1. If A∩B = Ï• , prove that

a) A⊆ B̅

Solution:

We have

A∩B = Ï•

Now,

x∈ A ⇒ x∉ B       [∵ A∩B = Ï• ]

          ⇒ x∈ B

since x∈ A ⇒ x∈ B

∴  A⊆ B̅    proved

 

b) B∩A̅ = B

Solution:

We have,

A∩B = Ï•

B∩A̅ = {x: x∈ B and x∈ A̅ }

          = {x: x∈ B and x∉ A}

          = {x: x∈ B}    [∵ A∩B = Ï• ]

          = B

∴ B∩A̅ = B


c) A∪B̅ = B̅

Solution:

Since A∩B = Ï• ⇒ x∈ A ⇒ x∉ B

Now,

A∪B̅ = {x: x∈ A or x∈ B̅}

         = {x: x∉ B or x∈ B̅}

         = {x : x∈ B̅  or x∈ B̅}

         = { x : x∈ B̅ }

         = B

∴ A∪B̅ = B

2. If A⊆B, prove that

a) B̅ ⊆ A̅

Solution:

Let x∈ B̅ then

x∈ B̅ ⇒ x∉ B

          ⇒ x∉ A     [∵ A⊆ B]

since x∈ B̅ ⇒ x∉ A

∴ B̅ ⊆ A̅    proved.

 

b) A∪B = B

Solution:

Since A⊆B ⇒ x∈ A ⇒ x∈ B

Now,

A ∪ B = {x: x∈ A or x∈ B}

         = {x: x∈ B or x ∈ B}

         = {x: x∈ B}

         = B

∴ A∪B = B


c) A∩B = A

Solution:

Since A⊆B ⇒ x∈ A ⇒ x∈ B

Now,

A      = {x: x∈ A }

= {x: x∈ A and x ∈ A}

         = {x: x∈ A and x∈ B}

         = A∩B

∴ A∩B = A        Proved.

 

3. Prove that

    a) B – A = B ∩ A̅

         Proof :

         B – A = {x: x∈ B and x∉ A}

                  = {x: x∈ B and x∈ A̅ }

                  = B∩ A̅

     ∴ B – A = B∩ A̅     Proved.

   

    b) A – B̅ = A∩B

         Proof:

         A – B̅ = {x: x∈ A and x∉ B̅}

                  = {x: x∈ A and x∈ B}

                  = A∩B

         ∴ A – B̅ = A∩B          Proved.

   

   c) A – B = B̅ – A̅

         Proof:

         A – B = {x: x∈ A and x∉ B}

                  = {x : x∉  A̅ and x∈ B̅}

                  = {x: x∈ B̅ and x∉ A̅}

                  = B̅ – A̅

         ∴ A – B = B̅ – A̅        Proved.