Question

Let a relation $r$ in the set $N$ of natural numbers be defined as $(x,y)\iff x^2–4xy+3y^2=0,\forall x,y\in N$. The relation $R$ is

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

Answer 1

The correct option is A

Reflexive

Explanation for the correct option:

Checking reflexivity of relation:

The relation $R$ in the set $N$ of natural numbers be defined as $(x,y)\iff x^2–4xy+3y^2=0,\forall x,y\in N$

$$\begin{gathered} \Rightarrow \left(x,y\right)\iff x^2–3xy-xy+3y^2=0,\forall x,y\in N....\left(i\right) \\ \Rightarrow \left(x,y\right)\iff x\left(x–3y\right)-y\left(x-3y\right)=0,\forall x,y\in N \\ \Rightarrow \left(x,y\right)\iff \left(x–3y\right)\left(x-y\right)=0,\forall x,y\in N....\left(ii\right) \end{gathered}$$

Considering $\left(x,y\right)=\left(a,a\right),\forall a\in N$

$$\begin{gathered} \Rightarrow \left(a–3a\right)\left(a-a\right)\left[\text{from}\left(ii\right)\right] \\ =\left(-2a\right)\left(0\right) \\ =0 \end{gathered}$$

Thus, $\left(a,a\right)\in R$ so it is reflexive relation.

Hence, option A is the correct answer.

Explanation for incorrect options:

Option(B):

Checking symmetricity of relation:

Considering $\left(x,y\right)=\left(3,1\right)$

$$\begin{gathered} \Rightarrow (3–3\times 1)(3–1)\left[\text{from}\left(ii\right)\right] \\ =\left(0\right)\left(2\right) \\ =0 \end{gathered}$$

Thus, $(3,1)\in R$

Again Considering $(x,y)=(1,3)$

$$\begin{gathered} \Rightarrow (1–3\times 3)(1–3)\left[\text{from}\left(ii\right)\right] \\ =(-8)(-2) \\ =16 \end{gathered}$$

Therefore, $(1,3)\notin R$

Thus it is not symmetric.

Hence Option (B) is incorrect.

Option(C):

Checking Transitivity of relation:

Considering $(x,y)=(9,3)$

$$\begin{gathered} \Rightarrow (9,3)\iff 9^2-4\left(9\right)\left(3\right)+3{\left(3\right)}^2=0\left[\text{from}\left(ii\right)\right] \\ \Rightarrow (9,3)\iff 81-108+27=0 \\ \Rightarrow (9,3)\iff 0=0 \end{gathered}$$

Thus, $(9,3)\in R$

Again Considering $(x,y)=(3,1)$

$$\begin{gathered} \Rightarrow (3,1)\iff 3^2-4\left(3\right)\left(1\right)+3{\left(1\right)}^2=0\left[\text{from}\left(ii\right)\right] \\ \Rightarrow (3,1)\iff 9-12+3=0 \\ \Rightarrow (3,1)\iff 0=0 \end{gathered}$$

$(3,1)\in R$

Now Considering $Let(x,y)=(9,1)$

$$\begin{gathered} \Rightarrow (9,1)\iff 9^2-4\left(9\right)\left(1\right)+3{\left(1\right)}^2=0\left[\text{from}\left(ii\right)\right] \\ \Rightarrow (9,1)\iff 81-36+3=0 \\ \Rightarrow (9,1)\iff 48\ne 0 \end{gathered}$$

Thus, $(9,1)\notin R$

As $(3,1)\in R$ and $(3,1)\in R$ but $(9,1)\notin R$

Thus, it is not transitive.

Hence Option (C) is incorrect.

Option(D):

From above explanation

The given relation is not symmetric and not transitive.

Thus it is not an equivalence relation.

Hence Option (D) is incorrect.

Therefore, the correct answer is option (A).