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Inverse of a relation: If (a,b) ϵ R, then (b,a) ϵ R-1 for all a,b ϵ X.

PART-1

We know '∩' of two equivalence relations is equivalence. So,

we just need to prove R2-1 is an equivalence relation.

1. If (a,a) ϵ R2, then (a,a) ϵ R2-1 also for all a ϵ X

So R2-1 is reflexive.

2. If symmetric pairs (a,b) & (b,a) ϵ R2, then corresponding elements (b,a) & (a,b) also ϵ R2-1.

So R2-1 is symmetric.

3. If (a,b), (b,c) & (a,c) ϵ R2, then (b,a), (c,b) & (c,a) ϵ R2-1 satisfying transitivity property, ((c,b) & (b,a) --> (c,a))

So R2-1 is transitive.

Thus R2-1 is an equivalence relation.

So R1 ∩ R2-1 is also equivalence.

PART-2

1. Since R is reflexive, so R-1 will also contain reflexive elements.

So R ∩ R-1 is reflexive.

2. Since R is not symmetric, then

only some but not every symmetric pairs (a,b) & (b,a) are present in R.

R contains only (a,b) but not (b,a) for some a,b ϵ X, then

R-1 will contain only (b,a) but not (a,b). So R ∩ R-1 contains neither (a,b) nor (b,a) thus ensuring no asymmetric elements.

R contains both (a,b) & (b,a) for some a,b ϵ X, then

R-1 will also contain both (b,a) & (a,b). So R ∩ R-1 contains both (a,b) & (b,a) thus ensuring both symmetric pairs are present.

So R ∩ R-1 will either contain both symmetric pairs or none at all making it symmetric.

3. Since R is transitive, so

if (a,b), (b,c) & (a,c) ϵ R for some a,b,c ϵ X, then (b,a), (c,b) & (c,a) ϵ R-1.

This means R ∩ R-1 doesn't contain any of (a,b), (b,c) & (a,c) or (b,a), (c,b) & (c,a) elements, so no need to check transitivity property here.

Finally R ∩ R-1 is equivalence relation as it is reflexive, symmetric and transitive!

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