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Solution (i) if r is reflexive, then r-1 is reflexive.. Assume that the relation r on a is reflexive. By using definition, ∀ a ∈ a, (a, a) ∈r. In view that r-1 consists of precisely those ordered pairs which can be obtained with the aid of  interchanging the first and second detail of ordered pairs in r, consequently, if (a, a) ∈ r then (a, a) ∈ r-1. Thus, ∀ a ∈ a, (a, a) ∈ r-1. As a result r-1is reflexive as properly. Answer (ii) assume that the relation r on a is symmetric. Let (a, b) ∈ r-1for a,b ∈a. By means of definition of r-1, (b, a) ∈r. Since r is symmetric.


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Therefore (a, b) ∈r. However then by way of definition of r-1, (b, a) ∈r. We have as a result shown that for all a, b ∈ a, if (a, b) ∈r-1 then (b, a) ∈ r-1. Therefore r-1is symmetric. Solution (iii) show that if r is transitive, then r-1 is transitive. Assume that the relation r on a is transitive. Permit (a, b) ∈ r-1and (b, c) ∈ r-1. Then via definition of r-1, (b, a) ∈r and (c, b) ∈r.

Now r is transitive, consequently If (c, b) ∈r and (b, a) ∈r then (c, a) ∈r. Once more through definition of r-1, we’ve got (a, c) ∈ r-1. We’ve got for this reason proven that for all a, b, c ∈ a, if (a, b) ∈ r-1and (b, c) ∈ r-1then (a, c) ∈ r-1. Thus r-1 is transitive. Answer (iv) show that if r is anti-symmetric. Then r-1 is anti-symmetric. Think that relation r on a is anti-symmetric. Permit (a,b)∈ r-1 1 and (b,a)∈ r-1 then through definition of r-1(b,a) ∈r and (a,b) ∈r. Due to the fact that r is antisymmetric, so if (a,b)∈r and (b,a)∈r then a = b.As a result we have shown that if (a,b) ∈ r-1 and (b,a) ∈ r-1 then a=b. For that reason r-1 is antisymmetric.