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Question: Let T be the set of all triangles in the Euclidean plane, and let a relation R and T be defined as a...

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

Explanation

Solution

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, (a,a)R\left( a,a \right)\in R.
2)For a relation to be symmetric, (a,b)R(b,a)R\left( a,b \right)\in R\Rightarrow \left( b,a \right)\in R.
3)For a relation to be transitive, (a,b)R,(b,c)R(a,c)R\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,
T1T1(a,a)R{{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, T1 and T2{{T}_{1}}\text{ and }{{T}_{2}} are congruent two each other and similarly T2 and T1{{T}_{2}}\text{ and }{{T}_{1}} are congruent to each other.
T1T2 T2T1 (a,b)R(b,a)R \begin{aligned} & {{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{aligned}
Hence, the relation R is symmetric in nature.
We have to prove that T is transitive in nature.
Since, we know that the T1{{T}_{1}} and T2{{T}_{2}} are congruent to each other and similarly T2{{T}_{2}} and T3{{T}_{3}} are congruent to each other.
T1T2 and T2T3 T1T3 \begin{aligned} & \because {{T}_{1}}\cong {{T}_{2}}\text{ and }{{T}_{2}}\cong {{T}_{3}} \\\ & \therefore {{T}_{1}}\cong {{T}_{3}} \\\ \end{aligned}
So, T1T3R{{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.