Question

Let A={1,2,3,4,5,6}, B={2,4,5,6}, and C={1,6,7}, find:
a. A U B
b. A n C
c. A-C
d. A U(B n C)
e. n(A-C)​

Answer 1

Step-by-step explanation:

(i) To verify : A×(B∩C)=(A×B)∩(A×C)

(i) To verify : A×(B∩C)=(A×B)∩(A×C)We have B∩C={1,2,3,4}∩{5,6}=ϕ

(i) To verify : A×(B∩C)=(A×B)∩(A×C)We have B∩C={1,2,3,4}∩{5,6}=ϕ∴ L.H.S = A×(B∩C)=A×ϕ=ϕ

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)}

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ∴L.H.S=R.H.S

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ∴L.H.S=R.H.SHence A×(B∩C)=(A×B)∩(A×C)

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ∴L.H.S=R.H.SHence A×(B∩C)=(A×B)∩(A×C)(ii) To verify: A×C is a subset of B×D

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ∴L.H.S=R.H.SHence A×(B∩C)=(A×B)∩(A×C)(ii) To verify: A×C is a subset of B×DA×C={(1,5),(1,6),(2,5),(2,6)}

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ∴L.H.S=R.H.SHence A×(B∩C)=(A×B)∩(A×C)(ii) To verify: A×C is a subset of B×DA×C={(1,5),(1,6),(2,5),(2,6)}B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7),

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ∴L.H.S=R.H.SHence A×(B∩C)=(A×B)∩(A×C)(ii) To verify: A×C is a subset of B×DA×C={(1,5),(1,6),(2,5),(2,6)}B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7), (3,8),(4,5),(4,6),(4,7),(4,8)}

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ∴L.H.S=R.H.SHence A×(B∩C)=(A×B)∩(A×C)(ii) To verify: A×C is a subset of B×DA×C={(1,5),(1,6),(2,5),(2,6)}B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7), (3,8),(4,5),(4,6),(4,7),(4,8)}We can observe that all the elements of set A×C are the elements of set B×D

A×B={(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4)}A×C={(1,5),(1,6),(2,5),(2,6)} ∴R.H.S.=(A×B)∩(A×C)=ϕ∴L.H.S=R.H.SHence A×(B∩C)=(A×B)∩(A×C)(ii) To verify: A×C is a subset of B×DA×C={(1,5),(1,6),(2,5),(2,6)}B×D={(1,5),(1,6),(1,7),(1,8),(2,5),(2,6),(2,7),(2,8),(3,5),(3,6),(3,7), (3,8),(4,5),(4,6),(4,7),(4,8)}We can observe that all the elements of set A×C are the elements of set B×D Therefore A×C is a subset of B×D