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Hint: Here we have been given that $n$ is a factor of $m$. So use $(a,a)\in R\forall a\in A$, $\forall a,b,c\in X:(aRb\wedge bRc)\Rightarrow aRc$ and $(a,b)\in R\Rightarrow (b,a)\in R$. You will get the answer.

Complete step-by-step answer:

In maths, a binary relation $R$ across a set $X$ is reflexive if each element of set $X$ is related or linked to itself. In terms of relations, this can be defined with $(a,a)\in R\forall a\in X$ or $I\in R$ where I is the identity relation on A. It has a reflexive property and is said to hold reflexivity. Symmetry, transitivity and reflexivity are the three properties representing equivalence relations.

In relation and functions, a reflexive relation is the one in which every element maps to itself. For example, let us consider a set $A=\{1,2\}$. Now here the reflexive relation will be $R=\{(1,1),(2,2),(1,2),(2.1)\}$. Hence, a relation is reflexive if:

$(a,a)\in R\forall a\in A$

where $a$ is the element, $A$ is the set and $R$ is the relation.

A binary relation $R$ over a set $X$ is transitive if whenever an element $a$ is related to an element $b$, and $b$ is in turn related to an element $c$, then a is also related to $c$.

$\forall a,b,c\in X:(aRb\wedge bRc)\Rightarrow aRc$

On the other hand, "is the mother of" is not a transitive relation, because if Alice is the mother of Brenda, and Brenda is the mother of Claire, then Alice is not the mother of Claire. What is more, it is anti transitive: Alice can never be the mother of Claire.

Let $A$ be a set in which the relation $R$ defined. Then $R$ is said to be a symmetric relation, if $(a,b)\in R\Rightarrow (b,a)\in R$, that is, $aRb\Rightarrow bRa$ for all $(a,b)\in R$.

Consider, for example, the set $A$ of natural numbers. If a relation $A$ be defined by $''(x+y)''$, then this relation is symmetric in $A$, for $a+b=5\Rightarrow b+a=5$.

But in the set $A$ of natural numbers if the relation $R$ be defined as ‘$x$ is a divisor of $y$’, then the relation $R$ is not symmetric as $3R9$ does not imply $9R3$; for, $3$ divides $9$ but $9$ does not divide $3$.

For a symmetric relation $R$, ${{R}^{-1}}=R$.

Since $n$ is a factor of $n$, since every natural number is a factor of itself so the relation is reflexive.

If $n$ is a factor of $m$ and $m$ is a factor of $p$, then $n$ is surely a factor of $p$, so the relation is transitive.

If however $n$ is a factor of $m$, $m$ is not necessarily a factor of $n$ so the relation is not symmetric.

Hence, the answer is option D.

Note: Read the question carefully. You should know the concept of reflexive, transitive and symmetric. Also, you should know the basics of these and their properties. You must also know the types of properties.

Complete step-by-step answer:

In maths, a binary relation $R$ across a set $X$ is reflexive if each element of set $X$ is related or linked to itself. In terms of relations, this can be defined with $(a,a)\in R\forall a\in X$ or $I\in R$ where I is the identity relation on A. It has a reflexive property and is said to hold reflexivity. Symmetry, transitivity and reflexivity are the three properties representing equivalence relations.

In relation and functions, a reflexive relation is the one in which every element maps to itself. For example, let us consider a set $A=\{1,2\}$. Now here the reflexive relation will be $R=\{(1,1),(2,2),(1,2),(2.1)\}$. Hence, a relation is reflexive if:

$(a,a)\in R\forall a\in A$

where $a$ is the element, $A$ is the set and $R$ is the relation.

A binary relation $R$ over a set $X$ is transitive if whenever an element $a$ is related to an element $b$, and $b$ is in turn related to an element $c$, then a is also related to $c$.

$\forall a,b,c\in X:(aRb\wedge bRc)\Rightarrow aRc$

On the other hand, "is the mother of" is not a transitive relation, because if Alice is the mother of Brenda, and Brenda is the mother of Claire, then Alice is not the mother of Claire. What is more, it is anti transitive: Alice can never be the mother of Claire.

Let $A$ be a set in which the relation $R$ defined. Then $R$ is said to be a symmetric relation, if $(a,b)\in R\Rightarrow (b,a)\in R$, that is, $aRb\Rightarrow bRa$ for all $(a,b)\in R$.

Consider, for example, the set $A$ of natural numbers. If a relation $A$ be defined by $''(x+y)''$, then this relation is symmetric in $A$, for $a+b=5\Rightarrow b+a=5$.

But in the set $A$ of natural numbers if the relation $R$ be defined as ‘$x$ is a divisor of $y$’, then the relation $R$ is not symmetric as $3R9$ does not imply $9R3$; for, $3$ divides $9$ but $9$ does not divide $3$.

For a symmetric relation $R$, ${{R}^{-1}}=R$.

Since $n$ is a factor of $n$, since every natural number is a factor of itself so the relation is reflexive.

If $n$ is a factor of $m$ and $m$ is a factor of $p$, then $n$ is surely a factor of $p$, so the relation is transitive.

If however $n$ is a factor of $m$, $m$ is not necessarily a factor of $n$ so the relation is not symmetric.

Hence, the answer is option D.

Note: Read the question carefully. You should know the concept of reflexive, transitive and symmetric. Also, you should know the basics of these and their properties. You must also know the types of properties.

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