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 ascribes what 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 facilitates our understanding of the concept of a set by providing a clear demarcation 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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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.

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…...

Complement of a set

Let $U$ be the universal set and $A$ is 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 use $A^{\prime}$ or $A^C$ to denote the complement of $A$ with respect to $U$.
$A^{\prime}=\{\mathrm{x}: \mathrm{x} \in \mathrm{U}$ and $\mathrm{x} \notin \mathrm{A}\}$. Obviously, $\mathrm{A}^{\prime}=\mathrm{U}-\mathrm{A}$

Law of Complement

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

Properties of Compliment
- When we take a union of a set with its complement then the result becomes a universal set.

$$
A \cup A^{\prime}=U
$$

- When we take an intersection of a set with its complement then the result becomes a null set.

$$
A \cap A^{\prime}=\varphi
$$

- When we take the double complement of a set then the result becomes an original set.

$$
\left(\mathrm{A}^{\prime}\right)^{\prime}=\mathrm{A}
$$

- The complement of a universal set is null and the complement of a null set is an universal set.

$$
\mathrm{U}^{\prime}=\varphi \text { and } \varphi^{\prime}=\mathrm{U}
$$

Note:

$A-B=A \cap B^{\prime}$

Summary

The complement of a set is also crucial in finding out elements that exist in the broader envelopes but not within a certain set. Thus let us use the Law of Complement to identify all the elements that can be included in this set as well as to state all the elements that are not part of it. However, applying this understanding is useful in managing the clubs’ memberships and subscriptions and also in organizing data and conducting in-depth analyses. A strong understanding of set complements increases a person’s potential to logically and effectively evaluate and organize various sets of objects, which is evident from this method’s theoretical and applied aspects.

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Solved Examples Based On the complement of sets:

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.

1. What is set?

Ans: A set is simply a collection of distinct objects, considered as a whole.
2. What is a complement of a set?

Ans: The complement of $A$ is the set of all elements of $U$ which are not the elements of $A$.
3. What is complementation law?

Ans: The complement of the complement of a set is the set itself.
4. What is universal complement law?

Ans: The complement of the universal set is the empty set.
5. What is the empty set complement law?

Ans: The complement of the empty set is the universal set.