Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x – y) is divisible by 3 is an equivalence relation.ORA binary operation on the set A = {0, 1, 2, 3, 4, 5} is defined...

Question

Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x – y) is divisible by 3 is an equivalence relation.
OR

A binary operation on the set A = {0, 1, 2, 3, 4, 5} is defined as

.

Write the operation table for a * b in A.

Show that zero is the identity for this operation * and each element ‘a’ ≠ 0 of the set is invertible with 6 – a, being the inverse of ‘a’.

Philoid · Accepted Answer

(x–x) = 0 is divisible by 3 for all x ∈ z. So, (x, x) ∈ R

∴ R is reflexive

(x – y) is divisible by 3 implies (y – x) is divisible by 3

So (x, y) ∈ R implies (y, x) ∈ R, x, y ∈ z

⇒ R is symmetric

(x – y) is divisible by 3 and (y – z) is divisible by 3

So (x – z) = (x – y) + (y – z) is divisible by 3

Hence (x, z) ∈ R ⇒ R is transitive

⇒ R is an equivalence relation

OR

Operation table a*b in A: –

Now for all a in A

∴a*0 = a + 0 = a ∀ a in A⇒0 is the identity element for*.

Now, for a’ ≠ 0 let, b = 6–a.

Since, a + b = a + 6–a = 6≥6

∴a*b = a + 6–a–6 = b*a = 0.

Hence, b = 6–a is the inverse of a.