Question

Let \[R\] be relation on the set \[N\] be defined by \[\left\{ {\left( {x,y} \right)|x,y \in N, 2x + y = 41} \right\}\]. Then find the type of the relation \[R\].
A. Reflexive
B. Symmetric
C. Transitive
D. None of these

Answer 1

Hint In the given question, the relation on a set of natural numbers is given. By using the definitions of reflexive, symmetric, and transitive relation, 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 on set \[N\] is, \[R = \left\{ {\left( {x,y} \right)|x,y \in N, 2x + y = 41} \right\}\].
Putting \[x = 1\] in the equation \[2x + y = 41\]
\[2 \cdot 1 + y = 41\]
\[ \Rightarrow y = 39\]

Putting \[x = 2\] in the equation \[2x + y = 41\]
\[2 \cdot 2 + y = 41\]
\[ \Rightarrow y = 37\]

Putting \[x = 3\] in the equation \[2x + y = 41\]
\[2 \cdot 3 + y = 41\]
\[ \Rightarrow y = 35\]

So on.
The relation can be written as:
\[R = \left\{ {\left( {1,39} \right),\left( {2,37} \right),\left( {3,35} \right),....,\left( {10,21} \right),\left( {11,19} \right),....,\left( {20,1} \right)} \right\}\]
Let’s check whether the relation is reflexive, symmetric, or transitive.
Reflexive:
Let \[\left( {x,x} \right) \in R\].
Putting \[y = x\] in the equation \[2x + y = 41\].
\[2x + x = 41\]
\[ \Rightarrow \]\[3x = 41\]
\[ \Rightarrow \]\[x = \dfrac{{41}}{3}\]
But \[x \notin N\]
Hence, the relation is not reflexive.

Symmetric:
If \[\left( {x,y} \right) \in R\], then \[\left( {y,x} \right) \in R\]
For \[\left( {1,39} \right)\]:
\[2\left( 1 \right) + 39 = 41\]
\[ \Rightarrow \]\[2 + 39 = 41\]
\[ \Rightarrow \]\[41 = 41\]
\[ \Rightarrow \]\[\left( {1,39} \right) \in R\]

For \[\left( {39,1} \right)\]:
\[2\left( {39} \right) + 1 = 41\]
\[ \Rightarrow \]\[78 + 1 = 41\]
\[ \Rightarrow \]\[79 \ne 41\]
\[ \Rightarrow \]\[\left( {39,1} \right) \notin R\]
Since \[\left( {1,39} \right) \in R\] does not imply \[\left( {39,1} \right) \in R\]
Hence, the relation is not symmetric.

Transitive:
If \[\left( {x,y} \right) \in R, \left( {y,z} \right) \in R\], then \[\left( {x,z} \right) \in R\]
Check the equation \[2x + y = 41\] for the point \[\left( {20,1} \right)\],
\[2\left( {20} \right) + 1 = 41\]
\[ \Rightarrow \]\[40+1=41\]
\[ \Rightarrow \]\[41 = 41\]
\[ \Rightarrow \]\[\left( {20,1} \right) \in R\]

Check the equation \[2x + y = 41\] for the point \[\left( {1,39} \right)\],
\[2\left( 1 \right) + 39 = 41\]
\[ \Rightarrow \]\[2 + 39 = 41\]
\[ \Rightarrow \]\[41 = 41\]
\[ \Rightarrow \]\[\left( {1,39} \right) \in R\]

Check the equation \[2x + y = 41\] for the point \[\left( {20,39} \right)\],
\[2\left( {20} \right) + 39 = 41\]
\[ \Rightarrow \]\[40 + 39 = 41\]
\[ \Rightarrow \]\[79 \ne 41\]
\[ \Rightarrow \]\[\left( {20,39} \right) \notin R\]
Since \[\left( {20,1} \right),\left( {1,39} \right) \in R\] does not imply \[\left( {20,39} \right) \in R\]
Hence, the relation is not transitive.
Hence the correct option is D.

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 \[a \ne b\].