Question

Let \[S\] be the set of real numbers. A relation \[R = \left\{ {\left( {a,b} \right):1 + ab > 0} \right\}\] on \[S\]. Then which of the following option is correct for the relation.
A. Reflexive and symmetric but not transitive
B. Reflexive and transitive but not symmetric
C. Symmetric, transitive but not reflexive
D. Reflexive, transitive and symmetric
E. None of the above is true

Answer 1

Hint:To check reflexive property, we will check whether \[\left( {a,a} \right) \in R\]. To check the symmetric property, we will check whether \[\left( {a,b} \right) \in R\] implies \[\left( {b,a} \right) \in R\]. To check the transitive property, we will check whether \[\left( {a,b} \right) \in R\], \[\left( {b,c} \right) \in R\] implies \[\left( {a,c} \right) \in R\].

Formula Used:
If a relation reflexive then \[\left( {a,a} \right) \in R\].
If a relation symmetric, then \[\left( {a,b} \right) \in R\] implies \[\left( {b,a} \right) \in R\].
If a relation transitive, then \[\left( {a,b} \right) \in R\], \[\left( {b,c} \right) \in R\] implies \[\left( {a,c} \right) \in R\].

Complete step by step solution:
Given relation is \[R = \left\{ {\left( {a,b} \right):1 + ab > 0} \right\}\].
We know that the square of a real number is always greater than 0.
Let \[a\] be a real number.
So \[{a^2} > 0\]
Add 1 both sides of the inequality.
\[1 + {a^2} > 1 + 0\]
\[ \Rightarrow 1 + a \cdot a > 1\]
\[ \Rightarrow 1 + a \cdot a > 0\]
So, \[\left( {a,a} \right) \in R\]. Thus \[R\] is reflexive.
Let \[a\] and \[b\] be any two real numbers such that \[\left( {a,b} \right) \in R\].
Since \[\left( {a,b} \right) \in R\], so \[1 + ab > 0\].
We can rewrite the inequality as \[1 + ba > 0\], since multiplication follows commutative property.
The inequality \[1 + ba > 0\] implies \[\left( {b,a} \right) \in R\].
Thus \[R\] is symmetric.

Let \[a\], \[b\], and \[c\] be any three real numbers such that \[\left( {a,b} \right) \in R\] and \[\left( {b,c} \right) \in R\].
Let \[a = - 12\], \[b = - 3\], and \[c = \frac{1}{6}\]
\[1 + ab = 1 + \left( { - 12} \right)\left( { - 3} \right) = 37 > 0\]
\[1 + bc = 1 + \left( { - 3} \right)\left( {\frac{1}{6}} \right) = 0.5 > 0\]
\[1 + ac = 1 + \left( { - 12} \right)\left( {\frac{1}{6}} \right) = - 2\ngtr{ 0}\]
Here \[1 + ac\ngtr{ 0 }\], so \[\left( {a,c} \right) \notin R\].
Hence \[R\] is not transitive.

Hence option A is the correct option.

Note:Student often use the inequalities \[1 + ab > 0\], \[1 + bc > 0\] to reach the inequality \[1 + ac > 0\]. But in this process, it impossible to reach the result. The easiest way to check transitive property is by letting the value of \[a\], \[b\] and \[c\]and then put the value of it’s in the inequality.