Question

Let a relation $R$ in 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 equivalent relation

Answer 1

Answer: (A)

reflexive

We have $R=\left\{\right(x,y):x^2-4xy+3y^2=0,x,y\in N\}$.

Let $x\in N.x^2-4xx+3x^2=4x^2-4x^2=0$

$\therefore (x,x)\in R.$

$\therefore R$ is reflexive

We have $(3{)}^2-4(3\left)\right(1)+3(1{)}^2=9-12+3=0$

$\therefore (3,1)\in R$.

Also $(1{)}^2-4(1\left)\right(3)+3(3{)}^2=1-12+27=16\ne 0$

$\therefore (1,3)\notin R.$

$\therefore R$ is not symmetric.

$(9,3)\in R$ because

$(9{)}^2-4(9\left)\right(3)+3(3{)}^2=81-108+27=0$

Also $(3,1)\in R$ because

$(3{)}^2-4(3\left)\right(1)+3(1{)}^2=9-12+3=0$

Now, $(9,1)\in R$ if $(9{)}^2-4(9\left)\right(1)+3(1{)}^2=0$

if $81-36+3=48\ne 0$,

which is not so.

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

$\therefore R$ is not transitive.