Let R be relation defined on the set of natural number N as follows:

    \[\mathbf{R}\text{ }=\text{ }\{\left( \mathbf{x},\text{ }\mathbf{y} \right):\text{ }\mathbf{x}\in \mathbf{N},\text{ }\mathbf{y}\in \mathbf{N},\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{41}\}\]

. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive.
Let R be relation defined on the set of natural number N as follows:

    \[\mathbf{R}\text{ }=\text{ }\{\left( \mathbf{x},\text{ }\mathbf{y} \right):\text{ }\mathbf{x}\in \mathbf{N},\text{ }\mathbf{y}\in \mathbf{N},\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{41}\}\]

. Find the domain and range of the relation R. Also verify whether R is reflexive, symmetric and transitive.

Given function:

    \[\mathbf{R}\text{ }=\text{ }\{\left( \mathbf{x},\text{ }\mathbf{y} \right):\text{ }\mathbf{x}\in \mathbf{N},\text{ }\mathbf{y}\in \mathbf{N},\text{ }\mathbf{2x}\text{ }+\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{41}\}\]

.

So, the domain =

    \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots ..,\text{ }20 \right\}\text{ }[Since,\text{ }y\in N\text{ }]\]

Finding the range, we have

    \[R\text{ }=\text{ }\left\{ \left( 1,\text{ }39 \right),\text{ }\left( 2,\text{ }37 \right),\text{ }\left( 3,\text{ }35 \right),\text{ }\ldots .,\text{ }\left( 19,\text{ }3 \right),\text{ }\left( 20,\text{ }1 \right) \right\}\]

Thus, Range of the function =

    \[\left\{ 1,\text{ }3,\text{ }5,\text{ }\ldots ..,\text{ }39 \right\}\]

R is not reflexive as (2,2) \notin R (as 2 x 2+2 \neq 41)

Also, R is not symmetric as

    \[\left( 1,\text{ }39 \right)\in R\]

but

    \[\left( 39,\text{ }1 \right)\notin R\]

Further R is not transitive as

    \[\left( 11,\text{ }19 \right)\notin R,\text{ }\left( 19,\text{ }3 \right)\notin R\]

; but

    \[\left( 11,\text{ }3 \right)\notin R\]

.

Therefore, R is neither reflexive nor symmetric and nor transitive.