Question

Let a relation $R$ is the set $N$ of natural numbers be defined as $(x,y)\in R$ if and only if $x^2-4xy+3y^2=0$ for all $x,y\in N$. The relation $R$ is

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

Answer 1

Answer: (A)

Reflexive

We have $R=\{(x,y):x^2-4xy+3y^2=0,x,y\in N\}$
Let $x=N$
$x^2-4xx+3x^2=4x^2-4x^2=0$
$\therefore (x,x)\in R$. Hence, $R$ is reflexive.
We have ${\left(3\right)}^2-4\left(3\right)\left(1\right)+3{\left(1\right)}^2=9-12+3=0$
$\therefore (3,1)\in R$
Also, ${\left(1\right)}^2-4\left(1\right)\left(3\right)+3{\left(3\right)}^2=1-12+27=16\ne 0$
$\therefore (1,3)\notin R$. Hence, $R$ is not symmetric.
Now, $(9,3)\in R$ because ${\left(9\right)}^2-4\left(9\right)\left(3\right)+3{\left(3\right)}^2=81-108+27=0$
Also, $(3,1)\in R$ because ${\left(3\right)}^2-4\left(3\right)\left(1\right)+3{\left(1\right)}^2=9-12+3=0$
Now, $(9,1)\in R$ but ${\left(9\right)}^2-4\left(9\right)\left(1\right)+3{\left(1\right)}^2=81-36+3=48\ne 0$
$\therefore (9,3),(3,1)\in R$ and $(9,1)\in R$
$\therefore R$ is not transitive.