Question

Let `R = {(x, y) : x, y in N and x^2-4xy+3y^2 = 0}`, where `N` is the set of all natural numbers. Then the relation R is
A. R is reflexive and symmetric, but not transitive
B. R is reflexive and transitive, but not symmetric
C. R is reflexive, symmetric and transitive
D. R is reflexive, but neither symmetric nor transitive

Answer 1

Correct Answer - D
Given, `x R Yimpliesx^(2)-4xy+3y^(2)=0`
`impliesx^(2)-xy-3xy+3y^(2)=0`
`impliesx(x-y)-3y(x-y)=0`
`implies(x-y)(x-3y)=0`
Reflexive property :
x R x implies (x-x)(-3x)=0
So, R is reflexive `" ….(1)"`
Symmetric property:
Let us check using an example (1, 2) and (2, 1)
for `(1, 2)implies(1-2)(1-6)=(-1)(-5)=10`
For (2, 1)implies(2-1)(2-3)=(1)(-1)=-1
So, R is not symmetric `" ....(2)"`
Transitive property :
For `(9x,3x)implies(9x-3x)(9x-9x)=0`
`"for "(3x, x)implies(3x-3x)(3x-9x)=0`
`"For "(9x, x)implies(9x-x)(9x-3x)0`
`"So, "(9x, 3x)inR,(3x, x)inR" but "(9x,x)cancelin R`
So, R is not transitive `" ....(3)"`
From (1), (2), (3), R is reflexive, but not symmetric and transitive.