Complement of a set, Law of Complement, Property of Complement

As for the complement of a set, it widely applies to real-life situations and lists out what all elements are not contained within a given set. For instance, if the frame concerns a university course where a given number of students were registered to take a math class. The complement of this set would include all other students within the university, who aren’t enrolled in that particular math class. It helps our understanding of the concept of a set by providing a clear boundary between elements that belong to it and those that do not. The Law of Complement specifically holds that the combination of a set and its complement is equal to the universal set and the combined set of a set and its complement has no element in common hence, the result is an empty set.

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1. Complement of set
2. Laws of Complement of set
3. Properties of Complement of Sets
4. Solved Examples Based on Complement of Set

In this article, we will cover the concept of the complement of a set, the law of complement, and its properties. This concept falls under the broader category of sets relation and function, a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of one question has been asked on this concept, including one in 2019.

Complement of set

Complement of sets is one of the fundamental operations in the concept of sets. Before looking into the concept of complement of sets, let us see what are sets.

Set

Sets are a foundational concept in mathematics, central to various fields such as statistics, geometry, and algebra. A set is simply a collection of distinct objects, considered as a whole. These objects, called elements or members of the set, can be anything: numbers, people, letters, etc. Sets are particularly useful in defining and working with groups of objects that share common properties.

It is a well-defined collection of distinct objects and it is usually denoted by capital letters A, B, C, S, U,V...

Now, let us look into the definition that explain complement of set in detail.

Complement of a set Definition

Let $U$ be the universal set and $B$ is a subset of $U$. Then the complement of B is the set of all elements of $U$ which are not the elements of $B$.

Symbolically, The complement of set B is denoted by $B^{\prime}$ or $B^C$ with respect to $U$.
$B^{\prime}=\{\mathrm{x}: \mathrm{x} \in \mathrm{U}$ and $\mathrm{x} \notin \mathrm{B}\}$. Obviously, $\mathrm{B}^{\prime}=\mathrm{U}-\mathrm{B}$

Complement of a Set Examples

The complement of a set examples are

1. Let $U = \{1,2,3,4,5,6,7,8,9,10\} and A=\{1,4,6,7,3,8\}$. Then, the complement of set A is $A^{\prime} = \{2,5,9,10\}$.

2. Let $U = \{blue, violet, green, yellow, grey, brown, black, white, red\}, A = \{violet, green, yellow, white\} and B = \{grey, brown\}$. Then, the complement of set A and B is $A^{\prime} = \{blue, grey, brown, black, red\}$ and $B^{\prime} = \{blue, violet, green, yellow, black, white, red \}$.

3. $U = \{x: x \in \mathbf N$ and $x \leq 10\}$ and $B=$ set of all natural numbers less than or equal to $5$.
Now, $U = \{1,2,3,4,5,6,7,8,9\}$ and $A = \{1,2,3,4,5\}$, then $A^\prime = \{6,7,8,9\}$

Complement of Set Venn Diagram

The venn diagram of complement of set is

Laws of Complement of set

The laws of complement are fundamental properties that relate a set to its complement. These include the following:

1. Complementation law: The complement of the complement of a set is the set itself. $\left(A^{\prime}\right)^{\prime}=A$

2. Universal complement law: The complement of the universal set is the empty set. $U^{\prime}=\phi$3

3. Empty set complement law: The complement of the empty set is the universal set. $\phi^{\prime}=U$

Properties of Complement of Sets

The properties of the complement of sets are

Complement Laws

• If $A$ is a subset of the universal set then $A' $ is also a subset of the universal set, therefore the union of $A$ and $A' $ is the universal set, represented as $A ∪ A’ = U$
• The intersection of Set $A$ and $A' $ provides the “$∅$”, represented as $A ∩ A’ = ∅$

For example, If $U = \{1, 2, 3, 4, 5\}$ and $A = \{4 , 5\}$, then $A' = \{1, 2, 3\}$. Now, notice that $A ∪ A’ = U = \{1, 2, 3, 4, 5\}$. Also, $A ∩ A’ = ∅$

Law of Double Complementation

• In this law, the complement of the complemented set is the original set, $(A')' = A$
• The complement of the set $ A^\prime$, where $A^\prime$ itself is the complement of $A$, the double complement of $A$ is thus $ A$ itself.

In earlier example, $U = \{1, 2, 3, 4, 5\}$ and $A = \{4 , 5\}$ then $A' = \{1 , 2 , 3\}$.
The complement of $A' = (A')' = \{4, 5\}$, which is equal to set $A$.

Law of Empty set and Universal Set

• The complement of the universal set is an empty set or null set ($∅$) and the complement of the empty set is the universal set.
• Since the universal set contains all elements and the empty set contains no elements, therefore, their complement is just opposite to each other, represented as $∅' = U And U' = ∅$

In the above-given example set $U = \{1, 2, 3, 4, 5\}$, we can observe that $U' = ∅$ (empty set) and $∅' = \{1, 2, 3, 4, 5\}$.

De Morgan’s law

The complement of the union of two sets is equal to the complement of sets and their intersection. $(A U B)’ = A’ ∩ B’ $(De Morgan’s Law of Union).

The complement of the intersection of two sets is equal to the complement of sets and their union. $(A ∩ B)’ = A’ U B’ $ (De Morgan’s Law of Intersection).

Important Notes on the Complement of a Set

• The complement of a set $A$ is denoted by $A' $ and is obtained by subtracting $A$ from the universal set $U$. i.e., $A' = U - A$.
• A set and its complement are always disjoint.
• The complement of the universal set is the empty set or null set.

Recommended Video Based on Complement of Set

Solved Examples Based on Complement of Set

Example 1: Given $\mathrm{n}(\mathrm{U})=10, \mathrm{n}(\mathrm{A})=5, \mathrm{n}(\mathrm{B})=3$ and $n(A \cap B)=2$. A and B are subsets of $U$, then $n(A \cup B)^{\prime}=$

Solution:
As we learned
COMPLEMENT OF A SET:
Let $U$ be the universal set and $A$ a subset of $U$. Then the complement of $A$ is the set of all elements of $U$ which are not the elements of A. Symbolically, we write A' to denote the complement of $A$ with respect to $U$.
where $A^{\prime}=\{x: x \in U$ and $x \notin A\}$.Obviously $A^{\prime}=U-A$

$
\begin{aligned}
& n(A \cup B)=5+3-2=6 \\
& n(A \cup B)^{\prime}=n(U)-n(A \cup B)=10-6=4
\end{aligned}
$

Hence, the answer is 4 .

Example 2: Two newspapers A and B are published in a city. It is known that 25% of the city population reads A and 20% reads B while 8% reads both A and B. Further, 30% of those who read A but not B look into advertisements, and 40% of those who read B but not A also look into advertisements, while 50% of those who read both A and B look into advertisements. then the percentage of the population who look into advertisements is:

Solution:

Let $P(A)$ and $P(B)$ denote respectively the percentage of the city population that reads newspapers $A$ and $B$.

Let us consider the total percentage to be 100 . Then from the given data, we have

$
P(A)=25, \quad P(B)=20, P(A \cap B)=8
$

$\therefore$ Percentage of those who read $A$ but not $B$

$
P(A \cap \bar{B})=P(A)-P(A \cap B)=25-8=17 \%
$

And,
Percentage of those who read $B$ but not $A$

$
P(\bar{A} \cap B)=P(B)-P(A \cap B)=20-8=12 \%
$

If $\mathrm{P}(\mathrm{C})$ denotes the percentage of those who look into an advertisement, then from the given data we obtain

$
\begin{aligned}
& \therefore P(C)=30 \% \text { of } P(A \cap \bar{B})+40 \% \text { of } P(\bar{A} \cap B)+50 \% \text { of } P(A \cap B) \\
& \Rightarrow P(C)=\frac{3}{10} \times 17+\frac{2}{5} \times 12+\frac{1}{2} \times 8 \\
& \Rightarrow P(C)=13.9 \%
\end{aligned}
$
Hence, the answer is 13.9%.

Example 3: If $U=\{1,2,3,4,5\}, A=\{3,4,5\}$ and $B=\{1,2\}$. Then which of the following is true, if $U$ is a universal set of $A$ and $B$ ?
1) $A \subset B$
2) $A=B$
3) $A=B^{\prime}$
4) None of these

Solution:
Clearly B $=\mathrm{U}-\mathrm{A}$
Hence, $B=A^{\prime}$ and $A=B^{\prime}$
Hence, the answer is the option 3.
Example 4: If A and B are such sets that $A \cup B=U$ is the universal set. Which of the following must be true?
1) $A \cap B=\phi$
2) $A \cup B=A \cap B$
3) $A=B^c$
4) $A \cap U=A$

Solution:

$A \cup A^{\prime}=U$
A and B don't need to be compliment sets. It is only possible that
$A \cap U=A$

Hence, the answer is the option 4.

Example 5: I $f A \cup B=U$ and $A \cap B=\phi$, then which of the following is not true?
1) $A^{\prime}=B$
2) $A=B^{\prime}$
3) $A \cap B=B \cap A$
4) $A \cup B=A \cap B$

Solution:
Clearly, $A$ and $B$ are complements of each other.
$\mathrm{A}=\mathrm{B}^{\prime}$ and $\mathrm{A}^{\prime}=\mathrm{B}$, so options (1) and (2) are correct.
Now option (3) is always correct as it is the commutative law.
In option (4), $A \cup B=U$ and $A \cap B=\phi$, so they are not equal.

Hence, the answer is the option 4.

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