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Any relation P is defined in set R0 of non zero real numbers by following ways: 

(i) xPy ⇔ x2 + y2 = 1 

(ii) xPy ⇔ xy = 1 

(iii) xPy ⇔ (x + y) is a rational number 

(iv) xPy ⇔ x/y is a rational number 

Test the reflexivity, symmetricity and transitivity of these relations.

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(i) Given set 

R0 = Set of non-zero real number 

Relation P in R0 is defined as 

xPy = x2 + y2 = 1 ∀ x, y ∈ R0 

Reflexivity: P is not reflexivie because when 2 ∈ R0 

Then (2)2 + (2)2 ≠ 1 

so, (2, 2) ∉ R

Similarly a2 + a2 ≠ 1 

so, (a, a) ∉ R0 

Hence, P is not reflexive. 

Symmetricity: 

Let a, b ∈ R0 

In this way (a, b) ∈ P 

(a, b) ∈ P ⇒ a2 + b2 = 1 

⇒ b2 + a2 = 1 

⇒ (b, a) ∈ P 

So, (a, b) ∈ P 

⇒ (b, a) ∈ P ∀ a, b ∈ R

P is symmetric relation. 

Transitivity: 

P is not transitive relation because

P is not transitive relation. 

(ii) Given 

set R0 = {Set of real numbers} 

Relation P in R0 is defined as 

xPy ⇔ xy = 1 ∀ x, y ∈ P 

Reflexivity: 

Let 2 ∈ R0 

But 2 × 2 = 4 ≠ 1 

⇒ (2, 2) ∉ P 

Hence, P is not reflexive 

Symmetricity: 

In this way (a, b) ∈ R0 

In this way (a, b) ∈ P 

(a, b) ∈ P 

⇒ ab = 1 

⇒ ba = 1 

[∵ Multiplication of real number is commutative] 

⇒ b . a ∈ P 

⇒ (a, b) ∈ P 

⇒ (b, a) ∈ P ∀ a, b ∈ R0 

P is a symmetric relation. 

Transitivity: 

Let a, b, c ∈ R0 is in this way 

a, b ∈ P and b, c ∈ P 

(a, b) ∈ P ⇒ a.b = 1 ……(1) 

(b, c) ∈ P ⇒ b.c = 1 …..(2) 

From equation (1) and (2), 

ab/bc = 1/1 

⇒ a/c = 1 

⇒ (a, c) ≠P ……..(3) 

So, (a, b) ∈ P, (b, c) ∈ P 

but (a, c) ∉ P 

Hence, P is not transitive relation. 

(iii) Given 

set R0 = Set of real number 

Relation P in R0 is defined as

xPy ⇔ x + y is a rational number ∀ x, y ∈ R0

Reflexivity : 

Let x ∈ R0 x ∈ R0 

⇒ x + x is need not to be rational

For example √3 ∈ R0 

⇒ √3 + √3 = 2√3 is an irrational number 

So, (x, x) ∉ P 

P is not reflexive relation. 

Symmetricity: 

Let a, b ∈ R0 then (a, b) ∈ P (a, b) ∈ P 

⇒ a + b is a rational number 

⇒ (b + a) is also a rational number 

⇒ (b, a) ∈ P 

So, (a, b) ∈P 

⇒ (b, a) ∈ P ∀ a, b ∈ R0 

P is a symmetric relation. 

Transitivity: 

Let a, b, c ∈ R0 are in this way (a, b) ∈ P 

⇒ a + b is a rational number 

(b, c) ∈ P ⇒ b + c is a rational number 

⇒ a + c is not necessary a rational number 

For example: 2 + √3, -√3 + 6, √3 + √7 ∈ R0 and (2 + √3, -√3 + 6) ∈ P 

because 2 + √3 – √3 + 6 = 8 is a rational number 

(-√3 + 6, √3 + 7) ∈ P 

because -√3 + 6 + √3 + 7 = 13 is a rational number

But (2 + √3, √3 + 7) ∈ P 

because 2 + √3 + √3 + 7 = 2√3 + 9 is an irrational number 

Hence, P is not a transitive relation. 

(iv) Given: 

Set : R0 = Set of real numbers 

Relation P in R0 is defined as a rational number

xPy = x/y is a rational number ∀ x, y ∈ R0 

Reflexivity: Let a ∈ R0 

a ∈ R0 ⇒ a/a = 1 is a rational number 

⇒ (a, a) ∈ P 

P is a reflexive number. 

Symmetricity: Let a, b ∈ R0 is in this way (a, b) ∈ P 

(a, b) ∈ P 

⇒ a/b is a rational number 

⇒ b/a is a rational number 

(From infination of set of rational numbers Q) 

⇒ (b, a) ∈ P 

(a, b) ∈ P 

⇒ (b, a) ∈ P ∀ a, b ∈ R0 

P is a symmetric relation 

Transitivity: 

Let a, b, c ∈ R0 is in this way 

(a, b) ∈ P and (b, c) ∈ P 

(a, b) ∈ P ⇒ a/b is a rational number 

(b, c) ∈ P ⇒ b/c is a rational number 

⇒ (a/b)(b/c)is also a rational number. 

[∵ Multiplication of rational number is also a rational number] 

⇒ a/c is a rational number 

⇒ (a, c) ∈ P 

(a, b) ∈ P, (b, c) ∈ P 

⇒ (a, c) ∈ P ∀ a, b, c ∈ R0 

P is a transitive relation.

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