Question

Let $R=\left\{\right(x,y):x,y\in Nand{x}^2-4xy+3y^2=0\}$, where $N$ is the set of all natural numbers. Then the relation $R$ is

• A. Reflexive but neither symmetry nor transistive
• B. symmetric and transistive
• C. reflexive and symmetric
• D. reflexive and transistive

Answer 1

The correct option is D reflexive and transistive

$R=(x,y):x,yENand{x}^2-4xy+3y^2=0$

For reflxurty

Let $a\in N$

The $a^2-4a.a+3a^2=a^2-4a^2+3a^2=0$

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

Thus $R$ is reflexive

For symmetry

Here, $(3,1)\in R$ is $3^2-4\left(3\right)\left(1\right)+3(1{)}^2=9-12+3=0$

For $(1,3)$

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

$=16\ne 0$

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

Thus $R$ is not symmetric.

For transiting

Let $(a,b),(b,c)\in R$ is $a^2-4ab+3b^2=0---------\left(1\right)$

and $b^2-4bc+3e^2=-----------\left(2\right)$

Subtracting $c\times \left(1\right)$ and $a\times \left(2\right)$ we get

$c{a}^2-4abc+3c{b}^2-a{b}^2+4abc-3a{c}^2=0$

$\Rightarrow c{a}^2-a{b}^2+3c{b}^2-3a{c}^2=0$

$\Rightarrow a(ca-b^2)+3c(b^2-ac)=0$

$\Rightarrow a(ca-b^2)-3c(ac-b^2)=0$

$\Rightarrow (a-3c)(ca-b^2)=0$

$\Rightarrow a-3x=0$ or $ca-b^2=0$

$\Rightarrow a-3c$ or $ca=b^2$

$a=\frac{b^2}{c}$ which is rejected since discussion of natural no. is not always a natural no. Eq $\frac{2}{3}$ where $2,3\in N$

Now, $a^2-4ac+3c^2=(3c{)}^2-4(3c\left)\right(c)+3c^2$

$=9c^2-12c^2+3c^2$

$=12c^2-12c^2$

$=0$ & $(a,c)\in R$.

$\therefore R$ is transitive