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Sets - Set Operations-advanced

Grade 9CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Union of Sets (A∪BA \cup B): The set of all elements that are in AA or in BB or in both. Formally, A∪B={x:x∈A or x∈B}A \cup B = \{x : x \in A \text{ or } x \in B\}.

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Intersection of Sets (A∩BA \cap B): The set of all elements that are common to both AA and BB. Formally, A∩B={x:x∈A and x∈B}A \cap B = \{x : x \in A \text{ and } x \in B\}.

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Difference of Sets (A−BA - B): The set of elements that belong to AA but not to BB. Formally, A−B={x:x∈A and x∉B}A - B = \{x : x \in A \text{ and } x \notin B\}.

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Complement of a Set (A′A'): The set of all elements in the Universal set UU that are not in AA. It is given by A′=U−AA' = U - A.

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Symmetric Difference (AΔBA \Delta B): The set of elements that belong to either AA or BB, but not to their intersection. AΔB=(A−B)∪(B−A)A \Delta B = (A - B) \cup (B - A).

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Disjoint Sets: Two sets AA and BB are disjoint if their intersection is an empty set, i.e., A∩B=∅A \cap B = \emptyset.

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De Morgan's Laws: These relate the complement of unions and intersections. They state that (A∪B)′=A′∩B′(A \cup B)' = A' \cap B' and (A∩B)′=A′∪B′(A \cap B)' = A' \cup B'.

📐Formulae

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B) samples

n(A∪B)=n(A)+n(B) (if A and B are disjoint)n(A \cup B) = n(A) + n(B) \text{ (if A and B are disjoint)}

n(A−B)=n(A)−n(A∩B)n(A - B) = n(A) - n(A \cap B)

n(AΔB)=n(A∪B)−n(A∩B)n(A \Delta B) = n(A \cup B) - n(A \cap B)

n(A∪B∪C)=n(A)+n(B)+n(C)−n(A∩B)−n(B∩C)−n(C∩A)+n(A∩B∩C)n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)

A∩(B∪C)=(A∩B)∪(A∩C) (Distributive Law)A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \text{ (Distributive Law)}

A∪(B∩C)=(A∪B)∩(A∪C) (Distributive Law)A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \text{ (Distributive Law)}

💡Examples

Problem 1:

In a class of 5050 students, 3030 students like Mathematics and 2525 like Science. If 1212 students like both subjects, find the number of students who like neither Mathematics nor Science.

Solution:

Let MM be the set of students who like Mathematics and SS be the set of students who like Science. Given: Total students n(U)=50n(U) = 50 n(M)=30n(M) = 30 n(S)=25n(S) = 25 n(M∩S)=12n(M \cap S) = 12 Step 1: Calculate the number of students who like at least one subject: n(M∪S)=n(M)+n(S)−n(M∩S)n(M \cup S) = n(M) + n(S) - n(M \cap S) n(M∪S)=30+25−12=43n(M \cup S) = 30 + 25 - 12 = 43 Step 2: Calculate the number of students who like neither subject: Neither=n(U)−n(M∪S)\text{Neither} = n(U) - n(M \cup S) Neither=50−43=7\text{Neither} = 50 - 43 = 7

Explanation:

We first use the principle of inclusion-exclusion to find the union of the two sets, which represents students who like at least one subject. Then, we subtract this from the total students to find those who like neither.

Problem 2:

If A={1,2,3,4,5,6}A = \{1, 2, 3, 4, 5, 6\} and B={4,5,6,7,8}B = \{4, 5, 6, 7, 8\}, find the symmetric difference AΔBA \Delta B.

Solution:

Step 1: Find the difference A−BA - B: A−B={x:x∈A and x∉B}A - B = \{x : x \in A \text{ and } x \notin B\} A−B={1,2,3}A - B = \{1, 2, 3\} Step 2: Find the difference B−AB - A: B−A={x:x∈B and x∉A}B - A = \{x : x \in B \text{ and } x \notin A\} B−A={7,8}B - A = \{7, 8\} Step 3: Find the union of these two differences: AΔB=(A−B)∪(B−A)A \Delta B = (A - B) \cup (B - A) AΔB={1,2,3}∪{7,8}={1,2,3,7,8}A \Delta B = \{1, 2, 3\} \cup \{7, 8\} = \{1, 2, 3, 7, 8\}

Explanation:

The symmetric difference consists of elements that are in set AA only and elements that are in set BB only, excluding the common elements.

Problem 3:

Verify De Morgan's First Law (A∪B)′=A′∩B′(A \cup B)' = A' \cap B' for the universal set U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, A={2,4,6,8}A = \{2, 4, 6, 8\}, and B={2,3,5,7,8}B = \{2, 3, 5, 7, 8\}.

Solution:

Step 1: Find A∪BA \cup B: A∪B={2,3,4,5,6,7,8}A \cup B = \{2, 3, 4, 5, 6, 7, 8\} Step 2: Find the complement (A∪B)′(A \cup B)': (A∪B)′=U−(A∪B)={1,9,10}(A \cup B)' = U - (A \cup B) = \{1, 9, 10\} Step 3: Find A′A' and B′B': A′=U−A={1,3,5,7,9,10}A' = U - A = \{1, 3, 5, 7, 9, 10\} B′=U−B={1,4,6,9,10}B' = U - B = \{1, 4, 6, 9, 10\} Step 4: Find A′∩B′A' \cap B': A′∩B′={1,3,5,7,9,10}∩{1,4,6,9,10}={1,9,10}A' \cap B' = \{1, 3, 5, 7, 9, 10\} \cap \{1, 4, 6, 9, 10\} = \{1, 9, 10\} Since (A∪B)′=A′∩B′={1,9,10}(A \cup B)' = A' \cap B' = \{1, 9, 10\}, the law is verified.

Explanation:

De Morgan's law shows that the complement of a union is the intersection of the complements.