Question

Show that the relation R in the set R of a real number defined as $R=\{\left( a,b \right):a < {{b}^{2}}\}$ is neither reflexive nor symmetric nor transitive.

Answer 1

Hint: Think of the basic definition of the types of relations given in the question and try to check whether the relation mentioned in the questions satisfies the condition for any type of relation or not.

Complete step by step answer:
Before starting with the solution, let us discuss different types of relations. There are a total of 8 types of relations that we study, out of which the major one’s are reflexive, symmetric, transitive, and equivalence relations.
Reflexive relations are those in which each and every element is mapped to itself, i.e., $\left( a,a \right)\in R$ . Symmetric relations are those for which, if \[\left( a,b \right)\in R\text{ }\] then $\left( b,a \right)$ must also belong to R. This can be represented as $aRb\Rightarrow bRa$ . Now, transitive relations are those for which, if $\left( a,b \right)\text{ and }\left( b,c \right)\in R$ then $\left( a,c \right)$ must also belong to R, i.e., $\left( a,b \right)\text{ and }\left( b,c \right)\in R\Rightarrow \left( a,c \right)\in R$ .
Now, if there exists a relation, which is reflexive, symmetric, and transitive at the same time, then the relation is said to be an equivalence relation. For example: let us consider a set A=(1,2). Then the relation {(1,2),(2,1),(1,1),(2,2)} is an equivalence relation.
Now let us start with the solution to the above question. The given set is a set of all real numbers, and the relation is $R=\{\left( a,b \right):a < {{b}^{2}}\}$ . If we take a to be a real number lying between 0 and 1, we will find that a is not less than ${{a}^{2}}$ . For example: if $a=\dfrac{1}{2}$ then ${{a}^{2}}=\dfrac{1}{4}$ , which doesn’t satisfy the condition of a reflexive relation as for all a belonging to the set cannot have the relation (a,a). Now, if we observe we will find that the relation is not symmetric as well. As for being a symmetric relation if $a < {{b}^{2}}$ then $b$ must also be less than square of a, but if we take a=2 and b=5 the condition is violated. Further, the relation is not transitive too, as for being transitive if $a < {{b}^{2}}\text{ and }b < {{c}^{2}}\text{ }$ then a must be less than the square of c, but if we take a=2, b=-2 and c=-1 the condition is violated, so the relation is not a transitive relation as well.

Hence, we have shown that the relation $R=\{\left( a,b \right):a < {{b}^{2}}\}$ for set of real numbers, is neither reflexive nor symmetric nor transitive.

Note: Remember a relation can also be called a transitive relation if there exists $a < {{b}^{2}}$ , but there doesn’t exist any relation of the type $b < {{c}^{2}}$ . Also, most of the questions as above are either solved by using statements based on observation or taking examples, as we did in the above question.