Question

Let a relation R be defined on the set N of natural numbers as-
$\left( {{\text{x}},\;{\text{y}}} \right) \Leftrightarrow {{\text{x}}^2} - 4xy + 3{{\text{y}}^2} = 0\;\forall \;{\text{x}},\;{\text{y}} \in {\text{N}}$. The relation R is-
A. Reflexive
B. Symmetric
C. Transitive
D. Equivalence

Answer 1

Hint: Any relation can be classified as reflexive, symmetric and transitive. If aRa exists in the relation, then it is said to be reflexive. If aRb and bRa both exist in the relation, then it is said to be symmetric. If aRb and bRc exist implies that aRc also exists, the relation is transitive.

Complete step by step solution:
A relation R on the set A is considered to be reflexive if an element in set A is related to itself.
For R to be reflexive, xRx should exist in R. That is,
${{\text{x}}^2} - 4{\text{x}}\left( {\text{x}} \right) + 3{\left( {\text{x}} \right)^2} = 0$
${{\text{x}}^2} - 4{{\text{x}}^2} + 3{{\text{x}}^2} = 0$
Hence, R is a reflexive relation.
A relation R on the set A is considered to be symmetric if an element in set A is related to another element then the reverse should also be true.
For R to be symmetric, if xRy exists then yRx also exists in R.
${{\text{x}}^2} - 4xy + 3{{\text{y}}^2} = 0$
${{\text{x}}^2} - xy - 3xy + 3{{\text{y}}^2} = 0$
(x - y)(x - 3y) = 0
On substituting x = 3 and y = 1,
(3 - 1)(3 - 3) = 0
On substituting x = 1 and y = 3,
(1 - 3)(1 - 9) = 16
Hence, we can see that (3, 1) belongs to R but (1, 3) does not. Hence, the relation is not symmetric.
For the relation R on a set A to be transitive, if an element a is related to b, and element b is related to c, then element a should be related to element c.
For R to be transitive, if xRy and yRz exist then xRz should also exist.
${{\text{x}}^2} - 4xy + 3{{\text{y}}^2} = 0$
${{\text{x}}^2} - xy - 3xy + 3{{\text{y}}^2} = 0$
(x - y)(x - 3y) = 0....(1)
Similarly,
(y - z)(y - 3z) = 0....(2)
On substituting x = 9 and y = 3 in equation (1),
(9 - 3)(9 - 9) = 0
On substituting y = 3 and z = 1 in equation (2),
(3 - 1)(3 - 3) = 0
Now, we can see that xRy and yRz exist, so we will check for xRz as-
On substituting x = 9 and z = 1,
(x - z)(x - 3z) = 0
(9 - 1)(9 - 3) = 8(6) = 48
This is not equal to 0, hence, xRz does not exist and R is not transitive.
Since, R is reflexive, not symmetric and not transitive. So, it cannot be an equivalence relation. So options B, C and D are eliminated. Option A is the correct choice.

Note: It is important to check carefully for each condition. It is also recommended to check and verify each condition using a suitable example. Even if one case is false, the condition is not verified. Also if it is not possible to prove that relation is symmetric, reflexive or transitive, then use a suitable example to show that it is not.