Question

If a relation $R$ on the set $N$ of natural number is defined as $(x,y)\Leftrightarrow\, x^2-4xy+3y^2=0$, $x,y$ is the element of $N$, Then the relation $R$ is
(1) reflexive
(2) symmetric
(3) transitive
(4) an equivalence relation

Answer 1

Hint: Here we have given that $(x,y)<{{x}^{\wedge }}2-4xy+3{{y}^{\wedge }}2=0,x,y$ is an element of natural number. We have to find whether $R$ is symmetric, reflexive, or transitive. By finding the value of the given expression and by comparing it with the general form, we can find the solution.

Formula used:
A relation $R$ is symmetric on a set $S$ if for all $x \in S$ and for all $y \in S$, if $(x, y) \in R$ then $(y, x) \in R .
A$ relation $R$ is transitive on a set $S$ if for all $x, y, z \in S$, if $(x, y) \in R$ and $(y, z) \in R$, then $(x, z) \in R$.
In a graph of a reflexive relation, every node will have an arc back to itself.

Complete step by step Solution:
Every element in a reflexive relation maps to another element. Consider a set $A = 1, 2$, for instance. Now consider the reflexive relation $R = "(1, 1), (2, 2), (1, 2), (2, 1)"$ as an example. The reciprocal relationship is provided by:
$(a, a) \in R$
We have $R=\left\{(x, y): x^{2}-4 x y+3 y^{2}=0, x, y \in N\right\}$
Let $x \in N \cdot x^{2}-4 x x+3 x^{2}=4 x^{2}-4 x^{2}=0$
$\therefore(\mathrm{x}, \mathrm{x}) \in \mathrm{R}$.
$\therefore \mathrm{R}$ is reflexive
If $a=b$ is true in a symmetric relationship, then $b=a$ is also true. To put it another way, a relationship $R$ is only symmetric if $(b,a)\in R$is true when $(a, b) \in R$
 $R=\{(1,2),(2,1)\}$for a set $A=\{1,2\}$ will serve as an illustration of a symmetric relation. Consequently, for a symmetric relation,
$a R b \Rightarrow b R a, \forall a, b \in A$
We have $(3)^{2}-4(3)(1)+3(1)^{2}=9-12+3=0$
$\therefore(3,1) \in \mathrm{R}$.
Also $(1)^{2}-4(1)(3)+3(3)^{2}=1-12+27=16 \equiv 0$
$\therefore(1,3) \notin \mathbf{R}$.
$\therefore \mathrm{R}$ is not symmetric.

If a relation is transitive, $(x, y) \in R,(y, z) \in R$and$(x, z) \in R$
 If a relationship is transitive,
 $aRb$and$b R c \Rightarrow a R c \quad \forall a, b, c \in A$
$(9,3) \in R$ because
$(9)^{2}-4(9)(3)+3(3)^{2}=81-108+27=0$
Also $(3,1) \in \mathrm{R}$ because
$(3)^{2}-4(3)(1)+3(1)^{2}=9-12+3=0$
Now, $(9,1) \in \mathrm{R}$ if $(9)^{2}-4(9)(1)+3(1)^{2}=0$
if $81-36+3=48 \equiv 0$
which is not so.
$\therefore(9,3),(3,1) \in \mathrm{R}$ and $(9,1) \notin \mathrm{R}$
$\therefore \mathrm{R}$ is not transitive.
An equivalency relation is one that is transitive, symmetric, and reflexive all at once.

Hence, the correct option is 1.

Note: A relationship between elements of a set A in which each element is related to the others in the set is known as a reflexive relation. As the name suggests, each piece of the set features an image that is a reflection of itself. The reflexive link is a key concept in set theory.