Suppose R and S are relations on the set X Provide a proof o
Solution
(a) Let x, y, z X and let R be transitive and let x R y and y R z. Then, y R-1x and z R-1y. Since R is transitive, therefore, x R z, therefore, z R-1x. Thus , if z R-1y and y R-1x, then zR-1x . This means that R-1 is transitive
(b) Let ( x , y ) R S , then ( x , y) R and (x , y) S so that x R y and x S y. Since R and S are symmetric, we have y R x and y S x. This implies that y (R S ) x which proves that R S is symmetric.
(c) Let X = { a,b, c} and let R = { (a,b), (b,a)} and let a b , a c and b c. Then R ‘ = { (a,a), (a,c), (b,b), (b,c), (c,a), (c,b), (c,c)} Now, obviously, R is symmetric. However R’ is not antisymmetric as ( a,c), (c,a) R’ , but c a.
(d) Let A={x,y,z}. Suppose that R={(x,y)} and S={(y,z)}.R and S are both transitive by default.However, RS is not transitive because (x,z) is not a member of RS.
(e) Let R(a,b) if a < b and S(a,b) if a > b Both R and S are anti symmetric but R U S = {(a,b): a = b} which is symmetric
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