Question

Let universal set \[u=\left\{ 1,2,3,4,5,6,7,8,9 \right\},A=\left\{ 1,2,3,4 \right\},B=\left\{ 2,4,6,8 \right\}\ and\ C=\left\{ 3,4,5,6 \right\}\]. Find \[\left( B-C \right)'\].

Answer 1

Hint: Subtracting a set B for a set C means removing all the elements which are common in B and C from C.
Complement of a set X denoted by X’ is the set which contains all the elements that occur in the universal set (u) but not in X.

Complete step-by-step answer:
A set is a collection of well – defined objects.
We have to find \[\left( B-C \right)'\].
Set \[\left( B-C \right)'\] is the complement of the set \[\left( B-C \right)\]. For finding the complement of \[\left( B-C \right)\], first we need to find the set \[B-C\].
As we know \[B-C\] will be set B minus all the elements which are present in both B and C.
$i.e.\ \ B-C=B-\left( B\cap C \right)$
$\left( B\cap C \right)=$ Set of the elements which are common in B and C.
Given: \[B=\left\{ 2,4,6,8 \right\}\ and\ C=\left\{ 3,4,5,6 \right\}\]
Elements which are common in B and C are 4 and 6.
$\Rightarrow B\cap C=\left\{ 4,6 \right\}$
By removing these elements (i.e. 4 and 6) from B, we will get \[B-C\].
$\begin{align}
  & \Rightarrow B-C=\left\{ 2,4,6,8 \right\}-\left\{ 4,6 \right\} \\
 & \Rightarrow B-C=\left\{ 2,8 \right\} \\
\end{align}$
We have found \[\left( B-C \right)\]. Now, we have to find the complement of \[\left( B-C \right)\] i.e. \[\left( B-C \right)'\].
We know, complement of a set X = universal set – X .
We have universal set (u) $=\left\{ 1,2,3,4,5,6,7,8,9 \right\}$
$\left( B-C \right)=\left\{ 2,8 \right\}$
And we have to find \[\left( B-C \right)'\].
And we know, \[\left( B-C \right)'=union-B-C\].
To find \[\left( union-\left( B-C \right) \right)\], we need to find elements which are common in union and \[B-C\].
$\begin{align}
  & universal set \left( u \right)=\left\{ 1,2,3,4,5,6,7,8,9 \right\} \\
 & B-C=\left\{ 2,8 \right\} \\
\end{align}$
Common elements: 2 and 8
\[\begin{align}
  & \Rightarrow u-\left( B-C \right)==\left\{ 1,2,3,4,5,6,7,8,9 \right\}-\left\{ 2,8 \right\} \\
 & \Rightarrow \left( B-C \right)'=\left\{ 1,3,5,6,7,9 \right\} \\
\end{align}\]
Hence, the required \[\left( B-C \right)'=\left\{ 1,3,5,6,7,9 \right\}\]

Note: We can also find \[\left( B-C \right)'\] by using Venn diagrams.

We will get the dotted area if we subtract B from ‘C’ and if we will subtract this dotted area from the union, we will get the entire un-dotted area.