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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(BC)=(AB)(AC) b)A(BC)=(AB)(AC) c)(AB)C=(AC)(BC) d)(AB)C=(AC)(BC) \begin{aligned} & a)A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right) \\\ & b)A\cap \left( B\cap C \right)=\left( A\cap B \right)\cap \left( A\cap C \right) \\\ & c)\left( A\cup B \right)\cup C=\left( A\cap C \right)\cup \left( B\cap C \right) \\\ & d)\left( A\cap B \right)\cup C=\left( A\cap C \right)\cap \left( B\cup C \right) \\\ \end{aligned}

Explanation

Solution

Now we know that the union intersection of sets is distributive over. Union is showed as A(BC)=(AB)(AC)A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right) . 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(BC)=(AB)(AC)A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right)
Hence we have Distributive property of intersection over union is given by A(BC)=(AB)(AC)A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right)
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(BC)A\cap \left( B\cup C \right)
Let us first find (BC)\left( B\cup C \right)
Now we have B ={3, 4, 5, 6, 7} and C = {5, 6, 7, 8, 9, 10}
Hence BC=3,4,5,6,7,8,9,10B\cup C=\\{3,4,5,6,7,8,9,10\\}
Now we have BC=3,4,5,6,7,8,9,10B\cup C=\\{3,4,5,6,7,8,9,10\\} and A = {1, 2, 3, 4, 5}
Hence we have A(BC)=3,4,5A\cap \left( B\cup C \right)=\\{3,4,5\\}
Let this be called equation (1)
A(BC)=3,4,5...........................(1)A\cap \left( B\cup C \right)=\\{3,4,5\\}...........................\left( 1 \right)
Now consider (AB)(AC)\left( A\cap B \right)\cup \left( A\cap C \right)
Now first let us find ABA\cap B
Now we have A = {1, 2, 3, 4, 5} and B ={3, 4, 5, 6, 7}
Hence we have AB=3,4,5.............(2)A\cap B=\\{3,4,5\\}.............(2)
Now consider ACA\cap C
We have A ={1,2,3, 4, 5} and C = {5, 6, 7, 8, 9, 10}
Hence we get AC=5............(3)A\cap C=\\{5\\}............(3)
Now let us take union of sets obtained in equation (2) and equation (3)
Hence we get
(AB)(AC)=3,4,5...........(4)\left( A\cap B \right)\cup \left( A\cap C \right)=\\{3,4,5\\}...........(4)
Hence from equation (4) and equation (1) we can say that A(BC)=(AB)(AC)A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right)

Note: Note that distributive property of intersection over union is A(BC)=(AB)(AC)A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( A\cap C \right) and distributive property of union over intersection is A(BC)=(AB)(AC)A\cup \left( B\cap C \right)=\left( A\cup B \right)\cap \left( A\cup C \right) . We can also verify this property by showing sets in the form of a venn diagram.