Question

Three relations ${{R}_{1}}$, ${{R}_{2}}$ and ${{R}_{3}}$ are defined on a set $A=\\{a,\,b,c\\}$ as follows:
${{R}_{1}}=\\{(a,\,a),\,(a,b),(a,c),(b,b),(b,c),(c,a)(c,b),(c,c)\\}$
${{R}_{2}}=\\{(a,\,a)\\}$
${{R}_{3}}=\\{(b,\,c)\\}$
${{R}_{4}}=\\{(a,b),(b,c),(c,a)\\}$
Find whether or not each of the relations ${{R}_{1}}$, ${{R}_{2}}$, ${{R}_{3}}$, ${{R}_{4}}$ on A is (i) reflexive (ii) symmetric (iii) transitive.

Answer 1

Hint: We will use the definitions of reflexive, symmetric and transitive relations to solve this question. A relation is a reflexive relation If every element of set A maps to itself. A relation in a set A is a symmetric relation if $({{a}_{1}},{{a}_{2}})\in R$ implies that $({{a}_{2}},{{a}_{1}})\in R$, for all ${{a}_{1}},{{a}_{2}}\in A$. A relation in a set A is a transitive relation if $({{a}_{1}},{{a}_{2}})\in R$ and $({{a}_{2}},{{a}_{1}})\in R$ implies that $({{a}_{1}},{{a}_{3}})\in R$ for all ${{a}_{1}},{{a}_{2}},{{a}_{3}}\in A$.

Complete step-by-step answer:
(i) Reflexive: A relation is a reflexive relation If every element of set A maps to itself.
We will see ${{R}_{1}}=\\{(a,\,a),\,(a,b),(a,c),(b,b),(b,c),(c,a)(c,b),(c,c)\\}$ first and here we can clearly see that (a, a), (b, b) and (c, c) belongs to ${{R}_{1}}$. Hence ${{R}_{1}}$ is reflexive.
Now we move on to ${{R}_{2}}=\\{(a,\,a)\\}$ and here we can clearly see that (a, a) belongs to ${{R}_{2}}$. Hence ${{R}_{2}}$ is Reflexive.
Now we move on to ${{R}_{3}}=\\{(b,\,c)\\}$ and here we can clearly see that (b, b), (c, c) does not belongs to ${{R}_{3}}$. Hence ${{R}_{3}}$ is not reflexive.
Now we move on to ${{R}_{4}}=\\{(a,b),(b,c),(c,a)\\}$ and here we can clearly see that (a, a), (b, b), (c, c) does not belongs to ${{R}_{4}}$. Hence ${{R}_{4}}$ is not reflexive.

(ii) Symmetric: A relation in a set A is a symmetric relation if $({{a}_{1}},{{a}_{2}})\in R$ implies that $({{a}_{2}},{{a}_{1}})\in R$, for all ${{a}_{1}},{{a}_{2}}\in A$.
We will see ${{R}_{1}}=\\{(a,\,a),\,(a,b),(a,c),(b,b),(b,c),(c,a)(c,b),(c,c)\\}$ first and here we can clearly see that (b, a) does not belongs to ${{R}_{1}}$. Hence ${{R}_{1}}$ is not symmetric.
Now we move on to ${{R}_{2}}=\\{(a,\,a)\\}$ and here we can clearly see that (a, a) belongs to ${{R}_{2}}$. Hence ${{R}_{2}}$ is Symmetric.
Now we move on to ${{R}_{3}}=\\{(b,\,c)\\}$ and here we can clearly see that (c, b) does not belongs to ${{R}_{3}}$. Hence ${{R}_{3}}$ is not symmetric.
Now we move on to ${{R}_{4}}=\\{(a,b),(b,c),(c,a)\\}$ and here we can clearly see that (b, a) does not belongs to ${{R}_{4}}$. Hence ${{R}_{4}}$ is not symmetric.

(iii) Transitive: A relation in a set A is a transitive relation if $({{a}_{1}},{{a}_{2}})\in R$ and $({{a}_{2}},{{a}_{1}})\in R$ implies that $({{a}_{1}},{{a}_{3}})\in R$ for all ${{a}_{1}},{{a}_{2}},{{a}_{3}}\in A$.
We will see ${{R}_{1}}=\\{(a,\,a),\,(a,b),(a,c),(b,b),(b,c),(c,a)(c,b),(c,c)\\}$ first and here we can clearly see that (a, b), (b, c) and also (a, c) belongs to ${{R}_{1}}$. Hence ${{R}_{1}}$ is not Transitive.
Now we move on to ${{R}_{2}}=\\{(a,\,a)\\}$ and it is clearly a transitive relation since there is only one element in it.
Now we move on to ${{R}_{3}}=\\{(b,\,c)\\}$ and here it has only two elements. Hence it is Transitive.
Now we move on to ${{R}_{4}}=\\{(a,b),(b,c),(c,a)\\}$ and here we can clearly see that (a, c) does not belongs to ${{R}_{4}}$. Hence ${{R}_{4}}$ is not Transitive.

Note: Remembering the definition of relations and the types of relations is the key here. We in a hurry can make a mistake in thinking ${{R}_{3}}$ as a symmetric set but we have to check the definition by taking subsets of the given set A.