Solution:
We can solve this problem using operation on sets.(union of two sets and Difference of two sets)
According to the operation on sets
Union of two sets: the union of two sets A and B, written as A ∪ B, is the set consisting of all the elements which are either in A or in B or in both, thus,
A ∪ B = {x: x∈ A or x ∈ B}
Clearly,
x ∊ A ∪ B ⇒ x ∊ A or x ∊ B, and
x ∊ A ∪ B ⇒ x ∊ A and x ∊ B.
Difference of two sets : If A and B are two sets, then their difference A - B is defined as :
A - B = {x : x ∊ A and x ∉ B}
Similarly, B - A = {x : x ∊ B and x ∉ A}.
By using these definitions we can find the answer
We have
A ∩ (B ∪ C)
x ∊ A ∩ (B ∪ C)
⇒ x ∊ A and (x ∊ B or x ∊ C)
⇒ (x ∊ A and x ∊ B) or (x ∊ A and x ∊ C)
(Since 'and' distributes 'or')
⇒ x ∊ (A ∩ B) or x ∊ (A ∩ C)
∴ A ∩ (B ∪ C) ⊆ (A ∩ B) ∪ (A ∩ C)
similarly , (A ∩ B) ∪ (A ∩ C) ⊆ A ∩ (B ∪ C)
Hence, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
It is one of the Distributive law.
For example :
Let A = {2, 3, 5, 7}, B = {2, 4, 6, 8} and C = {1, 3, 6, 8} then
(B∪C) = {2, 4, 6, 8} ∪ {1, 3, 6, 8}
= {1, 2, 3, 4, 6, 8}
LHS = A∩(B∪C)
= {2, 3, 5, 7} ∩ {1, 2, 3, 4, 6, 8}
= {2,3}
RHS = (A∩B) )∪(A∩C)
(A∩B)= {2, 3, 5, 7} ∩ {2, 4, 6, 8}
= {2}
(A∩C) = {2, 3, 5, 7} ∩ {1, 3, 6, 8}
= {3}
∴(A∩B)∪(A∩C) = {2, 3}
∴ LHS = RHS
i.e., A∩(B∪C) = (A∩B) ∪ (A∩C)
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