Question

Show that relation R in set A = {x€z:0<_x<_12} given as R = {(a, b):a=b} is reflexive, symmetric and transitive

Answer 1

$\huge\bold{Solution}$

• R = {(a, b) }:a=b}

• For any element a€A, we have (a,a) €R,

since a=a

So, R is reflexive .

• Now let (a, b) €R

a = b

b = a

(b, a) €R

Since R is symmetric.

• Now let (a, b) €R and (b, c) €R

a = b

b = c

a = c

(a, c) €R

Since R is transitive.

Answer 2

Hey mate,.

A = { x ∈ Z : 0 ≤ x ≤ 12} = {0,1,2,3,4,5,6,7,8,9,10,11,12}

(i) R = { (a,b) : |a - b| is a multiple of 4}

For any element a ∈A, we have (a, a) ∈ R as |a - a = 0|is a multiple of 4.

∴R is reflexive.

Now, let (a, b) ∈ R ⇒ |a - b| is a multiple of 4.

⇒ |-(a - b)| = ⇒ |b - a| is a multiple of 4.

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b), (b, c) ∈ R.

⇒ |(a - b)| is a multiple of 4 and |(b - c)| is a multiple of 4.

⇒ (a - b) is a multiple of 4 and (b - c) is a multiple of 4.

⇒ (a - c) = (a – b) + (b – c) is a multiple of 4.

⇒ |a - c| is a multiple of 4.

⇒ (a, c) ∈R

∴ R is transitive.

Hence, R is an equivalence relation.

The set of elements related to 1 is {1, 5, 9} since

|1 - 1| = 0 is a multiple of 4,

|5 - 1| = 4 is a multiple of 4, and

|9 - 1| = 8 is a multiple of 4.

(ii) R = {(a, b): a = b}

For any element a ∈A, we have (a, a) ∈ R, since a = a.

∴R is reflexive.

Now, let (a, b) ∈ R.

⇒ a = b

⇒ b = a

⇒ (b, a) ∈ R

∴R is symmetric.

Now, let (a, b) ∈ R and (b, c) ∈ R.

⇒ a = b and b = c

⇒ a = c

⇒ (a, c) ∈ R

∴ R is transitive.

Hope it will help you

✌️sai