(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) ∉ R0
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 ∈ R0
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.