Show that for any sets A and B , A = ( A ∩ B ) ∪ ( A - B )

Prove that, A = ( A ∩ B ) ∪ ( A - B ) RHS = ( A ∩ B ) ∪ ( A - B ) Using the result A - B = A - ( A ∩ B ) = A ∩ B ' ⇒ RHS = A ∩ B ∪ A ∩ B ' ⇒ RHS = A ∩ B ∪ B ' (using distributive property) ⇒ RHS = A ∩ U ( ∵ B ∪ B ' = U ) ⇒ RHS = A = LHS

Show that for any sets A and B , A = ( A ∩ B ) ∪ ( A - B )

Prove that, A = ( A ∩ B ) ∪ ( A - B ) RHS = ( A ∩ B ) ∪ ( A - B ) Using the result A - B = A - ( A ∩ B ) = A ∩ B ' ⇒ RHS = A ∩ B ∪ A ∩ B ' ⇒ RHS = A ∩ B ∪ B ' (using distributive property) ⇒ RHS = A ∩ U ( ∵ B ∪ B ' = U ) ⇒ RHS = A = LHS

Show that for any sets A and B , A ∪ ( B – A ) = A ∪ B .

L H S = A ∪ ( B - A ) Using B - A = B - ( A ∩ B ) = B ∩ A ' ⇒ L H S = A ∪ ( B ∩ A ' ) ⇒ L H S = ( A ∪ B ) ∩ ( A ∪ A ' ) ⇒ L H S = ( A ∪ B ) ∩ U ∵ A ∪ A ' = U ⇒ L H S = A ∪ B = R H S

For any two sets A and B , prove that following: A ∩ ( A ∪ B ) ' = ϕ

L H S = A ∩ ( A ∪ B ) ' = A ∩ A ' ∩ B ' (Using De-morgan's law) = ( A ∩ A ' ) ∩ B ' = ϕ ∩ B ' ∵ A ∩ A ' = ϕ = ϕ = R H S ∴ L H S = R H S Hence Proved

If A , B , C are three sets such that A ⊂ B , then prove that C – B ⊂ C – A .

Given A ⊂ B To prove: C - B ⊂ C - A L e t a ∈ C - B ⇒ a ∈ C and a ∉ B ⇒ a ∈ C and a ∉ A ∵ A ⊂ B ⇒ a ∈ C - A Thus, a ∈ C - B ⇒ a ∈ C - A This is true for all x ∈ C - B Hence C - B ⊂ C - A