prove that inverse of equivalence relation is also an equivalence relation

Asked by  | 15th Jul, 2008, 06:33: PM

Expert Answer:

Let R be an equivalence relation.

Reflexive (a,a) aA

Symmetric (a,b) R

                  ⇒ (b,a) R

Transitive (a,b), (b,c) R

                 ⇒ ac R

Let R-1 is the inverse relation of R

Now R-1 is reflexive

(a,a) R

⇒ (a,a) R-1

Symmetric

(b,a) R ⇒ (a,b) R-1

(a,b)  R ⇒ (b,a)  R-1

(a,b), (b,a) R-1

So R-1 is symmetric.

Transitive

(a,b), (b,c), (a,c) R

Now, (b,a) R-1 (a,c) R-1

(c,b) R-1

Since R-1 is symmetric (b,c) R-1

So, R-1 is transitive.

Hence R-1 is an equivalence relation.

 

 

Note:

Domain of R = Range of R-1

Range of R-1 = Domain of R

 

 

 

Answered by  | 31st Jan, 2009, 03:47: PM

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