Question
Question: If A, B and C are non-empty sets then the intersection of sets is distributive over union is sets is...
If A, B and C are non-empty sets then the intersection of sets is distributive over union is sets is represented as
a)A∩(B∪C)=(A∩B)∪(A∩C)b)A∩(B∩C)=(A∩B)∩(A∩C)c)(A∪B)∪C=(A∩C)∪(B∩C)d)(A∩B)∪C=(A∩C)∩(B∪C)
Solution
Now we know that the union intersection of sets is distributive over. Union is showed as A∩(B∪C)=(A∩B)∪(A∩C) . We will take set A, B and C of our choice and check the property.
Complete step by step answer:
Now consider 3 sets A, B and C.
Now we know that according to Algebra of Sets A∩(B∪C)=(A∩B)∪(A∩C)
Hence we have Distributive property of intersection over union is given by A∩(B∪C)=(A∩B)∪(A∩C)
Now let us take an example for the same and verify the distributive property.
Let us consider set A, B and C such that A = {1, 2, 3, 4, 5}, B ={3, 4, 5, 6, 7} and
C = {5, 6, 7, 8, 9, 10}
Now consider A∩(B∪C)
Let us first find (B∪C)
Now we have B ={3, 4, 5, 6, 7} and C = {5, 6, 7, 8, 9, 10}
Hence B∪C=3,4,5,6,7,8,9,10
Now we have B∪C=3,4,5,6,7,8,9,10 and A = {1, 2, 3, 4, 5}
Hence we have A∩(B∪C)=3,4,5
Let this be called equation (1)
A∩(B∪C)=3,4,5...........................(1)
Now consider (A∩B)∪(A∩C)
Now first let us find A∩B
Now we have A = {1, 2, 3, 4, 5} and B ={3, 4, 5, 6, 7}
Hence we have A∩B=3,4,5.............(2)
Now consider A∩C
We have A ={1,2,3, 4, 5} and C = {5, 6, 7, 8, 9, 10}
Hence we get A∩C=5............(3)
Now let us take union of sets obtained in equation (2) and equation (3)
Hence we get
(A∩B)∪(A∩C)=3,4,5...........(4)
Hence from equation (4) and equation (1) we can say that A∩(B∪C)=(A∩B)∪(A∩C)
Note: Note that distributive property of intersection over union is A∩(B∪C)=(A∩B)∪(A∩C) and distributive property of union over intersection is A∪(B∩C)=(A∪B)∩(A∪C) . We can also verify this property by showing sets in the form of a venn diagram.