In mathematics, a set is a collection or grouping of well-defined objects. All such objects, when grouped in a set, are called elements. Sets are represented by capital letter symbols, and the elements are placed together in a curly bracket {}.
For example, if W is the set of whole numbers, then W = {0, 1, 2, 3, 4, 5,....,∞}.
In this article, we will learn about a particular aspect of set theory that is the complement of a set, its meaning, symbol, properties, and Venn diagram.
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The complement of a set is the set that consists of all the elements from the universal set that are not already included in the given set. In other words, the difference between the universal set and the given set represents the complement of that set. The complement of a set S is mathematically expressed as:
S' = {x ∈ U and x ∉ S}
Where x denotes the elements in Universal Set, U but not in S.
For example, let S be the set of all Android mobiles, which is a subset of the universal set of all mobiles in the world. In this case, the complement of set S would be the set of all non-Android mobile phones (which would include iOS, Windows, and other operating systems), as these do not consist of Android mobile phones.
Mathematical Definition- Complement of a Set
If U is a universal set and A be any subset of U then the complement of A is the set of all members of the universal set U which are not the elements of A.
Mathematically Complement of a set can be expressed in the following form:
- S' = U - S
- S' = {x ∈ U : x ∉ S}
Symbol of Complement of a Set
The complements of sets A, B, C... are denoted as A', B', C',... and so on. Excluding all the elements of a given set from the universal set gives us the complement of that set.
Venn Diagram of Complement of a Set
The following Venn diagram shows the universal set U and its two subsets- A and A'. The green-shaded portion represents set A and the red portion depicts set A'. As is clear from the diagram, set A is not part of A' and vice-versa. This implies that A and A' are both disjoint sets or complements of each-other.
How to Find the Complement of a Set?
The complement of a set is simply found by excluding the elements of the given set from the universal set. This is shown in the example below.
Example: Find the complement of set S = {4, 8, 12, 16}, where the universal set is all multiples of 4 that are smaller than 50.
Solution:
Step 1: Identify the universal set (U) and the set (S) for which you want to find the complement.
Universal Set(U) = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48}
Also, given: S = {4, 8, 12, 16}
Step 2: Calculate the complement of set S by excluding the elements of S from the universal set U i.e., S' = U - S.
S' = U - S = {4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48} - {4, 8, 12, 16}
Step 3: Simplify the resulting set to obtain the complement.
Thus, S' = {20, 24, 28, 32, 36, 40, 44, 48}
Properties of Complement of a Set
There are various properties of the Complement of a Set, some of these properties are discussed as follows:
Complement Laws
- The union of set S and its complement set S' is the universal set, i.e., S ∪ S’ = U.
Example: If U = {10, 11, 12, 13, 14} and S = {10, 11, 12}; then S' = {13, 14}.
- The intersection of set S and its complement set S' is the null or empty set, i.e., S ∩ S’ = ∅.
Example: If U = {10, 11, 12, 13, 14} and S = {10, 11, 12}; then S' = {13, 14}.
Now, S ∩ S’ = {10, 11, 12} ∩ {13, 14} = ∅.
Law of Double Complementation
- The complement of the complement of a given set is the set itself, i.e., (S')' = S.
Example: If U = {10, 11, 12, 13, 14} and S = {10, 11, 12}; then S' = {13, 14}.
The complement of S' = U - S' = {10, 11, 12} = set S.
Law of Null and Universal Sets
Example: If U = {10, 11, 12, 13, 14}; then U' = ∅ and ∅' = U.
De Morgan’s laws
- The complement of the union of two sets is equal to the intersection of the complements of the two sets, i.e., (S U T)’ = S’ ∩ T’.
Example: If U = {10, 11, 12, 13, 14} and S = {13, 14} and T = {10, 11}.
Now, S ∪ T = {10, 11, 13, 14} and (S ∪ T)' = {12}.
Thus, S’ ∩ T’ = {10, 11, 12} ∩ {12, 13, 14} = {12} = (S ∪ T)'.
- The complement of the intersection of two sets is equal to the union of the complements of the two sets, i.e., (S ∩ T)’ = S’ U T’.
Example: If U = {10, 11, 12, 13, 14} and S = {13, 14} and T = {10, 11}.
Now, S ∩ T = ∅ and (S ∩ T)’ = {10, 11, 12, 13, 14}.
Thus, S’ ∪ T’ = {10, 11, 12} ∪ {12, 13, 14} = {10, 11, 12, 13, 14} = S’ ∪ T’.
Solved Examples on Complement of a Set
Learn various Practice Questions on Complement of Set.
Question 1. If D = { x | x is a multiple of 10, x ∈ N }, find D'.
Solution:
Given: the universal set U = set of all natural numbers = N = {1, 2, 3, 4, 5, 6,....}
Also it is given that: D = { x | x is a multiple of 10, x ∈ N }
⇒ D = {10, 20, 30, 40, 50, 60,....}
⇒ D' = U - D
D' = {1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12,...19, 21, 22,..., 29, 31,.....}.
Question 2. If S and T are any two sets, prove that S' – T' = T – S.
Solution:
S' – T' = T – S is only possible when S' – T' ⊆ S – T.
Let x ∈ S' – T'
⇒ x ∈ S' and x ∉ T'
⇒ x ∉ S and x ∈ T (since, S ∩ S' = ϕ )
⇒ x ∈ S – T
The above result would be true for all x ∈ S' – T'
∴ S' – T' = T – S
Question 3. How many elements are there in the complement of the set of all vowels in the English alphabet?
Solution:
The English alphabet consists of 26 alphabets of which 5 are vowels and 21 are consonants. The difference between the universal set(U) and the set of vowels(say, V) is the complement of the set of all vowels.
Now, n(U) = 26 and n(V) = 5.
Thus, n(V') = n(U) - n(V)
⇒ n(V') = 26 - 5
⇒ n(V') = 21
Question 4. If P and Q are two sets, then prove P ∩ (P' ∪ Q') = P ∩ Q.
Solution:
LHS = P ∩ (P' ∪ Q')
Let us expand the LHS to get: (P ∩ P’) ∪ (P ∩ Q)
Since, P ∩ P’ = ϕ; our LHS becomes:
LHS = ϕ ∪ (P ∩ Q)
⇒ LHS = P ∩ Q
Thus, LHS = RHS
Hence Proved.
Question 5. If U = {12, 13, 15, 17, 19} and S = {13, 17}, T = {12, 15, 17, 19}, then prove that (S ∪ T)' = S' ∩ T'.
Solution:
LHS = (S ∪ T)'. Let us first find S ∪ T.
S ∪ T = {12, 13, 15, 17, 19} = U
⇒ (S ∪ T)' = U – (S ∪ T) = U – U = ϕ.
So, LHS = ϕ.
RHS = S' ∩ T'
S' = U – S = {12, 15, 19}
and T' = U – T = {13}
⇒ S' ∩ T' = {12, 15, 19} ∩ {13} = ϕ
So, RHS = ϕ
Thus, LHS = RHS
Hence Proved.
Question 6. If X and Y are two sets, prove that: X – (X – Y) = X ∩ Y.
Solution:
According to the De Morgan's Laws, (A ∪ B)’ = A’ ∩ B’ and (A ∩ B) ‘ = A’ ∪ B’.
LHS = X – (X – Y)
⇒ LHS = X ∩ (X – Y)'
⇒ LHS = X ∩ (X ∩ Y')'
⇒ LHS = X ∩ (X' ∪ Y')') ((Y')' = Y)
⇒ LHS = X ∩ (X' ∪ Y)
⇒ LHS = (X ∩ X') ∪ (X ∩ Y)
⇒ LHS = ϕ ∪ (X ∩ Y)
⇒ LHS = (X ∩ Y) [As ϕ ∪ A = A]
But, RHS = (X ∩ Y)
∴ LHS = RHS
Hence Proved.
Practice Questions on Complement of a Set
Question 1: Let A = {1, 2, 3, 4, 5} and the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Find the complement of A, i.e., A'.
Question 2: If M = {a, b, c, d} and N = {c, d, e, f}, prove that: (M ∪ N)' = M' ∩ N'.
Question 3: The universal set U consists of all positive integers less than or equal to 20. If A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4, 5, 6}, find the complement of A ∩ B.
Question 4: If X = {3, 5, 7, 9}, Y = {1, 3, 5, 7}, and the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, find X' – Y.
Question 5: Let U = {10, 20, 30, 40, 50}, P = {20, 40}, and Q = {30, 50}. Find (P ∪ Q)'.
Question 6: If A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4, 5, 6}, find the number of elements in the complement of the union of A and B, i.e., (A ∪ B)'.
Question 7: If X = {a, b, c} and Y = {b, c, d}, and the universal set U = {a, b, c, d, e}, find the complement of (X ∩ Y).
Question 8: Prove that: (X ∩ Y)' = X' ∪ Y'
Question 9: If A = {1, 3, 5, 7, 9}, B = {2, 4, 6, 8, 10}, and the universal set U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, find A' ∩ B'.
Question 10: If the universal set U = {1, 2, 3, 4, 5, 6}, and A = {2, 3, 4}, find the number of elements in A' and B = {1, 5, 6}.