krit.club logo

Sets - Complement of a Set

Grade 11CBSE

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

πŸ”‘Concepts

β€’

Definition of Complement: The complement of a set AA, denoted by Aβ€²A', is the set of all elements of the Universal set UU which are not elements of AA. Mathematically, Aβ€²={x:x∈UΒ andΒ xβˆ‰A}A' = \{x : x \in U \text{ and } x \notin A\}. Visually, if the Universal set is a rectangle and set AA is a circle inside it, Aβ€²A' is represented by the entire shaded region inside the rectangle but outside the circle.

β€’

Complement Relative to Universal Set: The complement of a set depends entirely on the Universal set UU provided. For instance, if AA is the set of even numbers, its complement Aβ€²A' would be the set of odd numbers if UU is the set of natural numbers, but it could be different if UU is the set of integers.

β€’

Complement Laws: The union of a set AA and its complement Aβ€²A' always results in the Universal set UU (AβˆͺAβ€²=UA \cup A' = U). Conversely, the intersection of a set and its complement is always an empty set (A∩Aβ€²=βˆ…A \cap A' = \emptyset), meaning they are disjoint. Visually, this shows that the circle AA and its exterior Aβ€²A' share no space and together fill the bounding rectangle.

β€’

Law of Double Complementation: If you take the complement of a complement, you return to the original set, expressed as (Aβ€²)β€²=A(A')' = A. In a Venn diagram, this is equivalent to 'un-shading' the exterior and 're-shading' the interior circle.

β€’

Laws of Empty Set and Universal Set: The complement of the Universal set is the empty set (Uβ€²=βˆ…U' = \emptyset), and the complement of the empty set is the Universal set (βˆ…β€²=U\emptyset' = U). This indicates that nothing exists outside the boundary of UU, and everything in UU exists outside the boundary of nothingness.

β€’

De Morgan's First Law: The complement of the union of two sets is equal to the intersection of their complements, written as (AβˆͺB)β€²=Aβ€²βˆ©Bβ€²(A \cup B)' = A' \cap B'. Visually, if you take two overlapping circles and look at the area outside their combined shape, it is identical to finding the area that is simultaneously outside circle AA and outside circle BB.

β€’

De Morgan's Second Law: The complement of the intersection of two sets is equal to the union of their complements, written as (A∩B)β€²=Aβ€²βˆͺBβ€²(A \cap B)' = A' \cup B'. Visually, the region representing everything except the small overlapping 'almond' shape between two circles is the same as combining all area outside AA with all area outside BB.

πŸ“Formulae

Aβ€²=Uβˆ’AA' = U - A

AβˆͺAβ€²=UA \cup A' = U

A∩Aβ€²=βˆ…A \cap A' = \emptyset

(Aβ€²)β€²=A(A')' = A

Uβ€²=βˆ…U' = \emptyset

βˆ…β€²=U\emptyset' = U

(AβˆͺB)β€²=Aβ€²βˆ©Bβ€²(A \cup B)' = A' \cap B' (De Morgan's First Law)

(A∩B)β€²=Aβ€²βˆͺBβ€²(A \cap B)' = A' \cup B' (De Morgan's Second Law)

πŸ’‘Examples

Problem 1:

Let U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}, A={1,3,5,7,9}A = \{1, 3, 5, 7, 9\}, and B={2,3,5,7}B = \{2, 3, 5, 7\}. Find (AβˆͺB)β€²(A \cup B)'.

Solution:

Step 1: Find AβˆͺBA \cup B. Combine all elements from both sets without repetition. AβˆͺB={1,2,3,5,7,9}A \cup B = \{1, 2, 3, 5, 7, 9\} Step 2: Find the complement (AβˆͺB)β€²(A \cup B)' by identifying elements in UU that are NOT in AβˆͺBA \cup B. (AβˆͺB)β€²=Uβˆ’{1,2,3,5,7,9}(A \cup B)' = U - \{1, 2, 3, 5, 7, 9\} (AβˆͺB)β€²={4,6,8,10}(A \cup B)' = \{4, 6, 8, 10\}

Explanation:

To find the complement of a union, first determine the union of the two sets and then subtract those elements from the Universal set.

Problem 2:

Verify De Morgan's Second Law (A∩B)β€²=Aβ€²βˆͺBβ€²(A \cap B)' = A' \cup B' for U={a,b,c,d,e}U = \{a, b, c, d, e\}, A={a,b,c}A = \{a, b, c\}, and B={c,d}B = \{c, d\}.

Solution:

Step 1: Calculate the Left Hand Side (LHS). Find A∩BA \cap B: A∩B={c}A \cap B = \{c\}. Find (A∩B)β€²(A \cap B)': (A∩B)β€²=Uβˆ’{c}={a,b,d,e}(A \cap B)' = U - \{c\} = \{a, b, d, e\}.

Step 2: Calculate the Right Hand Side (RHS). Find Aβ€²A': Aβ€²=Uβˆ’{a,b,c}={d,e}A' = U - \{a, b, c\} = \{d, e\}. Find Bβ€²B': Bβ€²=Uβˆ’{c,d}={a,b,e}B' = U - \{c, d\} = \{a, b, e\}. Find Aβ€²βˆͺBβ€²A' \cup B': Aβ€²βˆͺBβ€²={d,e}βˆͺ{a,b,e}={a,b,d,e}A' \cup B' = \{d, e\} \cup \{a, b, e\} = \{a, b, d, e\}.

Step 3: Compare LHS and RHS. Since LHS={a,b,d,e}LHS = \{a, b, d, e\} and RHS={a,b,d,e}RHS = \{a, b, d, e\}, then (A∩B)β€²=Aβ€²βˆͺBβ€²(A \cap B)' = A' \cup B' is verified.

Explanation:

This step-by-step verification compares the result of taking the complement of an intersection against the union of the individual complements of the sets.