Question

Let a relation R in the set R of real numbers be defined as \[\left( {a,{\rm{ }}b} \right)\; \in R\] if and only if \[1 + ab > 0\] for all \[a,b \in R\]. Which of the following true for the relation R?
A. Reflexive and symmetric
B. Symmetric and transitive
C. Only transitive
D. An equivalence relation

Answer 1

Hint: First we will check whether the given relation is reflexive by putting b = a in the given inequality. Then we swap the value of a and b to check whether the given relation is a symmetric or not. To check whether the given relation transitive we will take an example.

Complete step by step solution: Given that a relation R in the set R of real numbers be defined as \[\left( {a,{\rm{ }}b} \right)\; \in R\] if and only if \[1 + ab > 0\] for all \[a,b \in R\].
Now we will put b = a in the given inequality:
\[1 + a \cdot a > 0\]
Subtract 1 from both sides:
\[{a^2} > - 1\]
We know that the square of a number is always greater than or equal to zero. In other words, the square of a number is always greater than -1.
Thus \[{a^2} > - 1\] is true.
Thus \[\left( {a,a} \right) \in R\], Hence R is reflexive.

 We know that, multiplication follows the commutative law.
The given inequality can be written in the form \[1 + ba > 0\]. This implies \[\left( {b,a} \right) \in R\]. Hence R is symmetric.
For transitive we will use an example:
Putting \[a = 2\] and \[b = - \dfrac{1}{4}\] in the given inequality:
\[1 + 2 \cdot \left( { - \dfrac{1}{4}} \right) > 0\]
\[ \Rightarrow 1 - \dfrac{1}{2} > 0\]
\[ \Rightarrow \dfrac{1}{2} > 0\]
Hence \[\left( {2, - \dfrac{1}{4}} \right) \in R\]
Putting \[a = - \dfrac{1}{4}\]and \[b = - 1\] in the given inequality:
\[1 + \left( { - \dfrac{1}{4}} \right) \cdot \left( { - 1} \right) > 0\]
\[ \Rightarrow 1 + \dfrac{1}{4} > 0\]
\[ \Rightarrow \dfrac{5}{4} > 0\]
Hence \[\left( { - \dfrac{1}{4}, - 1} \right) \in R\]
According to the transitive, \[\left( {2, - 1} \right) \in R\]
But \[1 + 2 \cdot \left( { - 1} \right) = - 1 < 0\]. This implies \[\left( {2, - 1} \right) \notin R\]
Hence R is not transitive.

Option ‘A’ is correct

Note: Students often do mistake to check transitive property. They used the inequality to check it which is incorrect way. Here we have to use counter example to check it.