
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
163.2k+ views
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.
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.
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