
In the set \[A = \left\{ {1,2,3,4,5} \right\}\] a relation \[R\] is defined by \[R = \left\{ {\left( {x,y} \right)|x,y \in A{\text{ and x < y}}} \right\}\] . then \[R\] is
A.Reflexive
B.Symmetric
C.Transitive
D.None of these
Answer
490.2k+ views
Hint: The reflexivity , symmetric and transitive nature of \[R\] should be checked with the elements of \[A\] only as it is defined that \[x,y \in A\] and when checking the above conditions also considered the condition \[{\text{x < y}}\] .
Complete step-by-step answer:
Given : \[A = \left\{ {1,2,3,4,5} \right\}\] and \[R = \left\{ {\left( {x,y} \right)|x,y \in A{\text{ and x < y}}} \right\}\]
First we will check the reflexivity of relation \[R\] .
Reflexive:
For that lets say \[R:(x,y)\] . Now , let \[a\] be a element such that
\[a \in x\] and \[a \in y\] , for reflexive \[\left( {a,a} \right) \in R\]
But the condition \[{\text{x < y}}\] is not satisfied here as \[a \nless.a\]
Therefore the relation \[R\] is not Reflexive in nature .
Symmetric:
lets say \[R:(x,y)\] . Now , let \[a,b\] be elements of \[A\] such that \[a \in x\] and \[b \in y\]
Now , for symmetric \[(x,y) \in R\] and \[(y,x) \in R\] .
But the condition \[{\text{x < y}}\] is not satisfied here as if \[{\text{a < b}}\] then it should be \[b > a\] .
Therefore , the given relation is symmetric in nature .
Transitive:
Let us take an element \[z\] such that \[xRy\] and \[yRz\] which means that \[x\] is related to \[y\] and the element \[z\] is related to \[y\] .
Therefore , \[x < y\] and \[y < z\], which results in
\[x < z\] .
Therefore , \[x\] is related to \[z\] (\[x < z\]) , which fulfills the condition for the transitive nature of relation \[R\].
For example :
\[1 < 2\] and \[2 < 4\]
therefore , \[1 < 4\]
Hence \[R\] is transitive in nature .
Therefore , option ( C ) is the correct answer for the given question .
So, the correct answer is “Option C”.
Note: Let \[R\] be a relation from set \[A\] to set \[B\] . Then the set of all first components of the ordered pairs belonging to \[R\] is called the DOMAIN of \[R\] , while the set of all the components of ordered pairs in \[R\] is called the RANGE of \[R\] .
Complete step-by-step answer:
Given : \[A = \left\{ {1,2,3,4,5} \right\}\] and \[R = \left\{ {\left( {x,y} \right)|x,y \in A{\text{ and x < y}}} \right\}\]
First we will check the reflexivity of relation \[R\] .
Reflexive:
For that lets say \[R:(x,y)\] . Now , let \[a\] be a element such that
\[a \in x\] and \[a \in y\] , for reflexive \[\left( {a,a} \right) \in R\]
But the condition \[{\text{x < y}}\] is not satisfied here as \[a \nless.a\]
Therefore the relation \[R\] is not Reflexive in nature .
Symmetric:
lets say \[R:(x,y)\] . Now , let \[a,b\] be elements of \[A\] such that \[a \in x\] and \[b \in y\]
Now , for symmetric \[(x,y) \in R\] and \[(y,x) \in R\] .
But the condition \[{\text{x < y}}\] is not satisfied here as if \[{\text{a < b}}\] then it should be \[b > a\] .
Therefore , the given relation is symmetric in nature .
Transitive:
Let us take an element \[z\] such that \[xRy\] and \[yRz\] which means that \[x\] is related to \[y\] and the element \[z\] is related to \[y\] .
Therefore , \[x < y\] and \[y < z\], which results in
\[x < z\] .
Therefore , \[x\] is related to \[z\] (\[x < z\]) , which fulfills the condition for the transitive nature of relation \[R\].
For example :
\[1 < 2\] and \[2 < 4\]
therefore , \[1 < 4\]
Hence \[R\] is transitive in nature .
Therefore , option ( C ) is the correct answer for the given question .
So, the correct answer is “Option C”.
Note: Let \[R\] be a relation from set \[A\] to set \[B\] . Then the set of all first components of the ordered pairs belonging to \[R\] is called the DOMAIN of \[R\] , while the set of all the components of ordered pairs in \[R\] is called the RANGE of \[R\] .
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