If A=\ 1,2,3,4\ B=\ 3,4,5,6,\ X=\ 1,2,....10\ then verify . AU(B cap C) equiv(AUB) cap(AUC)
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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×C={(1,5),(1,6),(2,5),(2,6)}
∴R.H.S.=(A×B)∩(A×C)=ϕ
∴L.H.S=R.H.S
Hence A×(B∩C)=(A×B)∩(A×C)
(ii) To verify: A×C is a subset of B×D
A×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
Step-by-step explanation:
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