Question

The following relation is defined on the set of real numbers: aRb if $1+ab>0$. Find whether the relation is reflexive, symmetric or transitive.

Answer 1

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 :
Before proceeding with the question we should know about the concept of relations and different types of relations that are reflexive, symmetric and transitive relations.
A relation in set A is a subset of $A\times A$. Thus, $A\times A$ is two extreme relations.
A relation in a set A is a reflexive relation if $(a,a)\in R$, for every $a\in A$.
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 ${{a}_{1}},{{a}_{2}},{{a}_{3}}\in A$.
A relation in a set A is an equivalence relation if R is reflexive, symmetric and transitive.
We will first check reflexivity. Now let a be an arbitrary element of R. Then,

& \Rightarrow 1+a\times a>0 \\\ & \Rightarrow 1+{{a}^{2}}>0 \\\ \end{aligned}$$ Now we know that the square of any number is positive. So the given relation is reflexive. Now moving on to symmetry. Let $$(a,b)\in R$$. $$\begin{aligned} & \Rightarrow 1+ab>0 \\\ & \Rightarrow 1+ba>0 \\\ \end{aligned}$$ This implies that $$(b,a)\in R$$. Hence the given relation is symmetric. Finally, we will check for transitivity. Let $$(a,b)\in R$$ and $$(b,c)\in R$$. $$\Rightarrow 1+ab>0$$ and $$\Rightarrow 1+bc>0$$ but 1+ac is not greater than 0. This implies that $$(a,c)\notin R$$ and hence the given relation is not transitive. So the given relation is reflexive and symmetric but not transitive. **Note** : We in a hurry can make a mistake in thinking it as a transitive set but we have to check the definition by taking subsets of the given set A. A relation in a set A is a transitive relation if $$({{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$$.