Question

If $A=\{a,b,c,d,e\}$, $B=\{a,c,e,g\}$ and $C=\{b,e,f,g\}$, verify that:

1)$A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$
2)$A-\left( B\cap C \right)=\left( A-B \right)\cup \left( A-C \right)$

Answer 1

Hint: Here, for both the proofs it is sufficient to show that LHS = RHS. For the proof (i) first consider LHS and find B – C then find $A\cap \left( B-C \right)$. Now, for RHS, first find $A\cap B$ and $A\cap C$, then from this one find $\left( A\cap B \right)-\left( A\cap C \right)$. Then, we will get LHS = RHS. Similarly for proof (ii), first find $B\cap C$, then consider $A-\left( B\cap C \right)$. For RHS, first find $A-B$ and $A-C$, then at last find $\left( A-B \right)\cup \left( A-C \right)$. Here, also we will get LHS =RHS.

Complete step-by-step answer:
Here, we are given that $A=\{a,b,c,d,e\}$, $B=\{a,c,e,g\}$ and $C=\{b,e,f,g\}$.

Now, we have to prove that $A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C

\right)$ and $A-\left( B\cap C \right)=\left( A-B \right)\cup \left( A-C \right)$.

Here, to prove this it is sufficient to show that LHS = RHS.

1) $A\cap \left( B-C \right)=\left( A\cap B \right)-\left( A\cap C \right)$

Here, first consider the LHS, $A\cap \left( B-C \right)$. First let us find B – C.

We know that B –C is the set of all elements in B that are not in C.

Here, $B=\{a,c,e,g\}$and $C=\{b,e,f,g\}$. Hence, we will get:

$B-C=\{a,c\}$

Now, let us find $A\cap \left( B-C \right)$, the elements that are common in both A and B – C.

$\Rightarrow A\cap \left( B-C \right)=\{a,c\}$

Now, consider RHS, $\left( A\cap B \right)-\left( A\cap C \right)$, first find $A\cap B$, that is

elements that are common in both A and B.

$\Rightarrow A\cap B=\{a,c,e\}$ …… (1)

Next, find the elements in $A\cap C$, the elements that are common in both A and C.

$\Rightarrow A\cap C=\{b,e\}$

Now, consider $\left( A\cap B \right)-\left( A\cap C \right)$, here we have to find the set of

all elements in $A\cap B$ that are not in $A\cap C$.

$\Rightarrow \left( A\cap B \right)-\left( A\cap C \right)=\{a,c\}$ ……. (2)

Form equation (1) and equation (2) we can say that, LHS = RHS, that is,

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


2) $A-\left( B\cap C \right)=\left( A-B \right)\cup \left( A-C \right)$

Now, consider the LHS $A-\left( B\cap C \right)$ , first we will find $B\cap C$ where we have

to find elements that are common in both B and C.

$\Rightarrow B\cap C=\{e,g\}$

Now, find $A-\left( B\cap C \right)$, to find elements which are in A but not in $B\cap C$.

$\Rightarrow A-\left( B\cap C \right)=\{a,b,c,d\}$ …… (3)

Next, consider RHS, $\left( A-B \right)\cup \left( A-C \right)$ first find $A-B$, the elements

that are in A but not in B.

$\Rightarrow A-B=\{b,d\}$

Next, we have to find $A-C$, the elements that are in A but not in C.

$\Rightarrow A-C=\{a,c,d\}$

Now, let us find $\left( A-B \right)\cup \left( A-C \right)$, that is to find elements that are

either in $A-B$ or $A-C$.

$\Rightarrow \left( A-B \right)\cup \left( A-C \right)=\{a,b,c,d\}$ …… (4)

Hence, from equation (3) and equation (4) we will get LHS = RHS.

Therefore, we can say that $A-\left( B\cap C \right)=\left( A-B \right)\cup \left( A-C \right)$.

Note: Here, we can also find this easily with the help of Venn diagram. A Venn diagram is a diagram that shows all possible logical relations between a finite collection of different sets. These diagrams depict elements as points in the plane and sets as regions inside closed curves.