The difference between the two collections is a concept that examines the presence of elements in one collection but not in another. For instance, consider two groups of employees in a company; In one group, some members participated in a mandatory training session, but the other group did not participate in it at all. Therefore, the possibilities in the second set that are not in the first set indicate the number of employees who have not experienced the training. This operation is essential for defining the specific problem and issue domains or domains that require a solution in specific sectors or segments within a broad classification.
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In this article, we will cover the concept of the difference of sets. 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 2021.
Difference of sets is one of the fundamental operations in the concept of sets. It has various applications in concepts involving a specific category. Before looking into the concept of difference of sets, let us see what are sets.
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…...
The difference of sets $A$ and $B$ in this order is the set of elements that belong to $A$ but not to B.
Symbolically, we write $A - B$ and read as "A minus B".
To find the difference of sets,
Now, let us look into difference of set example for better understanding.
1. $A=\{1,2,3,4\}$ and $B=\{4,5,6,8\}$,
then, $A-B=\{1,2,3\}$ and $B-A=\{5,6,8\}$
2. $A = \{blue, violet, green, yellow, black, white, red\}$ and $B = \{black, blue, red, grey, brown\}$, then $A-B=\{violet, green, yellow, white\}$ and $B-A = \{grey, brown\}$
3. $A = \{x: x \in \mathbf N$ and $x \leq 10\}$ and $ B=$ set of all integers greater than $-5$ and less than or equal to $5$.
Now, $A = \{1,2,3,4,5,6,7,8,9\}$ and $B = \{-4,-3,-2,-1,0,1,2,3,4,5\}$, then $A-B=\{6,7,8,9\}$ and $B-A =\{-4,-3,-2,-1,0\}$
4. $A=$ set of all even natural numbers less than $10$ and $B=\{2,4,6,8\}$. Now, $A = \{2,4,6,8\}$ and $B=\{2,4,6,8\}$, then $A-B=\phi$ and $B-A =\phi$
The difference of sets $A$ and $B$ is $A-B$. The difference of set Venn diagram is
For instance, The difference of sets venn diagram of $A=\{1,2,3,4\}$ and $B=\{4,5,6,8\}$ is
From the venn diagram, $A-B=\{1,2,3\}$ and $B-A=\{5,6,8\}$.
1. In general $A - B$ does not equal $B - A$. This means that the difference of sets are not interchangeable.
2. When we take the difference of the same set then the result is empty. (i.e) $\mathrm{A}-\mathrm{A}=\phi$
3. When we take the difference of a finite set with the null set then the result is a finite set. (i.e) $\mathrm{A}-\phi=\mathrm{A}$
4. When we take the difference of a finite set with a universal set then the result is empty. (i.e) $A-U=\phi$
5. The sets $A-B, A \cap B$ and $B-A$ are mutually disjoint sets, (i.e) the intersection of any two of these sets is the null set(empty set).
Symmetric difference of two sets $A$ and $B$ is defined as $A Δ B = ( A - B ) ∪ ( B - A )$. This can also be represented as $A Δ B = ( A ∪ B) - ( A ∩ B )$
The symmetric difference of sets can be found with the help of the following steps:
1. $A=\{1,2,3,4\}$ and $B=\{4,5,6,8\}$,
then, $A-B=\{1,2,3\}$ and $B-A=\{5,6,8\}$. The symmetric difference of set are $(A-B) \cup (B-A) = \{1,2,3\} \cup \{5,6,8\} = \{1,2,3,5,6,8\}$
2. $A = \{blue, violet, green, yellow, black, white, red\}$ and $B = \{black, blue, red, grey, brown\}$, then $A-B=\{violet, green, yellow, white\}$ and $B-A = \{grey, brown\}$. The symmetric difference of set are $(A-B) \cup (B-A) = \{violet, green, yellow, white\} \cup \{grey, brown\} = \{violet, green, yellow, white, grey, brown\} $
3. $A = \{x: x \in \mathbf N$ and $x \leq 10\}$ and $ B=$ set of all integers greater than $-5$ and less than or equal to $5$.
Now, $A = \{1,2,3,4,5,6,7,8,9\}$ and $B = \{-4,-3,-2,-1,0,1,2,3,4\}$, then $A-B=\{5,6,7,8,9\}$ and $B-A =\{-4,-3,-2,-1,0\}$. The symmetric difference of set are $(A-B) \cup (B-A) = \{5,6,7,8,9\} \cup \{-4,-3,-2,-1,0\} = \{-4,-3,-2,-1,0,5,6,7,8,9\}$
4. $A=$ set of all even natural numbers less than $10$ and $B=\{2,4,6,8\}$. Now, $A = \{2,4,6,8\}$ and $B=\{2,4,6,8\}$, then $A-B=\phi$ and $B-A =\phi$. The symmetric difference of set are $(A-B) \cup (B-A) = \phi \cup \phi = \phi$
The symmetric difference of sets $A$ and $B$ is $(A-B) \cup (B-A)$. The symmetric difference of set Venn diagram is
For instance, The difference of sets venn diagram of $A=\{1,2,3,4\}$ and $B=\{4,5,6,8\}$ is
From the venn diagram, $A Δ B = \{1,2,3,5,6,8\}$
Example 1: If $\mathbf{A}, \mathbf{B}$, and $\mathbf{C}$ are non-empty sets, then $(A \cup B)-(A \cap B)$
1) $(A \cup B)-B$
2) $A-(A \cap B)$
3) $(A-B) \cup(B-A)$
4) $(A \cap B) \cup(A \cup B)$
Solution:
Clearly, as the sets in the question and in the third option, both equal the symmetric difference of $A$ and $B$, so both these are equal.
$
(A-B) \cup(B-A)=(A \cup B)-(A \cap B)
$
Hence, the answer is the option 3.
Example 2: $
\begin{aligned}
& \mathrm{A}=\left\{\mathrm{n} \in \mathrm{N} \mid \mathrm{n}^2 \leq \mathrm{n}+10,000\right\}, \mathrm{B}=\{3 \mathrm{k}+1 \mid \mathrm{k} \in \mathrm{N}\} \\
& C=\{2 \mathrm{k} \mid \mathrm{k} \in \mathrm{N}\} A \cap(B-C)
\end{aligned}
$
$ \text { Let } \mathrm{A}=\left\{\mathrm{n} \in \mathrm{N} \mid \mathrm{n}^2 \leq \mathrm{n}+10,000\right\}, \mathrm{B}=\{3 \mathrm{k}+1 \mid \mathrm{k} \in \mathrm{N}\} \text { and } C=\{2 \mathrm{k} \mid \mathrm{k} \in \mathrm{N}\} \text {, then the sum of all the elements of the set } A \cap(B-C) \text { is equal to } $
Solution:
$\begin{aligned} & A: n^2-n \leq 10,000 \\ & \Rightarrow n(n-1) \leq 100 \cdot 100 \\ & \Rightarrow n=\{1,2, \ldots, 100\} \\ & B=\{4,7,10,13, \ldots\} \\ & C=\{2,4,6,8,10, \ldots\} \\ & B-C=\{7,13,19, \ldots\} \\ & \begin{aligned} A \cap(B-C)=\{7,13,19, \ldots, 97\} \\ \text { Sum }=\frac{16}{2}\{14+14.6\} \\ \quad=832\end{aligned}\end{aligned}$
Hence, the answer is 832.
Example 3:
Let $\mathrm{A}, \mathrm{B}$, and $\mathbf{C}$ be sets such that $\phi \neq A \cap B \subseteq C$. Then which of the following statements is not true?
1) $B \cap C \neq \phi$
2) If $(A-B) \subseteq C$, then $A \subseteq C$
3) $(C \cup A) \cap(C \cup B)=C$
4) If $(A-C) \subseteq B$, then $A \subseteq B$
Solution:
$
\begin{aligned}
& \text { As }(A \cap B) \subseteq C \\
& \Rightarrow(A \cap B) \subseteq(B \cap C) \\
& \text { as }(A \cap B) \neq \phi_{\Rightarrow}(B \cap C) \neq \phi
\end{aligned}
$
So, option (1) is true
Let $x \in A$ and $x \notin B \Rightarrow x \epsilon(A-B)_{\Rightarrow} \Rightarrow x \epsilon C$ let $x \in A$ and $x \epsilon B \Rightarrow x \epsilon(A \cap B) \Rightarrow x \epsilon C$ Hence $x \in A$ and $x \in C \Rightarrow A \subseteq C$ So, option (2) is true.
Let
$\begin{aligned} & x \in C, x \in(C \cup A) \cap(C \cup B)) \\ & =>x \epsilon(C \cup A) \text { and } x \epsilon(C \cup B)) \\ & (x \in C \text { or } x \in A) \text { and }(x \in C \text { or } x \in B) \\ & =>x \epsilon C \text { or } x \epsilon(A \cap B) \\ & =>x \epsilon C \quad A s, A \cup B \subseteq C\end{aligned}$
$\begin{aligned}
&(C \cup A) \cap(C \cup B) \subseteq C.......\text { (1) }
\end{aligned}$
Now,
$
\begin{aligned}
& =>x \epsilon C \\
& =>x \epsilon(C \cup A) \cap(C \cup B) \\
& =>C \subseteq(C \cup A) \cap(C \cup B) ......(2)
\end{aligned}
$
From (1) and (2)
$
(C \cup A) \cap(C \cup B)=C
$
=> Option (3) is correct.
For $\mathrm{A}=\mathrm{c}, A-C=\phi$
$
=>\phi \subseteq B \text { but }=>A \nsubseteq B
$
So, option (4) is not true.
Example 4:
If $A=(x \in \mathbf{R}:|x|<2)$ and $B=(x \in \mathbf{R}:|x-2| \geqslant 3)$; then:
1) $A-B=[-1,2)$
2) $B-A=\mathbf{R}-(-2,5)$
3) $A \cup B=\mathbf{R}-(2,5)$
4) $A \cap B=(-2,-1)$
Solution:
$\begin{aligned} & A=\{x: x \in(-2,2)\} \\ & B=\{x: x \in(-\infty,-1] \cup[5, \infty)\} \\ & A \cap B=\{x: x \in(-2,-1]\} \\ & A \cup B=\{x: x \in(-\infty, 2) \cup[5, \infty)\} \\ & A-B=\{x: x \in(-1,2)\} \\ & B-A=\{x: x \in(-\infty,-2] \cup[5, \infty)\}\end{aligned}$
Hence, the answer is the option 2.
The difference of set definition is, $A-B$ in this order is the set of elements that belong to $A$ but not to $B$.
The intersection of any two of these sets is the null set is called a disjoint set.
The symmetric difference of two sets $A$ and $B$ is defined as
$A \Delta B=(A-B) \cup(B-A)$. The symmetric difference of set A and B can also represented as $A Δ B = ( A ∪ B) - ( A ∩ B )$
The set difference of set A with null set is $A-\{\} = A$.
No, $A-B$ is not equal to $B-A$.
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