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Let $\mathrm{R}$ be a relation on the set $\mathrm{N}$ of natural numbers defined
by 'nRM $\Leftrightarrow \mathrm{n}$ is a factor of $\mathrm{m}$ '. Then which one of the following is correct?
Options:
by 'nRM $\Leftrightarrow \mathrm{n}$ is a factor of $\mathrm{m}$ '. Then which one of the following is correct?
Solution:
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Verified Answer
The correct answer is:
$\mathrm{R}$ is reflexive, transitive but not symmetric
$n R m \Leftrightarrow n$ is a factor of $m$ $\Rightarrow m$ is divisible by $n$. Reflexivity We know that
$n$ is divisible by $n \forall n \in N$
$(n, n) \in R \forall n \in N$
$R$ is reflexive. Symmetric
$n, m \in N$
Let $n=2, m=6$
$m$ is divisible by $n$ but $n$ is not divisible by $m$. Hence $R$ is not symmetric. Transitivity
Let $(n, m) \in R$ and $(m, p) \in R$ then $(n, m) \in R$ and
$(m, p) \in R \Rightarrow(\mathrm{n}, \mathrm{p}) \in \mathrm{R}$
or If $m$ is divisible by $n$ and $p$ is divisible by $m$. Hence $p$ is divisible by $n$.
$(n, p) \in R \forall n, p \in N$
$R$ is transitive relation on $N$. Hence $R$ is reflexive, transitive but not symmetric.
$\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x$ if $f(2 a-x)=f x$
$n$ is divisible by $n \forall n \in N$
$(n, n) \in R \forall n \in N$
$R$ is reflexive. Symmetric
$n, m \in N$
Let $n=2, m=6$
$m$ is divisible by $n$ but $n$ is not divisible by $m$. Hence $R$ is not symmetric. Transitivity
Let $(n, m) \in R$ and $(m, p) \in R$ then $(n, m) \in R$ and
$(m, p) \in R \Rightarrow(\mathrm{n}, \mathrm{p}) \in \mathrm{R}$
or If $m$ is divisible by $n$ and $p$ is divisible by $m$. Hence $p$ is divisible by $n$.
$(n, p) \in R \forall n, p \in N$
$R$ is transitive relation on $N$. Hence $R$ is reflexive, transitive but not symmetric.
$\int_{0}^{2 a} f(x) d x=2 \int_{0}^{a} f(x) d x$ if $f(2 a-x)=f x$
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