Question

Let \[R = \left\{ {\left( {x,y} \right):x,y \in N and {x^2} - 4xy + 3{y^2} = 0} \right\}\] , where \[N\] is the set of all natural numbers. Then what is the type of the relation \[R\]?
A. Reflexive but neither symmetric nor transitive
B. Symmetric and transitive
C. Reflexive and transitive
D. Reflexive and transitive

Answer 1

Hint: Here, a relation is given. First, simplify the equation given in the relation and find out the values of \[x\] and \[y\]. Then, by using the definitions of reflexive, symmetric, and transitive relations, we will find the type of relation \[R\].

Formula Used: Let \[R\] be the relation on a set \[A\]. Then,
\[R\] is reflexive if for all \[x \in A\], \[xRx\].
\[R\] is symmetric if for all \[x,y \in A\], if \[xRy\], then \[yRx\].
\[R\] is transitive if for all \[x,y,z \in A\], if \[xRy\]and \[yRz\], then \[xRz\].

Complete step by step solution: The given relation is \[R = \left\{ {\left( {x,y} \right):x,y \in N and {x^2} - 4xy + 3{y^2} = 0} \right\}\], where \[N\] is the set of all natural numbers.

Let’s check whether the relation is reflexive, symmetric, or transitive.
We have, \[R = \left\{ {\left( {x,y} \right):x,y \in N and {x^2} - 4xy + 3{y^2} = 0} \right\}\]
Now simplify the equation \[{x^2} - 4xy + 3{y^2} = 0\].
\[{x^2} - xy - 3xy + 3{y^2} = 0\]
\[ \Rightarrow x\left( {x - y} \right) - 3y\left( {x - y} \right) = 0\]
\[ \Rightarrow \left( {x - y} \right)\left( {x - 3y} \right) = 0\]
\[ \Rightarrow \left( {x - y} \right) = 0\] or \[\left( {x - 3y} \right) = 0\]
\[ \Rightarrow x = y\] or \[x = 3y\]
Therefore, the elements of the given relation are:
\[R = \left\{ {\left( {1,1} \right),\left( {3,1} \right),\left( {2,2} \right),\left( {6,2} \right),\left( {3,3} \right),\left( {9,3} \right),....} \right\}\]

Now, check the type of the given relation.
For reflexive:
We know that, \[R\] is reflexive if for all \[x \in A\], \[xRx\].
Since, \[\left( {1,1} \right),\left( {2,2} \right),\left( {3,3} \right),...\] are the elements of the given relation \[R\].
Therefore, \[R\] is a reflexive relation.

For symmetric:
We know that \[R\] is symmetric if for all \[x,y \in A\], if \[xRy\], then \[yRx\].
Here, \[\left( {3,1} \right)\] is an element of the relation \[R\], but \[\left( {1,3} \right)\] is not an element of \[R\].
Therefore, \[R\] is not a symmetric relation.

For transitive:
We know that \[R\] is transitive if for all \[x,y,z \in A\], if \[xRy\]and \[yRz\], then \[xRz\].
In the given relation \[R\]:
\[\left( {1,1} \right) \in R\] and \[\left( {3,1} \right) \in R\] \[ \Rightarrow \left( {1,1} \right) \in R\]
Similarly, \[\left( {2,2} \right) \in R\] and \[\left( {6,2} \right) \in R\] \[ \Rightarrow \left( {2,2} \right) \in R\]
This condition is true for all elements of the relation \[R\].
Therefore, \[R\] is a transitive relation.

From the above statements, we get
The given relation \[R\] is reflexive and transitive.

Option ‘D’ is correct

Note: There is another property of relation. The name of the property is antisymmetric property. It states that \[a,b\] belong to set \[A\] and the relation \[R\] is defined on \[A\], then \[\left( {a,b} \right) \in R\] does not imply that \[\left( {b,a} \right) \in R\] when \[a \ne b\] .