Question

Let S be the set of all real numbers. Then show that the relation $R=\{\left( a,b \right):1+ab>0\}$ on S is reflexive and symmetric but not 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. For checking the relation, use the relation that multiplication is commutative, i.e., ab=ba. For the sake of showing that the relation is not transitive, you need to think of an example which shows that the relation is not transitive.

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 ones are reflexive, symmetric, transitive, and equivalence relation.
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. See if we select any real number, we will find that the number multiplied by itself gives a positive number, and 1+(a positive number) is always greater than zero. Hence, the relation would be reflexive.
Also, as multiplication is commutative, then ab=ba, so when (1+ab) is greater than 0 then (1+ba) will also be greater than 0. So, we can say that if (a,b) satisfies the relation $R=\{\left( a,b \right):1+ab>0\}$ , then (b,a) will also satisfy this relation. Hence the relation is symmetric as well.
However, if we take a=1, $b=-\dfrac{1}{2}$ , $c=-5$ then (a,b) and (b,c) will satisfy the relation $R=\{\left( a,b \right):1+ab>0\}$ , but (a,c) will not satisfy the given relation. Hence, the relation is not transitive.

Note: Remember, a relation can also be called a transitive relation if there exists $aRb$ , but there doesn’t exist any relation $bRc$ . 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. Remember, if you prove that the relation is not of a type for certain assumed values it is acceptable, but if you show that the relation is of a particular type for some assumed values, you cannot say that it the relation has been proved to be of that particular type.