Question

Prove that $A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$.

Answer 1

Hint: Firstly, we will assume an arbitrary element x such that it belongs to the set $A\cap \left( B-C \right)$. Then, by using the properties of the sets we conclude that $A\cap \left( B-C \right)\subseteq \left( A\cap B \right)-\left( A\cap C \right)$. Then, again we will assume another element y such that it belongs to the set $\left( A\cap B \right)-\left( A\cap C \right)$.
Then , again by using the properties of sets we get a relation as $\left( A\cap B \right)-\left( A\cap C \right)\subseteq A\cap \left( B-C \right)$.
By using these two relations, we proved that $A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$.

Complete step by step answer:
In this question, we are supposed to find the relation between the set elements A, B and C which are arranged as follows:
$A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$
Before attempting the actual question we should know the following defines:
The union of A and B $\left( A\cup B \right)$is the total elements of A and B which are common or not all are included.
The intersection of A and B $\left( A\cap B \right)$is the elements of A and B which are common in both sets.
The set A is called a subset of B $A\subseteq B$if set A is totally contained by set B.
Now, let us assume an arbitrary element x such that it belongs to the set $A\cap \left( B-C \right)$.
So, it is represented as:
$x\in A\cap \left( B-C \right)......\left( i \right)$
Therefore, we can conclude from the above assumption that x belongs to whole set $A\cap \left( B-C \right)$the it also belongs to the set A and set (B-C) both as:
$x\in A$ and $x\in \left( B-C \right)$
Now, by again using the same property if x belongs to set (B-C), then it also belong to the set B and set C as:
$\begin{align}
  & x\in A \\
 & x\in B \\
 & x\in C \\
\end{align}$
These above statements implies that:
$\begin{align}
  & x\in \left( A\cap B \right) \\
 & x\notin \left( A\cap C \right) \\
\end{align}$
So, by using above two statements, we can conclude that:
$x\in \left( A\cap B \right)-\left( A\cap C \right).....\left( ii \right)$
From equation (i) and (ii), it is clear fact that $A\cap \left( B-C \right)$is the subset of the $\left( A\cap B \right)-\left( A\cap C \right)$.
$A\cap \left( B-C \right)\subseteq \left( A\cap B \right)-\left( A\cap C \right).....\left( iii \right)$
Now, let us assume another arbitrary element y such that it belongs to the set $\left( A\cap B \right)-\left( A\cap C \right)$.
So, it is represented as:
$x\in \left( A\cap B \right)-\left( A\cap C \right)......\left( iv \right)$
Therefore, we can conclude from the above assumption that y belongs to whole set $\left( A\cap B \right)-\left( A\cap C \right)$the it also belongs to the set $\left( A\cap B \right)$ and but doesn’t to belong to set $\left( B\cap C \right)$ as:
$y\in \left( A\cap B \right)$ and $y\notin \left( B\cap C \right)$
Now, by again using the same property if y belongs to set (A-B), then it also belong to the set A and set B as:
$\begin{align}
  & y\in A \\
 & y\in B \\
 & y\notin C \\
\end{align}$
These above statements implies that:
$\begin{align}
  & y\in A \\
 & y\in \left( B-C \right) \\
\end{align}$
So, by using above two statements, we can conclude that:
$y\in A\cap \left( B-C \right).....\left( v \right)$
From equation (iv) and (v), it is clear fact that $\left( A\cap B \right)-\left( A\cap C \right)$is the subset of $A\cap \left( B-C \right)$.
$\left( A\cap B \right)-\left( A\cap C \right)\subseteq A\cap \left( B-C \right).....\left( vi \right)$
Therefore, from equation (iii) and (vi), we get:
$A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$
Hence, it is proved that $A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$.

Note: Another approach to solve this kind of problem is as given below:
Simply we can use the property of the distributive law which states that:
$a\times \left( b+c \right)=\left( a\times b \right)+\left( a\times c \right)$
Similarly, by using the same property, we can conclude that:
$A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$
You can verify it also by using any non-empty or null sets.