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Sets - Introduction-advanced

Grade 9CBSE

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

🔑Concepts

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A set is a well-defined collection of objects. The objects are called elements or members of the set. If xx is an element of set AA, we write x∈Ax \in A. If not, we write x∉Ax \notin A.

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Sets can be represented in two ways: Roster (Tabular) Form, where elements are listed within braces like A={1,2,3,4,5}A = \{1, 2, 3, 4, 5\}, and Set-builder Form, which describes the common property of elements, like A={x:x∈N,x≤5}A = \{x : x \in \mathbb{N}, x \leq 5\}.

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A set with no elements is called an Empty Set or Null Set, denoted by ∅\emptyset or {}\{\}.

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A set AA is a Subset of BB (A⊆BA \subseteq B) if every element of AA is also an element of BB. The total number of subsets of a set with nn elements is 2n2^n.

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The Power Set of a set AA, denoted by P(A)P(A), is the set of all possible subsets of AA.

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The Union of two sets AA and BB (A∪BA \cup B) is the set of elements that are in AA, or in BB, or in both. The Intersection (A∩BA \cap B) is the set of elements common to both AA and BB.

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The Difference of sets A−BA - B consists of elements that belong to AA but not to BB.

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

📐Formulae

n(A∪B)=n(A)+n(B)−n(A∩B)n(A \cup B) = n(A) + n(B) - n(A \cap B) or for disjoint sets where n(A∩B)=0n(A \cap B) = 0: n(A∪B)=n(A)+n(B)n(A \cup B) = n(A) + n(B)

Number of subsets of a set with n elements=2n\text{Number of subsets of a set with } n \text{ elements} = 2^n

Number of non-empty subsets=2n−1\text{Number of non-empty subsets} = 2^n - 1

De Morgan’s Law 1: (A∪B)′=A′∩B′\text{De Morgan's Law 1: } (A \cup B)' = A' \cap B'

De Morgan’s Law 2: (A∩B)′=A′∪B′\text{De Morgan's Law 2: } (A \cap B)' = A' \cup 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)

💡Examples

Problem 1:

Write the set A={1,4,9,16,25,… }A = \{1, 4, 9, 16, 25, \dots\} in set-builder form.

Solution:

A={x:x=n2,n∈N}A = \{x : x = n^2, n \in \mathbb{N}\}

Explanation:

Each element in the given set is a perfect square. 1=121 = 1^2, 4=224 = 2^2, 9=329 = 3^2, and so on. Therefore, we define xx as the square of natural numbers nn.

Problem 2:

If set A={a,b,c}A = \{a, b, c\}, find the number of subsets and list the power set P(A)P(A).

Solution:

Number of subsets = 23=82^3 = 8. P(A)={∅,{a},{b},{c},{a,b},{b,c},{a,c},{a,b,c}}P(A) = \{\emptyset, \{a\}, \{b\}, \{c\}, \{a, b\}, \{b, c\}, \{a, c\}, \{a, b, c\}\}

Explanation:

A set with nn elements has 2n2^n subsets. Here n=3n=3, so there are 8 subsets, including the empty set and the set itself.

Problem 3:

In a class of 50 students, 30 like Mathematics, 25 like Science, and 15 like both. How many students like neither Mathematics nor Science?

Solution:

Let MM be the set of students liking Maths and SS be the set of students liking Science. n(U)=50n(U) = 50, n(M)=30n(M) = 30, n(S)=25n(S) = 25, n(M∩S)=15n(M \cap S) = 15. 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+25−15=40n(M \cup S) = 30 + 25 - 15 = 40 Students liking neither = n(U)−n(M∪S)n(U) - n(M \cup S) 50−40=1050 - 40 = 10

Explanation:

First, we find the number of students who like at least one subject using the Addition Principle. Then we subtract this from the total number of students to find those who like neither.