Question

Q: If R is an equivalence relation on a set A,
then show that the inverse relation R-l of R
on A is also an equivalence relation​

Answer 1

ANSWER
Let R be a relation on A i.e. R⊆A×A
R={(a,b)∣a,b∈A}
Also, given R is equivalence relation,

Now, let R
−1
={(b,a)∣(a,b)∈R}
We will check whether R
−1
is reflexive, symmetric, transitive or an equivalence relation.

Reflexive:
Since, R is reflexive
⇒(a,a)∈R
⇒(a,a)∈R
−1
(by def of R
−1
)
Hence, R
−1
is reflexive.

Symmetric: Let (b,a)∈R
−1

⇒(a,b)∈R (by def of R
−1
)
⇒(b,a)∈R (Since, R is symmetric)
⇒(a,b)∈R
−1
(by def of R
−1
)
Hence, R
−1
is symmetric.

Transitive : Let (b,a),(a,c)∈R
−1

⇒(a,b),(c,a)∈R (by def of R
−1
)
or (c,a)(a,b)∈R
⇒(c,b)∈R (since, R is transitive.)
⇒(b,c)∈R
−1

Hence, R
−1
is transitive.
Hence, R
−1
is an equivalence relation.