The symbol “∩” can be used to represent the intersection of two sets. The intersection of two sets A and B can be explained as the set of all elements that are common of both sets – A and B. A∩B is a symbol that represents the intersection of sets – A and B.
The intersection, A∩B (read as A intersection B) lists all the items that are present in both sets and are the common elements of A and B for any two sets A and B.
Example – Set A = {a, b, c, d, e} and Set B = {d, e, f, g}
Hence, A∩ B = {d, e}
The most considerable set containing all the elements common to A and B is the intersection of two given sets, say A and B. The intersection of two sets can be a set with at least one element or an empty set with no items. If A and B are two sets with the property A ∩ B = φ, they are referred to as disjoint sets. That is, at the intersection of A and B, there are no elements.
Finding the intersection of more than two sets is achievable. You’ll learn how to find the intersection of three sets in this section. If A, B, and C are three sets, then the set of all elements that are common to A, B, and C is the intersection of these three sets. A ∩B ∩C can be used to symbolise this.
The properties of the intersection of the sets are as follows:
Let’s take a look at each of these properties one by one:-
1.Commutative law: P∩Q = Q∩P
Consider two sets P = {2, 4, 6, 8} and Q = {2, 3, 6, 9}.
Now, P∩Q = {2, 4, 6, 8} ∩ {2, 3, 6, 9} = {2, 6}
Q∩P= {2, 3, 6, 9} ∩ {2, 4, 6, 8} = {2, 6}
Hence, P∩Q = Q∩P.
2. Associative law: (P∩Q) ∩ R = P ∩ (Q∩R)
Let P = {2, 3, 4, 5}, Q = {4, 5, 6, 7}, and R = {6, 7, 8, 9}.
Now, P∩Q = {2, 3, 4, 5} ∩ {4, 5, 6, 7} = {4, 5}
(P∩Q) ∩ R = {4, 5} ∩ {6, 7, 8, 9} = { } = φ
Similarly, Q ∩ R = {4, 5, 6, 7} ∩ {6, 7, 8, 9} = {6, 7}
P ∩ (Q ∩ R) = {2, 3, 4, 5} ∩ {6, 7} = { } = φ
Hence, (P∩Q) ∩ R = P ∩ (Q ∩ R)
3. Idempotent law: P∩P = P
Suppose P = {w, x, y, z} such that P ∩ P = {w, x, y, z} ∩ {w, x, y, z} = {w, x, y, z} = P
4. Law of φ and U: φ ∩ A = φ, U ∩ A = A
Consider φ = { } and A = {10, 11, 12}.
φ ∩ A = { } ∩ {10, 11, 12} = { } = φ
Let U = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} and A = {4, 8, 12, 16, 20}.
U ∩ A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} ∩ {4, 8, 12, 16, 20} = {4, 8, 12, 16, 20} = A
5. Distributive law: P ∩ (Q U R) = (P∩Q) U (P ∩ R)
Let us take three sets P = {1, 3, 6, 9}, Q = {2, 5, 7, 9} and R = {4, 5, 6, 9}.
Q U R = {2, 5, 7, 9} U {4, 5, 6, 9} = {2, 4, 5, 6, 7, 9}
P ∩ (Q U R) = {1, 3, 6, 9} ∩ {2, 4, 5, 6, 7, 9} = {6, 9}
And, P ∩ Q = {1, 3, 6, 9} ∩ {2, 5, 7, 9} = {9}
P ∩ R = {1, 3, 6, 9} ∩ {4, 5, 6, 9} = {6, 9}
(P∩Q) U (P ∩ R) = {9} U {6, 9} = {6, 9}
Hence, P ∩ (Q U R) = (P∩Q) U (P∩R)
The most extensive set containing all the elements common to P and Q is the intersection of two given sets, say P and Q. The intersection of two sets is represented by the symbol “∩”.
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