
R is a relation over the set of real numbers, and it is given by mn≥0. Then R is
A. Symmetric and transitive
B. Reflexive and symmetric
C. A partial order relation
D. An equivalence relation
Answer
233.1k+ views
Hint: Use the concept of relation to solve the problem which states that in mathematics, a relation describes the connection between two distinct collections of data. If more than two non-empty sets are taken into consideration, the relationship between them will be determined if there is a link between their items. A relation is described as a collection of ordered pairs. We will then use the properties of relation to prove its identity.
Formula Used:
The reflexive relation is established by: $\left( {{{a}},{{a}}} \right) \in {{R}}$
The symmetric relation is established by: ${{aRb}} \Rightarrow {{bRa}},\forall {{a}},{{b}} \in {{A}}$
The transitive relation is established by: ${{aRb}}{{and}}{{bRc}} \Rightarrow {{aRc}}\forall {{a}},{{b}},{{c}} \in {{A}}$
The relation is said to be equivalent if it is reflexive, transitive and symmetric at the same time.
Complete step by step solution:
Here, we are given that
m R n if $mn \ge 0$
Checking for Reflexivity:
Since, we have real numbers
${m^2} \ge 0$
${{mm}} \ge 0$
${{mRm}}$
This proves R is reflexive
Checking for Symmetry:
Assuming ${{mRm}}$
${{mn}} \ge 0$
Since, product of real numbers is commutative, we can say
${{nm}} \ge 0$
${{nRm}}$
This proves R is symmetric.
Checking for transitive
Assuming $mRn,nRp$
Since, $mn \ge 0;np \ge 0$ we can state that $m,n$ and $p$ are of same signs
Further,
$mp \ge 0$
$mRp$
This proves, R is transitive.
As we can see the given sets are reflexive, symmetric and transitive, so we can state that it’s an equivalence relation.
Option ‘D’ is correct
Note: Before attempting this question, one should have prior knowledge about the concept of relations and also remember that when a relation is reflexive, symmetric and transitive at the same time such relation is equivalence relation, use this information to approach the solution.
Formula Used:
The reflexive relation is established by: $\left( {{{a}},{{a}}} \right) \in {{R}}$
The symmetric relation is established by: ${{aRb}} \Rightarrow {{bRa}},\forall {{a}},{{b}} \in {{A}}$
The transitive relation is established by: ${{aRb}}{{and}}{{bRc}} \Rightarrow {{aRc}}\forall {{a}},{{b}},{{c}} \in {{A}}$
The relation is said to be equivalent if it is reflexive, transitive and symmetric at the same time.
Complete step by step solution:
Here, we are given that
m R n if $mn \ge 0$
Checking for Reflexivity:
Since, we have real numbers
${m^2} \ge 0$
${{mm}} \ge 0$
${{mRm}}$
This proves R is reflexive
Checking for Symmetry:
Assuming ${{mRm}}$
${{mn}} \ge 0$
Since, product of real numbers is commutative, we can say
${{nm}} \ge 0$
${{nRm}}$
This proves R is symmetric.
Checking for transitive
Assuming $mRn,nRp$
Since, $mn \ge 0;np \ge 0$ we can state that $m,n$ and $p$ are of same signs
Further,
$mp \ge 0$
$mRp$
This proves, R is transitive.
As we can see the given sets are reflexive, symmetric and transitive, so we can state that it’s an equivalence relation.
Option ‘D’ is correct
Note: Before attempting this question, one should have prior knowledge about the concept of relations and also remember that when a relation is reflexive, symmetric and transitive at the same time such relation is equivalence relation, use this information to approach the solution.
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