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Let $R$ be the relation defined on the set of natural number $\mathrm{N}$ as $a \mathrm{Rb} ; \mathrm{a}, \mathrm{b} \in \mathrm{N}$, if a divides $\mathrm{b}$. Then, which one of the following is correct?
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Verified Answer
The correct answer is:
$\mathrm{R}$ is reflexive and transitive
For reflexive: (d) For reflexive:
$\mathrm{aRa} \Rightarrow$ a divides a R is reflexive. For symmetric :
$\mathrm{aRb} \Rightarrow$ a divides $\mathrm{b}$
$\mathrm{bRa} \Rightarrow \mathrm{b}$ divides a
which may not be true $\Rightarrow \quad \mathrm{R}$ is not symmetric For transitive $\mathrm{aRB} \Rightarrow$ a divides $\mathrm{b} \Rightarrow \mathrm{b}=\mathrm{ka}$
$\mathrm{bRc} \Rightarrow \mathrm{b}$ divides $\mathrm{c} \Rightarrow \mathrm{c}=\mathrm{lb}$
Now, $\mathrm{c}=1 \mathrm{ka}$
$\Rightarrow$ a divides $\mathrm{c}$
$\Rightarrow$ a $R c$
$\Rightarrow \quad \mathrm{aRb}, \mathrm{bRc} \Rightarrow \mathrm{c} \mathrm{Ra}$
$\Rightarrow \mathrm{R}$ is transitive.
$\mathrm{aRa} \Rightarrow$ a divides a R is reflexive. For symmetric :
$\mathrm{aRb} \Rightarrow$ a divides $\mathrm{b}$
$\mathrm{bRa} \Rightarrow \mathrm{b}$ divides a
which may not be true $\Rightarrow \quad \mathrm{R}$ is not symmetric For transitive $\mathrm{aRB} \Rightarrow$ a divides $\mathrm{b} \Rightarrow \mathrm{b}=\mathrm{ka}$
$\mathrm{bRc} \Rightarrow \mathrm{b}$ divides $\mathrm{c} \Rightarrow \mathrm{c}=\mathrm{lb}$
Now, $\mathrm{c}=1 \mathrm{ka}$
$\Rightarrow$ a divides $\mathrm{c}$
$\Rightarrow$ a $R c$
$\Rightarrow \quad \mathrm{aRb}, \mathrm{bRc} \Rightarrow \mathrm{c} \mathrm{Ra}$
$\Rightarrow \mathrm{R}$ is transitive.
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