Question

Identify which of the following relations are reflexive, symmetric, and transitive.

R = {(a, b) : a, b ∈ Z, a - b is an integer}
R = {(a, b) : a, b ∈ N, a + b is even}	✓	✓	x
R = {(a, b) : a, b ∈ N, a divides b}
R = {(a, b) : a, b ∈ N, a2 - 4ab + 3b2 = 0}
R = {(a, b) : a is sister of b and a, b ∈ N = G = Set of girls}
R = {(a, b) : Line a is perpendicular to line b in a place}
R = {(a, b) : a, b ∈ R, a < b}
R = {(a, b) : a, b ∈ R, a \(\leq\) b3}

Answer 1

i.	R = {(a, b) : a, b ∈ Z, a - b is an integer}	✓	✓	✓
ii.	R = {(a, b) : a, b ∈ N, a + b is even}	✓	✓	✓
iii.	R = {(a, b) : a, b ∈ N, a divides b}	✓	x	✓
iv.	R = {(a, b) : a, b ∈ N, a2 - 4ab + 3b2 = 0}	✓	x	x
v.	R = {(a, b) : a is sister of b and a, b ∈ N = G = Set of girls}	x	✓	✓
vi.	R = {(a, b) : Line a is perpendicular to line b in a place}	x	✓	x
vii.	R = {(a, b) : a, b ∈ R, a < b}	x	x	✓
viii.	R = {(a, b) : a, b ∈ R, a \(\leq\) b3}	x	x	x

(i) Given, R = {(a, b): a, b ∈ Z, a – b is an integer}

Let a ∈ Z, then a – a ∈ Z ∴ (a, a) ∈ R

∴ R is reflexive.

Let (a, b) ∈ R

∴ a – b ∈ Z

∴ -(a – b) ∈ Z, i.e., b – a ∈ Z

∴ (b, a) ∈ R

∴ R is symmetric.

Let (a, b) and (b, c) ∈ R

∴ a – b ∈ Z and b – c ∈ Z

∴ (a – b) + (b – c) ∈ Z

∴ a – c ∈ Z

∴ (a, c) ∈ R

∴ R is transitive.

(ii) Given, R = {(a, b) : a, b ∈ N, a + b is even

Let a ∈ N, then a + a = 2a, which is even.

∴ (a, a) ∈ R

∴ R is reflexive.

Let (a, b) ∈ R

∴ a + b is even

∴ b + a is even

∴ (b, a) ∈ R

∴ R is symmetric.

Let (a, b) and (b, c) ∈ R

∴ a + b and b + c is even

Let a + b = 2x and b + c = 2y for x, y ∈ N

∴ (a + b) + (b + c) = 2x + 2y

∴ a + 2b + c = 2(x + y)

∴ a + c = 2(x + y) – 2b = 2(x + y – b)

∴ a + c is even ……..[∵ x, y, b ∈ N, x + y – b ∈ N]

∴ (a, c) ∈ R ∴ R is transitive.

(iii) Given, R = {(a, b) : a, b ∈ N, a divides b}

Let a ∈ N, then a divides a.

∴ (a, a) ∈ R ∴ R is reflexive.

Let a = 2 and b = 8, then 2 divides 8

∴ (a, b) ∈ R But 8 does not divide 2.

∴ (b, a) ∉ R

∴ R is not symmetric.

Let (a, b) and (b, c) ∈ R

∴ a divides b and b divides c.

Let b = ax and c = by for x, y ∈ N.

∴ c = (ax) y = a(xy)

i.e., a divides c.

∴ (a, c) ∈ R

∴ R is transitive.

(iv) Given, R = {(a, b) : a, b ∈ N, a – 4ab + 3b = 0}

Let a ∈ N, then a² – 4aa + 3a² = a² – 4a² + 3a² = 0

∴ (a, a) ∈ R

∴ R is reflexive.

Let a = 3 and b = 1,

then a² – 4ab + 3b² = 9 – 12 + 3 = 0

∴ (a, b) ∈ R

Consider, b² – 4ba + 3a² = 1 – 12 + 9 = -2 ≠ 0

∴ (b, a) ∉ R

∴ R is not symmetric.

Let a = 3, b = 1 and c = 1/3,

then a² – 4ab + 3b² = 9 – 12 + 3 = 0

and b² – 4bc + 3c² = 1 – 4/3 + 1/3 = 1 – 1 = 0

∴ we get (a, b) and (b, c) ∈ R.

Consider, a² – 4ac + 3c² = 9 – 4 +1/3 =16/3 ≠ 0

∴ (a, c) ∉ R

∴ R is not transitive.

(v) Given, R = {(a, b) : a is sister of b and a, b ∈ G = Set of girls}

Let a ∈ G, then ‘a’ cannot be a sister of herself.

∴ (a, a) ∉ R

∴ R is not reflexive.

Let (a, b) ∈ R ∴ ‘a’ is a sister of ‘b’.

∴ ‘b’ is a sister of ‘a’.

∴ (b, c) ∈ R

∴ R is symmetric.

Let (a, b) and (b, c) ∈ R

∴ ‘a’ is a sister of ‘b’ and ‘b’ is a sister of ‘c’

∴ ‘a’ is a sister of ‘c’.

∴ (a, c) ∈ R

∴ R is transitive.

(vi) Given, R = {(a, b) : Line a is perpendicular to line b in a plane}

Let a be any line in the plane, then a cannot be perpendicular to itself.

∴ (a, a) ∉ R ∴ R is not reflexive.

Let (a, b) ∈ R

∴ a is perpendicular to b.

∴ b is perpendicular to a.

∴ (b, a) ∈ R. ∴ R is symmetric.

Let (a, b) and (b, c) ∈ R.

∴ a is perpendicular to b and b is perpendicular to c.

∴ a is parallel to c.

∴ (a, c) ∉ R

∴ R is not transitive.

(vii) Given, R = {(a, b) : a, b ∈ R, a < b}

Let a ∈ R, then a ≮ a.

∴ (a, a) ∉ R ∴ R is not reflexive.

Let a = 1 and b = 2, then 1 < 2

∴ (a, b) ∈ R But 2 ≮ 1

∴ (b, a) ∉ R

∴ R is not symmetric.

Let (a, b) and (b, c) ∈ R

∴ a < b and b < c

∴ a < c

∴ (a, c) ∈ R

∴ R is transitive.

(viii) Given, R = {(a, b) : a, b ∈ R, a ≤ b³ }

Let a = -3, then a³ = -27.

Here, a ≮ a

∴ (a, a) ∉ R

∴ R is not reflexive.

Let a = 2 and b = 9, then b = 729

Here, a < b

∴ (a, b) ∈ R

Consider, a³ = 8

Here, b ≮ a³ ∴ (b, a) ∉ R

∴ R is not symmetric.

Let a = 10, b = 3, c = 2

 then b³ = 27 and c³ = 8

Here, a < b³ and b < c³ .

∴ (a, b) and (b, c) ∈ R

But a ≮ c³

∴ (a, c) ∉ R.

∴ R is not transitive.