Question

An integer $m$ is said to be related to another integer $n$ if $m$ is a multiple of $n$. Then find the type of relation.
A. Reflexive and symmetric
B. Reflexive and transitive
C. Symmetric and transitive
D. Equivalence relation

Answer 1

Hint: First we will assume a relation $R$ that is defined by an integer $m$ is said to be related to another integer $n$ if $m$ is a multiple of $n$. Then we will check whether it satisfies the reflexive property, symmetric property, and transitive property.

Formula Used:
Equivalent Relation:
If $\left( {x,x} \right) \in R$ then $R$ is reflexive.
If $\left( {x,y} \right) \in R$ and $\left( {y,x} \right) \in R$ then $R$ is symmetric.
If $\left( {x,y} \right) \in R$ and $\left( {y,z} \right) \in R$ implies $\left( {x,z} \right) \in R$, then $R$ is transitive.

Complete step by step solution
Given that, an integer $m$ is said to be related to another integer $n$ if $m$ is a multiple of $n$. This means $n$ divisible by $m$.
We know that a number is divisible by itself.
Let $m$ be a number. Then $m$ is divisible by $m$.
Thus, $\left( {m,m} \right) \in R$. Hence the relation is reflexive.

Let $m$ and $n$ be two numbers and $\left( {m,n} \right) \in R$. Since $\left( {m,n} \right) \in R$, so $m$ is divisible by $n$.
Therefore, $\dfrac{m}{n} = k$ where $k$ is an integer.
$ \Rightarrow \dfrac{n}{m} = \dfrac{1}{k}$, but $\dfrac{1}{k}$ is not an integer.
Thus, $n$ is not divisible by $m$. So, $\left( {n,m} \right) \notin R$.
Hence $R$ is not symmetric.

Let $m$, $n$, and $p$ be two numbers and $\left( {m,n} \right) \in R$ and $\left( {n,p} \right) \in R$.
Since $\left( {m,n} \right) \in R$, so $m$ is divisible by $n$. This implies $\dfrac{m}{n} = {k_1}$ where ${k_1}$ is an integer.
Since $\left( {n,p} \right) \in R$, so $n$ is divisible by $p$. This implies $\dfrac{n}{p} = {k_2}$ where ${k_2}$ is an integer.
Multiply $\dfrac{m}{n} = {k_1}$ and $\dfrac{n}{p} = {k_2}$
$\dfrac{m}{n} \cdot \dfrac{n}{p} = {k_1} \cdot {k_2}$
$ \Rightarrow \dfrac{m}{p} = {k_1}{k_2}$
Since ${k_1}{k_2}$ is an integer. So $m$ is divisible by $p$. This implies $\left( {m,p} \right) \in R$.
Hence $R$ is transitive.

Option ‘C’ is correct

Note: Whenever you get this type of question the key concept of solving is you have first knowledge of all the relations like reflexive, symmetric, and transitive. A relation is equivalent if and only if the relation is reflexive, symmetric, and transitive. Since the relation is not symmetric, the relation never is an equivalence relation.