a relation to be transitive, $\left( a,b \right)\in R,\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$.
4)For a relation to be equivalence, it should be reflexive, symmetric and transitive.
Let T be all triangles in the Euclidean plane.
T = {all triangles in the Euclidean plane}
R = {(a, b); a is congruent to b}
First, we have to prove that R is reflexive.
We know that every triangle is congruent to itself. Using this,
${{T}_{1}}\cong {{T}_{1}}\Rightarrow (a,a)\in R$
Hence, the relation R is reflexive.
Now, we have to prove that R is symmetric.
So, ${{T}_{1}}\text{ and }{{T}_{2}}$ are congruent two each other and similarly ${{T}_{2}}\text{ and }{{T}_{1}}$ are congruent to each other.
$\begin{align}
  & {{T}_{1}}\cong {{T}_{2}} \\
 & {{T}_{2}}\cong {{T}_{1}} \\
 & \therefore \left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R \\
\end{align}$
Hence, the relation R is symmetric in nature.
We have to prove that T is transitive in nature.
Since, we know that the ${{T}_{1}}$ and ${{T}_{2}}$ are congruent to each other and similarly ${{T}_{2}}$ and ${{T}_{3}}$ are congruent to each other.
$\begin{align}
  & \because {{T}_{1}}\cong {{T}_{2}}\text{ and }{{T}_{2}}\cong {{T}_{3}} \\
 & \therefore {{T}_{1}}\cong {{T}_{3}} \\
\end{align}$
So, ${{T}_{1}}\cong {{T}_{3}}\in R$
Hence, the relation R is transitive in nature.
Therefore, R is equivalence in nature.
Hence, option (c) is correct.

Note: The knowledge of equivalence of a relation is must for solving this problem. Students must remember all the necessary conditions for proving a set is reflexive, symmetric and transitive. All three relations must be verified for equivalence.