Question

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

Relation	Reflexive	Symmetric	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 ∈ G = Set of girls}
R = {(a, b) : Line a is perpendicular to line b in a plane}
R = {(a, b) : a, b ∈ R, a < b}
R = {(a, b) : a, b ∈ R, a ≤ b3}

Answer 1

i. R = {(a, b)/a, b ∈ Z, a - b is an integer}

Let a, b, c ∈ Z

∵ a − a = 0 ∈ Z

∴ aRa ∀ a ∈ Z

∴ R is reflective

Let aRb ∴ a − b is an integer

∴ −(a − b) = b − a is also an integer

∴ bRa

∴ aRb ⇒ bRa ∀ a, b ∈ Z

∴ R is symmetric

Let aRb, bRc

∴ a − b, b − c are integers

∴ (a − b) + (b − c) = a − c is an integer

∴ aRb, bRc ⇒ aRc ∀ a, b, c ∈ Z

∴ R is transitive.

ii. R = {(a, b) / a, b ∈ N, a + b is even}

Let a, b, c ∈ N

a + a = 2a is even

∴ aRa ∀ a ∈ N

R is reflexive

Let aRb . ·. a + b is even

∴ b + a is even

∴ bRa

∴ aRb ⇒ bRa ∀ a, b ∈ N

∴ R is symmetric

Let aRb, bRc

∴ a + b, b + c are even

Let a + b = 2m, b + c = 2n

∴ (a + b) + (b + c) = 2m + 2n

∴ a + c = 2m + 2n -− 2b = 2 (m + n − b) is even

∴ aRc

∴ aRb, bRc ⇒ aRc ∀ a, b, c ∈ N

∴ R is transitive.

iii. R = {(a, b) / a, b ∈ N, a divides b}

∵ a divides a   aRa ∀ a ∈ N

∵ R is reflexive

Let a = 2, b = 4

∴ 2 divides 4 so that aRb

But 4 does not divide 2 ∴ `bcancelRa`

∴ aRb `cancel=> bRc`

∴ R is not symmetric

Let aRb, bRc

∴ a divides b, b divides c

∴ b = am, c = bn, m, n ∈ N

∴ c = bn = (am)n = a(mn)

∴ a divides c  ∴ aRc

∴ aRb, bRc ⇒ aRc ∀ a, b, c ∈ N

∴ R is transitive.

iv. R = {(a, b) / a, b ∈ N, a² − 4ab + 3b² = 0}

aRb if a2 − 4ab + 3b² = 0

i.e., if (a − b)(a − 3b) = 0

i.e., if a = b or a = 3b

a = a ∴ aRa ∀ a ∈ N

R is reflexive

Let a = 27, b = 9

∴ a = 3b ∴ aRb

`b cancel=a and b cancel= 3a`

`b cancelRa`

`bRb cancel=> bRa`

R is not symmetric

Let a = 27, b = 9, c = 3

a = 3b  ∴ aRb

Also, b = 3c ∴ bRc

But `a cancel= c and a cancel= 3c`

`a cancelR c`

`aRb, bRc cancel=> aRc`

R is not transitive.

v. R = {(a, b) / a is a sister of b,

a, b ∈ G = Set of girls}

No girl is her own sister

`a cancelR a` for any a ∈ G

R is not reflexive

Let aRb

a is a sister of b

b is a sister of a

bRa

aRb ⇒ bRa ∀ a, b ∈ G

R is symmetric

Let aRb, bRc

a is a sister of b and b is a sister of c

aRc

aRb, bRc ⇒ aRc ∀ a, b, c ∈ G

R is transitive. 

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

No line is perpendicular to itself

`a cancelR a` for any line

R is not reflexive

Let aRb

a is perpendicular to b

b is perpendicular to a

bRa

aRb ⇒ bRa ∀ a, b

R is symmetric

If a is perpendicular to b and b is perpendicular to c, then a is parallel to c

aRb, bRc `cancel=>` aRc

R is not transitive.

vii. R = {(a, b) / a, be R, a < b}

a ≮  a ∀ a ∈ R

R is not reflexive

Let a = 2, b = 4

a< b

aRb

But b ≮  a

b `cancelR` a

`aRb cancel=> bRa`

R is not symmetric

Let aRb, bRc

a < b, b < c

a < b < c i.e., a < c

aRc

aRb, bRc ⇒ aRc ∀ a, b, c ∈ R

R is transitive.

viii. R = {(a, b) / a, b ∈ R, a ≤ b³ }

Let a = −2 

a³ = −8

But −2 > −8

`a cancel≤ a^3` for all a ∈ R

R is not reflexive

Let a = 1, b = 2 so that a³ = 1, b³ = 8

a < b³      ...aRb

But b > a³  `bcancelRa`

aRb ~ bRa

R is not symmetric

Let a = 8, b = 2, c = 1.5

a = b³   ...aRb

c³ = (1.5)³ = 3.375

b < c³

bRc

But a < c³

aRb, bRc `cancel=>` aRc

R is not transitive.

Relation	Reflexive	Symmetric	Transitive
R = {(a, b) : a, b ∈ Z, a – b is an integer}	√	√	√
R = {(a, b) : a, b ∈ N, a + b is even}	√	√	√
R = {(a, b) : a, b ∈ N, a divides b}	√	x	√
R = {(a, b) : a, b ∈ N, a2 – 4ab + 3b2 = 0}	√	x	x
R = {(a, b) : a is sister of b and a, b ∈ G = Set of girls}	x	√	√
R = {(a, b) : Line a is perpendicular to line b in a plane}	x	√	x
R = {(a, b) : a, b ∈ R, a < b}	x	x	√
R = {(a, b) : a, b ∈ R, a ≤ b3}	x	x	√