Question

Let $N$ denote the set of all-natural numbers and $R$ be the relation on $N\times N$ defined by $(a,b)R(c,d)$ if $ad(b+c)=bc(a+d)$, then $R$ is

• A. Symmetric only
• B. Reflexive only
• C. Transitive only
• D. An equivalence relation

Answer 1

The correct option is D

An equivalence relation

Explanation for the correct option:

We know that a relation $R$ on a set $A$ is said to an equivalence relation it is transitive, symmetric and reflexive.

Step 1: Checking if $R$ is reflexive

Given $N$ is a set of natural numbers and $R$ is a relation on $N\times N$ defined by $(a,b)R(c,d)$ if $ad(b+c)=bc(a+d)$

Definition of a reflexive relation: We know that a relation $R$ on a set $A$ is said to be reflexive if $\left(\mathrm{a},\mathrm{a}\right)\in \mathrm{R},\forall \mathrm{a}\in \mathrm{A}$

Let $\left(a,b\right)R\left(a,b\right)$

Therefore,

$$\begin{gathered} ab\left(b+a\right)=ba\left(a+b\right) \\ \Rightarrow ab\left(b+a\right)=ab\left(b+a\right) \end{gathered}$$

This implies that $R$ is reflexive.

Step 2: Checking if $R$ is symmetric

Definition of a symmetric relation: We know that a relation $R$ on a set $A$ is said to be symmetric if $\left(\mathrm{a},\mathrm{b}\right)\in \mathrm{R}$, then $\left(\mathrm{b},\mathrm{a}\right)\in \mathrm{R}$, $\forall a,b\in A$.

Let $\left(a,b\right)R\left(c,d\right)$

Therefore,

$$\begin{gathered} ad\left(b+c\right)=bc\left(a+d\right) \\ \Rightarrow bc\left(a+d\right)=ad\left(b+c\right) \\ \Rightarrow cb\left(d+a\right)=da\left(c+b\right) \\ \Rightarrow \left(c,d\right)R\left(a,b\right) \end{gathered}$$

This implies that $R$ is symmetric.

Step 3: Checking if $R$ is transitive

Definition of a transitive relation: We know that a relation $R$ on a set $A$ is said to be transitive if $\left(\mathrm{a},\mathrm{b}\right)\in \mathrm{R}\&\left(\mathrm{b},\mathrm{c}\right)\in \mathrm{R}$ then $\left(\mathrm{a},\mathrm{c}\right)\in \mathrm{R}$, $\forall a,b,c\in A$.

Let $\left(a,b\right)R\left(c,d\right)$

Therefore,

$$\begin{gathered} ad\left(b+c\right)=bc\left(a+d\right) \\ \Rightarrow adb+adc=abc+bcd \\ \Rightarrow abd-abc=bcd-acd \\ \Rightarrow ab\left(d-c\right)=cd\left(b-a\right) \\ \Rightarrow \frac{ab}{b-a}=\frac{cd}{d-c}\left(1\right) \end{gathered}$$

And let $\left(c,d\right)R\left(e,f\right)$.

Therefore,

$$\begin{gathered} cf\left(d+e\right)=de\left(c+f\right) \\ \Rightarrow cfd+cef=ced+edf \\ \Rightarrow cfd-ced=edf-cef \\ \Rightarrow cd\left(f-e\right)=ef\left(d-c\right) \\ \Rightarrow \frac{cd}{d-c}=\frac{ef}{f-e}\left(2\right) \end{gathered}$$

From equations $\left(1\right)$ and$\left(2\right)$ we get,

$$\begin{gathered} \frac{ab}{b-a}=\frac{ef}{f-e} \\ \Rightarrow abf-abe=efb-efa \\ \Rightarrow abf+efa=efb+abe \\ \Rightarrow af\left(b+e\right)=be\left(a+f\right) \\ \Rightarrow \left(a,b\right)R\left(e,f\right) \end{gathered}$$

This implies that $R$ is transitive.

Therefore, the given relation is reflexive, symmetric and transitive. So, it is an equivalence relation.

Hence, option (D) is the correct answer.