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Let $\mathrm{R}$ be relation defined on the set of natural number $\mathrm{N}$ as follows, $R=\{(x, y): x \in N, y \in N, 2 x+y=41\}$. Find the domain and range of the relation $R$. Also verify whether $R$ is reflexive, symmetric and transitive.
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We have,
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x} \in \mathrm{N}, \mathrm{y} \in \mathrm{N}, 2 \mathrm{x}+\mathrm{y}=41\}$
Domain $=\{1,2,3, \ldots, 20\}$
Range $=\{1,3,5,7, \ldots \ldots ., 39\}$
$\mathrm{R}=\{(1,39),(2,37),(3,35), \ldots .,(19,3),(20,1)\}$
$\mathrm{R}$ is not reflexive as $(2,2) \notin \mathrm{R}[\because \quad 2 \times 2+2 \neq 41]$
Also, $R$ is not symmetric.
As $\quad(1,39) \in \operatorname{R}$ but $(39,1) \notin \mathrm{R}$
$R$ is not transitive.
As $(11,19) \in \mathrm{R},(19,3) \in \mathrm{R}$
But $\quad(11,3) \notin \mathrm{R}$
Hence, $\mathrm{R}$ is neither reflexive, nor symmetric and nor transitive.
$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}): \mathrm{x} \in \mathrm{N}, \mathrm{y} \in \mathrm{N}, 2 \mathrm{x}+\mathrm{y}=41\}$
Domain $=\{1,2,3, \ldots, 20\}$
Range $=\{1,3,5,7, \ldots \ldots ., 39\}$
$\mathrm{R}=\{(1,39),(2,37),(3,35), \ldots .,(19,3),(20,1)\}$
$\mathrm{R}$ is not reflexive as $(2,2) \notin \mathrm{R}[\because \quad 2 \times 2+2 \neq 41]$
Also, $R$ is not symmetric.
As $\quad(1,39) \in \operatorname{R}$ but $(39,1) \notin \mathrm{R}$
$R$ is not transitive.
As $(11,19) \in \mathrm{R},(19,3) \in \mathrm{R}$
But $\quad(11,3) \notin \mathrm{R}$
Hence, $\mathrm{R}$ is neither reflexive, nor symmetric and nor transitive.
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