**Proof:** Let, R1 and R2 be two equivalence relation

For all a∊ A, (a, a) ∊ R1 and (a, a) ∊ R2. Since R1 and R2 are reflexive.

∴ (a, a) ∊ R1 ∪ R2

So R1 ∪ R2 is reflexive.

If (a, b) ∊ R1 ∪ R2, then (a, b) ∊ R1 and (a, b) ∊ R2

Therefore (b, a) ∊ R1 or (b, a) ∊ R2.

Since R1 and R2 are symmetric.

So R1 ∪ R2 is symmetric.

Again (a, b) ∊ R1 ∪ R2 and (b, c) ∊ R1 ∪ R2

⇒ (a, b) ∊ R1 or (a, b) ∊ R2 and (b, c) ∊ R1 or (b, c) ∊ R2

Since both R1 and R2 are equivalence relation. It is clear R1 and R2 are both transitive.

Since R1 is transitive, i.e., (a, b) ∊ R1 and (b, c) ∊ R1 then (a, c) ∊ R1.

and Since R2 is transitive, i.e., (a, b) ∊ R2 and (b, c) ∊ R2 then (a, c) ∊ R2.

Now (a, c) ∊ R1 and (a, c) ∊ R2 implies that (a, c) ∊ R1 ∪ R2.

Home / Discrete mathematics / Show that union of equivalence relation is always a equivalence relation.

Tags Discrete mathematics Equivalence relation mathematics Reflexive Transitive relation

### Check Also

## Procedure for computing shortest distance| Discrete Mathematics

Solution: The procedure for computing the shortest distance/path from a to any vertex G. Initially, ...