Let U 1 2 34 and S 1 2We define the biliary relation R on t

Let U = {1, 2, 3,4} and S= {1, 2}.We define the biliary relation R on the nonempty subsets of U ftilovvs.R = {(a, b):A-S= B;A, B U} (a)Is R refelxive? If ypu belive it is true, then prove your answer wise, give counter examples. (b) Is R symmetric? Again Prove or disprove. (c) Is R teansitive? Again prove or disprove. (d) Is R antisymmetric? Again, Prove or disprove.

Solution

Here A and B may be any set out of {1}{2}{3}{4}{1,2}{1,3}{1,4}{2,3}{2,4}{1,2,3},{2,3,4},{1,2,3,4} etc.

Now in case of Reflexive, we must obtaine an ordered pair of form (A,A) so that

A-S= A

that is only possible if S is a NULL set that is not as S ={1,2} given here. So R is not reflexive.

Counter example : Now here if we take A={1,2,3} then

A-S = {1,2,3}-{1,2}={3} that is not A.
So R is not reflexive.

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Now for symmetry, if R= {A,B} then R should have (B,A) also.

Now if A - S = B

for example if A = {1,2,3,4 }

then A-S = {1,2,3,4} -{1,2} = {3,4} =B

then B-S = {3,4} - {1,2 }= a NULL set that is not equal to A.

So if R=(A,B) is true then R=(B,A) is not true. Thus R is not symmetric using above counter example.

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Now for transitive, if R=(A,B) and R=(B,C) then R=(A,C)

So let A ={1,2,3,4}, then

A-S = {3,4} =B

then B-S = {3,4} - {1,2 } = a NULL SET so C must be a NULL SET.

So even initial conditions for transitivity are not being fullfilled here that is why R is not transitive even.

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Now by rule if a,b belongs to set X then if aRb and bRa then b=a or if ARB such that A is not equal to B then BRA does not hold.

Now here ARB : A - S = B   then BRA : B -S =A ( that is not true by above examples also)

as if A={1,2,3,4 } then A - S = {3,4}=B

and accordingly B-S = {3,4}-{1,2}= NULL set.

so clearly as
ARB such that A is not equal B, then BRA does not hold.

So surely R is an antisymmetric relation.