You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equivalence Relation

Prove that if...

Question

Prove that if $R$ is an equivalence relation then $R^{- 1}$ is also.

Open in App

Solution

R be an equilence relation
$a R^{- 1} b = b R a$
R is an equilance relation
$a R a a R^{- 1} a R^{- 1}$ is reflexive
$\begin{matrix} a R^{- 1} b b R a a R b \end{matrix}$ systematic
$a R^{- 1} b b R^{- 1} c$
R is transitive
$c R a a R^{- 1} c$
$R^{- 1}$ is transitive
combining we say $R^{- 1}$ is equivalence

Suggest Corrections

Similar questions

R \cap S

A = {1, 2, 3, . . . ., 9}

R

A \times A

(a, b) R (c, d)

a + d = b + c

(a, b), (c, d)

A \times A

R

[(2, 5)]

R

A

R^{- 1}

A = {1, 2, 3, . . . . 9}

R

A \times A

(a, b) R (c, d)

a + d = b + c

(a, b), (c, d)

A \times A

R

[(2, 5)] .

R

A

R^{- 1}

Join BYJU'S Learning Program