Question

If $A=\{x\in N,x\le 7\}$, $B=\{x:x\text{ }is\text{ }prime,\text{ }x<8\}$ and $C=\{x:x\in N,\text{ }x\text{ }is\text{ }odd\text{ }and\text{ }x<10\}$, verify that:
(i) $A\cup \left( B\cap C \right)=\left( A\cup B \right)\cap \left( A\cup C \right)$
(ii) $A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( B\cap C \right)$

Answer 1

Hint: Here, first we have to find the elements of A, B and C. Then, for both the proofs first we have to consider the LHS part and then the RHS part and we have to show that LHS = RHS. For the first proof consider LHS and find the elements of $B\cap C$ and then find the elements of $A\cup \left( B\cap C \right)$. In RHS, first find $A\cup B$ and then $A\cup C$. Then, find the members of $\left( A\cup B \right)\cap \left( A\cup C \right)$, we will get LHS = RHS. Similarly, for the second proof first consider LHS, find the members of $B\cup C$, next find $A\cup \left( B\cap C \right).$Then, consider RHS, first find members of $A\cap B$ and $B\cap C$, then at last find $\left( A\cap B \right)\cup \left( B\cap C \right)$, we will get LHS = RHS.

Complete step-by-step answer:
Here, we are given that $A=\{x\in N,x\le 7\}$, $B=\{x:x\text{ }is\text{ }prime,\text{ }x<8\}$ and $C=\{x:x\in N,\text{ }x\text{ }is\text{ }odd\text{ }and\text{ }x<10\}$.
Now, we have to prove that $A\cup \left( B\cap C \right)=\left( A\cup B \right)\cap \left( A\cup C \right)$ and $A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( B\cap C \right)$
Now, let us write the elements in A, B and C. They are:
$\begin{align}
  & A=\{x\in N,x\le 7\}=\{1,2,3,4,5,6,7\} \\
 & B=\{x:x\text{ }is\text{ }prime,\text{ }x<8\}=\{2,3,5,7\} \\
 & C=\{x:x\in N,\text{ }x\text{ }is\text{ }odd\text{ }and\text{ }x<10\}=\{1,3,5,7,9\} \\
\end{align}$
(i) $A\cup \left( B\cap C \right)=\left( A\cup B \right)\cap \left( A\cup C \right)$
To prove $A\cup \left( B\cap C \right)=\left( A\cup B \right)\cap \left( A\cup C \right)$, it is sufficient to show that LHS = RHS.
First let us consider the LHS.
Let us find $B\cap C$, that is, we have to find the elements that are common in both B and C.
$\Rightarrow B\cap C=\{3,5,7\}$
Now, consider $A\cup \left( B\cap C \right)$, that is, elements that are in either A or $B\cap C$. We will get:
$\Rightarrow A\cup \left( B\cap C \right)=\{1,2,3,4,5,6,7\}$ …….. (1)
Now, take the RHS part. First, find $A\cup B$, that is, we have to find elements that are either in A or B.:
$\Rightarrow A\cup B=\{1,2,3,4,5,6,7\}$
Next, find $A\cup C$, that is, we have to find elements that are either in A or C. Therefore, we will get it as:
$\Rightarrow A\cup C=\{1,2,3,4,5,6,7,9\}$
Now, find $\left( A\cup B \right)\cap \left( A\cup C \right)$, that is we, have to find elements that are common in both $A\cup B$ and $A\cup C$. Hence,
$\Rightarrow \left( A\cup B \right)\cap \left( A\cup C \right)=\{1,2,3,4,5,6,7\}$ …… (2)
Hence, from equation (1) and equation (2), we have:
$A\cup \left( B\cap C \right)=\left( A\cup B \right)\cap \left( A\cup C \right)$

(ii) $A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( B\cap C \right)$
Here, first we will find the LHS part then RHS .Then we have to prove that LHS = RHS.
First find, $B\cup C$, that is, find elements that are either in B or C. We will get it as:
$\Rightarrow B\cup C=\{1,2,3,5,7,9\}$
Next we can find $A\cap \left( B\cup C \right)$, that is, to find elements that are common in both A and $B\cup C$.
Thus,
$\Rightarrow A\cap \left( B\cup C \right)=\{1,2,3,5,7\}$ ……. (3)
Now, consider the RHS, $\left( A\cap B \right)\cup \left( B\cap C \right)$. First, let us find $A\cap B$, that is to find elements that are common in A and B.
$\Rightarrow A\cap B=\{1,2,3,5,7\}$
Now, we can find $B\cap C$, that is, we have to find elements that are common in B and C.
$\Rightarrow B\cap C=\{3,5,7\}$
Now, consider $\left( A\cap B \right)\cup \left( B\cap C \right)$, that is, find elements that are either in $A\cap B$ or $A\cap C$.
$\Rightarrow \left( A\cap B \right)\cup \left( B\cap C \right)=\{1,2,3,5,7\}$ ……. (4)
From equation (3) and equation (4) we can say that:
$A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( B\cap C \right)$

Note: $A\cup \left( B\cap C \right)=\left( A\cup B \right)\cap \left( A\cup C \right)$ and $A\cap \left( B\cup C \right)=\left( A\cap B \right)\cup \left( B\cap C \right)$ is the distributive law of sets. We can also prove this with the help of Venn diagram by taking three sets A, B and C.