BIGtheme.net http://bigtheme.net/ecommerce/opencart OpenCart Templates
Saturday , July 22 2017
Home / Discrete mathematics / Show that union of equivalence relation is always a equivalence relation.

Show that union of equivalence relation is always a equivalence relation.

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.

So R1∪R2 is transitive. Hence R1 ∪ R2 is an equivalence relation. (Proved)

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, ...

Leave a Reply

Your email address will not be published. Required fields are marked *